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Math Forum / Mathematics / Mathematical Logic / July 2009Who introduced false logic?
... classical logic was abstracted from the mathematics of finite sets and their subsets .... Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and finally applied it, without justification, to the mathematics of infinite sets. [H. Weyl]. One example is this. For infinite sets with a linear ordering “=<” the implication En Am: m =< n ==> Am En m =< n (1) is held correct, but not the equivalence En Am: m =< n <==> Am En m =< n (2) For any finite set with linear ordering however, (2) is true. The reason is that all elements of a finite set are subject to investigation. Therefore, if in “completed”, i.e., “actual” infinity, as used in set theory, all elements are available, then also (2) should be used. (2) can only be false, if potentially infinite sets are considered, i.e., non-static sets which are never complete but allow that elements can be added. Not necessary to mention, that (2) excludes infinity. Hence actual infinity is a contradiction per se. There is not “a set” that contains { } and with A also {A}. Who introduced the false logical law into set theory? Regards, WM > ... classical logic was abstracted from the mathematics of finite sets > and their subsets .... Forgetful of this limited origin, one > afterwards mistook that logic for something above and prior to all > mathematics, and finally applied it, without justification, to the > mathematics of infinite sets. [H. Weyl]. This is an ancient and discredited take. Logic is logic. It IS NOT connected to anything ELSE OUT THERE that it COULD NEED "justification" for getting "applied" to. > One example is this. For infinite sets with a linear ordering “=<” the > implication [quoted text clipped - 6 lines] > as used in set theory, all elements are available, then also (2) > should be used. This is just bullshit. You are basically saying that N cannot exist. Since everybody VERY MUCH INCLUDING WEYL has been playing with N for a long time, we are just going to keep playing. There is nothing you can do to stop us, because our play does NOT produce contradictions. YOURS DOES. > (2) can only be false, If accepting it as true produces contradictions, i.e., IF IT CAN BE PROVEN false. OTHERWise, ANY AND EVERYbody who WANTS to gets to KEEP USING IT AS AN AXIOM. > if potentially infinite sets > are considered, This is just bullshit. > i.e., non-static sets EVERYTHING in this paradigm is static. NOTHING MAY EVER be added TO ANYthing. Unlike you, WE KNOW THE DIFFERENCE between a constant and a variable. >> ... classical logic was abstracted from the mathematics of finite sets >> and their subsets .... Forgetful of this limited origin, one [quoted text clipped - 5 lines] > Logic is logic. It IS NOT connected to anything ELSE OUT THERE > that it COULD NEED "justification" for getting "applied" to. Somewhere Weyl also asserted that “Mathematik ... ist die Wissenschaft vom Unendlichen.” ("Mathematics is the science of the infinite"). I don't know where he wrote that, but Mueckenheim doesn't need to give a source of his quotation either. And what Weyl refers to in Mueckenheim's quote is classical vs. intuitionistic logic. That makes perfect sense, but it has absolutely nothing to do with Mueckenheim's idiotic crap.  SignatureW. Hughes, in sci.math.: "No set of natural numbers without a last element [is finite]" Prof. Dr. W. Mückenheim, mathematical mastermind of "Augsburg University of Applied Science": "There is no natural number called "out a last element". Am Wed, 27 May 2009 00:37:32 +0200 schrieb Ralf Bader: > Somewhere Weyl also asserted that > “Mathematik ... ist die Wissenschaft vom Unendlichen.” > ("Mathematics is the science of the infinite"). Maybe with "infinite" he referred to the /potentially infinite/ here? ;-) SCNR Herb > Am Wed, 27 May 2009 00:37:32 +0200 schrieb Ralf Bader: > [quoted text clipped - 3 lines] > > Maybe with "infinite" he referred to the /potentially infinite/ here? ;-) No. I am applying Mueckenheim's rules of quotation slandering and so it means what I like it to mean. Anyway it wouldn't change the idiocy of Mueckenheim's example in the initial posting of this thread. Ralf > Anyway it wouldn't change the idiocy of Mueckenheim's > example in the initial posting of this thread. Completely agree with you. (Do you still read his crap?) Herb >> Anyway it wouldn't change the idiocy of Mueckenheim's >> example in the initial posting of this thread. > > Completely agree with you. (Do you still read his crap?) Yes, I read it, selectively for a long time. It is still mildly interesting, psychologically. I think one must have some kind of mental representation (in Brouwer's language an intuition) of a structure like the natural numbers to be able to deal with it, and Mueckenheim doesn't have such a representation. He is constantly mixing up the notions of actual and potential infinity (whatever they precisely mean) with the presence or absence of a maximal element in an ordered countably infinite set. In all his pseudoproofs of the "nonexistence" or "inconsistency" of the "actually infinite" naturals an alleged maximal element slips in. Concerning the "binary tree", another devastating attitude is visible: The preoccupation with the "existence" of objects. But mathematical objects are uninteresting in themselves; they are property-less carriers of relations to other mathematical objects. Mueckenheim asserts two things: a relationship between that tree and the reals, and a presentation of the full tree as a limit of finite partial trees. Now the question is, what is the data necessary to obtain a limit? That is not just a sequence of objects. E.g., to get a limit of a sequence of reals a topology is required. Mueckenheims attempted formation of the full tree by a sequence of subtrees is, basically, a categorical limit. Such limit can be taken from a diagram of objects _and_ _morphisms_; for example, the limit of an inclusion chain of sets: A_1 c A_2 c A_3... is the union of all A_i. In a topological category one would call this a tower of cofibrations, in another context it would be called a sum, and in Mueckenheim's case A_i would be the tree with i stages. It should be possible to set up a category TREE and a functor path:TREE -> SET which associates with any tree its set of paths. And Mueckenheim's limit formation should work in TREE. Mueckenhiem's "argument" in principle is that his kind of limit is the only one conceivable. But that is just a limitation of Grömaz Mueckenheim's imagination or, turned positively, a manifestation of his stupidity. Instead of the cofibration tower we can take a fibration tower: A_1 <- A_2 <- A_3...; in going the reverse arrow from A_i to A_i+1, each element of A_i is split up in several elements of A_i+1; for example, the path p in the i-stage binary tree is split up in the two paths starting like p and finally turning right or left in the (i+1)-stage. Here the categorical limit is a kind of product. Now, the path functor seems well-behaved w.r.t. limits of fibration towers, but not w.r.t. limits of cofibration towers. That is in coincidence with the observation that we get the reals as a limit of chopping up the "line" into smaller and smaller pieces (a fibration-like procedure) and not by accumulating more and more pieces (a cofibration-like procedure). This is, in loose talk, what I view as the background of the unending binary bonsai climbing, and it is kind of funny to see what is going on in these threads, from both sides. Ralf Ralf Bader wrote: > >> ... classical logic was abstracted from the mathematics of finite sets > >> and their subsets .... Forgetful of this limited origin, one [quoted text clipped - 13 lines] > intuitionistic logic. That makes perfect sense, but it has absolutely > nothing to do with Mueckenheim's idiotic crap. Weyl underwent a number of changes in his mathematical philosophy, as mentioned in http://en.wikipedia.org/wiki/Hermann_Weyl#Foundations_of_mathematics . Based on other stuff I've read but only vaguely recall, it may be that the wiki article understates the degree of Weyl's philosophical evolution. -- hz My, cranks are cute when they try to order quantifiers. > En Am: m =< n ==> Am En m =< n (1) > is held correct, This is not MERELY "held correct": THIS IS A LAW of first-order logic: THIS IS A VALIDITY of first-order logic. THIS DOES NOT HAVE A DAMN thing to do with n, with =<, with ordering, OR WITH NUMBERS!! "=<" could be replaced WITH ANY BINARY PREDICATE WHAT*SO*f.cking*EVER and this sentence WOULD STILL BE A THEOREM of first-order logic! In particular, m and n could be men and "=<" could mean "is shaved by, AND THIS WOULD STILL BE TRUE: If there is an n such that for all men m, m is shaved by n (i.e., if there is a barber), then it must also be the case that for every man, SOMEman (the barber, n, of course!) shaves him. but not the equivalence > En Am: m =< n <==> Am En m =< n Because the BACK arrow is obviously NOT valid, DUMBASS. If you want to actually HAVE this argument, you have to REPLACE your idiotic use of <==> in the middle of (2) with <== , i.e., you HAVE TO REVERSE THE ORDER of the halves of the conditional! The FORWARD arrow IS ALREADY (1)! > One example is this. This is not an example of anything except your complete ignorance of logic. > For infinite sets with a linear ordering “=<” Finite sets HAVE NOTHING TO DO with this. SETS have NOthing to do with this. ORDERING HAS NOTHING to do with this. > the > implication > En Am: m =< n ==> Am En m =< n (1) > is held correct, This implication IS NOT merely HELD correct! This implication IS VALID! This implication is correct WHETHER YOU LIKE IT OR NOT! WE CAN PROVE this implication! FROM NOTHING! We don't even need any premises! ALL we need is the relevant INFERENCE RULES about quantifiers! No special property of ordering or sets is used in the proof of this implication! The fact that you spelled the predicate "=<" IS NOT relevant! This theorem holds FOR ANY BINARY PREDICATE WHATSOEVER! And any cardinality of domain whatsoever AS WELL. > but not the equivalence > En Am: m =< n <==> Am En m =< n (2) You have completely misspelled this. You cannot rationally compare P --> Q to P <--> Q because P <--> Q is nothing BUT P-->Q (which you've ALREADY dealt with) CONJOINED WITH P <-- Q ! THE BACK arrow is the only thing that is relevant! Given that the forward-arrow is valid, the ONLY way the conjunction of both arrows can be valid IS IF THE BACK-ARROW is valid as well! So THE ONLY question you have left to investigate is whether Q --> P is valid! So just write it that way. > En Am: m =< n <==> Am En m =< n (2) > For any finite set with linear ordering however, (2) is true. The > reason is that all elements of a finite set are subject to > investigation. This is just bullshit. If the finite set is so finite that it has ONLY TWO real numbers as elements, namely, pi and e, then DESPITE the fact that it is finite, its elements are NOT (not completely, anyway) subject to investigation, because you will not live long enough to learn (or investigate, or anything else) the 10^(googolplex)th digit of pi, OR of e. Pi and e, are themselves, DESPITE being computable, NOT subject to (sufficient) investigation. Even if they were, it is easy to show the existence of real numbers that are NOT definable or computable. The SET may be finite but individual elements WITHIN it may NOT be. Of course, you, being you, would probably insist that these elements as well were "incomplete" and only "potentially" extant. Really? Does a finite circle with a finite diameter have a NON-finite RATIO of its circumference to its diameter?? All you have to do to refute your position is turn this set on its side, is consider its transpose. Consider the INFINITE set of two-element sets with the property that each of its elements is a (very finite) pair: the pair of numbers you get by taking the next few digits of pi and e. You know, {3,2} {14,71} {159,828} ad naus. The nth pair of this set is 2 n-digit numbers (maybe one of the sets will 1 day only have 1 number, since the 2 will coincide, but my point is, pi and e ARE NOT sufficiently amenable to "investigation" FOR ANYBODY TO EVER answer THAT question). Every ELEMENT of this set is a 2 element set. Every one of THOSE 2 elements is finite and is a natural number. Yet you say that one set is legitimate (or satisfies (2)) and the other does not. > ... classical logic was abstracted from the mathematics of finite sets > and their subsets .... Forgetful of this limited origin, one [quoted text clipped - 21 lines] > > Regards, WM Does nobody know who introduced this illegal logical rule into set theory? Or does anybody have a an example of a finite set that makes this rule legal? Regards, WM In article <b54a16b3-914d-47eb-bd89-967cc10d2cac@u10g2000vbd.googlegroups.com>, > > ... classical logic was abstracted from the mathematics of finite sets > > and their subsets .... Forgetful of this limited origin, one [quoted text clipped - 14 lines] > > are considered, i.e., non-static sets which are never complete but > > allow that elements can be added. But non-static sets do not exist in any mathematical set theory. only in such anti-mathematical theories as anti-mathematicians like WM dream up. SignatureVirgil > > If we define: > [quoted text clipped - 10 lines] > > them, names them, or makes models of them, but because the natural > > numbers are simply existing and mathematics is built upon them. > But according to WM, no such thing as N can exist. > So WM wold throw out the naturals on which so much is built. The question is not whether the complete set of all naturals exists. That question alrady is nearly as ridiculous as any affirmative answer. The question is whether we could inform someone who does not yet know, what we understand by the sequence of natural numbers. In order to answer this question, we need not wait until SETI gets contact. We can answer it in every first class of every elementary school. Of course we can inform any child with average intelligence what we understand by this sequence 1, 2, 3, ... Only logicians seem to see problems where no problems are. (As some kind of compensation they see no problems where problems are.) Of course this sequence 1, 2, 3, ... is uniquely defined, because it is possible to inform any intelligent being about it. There is no the slightest difference between the idea of N and the idea of a sonata or pi. Regards, WM > In article > <b54a16b3-914d-47eb-bd89-967cc10d2...@u10g2000vbd.googlegroups.com>, [quoted text clipped - 19 lines] > > But non-static sets do not exist in any mathematical set theory. They do not exist in what is commonly called ste theory and what is eternally false mathematics. Regards, WM > > > If we define: > > [quoted text clipped - 64 lines] > > Regards, WM WM: I offer this text as further proof of polynomial time achievement: Search: Login 1. Create Account 2. Login 1. File Exchange 2. Newsgroup 3. Link Exchange 4. Blogs 5. Contest 6. MathWorks.com • About the MATLAB Newsgroup • Post A New Message • E-mail this page MATLAB Central > MATLAB Newsreader > An exact simplification challenge - 91 (eerie P... Add thread to My Watch List What is a Watch List? Thread Subject: An exact simplification challenge - 91 (eerie PolyGamma) Subject: An exact simplification challenge - 91 (eerie PolyGamma) From: Vladimir Bondarenko Date: 25 May, 2009 03:23:31 Message: 1 of 6 Reply to this message Add author to My Watch List View original format Flag as spam Hello, - PolyGamma[0, 1 + I] - PolyGamma[0, 1 - I] - PolyGamma[0, 3/2 + I] - PolyGamma[0, 3/2 - I] + 2 PolyGamma[0, 2 + 2 I] + 2 PolyGamma[0, 2 - 2 I] + (1/2 + I) PolyGamma[1, 1 + I] + (1/2 - I) PolyGamma[1, 1 - I] - (1/2 + I) PolyGamma[1, 3/2 + I] - (1/2 - I) PolyGamma[1, 3/2 - I] + (2 + 4 I) PolyGamma[1, 2 + 2 I] + (2 - 4 I) PolyGamma[1, 2 - 2 I] - 2 Re [I PolyGamma[1, 1 + I]] + 2 Re[I PolyGamma[1, 1 - I]] ?Folks, please give not just the answer but the processing.Cheers,Vladimir BondarenkoCo-founder, CEO, Mathematical Directorhttp://www.cybertester.com/ Cyber Tester Ltd.----------------------------------------------------------"We must understand that technologieslike these are the way of the future."---------------------------------------------------------- Subject: An exact simplification challenge - 91 (eerie PolyGamma) From: Martin Musatov Date: 25 May, 2009 03:36:14 Message: 2 of 6 Reply to this message Add author to My Watch List View original format Flag as spam Vladimir Bondarenko wrote:> Hello,>> - PolyGamma[0, 1 + I]> - PolyGamma[0, 1 - I]> - PolyGamma[0, 3/2 + I]> - PolyGamma[0, 3/2 - I]> + 2 PolyGamma[0, 2 + 2 I]> + 2 PolyGamma[0, 2 - 2 I]> + (1/2 + I) PolyGamma[1, 1 + I]> + (1/2 - I) PolyGamma[1, 1 - I] > - (1/2 + I) PolyGamma[1, 3/2 + I]> - (1/2 - I) PolyGamma[1, 3/2 - I] > + (2 + 4 I) PolyGamma[1, 2 + 2 I]> + (2 - 4 I) PolyGamma[1, 2 - 2 I] > - 2 Re[I PolyGamma[1, 1 + I]]> + 2 Re[I PolyGamma[1, 1 - I]]>> ?>> Folks, please give not just the answer but the processing.>> Cheers,>> Vladimir Bondarenko>> Co-founder, CEO, Mathematical Director>> http://www.cybertester.com/ Cyber Tester Ltd.>> ---------------------------------------------------------->> "We must understand that technologies> like these are the way of the future.">> ----------------------------------------------------------People, did I or did I not call it: "eerie effectiveness". Googlesearch sci.math: "Musatov" and "Eerie". Want to keep going? I enjoyit. It keeps getting more gratifying each passing day, like puttingpennies into a piggy bank. Subject: An exact simplification challenge - 91 (eerie PolyGamma) From: Martin Musatov Date: 25 May, 2009 03:40:46 Message: 3 of 6 Reply to this message Add author to My Watch List View original format Flag as spam (C)2009:Here you go folks, call me the information prophet what youare witnessing is simple P=NP mathematics over large data sets:"Eerie" isn't it?...The "Effectiveness"...Martin Musatov wrote:> Vladimir Bondarenko wrote:> > Hello,> >> > - PolyGamma[0, 1 + I]> > - PolyGamma[0, 1 - I]> > - PolyGamma[0, 3/2 + I]> > - PolyGamma[0, 3/2 - I]> > + 2 PolyGamma[0, 2 + 2 I]> > + 2 PolyGamma[0, 2 - 2 I]> > + (1/2 + I) PolyGamma[1, 1 + I] > > + (1/2 - I) PolyGamma[1, 1 - I]> > - (1/2 + I) PolyGamma[1, 3/2 + I]> > - (1/2 - I) PolyGamma[1, 3/2 - I]> > + (2 + 4 I) PolyGamma[1, 2 + 2 I]> > + (2 - 4 I) PolyGamma[1, 2 - 2 I]> > - 2 Re[I PolyGamma[1, 1 + I]]> > + 2 Re[I PolyGamma[1, 1 - I]]> >> > ?> >> > Folks, please give not just the answer but the processing.> >> > Cheers,> >> > Vladimir Bondarenko> >> > Co-founder, CEO, Mathematical Director> >> > http://www.cybertester.com/ Cyber Tester Ltd.> >> > ----------------------------------------------------------> >> > "We must understand that technologies> > like these are the way of the future."> >> > ----------------------------------------------------------> People, did I or did I not call it: "eerie effectiveness". Google> search sci.math: "Musatov" and "Eerie". Want to keep going? I enjoy> it. It keeps getting more gratifying each passing day, like putting> pennies into a piggy bank.Martin MusatovFounderMeAmI.org Subject: An exact simplification challenge - 91 (eerie PolyGamma) From: clicliclic@freenet.de Date: 25 May, 2009 18:32:02 Message: 4 of 6 Reply to this message Add author to My Watch List View original format Flag as spam Vladimir Bondarenko schrieb:>> - PolyGamma[0, 1 + I]> - PolyGamma[0, 1 - I]> - PolyGamma[0, 3/2 + I]> - PolyGamma[0, 3/2 - I]> + 2 PolyGamma[0, 2 + 2 I]> + 2 PolyGamma[0, 2 - 2 I]> + (1/2 + I) PolyGamma[1, 1 + I]> + (1/2 - I) PolyGamma[1, 1 - I]> - (1/2 + I) PolyGamma[1, 3/2 + I]> - (1/2 - I) PolyGamma[1, 3/2 - I]> + (2 + 4 I) PolyGamma[1, 2 + 2 I]> + (2 - 4 I) PolyGamma[1, 2 - 2 I]> - 2 Re[I PolyGamma[1, 1 + I]]> + 2 Re[I PolyGamma[1, 1 - I]]>> ?>This simple challenge seems to be specifically made for Derive 6.10:-POLYGAMMA (0,1+#i)-POLYGAMMA(0,1-#i)-POLYGAMMA(0,3/2+#i)-POLYGAM~MA(0,3/2-#i) +2*POLYGAMMA(0,2+2*#i)+2*POLYGAMMA(0,2-2*#i)+(1/2+#i~)*POLYGAMMA (1,1+#i)+(1/2-#i)*POLYGAMMA(1,1-#i)-(1/2+#i)*POLYGAMM~A(1,3/2+#i)-(1/2- #i)*POLYGAMMA(1,3/2-#i)+(2+4*#i)*POLYGAMMA(1,2+~2*#i)+(2-4*#i) *POLYGAMMA(1,2-2*#i)-2*RE(#i*POLYGAMMA(1,1+#i))+2*~RE(#i*POLYGAMMA(1,1- #i))" ... is automatically rewritten to ... "-DIGAMMA(3/2-#i)-DIGAMMA (3/2+#i)+2*DIGAMMA(2-2*#i)+2*DIGAMMA(2+2~*#i)-DIGAMMA(1-#i)-DIGAMMA (1+#i)-ZETA(2,1/2-#i)/2-ZETA(2,1/2+#i)~/2+2*ZETA(2,1-2*#i)+2*ZETA (2,1+2*#i)+ZETA(2,1-#i)/2+ZETA(2,1+#i)~/2+#i*(-CONJ(ZETA(2,1-#i))+CONJ (ZETA(2,1+#i))+ZETA(2,1/2-#i)-ZET~A(2,1/2+#i)-4*ZETA(2,1-2*#i)+4*ZETA (2,1+2*#i))" judicious substitution of the helper functions ... "mpsi (z,m):=1/m*SUM(DIGAMMA((z+k)/m),k,0,m-1)+LN(m)mzeta(s,z,m):=1/m^s*SUM (ZETA(s,(z+k)/m),k,0,m-1)" ... along with manual help for CONJ (ZETA) ... "-DIGAMMA(3/2-#i)-DIGAMMA(3/2+#i)+2*mpsi(2-2*#i,2)+2*mpsi (2+2*#i,~2)-DIGAMMA(1-#i)-DIGAMMA(1+#i)-ZETA(2,1/2-#i)/2-ZETA (2,1/2+#i)/2~+2*mzeta(2,1-2*#i,2)+2*mzeta(2,1+2*#i,2)+ZETA(2,1-#i)/ 2+ZETA(2,1~+#i)/2+#i*(-ZETA(CONJ(2),CONJ(1-#i))+ZETA(CONJ(2),CONJ (1+#i))+ZE~TA(2,1/2-#i)-ZETA(2,1/2+#i)-4*mzeta(2,1-2*#i,2)+4*mzeta (2,1+2*#i~,2))ZETA(2,1-#i)+ZETA(2,1+#i)+4*LN(2)2*RE(ZETA(2,1+#i))+4*LN (2)3.698588915Martin. Subject: An exact simplification challenge - 91 (eerie PolyGamma) From: Vladimir Bondarenko Date: 25 May, 2009 19:41:14 Message: 5 of 6 Reply to this message Add author to My Watch List View original format Flag as spam On May 25, 9:32 pm, cliclic...@freenet.de wrote:> Vladimir Bondarenko schrieb:>>>>>> > - PolyGamma[0, 1 + I]> > - PolyGamma[0, 1 - I] > > - PolyGamma[0, 3/2 + I]> > - PolyGamma [0, 3/2 - I]> > + 2 PolyGamma[0, 2 + 2 I]> > + 2 PolyGamma[0, 2 - 2 I]> > + (1/2 + I) PolyGamma[1, 1 + I]> > + (1/2 - I) PolyGamma[1, 1 - I]> > - (1/2 + I) PolyGamma[1, 3/2 + I]> > - (1/2 - I) PolyGamma[1, 3/2 - I]> > + (2 + 4 I) PolyGamma[1, 2 + 2 I]> > + (2 - 4 I) PolyGamma[1, 2 - 2 I]> > - 2 Re[I PolyGamma[1, 1 + I]]> > + 2 Re[I PolyGamma[1, 1 - I]]>> > ?>> This simple challenge seems to be specifically made for Derive 6.10:>> -POLYGAMMA(0,1+#i)- POLYGAMMA(0,1-#i)-POLYGAMMA(0,3/2+#i)-POLYGAM~> MA(0,3/2-#i) +2*POLYGAMMA(0,2+2*#i)+2*POLYGAMMA(0,2-2*#i)+(1/2+#i~> )*POLYGAMMA (1,1+#i)+(1/2-#i)*POLYGAMMA(1,1-#i)-(1/2+#i)*POLYGAMM~> A(1,3/2+#i)- (1/2-#i)*POLYGAMMA(1,3/2-#i)+(2+4*#i)*POLYGAMMA(1,2+~> 2*#i)+(2-4*#i) *POLYGAMMA(1,2-2*#i)-2*RE(#i*POLYGAMMA(1,1+#i))+2*~> RE(#i*POLYGAMMA (1,1-#i))>> " ... is automatically rewritten to ... ">> -DIGAMMA(3/2- #i)-DIGAMMA(3/2+#i)+2*DIGAMMA(2-2*#i)+2*DIGAMMA(2+2~> *#i)-DIGAMMA(1- #i)-DIGAMMA(1+#i)-ZETA(2,1/2-#i)/2-ZETA(2,1/2+#i)~> /2+2*ZETA(2,1-2*#i) +2*ZETA(2,1+2*#i)+ZETA(2,1-#i)/2+ZETA(2,1+#i)~> /2+#i*(-CONJ(ZETA(2,1- #i))+CONJ(ZETA(2,1+#i))+ZETA(2,1/2-#i)-ZET~> A(2,1/2+#i)-4*ZETA (2,1-2*#i)+4*ZETA(2,1+2*#i))>> " judicious substitution of the helper functions ... ">> mpsi(z,m):=1/m*SUM(DIGAMMA((z+k)/m),k,0,m-1)+LN(m)>> mzeta(s,z,m):=1/m^s*SUM(ZETA(s,(z+k)/m),k,0,m-1)>> " ... along with manual help for CONJ(ZETA) ... ">> -DIGAMMA(3/2-#i)-DIGAMMA(3/2+#i) +2*mpsi(2-2*#i,2)+2*mpsi(2+2*#i,~> 2)-DIGAMMA(1-#i)-DIGAMMA(1+#i)-ZETA (2,1/2-#i)/2-ZETA(2,1/2+#i)/2~> +2*mzeta(2,1-2*#i,2)+2*mzeta(2,1+2*#i, 2)+ZETA(2,1-#i)/2+ZETA(2,1~> +#i)/2+#i*(-ZETA(CONJ(2),CONJ(1-#i))+ZETA (CONJ(2),CONJ(1+#i))+ZE~> TA(2,1/2-#i)-ZETA(2,1/2+#i)-4*mzeta(2,1-2*#i, 2)+4*mzeta(2,1+2*#i~> ,2))>> ZETA(2,1-#i)+ZETA(2,1+#i)+4*LN(2)>> 2*RE (ZETA(2,1+#i))+4*LN(2)>> 3.698588915>> Martin.Great!I'd only offer a bit simpler answer1-pi^2*CSCH(pi)^2+4*LOG(2)3.698588915But there's still something about this andmany other challenges untold :) Cheers,Vladimir BondarenkoCo-founder, CEO, Mathematical Directorhttp://www.cybertester.com/ Cyber Tester Ltd.----------------------------------------------------------"We must understand that technologieslike these are the way of the future."---------------------------------------------------------- Subject: An exact simplification challenge - 91 (eerie PolyGamma) From: clicliclic@freenet.de Date: 25 May, 2009 21:34:03 Message: 6 of 6 Reply to this message Add author to My Watch List View original format Flag as spam Vladimir Bondarenko schrieb:> On May 25, 9:32 pm, cliclic...@freenet.de wrote:> > Vladimir Bondarenko schrieb:> >> > > - PolyGamma[0, 1 + I]> > > - PolyGamma[0, 1 - I]> > > - PolyGamma[0, 3/2 + I]> > > - PolyGamma[0, 3/2 - I]> > > + 2 PolyGamma[0, 2 + 2 I]> > > + 2 PolyGamma[0, 2 - 2 I]> > > + (1/2 + I) PolyGamma[1, 1 + I]> > > + (1/2 - I) PolyGamma[1, 1 - I]> > > - (1/2 + I) PolyGamma[1, 3/2 + I]> > > - (1/2 - I) PolyGamma[1, 3/2 - I]> > > + (2 + 4 I) PolyGamma [1, 2 + 2 I]> > > + (2 - 4 I) PolyGamma[1, 2 - 2 I]> > > - 2 Re[I PolyGamma[1, 1 + I]]> > > + 2 Re[I PolyGamma[1, 1 - I]]> >> > > ?> >> > This simple challenge seems to be specifically made for Derive 6.10:> >> > -POLYGAMMA(0,1+#i)-POLYGAMMA(0,1-#i)-POLYGAMMA(0,3/2+#i)- POLYGAM~> > MA(0,3/2-#i)+2*POLYGAMMA(0,2+2*#i)+2*POLYGAMMA(0,2-2*#i) +(1/2+#i~> > )*POLYGAMMA(1,1+#i)+(1/2-#i)*POLYGAMMA(1,1-#i)-(1/2+#i) *POLYGAMM~> > A(1,3/2+#i)-(1/2-#i)*POLYGAMMA(1,3/2-#i)+(2+4*#i) *POLYGAMMA(1,2+~> > 2*#i)+(2-4*#i)*POLYGAMMA(1,2-2*#i)-2*RE (#i*POLYGAMMA(1,1+#i))+2*~> > RE(#i*POLYGAMMA(1,1-#i))> >> > " ... is automatically rewritten to ... "> >> > -DIGAMMA(3/2-#i)-DIGAMMA(3/2+#i) +2*DIGAMMA(2-2*#i)+2*DIGAMMA(2+2~> > *#i)-DIGAMMA(1-#i)-DIGAMMA(1+#i)- ZETA(2,1/2-#i)/2-ZETA(2,1/2+#i)~> > /2+2*ZETA(2,1-2*#i)+2*ZETA (2,1+2*#i)+ZETA(2,1-#i)/2+ZETA(2,1+#i)~> > /2+#i*(-CONJ(ZETA(2,1-#i)) +CONJ(ZETA(2,1+#i))+ZETA(2,1/2-#i)-ZET~> > A(2,1/2+#i)-4*ZETA(2,1-2*#i) +4*ZETA(2,1+2*#i))> >> > " judicious substitution of the helper functions ... "> >> > mpsi(z,m):=1/m*SUM(DIGAMMA((z+k)/m),k,0,m-1)+LN (m)> >> > mzeta(s,z,m):=1/m^s*SUM(ZETA(s,(z+k)/m),k,0,m-1)> >> > " ... along with manual help for CONJ(ZETA) ... "> >> > -DIGAMMA(3/2-#i)- DIGAMMA(3/2+#i)+2*mpsi(2-2*#i,2)+2*mpsi(2+2*#i,~> > 2)-DIGAMMA(1-#i)- DIGAMMA(1+#i)-ZETA(2,1/2-#i)/2-ZETA(2,1/2+#i)/2~> > +2*mzeta(2,1-2*#i, 2)+2*mzeta(2,1+2*#i,2)+ZETA(2,1-#i)/2+ZETA(2,1~> > +#i)/2+#i*(-ZETA (CONJ(2),CONJ(1-#i))+ZETA(CONJ(2),CONJ(1+#i))+ZE~> > TA(2,1/2-#i)-ZETA (2,1/2+#i)-4*mzeta(2,1-2*#i,2)+4*mzeta(2,1+2*#i~> > ,2))> >> > ZETA (2,1-#i)+ZETA(2,1+#i)+4*LN(2)> >> > 2*RE(ZETA(2,1+#i))+4*LN(2)> >> > 3.698588915> >>> Great!>> I'd only offer a bit simpler answer>> 1- pi^2*CSCH(pi)^2+4*LOG(2)>> 3.698588915>Sorry, I didn't try to "elementarize" the ZETA's.ZETA(2, 1+#i) + ZETA(2, 1-#i) =ZETA(2, 1+#i) + ZETA(2, -#i) + 1 =-(2 pi)^2 LI(-1, #e^(-2 pi)) + 1which is equivalent to your expression since LI(-1,z) = z/(1-z)^2.Can Maple or Mathematica perform the conversion automatically?Martin.Tags for this Thread Everyone's Tags: spam(2) Add a New Tag: Separated by commasEx.: root locus, bode What are tags? A tag is like a keyword or category label associated with each thread. Tags make it easier for you to find threads of interest. Anyone can tag a thread. Tags are public and visible to everyone. Tag Activity for This Thread Tag Applied By Date/Time spam per isakson 25 May, 2009 00:25:24 spam John D'Errico 24 May, 2009 23:32:10 Feed for this Thread Public Submission Policy NOTICE: Any content you submit to MATLAB Central, including personal information, is not subject to the protections which may be afforded information collected under other sections of The MathWorks, Inc. Web site. You are entirely responsible for all content that you upload, post, e-mail, transmit or otherwise make available via MATLAB Central. The MathWorks does not control the content posted by visitors to MATLAB Central and, does not guarantee the accuracy, integrity, or quality of such content. Under no circumstances will The MathWorks be liable in any way for any content not authored by The MathWorks, or any loss or damage of any kind incurred as a result of the use of any content posted, e-mailed, transmitted or otherwise made available via MATLAB Central. Read the complete Disclaimer prior to use. Contact us at files@mathworks.com • Featured MathWorks.com Topics: • New Products • Support • Documentation • Training • Webinars • Newsletters • MATLAB Trials • Careers • © 1994-2009 The MathWorks, Inc. • Site Help • Patents • Trademarks • Privacy Policy • Preventing Piracy Signed, Martin Musatov __ The unpopular opinion is not always wrong. Sometimes it is simply perfect. >> > If we define: >> [quoted text clipped - 20 lines] >The question is whether we could inform someone who does not yet know, >what we understand by the sequence of natural numbers. Sorry, I can't follow any of this, because I don't already know what you mean by the words "question", "is", "whether", "we". "could", "inform", "someone", "who", "does", "not", "yet", and "understand". >In order to answer this question, we need not wait until SETI gets >contact. We can answer it in every first class of every elementary [quoted text clipped - 39 lines] > >Regards, WM David C. Ullrich "Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to." (John Jones, "My talk about Godel to the post-grads." in sci.logic.) > On Wed, 27 May 2009 01:04:52 -0700 (PDT), WM > [quoted text clipped - 29 lines] > "could", "inform", "someone", "who", "does", "not", "yet", > and "understand". 1. Then let us start with the meaning of your sentence "because I don't already know what you mean by the words" Obviously you can understand that sentence, because you even corrected a typo (alrady) of mine. "does" is closely related to "do", "don't" is the negation of "do", apending a truncated version of "not". So much for the first lesson. "First", abbreviated by "1." or "1^st" points to the last member of an ordered set of cardinality 1. Any open questions? Regards, WM In article <1d5fcfb8-e4f6-496f-8fc9-1ff6b78184a0@k8g2000yqn.googlegroups.com>, > > > If we define: > > [quoted text clipped - 15 lines] > > The question is not whether the complete set of all naturals exists. That is a question on which WM differs from the mainstream. But WM cannot grant its existence in a given argument and then deny it in the same argument and expect anyone to accept that sort of argument. > That question alrady is nearly as ridiculous as any affirmative > answer. Wrong! > The question is whether we could inform someone who does not yet know, > what we understand by the sequence of natural numbers. Children know from a very early age. Whom does WM think is left to inform? But non-static sets do not exist in any mathematical set theory. > They do not exist in what is commonly called ste theory and what is > eternally false mathematics. Only in WM's MathUnrealism is it false. Everywhere else it is the true math. > Regards, WM SignatureVirgil > > > > Of course there can be different models for N and there can be > > > > different names for the elements of N. But that does not matter. The [quoted text clipped - 10 lines] > But WM cannot grant its existence in a given argument and then deny it > in the same argument and expect anyone to accept that sort of argument. Grant the existence of two natural numbers m and n such that m/n = sqrt (2). Then falsify it. Grant the existence of a largest natural. Then falsify it. Grant the existence of a largest prime number. Then falsify it. Grant the existence of all natural numbers. Then falsify it. All these proofs are proofs by contradiction. > > That question alrady is nearly as ridiculous as any affirmative > > answer. > > Wrong! I said "nearly"! This is a fuzzy quantification, not a sharp one. It cannot be wrong. One might qualify it as nearly wrong or nearly right, at most. > > The question is whether we could inform someone who does not yet know, > > what we understand by the sequence of natural numbers. > > Children know from a very early age. Whom does WM think is left to > inform? How do children get to know what the real numbers are? Do they have Peano's axioms in their genes or in their strollers? > But non-static sets do not exist in any mathematical set theory. That is orthodox nonsense. The natural numbers as a potetntially infinite set exist in every mathematical theory before BC (before Cantor) and in many mathematical theories AC, for instance in every theory that is taught in elementary schools as well as in universities outside the departments of "logics" and mathematics. Regards, WM In article <ee23e72e-206f-48c9-ac34-ad7b2046f299@c9g2000yqm.googlegroups.com>, > > > > > Of course there can be different models for N and there can be > > > > > different names for the elements of N. But that does not matter. The [quoted text clipped - 18 lines] > > All these proofs are proofs by contradiction. I have never seen a logically correct or complete falsification of the existence of all naturals. Note that an argument against the existence of numerals (names) for all naturals does not suffice. > > > That question alrady is nearly as ridiculous as any affirmative > > > answer. [quoted text clipped - 4 lines] > cannot be wrong. One might qualify it as nearly wrong or nearly right, > at most. All of WM's logic is fuzzy, much to fuzzy for serious mathematics. > > > The question is whether we could inform someone who does not yet know, > > > what we understand by the sequence of natural numbers. [quoted text clipped - 4 lines] > How do children get to know what the real numbers are? Do they have > Peano's axioms in their genes or in their strollers? Quite possible in their genes. > > But non-static sets do not exist in any mathematical set theory. > > That is orthodox nonsense. Then produce an example of a mathematical set theory is which there are non-static sets. > The natural numbers as a potetntially > infinite set exist in every mathematical theory before BC (before > Cantor) and in many mathematical theories AC, for instance in every > theory that is taught in elementary schools as well as in universities > outside the departments of "logics" and mathematics. Balls! Set theories as taught in elementary schools almost all require a fixed universal set in which all other sets are fixed proper subsets, a la Venn diagrams. And this approach even carries over into some introductory math classes in universities. Certainly I have never seen such variable sets anywhere in either England or the United states at any level from kindergarten through postgraduate. > Regards, WM SignatureVirgil On 27 Mai, 23:15, Virgil <virg...@nowhere.com> wrote: > > > > Of course there can be different models for N and there can be > > > > different names for the elements of N. But that does not matter. The [quoted text clipped - 10 lines] > But WM cannot grant its existence in a given argument and then deny it > in the same argument and expect anyone to accept that sort of argument. Grant the existence of two natural numbers m and n such that m/n = sqrt (2). Then falsify it. Grant the existence of a largest natural. Then falsify it. Grant the existence of a largest prime number. Then falsify it. Grant the existence of all natural numbers. Then falsify it. All these proofs are proofs by contradiction. ************************* How do you falsify the existence of the set of all Natural numbers in ZF ? ZF includes an axiom of infinity, which pretty much directly guarantees that there is an infinite set of all finite ordinals. Assuming ZF is consistent, this is also known to be independent of the other ZF axioms. So unless you are arguing that ZF is inconsistent, I think you are going to have a lot of problems showing that there cannot be a set of all natural numbers. > How do you falsify the existence of the set of all Natural numbers in ZF ? > ZF includes an axiom of infinity, which pretty much directly guarantees that > there is an infinite set of all finite ordinals. Assuming ZF is consistent, > this is also known to be independent of the other ZF axioms. So unless you > are arguing that ZF is inconsistent, I think you are going to have a lot of > problems showing that there cannot be a set of all natural numbers. WM keeps claiming that any system which allows a set of ALL natural number to exist is necessarily inconsistent, though he has not produced anything that qualifies as mathematically or logically valid proof. WM must have skipped plane geometry as a child, and ever since, as he has no notion of what constitutes a mathematically valid or logically valid proof, but argues more like a politician. SignatureVirgil On 29 Mai, 06:44, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote: > On 27 Mai, 23:15, Virgil <virg...@nowhere.com> wrote: > [quoted text clipped - 26 lines] > ZF includes an axiom of infinity, which pretty much directly guarantees that > there is an infinite set of all finite ordinals. What about classical arithmetics with an axiom that sqrt(2) is a rational number? Regards, WM In article <ea1c1694-68e6-44a3-86b6-c33020c8e511@a36g2000yqc.googlegroups.com>, > On 29 Mai, 06:44, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> > wrote: [quoted text clipped - 31 lines] > What about classical arithmetics with an axiom that sqrt(2) is a > rational number? WM's suggested system is provably inconsistent so his suggestion is about as useful as tits on a bull, whereas ZF has no known inconsistencies that are logically derivable without imposing other assumptions. SignatureVirgil >> Grant the existence of two natural numbers m and n such that m/n = sqrt >> (2). Then falsify it. [quoted text clipped - 14 lines] > What about classical arithmetics with an axiom that sqrt(2) is a > rational number? This does not answer my question. But I will answer yours for you. If you create a version of "classical" (= "standard" ?) arithmetic with an axiom that sqrt(2) is rational, it would be inconsistent, and hence useless. Now how about answering my question. How do you falsify the existence of the set of all Natural numbers in ZF, as you claimed you could? > Regards, WM On 30 Mai, 02:56, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote: > >> Grant the existence of two natural numbers m and n such that m/n = sqrt > >> (2). Then falsify it. [quoted text clipped - 20 lines] > "standard" ?) arithmetic with an axiom that sqrt(2) is rational, it would > be > inconsistent, and hence useless. Same with the axiom of infinity: “There exists a complete linear infinite set” is a self-contradictory similar to “there exists a pair of natural numbers, a and b, such that b^2 = 2a^2. > Now how about answering my question. How do you falsify the existence of the > set of all Natural numbers in ZF, as you claimed you could? It is simple: ... classical logic was abstracted from the mathematics of finite sets and their subsets .... Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and finally applied it, without justification, to the mathematics of infinite sets. [Hermann Weyl, "Mathematics and logic: A brief survey serving as a preface to a review of The Philosophy of Bertrand Russell", American Mathematical Monthly 53: 2–13] Show me a complete finite linear set that does not allow for quantifier reversal. En Am: m =< n <==> Am En m =< n [*] . Therefore: Either [*] holds or the set is not complete but allows for extension. Then we have only En Am: m =< n ==> Am En m =< n [**] because not all elements are readily available. That is called a potentially infinite set. But in this case there is no chance to prove uncountability. This had already been recognized by the late Alexander Zenkin, one of the brave scientists who dared to condemn this hypocritical behaviour: Cantor's 'paradise' as well as all modern axiomatic set theory is based on the (self-contradictory) concept of actual infinity. Cantor emphasized plainly and constantly that all transfinite objects of his set theory are based on the actual infinity. Modern AST-people try to persuade us to believe that the AST does not use actual infinity. It is an intentional and blatant lie, since if infinite sets, X and N, are potential, then the uncountability of the continuum becomes unprovable, but without the notorious uncountablity of continuum the modern AST as a whole transforms into a long twaddle about nothing. Resume: The internal contradiction in set theory is veiled by mixing up potential and actual infinity. That is the reason why set theorists usually refuse to specify which infinity they apply. Most even pretend (or profess) not to know the difference. Regards, WM In article <4f20ac0a-a6b7-4a20-8eca-84a63b1f69b6@c19g2000yqc.googlegroups.com>, > > But I will answer yours for you. If you create a version of "classical" (= > > "standard" ?) arithmetic with an axiom that sqrt(2) is rational, it would > [quoted text clipped - 4 lines] > infinite set² is a self-contradictory similar to ³there exists a pair > of natural numbers, a and b, such that b^2 = 2a^2. WM keeps claiming this but never manages to prove it. All of his attempts at proofs require assuming a priori and without proof that no infinite set can exist. > > Now how about answering my question. How do you falsify the existence of > > the [quoted text clipped - 8 lines] > review of The Philosophy of Bertrand Russell", American Mathematical > Monthly 53: 213] So from what cathedra was Weyl speaking? Arguing from citing authorities may work in law courts, but carries little weight mathematics. > Show me a complete finite linear set that does not allow for > quantifier reversal. > En Am: m =< n <==> Am En m =< n [*] . Show me that every infinite well ordered set does allow it. > Therefore: Either [*] holds or the set is not complete but allows for > extension. Or the set is infintie, well-ordered, and has no maximal member. > Then we have only > En Am: m =< n ==> Am En m =< n [**] > because not all elements are readily available. Any set in a sane set theory has all members equally "available". > That is called a potentially infinite set. That is called nonsense in set theories, as is much of WM's MathUnrealism. SignatureVirgil > > Then we have only > > En Am: m =< n ==> Am En m =< n [**] > > because not all elements are readily available. > > Any set in a sane set theory has all members equally "available". Then a complete set with linear order sould have a last element. Regards, WM In article <e705c332-c068-4e0f-ade4-d3c25f33b380@l12g2000yqo.googlegroups.com>, > > > Then we have only > > > En Am: m =< n ==> Am En m =< n [**] [quoted text clipped - 3 lines] > > Then a complete set with linear order sould have a last element. Non sequitur, as usual. There is no reason why having n+1 available whenever n is available requires existence of an n with no n+1. SignatureVirgil > In article > <e705c332-c068-4e0f-ade4-d3c25f33b...@l12g2000yqo.googlegroups.com>, [quoted text clipped - 8 lines] > > Non sequitur, as usual. This is a logical truth obtained from observation of sets --- finite sets of course, because actually infinite sets are not observable. > There is no reason why having n+1 available whenever n is available > requires existence of an n with no n+1. Not for an infinite set that is not complete. But for every complete set. Regards, WM In article <6ef6cae8-13cf-4983-a384-1be43b0cfa50@q16g2000yqg.googlegroups.com>, > > In article > > <e705c332-c068-4e0f-ade4-d3c25f33b...@l12g2000yqo.googlegroups.com>, [quoted text clipped - 11 lines] > This is a logical truth obtained from observation of sets --- finite > sets of course, because actually infinite sets are not observable. At least not observable by WM, though they are as observable to others as large finite sets. > > There is no reason why having n+1 available whenever n is available > > requires existence of an n with no n+1. > > Not for an infinite set that is not complete. But for every complete > set. Then whatever WM's special notion of "completeness" may be in his world of MathUnrealism, it is not universal among sets outside that world. SignatureVirgil > This is a logical truth obtained from observation That is a contradiction. Observation can INSPIRE you to adopt some AXIOMS that will thereAFTER constitute "logical truth", but observation of the physical world is simply irrelevant to logic in general. Logic is about DEFINITIONS, NOT observations. > > > > Any set in a sane set theory has all members equally "available". > > > Then a complete set with linear order should have a last element. "should" is NOT a logical notion. NEITHER is "complete", as YOU use it, SINCE YOU FAIL TO DEFINE IT. It is, moreover, fairly comical that you say "set with linear order" as though that were somehow ONE thing. ANY set can be put into ANY order. ANY ordering on ANY superset of ANY set can be RESTRICTED to a smaller set. The point is that the set and the ordering SIMPLY HAVE NOTHING TO DO WITH each other. If a set is infinite then YOU MAY ORDER IT ANY WAY YOU LIKE. Insisting that the order be linear DOES NOT AFFECT the question of whether there is a FIRST OR a last element! Since the reverse/converse of any linear order IS ALSO A LINEAR order, if it had to have a last element then it would also have to have a first one. The set of the integers HAS NEITHER. > > There is no reason why having n+1 available whenever n is available > > requires existence of an n with no n+1. > > Not for an infinite set that is not complete. But for every complete > set. YOU CAN'T *DEFINE* "complete", MORON!!! If you want to define it AS having a last element, then you will win BY CIRCULARITY! And lose thereby as well! On 30 Mai, 02:56, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote: > >> Grant the existence of two natural numbers m and n such that m/n = sqrt > >> (2). Then falsify it. [quoted text clipped - 22 lines] > > be > inconsistent, and hence useless. Same with the axiom of infinity: “There exists a complete linear infinite set” is a self-contradictory similar to “there exists a pair of natural numbers, a and b, such that b^2 = 2a^2. ***************************** I understand your claim. I was asking fir a proof of it, not you restating it. > Now how about answering my question. How do you falsify the existence of > the > set of all Natural numbers in ZF, as you claimed you could? It is simple: ... classical logic was abstracted from the mathematics of finite sets and their subsets .... Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and finally applied it, without justification, to the mathematics of infinite sets. [Hermann Weyl, "Mathematics and logic: A brief survey serving as a preface to a review of The Philosophy of Bertrand Russell", American Mathematical Monthly 53: 2–13] ******************************* Well, Weyl never claimed or proved your assertion, so starting off with a quote from him is a piss-poor starting point. I assume you have no proof, which is why you are just in random quote mode. Show me a complete finite linear set that does not allow for quantifier reversal. En Am: m =< n <==> Am En m =< n [*] . ************************ Its your proof. Either provide such a set yourself, or prove that none exists. Therefore: Either [*] holds or the set is not complete but allows for extension. ***************************** You better define "extension", and show its relationship to "complete". Just in case, you better define "complete" as well, in case you are using it differently to everybody else. Then we have only En Am: m =< n ==> Am En m =< n [**] because not all elements are readily available. ************************** You better define "readily available" while you are at it. That is called a potentially infinite set. But in this case there is no chance to prove uncountability. ************************* Define "potentially infinite" as opposed to "infinite". Show that this does not allow uncountability to be proved. This had already been recognized by the late Alexander Zenkin, one of the brave scientists who dared to condemn this hypocritical behaviour: Cantor's 'paradise' as well as all modern axiomatic set theory is based on the (self-contradictory) concept of actual infinity. Cantor emphasized plainly and constantly that all transfinite objects of his set theory are based on the actual infinity. Modern AST-people try to persuade us to believe that the AST does not use actual infinity. It is an intentional and blatant lie, since if infinite sets, X and N, are potential, then the uncountability of the continuum becomes unprovable, but without the notorious uncountablity of continuum the modern AST as a whole transforms into a long twaddle about nothing. Resume: The internal contradiction in set theory is veiled by mixing up potential and actual infinity. That is the reason why set theorists usually refuse to specify which infinity they apply. Most even pretend (or profess) not to know the difference. ************************* What *is* the difference? Regards, WM ************************** Gee, such a big claim you make. Yet no proof. Not even a lame attempt. Even cranks generally try and produce bad proofs, you don't even bother trying to do that. You are actually sub-crank. To become merely crank, define some terms, and try and make your rant at least look like a proof. On 31 Mai, 03:12, "Peter Webb" > > How do you falsify the existence of the > > set of all Natural numbers in ZF, as you claimed you could? [quoted text clipped - 9 lines] > > ******************************* > Show me a complete finite linear set that does not allow for > quantifier reversal. [quoted text clipped - 3 lines] > Its your proof. Either provide such a set yourself, or prove that none > exists. [*] belongs to the basic logic of finite sets. As this logic has been obtained by observing the behaviour of finite linear sets, there is no further proof except that a finite linear set violating [*] has never been observed. > Therefore: Either [*] holds or the set is not complete but allows for > extension. [quoted text clipped - 3 lines] > in case, you better define "complete" as well, in case you are using it > differently to everybody else. Complete means that every element of a set exists. In case of a linear set, complete means that also the last element of the linear order does exist. > Then we have only > En Am: m =< n ==> Am En m =< n [**] > because not all elements are readily available. > > ************************** > You better define "readily available" while you are at it. Every element of a complete set is readily available. If not eery element is readily available, then the set is incomplete, that mean not complete. > That is called a potentially infinite set. But in this case there is > no chance to prove uncountability. > > ************************* > Define "potentially infinite" as opposed to "infinite". Show that this does > not allow uncountability to be proved. Potentially infinite is not opposed to infinite but is one kind of infinity. The other one is actually infinite. > This had already been recognized by the late Alexander Zenkin, one of > the brave scientists who dared to condemn this hypocritical [quoted text clipped - 16 lines] > ************************* > What *is* the difference? Potential infinity is possible, actual infinity is not. > ************************** > Gee, such a big claim you make. Yet no proof. Not even a lame attempt. Sorry, you are only unable to understand it. Not everybody is able to understand everthing. That is an odl wisdom. > Even > cranks generally try and produce bad proofs, Yes, set theorists, for instance. But I have shown that they need to use logic of potential infinity to "prove" their actual infinity. Regards, WM I see. So your theory really has two main components. One component is a sort of psuedo-mystical word salad of undefined terms and fragments of philosophy, and the other component is the occasional use of mathematical looking equations and terms. But is it art? On 31 Mai, 16:56, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote: > I see. No, you don't. You intermingle mathematical theorems and logical foundations. Mathematical theorems can be proven using axioms and logic. Logical foundations cannot be proven (how should they? By some pre- logic laws?). Logical foundations can only be obtained from observing the behaviour of sets --- necessarily finite sets, because there are no infinite sets in reality that could be observed. This observation results in a logical law that states: Every linear complete set has a last element. Think a while about that. Then you may come back and tell us the result (not about the truth of what I just said, but about your level of having understood it). Regards, WM > On 31 Mai, 16:56, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> > wrote: [quoted text clipped - 17 lines] > > Regards, WM How do you define "Parallel lines" ? Regards, David Bernier > How do you define "Parallel lines" ? A music album released in 1978 by the new wave rock band Blondie. MoeBlee In article <70da2ace-75b1-46be-8b0a-209e79a0e839@y9g2000yqg.googlegroups.com>, > On 31 Mai, 16:56, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> > wrote: [quoted text clipped - 7 lines] > logic laws?). Logical foundations can only be obtained from observing > the behaviour of sets That could only be the case if one assumes a priori that nothing but sets can exist,and not necessarily then. > This observation results in a logical law that states: Every linear > complete set has a last element. Then that means there are non-complete linear complete sets which, though ordered, do not have last elements. > Think a while about that. Did so, and, as usual, found that WM's arguments do not hold water. SignatureVirgil > Logical foundations cannot be proven (how should they? By some pre- > logic laws?). Logical foundations can only be obtained from observing > the behaviour of sets --- necessarily finite sets, because there are > no infinite sets in reality that could be observed. Most finite sets cannot be observed EITHER, dumbass. They are WAY TOO BIG. In article <eb95bd31-1206-411a-a936-f93a8ec5205e@g37g2000yqn.googlegroups.com>, > On 31 Mai, 03:12, "Peter Webb" > [quoted text clipped - 10 lines] > further proof except that a finite linear set violating [*] has never > been observed. It still requires that which WM is incapable of providing, namely a proof. > > Therefore: Either [*] holds or the set is not complete but allows for > > extension. It still requires that which WM is incapable of providing, namely a proof. > > ***************************** > > You better define "extension", and show its relationship to "complete". Just > > in case, you better define "complete" as well, in case you are using it > > differently to everybody else. > > Complete means that every element of a set exists. Then N is complete. > In case of a linear > set, complete means that also the last element of the linear order > does exist. If that is a part of WM's definition, then WM need to prove that no "incomplete" set can exist, something that he has been failing to do for years, at least without assumes it a priori. . > > Then we have only > > En Am: m =< n ==> Am En m =< n [**] [quoted text clipped - 4 lines] > > Every element of a complete set is readily available. That does not define anything. > If not eery element is readily available, then the set is incomplete, > that mean not complete. > > > > That is called a potentially infinite set. Except that there are no such things. > > ************************* > > Define "potentially infinite" as opposed to "infinite". Show that this does [quoted text clipped - 7 lines] > > usually refuse to specify which infinity they apply. Most even pretend > > (or profess) not to know the difference. Sure they do. One major difference is that otentially infinite sets do not ever exist but actually infinite sets exist in many set theories. > > ************************* > > What *is* the difference? > > > Potential infinity is possible, actual infinity is not. Backwards as usual. > > ************************** > > Gee, such a big claim you make. Yet no proof. Not even a lame attempt. [quoted text clipped - 7 lines] > Yes, set theorists, for instance. But I have shown that they need to > use logic of potential infinity to "prove" their actual infinity. Wm keeps claiming to have "shown" all sorts of things, but until the recipients of his "showings" acknowledge the validity of such "showings", WM has shown nothing but his ignorance to anyone. It was not Wiles "showing" a proof of FLT so much as the rest of the world's acknowledging its validity that was important, since he was nowhere near the first to claim a proof of FLT. When WM can successfully square a circle, trisect an angle and duplicate a cube using only the classical methods, only then will anyone here be liable to accept WM's proofs that all sets are necessarily finite in any and every set theory. SignatureVirgil > The question is whether we could inform someone who does not yet know, > what we understand by the sequence of natural numbers. Could we inform them COMPLETELY? Could we ever FINISH informing them? Wouldn't they just POTENTIALLY informed?? > > But non-static sets do not exist in any mathematical set theory. Of COURSE not. THE PARADIGM is static! IF you want something variable as OPPOSED to static then you CALL that something A VARIABLE or A PROCESS AS OPPOSED to a static constant! The theory is INHERENTLY ABOUT static, constant things! AND some of these static constant things are statically constantly LACKING first and last elements (like the set of all&only the integers, for example). Or at least they CAN be ordered that way -- order is NOT an INHERENT property OF ANY set -- indeed, the fact that ALL possible orders are EQUALLY legitimate -- STATICALLY -- is what MAKES a thing a set! > They do not exist in what is commonly called set theory and what is > eternally false mathematics. Because the paradigm IS STATIC, it does NOT do TIME, EITHER, which makes your calling it "eternally" anything, well, even stupider than usual. The mathematics you are calling eternally false is in fact PERFECTLY capable of dealing with variables and processes. It's just that N *IS NOT ONE* of them. N *is* static, PRECISELY AS your contention that we can finish communicating it clearly SHOWS. > ... classical logic was abstracted from the mathematics of finite sets > and their subsets .... Forgetful of this limited origin, one [quoted text clipped - 19 lines] > > Who introduced the false logical law into set theory? HA-HA-HA! You *idiot*! You attempt to support Weyl with an example that directly undercuts his argument! If classical logic was abstracted from the mathematics of finite sets, then (2) would be true in classical logic. But, as you point out, it ain't, so it wasn't. Lol! What a maroon! -- hz On 26 Mai, 04:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > In article <15af9b47-eac6-4d22-8720-0d6e7ebba...@e21g2000yqb.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > The problem boils down to the following: [quoted text clipped - 7 lines] > The above is not contested: [**] implies > [*]. I said: For complete linear sets [*] is true. You said [*] is not true, but [**] is true. Weyl said: Classical logic was obtained from finite sets. Therefore I asked you: Show me a linear complete finite set, that makes your claim [**] right and my claim [*] wrong. This is so simple that it should be understandable even for someone who is not "very deep in logic". > What is contested is that: > En Am: m =< n <== Am En m =< n [***] > implies [*]. And *that* is the form you do use. No. I do not use the implication only, I use the full equivalence. Of course the equivalence includes the implication [***] as well as the implication [**] There is a trivial finite > counter-example. Take three dice where on each of the sides one of the > numbers one to nine is printed (some of them repeated). Say the set is {d1, [quoted text clipped - 7 lines] > but not > En Am d_m < d_n (there is no best die). Of course there is no best die. Therefore this set is not linear. Every finite set of natural numbers has a "best" number. Why do you bother with such nonsense examples? But the answer is easy: Because you have no other examples. _____________________ > Doesn't it bother you that he gets letters from > other mathematicians in Germany complaining > about it, and that he is proud about that fact? I have never got a letter from a mathematician complaining about that. I would never publish a letter of my private correspondence without consent of the correspondent. My university of applied sciences got a letter from a greasy informer who may be whatever but certainly is not a mathematician. > Do you not think that it might lower the value of other > degrees in Germany as well? But you think it would increase this value if I taught, as you propose, that the sum of all natural numbers can be zero? Or if I taught, contrary to fact, Cantor's claim that a real number is the limit of its finite initial segments but all real numbers are not the limits of all their finite initial segments? Regards, WM > On 26 Mai, 04:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > In article <15af9b47-eac6-4d22-8720-0d6e7ebba...@e21g2000yqb.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: [quoted text clipped - 10 lines] > > I said: For complete linear sets [*] is true. Not in the article to which I responded. > You said [*] is not true, but [**] is true. That is not what I said. I said that for the case involved you have to *prove* that it is true, because it is not generally true. > Weyl said: Classical logic was obtained from finite sets. Right. > Therefore I asked you: Show me a linear complete finite set, that > makes your claim [**] right and my claim [*] wrong. But now you include the word "linear". Where did "Weyl" include the word "linear"? > This is so simple that it should be understandable even for someone > who is not "very deep in logic". [quoted text clipped - 6 lines] > course the equivalence includes the implication [***] as well as the > implication [**] Because the implication [**] is always true, the only part of the equivalence that is new is the implication [***]. > > There is a trivial finite > > counter-example. Take three dice where on each of the sides one of the [quoted text clipped - 14 lines] > Why do you bother with such nonsense examples? > But the answer is easy: Because you have no other examples. Because they are answers to what you actually ask. It proves, in general, that [***] does *not* imply [*], even not in finite cases. So in order to use such implication you have to *prove* it. Now you state that classical logic is derived from the finite case. But in the finite case that implication is not generally available, so it is not part of classical logic, I would say. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <351d0bd7-fe7d-4005-a10a-42b63a903...@q16g2000yqg.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 26 Mai, 04:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 12 lines] > > Not in the article to which I responded. But frequently I made use of what you call quatifier exchange and what is allowed in case of complete linear sets. > > You said [*] is not true, but [**] is true. > > That is not what I said. You said: The only thing that can be stated is (symbolically): E n A m P(m, n) -> A m E n P(m, n) not the reverse, this is just basic logic. > I said that for the case involved you have to > *prove* that it is true, because it is not generally true. It is generally true for complete linear sets. You have to prove that it is not. > > Weyl said: Classical logic was obtained from finite sets. > [quoted text clipped - 5 lines] > But now you include the word "linear". Where did "Weyl" include the > word "linear"? I did never claim that quantifier exchange is allowed in case of non- linear sets, like cyclic sets as, for instance, your dice. That would be nonsense. A simple example: Every country has a country that lies west of it. But there is no country that lies west of all countries. > > This is so simple that it should be understandable even for someone > > who is not "very deep in logic". [quoted text clipped - 9 lines] > Because the implication [**] is always true, the only part of the equivalence > that is new is the implication [***]. For complete linear sets both are true, therefore [*] holds. And you should recognize that actually complete linear sets obey [*]. Only potentially infinite sets do not. But you mix up things. You claim the existence of a complete linear set but disregard the necessary consequence of completeness or linearity, namely the validity of [*]. Regards, WM I have seen MatheRealism being discussed in other threads. As I don't want to get involved in too many threads, I answer here: -possums.net> wrote: > > The "is" of "there is" is not the exists() predicate, so (I suppose!) > > if you could formalise this it would not be a contradiction "as such". That is correct! > As I recall, he actually used the word "exist" for both aspects: both > that the number exists and that it cannot exist. He may well have > intended different meanings for each use without clarification or > distinction. Due to the lack of tools for representing numbers with large information contents (larger than 2^80 bits, or 2^365 bits, or 2^X bits, where X is a number that may depend on the progress of physics but in any case is finite) we must accept that numbers with larger information contents (that canot be reduced) do not exist. That means, there exist numbers (namely according to current mathematics, which calls itself realism but is simply a form of idealism) that do not exist according to fact. "to exist" is used here in two different meanings such that "there exist(1) numbers that do not exist(2)" is not a self-contradiction. Regards, WM In article <5f791699-355d-41db-80ec-208739da808f@s20g2000vbp.googlegroups.com>, > I have seen MatheRealism being discussed in other threads. As I don't > want to get involved in too many threads, I answer here: [quoted text clipped - 16 lines] > but in any case is finite) we must accept that numbers with larger > information contents (that canot be reduced) do not exist. WM may have to accept if, but o one else need do so. WM conflates our ability to name a number with its existence. There is nothing inconsistent or illogical in the existence of things for which we do not have names. > That means, there exist numbers (namely according to current > mathematics, which calls itself realism but is simply a form of > idealism) that do not exist according to fact. It only means that there are numbers for which specific forms of names do not exist, but that is only a fault in those specific naming mechanisms. > "to exist" is used here in two different meanings such that "there > exist(1) numbers that do not exist(2)" is not a self-contradiction. It is a contradiction in English. It probably is a contradiction in the German of scholars. But WM would not know about that. SignatureVirgil > In article > <5f791699-355d-41db-80ec-208739da8...@s20g2000vbp.googlegroups.com>, [quoted text clipped - 24 lines] > There is nothing inconsistent or illogical in the existence of things > for which we do not have names. Unless these things are names only. > > That means, there exist numbers (namely according to current > > mathematics, which calls itself realism but is simply a form of [quoted text clipped - 3 lines] > do not exist, but that is only a fault in those specific naming > mechanisms. Replace "do not exist" by "cannot exist" and you see that such numbers cannot be used for enumerating anything, hence are not numbers. > > "to exist" is used here in two different meanings such that "there > > exist(1) numbers that do not exist(2)" is not a self-contradiction. > > It is a contradiction in English. It probably is a contradiction in the > German of scholars. But WM would not know about that. In philosophical texts you can find frequently words with different meanings that are distinguished by numbers as I did. To be(1) or to be (2) or to be(14) that is the question. But you seem not to be(0) aware of such scholarly texts. Regards, WM In article <6590202d-598a-41fd-acde-d000fddd2db5@x6g2000vbg.googlegroups.com>, > > In article > > <5f791699-355d-41db-80ec-208739da8...@s20g2000vbp.googlegroups.com>, [quoted text clipped - 26 lines] > > Unless these things are names only. Then different names, being names only, would necessarily be different things and then the name 1/2 and the name 2/4, being different things, could not be equal. > > > That means, there exist numbers (namely according to current > > > mathematics, which calls itself realism but is simply a form of [quoted text clipped - 6 lines] > Replace "do not exist" by "cannot exist" and you see that such numbers > cannot be used for enumerating anything, hence are not numbers. Only cardinals can ennumerate anything, so WM would have us eliminate all rationals, ordinals, algebraics, etc. > > > "to exist" is used here in two different meanings such that "there > > > exist(1) numbers that do not exist(2)" is not a self-contradiction. [quoted text clipped - 6 lines] > (2) or to be(14) that is the question. But you seem not to be(0) aware > of such scholarly texts. If WM had indexed those usages of "exist" in his original claim, he might be able to justify his present claim, but he didn't and thus can't. SignatureVirgil > > > There is nothing inconsistent or illogical in the existence of things > > > for which we do not have names. [quoted text clipped - 4 lines] > things and then the name 1/2 and the name 2/4, being different things, > could not be equal. Why not? There are laws, i.e., other abstract things, that allow to find out whether two names are the same or not. There are many names identical to the name pi. > > > > "to exist" is used here in two different meanings such that "there > > > > exist(1) numbers that do not exist(2)" is not a self-contradiction. [quoted text clipped - 9 lines] > If WM had indexed those usages of "exist" in his original claim, he > might be able to justify his present claim, but he didn't and thus can't. If I had written a philosophical text, I would have explained this, and in fact I did it in many of my contributions here, but nobody can expect that I do so in every contribution. A minimum of brainwork is required to understand my texts, but really, it is a very small amount. Regards, WM In article <f8cdb05d-9469-4247-9c76-d431508ea9d2@3g2000yqk.googlegroups.com>, > > > > There is nothing inconsistent or illogical in the existence of things > > > > for which we do not have names. [quoted text clipped - 8 lines] > find out whether two names are the same or not. There are many names > identical to the name pi. False. There is only one *name* that is identical to the name "pi", and that is "pi" itself, but there are lots of names which name the same number that "pi" names. WM still does not understand the differnce between a name and the thing named. That is only one of his major errors in logical comprehension. > A minimum of brainwork is > required to understand my texts, but really, it is a very small > amount. Actually, there is a major absence of brain work that is required to accept WM's theses. SignatureVirgil In article <37a98f77-6bc9-4cae-88e5-fcbadd1d0a75@q2g2000vbr.googlegroups.com>, > > Not in the article to which I responded. > > But frequently I made use of what you call quatifier exchange and what > is allowed in case of complete linear sets. You must prove it is allowed for the special case of what you call "complete linear sets" as it is invalid in general. > > > You said [*] is not true, but [**] is true. > > [quoted text clipped - 9 lines] > > It is generally true for complete linear sets. Not until you prove it to be so, which you have not done. You have not even given an adequate definition of what you mean by "complete linear sets", unless you mean only sets for which what you claim is true. > > > Weyl said: Classical logic was obtained from finite sets. > > [quoted text clipped - 10 lines] > be nonsense. A simple example: Every country has a country that lies > west of it. But there is no country that lies west of all countries. If you concede that something does not hold for arbitrary cases and you claim it does holds for certain special cases, then it is your obligation to prove it for those special cases. Which anyone competent in pure mathematics would have known, but which such professors of impure math as WM does not know. > For complete linear sets both are true Often claimed but never proven (at least by WM). SignatureVirgil ... > > > > > The problem boils down to the following: > > > > > [quoted text clipped - 12 lines] > But frequently I made use of what you call quatifier exchange and what > is allowed in case of complete linear sets. You think so, but you have to prove that it is valid for infinite complete linear sets. Note that "classical logic is obtained from finite sets". Nowhere in that quote the word linear is mentioned. > > > You said [*] is not true, but [**] is true. > > [quoted text clipped - 4 lines] > E n A m P(m, n) -> A m E n P(m, n) > not the reverse, this is just basic logic. The reverse of E n A m P(m, n) -> A m E n P(m, n) is E n A m P(m, n) <- A m E n P(m, n) which is [***], neither [*] nor [**]. > > I said that for the case involved you have to > > *prove* that it is true, because it is not generally true. > > It is generally true for complete linear sets. You have to prove that > it is not. It is not true for the infinite set of naturals. (1) define FISON(n) be the set of naturals from 1 to n, that is: {1, ..., n}. (2) A{m in N} E(n in N} such that FISON(m) subset FISON(n), trivial, take n = m + 1. (3) E{n in N} A{m in N} such that FISON(m) subset FISON(n), trivially false, take m = n + 1. Which part of this proof is wrong? It clearly shows that E n A m P(m, n) <- A m En P(m, n) is false. Here with: E n meaning E{n in N} A n meaning A{n in N} P(m, n) meaning FISON(m) subset FISON(n). The first part of the implication is false while the second part of the implication is true, and so the implication is false (all by classical logic). > > But now you include the word "linear". Where did "Weyl" include the > > word "linear"? [quoted text clipped - 3 lines] > be nonsense. A simple example: Every country has a country that lies > west of it. But there is no country that lies west of all countries. But as Weyl did not include "linear" in his words, how can that quote support your claim? > > > > What is contested is that: > > > > En Am: m =< n <== Am En m =< n [***] > > > > implies [*]. And *that* is the form you do use. ... > > Because the implication [**] is always true, the only part of the > > equivalence that is new is the implication [***]. > > For complete linear sets both are true, therefore [*] holds. You just state without proof. Where in my proof above that it is false did I go wrong? > And you > should recognize that actually complete linear sets obey [*]. Strange, I give above a proof that it does not hold. I did not use "actual infinity" nor "potentially infinity", only logic and the axioms of ZF. > Only > potentially infinite sets do not. But you mix up things. You claim the > existence of a complete linear set but disregard the necessary > consequence of completeness or linearity, namely the validity of [*]. Why is that a necessary consequence? Because you want it to be so? Can you give a *mathematical* reason? Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > Why is that a necessary consequence? Because you want it to be so? Can > you give a *mathematical* reason? Oh, you REALLY can't stop "arguing" with WM, right? Is this some sort of compulsive act? Herb >> Why is that a necessary consequence? Because you want it to be so? Can >> you give a *mathematical* reason? > > Oh, you REALLY can't stop "arguing" with WM, right? Is this some sort of > compulsive act? Why don't you mind your own business? What compels you to these confrontational posts? Just a natural a.shole? SignatureJesse F. Hughes "Wiles made somewhere around half a million dollars U.S. that I heard about, and I know he didn't take major endorsements." --JSH on the rewards of proving Fermat's last theorem. > > Why is that a necessary consequence? Because you want it to be so? Can > > you give a *mathematical* reason? > > Oh, you REALLY can't stop "arguing" with WM, right? Is this some sort of > compulsive act? I think it is justified if my name is in the Subject. But if you do not like it, please kilfile me. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > I think it is justified if my name is in the Subject. Yes, you are right. But I guess you see now the dependence between a crank and its codependent partners. The crank (or at least this one) is craving for attention. If you stop responding he is suffering (from this lack of attention). [Imho one of the MAIN reasons why this loon stil resides in this NG is the _constant_ attention he gets from _certain_ posters...] > But if you do not like it, please killfile me. Hell, no! You are a competent and intelligent participant in this NG. It would be a shame to ignore your (more intelligent) contributions. [Though this does in no way relativize the regrettable effects of your codependent behavior concerning WM, if so.] Herb > > I think it is justified if my name is in the Subject. > [quoted text clipped - 3 lines] > attention). [Imho one of the MAIN reasons why this loon stil resides in > this NG is the _constant_ attention he gets from _certain_ posters...] I do not think so. WM has now two books about mathematics on his name. The first one he got published some time ago by some vanity press (Shaker Verlag). (It is vanity press because you have to pay for your book to get published, for what you pay you get a number of copies, everything further is handled by print-on-demand.) I did a review of it that is available through my web-pages. You can find it at: <http://homepages.cwi.nl/~dik/english/mathematics/mueck/index.html> The second (recently published, 2008 I think) is however by a reputable publisher. It is intended as textbook for students, and used by WM at his university. You may expect an influx of German students that think what WM writes is gospel. There are already some around here: Albrecht (if I remember well he is connected to a major German technical institution) and also Eckard Blumschein (emiritus professor of the Magdeburg University in some technical subject). Han de Bruijn was also one of them, but recently he seems to have changed opinion. What i find is a growing clash between technical physists and mathematicians. The technical physists do not understand what mathematicians are doing, and think they are subordinate to their needs, but on the other hand do not understand numerical mathematics either (which caters for their needs). They just think that numbers are absolute things. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ >> ... you are right. But I guess you see now the dependence between a crank >> and its codependent partners. The crank (or at least this one) is craving [quoted text clipped - 3 lines] >> > I do not think so. [...] Of course not, since you are suffering from co-dependence. *sigh* Anyway, WM was succesfully driven out from de.sci.mathematic since no one was willing to "argue" with him the way _certain_ poster in this NG do (for such a long time). EOD Herb > The second (recently published, 2008 I think) is however by a reputable > publisher. It is intended as textbook for students, and used by WM at his > university. That is horrible news indeed. WM must be the most successful crank I've ever seen. He *teaches* and *publishes* his delusions! How remarkable and how regrettable. (Yes, Abian was well-respected as a mathematician, but did not teach or publish his cranky "discoveries", as far as I know.) So who the hell published a WM textbook? SignatureJesse F. Hughes "Just goes to tell you. If you make a major discovery, and some stupid interviewer asks you if you're the greatest mathematician of all time, just say no." -- practical advice from James S. Harris > > The second (recently published, 2008 I think) is however by a reputable > > publisher. It is intended as textbook for students, and used by WM at his > > university. > > So who the hell published a WM textbook? Not so difficult to find. Oldenbourg Wissenschaftsverlag. The title is "Mathematik für die ersten Semester". Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Mueckenheim is also cited as an authority on infinity at this educational site: http://www.learner.org/courses/mathilluminated/units/3/resources/index.php I didn't see any evidence that WM's horrid arguments actually influenced the text of the site. Nonetheless, students who want to learn more about Cantor are directed to WM's illogical blatherings. Still, I blame most WM's employer for putting him in a position of authority to educate students on exactly that material he has shown no capacity to understand. I can't comprehend how that situation has remained. I'm sure that WM is tenured, but that doesn't entail that he can teach bad mathematical reasoning in the classroom, does it? Signature"It's one of the easiest tickets to true fame--not this silly stuff where people cheer you for a few years and then forget about you--but the kind of fame where school kids have to read your biography and do reports on you." -- Another reason to support James S. Harris. > Mueckenheim is also cited as an authority on infinity at this > educational site: > > http://www.learner.org/courses/mathilluminated/units/3/resources/index.php > I didn't see any evidence that WM's horrid arguments actually > influenced the text of the site. Nonetheless, students who want to > learn more about Cantor are directed to WM's illogical blatherings. Did you consider writing to a person on the advisory board http://www.learner.org/courses/mathilluminated/about/advisors.php about this? That might have more effect than another round of discussion with Mueckenheim in this group (but there could be an exercise in the main text asking to explain why those papers are scrap) > Still, I blame most WM's employer for putting him in a position of > authority to educate students on exactly that material he has shown no > capacity to understand. I can't comprehend how that situation has > remained. I'm sure that WM is tenured, but that doesn't entail that > he can teach bad mathematical reasoning in the classroom, does it? That is a school for engineers and the like, not for science or mathematics majors. Mueckenheim obviously came there as a physicist, and what he does is to teach physics and mathematics to engineering students. I recently had a short look into his textbook, and while the preface is kind of crazy, the body of the book makes the impression of being fairly standard, on a first sight. Moreover, the students at that school seem to be obliged to take some courses of a general interst nature which they can select according to personal preference, and Mueckenheim's lecture on the "history of the infinite" is among these courses. As it is a state school one could send a formal complaint ("Dienstaufsichtsbeschwerde") to that school or the ministry responsible for the state school system, of course in German language, and they would have to answer it. If Mueckenheim gets to know this then he will blather around that he had been denounced like Zermelo who had been driven out of Freiburg university because he didn't perform the "Hitler greeting" well enough. A couple of years ago I stumbled in a public library across this book http://www.amazon.de/Götzen-Computer-Kritik-unreinen-Vernunft/dp/3879592942/ref =sr_1_7?ie=UTF8&s=books&qid=1243715629&sr=1-7 "Gödel, Götzen und Computer", by Max Woitschach. The author was primarily working for the german branch of IBM, and as a secondary job he lectured at a school like Mueckenheim's. And that book contained his collected misunderstandings (different from Mueckenheim's) about mathematics. So Mueckenheim has predecessors. Woitschach deceased in 1993, and now there is an endowment for "ideology-free science" with an annual award, see http://www.woitschach-stiftung.de/ Ralf > Mueckenheim is also cited as an authority on infinity at this > educational site: [quoted text clipped - 3 lines] > I didn't see any evidence that WM's horrid arguments actually > influenced the text of the site. You seem to have overlooked: “His polarizing results generated much controversy that, to this day, is not completely resolved.” > Nonetheless, students who want to > learn more about Cantor are directed to WM's illogical blatherings. Illogical blatherings you may find when reading Fools Of Mathematics or something like that. > Still, I blame most WM's employer for putting him in a position of > authority to educate students on exactly that material he has shown no > capacity to understand. Wow, do you really belong to that small elite group of scholars who understand cardinal and ordinal exponentiation? How did you manage that task? Certainly you also understand the ordinary mistakes of set theory. (Only the cardinal mistakes may have escaped you.) How can someone believe that set theory is too difficult to understand for an average intelligence? Are you so proud to have managed it that you have lost all measure? > I can't comprehend how that situation has > remained. I'm sure that WM is tenured, but that doesn't entail that > he can teach bad mathematical reasoning in the classroom, does it? Therefore I don’t do so, but teach good mathematics, namely mathematics that is free of confusing the different kinds of infinity and free of the due silly results. Regards, WM > How can someone believe that set theory is too difficult to understand > for an average intelligence? Are you so proud to have managed it that > you have lost all measure? On the contrary, I *don't* believe that Cantor's theorem is too difficult for the average person, given appropriate background knowledge and skills. For this reason, I find you exceptional, not average. SignatureJesse F. Hughes "If you are a consumer that's taking advantage of the technologies that exist ... then the spam problem for you is solved." --MS spokesman verifying that the spam problem has been solved. > On the contrary, I *don't* believe that Cantor's theorem is too > difficult for the average person, given appropriate background > knowledge and skills. This should have been followed by "You're just REALLY Stupid". >> On the contrary, I *don't* believe that Cantor's theorem is too >> difficult for the average person, given appropriate background >> knowledge and skills. > > This should have been followed by > "You're just REALLY Stupid". I more or less said that in the part you snipped. "For this reason, I find [WM] exceptional, not average." SignatureJesse F. Hughes Baba: Spell checkers are bad. Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D. > > This should have been followed by > > "You're just REALLY Stupid". > I more or less said that in the part you snipped. > > "For this reason, I find [WM] exceptional, not average." Well, more less than more, but sometimes, less is more. In article <6f02bd25-bb41-45f7-b2dc-7010c10260eb@s12g2000yqi.googlegroups.com>, > > Mueckenheim is also cited as an authority on infinity at this > > educational site: [quoted text clipped - 20 lines] > Wow, do you really belong to that small elite group of scholars who > understand cardinal and ordinal exponentiation? One does not have to understand all of it to undersatnd that it is both valid mathematics and often quite interesting, whereas what WM is trying to sell is neither. > How did you manage > that task? By not taking WM's claims on faith. At least one can spot many of the errors of WM's perverted version of set theory. > How can someone believe that set theory is too difficult to > understand for an average intelligence? WM is a true believer in his faith (see Eric Hoffer on true believers) and is untouchable by any fact or logical argument not agreeing with that faith. But WM's claims require that names of numbers and the numbers they name are identical, which as anyone can plainly see thy are not. > Are you so proud to have managed it that you have lost all measure? Mathematical measure theory requires infinite sets, so it is WM who has lost all measure by denying its basis. > > I can't comprehend how that situation has > > remained. I'm sure that WM is tenured, but that doesn't entail [quoted text clipped - 6 lines] > > Regards, WM SignatureVirgil > Therefore I don’t do so, but teach good mathematics, namely > mathematics that is free of confusing the different kinds of infinity > and free of the due silly results. If you were posting this crap in German, you would've been fired long ago. George Greene a écrit : >> Therefore I don’t do so, but teach good mathematics, namely >> mathematics that is free of confusing the different kinds of infinity >> and free of the due silly results. > > If you were posting this crap in German, you would've > been fired long ago. If there were mathematicians in this newsgroup with the JSH mentality, they would have tranlated it and send it to Germans "authorities" long ago. The logical conclusion is that true mathematicians are (relatively) nice people. On the other hand, the damage done to some young Germans should have been quite noticeable by now... > > Therefore I don=92t do so, but teach good mathematics, namely > > mathematics that is free of confusing the different kinds of infinity > > and free of the due silly results. > > If you were posting this crap in German, you would've > been fired long ago. You apparently do not understand it at all. WM has posted this stuff for years in the German newsgroup on mathematics, in German. Moreover, he has two books about mathematics on his name, in German. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <db512ef4-fbff-48ee-af30-ed25dc9b7...@s16g2000vbp.googlegroups.com> George Greene <gree...@email.unc.edu> writes: > > On Jun 1, 6:38=A0am, WM <mueck...@rz.fh-augsburg.de> wrote: [quoted text clipped - 8 lines] > years in the German newsgroup on mathematics, in German. Moreover, he > has two books about mathematics on his name, in German. And both say, in effect, about the existence of actual infinity, what Kant said about the proof of the existence of God: These assumptions (proof of God, axiom of infinity) are as ridiculous as a merchant who would try to improve his balance by adding some zeros behind his result. I would be glad if a court would have to decide about my position based on the binary tree and my further arguments. Kant, at his time, had to stop his lectures about religious topics by order of his Prussian King. But I am sure that all this Cantor-nonsense would burst like a bubble filled with foul gas once the public got to know what crap has to be paid by the taxpayer. Regards, WM Regards, WM > > You apparently do not understand it at all. WM has posted this stuff for > > years in the German newsgroup on mathematics, in German. Moreover, he > > has two books about mathematics on his name, in German. > And both say, in effect, about the existence of actual infinity, what > Kant said about the proof of the existence of God: These assumptions > (proof of God, axiom of infinity) are as ridiculous as a merchant who > would try to improve his balance by adding some zeros behind his > result. That is not something that can legitimately be said in any mathematical context. In the first place, WE DON'T prove that infinity exists. WE JUST HAVE AN AXIOM of infinity. YOU CAN USE THE DENIAL OF THAT AXIOM and talk about A FINITE universe IF YOU LIKE! NOTHING IS STOPPING you! Except maybe for the fact that IF you allow EVERYthing to have a successor that is distinct FROM ALL SMALLER things, then THERE FACTUALLY ARE an infinite number of things and that assuming otherwise LEADS TO CONTRADICTIONS, EVEN IN YOUR impoverished logic. > I would be glad if a court would have to decide about my position > based on the binary tree and my further arguments. It has. You've been convicted. Unanimously. > Kant, at his time, > had to stop his lectures about religious topics by order of his > Prussian King. But I am sure that all this Cantor-nonsense would burst > like a bubble filled with foul gas once the public got to know what > crap has to be paid by the taxpayer. The public doesn't know sh.t about math, and if you are at a public university and taxpayers are paying YOUR salary, well, be careful what you wish for. Come the day. In article <103a00da-91c5-47fc-b40e-79f3303930bc@e24g2000vbe.googlegroups.com>, > > In article > > <db512ef4-fbff-48ee-af30-ed25dc9b7...@s16g2000vbp.googlegroups.com> George [quoted text clipped - 16 lines] > would try to improve his balance by adding some zeros behind his > result. If I recall correctly, Kant also claimed that only Eulidean geometry could be consistent. So perhaps one ought not rely on Kant's opinions about mathematics. > I would be glad if a court would have to decide about my position > based on the binary tree and my further arguments. Anyone who would submit the validity of a mathematical argument to a court of law, knows very little about mathematics and very little about the law. > Kant, at his time, > had to stop his lectures about religious topics by order of his > Prussian King. But I am sure that all this Cantor-nonsense would burst > like a bubble filled with foul gas once the public got to know what > crap has to be paid by the taxpayer. If the mathematicians of Germany knew what anti-mathematical nonsense WM was pouring into his students, they might very well act to burst that bubble filled with even fouler gas. SignatureVirgil Martin Musatov wrote: > In article > <103a00da-91c5-47fc-b40e-79f3303930bc@e24g2000vbe.googlegroups.com>, [quoted text clipped - 44 lines] > -- > Virgil I tell you the truth, I do not know Dik T. Winter. Now remove him from my sight. ++ Martin Musatov > > I would be glad if a court would have to decide about my position > > based on the binary tree and my further arguments. > > Anyone who would submit the validity of a mathematical argument to a > court of law, knows very little about mathematics and very little about > the law. My remark did not concern mathematics but only transfinite set theory. Anybody who has seen how the advocates of transfinite set theory react to counter proofs, of mine and of others, which are easily understood and accepted by anybody except those advocates of transfinite set theory --- anybody who has seen that knows that only judges and psychiatrists can help. Regards, WM In article <50b6dece-a0a1-4dc4-be24-c6474d1111b7@h23g2000vbc.googlegroups.com>, > > > I would be glad if a court would have to decide about my position > > > based on the binary tree and my further arguments. [quoted text clipped - 4 lines] > > My remark did not concern mathematics but only transfinite set theory. Which is, despite WM's petty objections, a well recognized part of mathematics. > Anybody who has seen how the advocates of transfinite set theory react > to counter proofs, of mine and of others, which are easily understood > and accepted by anybody except those advocates of transfinite set > theory I do not speak to everyone's objections to transfinite set theories, but to WM's, and both I and others note that WM's alleged counterproofs fail on purely logical grounds, as he reveals himself to be both logically and mathematically incompetent, being unable either to analyse or create logical arguments. There are objectors to infinite sets, like the supporters of IST, whose case is at last presented logically. Wm would do well to learn to emulate them if he ever wishes to impose his views on anyone other than is captive students. > --- anybody who has seen that knows that only judges and > psychiatrists can help. Psychiatry might help WM get his act together, but judges at law would have no more right to impose their judgements on mathematics than mathematicians would have to impose their judgements on the law. Since WM brings up the issue of psychiatry, perhaps his unconscious is aware of his need for its help. SignatureVirgil > I would be glad if a court would have to decide about my position > based on the binary tree and my further arguments. Kant, at his time, > had to stop his lectures about religious topics by order of his > Prussian King. But I am sure that all this Cantor-nonsense would burst > like a bubble filled with foul gas once the public got to know what > crap has to be paid by the taxpayer. Do we have any evidence that JSH and WM are different people? Have they ever been photographed together? Marshall j/k > Do we have any evidence that JSH and WM are different > people? Have they ever been photographed together? Yes they have been. See: http://www.ltn.lv/~podnieks/slides/goedel/Goedel_Einstein.jpg MoeBlee ... > > You apparently do not understand it at all. WM has posted this stuff for > > years in the German newsgroup on mathematics, in German. Moreover, he [quoted text clipped - 5 lines] > would try to improve his balance by adding some zeros behind his > result. And you are deluded. An axiom is a statement of something that can not be proven, neither disproven using the remainder of the theory. So comparing the "axiom of infinity" with a "proof of God" is pretty stupid. In mathematics a theory depends on the axioms used. There is nothing sacred about the axioms, but as long as you are discussing a theory you should use the axioms of that theory. So, if you are discussion Eucliedan geometry you should use the parallel axiom. Of course you can reject it, but in that case you are not discussing Euclidean geometry but something else. > But I am sure that all this Cantor-nonsense would burst > like a bubble filled with foul gas once the public got to know what > crap has to be paid by the taxpayer. Are you mimicking JSH on purpose? In that case do not forget your uprising as the best mathematician, hand-shaking with rulers of the world and raking in large amounts of dollars. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ By classical logic "Cantor" means "can't or" it is a recursive negative argument built on a cornerstone of limitation. It is a non productive axiom. Same as "Boolean" or "Boo-lean". The purpose of Boolean algebra is to scare or push a truly logical mind from a truly lean and defined set, and it does this by presenting a seemingly sound but bloated statement of limitation. It leans toward limitation instead of governing by an open statement. An open statement is one open to unknown (or "uncountable") possibilities but nonetheless firmly grounded in logic. "Elepants can be colored." is an example of an open statement. Reality: elephants are mostly grey or brown. If you painted an elephant green it could exist. But the notion "seems" absurd so it is blindly dismissed. > > On 3 Jun., 04:37, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > ... [quoted text clipped - 28 lines] > dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <103a00da-91c5-47fc-b40e-79f330393...@e24g2000vbe.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 3 Jun., 04:37, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 11 lines] > And you are deluded. An axiom is a statement of something that can not be > proven, neither disproven using the remainder of the theory. The axiom can be contradicted. Simple example: The axiom could be: The binary tree has uncountably many paths. I show that the end of each path p of the set P can be mapped on a node, and that all paths p of P cover all nodes of the tree. Therefore, after having completed the covering of the whole tree, there remains no node that could be used to construct a path that does not belong to P. This disproves the mentioned axiom. > So comparing > the "axiom of infinity" with a "proof of God" is pretty stupid. In > mathematics a theory depends on the axioms used. There is nothing sacred > about the axioms, but as long as you are discussing a theory you should > use the axioms of that theory. That is same as with proofs of God. The Vatican published an axiom (they call it a dogma, but it is of the same meaning) according to which it is possible to prove the existence of God. As long as you are discussing catholicism, you should use it, i.e., believe in that axiom. > So, if you are discussion Eucliedan > geometry you should use the parallel axiom. Of course you can reject it, > but in that case you are not discussing Euclidean geometry but something > else. That means, you are willing to believe in what the Vatican says? Probably they think that the Dutch better should have stayed within the Spanish Empire. Regards, WM In article <74f33c12-95fe-4b83-a7e3-941591defd34@c9g2000yqm.googlegroups.com>, > > In article > > <103a00da-91c5-47fc-b40e-79f330393...@e24g2000vbe.googlegroups.com> [quoted text clipped - 15 lines] > The binary tree has uncountably many paths. I show that the end of > each path p of the set P can be mapped on a node, In a maximal infinite binary tree a path can have only one "end" and all paths have the same end: the root node. But since one of WM's often hidden assumptions is that there re no infinite sets, in his world there is no such tree at all. > and that all paths p of P cover all nodes of the tree. Therefore, > after having completed the covering of the whole tree Which in WM's world is guaranteed to be finite anyway, > there remains no node that could be used to construct a path that > does not belong to P. That presumes that through each node passes only a less than uncountable infinity of paths, which is false in any system, such as ZF, in which a maximal infinite binary tree can exist. > This disproves the mentioned axiom. Not in ZF it doesn't. Or, at least, not without sneaking in some of WM's hidden assumptions it doesn't. > > So comparing the "axiom of infinity" with a "proof of God" is > > pretty stupid. In mathematics a theory depends on the axioms used. [quoted text clipped - 6 lines] > are discussing catholicism, you should use it, i.e., believe in that > axiom. Since mathematics is about what follows from a given system of axioms or assumptions without regard to whether those axioms or assumptions themselves are in any sense true, if one ACCEPTS, for purposes of argument, that Catholic axiom, one can see whether the proof their God is valid. That is where WM fails to see what axiomatic mathematics is about. It does not CARE whether an axiom in some system is true, but only cares what follows from assuming the whole axiom system. But the notion that WM has access to THE TRUTH, when he has such a profound lack of understanding of logic is ridiculous. > > So, if you are discussion Eucliedan geometry you should use the > > parallel axiom. Of course you can reject it, but in that case you > > are not discussing Euclidean geometry but something else. > > That means, you are willing to believe in what the Vatican says? Not at all, but the difference between what can be deduced from a set of axioms and whether the set of axioms represent what is actually true are quite different problems, neither of which WM is competent to discuss. > Probably they think that the Dutch better should have stayed within > the Spanish Empire. Possible for the Spanish, but almost certainly not for the Dutch. SignatureVirgil > In a maximal infinite binary tree a path can have only one "end" and > all paths have the same end: the root node. That is the beginning. If you dislike the name "end", then speak of X = "path without its first n nodes". There are infinitely many nodes mapped on every X, namely those which belong to that X. The X exhaust the complete binary tree, because there is no node that remains unmapped. The X of every real number you wish, including 1/3, 1/sqrt(2), 1/pi and so on, is among all paths constructed. Regards, WM In article <00852fd6-5479-4b70-9233-67f85b6e5961@o20g2000vbh.googlegroups.com>, > > In a maximal infinite binary tree a path can have only one "end" > > and all paths have the same end: the root node. > > That is the beginning. It is the front end, so that it is an end. > If you dislike the name "end", then speak of X = "path without its > first n nodes". I prefer to call what is left after removing the head of a path, its tail. > There are infinitely many nodes mapped on every X, And uncountably many X through every node. > namely those which > belong to that X. The X exhaust the complete binary tree, because > there is no node that remains unmapped. The X of every real number > you wish, including 1/3, 1/sqrt(2), 1/pi and so on, is among all > paths constructed. Not so, as sequential construction, a la WM, doesn't cut it. > Regards, WM SignatureVirgil > I prefer to call what is left after removing the head of a path, its > tail. No objection. X is a tail. > > There are infinitely many nodes mapped on every X, > > And uncountably many X through every node. Let us choose one of them and map it one that node. For the others we will find other nodes, below that one chosen. Then we have one tail mapped on one node, and therefore, one path mapped on one node. This procedure uses up a countable number of paths as it exhausts the set of nodes. Therefore there is no possibility to construct any further tail. All nodes are already used up. And, what is more important, also all combinations of nodes are used up. Regards, WM > > And uncountably many X through every node. > Let us choose one of them and map it one that node. For the others we > will find other nodes, below that one chosen. Nope. You cannot find enough other nodes. There are uncountably many paths going through the node and only countably many nodes below it. - William Hughes In article <dec4d13d-b292-4b86-9965-16c8fe0492af@n4g2000vba.googlegroups.com>, > > > And uncountably many X through every node. > [quoted text clipped - 6 lines] > > - William Hughes Does WM's "below" mean nearer to the root or futher from it? If "below it" mean closer to the root node, then there are only FINITELY many nodes "below" any node. SignatureVirgil > Does WM's "below" mean nearer to the root or futher from it? "Below" means further from the root. Regards, WM > > > And uncountably many X through every node. > > Let us choose one of them and map it one that node. For the others we [quoted text clipped - 3 lines] > paths > going through the node and only countably many nodes below it. That's just the proof saying why Cantor's proof is wrong! Cantor (and his disciples) assume that the complete diagonal of the famous list is the limit of an infinite process. But he and they deny that the complete binary tree is the limit of an infinite process. Cantor spelled that out in a letter to Letter to Vivanti of Dec. 3 1885: Introducing the expression "path" that Cantor did not use but mean, he said: "This apparent difficulty is solved as follows: The set of all "path" is not the limit of all "finite path" z_n for n = oo." This is an inconsistency. Either there is no limit at all (then Cantor's proof is wrong, because the list cannot be checked completely) or the limit holds also in case of the binary tree. Regards, WM In article <ae267554-bb7b-4185-8acf-7b034e268def@g19g2000yql.googlegroups.com>, > > > > And uncountably many X through every node. > > > Let us choose one of them and map it one that node. For the others we [quoted text clipped - 9 lines] > the > limit of an infinite process. That is the immediate result of a definition. That it may be produced as a limit is irrelevant. > But he and they deny that the complete > binary tree is the limit of an infinite process. Irrelevant. The union of a set of infinitely many sets is a set regardless. And if infinitely many "complete" finite binary trees are regarded as sets of nodes, their union is a set of nodes in which every maximal totally ordered under "ancestor" set of nodes is a path and there are uncountably many such paths. > Cantor spelled that > out in a letter to Letter to Vivanti of Dec. 3 1885: Introducing the > expression "path" that Cantor did not use but mean WM cannot claim Cantor is always wrong, as he so often does, and simultaneously cite him as an authority on what is right. SignatureVirgil > > > > And uncountably many X through every node. > > > Let us choose one of them and map it one that node. For the others we [quoted text clipped - 3 lines] > > paths > > going through the node and only countably many nodes below it. <snip non sequitur> We are discussing your statement: Let us choose one of them and map it one that node. For the others we will find other nodes, below that one chosen. Please indicate the first line you disagree. - There is more that one path through "that node" - There are countably many nodes below "that node" - Assuming that "For the others we will find other nodes, below that one chosen" is equivalent to assuming that there are only countably many paths through "that node" - You are trying to prove that there are countably many paths through the root node. - This is equivalent to proving that there are countably many paths through "that node" - William Hughes - > - There is more that one path through "that node" > [quoted text clipped - 4 lines] > is equivalent to assuming that there are only > countably many paths through "that node" No. It is not assuming this but *proving* this. Because paths cannot be distinguished without nodes. Equivalently: Assuming that that, in Cantor's proof, every line contains a digit that differs from the corresponding digit of the anti diagonal would be assuming that the anti diagonal is not in the list. > - You are trying to prove that there are countably many > paths through the root node. > > - This is equivalent to proving that there are countably many > paths through "that node" This is equivalent to proving that there are only countably many possibilities of distinguishing paths below "that node". Regards, WM > > - There is more that one path through "that node" > > > - There are countably many nodes below "that node" We are discussing WM's statement: Let us choose one of them and map it one that node. For the others we will find other nodes, below that one chosen. WM has agreed - There is more that one path through "that node" - There are countably many nodes below "that node" We are not at > > - Assuming that "For the others we > > will find other nodes, below that one chosen" > > is equivalent to assuming that there are only > > countably many paths through "that node" > No. It is not assuming this but *proving* this. Nope, "assuming" is not "proving". You give a putative proof that starts out by assuming what you want to prove. This is circular. > Because paths cannot be distinguished without nodes. Proves nothing. We know that a countable number of elements can distinguish an uncountable number of subsets. A countable number of nodes can distinguish an uncountable number of paths. - William Hughes > > > - There is more that one path through "that node" > [quoted text clipped - 23 lines] > proof that starts out by assuming what you want > to prove. This is circular. Not at all. I show that there are no nodes left over, after a countable number of infinite paths have been constructed. My only assumption is that in a tree, the nodes of which are completely covered by paths, no further path can be constructed. My proof rests upon the fact that after the set of all finite paths of the form 0.1, 0.11, 0.111, ... has been constructed, there is no chance to construct the path 0.111... in addition. Regards, WM > > Because paths cannot be distinguished without nodes. > [quoted text clipped - 4 lines] > > - William Hughes > My proof rests upon the fact that after the set of all finite paths of > the form 0.1, 0.11, 0.111, ... has been constructed, there is no > chance to construct the path 0.111... in addition. Why not start with path 0.111... and then add those other paths 0.1, 0.11, 0.111, ...? Cheers, Rainer Rosenthal r.rosenthal@web.de > > My proof rests upon the fact that after the set of all finite paths of > > the form 0.1, 0.11, 0.111, ... has been constructed, there is no > > chance to construct the path 0.111... in addition. > > Why not start with path 0.111... and then add those other paths > 0.1, 0.11, 0.111, ...? This is really a splendid idea! In this manner all real numbers of the unit interval can be inserted into the infinite binary tree, 1/3 and 3/137 and even 1/pi. In addidion the tree can be completed by all the "terminating paths" (those with tails 000...). Alas, we have used for construction a countable set of paths (because all real numbers you know or can construct by means of Cantor's diagonal method form a countable set). This leads to the result that there are not uncountably many paths in the tree, unless they sneak in during / after construction. Refusing such we magic, we find that there are no infinite paths at all. And this implies that we cannot start off with 0.111.... However, your contribution to this problem is of high value and I appreciate it very much. Regards, WM WM schrieb: >>> My proof rests upon the fact that after the set of all finite paths of >>> the form 0.1, 0.11, 0.111, ... has been constructed, there is no [quoted text clipped - 6 lines] > However, your contribution to this problem is of high value and I > appreciate it very much. Thank you, but it seems as if I wasn't clear in my question. Didn't you say: "If I am going to enumerate the real numbers in such and such a way then there will be some of them not in my list"? But then, that doesn't prove all too much, because it's obvious that there are many ways of not counting the reals. The point is that there is no way of counting the reals. Best regards, Rainer Rosenthal r.rosenthal@web.de > WM schrieb: > [quoted text clipped - 13 lines] > in such and such a way then there will be some of them not > in my list"? It is not enumerating all real numbers. What I do is to use actually infinite paths (at the end we will see that they do not exist, but I assume their existence) for constructing the complete binary tree. As a simple example I use only such paths that have tails of the form 000..., i.e., "terminating" paths. The result is: They exhaust the complete infinite binary tree. There is no node that remains uncovered. And there remains no node and no set of nodes that could be used to construct a path that is not yet in the tree. The tree is complete at the end. With the tip of your finger you can follow every path, even 1/pi or 1/3. > But then, that doesn't prove all too much, because it's obvious > that there are many ways of not counting the reals. > The point is that there is no way of counting the reals. As I said, I do not count the reals. I cover all possible infinite sequences of bits by means of a countable set of paths = a countable set of binary sequences. And your idea to start with some paths like 0.111... or 1/pi that do not belong to the countable set (of "terminating" paths, i.e., such with tails 000...) that I use for construction, shows us, that the result is in no way different. But how can we explain that the result is not influenced by those paths? In my opinion by recognizing that such paths do not exist. Regards, WM > > WM schrieb: > [quoted text clipped - 44 lines] > > - Show quoted text - Welcome to the "real" mathematics and not the fake "Matrix" movie sorry excuse for chess you've been playing for the last 20 years. Musatov (Founder, Inverse 19 Mathematics) Not the only, but one without it would not exist. QED Dear Simon, Efforts by mathematicians to determine the value of pi since have led to the conclusion that it ... Unlike rational numbers, pi cannot be represented by a common fraction, .... But what about an "uncommon fraction"? Will you look at my drawing and in particular how the twelve points are the outside points of the "sphere" are all "rationally" reached by "decimal" numbers? 6.2035884*2 and doubled an infinite number of times should yield a constant decimal binary expansion (what I mean by this is the numbers to the right of the decimal point should always end in "4, then 8, then 6, then 2, then 4, then 8, then 6, then 2...repeating for infinity). So in binary, by starting with this number we predict for n>2 the infinite binary tree will end with decimal progression: 2, 4, 8, 6 For n<2 the infinite binary tree will begin and end (depends on if you are increasing or decreasing the value) 1, 5, 5, 5 (repeating 5 infinitely) as we decrease to infinitely small. Musatov Inverse Counting Chain of Infinite Proportions (sample) 0.0000473372528076171875 0.000094674505615234375 0.00018934901123046875 0.0003786980224609375 0.000757396044921875 0.00151479208984375 0.0030295841796875 0.006059168359375 0.01211833671875 0.0242366734375 0.048473346875 0.09694669375 0.1938933875 0.387786775 0.77557355 1.5511471 3.1022942 6.2035884 12.4071768 24.8143536 49.6287072 99.2574144 198.5148288 397.0296576 794.0593152 1588.1186304 3176.2372608 6352.4745216 12704.9490432 25409.8980864 50819.7961728 101639.5923456 203279.1846912 406558.3693824 813116.7387648 1626233.4775296 3252466.9550592 6504933.9101184 13009867.8202368 26019735.6404736 52039471.2809472 104078942.5618944 208157885.1237888 416315770.2475776 832631540.4951552 1665263080.9903104 3330526161.9806208 6661052323.9612416 13322104647.9224832 26644209295.8449664 53288418591.6899328 106576837183.3798656 213153674366.7597312 426307348733.5194624 852614697467.0389248 1705229394934.0778496 3410458789868.1556992 6820917579736.3113984 13641835159472.6227968 27283670318945.2455936 54567340637890.4911872 109134681275780.9823744 218269362551561.9647488 436538725103123.9294976 873077450206247.8589952 1746154900412495.7179904 3492309800824991.4359808 6984619601649982.8719616 13969239203299965.7439232 27938478406599931.4878464 55876956813199862.9756928 111753913626399725.9513856 223507827252799451.9027712 447015654505598903.8055424 894031309011197807.6110848 1788062618022395615.2221696 3576125236044791230.4443392 7152250472089582460.8886784 14304500944179164921.7773568 28609001888358329843.5547136 57218003776716659687.1094272 114436007553433319374.2188544 228872015106866638748.4377088 457744030213733277496.8754176 915488060427466554993.7508352 1830976120854933109987.5016704 3661952241709866219975.0033408 7323904483419732439950.0066816 14647808966839464879900.0133632 29295617933678929759800.0267264 58591235867357859519600.0534528 117182471734715719039200.1069056 234364943469431438078400.2138112 468729886938862876156800.4276224 937459773877725752313600.8552448 1874919547755451504627201.7104896 3749839095510903009254403.4209792 7499678191021806018508806.8419584 (sample) If you would like to do the honors Mr. Plouffe, I endear you to calculate the angles and lengths of the number of ways, it is quite obviously possible to "square a circle" based on the geometry and numerical progression. As is self-evident to anyone with even a modest understanding of mathematics and geometry after visual inspection of the drawn proof (http://www.flickr.com/photos/24869933@N07/3619373444/sizes/l/) it is quite easily accomplished to connect and triangulate the outer points of the rectangles to the outer points of the square and circle and even by dimension sphere. Yes, the drawing clearly by definition of inverse symmetry (take the drawing cut it in half at the "equator" and turn the paper over and rotate it and compare the top and bottom "mirrored symmetry" and you will see it is an exact inverse of symmetry itself) the drawing explains how to properly and mathematically soundly represent a fourth dimension on a two dimensional plane. In plane language if you take the picture and separate the top half of the sphere by the bottom half of the sphere they are identical but only if the sphere existed on a complete three dimensional plane through expansion. Since the paper is flat, this is an impossibility, nonetheless it is completely expressed and retractable and countable infinitely. By the definition of our standard we have established any k-digit approximation to pi is rational. Which could have many practical scientific applications. For instance, search: ON THE THEOREM OF IVA$EV-MUSATOV IIIx e F(l) + F(2) + ... + F(p). This remark is the basis of our proof of Theorem .... -y(M + 1), x' + y(M + 1)] = 0 it is easy to check that Conditions (i), ...... Now set \i = pi x. *jj, v. Since p, x is positive,. (i)" fi is positive, ... plms.oxfordjournals.org/cgi/reprint/s3-53/1/143.pdf by TW Korner - 1986 - Cited by 7 - Related articles REPRESENTATION OF FUNCTIONSBY SERIES AND CLASSES ψ(1)with measure \E\> 2π -- ε and such that the trigonometric series for g(x) converges uniformly on [0, 2π]. .... trigonometric series are due to Ivashev-Musatov [40] who has shown that there are null series (in the sense of ..... function ψ(χ) e C(0, 1) with ψ(0) = ψ(1) = 0 such that ...... <*i<Pi(0= Σ Ψ;(0-4-a ... www.iop.org/EJ/article/0036-0279/27/2/R01/RMS_27_2_R01.pdf by PL Ul'yanov - 1972 - Cited by 24 - Related articles Full text of "Bulletin of the British Museum (Natural History)"X 20. FIG. 5. Well preserved large fragment showing the different aspect and ...... 0-1-0-5 P i n height) so giving the shell surface the appearance of ...... 94-95, i pi. Maseru. BRINK, A. S. 1963. Two cynodonts from the Ntawere ...... MUSATOV, D. L, NEMIROVSKAYA, V. N., SHIROKOVA, E. V. & ZHURAVLEVA, I. T. 1961. ... www.archive.org/stream/bulletinofbritis17geollond/bulletinofbritis17geollond_djvu.txt Trigonometric series with rapidly decreasing coefficients a i0, 5 ...Nt,+i - Pi = n,. N tl+j. -pi =N tl+j. -i +pi +1. ..... number 8 6 (0,1) for the product (32) such that for every point x 6 E there are ... www.iop.org/EJ/article/1064-5616/187/11/A01/MSB_187_11_A01.pdf by SS Galstyan - 1996 - Related articles - All 7 versions Plouffe's formula for pi - sci.math | Google Groups3 posts - 2 authors - Last post: 41 minutes ago x=pi e^(ipi)+1=0. Martin Musatov. Reply Forward. Martin Musatov to Robert show details May 3 Reply. Thank you, Sir. No problem. ... groups.google.com/group/sci.math/browse_thread/thread/65ba40fd9e5e8f46/ bf6d1ee13d87270b?lnk=raot - 41 minutes ago [PDF] -File Format: PDF/Adobe Acrobat Gli storici ci dicono che, nel X secolo, il principe Vladi- ...... i pi`u anziani e carismatici si evidenziavano i nomi di ...... tin Somov, Viktor Borisov-Musatov, Aleksandr Benois e Lev Bakst ... 5/0; Judovin 5/0; Korovin 2/1; Tukacov 1/0; Goli- cyn 1/1; Zacharov 1/1; Jurkunas 5/0; Krasauskas 2/0;. Bulaka 3/0. ... www.esamizdat.it/eSamizdat_2005_(III)_2-3.pdf I am testing these identities: (1/2) * ( - k ) * ( k - 1/2 ) * ( - 3 ) / k! This equation produces M = k! ( - 3 ) * ( k - 1/2) * ( - k ) * (1/2) Thank you for reading and writing the great comments back! Sincerely, Martin Musatov Reply Forward ---------- Forwarded message ---------- From: inverse 19 <hope9...@verizon.net> Date: Apr 11, 7:41 pm Subject: CHALLENGE TO MATHEMATICIANS OF THE WORLD To: sci.math On Apr 11, 11:40 am, "Harris Moran" <inva...@example.com> wrote: > L@@K! A caps-lock google-poster! > *plonk* > "inverse 19" <hope9...@verizon.net> wrote in message >news:f43770df-ad76-4968-af06-2c8979537339@u8g2000yqn.googlegroups.com... > > muh brane i needs one- Hide quoted text - > - Show quoted text - NOTE THE PROGRESSION OF NUMBERS PELLS PROPORTION X^2=3 or 4 = 3 as the numbers progression, WOW if this does not proof this what will 1. Every odd number , number as Y value till infinity is exactly solvable with a proportionate result using the base Pells equation i.e 3,5,7,9,etc etc till infinty and the answer is as in 2(2^2)+1=3^2 and so on till infinity with a whole number integer. 2.Every even number , without exception, using Pells equation, the integer always is a .75 till infinity consistant as 63.75(2^4)=1=4^4 till infinity every integer will end with .75. is that the same proportion as 3 over 4, Y=3 for the least with numbers proportion There is a definite pattern by pells to numbers, and I note that in that pattern the number 19 proportion is strikingly proprtionate, and I clearly surmise that that circluar progression is the same as numbers for vector 19 progression. > But how can we explain that the result > is not influenced by those paths? In my opinion by > recognizing that such paths do not exist. "result ... not influenced by those paths"? I disagree. Without those paths (like 0.111...) you are able tu enumerate, but taking care of all of them brings you into trouble. It's *them* which are the difficult part. Listing the finite subsets of the natural numbers is easy: just sort by their number of elements and within the sets of equal size order the sets lexicographical. Those sets influencing the result, I am not forced to disbelieve in the existence of them. Best regards, Rainer Rosenthal r.rosenthal@web.de In article <67306b57-4783-4850-a892-ee8064981566@d7g2000prl.googlegroups.com>, > > WM schrieb: > > [quoted text clipped - 25 lines] > the tree. The tree is complete at the end. With the tip of your finger > you can follow every path, even 1/pi or 1/3. While there is no node "uncovered" there are maximal totally ordered sets of nodes under the order induced by the the "parent of" relation (paths) which have not been accounted for. Namely all those which have infnitely many 1's in there binary representation. > > But then, that doesn't prove all too much, because it's obvious > > that there are many ways of not counting the reals. > > The point is that there is no way of counting the reals. > > As I said, I do not count the reals. Because you can't. > I cover all possible infinite > sequences of bits by means of a countable set of paths = a countable > set of binary sequences. You have deliberately overlooked every sequence of bits which contains as many 1 bits as 0 bits. WHich is most of them. > And your idea to start with some paths like 0.111... or 1/pi that do > not belong to the countable set (of "terminating" paths, i.e., such > with tails 000...) that I use for construction, shows us, that the > result is in no way different. You have deliberately overlooked every sequence of bits which contains as many 1 bits as 0 bits. WHich is most of them > But how can we explain that the result > is not influenced by those paths? In my opinion by recognizing that > such paths do not exist. So a path of infinitely many 0's does exist but a path of infinitely many 1's does not? Then your tree is not maximal, and omits some necessary nodes. SignatureVirgil In article <d8a677b7-f150-4392-b578-4d90aa10df82@a36g2000yqc.googlegroups.com>, > > > My proof rests upon the fact that after the set of all finite paths of > > > the form 0.1, 0.11, 0.111, ... has been constructed, there is no [quoted text clipped - 11 lines] > all real numbers you know or can construct by means of Cantor's > diagonal method form a countable set). But every such countable set is of paths, like every countable set of other sorts of binary sequences, is provably incomplete, so there is no ay to count all of them. > This leads to the result that there are not uncountably many paths in > the tree, unless they sneak in during / after construction. They do! it is in forming the union of all the node sets of finite trees that the node set of the infinite tree gets its uncountably many paths. > Refusing > such we magic, we find that there are no infinite paths at all. That is only because WM is incapable of looking at anything outside his MathUnrealistic world and seeing what is there. Once one has an infinite number of nodes making a complete infinite binary tree, the result that WM refuses to accept is inevitable. > And > this implies that we cannot start off with 0.111.... In an attempt to list all such binary sequences we can start with any of them. The "cannot" is in trying to finish it off. SignatureVirgil In article <2614a3b6-17bb-48b3-aa1d-37fe32a7aa45@o18g2000yqi.googlegroups.com>, > > > > - There is more that one path through "that node" > > [quoted text clipped - 26 lines] > Not at all. I show that there are no nodes left over, after a > countable number of infinite paths have been constructed. Since infinitely many paths use each node, and you have only eliminated one per node, you would then need to show that infinity minus one is zero for your argument to hold. > My only > assumption is that in a tree, the nodes of which are completely > covered by paths, no further path can be constructed. But that is not the case for a MAXIMAL infinite binary tree. For example, WM's own construction of a tree in which no path has more than finitely many 1 branchings and infinitely many 0 branchings By his own argument, correct for once, there is a path through every node of even a maximal infinite binary tree. But such a tree is far from maximal as there are more paths missing than present. > My proof rests upon the fact that after the set of all finite paths of > the form 0.1, 0.11, 0.111, ... has been constructed, there is no > chance to construct the path 0.111... in addition. Nevertheless, if a maximal tree that path must be present, so that WM's construction does not produce the desired maximal infinite binary tree. > Regards, WM > > [quoted text clipped - 6 lines] > > > > - William Hughes SignatureVirgil > In article > <2614a3b6-17bb-48b3-aa1d-37fe32a7a...@o18g2000yqi.googlegroups.com>, [quoted text clipped - 69 lines] > > - Show quoted text - P-versus-NP pageIt has been archived in June 2004, and it proves that P=NP. ... In October 2005, Moustapha Diaby also constructed a linear programming formulation of the ... www.win.tue.nl/~gwoegi/P-versus-NP.htm [PDF] The P versus NP ProblemFile Format: PDF/Adobe Acrobat - View as HTML search problem for every NP problem has a polynomial-time algorithm. For .... generators have been constructed assuming the hardness of integer factor- ... www.claymath.org/millennium/P_vs_NP/Official_Problem_Description.pdf ++ Musatov > > Because paths cannot be distinguished without nodes. > > Proves nothing. We know that a countable > number of elements can distinguish an uncountable > number of subsets. A countable number of nodes can > distinguish an uncountable number of paths. No that is provably wrong. All nodes are used up by a countable number of paths, e.g., all paths ending in a tail of zeros. Therefore no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path. All combinations of nodes that are possible in the tree have already been occupied. Regards, WM > > > Because paths cannot be distinguished without nodes. > [quoted text clipped - 7 lines] > possibility exists to construct or to distinguish by one or many or > infinitely many nodes of the tree another path. Your claim is that "no possibility exists" Nope. In any tree, any node that is not a leaf node can contribute to more than one path. The possibility exists to construct another path using nodes which are not leaf nodes. - William Hughes > > > > Because paths cannot be distinguished without nodes. > [quoted text clipped - 14 lines] > exists to construct another path using nodes > which are not leaf nodes. If you have a tree that is (the nodes of which are) completely covered by a set of paths then you cannot show any node that will distinguish a further path from that given set of paths. But if you cannot distinguish a path from a given set of paths, then it belongs to that set. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path. > If you have a tree that is (the nodes of which are) completely covered > by a set of paths then you cannot show any node that will distinguish > a further path from that given set of paths. You can, however, show a subset of nodes that will distinguish a further path from that given set of paths. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish [quoted text clipped - 6 lines] > You can, however, show a subset of nodes that will distinguish > a further path from that given set of paths. But only before those paths have occupied the tree. Because after construction the given set of paths has occupied the whole tree in that it includes all possible combinations of nodes, i.e., every subset that you might think of is already represented in the tree, because it is constructed such that no subset remains. Calvin Ostrum said: They sneak in as sets of joined edges and there is nothing you can do about it. But the contrary is true. All actually infinite paths sneak out. There is no path 0.111... There is only the set of paths 0.1 0.11 0.111 ... That set does not contain an actually infinite path. How should it? Should an infinite union of paths of the form 0.1 yield an infinite path? No The union of all finite paths does not supply an actually infinite path, no matter how large they are. All those numbers which would make R uncountable do not exist! Alas, in Cantor's list that is not so sharply visible. It's getting time to recognize that actual infinity is actually absent. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish [quoted text clipped - 8 lines] > > But only before those paths have occupied the tree. Nope. Let the given set of paths be P. Please acknowledge There is a subset of nodes that cannot be found in a single element of P. A subset of nodes is distinguished from an element p of P if and only if it is not contained in p. A subset of nodes is distinguished from every element of P if and only if it is not contained in a single element of P. There is a subset of nodes which forms a path that is not contained in a single element of P. This subset of nodes is distinguished from every element of P. The subset of nodes can be found in the tree, however it cannot be found in a single element of P. - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish [quoted text clipped - 30 lines] > The subset of nodes can be found in the tree, however > it cannot be found in a single element of P. All right as long as P is not used to construct the binary tree. Afterwards, however, we can see that every p is present in the tree. Either sneaking in at _some_ (which?) stage, based upon unmathematical superstition, or recognizing that there are no actually infinite paths. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." > > The subset of nodes can be found in the tree, however > > it cannot be found in a single element of P. > > All right as long as P is not used to construct the binary tree. Please acknowledge Before P is used to construct the binary tree, the binary tree contains a path p that can be distinguished from every element of P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." After the tree has been constructed. Yes, that is true. > > > The subset of nodes can be found in the tree, however > > > it cannot be found in a single element of P. [quoted text clipped - 5 lines] > p that can be distinguished from > every element of P. Yes, I just answered a contribution by Rainer Rosenthal, who essentially had the same idea. You can start off with the path 1/e and then add all paths with tails 000... You can start off with all paths you like or can construct by means of Cantor's diagonal method. Then complete the construction by inserting those paths with tails 000...which are yet missing. As a result you get a binary tree that is complete in that you cannot apply any diagonalization, because there is no node or set of nodes that is not already belonging to a path. If you insist in an uncountable set of paths, you must believe the "sneaking in" during or after completion. I do not believe in "sneaking in" in mathematics. Therefore, Cantor's diagonal argument, if correct, forces us to deny the existence of actually infinite paths at all. Therefore you cannot start with 1/e. It does not exist as a path or as a complete digit sequence (it does exist as 1/e though). Regards, WM In article <02e75db8-d2eb-4cb9-9072-3f68ff3c167b@q16g2000yqg.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." > > After the tree has been constructed. Yes, that is true. After the tree is constructed, every path (maximal totally ordered set of nodes) can be distinguished from any or all others by any infinite subset of its set of nodes. But the set of such paths cannot be ennumerated. > > > > The subset of nodes can be found in the tree, however > > > > it cannot be found in a single element of P. [quoted text clipped - 20 lines] > at all. Therefore you cannot start with 1/e. It does not exist as a > path or as a complete digit sequence (it does exist as 1/e though). It doesn't matter how you start or how you go on, as soon as you have included all the nodes from all possible finite subtrees, you automatically have all the uncountable set of paths. At least, you have uncountably many subsets of the set of nodes which are totally ordered by the transitive closure of the "parent of" relation. Wm seems to think that most of such maximal subsets are not paths because he doesn't like them, but they are my definition of paths, and any maximal infinite binary tree has uncountably many of them. SignatureVirgil Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." > > Before P is used to construct the binary tree, > > the binary tree contains a path > > p that can be distinguished from > > every element of P. > > Yes Using P to construct the binary tree does not change any of the elements of P or the path p. Please acknowledge After P is used to construct the binary tree, the binary tree contains a path p that can be distinguished from every element of P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 17 lines] > p that can be distinguished from > every element of P. After construction, the tree contains P (and every other path of the unit interval you wish) in same same form, whether or not P is constructed first, before the construction by terminating paths has been done. That suggests that P is somewhat decoupled from the digit sequences represented in the tree. It has nothing to do with a digit sequence. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 20 lines] > After construction, the tree contains P (and every other path of the > unit interval you wish) in same same form So after construction the tree contains every element of P and the path p. Since they have not changed form it is still possible to distinguish p from every element of P So it is possible to contruct another path, p, which can be distinguished from every element of P. This directly contradicts your claim "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 24 lines] > P and the path p. Since they have not changed form it > is still possible to distinguish p from every element of P. Sorry, that is impossible. Every node of path p and every set of nodes of path p is covered by one or more paths of P. > So it is possible to contruct another path, p, which can be > distinguished from every element of P. That is impossible. The path p_0 = 0.111... for instance is completely covered by terminating paths, whether or not it had been inserted originally. > This directly contradicts your claim > > "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." Sorry, this claim still stands. The reason is, that the path 0.111... does not exist. It is nothing but a union of terminating paths (potential infinity). The explanation is as follows: In order to define p_0 as a binary sequence, you say that for every terminating path p_n there is an m > n such that the node m is covered by p_0 but not by p_n. In the complete binary tree however, we have all nodes already covered by terminating paths. Therefore this bad trick of logic (using potential infinity for a set of nodes that should have actual existence) fails. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." > > the tree contains every element of > > P and the path p. Since they have not changed form it > > is still possible to distinguish p from every element of P. > > Sorry, that is impossible. Note that you have agreed the binary tree contains a path p that can be distinguished from every element of P. this directly contradicts your claim that no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path. > Every node of path p and every set of nodes > of path p is covered by one or more paths of P. Indeed, one *or more* paths of P. The set of nodes in path p is not covered by one path in P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." So it is. Proof: Let me construct a tree. I will not tell you what kind of paths I use. For instance I could use terminating paths or paths with tails 111... or paths with tails pi-3 or paths with tails e-2 or a mixture of many kinds of paths. I will present you the tree when it is ready. Would you bet to be able to distinuish your path from the set that I have used? > > > the tree contains every element of > > > P and the path p. Since they have not changed form it [quoted text clipped - 13 lines] > distinguish by one or many or infinitely many nodes > of the tree another path. So let us play the game. Choose a path and try to distinguish it from the set of paths that completely cover my tree. > > Every node of path p and every set of nodes > > of path p is covered by one or more paths of P. > > Indeed, one *or more* paths of P. > The set of nodes in path p is > not covered by one path in P. Ok. Let us start. I often get spam mails offering some millions of dollars. I will risk one of them, say 10 million dollars on the claim that you will not be able to distinguish your path from the set of paths that I have used for construction. What about your stakes? Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." <snip attempt to change P after p has been chosen> Note that you have agreed the binary tree contains a path p that can be distinguished from every element of P. This directly contardicts your claim. - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes > > > of the tree another path." > > <snip attempt to change P > after p has been chosen> You should give up. There is no attempt to change P. There is a proof that you cannot distinguish p from the set of paths P that has been used to construct the tree. > Note that you have agreed > > the binary tree contains a path > p that can be distinguished from > every element of P. If I have agreed, then I was in error. The binary tree contains no path that can be distinguished from the countable set P that has been used to construct the tree. Or let me put is this way: Every path that is present in the tree belongs to a countable set that could have been unsed to construct the tree. Wanna raise the stakes? Regards, WM In article <ab2eed5f-d45a-42bf-8293-d6b4db344e6f@s12g2000yqi.googlegroups.com>, > > > > > > > > > On 11 Jun., 23:06, William Hughes <wpihug...@hotmail.com> > > > > > > > > > wrote: [quoted text clipped - 12 lines] > that you cannot distinguish p from the set of paths P that has been > used to construct the tree. If that list of paths is finite or only countably infinite, then its anti-diagonal is not in it. So that unless WM's set of paths is uncountable, one can, quite mechanically, find a non-member path. > > Note that you have agreed > > [quoted text clipped - 5 lines] > path that can be distinguished from the countable set P that has been > used to construct the tree. If P is countable then for any counting of its members there is an anti-diagonal path NOT included in P. > Or let me put is this way: Every path that > is present in the tree belongs to a countable set that could have been > unsed to construct the tree. Which is, as usual, not the same thing at all. Every path is a member of many countable sets that could be used to "construct" the tree by WM's method. But that toes not imply that there is any one countable set which contains all these paths. And for every countable set roving it to be countable provides a direct method for proving it incomplete. > Wanna raise the stakes? Whatever stakes you like, since as soon as your list is proved countable it is also proved incomplete. SignatureVirgil > > <snip attempt to change P > > after p has been chosen> [quoted text clipped - 10 lines] > > If I have agreed, then I was in error. OK. I will give you the list of things you agreed with. Point out where you made your error. There is a subset of nodes that cannot be found in a single element of P. A subset of nodes is distinguished from an element q of P if and only if it is not contained in q. A subset of nodes is distinguished from every element of P if and only if it is not contained in a single element of P. There is a subset of nodes which forms a path, call it p, that is not contained in a single element of P. The path p is distinguished from every element of P. All of the nodes of path p are in the tree. - William Hughes > > You should give up. There is no attempt to change P. There is a proof > > that you cannot distinguish p from the set of paths P that has been [quoted text clipped - 13 lines] > There is a subset of nodes that cannot be found in a single > element of P. That is correct --- iff actually infinitely long sequences of nodes exist! And that is the assumption we start off with (and that we disprove). Consider the set P of terminating paths and p = 1/pi. The nodes of 1/pi are not in a single element of P. > A subset of nodes is distinguished from an element > q of P if and only if it is not contained in q. 1/pi is not contained in any element of P --- if 1/pi exists! > A subset of nodes is distinguished from every > element of P if and only if it is not contained in > a single element of P. Same as above. No error yet. > There is a subset of nodes > which forms a path, call it p, > that is not contained in a single element of P. No error detected so far. > The path p is distinguished from every element > of P. Ok. > All of the nodes of path p are in the tree. That is correct because the tree contains all subsets of nodes. All information that is available about a number in binary representation is in (one path of) the tree. If I present you the complete tree without saying how I constructed it, then you have exactly the same information that can be applied in Cantor's list argument. You cannot distinguish your favourite path from the tree constructed by means of a countable set of paths. What's now? Play the game in order to prove that? Thou shalt not belittle my binary tree! Regards, WM In article <84299d67-0445-495d-9da3-55fa8ea2a828@r10g2000yqa.googlegroups.com>, > > > You should give up. There is no attempt to change P. There is a proof > > > that you cannot distinguish p from the set of paths P that has been [quoted text clipped - 16 lines] > That is correct --- iff actually infinitely long sequences of nodes > exist! WM has already said that HE can construct a tree with infinitely infinitely long sequences of nodes in it. So they must exist at least for him. But now he does not want to allow anyone else to be able touse them? > And that is the assumption we start off with (and that we > disprove). WM's perpetual claims of his own proofs and disproofs have yet to refer to any actual proof or disproof that is either mathematically or logically valid. > Consider the set P of terminating paths and p = 1/pi. The > nodes of 1/pi are not in a single element of P. [quoted text clipped - 9 lines] > > Same as above. No error yet. Given a maximal infinite binary tree and any finite or countably infinite set of subsets of its node set then there will be paths as node sets not members of that given set. > > There is a subset of nodes > > which forms a path, call it p, [quoted text clipped - 16 lines] > exactly the same information that can be applied in Cantor's list > argument. If you, or anyone else, claims that they can count (list) all the paths in that tree, then the Cantor diagonal argument refutes you all. > You cannot distinguish your favourite path from the tree constructed > by means of a countable set of paths. Sure I can! At least in binary trees with more than two nodes. The first node of a path in such a tree has at most one child whereas in the tree it has two. That counts as distinguishing a path from a tree. > What's now? Play the game in order to prove that? > > Thou shalt not belittle my binary tree! Your form of allegedly maximal infinite binary tree belittles itself by being too little to be a maximal infinite binary tree. SignatureVirgil Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." In order to claim you do not contradict this you resort to Wolkenmuekenheim logic. The path p is distinguished from every element of P. All of the nodes of path p are in the tree. The binary tree does not contain a path p that can be distinguished from every element of P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." Yes. > The path p is distinguished > from every element of P. 1/pi is distinct from any terminating path. > All of the nodes of path p > are in the tree. All nodes of the path 1/pi are in the tree (together with all nodes of any other path of the unit interval). > The binary tree does not contain a path > p that can be distinguished from > every element of P. The binary tree does not contain any path that can be distinguished from every element of the set P of paths by which it was constructed. If this were no true, then you could determine whether a given path p differs from every element of P without knowing P. Note: Cantor's diagonal method uses only digits respective nodes, no additional information like the age of the writer or so. Same holds for my tree. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." > > Yes. Wolkenmuekenheim logic. You agree that there is a path p in the tree You agree that path p can be distinguished from every element of P. You do not agree that the tree contains a path that can be distinguished from every element of P. - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes > > > of the tree another path." > > > Yes. > You agree that there is a path p in the tree > > You agree that path p can be distinguished > from every element of P. If actually infinite paths exist! That is obviously a prerequisite. Otherwise the construction using terminating paths covers everything. > You do not agree that > the tree contains a path that > can be distinguished from every element of P. I can prove it. You only have to accept the game. You could see it even by yourself without any game. Regards, WM > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 8 lines] > > If actually infinite paths exist! Ok. We make the assumption explicit. Wolkenmuekenheim logic You agree that if actually infinite paths exist, there is a path p in the tree You agree that if actually infinite paths exist, path p can be distinguished from every element of P. You do not agree that if actually infinite paths exist, the tree contains a path that can be distinguished from every element of P. - William Hughes > > > > > Your claim is that "no possibility exists to construct or to > > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 10 lines] > > Ok. We make the assumption explicit. > You agree that if actually infinite paths > exist, there is a path p in the tree [quoted text clipped - 3 lines] > can be distinguished > from every element of P. by using nodes of the tree only without additional knowledge. > You do not agree that > if actually infinite paths exist, > the tree contains a path that > can be distinguished from every element of P. I prove that it is impossible to distinguish a give path p from the paths in P using only the information stored in the tree, i.e. using only digit sequences to identify numbers, thereby proving that no actual infinity exists. Instead of grumbling about this fact, you either should accept it and confess that I am right and that set theory is wrong, or you should be able to distinguish your path p from the set P of paths that I have used for construction, but without any other knowledge than the final result, namely the digits, i.e., the nodes of the binary tree. That are the two possible logical alternatives. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." You claim you do not contradict this by using Wolkenmuekenheim logic. Assume A: I agree that B is true I agree that C is true I do not agree that (B and C) is true. - William Hughes We agree that in Cantor's diagonal argument, applied to real numbers, the numbers are represented and identified solely by their digits. No further information is available. We assume that a real number p can be distinguished from a set Q of real numbers q by general considerations, for instance, if p is a transcendental number and Q consists of rational numbers q only. Of course it would be impossible to distinguish p from all q, because for every digit d_n of p, there is a number q that shares all digits up to d_n with p. It is possible to construct the complete binary tree from the numbers in Q. That means, it is possible to construct all possible digit sequences from the numbers in Q. It is also possible to construct the complete binary tree from the set Tp that consists of all termintaing rationals which are appended by a tail consisting of p. Therefore it is impossible to distinguish p from the binary tree. But the binary tree has been errected by using a countable set of paths only. Therefore it is impossible to distinguish, by means of digits, any real number from a countable set of paths. Conclusion: Cantor's diagonal argument is contradicted. Comment: This is not surprising, because it is impossible to identify any irrational number by means of its decimal representation, but just this is claimed by the diagonal argument. Regards, WM <snip evasion> Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." You claim you do not contradict this by using Wolkenmuekenheim logic. Assume A: I agree that B is true I agree that C is true I do not agree that (B and C) is true. - William Hughes > <snip evasion> > > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." Choose a path and try it. I then will tell you whether you are right. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." <snip attempt to provide a new putative proof> You claim you do not contradict this by using Wolkenmuekenheim logic. Assume A: I agree that B is true I agree that C is true I do not agree that (B and C) is true. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." snip attempt to evade the topic. Topic restated and elaborated: We agree that in Cantor's diagonal argument, applied to real numbers, the numbers are represented and identified solely by their digits. No further information is available. We assume that a real number p can be distinguished from all elements of a set Q of real numbers q by general considerations, for instance, if p is a transcendental number and Q consists of rational numbers q only. (This implies the existence of such a number p, but does not imply the existence of a decimal expansion of p.) Of course it would be impossible to distinguish p from all q, because for every digit d_n of p, there is a number q that shares all digits up to d_n with p. It is possible to construct the complete binary tree from the numbers in Q. That means, it is possible to construct all possible digit sequences from the numbers in Q. It is also possible to construct the complete binary tree from the set Tp that consists of all termintaing rationals which are appended by a tail consisting of p. Therefore it is impossible to distinguish p from the binary tree. But the binary tree has been errected by using a countable set of paths only. Therefore it is impossible to distinguish, by means of digits, any real number from all elements a countable set of paths. Conclusion: Cantor's diagonal argument is contradicted. Comment: This is not surprising, because it is impossible to identify any irrational number by means of its decimal representation, but just this is claimed by the diagonal argument. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." <snip continuing evasion> You have agreed to Assuming infinite paths exist: The path p is in the tree The path p can be distinguished from every element of P. You refuse to agree to Assuming infinite paths exist: There is a path p in the tree that can be distinguished from every element of P. Until you explain this, no other argument as to why you refuse to accept this statment will be considered. - William Hughes > Assuming infinite paths exist: > > The path p is in the tree > > The path p can be distinguished from > every element of P. For every path p_n of P there is a digit such that p differs from p_n. But there is no digit of p that differs from every path p_n of P because, if p exists as binary representation, then p is the union of all p_n and as such cannot differ from the union. > You refuse to agree to > [quoted text clipped - 6 lines] > to why you refuse to accept this statment > will be considered. It is obviously impossible to distinguish p from all paths p_n that are in the binary tree after construction has been completed. If you don't believe me, then play the game. Minimum stake 10^6 dollars or (if you come from Europe) Euros or Pounds. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." > > Assuming infinite paths exist: > > > The path p is in the tree > > > The path p can be distinguished from > > every element of P. > For every path p_n of P there is a digit such that p differs from p_n. > But there is no digit of p that differs from every path p_n of P > because, if p exists as binary representation, then p is the union of > all p_n and as such cannot differ from the union. The statement says nothing about p being distinguished from the union of all p_n, the statment says that p can be distinguished from every *element* of the union of all p_n. Do you wish to repudiate your repeated agreement to The path p can be distinguished from every element of P ? > > You refuse to agree to > [quoted text clipped - 9 lines] > It is obviously impossible to distinguish p from all paths p_n that > are in the binary tree after construction has been completed. Indeed. However, the statement is not about distinguishing p from paths in the tree, it is about distinguishing p from paths in P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 22 lines] > > ? No. The union of paths is not larger than every element - unless we agree to magic. In particular p cannot be distinguished from the paths of P, used to construct the tree, if I had included p in P. I agree that p was not included. But that does not play a role unless you are able to obtain that from the complete tree. But you are not! > > > You refuse to agree to > [quoted text clipped - 13 lines] > distinguishing p from paths in the > tree, it is about distinguishing p from paths in P. If this was possible before, then it would also be possible post festum, because only P was used to construct the tree. But it is not. Therefore it turns out that the original assumption "p can be distuinguished from every path of P" is contradicted. A classical proof by conradiction. By the way, one of the finest proofs that exists in mathematics. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." Your answer to "Do you wish to repudiate your repeated agreement to" The path p can be distinguished from every element of P. is "No". Then you claim p cannot be distinguished from the paths of P, used to construct the tree, You agree to the stamentemt The path p is in the tree. then you say I agree that p was not included. But that does not play a role unless you are able to obtain that from the complete tree. But you are not! You are becoming incoherent. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 23 lines] > > You are becoming incoherent. There are two statements: 1) Path p can be distinguished from every path of P. 2) Path p cannot be distinguished from every path of P. The first is assumed to be correct before P was used to construct the tree. The second is assumed to be correct after P was used to construct the tree. One of them is false, unless it is a magic tree where something happens during construction. But I do not believe in magic, least in mathematics. The second statement can be proved to be correct, because in fact you are not able to distinguish p from P (by means of digits). Therefore the first statement is falsified. This means that there are no actually infinite digit sequences, but only potentially infinite digit sequences. It is not that difficult to understand. Regards, WM In article <8c1af2fd-8882-457f-a2c7-ec6ccb52d7db@y17g2000yqn.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 32 lines] > The second is assumed to be correct after P was used to construct the > tree. BOTH WRONG! At least if P is a countable and fixed set of paths. Unless P is allowed to change as it is used, every p that exists before the construction must be in P (as there is as yet no tree from which to get others), but after the construction is complete, the tree necessarily contains uncountably many paths some of which could not have been in the original P. > One of them is false, unless it is a magic tree where something > happens during construction. But I do not believe in magic, least in > mathematics. Something happens during construction, but it is only magic to those too thick to understand it. > The second statement can be proved to be correct, because in fact you > are not able to distinguish p from P (by means of digits). Cantor could and did. And those using his methods still can anddo. > Therefore the first statement is falsified. This means that there are > no actually infinite digit sequences, but only potentially infinite > digit sequences. Then WMcan never build a maximal infinite binary tree at all, even though others can still build them. But since WM conceded complete paths at the onset of this discussion, he is not now allowed to renege now. SignatureVirgil Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." > There are two statements: > 1) Path p can be distinguished from every path of P. > 2) Path p cannot be distinguished from every path of P. > > The first is assumed to be correct before P was used to construct the > tree. Nope. No assumption. You agreed that statement 1 is correct before P was used to construct the tree. You agreed that constucting the tree does not change path p or any element of P You are now trying to claim statement 2 is correct after P is used to construct the tree. (Note P is used to construct the nodes of the tree, the nodes of the tree are used to construct the paths of the tree. The set of paths in P and the set of paths in the tree are not the same. The fact that p cannot be distinguished from all paths in the tree does not mean that p cannot be distinguished from all paths in P) > The second statement can be proved to be correct, because in fact you > are not able to distinguish p from P (by means of digits). P contains every digit in p, so you are not able to distinguish p from P by digits. P does not contain every subset of digits in p, so you are able to distinguish p from P by subsets of digits. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 9 lines] > Nope. No assumption. You agreed that statement 1 > is correct before P was used to construct the tree. I agreed under the assumption that actual infinity exists, in order to contradict this assumption. > You agreed that constucting the tree does not change > path p or any element of P That is correct. It is a basic of mathematics. > You are now trying to claim statement 2 is correct > after P is used to construct the tree. I can prove that statement 2 is correct. It is proved by your inability to distinguish p from all paths of the tree. > (Note P is used to construct the nodes of the > tree, the nodes of the tree are used to construct > the paths of the tree. The set of paths in P > and the set of paths in the tree are not the same. There is nothing added. So you are trying to invent magic. > The fact that p cannot be distinguished from > all paths in the tree does not mean that p cannot > be distinguished from all paths in P) It does. > > The second statement can be proved to be correct, because in fact you > > are not able to distinguish p from P (by means of digits). > > P contains every digit in p, so you are not able to distinguish > p from P by digits. P does not contain every subset of digits > in p, so you are able to distinguish p from P by subsets of digits. Cantor's list argument, applied to sequences of digits, is applied to single digits only. If this is not justified, then his proof is false. Regards, WM In article <0436222c-24dd-46c7-8d87-45f3e95a12d3@y17g2000yqn.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 23 lines] > I can prove that statement 2 is correct. It is proved by your > inability to distinguish p from all paths of the tree. WRONG! Unless WM can also prove that every path in the constructed tree WA already there in his set P, he lies. > > (Note P is used to construct the nodes of the > > tree, the nodes of the tree are used to construct > > the paths of the tree. The set of paths in P > > and the set of paths in the tree are not the same. > > There is nothing added. So you are trying to invent magic. Wile there are no nodes added, there now exist all sorts of sets of nodes that were not present in any member of P. WM's inability to distinguish between nodes and sets of nodes, among a number of other important distinctions that he seems incapable of making, repeatedly destroys the validity of his arguments. > > The fact that p cannot be distinguished from > > all paths in the tree does not mean that p cannot > > be distinguished from all paths in P) > > It does. Not outside of WM's world of MathUnrealism. If P is a countable set of paths(as sets of nodes), even though the union of P is the set of all nodes, there are sets of nodes which are maximal totally ordered subsets under the 'ancestor of" partial ordering of the set of all nodes ordering which are not in P. > Cantor's list argument, applied to sequences of digits, is applied to > single digits only. WM cannot have read the same proof I read then. In the proof I read, Cantor is presented with a list of binary digit sequences, say s_0, s_1, s_2, ..., and creates an entire sequence t_0 such that for all n, t_0(n) = 1 - s_n(n). But One can as easily create for each k a sequence of sequences t_k such that for all k and all n, t_k(n) = 1- s_n(n+k). We now have at least as many nonmembers of that list as we have members. And this holds for any such list of lists of binary digits. Thus if any of WM's outre claims should turn out to be true, it is clear that WM is not competent to prove them. And many of them admit to valid counterproofs. SignatureVirgil Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." > > > There are two statements: > > > 1) Path p can be distinguished from every path of P. [quoted text clipped - 8 lines] > I agreed under the assumption that actual infinity exists, in order to > contradict this assumption. So we are talking about a proof by contradiction. It needs two parts If A then B If A then not B Consider the first part > > You agreed that constucting the tree does not change > > path p or any element of P > > That is correct. It is a basic of mathematics. So either you are completly incoherent or you now agree that under the assumption that actual infinity exists path p can be distinguished from every path of P. We now proceed to the second part of the putative proof by contradiction. > > > The second statement can be proved to be correct, because in fact you > > > are not able to distinguish p from P (by means of digits). > > > P contains every digit in p, so you are not able to distinguish > > p from P by digits. P does not contain every subset of digits > > in p, so you are able to distinguish p from P by subsets of digits. <snip evasion> Your claim explicitly mentions distinguishing p from P by "one or many or infinitely many nodes". Answer yes or no you are able to distinguish p from P by subsets of digits - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." In fact that is not possible. > > > > There are two statements: > > > > 1) Path p can be distinguished from every path of P. > > > > 2) Path p cannot be distinguished from every path of P. > > > You agreed that constucting the tree does not change > > > path p or any element of P [quoted text clipped - 5 lines] > actual infinity exists path p can be distinguished > from every path of P. But it turns out that this assumption cannot be satisfied. > Your claim explicitly mentions distinguishing p from P by > "one or many or infinitely many nodes". > Answer yes or no > > you are able to distinguish p from P by subsets of digits I will try to explain it for you again: We have a list of a countable set P of terminating paths p_n. By the diagonal method we can distinguish p from every p_n (if actual infinity exists). Now we write the paths p_n in slightly different form: The beginning zeros of all paths are written only once, the following 1 or 0 also are written only once each and so on. Note, we have not done anything else but writing the list in slightly different form, saving some ink. In particular we have not added any new path. This yields the complete binary tree. You know that you are unable to distinguish any binary sequence representing a real of the unit interval from every path of the tree. This proves that you, contrary to the assumption, have been unable at the beginning too, because there the same set of paths was presented to you. You only believed that you were able. Cantor's diagonal method is falsified. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." You are trying a proof by contradiction of "Infinite paths exist" We have If infinite paths exist there is a path p that can be distinguished from every path of P. We need If infinite paths exist path there is no path p that can be distinguished from every path of P. Your putative proof you are not able to distinguish p from P (by means of digits). (Note that you have agreed that distinguish p from P means distinguish p from every element of P) I agreed, and pointed out that you distinguish p from every elemnt of P by subsets of digits. > > Answer yes or no > > > you are able to distinguish p from P by subsets of digits <keep evasion> > I will try to explain it for you again: > [quoted text clipped - 9 lines] > > This yields the complete binary tree. Correct, you can get every node in the binary tree without using every subset of nodes in the binary tree. > You know that you are unable to > distinguish any binary sequence representing a real of the unit > interval from every path of the tree. Correct the tree contains every subset of nodes, however P ( a list of subsets of nodes) does not contain every subset of nodes You can distinguish p from every element of P by subsets of nodes. William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 4 lines] > > We have > If [actually]infinite paths exist > there is a path p that can be distinguished [quoted text clipped - 5 lines] > there is no path p that can be distinguished > from every path of P. No we need and have in fact: There is no path p that can be distinguished from every path of P. > Your putative proof > > you are not able to distinguish p from P (by means of digits). > > (Note that you have agreed that distinguish p from P > means distinguish p from every element of P) only if actually infinite path exist. > I agreed, and pointed out that you distinguish p from every element of > P by > subsets of digits. That is wrong once P has been used to construct the tree. Proof: You cannot distinguish p from the tree, because the tree would not be different if p and only paths with tails consisting of p were used to construct it. > > I will try to explain it for you again: > [quoted text clipped - 12 lines] > Correct, you can get every node in the binary tree without > using every subset of nodes in the binary tree. Wrong. P contains every subset of nodes of the binary tree. There is nothing else. > > You know that you are unable to > > distinguish any binary sequence representing a real of the unit [quoted text clipped - 3 lines] > however P ( a list of subsets of nodes) > does not contain every subset of nodes As my proof shows, P contains every subset of nodes. > You can distinguish p from every element of P by > subsets of nodes. Wrong. The tree contains nothing more than the elements of P. This gets clear from the fact that you only write the list in somewhat compressed form. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." You are trying a proof by contradiction of " Actually infinite paths exist" We have If actually infinite paths exist there is a path p that can be distinguished from every path of P. We need If actually infinite paths exist there is no path p that can be distinguished from every path of P. <snip> > Proof: You > cannot distinguish p from the tree, Irrelevant. You do not distinguish p from the tree. You distinguish p from every element of P. > P contains every subset of nodes of the binary tree. The union of the paths in P contains every subset of nodes. However, p is a subset of nodes that is not contained in any single element of P. - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 13 lines] > there is no path p that can be distinguished > from every path of P. Please check that again. Compare: We want to prove by contradiction that pi is not rational. We have, if pi is rational, then pi is p/q. We need, according to you, if pi is rational, then pi is not p/q. My proof is as follows: If Actual then Distinct and not Distinct. > > Proof: You > > cannot distinguish p from the tree, > > Irrelevant. You do not distinguish p from the tree. > You distinguish p from every element of P. You cannot distinguish p from every path of the tree. Every path of the tree is is from P. > > P contains every subset of nodes of the binary tree. > > The union of the paths in P contains every subset of > nodes. However, p is a subset of nodes that is not > contained in any single element of P. Then you would be able to distinguish p from every path of the tree that is constructed by P. But you aren't. Regards, WM > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes > > > > of the tree another path." You agree that if actually infinite paths exist your claim is false. > > You are trying a proof by contradiction of > > " Actually infinite paths exist" [quoted text clipped - 9 lines] > > there is no path p that can be distinguished > > from every path of P. <snip> > > > Proof: You > > > cannot distinguish p from the tree, [quoted text clipped - 3 lines] > > You cannot distinguish p from every path of the tree. Irrelevant, you distinguish p from every element of P > Every path of the tree is is from P. Nope. Every *node* of the tree is from P. However, there is a *subset of nodes* in the tree that is not contained in one element of P. - William Hughes > > > > > Your claim is that "no possibility exists to construct or to > > > > > distinguish by one or many or infinitely many nodes > > > > > of the tree another path." > > You agree that if actually infinite paths exist > your claim is false. No. I see that my claim is correct under any circumstances. That fact is independent of whether or nor actually infinite paths exist. Proof: It is impossible to distinguish your path p from every path of the tree. > > > You are trying a proof by contradiction of > > > " Actually infinite paths exist" [quoted text clipped - 9 lines] > > > there is no path p that can be distinguished > > > from every path of P. If actually infinite paths exist, then there is a path p that can be distinguished from every path of P. And: There is no path p that can be distinguished from every path of P. This kind of proof is called proof by contradiction or modus tollens: ((A ==> B) & ~B) ==> ~A > > > > Proof: You > > > > cannot distinguish p from the tree, [quoted text clipped - 5 lines] > > Irrelevant, you distinguish p from every element of P Not after they have been written in form of tree. > > Every path of the tree is is from P. > > Nope. Every *node* of the tree is from P. Every path of the tree is from P. I explicitly forbid every other path to enter my tree. But, surprise, there is no other path willing to do so. What kind of path should apply, after I have every node of the tree covered with a path of P? > However, there is a *subset of nodes* in the tree that is not > contained in one element of P. Wrong. Every node and every subset of nodes is in a path of P. The ideas you talk about are not realized by digits or bits or nodes - as my proof shows. Regards, WM > > > > > > Your claim is that "no possibility exists to construct or to > > > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 5 lines] > No. I see that my claim is correct under any circumstances. That fact > is independent of whether or nor actually infinite paths exist. Nope. You have repeatedly agreed that if actually infinite paths exist, then a path p exists that can be distinguished from every element of P. WM: If actually infinite paths exist, then there is a path p that can be WM: distinguished from every path of P. You also clasim WM: And: WM: There is no path p that can be distinguished from every path of P. <snip> > > > Every path of the tree is is from P. > > > Nope. Every *node* of the tree is from P. > > Every path of the tree is from P. I explicitly forbid every other path > to enter my tree. Nope. You cannot forbid every other path to enter the tree. You add nodes to the tree. When you add a node to the tree you add subsets of nodes to the tree as well. You add a subset of nodes that is not in a single element of P. - William Hughes > > > > > > > Your claim is that "no possibility exists to construct or to > > > > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 13 lines] > be > WM: distinguished from every path of P. A ==> B > You also claim > > WM: And: > WM: There is no path p that can be distinguished from every path of > P. ~B > <snip> ((A ==> B) & ~B) ==> ~A > > > > Every path of the tree is is from P. > [quoted text clipped - 5 lines] > Nope. You cannot forbid every other path to enter the > tree. You add nodes to the tree. I add infinite paths. > When you add a node to the tree you add subsets of nodes > to the tree as well. You add a subset of nodes that is > not in a single element of P. Wrong. I add a path p_n like 0.himpidimpydowadididum000... That path p_n is different from any other path p_m and there is no path p_m that together with p_n produces anything different from all finite subsets that are already contained in p_n or p_m. You are dreaming of an ideal set theoretic world - without paradoxes and antinomies. But this world has come to an end in 2004 with the first construction, recognition and careful interpretation of the binary tree. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." You have repeatedly noted WM: If actually infinite paths exist, WM: then there is a path p that can be WM: distinguished from every path of P. You also claim WM: There is no path p that can WM: be distinguished from every path of P. > > > Every path of the tree is from P. I explicitly forbid every other path > > > to enter my tree. [quoted text clipped - 3 lines] > > I add infinite paths. Which consist of subsets of nodes. When you add subsets of nodes to the tree you create other subsets nodes in the tree that you did not add. > > When you add a node to the tree you add subsets of nodes > > to the tree as well. You add a subset of nodes that is [quoted text clipped - 5 lines] > path p_m that together with p_n produces anything different from all > finite subsets that are already contained in p_n or p_m. Indeed, you do not produce a path by adding a single path. You produce a path by adding a set of paths. -William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 5 lines] > WM: then there is a path p that can be > WM: distinguished from every path of P. A ==> B > You also claim > > WM: There is no path p that can > WM: be distinguished from every path of P. ~B This is not my claim but the proven impossibility to distinguish p from every element of P. ( ... ) ==> ~A > > > > Every path of the tree is from P. I explicitly forbid every other path > > > > to enter my tree. [quoted text clipped - 7 lines] > When you add subsets of nodes to the tree you create > other subsets nodes in the tree that you did not add. That is a bare lie. I add the paths p_n , that have a tail of zeros beginning at node n, one by one, in the following order: 0 -, 1 -, 2, -, 3 -, 4, ... What unadded subset is created when what path is added? Remember: A union of sets contains only elements that are contained in at least one of the united sets. I add singlets {p_n}. > > > When you add a node to the tree you add subsets of nodes > > > to the tree as well. You add a subset of nodes that is [quoted text clipped - 8 lines] > Indeed, you do not produce a path by adding a single path. > You produce a path by adding a set of paths. And Cantor's proof does not fail when considering a single line but when considering a set of lines. Why do you believe that we should stick to mathematics and logic in that case, if we believe in magic creation of paths in case of the tree? Regards, WM In article <f6d08279-d0cf-499f-8dc2-ebf332d4c877@a36g2000yqc.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 12 lines] > > WM: There is no path p that can > > WM: be distinguished from every path of P. Since that does not, and cannot, happen in ZF nor anywhere else outside WM's silly world, it is of no mathematical consequence. > This is not my claim but the proven impossibility to distinguish p > from every element of P. Except that everyone but WM can do it. > > > > > Every path of the tree is from P. I explicitly forbid every other path > > > > > to enter my tree. Maximal infinite binary trees are not yours to command, and yours, not being among them, are irrelevant. > > > > Nope. You cannot forbid every other path to enter the > > > > tree. You add nodes to the tree. [quoted text clipped - 6 lines] > > That is a bare lie. It is true even for finite sets: Given P = {{a,b}, {a,c}}, there are subsets of Union(P) that are not members of P. More of them than there are members of P. > I add the paths p_n , that have a tail of zeros beginning at node n, > one by one, in the following order: [quoted text clipped - 5 lines] > > What unadded subset is created when what path is added? All sorts of set which are not paths are added with each new node. But the new paths do not all appear until after WM's additions are completed. > Remember: A union of sets contains only elements that are contained in > at least one of the united sets. Which means that for any given fison, the union of all fisons also contains at least one element not in that given fison. > I add singlets {p_n}. Since none of them is infinite, no set of them contains, as a subset, any paths at all but the union of the set of all of them contains all paths as subsets. > Why do you believe that we should stick to mathematics and logic in > that case, if we believe in magic creation of paths in case of the > tree? WM may believe in such magic, but those of us who pay attention to definitions and logic only see such sight of hand trickery in WM's arguments, not in the actual meanings of actual definitions or in the implications of actual logic. SignatureVirgil > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 7 lines] > > A ==> B Yes, you agree A ==> B is true. > > You also claim > > > WM: There is no path p that can > > WM: be distinguished from every path of P. > > ~B Yes, you claim ~B is true. <snip> > > When you add subsets of nodes to the tree you create > > other subsets nodes in the tree that you did not add. > > That is a bare lie. We have to deal with this claim before we can move on. Note we are talking about subsets of nodes. > I add the paths p_n , that have a tail of zeros beginning at node n, > one by one, in the following order: [quoted text clipped - 3 lines] > -, 2, -, 3 > -, 4, ... > What unadded subset is created when what path is added? Note you ask for a subset of nodes not a path. I will give an example of a subset of nodes that is not a path. You add the paths 01000... 101000... At this point the subset of nodes q={01,010,10,101} is in the tree, but q is not in any single element of P. Since only elements of P are added to the tree, q is not added to the tree. - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 9 lines] > > Yes, you agree A ==> B is true. Yes. > > > You also claim > [quoted text clipped - 14 lines] > We have to deal with this claim before we can move > on. Note we are talking about subsets of nodes. Precisely: We are talking about subsets of nodes that can be paths or parts of paths. The subset of nodes consisting of all nodes of the 17th level of the tree is not under consideration. > > I add the paths p_n , that have a tail of zeros beginning at node n, > > one by one, in the following order: [quoted text clipped - 22 lines] > element of P. Since only elements of P are added > to the tree, q is not added to the tree. Making my question more explicite: What unadded subset, than can be (a part of) a path, is created when what path is added? Regards, WM > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 32 lines] > Precisely: We are talking about subsets of nodes that can be paths or > parts of paths. Nope, you just added the condition "that can be paths or parts of paths". <snip> > Making my question more explicite: > What unadded subset, than can be (a part of) a path, is created when > what path is added? Before I answer this question, please aanswer yes or no. When you add subsets of nodes to the tree you create other subsets of nodes in the tree that you did not add. - William Hughes > > > > > Your claim is that "no possibility exists to construct or to > > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 35 lines] > Nope, you just added the condition "that can be paths or > parts of paths". Other subsets of nodes are irrelevant. > <snip> > [quoted text clipped - 3 lines] > > Before I answer this question, please answer yes or no. Yes. But that question does belong to the topic. Subsets of nodes that are not connected by paths, i.e., nodes that do not stand in the ancestor relation are as irrelevant as the question of present day lunch. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) WH: When you add subsets of nodes to the tree you create WH: other subsets nodes in the tree that you did not add. WM: That is a bare lie. WH: please answer yes or no. WH: When you add subsets of nodes to the tree you create WH: other subsets nodes in the tree that you did not add. WM: Yes WM: What unadded subset, than can be WM: (a part of) a path, is created when WM: what path is added? As we have seen we can create a finite subset by adding a single path. However, as previously noted, to create a path requires creating an infinite subset. To do this we need to add a subset of paths. E.g. The subset of paths added is 1000... 11000... 111000... ... The subset of nodes that is now in the tree is t={1,11,111, ...} Note that t is not an element of P, so the subset t is never added to the tree. -William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 9 lines] > > So if A is true your claim is false. A ==> B should go undisputed, like: If pi is rational, then it can be represented by p/q. Nobody will claim that pi is rational but cannot be represented by p/ q. The only conceivable instance to lay claim to pi's rationality without its being representable as a fraction is, if Cantor had said that pi is rational. Well, then further arguing would be hopeless and in vain. > You want to show > [quoted text clipped - 3 lines] > > (Note that assuming ~A is circular) Of course. > As we have seen we can create a finite subset > by adding a single path. That is of no relevance because only allowed subsets can be of interest. Why do you want to raise the impression your arguing with respect to irrelevant subsets would be relevant? > However, > as previously noted, > to create a path requires creating an infinite subset. To create a path requires to add that path. > To do this we > need to add a subset of paths. [quoted text clipped - 12 lines] > Note that t is not an element of P, so the subset > t is never added to the tree. If it has not been added, then it cannot be in tree. Otherwise, if it could be in the tree without having been added, why then should it not be in the list without having been added? Why then should it not be in P without being in P? Can you explain that? Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) <snip> WM: What unadded subset, than can be WM: (a part of) a path, is created when WM: what path is added? <snip> > > as previously noted, > > to create a path requires creating an infinite subset. > > To create a path requires to add that path. Nope I give a counterexample. > > To do this [create a path] we > > need to add a subset of paths. [quoted text clipped - 12 lines] > > Note that t is not an element of P, so the subset > > t is never added to the tree. Which is the first statement you disagree with all nodes of t are in the tree t is in the tree all nodes of t are in P t is not an element of P Note: if all the nodes of a path h are in the tree then h is in the tree, however, if all the nodes of h are in P, h may or may not be an element of P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 9 lines] > > So if A is true your claim is false. The claim is false if and only if you can distingusih a path p from all pats from the set P when written in form of the tree. > You want to show > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > by proving ~B > > To create a path requires to add that path. > [quoted text clipped - 13 lines] > > > > t={1,11,111, ...} There is no sequence of 1's added that is longer than every such sequence of paths in P. > > > Note that t is not an element of P, so the subset > > > t is never added to the tree. [quoted text clipped - 5 lines] > all nodes of t are in P > t is not an element of P I disagree with the statement: More nodes of t are in tree than are in any single path of P. Proof: The tree is nothing but a another way to write the paths of P. So why do you think (or rather pretend to think) that there are longer sequences of 1's in P than in P? > Note: if all the nodes of a path h are in the > tree then h is in the tree, however, if all the > nodes of h are in P, h may or may not be an element > of P. Wrong. The list of paths of P is the tree in slightly different notation. When changing from list to tree, then several bits are mapped on one node, not the other way round. That number does decrease, not increase. Again you are trying to rape logic. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 9 lines] > > > So if A is true your claim is false. Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) <snip> > > Which is the first statement you disagree with > > > all nodes of t are in the tree > > t is in the tree > > all nodes of t are in P > > t is not an element of P <snip evasion> Please answer the question Note: The list of paths of P is not the tree in slightly different notation. If all the nodes of a path h are in the tree then h is in the tree, however, if all the nodes of h are in P, h may or may not be an element of P. - William Hughes > A: actually infinite paths exist, > B: the infinite tree contains a path p that can be [quoted text clipped - 5 lines] > > So if A is true your claim is false. That is nonsense. If (A) pi has a rational presentation, then (B) it can be written as p/ q. B is false. And A cannot be true without B. > You want to show > [quoted text clipped - 16 lines] > > Please answer the question What do you understand by "all nodes"? > Note: The list of paths of P is not > the tree in slightly different > notation. Note the list of paths P is the same as the tree in slightly different notation. Proof: Every pair of sequences is written in a form such that their common initial sequence is written only once. This is continued as far as possible. >If all the nodes of a path h are in the > tree then h is in the tree, however, if all the > nodes of h are in P, h may or may not be an element > of P. That is a self-contradiction, because the tree is not different from the list other than that some bits are written only once. In particular there cannot be more in the tree than is already in the list. (With exception of mathemagicians' tricks, of course.) Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true then B is true and your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) > > > > Which is the first statement you disagree with > [quoted text clipped - 8 lines] > > What do you understand by "all nodes"? "all nodes of t are in the tree" There is no node that is in t but is not in the tree "all nodes of t are in P" There is no node that is in t but is not in an element of P WM: the list of paths P is the same as the tree Nope. You have agreed that the tree contains a subset of nodes that is not contained in one element of the list of paths P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 10 lines] > So if A is true then B is true > and your claim is false. So it is. > > What do you understand by "all nodes"? > > "all nodes of t are in the tree" > There is no node that is in t but is not in the tree That is correct. > "all nodes of t are in P" > There is no node that is in t but is not in an element of P Also correct. Every node of t and all predecessor nodes are in one path of P. > WM: the list of paths P is the same as the tree > > Nope. You have agreed that the tree contains a subset of > nodes that is not contained in one element of the list of paths P. All subsets of nodes that are in the tree also are in the list of paths. Why should the subset belong to one element of the list of paths if it does not belong to a single path in the tree? Can you explain that? Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true then B is true and your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) > > > What do you understand by "all nodes"? > [quoted text clipped - 7 lines] > > Also correct. <snip evasion> Which is the first statement you disagree with all nodes of t are in the tree t is in the tree all nodes of t are in P t is not an element of P > > WM: the list of paths P is the same as the tree > > > Nope. You have agreed that the tree contains a subset of > > nodes that is not contained in one element of the list of paths P. > All subsets of nodes that are in the tree also are in the list of > paths. But, as you have agreed, not in a single path. > Why should the subset belong to one element of the list of > paths It doesn't. "the tree contains a subset of nodes that is *not* contained in one element of the list of paths" - William Hughes i > > Why should the subset belong to one element of the list of > > paths > > It doesn't. > "the tree contains a subset of nodes that is > *not* contained in one element of the list of paths" As it is neither contained in a path of the tree, this observation is completely irrelevant. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true then B is true and your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) <snip evasion> Please answer yes or no t is not an element of P - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 24 lines] > > t is not an element of P The actually infinite path t is not an element of P and not a path of T, because t does not exist. Proof by the fact that t cannot be distinguished from the set P after its elements have been ordered as paths of the tree. REgards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 26 lines] > > The actually infinite path t is not an element of P > and not a path of T, because t does not exist. Note it would be circular to simply assume that actually infinite paths do not exist. > Proof by the fact that t cannot be distinguished from the set P Nope: you have agreed that t can be distintuished from the every element of the set P (Recall, You have agreed. A subset of nodes is distinguished from every element of P if and only if it is not contained in a single element of P.) - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 30 lines] > Note it would be circular to simply assume that actually > infinite paths do not exist. It is not assumed. It is proved by the fact that t cannot be distinguished from the paths of P and T. > > Proof by the fact that t cannot be distinguished from the set P > [quoted text clipped - 7 lines] > element of P if and only if it is not contained in > a single element of P.) All of t, that does exist, is contained in a path of P. It is only your delusion that opposes to this fact. You say that for every path p of P there is a bit of t that does not belong to p. And you think this would exclude t from P. But that is wrong, because "every" always only concerns to paths of P that you can specifically address. The important fact is: Every bit of t and all its predecessors are in infinitely many paths of P. When P is ordered as a tree, your delusion can no longer deceive you. Regards, WM > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 52 lines] > wrong, because "every" always only concerns to paths of P that you can > specifically address. So. "every element of P" becomes a Wolkenmeukenheim special, something fixed that changes. However, even with this modification of the meaning of every we have. A subset of nodes is distinguished from every element of P that you can specifically address if and only if it is not contained in a single element of P that you can specifically address. and you have agreed that t can be distinguished from every element of P. -William Hughes > A subset of nodes is distinguished from every > element of P that you can specifically address [quoted text clipped - 4 lines] > and you have agreed that t can be distinguished from > every element of P. It does not matter whether I have agreed, it matters whether it is true. As you see from the tree, it is not true, unless the transformation from the list to the tree would introduce new paths. That, however, would be something between mathemagic and matheology. Under any circumstances that is rubbish. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true then B is true and your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) > > you have agreed that t can be distinguished from > > every element of P. > It does not matter whether I have agreed Outside of Wolkenmuekenheim it does. > it matters whether it is > true. > As you see from the tree, the subset of nodes t is in the tree An we agree that the subset of nodes t is not in a single element of the list P so it is true that the subset of nodes t is in the tree and can be distinguished from every element of P. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 27 lines] > > the subset of nodes t is in the tree You cannot distinguish it from all paths of the tree. > And we agree that > > the subset of nodes t is not in > a single element of the list P if actual infinity axists. > so it is true that the subset of nodes t is > in the tree and can be distinguished from > every element of P. If you were right. But that would imply the existence of actual infinity and the creation of paths by dropping bits. As the latter is wrong, you are not right. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 30 lines] > > > the subset of nodes t is in the tree <snip> > > And we agree that > > > the subset of nodes t is not in > > a single element of the list P > > if actual infinity axists. So you cannot prove ~B without making the assumption ~A which is what you are trying to prove. We are left we the result that you cannot prove A or ~A. - William Hughes > > > Your claim is that "no possibility exists to construct or to > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 39 lines] > > > if actual infinity axists. Otherwise your alleged "proof" is even more obviously wrong. > So you cannot prove ~B without making > the assumption ~A which is what you are trying > to prove. Wrong. ~B is proven by the impossibility to distinguish t from every path of the binary tree and the fact that the tree does not contain paths that are missing in P. Regards, WM > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 39 lines] > > > > if actual infinity axists. > > So you cannot prove ~B without making > > the assumption ~A which is what you are trying > > to prove. > > Wrong. ~B is proven by the impossibility to distinguish t from every > path of the binary tree Nope. You just said that if actual infinity exists, it is possible to distinguish t from every path of the binary tree. Hence you cannot prove ~B without assuming ~A. We are left we the result that you cannot prove A or ~A. - William Hughes > > Wrong. ~B is proven by the impossibility to distinguish t from every > > path of the binary tree > > Nope. You just said that if actual infinity exists, it is possible to > distinguish t from every path of the binary tree. Hence you cannot > prove ~B without assuming ~A. I prove ~B, for instance by your inability to distinguish t from T, with no regard to the truth of A. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true then B is true and your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) > I prove ~B, for instance by your inability to distinguish t from T, > with no regard to the truth of A. Nope. You need to distinguish the subset of nodes t from every element of P, not from the tree, and you have repeatedly agreed that you can distinguish t from every element of P if A is true. You have never shown that you cannot distinguish t from every element of P without regard to the truth of A. We are left we the result that you cannot prove A or ~A. - William Hughes > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes [quoted text clipped - 24 lines] > Nope. You need to distinguish the subset of nodes t from every > element of P, not from the tree, The paths of the tree are the same as the subsets of P. There is no reasonable argument that leads to a difference. Further we can derive from the axioms that the union of sets does not contain any element, that is not contained in at least one of the sets united. We unite the singlets {p_n} where p_n is in P. T = U{p_n}. There is no other path in the tree. > and you have > repeatedly agreed that you can distinguish t from every element of > P if A is true. You erroneously believed so. I could not contradict your error. Therefore I let it go. Now you can see that you were wrong. >You have never shown that you cannot distinguish > t from every element of P without regard to the truth of A. I cannot show that because I cannot disprove a dream unless the sleper is awaked. This is accomplished by the tree. > We are left we the result that you > cannot prove A or ~A. We are left with the result ~B. From this we obtain ~A. Regards, WM > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 26 lines] > > The paths of the tree are the same as the subsets of P. Nope. You have agreed that if infinite paths actually exist then t is not contained in any element of P. So you cannot argue "The paths of the tree are the same as the subsets of P" without assuming ~A. We are left we the result that you cannot prove A or ~A. - William Hughes > > The paths of the tree are the same as the subsets of P. > > Nope. You have agreed that if infinite paths > actually exist then t is not contained in any element > of P. So you cannot argue "The paths of the tree > are the same as the subsets of P" without assuming ~A. That is not a matter of A or not A. It is the same as x = x That is independent of A. Regards, WM Your claim is that "no possibility exists to construct or to distinguish by one or many or infinitely many nodes of the tree another path." A: actually infinite paths exist, B: the infinite tree contains a path p that can be distinguished from every path of P. You agree A ==> B So if A is true then B is true and your claim is false. You want to show ~A [Follows from ( A ==> B, ~B) ==> ~A ] by proving ~B (Note that assuming ~A is circular) Wolkenmuekenheim logic A: actually infinite paths exist, > > You have agreed that if infinite paths > > actually exist > That is not a matter of A or not A. - William Hughes In article <1ed82982-acee-457b-9a0b-ffaf45dfa8b6@f16g2000vbf.googlegroups.com>, > > > The paths of the tree are the same as the subsets of P. > > [quoted text clipped - 8 lines] > > That is independent of A. What it depends on is, in WM's view, the absence of any actually infinite sets, which implies the absence of the very maximal infinite binary tree that WM is pontificating about. In order to have such a tree at all, one must first have at least one actually infinite, and preferably well-ordered, set. Any actually infinite well -ordered set contains as a subset (possibly improper) a set that is order isomorphic to the infinite ordered set of all naturals, from which such a tree is easily built. And in such a tree, WM's claim are wrong. Specifically, the set of all paths each having only finitely many right branchings, while containing each node in at least one paths, does not contain any of those uncountably many paths which each contains infinitely many right branchings. SignatureVirgil In article <138f41db-b767-4ea6-87d7-3c37918ed63e@r16g2000vbn.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 26 lines] > > The paths of the tree are the same as the subsets of P. Not in the maximal infinite binary tree which one can so easily build from theactually infinite set of all 1-origin naturals by defining the left child of n to be 2*n+0 and the right child of n to be 2*n+1. > There is no > reasonable argument that leads to a difference. Cantor's argument re the incompleteness of any list of binary sequences is reasonable argument which leads to a difference. It is WM who has no reasonable argument prohibiting such a difference. > Further we can derive > from the axioms that the union of sets does not contain any element, > that is not contained in at least one of the sets united. It has been already granted that a countable set of paths can have every member of a countable set of nodes in at least one path. But such a fact does not in any way require that such a countable set of paths contain each possible path of a maximal infinite binary tree, and Cantor proved that it did not. We unite the > singlets {p_n} where p_n is in P. T = U{p_n}. There is no other path > in the tree. Every path is a countable set of nodes, all of which are members of members of P. But a family of sets whose union is a given set need not contain all subsets of that union, which is what WM's argument would require. So most paths of the tree are not in WM's set of paths, P. > > and you have > > repeatedly agreed that you can distinguish t from every element of > > P if A is true. > > You erroneously believed so. It is hardly erroneous to believe Cantor's diagonal proof. So given any COUNTABLE set of paths, P, in a maximal infinite binary tree, there are "more" paths in the tree but not in P than actually in P. > I could not contradict your error. But you did contradict his truth. > Therefore I let it go. Now you can see that you were wrong. On the contrary, we all still see where WM is still wrong. As soon as any actually infinite set exists, and one is necessary for any construction of a maximal infinite binary tree, then WM is wrong about such trees. And unless there is at least one such actually infinite set, there are no such trees for WM to pontificate about. > >You have never shown that you cannot distinguish > > t from every element of P without regard to the truth of A. > > I cannot show that because I cannot disprove a dream unless the sleper > is awaked. This is accomplished by the tree. Then it is WM who is doing all the dreaming, as his sort of maximal infinite binary tree can not ever exist anywhwere. Such trees can only exist in a world containing an actually infinite set. > > We are left we the result that you > > cannot prove A or ~A. > > We are left with the result ~B. WM is left with the result that his sort of tree cannot exist at all. In WM's world, a set of paths, P, with each path limited to finitely many right child nodes must contain a path containing infinitely many right child nodes. That is not allowed to happen in mathematical set theories. SignatureVirgil In article <9629db25-7da7-4675-8c24-44d176b9015d@o20g2000vbh.googlegroups.com>, > > > Wrong. ~B is proven by the impossibility to distinguish t from every > > > path of the binary tree [quoted text clipped - 7 lines] > > Regards, WM In the maximal infinite binary tree which I showed how to construct from the infinite set of all naturals, one can distinguish LOTS of other paths from any fixed countable set of paths. If it can't be done in WM's trees, then his trees aren't maximal infinite binary trees at all SignatureVirgil In article <d97be259-c60f-4c65-a49c-8efd08ddfabd@r34g2000vba.googlegroups.com>, > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 41 lines] > > Otherwise your alleged "proof" is even more obviously wrong. If actually infinite sets can exist, WM is wrong, and if they don't then all trees are finite and WM IS STILL WRONG. > > So you cannot prove ~B without making > > the assumption ~A which is what you are trying [quoted text clipped - 3 lines] > path of the binary tree and the fact that the tree does not contain > paths that are missing in P. Neither of which claimed "facts" is true if any actually infinite sets exist, and the alleged tree and set P themselves do not exist at all if no actually infinite sets exist, so the issue would then be moot. SignatureVirgil In article <f36b0001-fae5-4318-93a3-624f293016c4@n19g2000vba.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 36 lines] > > if actual infinity axists. Since no potentially infinite sets can exist, either there is no maximal infinite binary tree at all, in which case WM's speculating on its properties is foolish, or there is one, and its properties are those of actually infinite sets. > > so it is true that the subset of nodes t is > > in the tree and can be distinguished from > > every element of P. > > If you were right. And if such a tree exists at all, he is. > But that would imply the existence of actual > infinity and the creation of paths by dropping bits. Paths are "created" by the partial order on the set of nodes induced by the transitive closure of the"ancestor of" relation. In a maximal infinite binary tree, uncountably many of them are created. > As the latter is > wrong, you are not right. In set theory, WM is the one who is not right by being left out. > Regards, WM SignatureVirgil In article <0beaabb8-7024-453c-b23b-4c4030d8c63b@j12g2000vbl.googlegroups.com>, > > A subset of nodes is distinguished from every > > element of P that you can specifically address [quoted text clipped - 7 lines] > It does not matter whether I have agreed, it matters whether it is > true. And as WM has no access to truth, except by accident, and even then he may well reject it, one is better advised to go elsewhere to find it. > As you see from the tree, it is not true, unless the transformation > from the list to the tree would introduce new paths. Which it does. To get from a set of paths to a tree, one bmust take the union of the set of paths as the set of nodes of that tree, and that set of nodes contains subsets, including paths, that were not members of the original set of paths. WM has demonstrated this himself, though without realizing what he has shown: Wm starts with the set of all infinite binary sequences of 0 and 1 nodes which contain only finitely many one nodes. The the union of this set of sets is the set of all nodes of a maximal infinite binary tree, so is such a tree. But in that tree there are maximal totally ordered by "ancestor of" sets of nodes, i.e., paths, which contain infinitely many 1 nodes, so are NOT members of WM's original set of paths. That, however, > would be something between mathemagic and matheology. Under any > circumstances that is rubbish. It is WM's own personal rubbish then, since WM is the one who presented us with that set of paths and that tree. > Regards, WM SignatureVirgil In article <b29de6a2-fbc3-4c40-80ad-f94d52add184@l28g2000vba.googlegroups.com>, > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 27 lines] > > > The actually infinite path t is not an element of P > > > and not a path of T, because t does not exist. If t does not exist it CERTAINLY can be distinguished from things that do exist. > > Note it would be circular to simply assume that actually > > infinite paths do not exist. > > It is not assumed. WM repeatedly asumes it, but has never proved it. > It is proved by the fact that t cannot be > distinguished from the paths of P and T. But t CAN be distinguished from them,, particularly if the paths of P and T exist but t does not. > > > Proof by the fact that t cannot be distinguished from the set P > > [quoted text clipped - 10 lines] > All of t, that does exist, is contained in a path of P. It is only > your delusion that opposes to this fact. WM is not in a position to accuse others of having delusions when he has so many of his own that he is quite unable to distinguish from reality. > You say that for every path p of P there is a bit of t that does not > belong to p. That is not what anyone but WM says. What we do say is that if P is a countably infinite set of paths in some maximal infinite binary tree, then there are paths in that tree which are not members of P, even though the union of P may contain all nodes of that tree. In fact, WM's own example of the set of all paths with only finitely many right branches is such a P, since in the resulting tree there are paths with infinitely many right branches. So that WM has managed to disprove his own thesis. SignatureVirgil In article <556f71b5-452d-41d1-a00a-b6aef151bd36@l8g2000vbp.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 27 lines] > The actually infinite path t is not an element of P and not a path of > T, because t does not exist. WE have an actually infinite set of naturals N = {1, 2, 3, ...}. WE can make that set into the set of nodes of a maximal infinite binary tree by assigning to each natural N a left child, 2+n + 0 ,and a right child, 2*n+1. Note that we have, a priori, no paths at all but, a postiori, uncountably many of them > Proof by the fact that t cannot be distinguished from the set P after > its elements have been ordered as paths of the tree. In my tree, if P is only countably infinite, as a set of subsets of N, then there are more paths in the tree not in P than in it, even if the union of P is the set of all nodes. One of WM's models of P would be, in my tree as defined above, the set of paths having at most finitely many odd naturals as nodes. And WM tries to claim that among a family sets, P, with each set limited to finitely many odd naturals there must be a set containing infinitely many odd naturals. Sorry, WM, but that claim fails. SignatureVirgil In article <85db7468-1f62-407e-8232-aa31838b52d7@o20g2000vbh.googlegroups.com>, > > > Why should the subset belong to one element of the list of > > > paths [quoted text clipped - 5 lines] > As it is neither contained in a path of the tree, this observation is > completely irrelevant. If the original list of paths is no more than countable and the tree is a maxima; infinite binary tree, then there are necessarily paths in that tree not in the list. This is so because countably many paths can cover all nodes but cannot cover all paths, as Cantor proved. SignatureVirgil > In article > <85db7468-1f62-407e-8232-aa31838b5...@o20g2000vbh.googlegroups.com>, [quoted text clipped - 14 lines] > This is so because countably many paths can cover all nodes but cannot > cover all paths but they do, nevertheless > as Cantor proved. Therefore he was wrong. During the next 100 years we will laugh about his giant joke - and then he will be forgotten by mathematicians, except by some historians. Regards, WM In article <27965221-60ba-4020-a530-c89093f9f60a@e20g2000vbc.googlegroups.com>, > > In article > > <85db7468-1f62-407e-8232-aa31838b5...@o20g2000vbh.googlegroups.com>, [quoted text clipped - 16 lines] > > but they do, nevertheless They can cover all nodes, but Cantor proved that for any countable set of paths there are other paths. > > as Cantor proved. > > Therefore he was wrong. In a war between WM and Cantor, WM will always lose. > During the next 100 years we will laugh about his giant joke We do not even have to wait to laugh at WM. He is laughable now, and has been for years. - and > then he will be forgotten by mathematicians, except by some > historians. WM will be forgotten long before Cantor is. If it were not for WM's continual postings here, he would have been forgotten already. SignatureVirgil In article <ac5850a7-9c4c-4cd2-9ff7-cb56e419bcf3@y17g2000yqn.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 25 lines] > Also correct. Every node of t and all predecessor nodes are in one > path of P. But unless every chain of successor nodes to any node in t is in P, there are paths not in P. > > WM: the list of paths P is the same as the tree Nope: A set of sets of nodes is not a set of nodes. > > Nope. You have agreed that the tree contains a subset of > > nodes that is not contained in one element of the list of paths P. > > All subsets of nodes that are in the tree also are in the list of > paths. It is not even true that all totally-ordered-by-'ancestor of' sets of nodes are in P, at least if P is countable. And no set that is not thus totally ordered can possibly be in P. SignatureVirgil In article <c0fe63b1-27fd-493f-85df-1140b59a4c13@c36g2000yqn.googlegroups.com>, > > A: actually infinite paths exist, > > B: the infinite tree contains a path p that can be [quoted text clipped - 10 lines] > q. > B is false. And A cannot be true without B. That depends on P. If A. and (P is any countable set of paths), then B If A and (P is the uncountable set of all paths), then ~B > > You want to show > > [quoted text clipped - 18 lines] > > What do you understand by "all nodes"? In what context? All nodes of the tree, or all nodes of a path, or all nodes of a set of paths or something else? > > Note: The list of paths of P is not > > the tree in slightly different > > notation. > > Note the list of paths P is the same as the tree in slightly different > notation. Not at all. No node is a member of any set of paths, but every node is a member of the tree. > >If all the nodes of a path h are in the > > tree then h is in the tree, however, if all the > > nodes of h are in P, h may or may not be an element > > of P. > > That is a self-contradiction Not to those who understand the difference between being a member of a set and being a subset of a set. (With exception of mathemagicians' tricks, of course.) The only "trick" is noting that a subset of set need no be a member of that set and a member of a set need not be a subset of it. In dealing with sets and their members and their subsets, the distinction between subsets and members is important, even if WM cannot keep it straight in his head. Assuming that each node in a tree contains a pointer to any parent and pointers to any children, then a tree is merely a suitable pointer-connected set of nodes. A path is also a suitably pointer-connected set of nodes, though infinite paths are not binary trees and infinite binary trees are not paths. Furthermore, no set of paths IS an infinite binary tree, though its union may be. It is possible to have a set of paths whose union is a maximal infinite binary tree without its being the set of all paths of a maximal infinite binary tree. And any countable set of paths is like this. SignatureVirgil In article <1f069edc-4bee-4740-a0d5-4756d70b2738@j32g2000yqh.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 12 lines] > The claim is false if and only if you can distingusih a path p from > all pats from the set P when written in form of the tree. Since P is NOT a set of nodes, it cannot be "written in the form of a tree". Its union can be so written, provided that that union contains all nodes of the tree, but then one has lost the original path structure. > > > > The subset of nodes that is now in the tree is > > > > > > t={1,11,111, ...} > > There is no sequence of 1's added that is longer than every such > sequence of paths in P. Sequences of paths are irrelevant. No such "sequence" of paths can contain its "limit" unless the sequence is eventually constant. In which case it essentially contains only one path. > > > > Note that t is not an element of P, so the subset > > > > t is never added to the tree. [quoted text clipped - 8 lines] > I disagree with the statement: More nodes of t are in tree than are in > any single path of P. Since no one has said it, except you, why are you bothering disagree? > > Note: if all the nodes of a path h are in the > > tree then h is in the tree, however, if all the [quoted text clipped - 3 lines] > Wrong. The list of paths of P is the tree in slightly different > notation. Notation does not affect membership. If there are paths not in P according to one notation, which there are, then no change of notation can change that. > When changing from list to tree, then several bits are > mapped on one node, not the other way round. That number does > decrease, not increase. Again you are trying to rape logic. On the contrary. WE are not the ones who claim that change of notation induces a change in membership. If anyone is trying to rape logic, it is WM, with us defending her honor. SignatureVirgil In article <ddef1a81-ce6e-4886-94ad-793026852859@g19g2000yql.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 13 lines] > > should go undisputed Provided that A includes existence of infinite trees and WM's infinite trees are allowed to be a MAXIMAL infinite binary trees. > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > [quoted text clipped - 16 lines] > > To create a path requires to add that path. To create a maximal infinite binary tree does not require adding any path to anything. We can start with any actually infinite countable set, indexed by N, the set of 1-origin naturals, 1 is then the index of the root node, 2*n + 0 is the index of the LEFT child of the node indexed by n 2*n + 1 is the index of the right child of the node indexed by n One now has a maximal infinite binary tree without having even mentioned paths. And in this tree, there are too many paths to count, one path for every infinite binary sequence starting with a 1 and having 1's or 0's thereafter endlessly. SignatureVirgil In article <61b968d6-446d-4592-ba13-62335ce9dc5b@j18g2000yql.googlegroups.com>, > > > > > > > > > On 15 Jun., 22:04, William Hughes <wpihug...@hotmail.com> > > > > > > > > > wrote: [quoted text clipped - 40 lines] > > Other subsets of nodes are irrelevant. There is a very relevant 'subset of nodes', namely the set of all nodes. If this set has, as it must have, two children for each and every node then it must also have subsets of nodes ordered by "ancestor" in which each child node is a 1 node. The maximal such set is a path but is not in the P WM described as having all but finitely many 0 child nodes. > > <snip> > > [quoted text clipped - 8 lines] > ancestor relation are as irrelevant as the question of present day > lunch. Not at all. The most important set of nodes is the set of ALL nodes, which defines the tree. Since the set of all paths derives from the set of all nodes and the partial order induced by the "parent of" relation on that set of all nodes, nothing more is needed than the set of all nodes and the "parent of" relation. And paths only exist as maximal totally ordered subsets of that partially ordered set of nodes. And there are in any maximal infinite binary tree paths that are not members of any countable set of paths, as Cantor proved. SignatureVirgil > And there are in any maximal infinite binary tree paths that are not > members of any countable set of paths, as Cantor proved. Then let me know one of those paths, please. I will tell you, whether it belongs to a counatble set. Regards, WM >> And there are in any maximal infinite binary tree paths that are not >> members of any countable set of paths, as Cantor proved. [quoted text clipped - 3 lines] > > Regards, WM Your quantifier confusion is showing again. For any path in the complete maximal binary tree, there is at least one countable set of paths that contains it. (Trivially, the set {p} contains the path p for every path p. The union of {p} and the set of terminating paths is a countably infinite set which contains p.) For any countable set of paths, there is at least one path in the tree that is not in that set. Consider the set of terminating rooted paths: it is a countable infinite set (its elements can be put in 1-to-1 correspondence with the natural numbers), and missing every non-terminating rooted path, as well as every (terminating or non-terminating) non-rooted path. Describe a countable set of paths that you believe contains every path in the complete maximal binary tree. -o > >> And there are in any maximal infinite binary tree paths that are not > >> members of any countable set of paths, as Cantor proved. [quoted text clipped - 5 lines] > > Your quantifier confusion is showing again. Your belief in matheological nonsense is showing again. > For any path in the complete maximal binary tree, there is at least one > countable set of paths that contains it. (Trivially, the set {p} [quoted text clipped - 3 lines] > For any countable set of paths, there is at least one path in the tree > that is not in that set. Your belief in matheological nonsense is showing again. You claimn also: "For any list of binary sequences, there is at least one binary sequence that is not in that list"? The tree is constructed from the list of terminating sequences in the form of terminating paths. This list does not contain the path 0.111... Hence the union of terminating paths, namely the tree, does not contain that path 0.111... You cannot distinguish it from all paths of the tree because 0.111... does not contain any digit that is not in a "path" of the list, together with all its preceding digits. > Consider the set of terminating rooted paths: it is a countable > infinite set (its elements can be put in 1-to-1 correspondence with the [quoted text clipped - 3 lines] > Describe a countable set of paths that you believe contains every path > in the complete maximal binary tree. The set of all terminating paths extended by tails 000... contains all nodes and all combinations of nodes that can act as paths in the tree. Regards, WM In article <b10439d6-802c-4b0f-81d8-acf109037ca1@e20g2000vbc.googlegroups.com>, > > >> And there are in any maximal infinite binary tree paths that are not > > >> members of any countable set of paths, as Cantor proved. [quoted text clipped - 7 lines] > > Your belief in matheological nonsense is showing again. Fools like WM sneer at what they cannot understand. > > For any path in the complete maximal binary tree, there is at least one > > countable set of paths that contains it. (Trivially, the set {p} [quoted text clipped - 5 lines] > > Your belief in matheological nonsense is showing again. Cantor has proved, and WM has not disproved, what WM is calling nonsense. Such 'nonsense' is of much greater mathematical value than anything WM has ever done. > also: > "For any list of binary sequences, there is at least one binary > sequence > that is not in that list"? Cantor proved it, and none of WM's attempts to fault that proof are worth the time it takes to read them. > The tree is constructed from the list of terminating sequences in the > form of terminating paths. It can be, but need not be, constructed that way. One can construct the complete infinite binary tree directly from the 1-origin naturals by taking 1 as the root node and for any n taking 2*n + 0 and 2*n + 1 as, respectively, node n's left and right children. Note that for this construction, no paths at all exist until the construction is complete. > This list does not contain the path 0.111... > Hence the union of terminating paths, namely the tree, does not > contain that path 0.111... My tree above does. It is the minimal set of naturals containing 1 and for each of its members n containing also 2*n+1. > You cannot distinguish it from all paths of the tree because 0.111... > does not contain any digit that is not in a "path" of the list, > together with all its preceding digits. One can distinguish it in my tree from any path containing an even integer, which includes all other paths. > > Consider the set of terminating rooted paths One does not need to consider any paths to get the desired tree, as my construction above shows. > > Describe a countable set of paths that you believe contains every path > > in the complete maximal binary tree. > > The set of all terminating paths extended by tails 000... contains all > nodes and all combinations of nodes that can act as paths in the tree. In my tree, as defined above, that is flat out false. In my tree your paths all are limited to finitely many odd naturals, but there are paths with infinitely many odd naturals in them, including one which contains only odd naturals. > Regards, WM SignatureVirgil In article <eba4a799-5cf3-4a3a-a714-80c34b29e3ba@t21g2000yqi.googlegroups.com>, > > And there are in any maximal infinite binary tree paths that are not > > members of any countable set of paths, as Cantor proved. > > Then let me know one of those paths, please. I will tell you, whether > it belongs to a counatble set. Any path belongs to the countable set of which it is the sole member. The issue is whether there is any particular countable set of paths of which each and every path is member. Cantor proved the answer to be "no". SignatureVirgil > Then let me know one of those paths, please. I will tell you, whether > it belongs to a counatble set. EVERY path belongs to a countable set, DUMBASS. It belongs to the set of cardinality ONE, containing ONLY THAT path! This is NOT the same as "there is a countable set to which every path belongs". "Every path belongs to a countable set" DOES NOT MEAN "Every path belongs to some(fixed) countable set"!! In article <887a8991-e0b7-4fa3-8a5d-cfea7c5abfa7@n21g2000vba.googlegroups.com>, > > Then let me know one of those paths, please. I will tell you, whether > > it belongs to a counatble set. [quoted text clipped - 5 lines] > "Every path belongs to a countable set" DOES NOT MEAN > "Every path belongs to some(fixed) countable set"!! WM's quantifier dyslexia strikes again" Wm claims "For each path, p, there exists a countable set, S, with p e S" implies There exists a countable set S, such that for each path, p, p e S. But outside of WM's world of mathUnrealism, it doesn't. As Cantor proved, and WM has vainly tried to disprove. SignatureVirgil In article <937ab05c-8ff4-431d-8eb0-b361c703cd04@3g2000yqk.googlegroups.com>, > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 34 lines] > parts of paths. The subset of nodes consisting of all nodes of the > 17th level of the tree is not under consideration. Every set of odes is under consideration when WM only specifies "subsets of nodes". If WM wishes to limit the context specifically to subsets ordered by "ancestor of", he must explicitly indicate such a restriction. > > > I add the paths p_n , that have a tail of zeros beginning at node n, > > > one by one, in the following order: [quoted text clipped - 26 lines] > What unadded subset, than can be (a part of) a path, is created when > what path is added? When ALL your paths, as ordered node sets having a last 1, have been unioned into one large node set, that node set will contain paths with no last 1. Thus there are paths in the constructed tree which were not members of the set of paths from which it was constructed SignatureVirgil > Precisely: We are talking about subsets of nodes that can be paths or > parts of paths. NO subset of nodes can be a PATH, dumbass! PATHS have EDGES!! In article <6ac28556-4c38-46b5-8bb8-bc17447d740f@z16g2000prd.googlegroups.com>, > > > > > > > > Your claim is that "no possibility exists to construct or to > > > > > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 27 lines] > > ((A ==> B) & ~B) ==> ~A If no infinite sets can exist, as WM claims, then everything that WM claims about them is false.. > > > > > Every path of the tree is is from P. > > [quoted text clipped - 7 lines] > > I add infinite paths. Then WM is claiming that infinite paths both exist and do not exist, so that his set theory is self contradictory. > > When you add a node to the tree you add subsets of nodes > > to the tree as well. You add a subset of nodes that is > > not in a single element of P. > > Wrong. It is WM who is wrong, as there are all sorts of subsets of the set of nodes which are not paths, so cannot be in P, but are sets of nodes in that tree. > I add a path p_n like > 0.himpidimpydowadididum000... Then WM no longer has a binary tree at all, but only an irrelevant figment of his warped imagination. > That path p_n is different from any other path p_m and there is no > path p_m that together with p_n produces anything different from all > finite subsets that are already contained in p_n or p_m. No finite subset of the node set of a maximal infinite binary tree is a path in that tree. > You are dreaming of an ideal set theoretic world - without paradoxes > and antinomies. Why not? Such ideal worlds are what mathematics is all about. > But this world has come to an end in 2004 with the > first construction, recognition and careful interpretation of the > binary tree. Which binary tree or trees would that be? None of WM's quasi-trees can qualify as a maximal infinite binary tree. And as it is only in WM's misrepresentations of ZF, and not in ZF itself, that any anomalies appear, the maximal infinite binary trees in ZF behave quite nicely. SignatureVirgil In article <bf8603a8-ed62-48a7-afc3-6bfaeeb8bb91@y9g2000yqg.googlegroups.com>, > If actually infinite paths exist, then there is a path p that can be > distinguished from every path of P. [quoted text clipped - 3 lines] > This kind of proof is called proof by contradiction or modus tollens: > ((A ==> B) & ~B) ==> ~A But WM has not proved either. If the node set of a maximal infinite binary tree tree is actually infinite and P is any countable or finite set of paths (or other sets of nodes) of that tree (where a path is any maximal linear subset of the set of nodes under the ancestor relation) then there are paths not members of P. And any such non-member path can be distinguished from all members of P. > > Irrelevant, you distinguish p from every element of P > > Not after they have been written in form of tree. The set P does not contain, as a member, after a construction of a tree anything that P did not contain, as a member, before that construction. So that if P starts either finite countably infinite, it remains so, but the set of ALL paths in a maximal infinite binary tee, regardless of how it may have been "constructed", is uncountably infinite, a la Cantor. > > > Every path of the tree is is from P. > > > > Nope. Every *node* of the tree is from P. > > Every path of the tree is from P. I explicitly forbid every other path > to enter my tree. Then your tree is not maximal, and may not even be a tree in the sense that there will be in it linearly ordered sets of nodes which are not subsets of any path. > But, surprise, there is no other path willing to do > so. Maybe not in those quasi-trees that WM "constructs" but if WM's the member paths of WM's P are countable in cardinality, then there will be sets of nodes satisfying all the requirements of pathhood in the constructed tree that are not in WM's P. > What kind of path should apply, after I have every node of the > tree covered with a path of P? All those uncountably many maximal linearly-ordered sets of nodes that are not members of P. > > However, there is a *subset of nodes* in the tree that is not > > contained in one element of P. > > Wrong. Every node and every subset of nodes is in a path of P. It is WM who is dead wrong about that, apparently because he does not even understand what paths are. There are all kinds of "subsets of nodes" which cannot be members of any path. For example, no path can contain both children of any parent node. And if P is countable, then its member paths can be listed and the Cantor construction produces a path not in P > ideas you talk about are not realized by digits or bits or nodes They are "realized" in the imagination of those whose imaginations are not as crippled s WM's, which is all that is needed in mathematics. SignatureVirgil > > > You are trying a proof by contradiction of > > > " Actually infinite paths exist" [quoted text clipped - 9 lines] > > > there is no path p that can be distinguished > > > from every path of P. You know modus tollens? ((A ==> B) & ~B) ==> ~A A: actually infinite paths exist. B: p can be distinguished from every path of P. > > You cannot distinguish p from every path of the tree. > > Irrelevant, you distinguish p from every element of P The tree is nothing but every path of P written in some unconvential form. > > Every path of the tree is is from P. > > Nope. Every *node* of the tree is from P. > However, there is a *subset of nodes* in the tree that is not > contained in one element of P. Every subset of nodes, that can form a path of the tree, is in the tree. It stems from writing down the paths of P and no bit more. (Even if the tree is considered to be a union of sets {p_n} with p_n in P, then the union could not contain more than is contained in at least one of the united sets.) Regards, WM In article <f59dc977-e9eb-4bbc-bc62-42501f504a9a@r33g2000yqn.googlegroups.com>, > > > > You are trying a proof by contradiction of > > > > " Actually infinite paths exist" [quoted text clipped - 15 lines] > A: actually infinite paths exist. > B: p can be distinguished from every path of P. But you have not established your " ~"B anywhere outside of you own little world of MathUnrealism. > > > You cannot distinguish p from every path of the tree. > > > > Irrelevant, you distinguish p from every element of P > > The tree is nothing but every path of P written in some unconvential > form. Np maximal infinite binary tree is limited to a merely countable set of paths, so that WM's trees are all crippled. > > > Every path of the tree is is from P. > > [quoted text clipped - 4 lines] > Every subset of nodes, that can form a path of the tree, is in the > tree. It stems from writing down the paths of P and no bit more. P only provides the complete set of nodes for a maximal infinite binary tree, but, if only countable, not any *complete* set of paths, as is directly derivable from Cantor's diagonal proof. > (Even if the tree is considered to be a union of sets {p_n} with p_n > in P, then the union could not contain more than is contained in at > least one of the united sets.) WM falsely argues that what holds for finite trees must necessarily hold for infinite trees. Since the finiteness property does not carry over, it is clear that not ALL properties do. For a finite complete binary tree it is quite true that any set of paths whose union contains all nodes also contains all paths, but that is because of the easy bijection between leaf nodes and paths in such trees, but in trees without leaf nodes fails. In maximal infinite binary trees, each of countably many nodes is in all of the uncountably many paths passing through it. SignatureVirgil > In article > <f59dc977-e9eb-4bbc-bc62-42501f504...@r33g2000yqn.googlegroups.com>, [quoted text clipped - 20 lines] > > But you have not established your " ~"B Try to find out where the path 1/3 = 0.010101... or any other path differs from the complete, maximal, infinite binary tree. The tree is constructed by all terminating paths with tails 000... The same tree could be constructed by all terminating paths with tails 0.010101... There is no difference. That's ~B. Regards, WM In article <e1027530-bcc6-4a92-a3ea-f5e2bce14736@37g2000yqp.googlegroups.com>, > > In article > > <f59dc977-e9eb-4bbc-bc62-42501f504...@r33g2000yqn.googlegroups.com>, [quoted text clipped - 23 lines] > Try to find out where the path 1/3 = 0.010101... or any other path > differs from the complete, maximal, infinite binary tree. Your countable set of paths, P, does not contain as members all paths of any maximal infinite binary tree, though it may contain as members of members all the nodes. > The tree is constructed by all terminating paths with tails 000... Which is a countable set of paths, and therefore, by Cantor, does not include all paths. > same tree could be constructed by all terminating paths with tails > 0.010101... > There is no difference. Any set of paths, or other sets of nodes, whose union contains as members all nodes of a prospective maximal infinite binary tree tree can be used to construct that tree, whether that set of sets of nodes contains as members all the paths of that tree or not. Once one has the set of all its nodes, regardless of how they are come by, one has the tree, and all its paths derive from the node set and tha 'parent of' relation, not from any a priori set of paths. SignatureVirgil > > The tree is constructed by all terminating paths with tails 000... > > Which is a countable set of paths, and therefore, by Cantor, does not > include all paths. By Jove, it does! > > same tree could be constructed by all terminating paths with tails > > 0.010101... [quoted text clipped - 8 lines] > by, one has the tree, and all its paths derive from the node set and tha > 'parent of' relation, not from any a priori set of paths. Once one has the set of all binary sequences one has a list that contains all real numbers. This list is only written in a somewhat unconventional form. Regards, WM In article <faaecaf1-57f9-435a-aba9-f731cde7f5da@a36g2000yqc.googlegroups.com>, > > > The tree is constructed by all terminating paths with tails 000... > > [quoted text clipped - 18 lines] > Once one has the set of all binary sequences one has a list that > contains all real numbers. Wrong! One has a set which cannot be made into a list, as Cantor proved. > This list is only written in a somewhat > unconventional form. An incoherent form, since it cannot be written as a list at all. > Regards, WM SignatureVirgil > > > The tree is constructed by all terminating paths with tails 000... > > > Which is a countable set of paths, and therefore, by Cantor, does not > > include all paths. > > By Jove, it does! No, it doesn't. As you yourself have already pointed out, this set is easily bijectible with the set of paths terminating with ANY given tail, INCLUDING THE EMPTY tail (which would leave us with the set of all FINITE paths). The set of all FINITE paths will "cover" or "be able to construct" THE WHOLE tree, according to your (foolish) usage of these terms. This does NOT imply that actually infinite paths do not exist. The fact that one set of paths "covers or can construct" the tree goes NOWHERE toward proving that paths outside that set cannot be "contained in" the tree. > Try to find out where the path 1/3 = 0.010101... or any other path > differs from the complete, maximal, infinite binary tree. You don't know what you mean by "differ from". A proper subset differs from any superset BECAUSE there are elements in the superset that are not in the subset. These elements do NOT need to be anyWHERE. But if you insist on a location, every node and edge of the tree that IS NOT ON the path is where they (PLURAL, as TWO things) differ. > The tree is constructed by all terminating paths with tails 000... So what? That does NOT imply that no paths that end in something other than 000.... can be "in" the tree! You could LEAVE ALL THE TAILS OFF and still "construct" THE SAME tree! You can construct the ACTUALLY infinite maximal binary tree out of NOTHING BUT THE FINITE paths (as long as you use an actually infinite NUMBER OF these paths IN the construction). > The > same tree could be constructed by all terminating paths with tails > 0.010101... > There is no difference. The same tree could be constructed by all terminating paths WITH THE EMPTY tail, DUMBASS. The class of all FINITE paths is SUFFICIENT, *IF THIS IS WHAT YOU MEAN* by "construct". It is NOT what anybody else means. In article <8229ef12-6482-47df-9dd9-851dd82ab058@y17g2000yqn.googlegroups.com>, > > Try to find out where the path 1/3 = 0.010101... or any other path > > differs from the complete, maximal, infinite binary tree. [quoted text clipped - 16 lines] > long as you use an actually infinite NUMBER OF these paths IN the > construction). One can even construct a tree from nothing but its nodes, without any references to paths at all. But once constructed, certain sets of nodes will be paths while others will not, regardless of how, or from what, that tree was originally constructed. > > The > > same tree could be constructed by all terminating paths with tails [quoted text clipped - 5 lines] > *IF THIS IS WHAT YOU MEAN* by "construct". It is NOT what anybody > else means. SignatureVirgil In article <c3bd189f-fae5-490d-8c53-e7d516f74ebf@r31g2000prh.googlegroups.com>, > > Irrelevant. You do not distinguish p from the tree. > > You distinguish p from every element of P. > > You cannot distinguish p from every path of the tree. > Every path of the tree is is from P. Not if one can count the paths of P, since on cannot count the path of a maximal infinite binary tree. > > > P contains every subset of nodes of the binary tree. P, being a set of sets of nodes, would have to be the powerset of the set of all nodes to have as a member EVERY set of nodes, but since the set of nodes is countably infinite, that makes P, the set of all its subsets, uncountably infinite. SignatureVirgil In article <45a7a810-fac6-4e84-9bbe-759693328e53@j32g2000yqh.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 17 lines] > No we need and have in fact: There is no path p that can be > distinguished from every path of P. What WM claims he needs does not make those putative needs actual. > > Your putative proof > > > > you are not able to distinguish p from P (by means of digits). If one has any finite or countably infinite listing of infinite binary sequences (paths), then the Cantor construction finds one not in that listing. > Wrong. P contains every subset of nodes of the binary tree. There is > nothing else. If P is listable (countable) then every listing of its members produces a non-member by the Cantor construction. > > > You know that you are unable to > > > distinguish any binary sequence representing a real of the unit [quoted text clipped - 5 lines] > > As my proof shows, P contains every subset of nodes. WM has often claimed this, but never been able to prove it without imposing conditions contrary to, say, ZF, or whatever axiom system allows a maximal infintie binary tree to exist at all. In WM's "system", there is no such tree, but his is not the only system. There is nothing in mathematics that requires everyone to play by WM's rules. And until WM learns that, he will be forever an outsider. > > You can distinguish p from every element of P by > > subsets of nodes. > > > Wrong. That depends on P. If P, as a set of infinite binary sequences, is uncountably infinite, it may also be maximal, in the sense of containing every possible infinite binary seqeunce, but Cantor proved (or rather a corollary to his construction proved) once and for all that as soon as a set of such sequences can be listed (counted), it is missing at least as many members as it contains, The tree contains nothing more than the elements of P. The tree 'contains' all sorts of things not in P including many sets of nodes which are not paths at all. The tree, as a set of nodes, has no members which are paths, but only has subsets which are paths. So that WM is attempting to obfuscate things by using "contain" ambiguously. To be clear, one must distinguish between 'contain as a subset' and 'contain as a member'. While P as a set of paths may contain paths as members, it does not contain any Paths as subsets. So the fact that the union of P (the set of all elements of elements of P) may include all nodes does not in any way entail that P contains all paths. SignatureVirgil In article <0373342b-3226-4cd1-a89c-270e6b5a103a@a36g2000yqc.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." > > In fact that is not possible. That depends critically on what set of paths one is attempting to distinguish the given path from. If one takes any countable set of paths in the maximal infinite binary tree, then there are paths in that tree which can be "distinguished" from every path in that set. However if one takes the set of ALL paths in that tree , there are then no paths in the tree that can be distinguished from all paths in that set. > > > > > There are two statements: > > > > > 1) Path p can be distinguished from every path of P. [quoted text clipped - 11 lines] > > But it turns out that this assumption cannot be satisfied. Maybe not in WM's world, but there are worlds that WM apparently cannot access that are wide open to mathematicians. > > Your claim explicitly mentions distinguishing p from P by > > "one or many or infinitely many nodes". [quoted text clipped - 5 lines] > > We have a list of a countable set P of terminating paths p_n. Then any non-terminating path can be distinguished from every member of that set. > By the > diagonal method we can distinguish p from every p_n (if actual [quoted text clipped - 5 lines] > else but writing the list in slightly different form, saving some ink. > In particular we have not added any new path. You have made it impossible to distinguish between distinct paths even within your original set. > This yields the complete binary tree. No it doesn't. As soon as it combines any two paths into one, which it does en mass, whatever remains is incomplete. > You know that you are unable to > distinguish any binary sequence representing a real of the unit > interval from every path of the tree. If the tree is a maximal infinite binary tree, then every real in that interval has at least one binary representation. If the tree is, as WM would have it, less than maximal, then it will be missing some paths that the maximal tree contains. > This proves that you, contrary to the assumption, have been unable at > the beginning too, because there the same set of paths was presented > to you. WM is delusional again. Any maximal infinite binary tree will contain more that countably many paths because no countable set of paths can be maximal. There are always others not yet included. > You only believed that you were able. With much better justification that many of WM's beliefs. > Cantor's diagonal method > is falsified. Not by WM and not by anyone else. SignatureVirgil > In article > <0373342b-3226-4cd1-a89c-270e6b5a1...@a36g2000yqc.googlegroups.com>, [quoted text clipped - 7 lines] > That depends critically on what set of paths one is attempting to > distinguish the given path from. Every path that is in the tree belongs to this set. The tree is simply consructed by writing the paths in unconventional manner. > If one takes any countable set of paths in the maximal infinite binary > tree, then there are paths in that tree which can be "distinguished" [quoted text clipped - 3 lines] > no paths in the tree that can be distinguished from all paths in that > set. Even if we considered every path p_n as a set {p_n} and the tree as the union of these sets, then the tree should not contain a path that is absent from each of the sets. Nevertheless the complete tree contains all possible paths. (Possible means, paths tha can be written by bit sequences.) > > > > > > There are two statements: > > > > > > 1) Path p can be distinguished from every path of P. [quoted text clipped - 14 lines] > Maybe not in WM's world, but there are worlds that WM apparently cannot > access that are wide open to mathematicians. The world of matheology: spirituality, visionaries and delusions. > > I will try to explain it for you again: > > > We have a list of a countable set P of terminating paths p_n. > > Then any non-terminating path can be distinguished from every member of > that set. One is tempted to think so. And if there actually are non-terminating paths, you should be correct. > > By the > > diagonal method we can distinguish p from every p_n (if actual [quoted text clipped - 13 lines] > No it doesn't. As soon as it combines any two paths into one, which it > does en mass, whatever remains is incomplete. There is no node that remains unconstructed. > > You know that you are unable to > > distinguish any binary sequence representing a real of the unit [quoted text clipped - 4 lines] > would have it, less than maximal, then it will be missing some paths > that the maximal tree contains. The tree is as maximal as a tree can be, after construction by a countable set of paths. By the way: What more should be used tio construct the tree than a set of paths that reaches every node? What more than every node is the tree? Regards, WM In article <15d0007a-2eea-46cb-b40d-5a8ba266f138@d38g2000prn.googlegroups.com>, > > In article > > <0373342b-3226-4cd1-a89c-270e6b5a1...@a36g2000yqc.googlegroups.com>, [quoted text clipped - 9 lines] > > Every path that is in the tree belongs to this set. If the tree is a maximal infinite binary tree and every (maximal) path in it is a member of P, then P is uncountably infinite. > The tree is simply consructed by writing the paths in unconventional > manner. Since that "manner" conflates distinct paths, the membership of P is no longer deteminate. > > If one takes any countable set of paths in the maximal infinite binary > > tree, then there are paths in that tree which can be "distinguished" [quoted text clipped - 7 lines] > the union of these sets, then the tree should not contain a path that > is absent from each of the sets. No (maximal) path is a subset of any of your finite pathlets, WM, but infinitely many different finite pathlets of different lengths are subsets of every (maximal) path. Each of WM's pathlets determines and is determined by a unique node, its last node, but no (maximal) path has any last node. > Nevertheless the complete tree > contains all possible paths. (Possible means, paths tha can be written > by bit sequences.) Our "maximal infinite binary tree" certainly contains all possible paths, but unless WM's "complete" tree means the same as our "maximal infinite binary tree", his may well not contain all of them. > > > > > > > There are two statements: > > > > > > > 1) Path p can be distinguished from every path of P. [quoted text clipped - 16 lines] > > The world of matheology: spirituality, visionaries and delusions. Anything that WM cannot understand he equates with magic. > > > I will try to explain it for you again: > > [quoted text clipped - 5 lines] > One is tempted to think so. And if there actually are non-terminating > paths, you should be correct. In ZF, there are and he is. And WM has no set theory, only a bunch i=of incoherent assumptions. > > > By the > > > diagonal method we can distinguish p from every p_n (if actual [quoted text clipped - 15 lines] > > There is no node that remains unconstructed. Nodes are not paths in any binary tree with more than one node. > > > You know that you are unable to > > > distinguish any binary sequence representing a real of the unit [quoted text clipped - 7 lines] > The tree is as maximal as a tree can be, after construction by a > countable set of paths. But the completion of that construction can create paths not used in the construction itsef. The construction of a maximal infinite binary tree only requires existence of each of its NODES, with their appropriate connections to parent and children nodes. That construction does not require a priori any paths at all. Paths are a conseqeunce of the node set, not the cause of it. > By the way: What more should be used tio construct the tree than a set > of paths that reaches every node? Answered above: the set of nodes is all that is needed. Any paths that come into being then derive from the partial order induced by the "parent of" relation. > What more than every node is the tree? That's all a tree needs to exist, and all its paths need. So one can construct a tree without having a priori any paths at all. SignatureVirgil In article <e743cd44-0d28-4d21-a014-fce8422eba84@h28g2000yqd.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 25 lines] > No. The union of paths is not larger than every element - unless we > agree to magic. WM again loses track of what is going on. A union of paths, such as WM mentions, would be a set of nodes which is not, in general, a path at all. Hughes talks about unions of sets of paths, which would produce a new set of paths, each being a special sort of set of nodes, not merely a set of nodes which is not a path at all. > In particular p cannot be distinguished from the paths of P, used to > construct the tree, if I had included p in P. But there is no way to include all potential paths in such a construction as WM describes, so whatever ones he uses, others appear in his final tree. > I agree that p was not > included. But that does not play a role unless you are able to obtain > that from the complete tree. But you are not! Maybe WM is not able to do so. He seems self-crippled in so many ways, but those less handicapped can. > > > > You refuse to agree to > > [quoted text clipped - 16 lines] > If this was possible before, then it would also be possible post > festum, because only P was used to construct the tree. The point is that if only countably many paths are used in that construction, there will inevitably appear in the constructed tree other paths. At least so long as paths are defined as maximal totally ordered sets of nodes under the "ancestor" partial ordering of nodes. > But it is not. Wrong! > Therefore it turns out that the original assumption "p can be > distuinguished from every path of P" is contradicted. Wrong! > A classical proof by conradiction. Except that it is not a proof because it relies on assumptions contrary to fact. > By the way, one of the finest proofs that > exists in mathematics. WM will break his arm patting himself on the back so hard. But WM is doubly mistaken in his ego trip. His argument is neither a proof nor is it in mathematics. SignatureVirgil In article <ed7cfe4c-c486-43c0-bb1f-8ceb2652c77c@y17g2000yqn.googlegroups.com>, > > Assuming infinite paths exist: > > [quoted text clipped - 7 lines] > because, if p exists as binary representation, then p is the union of > all p_n and as such cannot differ from the union. Paths are not made up of digits, but of nodes. But in a binary tree, one can represent a path as an infinite sequence of binary digits with, say, 0 for each left child node and 1 for each right child node. But no such path p is then a union of other paths, as WM claims, as no union of a set of more than one path can be a path itself. > > You refuse to agree to > > [quoted text clipped - 9 lines] > It is obviously impossible to distinguish p from all paths p_n that > are in the binary tree after construction has been completed. If p_n refers to any finite or even countably infinite set of paths then every non-member of p_n, is distingishable from every member of p_n and every subset of p_n. > If you don't believe me, then play the game. Minimum stake 10^6 > dollars or (if you come from Europe) Euros or Pounds. Then I challenge WM to present his list of paths from which he claims no other path can be distinguished. Note that since any counterexample may be dependent on his choice of list, as Cantor's depends on the list being given, WM is constrained to publish his list first. SignatureVirgil > In article > <ed7cfe4c-c486-43c0-bb1f-8ceb2652c...@y17g2000yqn.googlegroups.com>, [quoted text clipped - 17 lines] > But no such path p is then a union of other paths, as WM claims, as no > union of a set of more than one path can be a path itself. In an actually infinite binary tree constructed of terminatings paths only there is/would be the non-terminating path 0.111... How does it come into being, if it comes into being? > > > You refuse to agree to > [quoted text clipped - 13 lines] > every non-member of p_n, is distingishable from every member of p_n and > every subset of p_n. Not from every path however that belongs to the binbary tree. Therefore your claim is not based upon digits or nodes but on superstition. > > If you don't believe me, then play the game. Minimum stake 10^6 > > dollars or (if you come from Europe) Euros or Pounds. > > Then I challenge WM to present his list of paths from which he claims no > other path can be distinguished. You must be able to to distinguish yout path from the complete binary tree constructed by my list of path by means of nodes only. What is a name? that which we call a rose by any other name would smell as sweet. Paths afre not identified by names in Cantor's lists. So we will not do it here. > Note that since any counterexample may be dependent on his choice of > list, as Cantor's depends on the list being given, WM is constrained to > publish his list first. I published the completed binary tree. It contains every sequence of nodes. Regards, WM In article <3b84e611-0e94-4d16-af3a-a568d149191f@n7g2000prc.googlegroups.com>, > > In article > > <ed7cfe4c-c486-43c0-bb1f-8ceb2652c...@y17g2000yqn.googlegroups.com>, [quoted text clipped - 22 lines] > > How does it come into being, if it comes into being? If one has all finite complete (all path of the same length and all non-leaf nodes having exactly two children) binary trees, then the union of the set of all their node sets is the node set of a maximal infinite binary tree having uncountably many actually infinite paths. > > If p_n refers to any finite or even countably infinite set of paths then > > every non-member of p_n, is distingishable from every member of p_n and > > every subset of p_n. > > Not from every path however that belongs to the binbary tree. But no one has claimed that. But since your p_n cannot simultaneously include all paths of the resulting maximal infinite binary tree and be countable, that is irrelevant. > Therefore your claim is not based upon digits or nodes but on > superstition. The only "superstition" involved is the definition of a maximal infinite binary tree, together with the definition of a path in such a tree. Which involves much less superstition than 'potentially infinite sets'. > > > If you don't believe me, then play the game. Minimum stake 10^6 > > > dollars or (if you come from Europe) Euros or Pounds. [quoted text clipped - 4 lines] > You must be able to to distinguish yout path from the complete binary > tree constructed by my list of path by means of nodes only. WM wants to change the game now. The original proposal was for WM to present a list of infinite binary paths and challenge me to find path not listed. Which game he would inevitably lose. But if WM wants to present an unlisted set of paths, I do not guarantee to find another path not in it. > What is a > name? that which we call a rose by any other name would smell as > sweet. The quote properly begins "What's in a name?" > Paths afre not identified by names in Cantor's lists. So we > will not do it here. > > [quoted text clipped - 4 lines] > I published the completed binary tree. It contains every sequence of > nodes. You have done no such thing. You may have referred to it, rather then your incomplete versions of it, but 'publishing' it would require at least explicitly listing ALL its nodes, which being infinite in number, cannot be done via newsnet. SignatureVirgil In article <25be3ee1-894f-4bcd-a5fa-b064b03dd3b6@g19g2000yql.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." > > snip attempt to evade the topic. Topic restated and elaborated: Snip WM's attempt to divert the discussion from issues raised by Hughes. Answer Hughes first. WM. SignatureVirgil > We agree that in Cantor's diagonal argument, applied to real numbers, NO, we DON'T agree to that, because Cantor's diagonal argument DOES NOT apply to real numbers. Cantor's diagonal argument IS NOT ABOUT ANY kind of numbers, with the possible exception of ordinals. Cantor's Theorem is ABOUT *SETS* and it applies TO ALL sets -- finite, infinite, ordered, ordinal, whatEVER. NUMBERS simply have NOTHING WHATSOEVER TO DO with Cantor's Theorem UNTIL AFTER numbers are ENCODED as certain kinds of sets. > the numbers are represented and identified solely by their digits. That is an artifact of the FACT that SETS are identified solely BY THEIR ELEMENTS. The digits are simply aliases for the presence or absence of an element (of the domain) FROM A SUBSET (of the domain). > No further information is available. > > We assume that a real number p can be distinguished from all elements > of a set Q of real numbers q by general considerations, NO, we don't. "Ability" to "distinguish" (or not) SIMPLY IS NOT A RELEVANT consideration AT ALL for this enterprise! FOR ANY two sets, they can be distinguished if and only if there exists some OTHER set that IS a member of one BUT NOT a member of the other! THAT IS ALL! NO OTHER MODE of "distinguishing" is relevant OR EVEN ALLOWED! If either of the sets is infinite then there can exist practical considerations regarding ACTUALLY FINDING or specifying a distinguishing element, but as far as the FACTUAL QUESTION of the distinctness of the sets is concerned, ALL THOSE CONSIDERATIONS ARE COMPLETELY irrelevant. > for instance, > if p is a transcendental number and Q consists of rational numbers q > only. > (This implies the existence of such a number p, but does not imply the > existence of a decimal expansion of p.) The existence of the decimal expansion of p IS ALREADY IMPLIED BY THE POWERSET AXIOM, DUMBASS. What you call "the decimal expansion of p" *IS* p, in THIS treatment! THIS treatment is about SUBSETS, NOT transcendental numbers, and ALL the relevant numbers, transcendental or otherwise, are being represented BY ENCODINGS which are ALREADY PROVEN to exist! > NO, we DON'T agree to that, because Cantor's diagonal argument DOES > NOT apply to real numbers. Cantor's diagonal argument IS NOT ABOUT > ANY kind of numbers, with the possible exception of ordinals. Why this exception? In any case, under discussion in this thread is the application of the diagonal argument used to establish the uncountability of reals. SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon mann nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus In article <b3c02082-904a-4829-91df-e1f47c8f9df9@g1g2000yqh.googlegroups.com>, > > <snip evasion> > > [quoted text clipped - 5 lines] > > Regards, WM O.K., I have my path. If you think you guessed it, tell me what you think it is. And I will tell you whether you are right. Or wrong! SignatureVirgil In article <9209f1b7-64eb-4486-9957-03349cd2c488@h28g2000yqd.googlegroups.com>, > We agree that in Cantor's diagonal argument, applied to real numbers, > the numbers are represented and identified solely by their digits. No [quoted text clipped - 15 lines] > is impossible to distinguish p from the binary tree. But the binary > tree has been errected by using a countable set of paths only. That ignores the fact that the result of such building automatically contains paths not required in the construction itself. > Therefore it is impossible to distinguish, by means of digits, any > real number from a countable set of paths. Not in ZF. > Conclusion: Cantor's diagonal argument is contradicted. Wrong outside of WM's world of MathUnrealism. > Comment: This is not surprising, because it is impossible to identify > any irrational number by means of its decimal representation, but just > this is claimed by the diagonal argument. Not at all. The diagonal argument, as applied to infinitely long strings of decimal digits, only says that any listing of such strings is incomplete. Which it is. SignatureVirgil > We agree that in Cantor's diagonal argument, applied to real numbers, > the numbers are represented and identified solely by their digits. No [quoted text clipped - 7 lines] > for every digit d_n of p, there is a number q that shares all digits > up to d_n with p. Wait, so you're arguing that you can't distinguish 1/pi (roughly 0.31830988...) from a set containing all rationals in [0, 1] because, for example, 3/10 is equal to 1/pi to the first decimal place, 31/100 is equal to the second, and so on? What about the non-zero difference between any one rational number and 1/pi? |1/pi - 3/10| is greater than the rational 1/50. |1/pi - 31/100| is greater than the rational 1/125. In fact, for each rational q in [0, 1], 1/pi differs from q by an amount greater than some other rational r in [0, 1], so we can distinguish 1/pi from every rational in [0, 1]. -o > > We agree that in Cantor's diagonal argument, applied to real numbers, > > the numbers are represented and identified solely by their digits. No [quoted text clipped - 18 lines] > 1], 1/pi differs from q by an amount greater than some other rational r > in [0, 1], so we can distinguish 1/pi from every rational in [0, 1]. If 1/pi exists as an actually infinite digit sequence. There are two statements: 1) Path p can be distinguished from every path of the countable set P that is used to construct the tree. 2) Path p cannot be distinguished from every path of P. The first is assumed to be correct before P was used to construct the tree. Above you are just arguing in favour of (1). The second is assumed to be correct after P was used to construct the tree. One of them is false, unless it is a magic tree where something happens during construction. But I do not believe in magic, least in mathematics. The second statement can be proved to be correct, because in fact you are not able to distinguish p from P (by means of digits) when the tree has been constructed. Therefore the first statement is falsified. This means that there are no actually infinite digit sequences, but only potentially infinite digit sequences. For *every* digit d_n of 1/pi there is a terminating path that is identical to the digit sequence d_1 to d_n and even until d-(n^n^n...^n). Therefore your argument above is not sufficient to save set theory. Regards, WM In article <5acbaafd-203e-4fef-8d81-be285ceca8e8@y9g2000yqg.googlegroups.com>, > > > We agree that in Cantor's diagonal argument, applied to real numbers, > > > the numbers are represented and identified solely by their digits. No [quoted text clipped - 24 lines] > 1) Path p can be distinguished from every path of the countable set P > that is used to construct the tree. Before the tree is built, every path that exists so far is, by necessity, one of those to be used to build the tree, so is in P. > 2) Path p cannot be distinguished from every path of P. Whatever countable set of paths, P, may be used to build a maximal infinite binary tree, there are too many paths in the resulting tree to be contained in the original P. SignatureVirgil > > 2) Path p cannot be distinguished from every path of P. > > Whatever countable set of paths, P, may be used to build a maximal > infinite binary tree, there are too many paths in the resulting tree to > be contained in the original P. I have built a complete maximal binary tree by means of a countable set P of path. You may choose a path and prove wether it is in the tree or not. I will tell you afterwards what P is. But this is of no relevance, because every countable set of terminating paths or paths withg a special tail like 31415... or 121212... will yield the same tree. Regards, WM In article <b1090bf0-4621-457f-9cd2-c6a617dba3fa@b9g2000yqm.googlegroups.com>, > > > 2) Path p cannot be distinguished from every path of P. > > [quoted text clipped - 4 lines] > I have built a complete maximal binary tree by means of a countable > set P of path. The issue is not whether my path is in the tree but whether it is in WM's original list of paths, P. And for any such list, Cantor provides an infallible method of producing an unlisted path. an with trivial variations, produce as many new paths as originally listed. > You may choose a path and prove wether it is in the > tree or not. I will tell you afterwards what P is. Unless you tell me in advance what P is, I would have to be guessing, but once you tell me, I can be quite sure that mine is not one of yours. > But this is of no > relevance, because every countable set of terminating paths or paths > withg a special tail like 31415... or 121212... will yield the same > tree. Including lots and lots of paths not used explicitly in the construction. Given any set of binary sequences, the very proof that it is countable, if it actually is, provides a method for finding sequences not included in it. SignatureVirgil >>> We agree that in Cantor's diagonal argument, applied to real numbers, >>> the numbers are represented and identified solely by their digits. No [quoted text clipped - 20 lines] > > If 1/pi exists as an actually infinite digit sequence. Since there is a function g from N to {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} that, for *every* element n of N, produces the n'th decimal place of 1/pi, it's safe to say that 1/pi has an infinite decimal representation. There's also a function g' from N to {0, 1} that produces, for *every* element n of N, the n'th bit of the binary representation of 1/pi. > There are two statements: > 1) Path p can be distinguished from every path of the countable set P > that is used to construct the tree. > 2) Path p cannot be distinguished from every path of P. Reading upstream a bit, I think you've misapprehended the point people are making. If P is a set of paths, and the union of all elements of P is the maximal complete binary tree, then there can be paths in the union of all elements of P that are not in P. If P is a countable set, then there are necessarily paths in the union of all elements of P that are not in P. In fact, one such set P is the set of finite rooted paths, which is isomorphic to the set of nodes in the maximal complete binary tree. We'll adopt a notation where each node is labelled with the set of left (denoted as 0) and right (denoted as 1) branchings necessary to reach it, prefixed by a dot (.): . is the root node. .0 is the left child of the root node. .1 is the right child of the root node. .00 is the left child of the left child of the root node. .01 is the right child of the left child of the root node. .10 is the left child of the right child of the root node. .11 is the right chind of the right child of the root node. (And so on.) Because every node can be reached in only finitely many branchings (as a consequence of the tree being representable as an infinite set of finite paths), every node has a finite label. We'll denote paths in the standard way, as sets of nodes: {., .0} is the path from the root to the leftmost rank 1 node. {., .0, .01} is the path from the root to the right-hand child of the left-hand child of the root node. Finite rooted paths can also be identified, for convenience, by their terminal node: there is one and only one path from the root node to the node .001101, so writing out the complete path is pointless. However, *infinite* paths do not have a terminal node, so they must be described in some other way - for example, by a function from N to the nodes of the tree. A path exists in the tree if and only if (every node on that path is in the tree) and (the path is a connected, linear subgraph of the tree). Two paths A and B are equal if and only if the set of nodes in path A is equal to the set of nodes on path B. There is a function f from N to the set of nodes that, for every natural number n, produces the node labelled by the first n binary places of 1/pi's binary expansion. This function describes a path. This path is distinct from every path in the set of finite paths used to describe the tree, because for every path in that set, the path described by this function contains more nodes. Informally, the path q from 1/pi under f is {., .0, .01, .010, .0101, .01010, .010100, .0101000, .01010001, .010100010, ...}. (Formally, it is only finitely describable as a function, not as an element-by-element list of nodes.) This path is distinct from each finite path: {.} is missing the node .0, which is in q. {., .0} is missing the node .01, which is in q. {., .0, .01} is missing the node .010, which is in q. In fact, we can continue this indefinitely; none of the finite rooted paths from which the tree is derived contains every node in q. However, every node in q is in the tree, since every one of its infinitely-many nodes is reachable from the root node after finitely many branchings. So q is a path over the maximal complete binary tree even though the maximal complete binary tree can be derived from a set that does not contain q. This is equivalent to the argument from arithmetic that there is, for every rational x, a rational y that is closer to 1/pi than x, presented above. > One of them is false, unless it is a magic tree where something > happens during construction. But I do not believe in magic, least in > mathematics. That a union of infinitely many distinct finite paths can contain infinite paths does not surprise me. In ZFC, one formulation of the axiom of infinity takes exactly the form of an infinite union of finite elements. Only a union of finitely many finite elements or a (possibly, infinite) union of finitely many distinct finite elements ({1} U {1} U {1} U {1} U {1} U ... = {1}, for example), neither of which is sufficient to produce the maximal complete binary tree, is a finite set. Conversely, in a set theory with no infinite sets, there is also no binary tree containing the path q (from above). Both kinds of theory can be internally consistent, which is as close to "true" as anything in mathematics gets. -o > >>> We agree that in Cantor's diagonal argument, applied to real numbers, > >>> the numbers are represented and identified solely by their digits. No [quoted text clipped - 27 lines] > produces, for *every* element n of N, the n'th bit of the binary > representation of 1/pi. That does not help, because *every* element of N is at a finite place, and for every element of N the sequence 1, ..., n is finite. > > There are two statements: > > 1) Path p can be distinguished from every path of the countable set P [quoted text clipped - 5 lines] > is the maximal complete binary tree, then there can be paths in the > union of all elements of P that are not in P. Magic? In fact there is no union. We simply write the paths p_n of P in slightly different form: The beginning zeros of all paths are written only once, the following 1 or 0 also are written only once each and so on. Note, we have not done anything else but writing the list in slightly different form, saving some ink. In particular we have not added any new path. This yields the complete binary tree. You are unable to distinguish any binary sequence representing a real of the unit interval from every path of that tree. >If P is a countable set, > then there are necessarily paths in the union of all elements of P that > are not in P. Why should they? "Necessarily" would be correct if actual infinity existed, but it necessarily doesn't. > In fact, one such set P is the set of finite rooted paths, which is > isomorphic to the set of nodes in the maximal complete binary tree. [quoted text clipped - 14 lines] > a consequence of the tree being representable as an infinite set of > finite paths), every node has a finite label. That is similar to every bit of 1/3 = 0.010101... in binary. Evere bit has a finite label. > We'll denote paths in the standard way, as sets of nodes: > [quoted text clipped - 20 lines] > describe the tree, because for every path in that set, the path > described by this function contains more nodes. And for *every* bit a_n of this infinite path there is a terminating path in the tree that contains n^n^n more nodes. So your argument is insufficient. > Informally, the path q from 1/pi under f is > {., .0, .01, .010, .0101, .01010, .010100, .0101000, .01010001, [quoted text clipped - 12 lines] > maximal complete binary tree can be derived from a set that does not > contain q. No it can't, unless magic plays a role. Every path in the tree is a finite path. > This is equivalent to the argument from arithmetic that there is, for > every rational x, a rational y that is closer to 1/pi than x, presented > above. And for each of these claoser rationals is far away from pi. > > One of them is false, unless it is a magic tree where something > > happens during construction. But I do not believe in magic, least in > > mathematics. > > That a union of infinitely many distinct finite paths can contain > infinite paths does not surprise me. There is no union. But even in set theory a union of sets should not contain elements that are in none of the united sets. > In ZFC, one formulation of the > axiom of infinity takes exactly the form of an infinite union of finite [quoted text clipped - 7 lines] > can be internally consistent, which is as close to "true" as anything > in mathematics gets. Actual or finished infinity as close to false as anything can get, within and outside of mathematics. Regards, WM >> In fact, one such set P is the set of finite rooted paths, which is >> isomorphic to the set of nodes in the maximal complete binary tree. [quoted text clipped - 17 lines] > That is similar to every bit of 1/3 = 0.010101... in binary. Evere bit > has a finite label. Similar to, but not the same. Under this labelling, there is no node labelled with the binary representation of 1/3, because that representation is not finite. There is a *path* which can be produced for 1/3, informally rendered as {., .0, .01, .010, .0101, ...} and formally rendered as a function h from N to the set of nodes such that h's image is a path. >> We'll denote paths in the standard way, as sets of nodes: >> [quoted text clipped - 22 lines] > > And for *every* bit a_n of this infinite path Paths do not contain bits. They contain nodes, and edges. Please be rigorous. > there is a terminating path in the tree that contains n^n^n more nodes. This is a red herring. There are, for any finite path x that is a subpath of the path described by f, infinitely many finite paths that contain x as a subpath. If the path x is rooted, then *every natural larger than the preimage of x under f* identifies the terminal node of a finite rooted path that contains x as a subpath. If paths were "constructed" in a stepwise fashion, then this would be a solid argument that such a stepwise process can never finish constructing the path described by f. However, paths are not "constructed". We give the function f, and its domain (N), and consider the complete path to have been described. It already existed; with those two entities, we have enough information to identify it and work with it. >> Informally, the path q from 1/pi under f is >> {., .0, .01, .010, .0101, .01010, .010100, .0101000, .01010001, [quoted text clipped - 15 lines] > No it can't, unless magic plays a role. Every path in the tree is a > finite path. Okay, wiseass. Which node in the path described by f is not in the tree? Every path in one of the sets from which the tree can be described is finite. That set is not the only way to describe the tree; we could, for example, describe it using a set of all of its infinitely-long paths, instead. Would you then argue that the tree contained no finite paths? >> This is equivalent to the argument from arithmetic that there is, for >> every rational x, a rational y that is closer to 1/pi than x, presented >> above. > > And for each of these claoser rationals is far away from pi. I cannot even guess at what you meant to say here. >>> One of them is false, unless it is a magic tree where something >>> happens during construction. But I do not believe in magic, least in [quoted text clipped - 5 lines] > There is no union. But even in set theory a union of sets should not > contain elements that are in none of the united sets. Once again: which element do you claim the path corresponding to 1/pi contains that is not contained in any of the finite paths used to describe the tree? If that path is not in the tree, then it must contain a node, or an edge, that is not in the tree. So which one is it? -o > >> In fact, one such set P is the set of finite rooted paths, which is > >> isomorphic to the set of nodes in the maximal complete binary tree. [quoted text clipped - 55 lines] > > Paths do not contain bits. They contain nodes, and edges. Please be rigorous. You are asking WM to be that which he is incapable of being. > > there is a terminating path in the tree that contains n^n^n more nodes. > [quoted text clipped - 66 lines] > > -o SignatureVirgil > >> In fact, one such set P is the set of finite rooted paths, which is > >> isomorphic to the set of nodes in the maximal complete binary tree. [quoted text clipped - 126 lines] > > -o Musatov produced: [PS] SIGACT News Complexity Theory Column 36 William Gasarch: (Univ of MD, 2100, P=NP) NP-completeness is important since (as we. tell our students) if a problem is NP-complete you look for other ways ... http://www.cs.umd.edu/~gasarch/papers/poll.ps Q: Can I get a witness? A: Yes. http://www.meami.org/?cx=000961116824240632825%3A5n3yth9xwbo&cof=FORID%3A10&ie=U TF-8&q=WM+P%3DNP#1005 > There is a *path* which can be produced for 1/3, informally rendered as > {., .0, .01, .010, .0101, ...} > and formally rendered as a function h from N to the set of nodes such > that h's image is a path. There is a path for 1/3 if there is actually infinity. We have the statement: If there is actual infinity, then the path for 1/3 is different from every terminating path. There is a second statement: 1/3 is not different from every terminating path. By modus tollens, we obtain: There is no actual infinity. The proof is as follows: The binary tree constructed from all terminating paths p_n of P does not contain more than these paths. Subproof: The tree is nothing else but the list of these paths written in a somewhat unconventional but correct form. But even if we consider the tree as a formal union of the sets {p_n}, then it is clear according to set theory, that the union can only contain an element if this element is contained in at least one of the united sets. Finally, if you consider the complete tree, then you cannot distinguish the path of 1/3 from every path of the tree. This is a very simple proof for the fact, that any further reasoning about actual infinity is useless. Regards, WM In article <f7b71b7d-d6a0-43a2-b5ff-5cf1e114d867@h11g2000yqb.googlegroups.com>, > > There is a *path* which can be produced for 1/3, informally rendered as > > {., .0, .01, .010, .0101, ...} [quoted text clipped - 10 lines] > By modus tollens, we obtain: > There is no actual infinity. Not until WM, or someone, shows how 1/3 is *the same* as some terminating path. > The proof is as follows: > > The binary tree constructed from all terminating paths p_n of P does > not contain more than these paths. False. The union of the nested sequence of node sets for *all* complete finite binary trees is not a finite set. > Subproof: The tree is nothing else but the list of these paths written > in a somewhat unconventional but correct form. The tree is the set of its nodes. Paths are merely a consequence of the partial order established by the "parent of" relation on that set of nodes. A apth is merely any maximal totally ordered subset > But even if we consider the tree as a formal union of the sets {p_n} It is the formal union of the node sets, not of the path sets, which allow building of a maximal infinite binary tree. Since no path in any one of those trees is path in any other tree, such a union of path sets produces only chaos. > Finally, if you consider the complete tree, then you cannot > distinguish the path of 1/3 from every path of the tree. If that tree is one of WM's construction we can distinguish it from all finitely many paths in that tree, and in our own maximal infinite binary trees, from all but 1 path. SignatureVirgil In article <30b4af79-0a18-464a-a2ac-a47933732167@x29g2000prf.googlegroups.com>, > > >>> We agree that in Cantor's diagonal argument, applied to real numbers, > > >>> the numbers are represented and identified solely by their digits. No [quoted text clipped - 30 lines] > That does not help, because *every* element of N is at a finite place, > and for every element of N the sequence 1, ..., n is finite. but for every n in N, there is a susequent finite place. > > > There are two statements: > > > 1) Path p can be distinguished from every path of the countable set P [quoted text clipped - 7 lines] > > Magic? Only to those without the wit to understand it. > In fact there is no union. There is in ZF. > We simply write the paths p_n of P in > slightly different form: The beginning > zeros of all paths are written only once, the following 1 or 0 also > are written only once each and so on. Note, we have not done anything > else but writing the list in slightly different form, saving some > ink. But it is now impossible to recover the original forms from their abbreviations, so it is no longer possible from those abbreviations to declare that a given path is or is not a member. > In particular we have not added any new path. But you have carefully obscured which ones are there. > This yields the complete binary tree. But not any maximal infinite binary tree. > You are unable to distinguish > any binary sequence representing a real of the unit interval from > every path of that tree. If the tree is a maximal infinite binary tree then every path is in it, but if it is any of your less than maximal trees, here may be paths missing. > >If P is a countable set, > > then there are necessarily paths in the union of all elements of P that > > are not in P. > > Why should they? "Necessarily" would be correct if actual infinity > existed, but it necessarily doesn't. It does in ZF, and as WM has no axiomatically satisfactory set theory to replace it with, not any internal proof of ZF's ncinsistency, I will continue to go with ZF. > > In fact, one such set P is the set of finite rooted paths, which is > > isomorphic to the set of nodes in the maximal complete binary tree. [quoted text clipped - 45 lines] > And for *every* bit a_n of this infinite path there is a terminating > path in the tree that contains n^n^n more nodes. Is this supposed to be relevant to anything? > So your argument is insufficient. Total non sequitur. It is WM's support of that claim which is insufficient. > > Informally, the path q from 1/pi under f is > > {., .0, .01, .010, .0101, .01010, .010100, .0101000, .01010001, [quoted text clipped - 14 lines] > > No it can't, unless magic plays a role. Whatever is sufficiently beyond the understanding of WM, he regards as magic. > Every path in the tree is a > finite path. Not by the standard definition of a (maximal) path. One can, in a maximal infinite binary tree, have all kinds of finite pathlets (finite linearly ordered rooted sets of nodes), but none of them are maximal so none of them are paths according to the definition requiring that they be maximal. > > This is equivalent to the argument from arithmetic that there is, for > > every rational x, a rational y that is closer to 1/pi than x, presented > > above. > > And for each of these claoser rationals is far away from pi. That does not parse in English. > > > One of them is false, unless it is a magic tree where something > > > happens during construction. But I do not believe in magic, least in > > > mathematics. WM declares magical whatever he does not comprehend. > > That a union of infinitely many distinct finite paths can contain > > infinite paths does not surprise me. > > There is no union. There is in ZF, and WM cannot fault ZF. > But even in set theory a union of sets should not > contain elements that are in none of the united sets. Nor does it. > > In ZFC, one formulation of the > > axiom of infinity takes exactly the form of an infinite union of finite [quoted text clipped - 10 lines] > Actual or finished infinity as close to false as anything can get, > within and outside of mathematics. Not as close to false as are "potentially infinite sets" in any standard set theory. And, as yet, WM has not produced any set theory in which they are not false. > Regards, WM SignatureVirgil > >>> We agree that in Cantor's diagonal argument, applied to real numbers, > >>> the numbers are represented and identified solely by their digits. No [quoted text clipped - 121 lines] > can be internally consistent, which is as close to "true" as anything > in mathematics gets. Very nice! SignatureVirgil > > We agree that in Cantor's diagonal argument, applied to real numbers, > > the numbers are represented and identified solely by their digits. No [quoted text clipped - 18 lines] > 1], 1/pi differs from q by an amount greater than some other rational r > in [0, 1], so we can distinguish 1/pi from every rational in [0, 1]. What everyone else can do, WM cannot. Which tells us all we need to know about WM. SignatureVirgil In article <406110a1-633d-43ff-8a7d-bc256aac5135@a36g2000yqc.googlegroups.com>, > > > > > > Your claim is that "no possibility exists to construct or to > > > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 20 lines] > > by using nodes of the tree only without additional knowledge. So that WM argues that if P is hidden then it somehow will contain paths that it does not contain when known? > > You do not agree that > > if actually infinite paths exist, [quoted text clipped - 4 lines] > paths in P using only the information stored in the tree, i.e. using > only digit sequences to identify numbers But it is eminently possible using the tree and knowing P. > thereby proving that no > actual infinity exists. Not a proof that will stand up to any scrutiny. > Instead of grumbling about this fact, you either should accept it and > confess that I am right and that set theory is wrong, or you should be > able to distinguish your path p from the set P of paths that I have > used for construction, but without any other knowledge than the final > result, namely the digits, i.e., the nodes of the binary tree. Or we can just show, as we have done so often, that there are more paths in the tree than in P (in the sense that there can be no surjection from P to the set of all paths or injection the other way). Theorem: if A and B are sets such that Card(A) > Card(B) then there is some a in A that is not in B. > That are the two possible logical alternatives. Except that WM carefully ignores the actuality. SignatureVirgil In article <222103da-c8b9-41d5-9461-689cdc417c59@f19g2000yqh.googlegroups.com>, > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 9 lines] > If actually infinite paths exist! That is obviously a prerequisite. > Otherwise the construction using terminating paths covers everything. You construction would require the existence of potentially infinite sets, which does not occur in any known set theory system. > > You do not agree that > > the tree contains a path that > > can be distinguished from every element of P. > > I can prove it. Between what WM claims to be able to prove and what he actually is able to prove lies an unbridgeable gulf. > You only have to accept the game. If it is the game WM offered before, he loses. SignatureVirgil In article <6296c995-102e-4c5e-95d4-30cbf4fa7420@j32g2000yqh.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 12 lines] > All nodes of the path 1/pi are in the tree (together with all nodes of > any other path of the unit interval). That is not the issue. WM has presented a countable list of paths, P, from which he claims to build a maximal infinite binary tree, T. While the union of P, as a set of nodes, can well be the set of all nodes of T, the members of P, being sets of nodes which for paths, d not exhaust the set of all paths, where a path is any maximal set of nodes totally ordered by the ancestor relation. > > The binary tree does not contain a path > > p that can be distinguished from > > every element of P. > > The binary tree does not contain any path that can be distinguished > from every element of the set P of paths by which it was constructed. On the contrary, the tree contains more of those paths than the ones from which it was constructed. > If this were no true, then you could determine whether a given path p > differs from every element of P without knowing P. Nonsense. There is difference between knowing that there ARE paths not in P and knowing WHICH paths are not in P. The cardinalities involved require that there be paths not in P, but do not identify which paths are not in P. If Card(A) > Card(B), then there are elements of A which are not members of B. At least outside of WM's world of MathUnrealism. > Note: Cantor's diagonal method uses only digits respective nodes, no > additional information like the age of the writer or so. Same holds > for my tree. How does the age of the writer of your tree affect its structure? > Regards, WM SignatureVirgil We agree that in Cantor's diagonal argument, applied to real numbers, the numbers are represented and identified solely by their digits. No further information is available. We assume that a real number p can be distinguished from a set Q of real numbers q by general considerations, for instance, if p is a transcendental number and Q consists of rational numbers q only. Of course it would be impossible to distinguish p from all q, because for every digit d_n of p, there is a number q that shares all digits up to d_n with p. It is possible to construct the complete binary tree from the numbers in Q. That means, it is possible to construct all possible digit sequences from the numbers in Q. It is also possible to construct the complete binary tree from the set Tp that consists of all termintaing rationals which are appended by a tail consisting of p. Therefore it is impossible to distinguish p from the binary tree. But the binary tree has been errected by using a countable set of paths only. Therefore it is impossible to distinguish, by means of digits, any real number from a countable set of paths. Conclusion: Cantor's diagonal argument is contradicted. Comment: This is not surprising, because it is impossible to identify any irrational number by means of its decimal representation, but just this is claimed by the diagonal argument. Regards, WM In article <b76b5101-f3f2-4b86-91b5-75792335690d@k8g2000yqn.googlegroups.com>, > We agree that in Cantor's diagonal argument, applied to real numbers, > the numbers are represented and identified solely by their digits. No [quoted text clipped - 3 lines] > real numbers q by general considerations, for instance, if p is a > transcendental number and Q consists of rational numbers q only. Then one can then distinguish between any real number whose binary or decimal or other base expansion is eventually periodic and those whose expansions are not eventually periodic. > Of course it would be impossible to distinguish p from all q, because > for every digit d_n of p, there is a number q that shares all digits > up to d_n with p. Thus one can distingish any number with a non-eventually-periodic expansion from all rationals. SignatureVirgil > In article > <b76b5101-f3f2-4b86-91b5-75792335690d@k8g2000yqn.googlegroups.com>, [quoted text clipped - 17 lines] > Thus one can distingish any number with a non-eventually-periodic > expansion from all rationals. What the hell has the diagonal argument to do with "distinguishing" numbers from one another? This is nothing but one more of Mueckenheim's red herrings. > > In article > > <b76b5101-f3f2-4b86-91b5-757923356...@k8g2000yqn.googlegroups.com>, [quoted text clipped - 23 lines] > > - Show quoted text - Consider these statements/results: (BE LOGICAL/LIKE SPOCK OR DO NOT POST A REPLY PLEASE) --I say this content with the fact you will do as you will. Simply read: Result (1): John 18:36 Jesus answered, "My kingdom is not of this world. If Bible in Basic English Jesus said in answer, My kingdom is not of this world: .... I If is it Jesus Jews kingdom My not now of over place prevent realm said ... http://bible.cc/john/18-36.htm Source (1): http://meami.org/?cx=000961116824240632825%3A5n3yth9xwbo&cof=FORID%3A... Result (2) Matthew 19 American King James Version 18 He said to him, Which? Jesus said, You shall do no murder, You shall not commit ... Truly I say to you, That a rich man shall hardly enter into the kingdom of heaven. ... You may not copyright it or prevent others from using it. ... http://kjv.us/matthew/19.htm Source (2) http://meami.org/?cx=000961116824240632825%3A5n3yth9xwbo&cof=FORID%3A... Result (3) CHAPTER 121 MAKE AMERICA KNOW HER SINS They refuse to go in themselves and seek to prevent those who are trying to enter. 16 I am sorry for my poor people who should be helping in the right way ... http://www.seventhfam.com/temple/books/black_man/blk121.htm Source (3) http://meami.org/?cx=000961116824240632825%3A5n3yth9xwbo&cof=FORID%3A... And finally *this*: (quoted from a physical "book" for my critics who claim I do not read) Title: Holy Bible (NIV) Publisher: Zondervan ISBN 0-310-92391-3 Matthew Chapter 23 13"Woe to you, teachers of the law and Pharisees, you hypocrites! You shut the kingdom of heaven in men's faces. You yourselves do not enter, nor will you let those who are trying to.^g 15"Woe to you, teachers of the law and Pharisees, you hypocrites! You travel over land and sea to win a single convert, and when he becomes one, you make him twice as much a son of hell as you are." ^g13 Some manuscripts <i>to</i> Questions: 1. Matthew 23:14 does not exist in this bible. Why? Conclusion (the verse is ommitted in the N.I.V): Proof: (Book) Title: Our African Heritage Holy Bible (Authorized King James Version) Publisher: Copyright (C) 2004 by World Publishing. All Rights Reserved. Matthew Chapter 23 14 Woe unto you, *scribes* and *Phar'-i-sees*, hypocrites!for ye devour widows' houses, and for a pretence make long prayer: therefore ye shall receive the greater damnation. Do you still not understand? In article <a9298759-d38b-46b5-b4c7-2c86c8c737ef@h11g2000yqb.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 7 lines] > Would you bet to be able to distinuish your path from the set that I > have used? If you hide the paths you have used, there is no way, without peeking, that anyone else can Tel wish paths yo have used, but that is immaterial. The point is that whatever paths you have used, by the time the maximal infinite binary tree finally exists, there are paths in it that you did not use. > So let us play the game. Choose a path and try to distinguish it from > the set of paths that completely cover my tree. Until I know which paths you used, I can only guess, but if I guess slyly enough, and your set of paths is countable, the odds are greatly in my favor. > > > Every node of path p and every set of nodes > > > of path p is covered by one or more paths of P. [quoted text clipped - 7 lines] > that you will not be able to distinguish your path from the set of > paths that I have used for construction. What about your stakes? Is your set of paths countable (and actually counted or listed, at least algorithmically), and recorded with a neutral referee? If 'yes' to both, you would almost certainly lose at any stakes. All your opponent would have to do is use the "Cantor antidiagonal" to your own list of paths, which you cannot ever include in that list. SignatureVirgil > In article > <a9298759-d38b-46b5-b4c7-2c86c8c73...@h11g2000yqb.googlegroups.com>, [quoted text clipped - 17 lines] > infinite binary tree finally exists, there are paths in it that you did > not use. How do they come in? The tree is constructed from a countable set P of terminating paths. If you have a Cantor list filled with those paths, then you claim to be able to construct another path p. So this path does not creaqpo intio the list. But it does creap into the tree? When does it creap? Where does it lurk and who tells it when to start? That is nonsense! There is no actualyy infinite path. So nothing needs to creap and superstision can be expelled from mathematics. > > So let us play the game. Choose a path and try to distinguish it from > > the set of paths that completely cover my tree. > > Until I know which paths you used, I can only guess, but if I guess > slyly enough, and your set of paths is countable, the odds are greatly > in my favor. But if I trick you out and decide only after you have guessed? Or if I am honest, as is my character, and use only every path that can be named? Then your odds are zero if not less. > > > > Every node of path p and every set of nodes > > > > of path p is covered by one or more paths of P. [quoted text clipped - 12 lines] > > If 'yes' to both, you would almost certainly lose at any stakes. Yes to both. Here is the algorithmus generating the words than label my paths including the dictionary and grammar of all languages that are able to identify paths: 0 1 00 01 10 11 .. > All your opponent would have to do is use the "Cantor antidiagonal" to > your own list of paths, which you cannot ever include in that list. There is only one problem: This list has no diagonal. No list of numbers that are given in a way such that they are uniquely defined has a diagonal. What has a diagonal is a list that contains blah-blah- blah like 0.9384791428759184... 0.2392847216773984... ... And the mathematics errected upon this blah-blah is like that. Regards, WM In article <2df84f81-1d47-45ea-b13e-c0e0025bd10d@h11g2000yqb.googlegroups.com>, > > In article > > <a9298759-d38b-46b5-b4c7-2c86c8c73...@h11g2000yqb.googlegroups.com>, [quoted text clipped - 19 lines] > > How do they come in? They are a natural and inevitable consequence of the tree becoming a maximal infinite binary tree. As soon as all the required nodes are there, so, automatically, are all those unused paths. As soon as one has the set of all nodes, every possible subset is a subset, and all those paths are merely paticular subsets. > The tree is constructed from a countable set P of > terminating paths. If the union of that set P contains every node of the maximal infinite binary tree as a member, then it contains every set of nodes as a subset, therefore contains every path as a subset. > If you have a Cantor list filled with those paths, > then you claim to be able to construct another path p. So this path > does not creaqpo intio the list. But it is in a new list. But P is not just a list, it must be a set of sets of nodes and the union of P (the set of all members of members of P) must then be the set of ALL nodes, at least if the tree is to be constructed from it. And once one has a set of all nodes, one has as subsets all those paths WM doesn't like hearing about. > But it does creap into the tree? Do you mean "creep" > When > does it creap? Where does it lurk and who tells it when to start? If the subsets of P cover all nodes then the union of P is the set of nodes and the subsets of that union include al those paths that upset WM so badly, so I guess one would have to say that when building your P, as soon as what you have built so far covers all nodes, you have all paths. > That is nonsense! There is no actualyy infinite path. So nothing needs > to creap and superstision can be expelled from mathematics. That is WM's nonsense. WM can hardly build his own infinite paths, as he has been doing here for some time, and then deny infinite paths to everyone else. > > > So let us play the game. Choose a path and try to distinguish it from > > > the set of paths that completely cover my tree. [quoted text clipped - 4 lines] > > But if I trick you out and decide only after you have guessed? That is not the game originally proposed, so that WM finds he needs to cheat. > Or if I > am honest, as is my character, and use only every path that can be > named? Then your odds are zero if not less. For any finite list of more than one name, the concatenation of all its names cannot be included within it as a name. And WM declares that absolutely no infinite lists of names are possible, so he loses. > > > > > Every node of path p and every set of nodes > > > > > of path p is covered by one or more paths of P. [quoted text clipped - 23 lines] > 10 > 11 None of the above are paths, or even sets of nodes, so are not allowable. Your own specification required a SET OF PATHS: "you will not be able to distinguish your path from the set of paths that I have used for construction" SignatureVirgil > > > The point is that whatever paths you have used, by the time the maximal > > > infinite binary tree finally exists, there are paths in it that you did [quoted text clipped - 7 lines] > the set of all nodes, every possible subset is a subset, and all those > paths are merely paticular subsets. All paths that I have used contain all possible subsets. > > The tree is constructed from a countable set P of > > terminating paths. > > If the union of that set P contains every node of the maximal infinite > binary tree as a member, then it contains every set of nodes as a > subset, therefore contains every path as a subset. It does. And it is a countable union.You remember a proof by Feferman and Levy? They showed that the statement that the set of all real numbers is the union of a denumerable set of denumerable sets cannot be refuted. > > If you have a Cantor list filled with those paths, > > then you claim to be able to construct another path p. So this path [quoted text clipped - 11 lines] > > Do you mean "creep" probably. Why do you English always speek such that one cannot distinguish between weak and week? > > When > > does it creap? Where does it lurk and who tells it when to start? [quoted text clipped - 3 lines] > so badly, so I guess one would have to say that when building your P, as > soon as what you have built so far covers all nodes, you have all paths. And if the right hand side of the tree is not yet ready, but the left hand side? Is there a border? > > That is nonsense! There is no actually infinite path. So nothing needs > > to creap and superstition can be expelled from mathematics. [quoted text clipped - 3 lines] > WM can hardly build his own infinite paths, as he has been doing here > for some time, and then deny infinite paths to everyone else. I can assume its existence and then find that this assumption was wrong. > > > > So let us play the game. Choose a path and try to distinguish it from > > > > the set of paths that completely cover my tree. [quoted text clipped - 15 lines] > names cannot be included within it as a name. And WM declares that > absolutely no infinite lists of names are possible, so he loses. The list of everything is shown below. > > > > > > Every node of path p and every set of nodes > > > > > > of path p is covered by one or more paths of P. [quoted text clipped - 25 lines] > > None of the above are paths, or even sets of nodes, so are not allowable. All paths that you can tink of and anything else is encoded in th elist above. Real numbers are not encoded as infinite sequences, though, because it is impossible to write infinite sequences. That is the main reason, why you and others never have written pi or 3.1415... as an infinite sequence. It is impossible. > Your own specification required a SET OF PATHS: > "you will not be able to distinguish your path from > the set of paths that I have used for construction after the construction has been completed." Yes, so it is. Regards, WM In article <28d86f4d-bfe2-4f37-9ebe-a563deb2e172@x3g2000yqa.googlegroups.com>, > > > > The point is that whatever paths you have used, by the time the maximal > > > > infinite binary tree finally exists, there are paths in it that you did [quoted text clipped - 9 lines] > > All paths that I have used contain all possible subsets. Outside of WM's world of MathUnrealism, if one has a maximal infinite binary tree and defines in it a path to be any maximal set of nodes totally ordered by the ancestor relation, then there are uncoutably many of paths. WM must be using some other definition of path in order to have so few of them. > > > The tree is constructed from a countable set P of > > > terminating paths. [quoted text clipped - 7 lines] > numbers is the union of a denumerable set of denumerable sets cannot > be refuted. Did they say it could be proved? And didn't their proof require denial of the axiom of choice? > > > If you have a Cantor list filled with those paths, > > > then you claim to be able to construct another path p. So this path [quoted text clipped - 25 lines] > And if the right hand side of the tree is not yet ready, but the left > hand side? Is there a border? The "border" appears to be between having all those nodes which precede and/or follow some particular node, and not having them. As soon as one has them, one also has uncountably many paths. > > > That is nonsense! There is no actually infinite path. So nothing needs > > > to creap and superstition can be expelled from mathematics. [quoted text clipped - 6 lines] > I can assume its existence and then find that this assumption was > wrong. You cannot find the assumption wrong without assuming that we do not grant. > the set of paths that completely cover my tree. > > [quoted text clipped - 49 lines] > All paths that you can tink of and anything else is encoded in th > elist above. Nope. Only nodes are encoded. To encode a path one would need an infinite binary sequence whose nodes are the finite initial sequences in your list. Either that or infinite subsets of members of your list. In either case, there are more that countably many of them. SignatureVirgil In article <63e6a071-5071-41b2-a0dd-2df7e53372cb@g1g2000yqh.googlegroups.com>, > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes [quoted text clipped - 27 lines] > Sorry, that is impossible. Every node of path p and every set of nodes > of path p is covered by one or more paths of P. Is P a set of paths or merely a set of nodes? If it is a set of nodes, it must be the set of all nodes or there will be paths wish are to subset of it. and the it will contain as subsets uncountably many paths If it is a set of paths, then it must contain all uncountably many paths or there will be a path which is not member of it. > > So it is possible to contruct another path, p, which can be > > distinguished from every element of P. > > That is impossible. The path p_0 = 0.111... for instance is completely > covered by terminating paths, whether or not it had been inserted > originally. If P is only a set of terminating paths then no non-terminating path will be a member of it. If P is a set of nodes, then it must contain every node of every finite subtree, and thus have every one of those uncountably many paths as a subset. Otherwise there will be paths that it does not contain as subsets. > > This directly contradicts your claim > > [quoted text clipped - 5 lines] > does not exist. It is nothing but a union of terminating paths > (potential infinity). Except that if the infinite tree exists at all, which has been conceded or we would not even be talking about it, then it is actually infinite and WM's finiteness restrictions no longer apply. Wm cannot concede the existence of an actually infinite tree, which he has been doing, and then insist that it does not have the properties of an actually infinite tree. SignatureVirgil > > Sorry, that is impossible. Every node of path p and every set of nodes > > of path p is covered by one or more paths of P. > > Is P a set of paths or merely a set of nodes? It is a set of paths. > If it is a set of paths, then it must contain all uncountably many paths > or there will be a path which is not member of it. Wrong. All countably many terminating paths are sufficient to cover all nodes of the tree. > > > So it is possible to contruct another path, p, which can be > > > distinguished from every element of P. [quoted text clipped - 5 lines] > If P is only a set of terminating paths then no non-terminating path > will be a member of it. All non-terminating paths are in the tree constructed by P. Regards, WM In article <257f6992-e801-43b8-a5df-30be71217db4@z9g2000yqi.googlegroups.com>, > > > Sorry, that is impossible. Every node of path p and every set of nodes > > > of path p is covered by one or more paths of P. [quoted text clipped - 8 lines] > Wrong. All countably many terminating paths are sufficient to cover > all nodes of the tree. But paths are not mere nodes, they are sets of nodes, and while every node many be a member of a member of P, that does not imply that every any particular subset of the set f nodes is a member of P. That does not even hold for finite sets. The set {{a},{b},{c}} contains every member of {a,b,c} as a member of a member but does NOT contain as a member every subset of {a,b,c}. For example {a,b} and {a,c} and {b,c} are not members nor subsets of {{a},{b},{c}}. WM needs to go back to set theory kindergarten, and brush up. > > > > So it is possible to contruct another path, p, which can be > > > > distinguished from every element of P. [quoted text clipped - 7 lines] > > All non-terminating paths are in the tree constructed by P. But not in P itself. Since P is countable it can be put into a list which necessarily has an anti-diagonal not in P. Actually, at least as many anti-diagonals as members. SignatureVirgil In article <3aa8c5c0-a906-46d1-9b0b-471f038dc3c7@b9g2000yqm.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes [quoted text clipped - 24 lines] > sequences represented in the tree. It has nothing to do with a digit > sequence. What happens in one of WM's non-maximal non-trees is irrelevant. In any maximal infinite binary tree, for every binary digit sequence there is a path and for every path there is a binary digit sequence. SignatureVirgil In article <9b068051-20e6-40e7-8fa5-8842974228e4@k38g2000yqh.googlegroups.com>, > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish [quoted text clipped - 8 lines] > > > > > But only before those paths have occupied the tree. Are they like an invading army? How do paths "occupy" a tree. If you mean that the set of nodes forming the path is subset of the set of nodes forming the tree, there must be a better way of expressing it. But if that is what you mean, Then every path comes into being as soon as the set of nodes of that tree comes into being. > > Nope. Let the given set of paths be P. > > [quoted text clipped - 24 lines] > superstition, or recognizing that there are no actually infinite > paths. The completion of the set of nodes of a tree automatically and immediately creates each subset of the set of nodes, thus creates each path in that tree. There is no deferred creation of such paths possible. SignatureVirgil In article <57880d13-7ab5-449c-af24-12129013f9c5@k8g2000yqn.googlegroups.com>, > > Your claim is that "no possibility exists to construct or to > > distinguish [quoted text clipped - 8 lines] > > But only before those paths have occupied the tree. Paths do not "occupy" trees. And if the definition of a path is a maximal set of nodes which is totally ordered by the transitive closure of the "parent of" relation, then every path that can ever exists in a tree always exists in that tree. So that WM must have some special definition of path which does not recognize all the above sort of pathsin order to have different sets of paths for the same set of nodes. > Because after > construction the given set of paths has occupied the whole tree in > that it includes all possible combinations of nodes, i.e., every > subset that you might think of is already represented in the tree, > because it is constructed such that no subset remains. Then Cantor's proof applies to every complete infinite binary tree, as the set of binary sequence bijects with the set of paths. > Calvin Ostrum said: They sneak in as sets of joined edges and there > is > nothing you can do about it. There is certainly nothing that WM can do about it, but it is the very definition of a path that necessarily "sneaks" them in. > But the contrary is true. All actually infinite paths sneak out. There > is no path 0.111... [quoted text clipped - 3 lines] > 0.111 > ... That may be true in WM's tiny world, but his horizons are too constrained. > That set does not contain an actually infinite path. How should it? Easily! > Should an infinite union of paths of the form 0.1 yield an infinite > path? There is only one path of the form 0.1, so no such "infinite union" is possible. The set {{0.1},{0.1},{0.1},...}, whose union WM wishes to form is actually just the set {{0.1}}. > No The union of all finite paths does not supply an actually > infinite path, no matter how large they are. Depends on one's definition of node and path. By my definition, the union of infinite sequence of node sets each totally ordered by the ancestor relation and each a proper subset of all its successors, is necessarily an infinite path, being a maximal infinite set of nodes totally ordered by the ancestor relation. > All those numbers which > would make R uncountable do not exist! What exists outside of WM's wee wee world is not under his control. > It's getting time to recognize that actual infinity is actually > absent. Not as notably absent as WM's ability to thing straight. SignatureVirgil In article <17677499-4a66-4a22-ab67-785214e395ca@r13g2000vbr.googlegroups.com>, > > > > > Because paths cannot be distinguished without nodes. > > [quoted text clipped - 18 lines] > by a set of paths then you cannot show any node that will distinguish > a further path from that given set of paths. In a maximal infinite binary tree, no one node is, not even any finite set of nodes, is enough to separate any one path from all others, it takes infinitely many nodes to do that. If WM's trees are not like that then they no maximal either, meaning that there are paths in maximal trees that WM's trees exclude. > But if you cannot > distinguish a path from a given set of paths, then it belongs to that > set. One can distinguish any path from any set of paths not containing it by a set of nodes if one is allowed sets with enough nodes to do the job, but not if, as WM would have it, one is restricted to too few nodes. In finite trees, the only single node that can separate any one path from all others is its leaf node, but in infinite trees there are no leaf nodes. In a maximal infinite binary tree, any infinite subset of any path's set of nodes suffices to separate it from any set of paths not containing it. But no finite set of its nodes, or any other nodes, suffices to separate a path from all sets of paths not containing it.. SignatureVirgil In article <42a79004-2290-4944-8636-04870303b7b0@g20g2000vba.googlegroups.com>, > > > Because paths cannot be distinguished without nodes. > > [quoted text clipped - 4 lines] > > No that is provably wrong. It may be considered wrong in WM's mythic world of MathUnrealism, but is not wrong in ZF, which WM s unable to show self-contradictory though it is easily show to contradict WM's mathUnrealism. > All nodes are used up by a countable number > of paths, e.g., all paths ending in a tail of zeros. But not all paths. And using the same set of nodes, one can easily add infinitely more paths. One can even add countably many more paths endlessly. > Therefore no > possibility exists to construct or to distinguish by one or many or > infinitely many nodes of the tree another path. Except that I can do what WM claims cannot be done merely by reversing all the 0's and 1's of every path to get as many new paths using only the same node set. > All combinations of > nodes that are possible in the tree have already been occupied. Except that there is an easy way to get as many new and supposedly impossible paths as WM already has by merely switching all 0's with 1's. WM really make an a.s of himself when he makes claims so easily and obviously falsifiable SignatureVirgil In article <4223f1bf-a04e-4e14-8513-c642212d4827@21g2000vbk.googlegroups.com>, > > - There is more that one path through "that node" > > [quoted text clipped - 6 lines] > > No. It is not assuming this but *proving* this. WM has not the least notion of how to distinguish valid proofs from invalid attempts, so that his claims of "proof" are all nonsense. > Because paths cannot > be distinguished without nodes. Non Sequitur. > Equivalently: Assuming that that, in Cantor's proof, every line > contains a digit that differs from the corresponding digit of the anti > diagonal would be assuming that the anti diagonal is not in the list. There is a rule by which just such a situation can be made to occur. And until WM can show why that rule does not achieve the claimed result, which he has often tried but universally failed to do, the proof remains valid. > > - You are trying to prove that there are countably many > > paths through the root node. [quoted text clipped - 4 lines] > This is equivalent to proving that there are only countably many > possibilities of distinguishing paths below "that node". Through through node in a maximal infinite binary tree there are as many paths as from the root node, and more than there are nodes, at least in cardinality. And that will remain the case until someone can show a surjection from the set of nodes to the set of paths. SignatureVirgil In article <fb1715f5-f578-444e-a1ec-915407281b08@x3g2000yqa.googlegroups.com>, > > I prefer to call what is left after removing the head of a path, its > > tail. [quoted text clipped - 11 lines] > of nodes. Therefore there is no possibility to construct any further > tail. There is no need to construct what are already there. > All nodes are already used up. From every node issue at least two tails (actually uncountably many tails in every maximal infinite binary tree), and your method only involves one of them, so is incomplete. > And, what is more important, also > all combinations of nodes are used up. Then paths must not be what WM calls "combinations of nodes" as not all paths are "used up". SignatureVirgil > > All nodes are already used up. > > From every node issue at least two tails (actually uncountably many > tails in every maximal infinite binary tree), and your method only > involves one of them, so is incomplete. One of them belongs to a tail that has already been mapped on a node at a higher level. The other is mapped on that actual node. That makes a complete 1 to 1 mapping. > > And, what is more important, also > > all combinations of nodes are used up. > > Then paths must not be what WM calls "combinations of nodes" as not all > paths are "used up". At least all the constructed paths cover all nodes and leave no node that could be used for another construction. Regards, WM In article <abc57a2e-ed45-4332-93e6-66c439beb9ef@k2g2000yql.googlegroups.com>, > > > All nodes are already used up. > > [quoted text clipped - 5 lines] > at a higher level. The other is mapped on that actual node. That makes > a complete 1 to 1 mapping. That accounts for only two paths out of uncountably many, so does not exhaust the paths. > > > And, what is more important, also > > > all combinations of nodes are used up. [quoted text clipped - 4 lines] > At least all the constructed paths cover all nodes and leave no node > that could be used for another construction. Only in infinite but not maximal binary trees in which some nodes have less than two children. > Regards, WM SignatureVirgil ... > > > And both say, in effect, about the existence of actual infinity, what > > > Kant said about the proof of the existence of God: These assumptions [quoted text clipped - 7 lines] > The axiom can be contradicted. Simple example: The axiom could be: The > binary tree has uncountably many paths. Perhaps, although in ZF it is not an axiom. > I show that the end of each > path p of the set P can be mapped on a node, and that all paths p of P > cover all nodes of the tree. Ignoring that in ZF the paths do not have an end. > Therefore, after having completed the > covering of the whole tree, there remains no node that could be used > to construct a path that does not belong to P. This is the wrong way around. You assume that you can cover this way the whole tree (I think with this you mean each path in the tree). But that is what you have to prove. > This disproves the > mentioned axiom. Indeed, when you assume it is false, it is easy to prove it is false. > > So comparing > > the "axiom of infinity" with a "proof of God" is pretty stupid. In [quoted text clipped - 5 lines] > (they call it a dogma, but it is of the same meaning) according to > which it is possible to prove the existence of God. I do not think such an axiom would be a valid axiom in mathematics. Axioms do not state what is or what is not possible to prove. Axioms states properties and existence of objects. > > So, if you are discussion Eucliedan > > geometry you should use the parallel axiom. Of course you can reject is > > but in that case you are not discussing Euclidean geometry but something > > else. > > That means, you are willing to believe in what the Vatican says? Well, no, because that "dogma" is not a valid "axiom". But can you tell me where that "dogma" actually is stated the way you say? > Probably they think that the Dutch better should have stayed within > the Spanish Empire. As, currently of the people that say they belong to a church, nearly 50% is Roman Catholic there may be some validity. But that nearly 50% is actually about 17% of the total population... Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <74f33c12-95fe-4b83-a7e3-941591def...@c9g2000yqm.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 4 Jun., 04:16, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 12 lines] > > Perhaps, although in ZF it is not an axiom. It is, because the paths of the tree are isomorphic with the real numbers in [0, 1] > > I show that the end of each > > path p of the set P can be mapped on a node, and that all paths p of P > > cover all nodes of the tree. > > Ignoring that in ZF the paths do not have an end. The paths of the tree have no end. But it can be shown for every node that it gets covered and that all nodes get covered by a countable set of paqths. > > Therefore, after having completed the > > covering of the whole tree, there remains no node that could be used [quoted text clipped - 3 lines] > whole tree (I think with this you mean each path in the tree). But that is > what you have to prove. There is not much to prove. Append a tail of a path to every node. Then every node is covered by at least one path, hence it does not remain uncovered. > > This disproves the > > mentioned axiom. > > Indeed, when you assume it is false, it is easy to prove it is false. I do not assume that the number of nodes is countable, but I count them. Here: http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#363,26,Folie 26 > > > So comparing > > > the "axiom of infinity" with a "proof of God" is pretty stupid. In [quoted text clipped - 9 lines] > do not state what is or what is not possible to prove. Axioms states > properties and existence of objects. Like God and the set of natural numbers. He knows all of them, according to Augustinus and Cantor. But I am sure he has no list of all the reals. > > > So, if you are discussion Eucliedan > > > geometry you should use the parallel axiom. Of course you can reject is [quoted text clipped - 5 lines] > Well, no, because that "dogma" is not a valid "axiom". But can you tell > me where that "dogma" actually is stated the way you say? Sorry, I only read it some time ago somwhere. But I think the set of dogmas must be in the net for those who are interested. I am not. Regards, WM In article <0f2e464e-9791-4085-a381-0e428be843fa@b9g2000yqm.googlegroups.com>, > > In article > > <74f33c12-95fe-4b83-a7e3-941591def...@c9g2000yqm.googlegroups.com> WM [quoted text clipped - 21 lines] > It is, because the paths of the tree are isomorphic with the real > numbers in [0, 1] No they are not. Not even order isomprphic. In the maximal binary tree, with the usual lexicographic order, there are adjacent paths, but there are no adjacent reals in binary, or any other, notation. > > > I show that the end of each > > > path p of the set P can be mapped on a node, and that all paths p of P [quoted text clipped - 5 lines] > that it gets covered and that all nodes get covered by a countable set > of paqths. But that is not at all the same as showing that all paths get "covered" but a countable set of nodes, with only one path per node. > > > Therefore, after having completed the > > > covering of the whole tree, there remains no node that could be used > > > to construct a path that does not belong to P. But there remain uncountably many paths in the maximal tree unaccounted for by that asociation. > > This is the wrong way around. You assume that you can cover this way the > > whole tree (I think with this you mean each path in the tree). But that is [quoted text clipped - 3 lines] > Then every node is covered by at least one path, hence it does not > remain uncovered. In fact every single node is covered by uncountably many paths this way. > > > This disproves the > > > mentioned axiom. [quoted text clipped - 3 lines] > I do not assume that the number of nodes is countable, but I count > them. Here: You assume the number of paths is countable despite you inability to count (list) them. > Like God and the set of natural numbers. He knows all of them, > according to Augustinus and Cantor. But I am sure he has no list of > all the reals. Are you questioning the ability of God? Why not? You question the ability of everyone else, except yourself. > > > > So, if you are discussion Eucliedan > > > > geometry you should use the parallel axiom. Of course you can reject [quoted text clipped - 10 lines] > Sorry, I only read it some time ago somwhere. But I think the set of > dogmas must be in the net for those who are interested. I am not. Then why bring it up? You have much more serious problems to face. SignatureVirgil ... > > > > And you are deluded. An axiom is a statement of something that can > > > > not be proven, neither disproven using the remainder of the theory. [quoted text clipped - 6 lines] > It is, because the paths of the tree are isomorphic with the real > numbers in [0, 1] First off, the paths of the tree are not isomorphic with the real number. There is no 1-1 mapping that is order preserving. Moreover, the statement "the binary tree has uncountably many paths" is a theorem, not an axiom. It can be proven. In ZF there is no axiom that uses the word "uncountable". It is defined and all statements using it are theorems. But apparently you still do not understand what an axiom is. > > > I show that the end of each > > > path p of the set P can be mapped on a node, and that all paths p of [quoted text clipped - 3 lines] > > The paths of the tree have no end. So your statment "I show that the end of each path p of the set P can be mapped on a node" is blatant nonsense. Why do you utter such contradictionary statements? > But it can be shown for every node > that it gets covered and that all nodes get covered by a countable set > of paqths. You have not shown it. You only show it by assuming that there are countably many paths. > > > Therefore after having completed the > > > covering of the whole tree, there remains no node that could be used [quoted text clipped - 7 lines] > Then every node is covered by at least one path, hence it does not > remain uncovered. Right. But do you use up *all* paths? *That* is what you have to prove. You are always going to it from the wrong way. It is right that all nodes can be covered by countably many paths. The set of paths where each path from some node always goes right is an example. It does indeed cover all nodes. But it is not the set of all paths, because there are paths that do not satisfy the condition that from some node onwards it always goes to the right. Or can you find a node that is *not* covered by a path that after some node to the right? Can you find a node that is covered by a path that alternates going left and right that is not covered by a path that after some node always goes to the right? > > > This disproves the > > > mentioned axiom. [quoted text clipped - 3 lines] > I do not assume that the number of nodes is countable, but I count > them. Darn. Misreeading *again*. The number of nodes *is* countable. You do assume the number of paths is countable. Why are you always misreading what people do write? > Here: > http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#363,26,Folie 26 Sorry, I am not able to look at that, no powerpoint on this system. > > > That is same as with proofs of God. The Vatican published an axiom > > > (they call it a dogma, but it is of the same meaning) according to [quoted text clipped - 7 lines] > according to Augustinus and Cantor. But I am sure he has no list of > all the reals. I am not interested in religious non-sequitors. > > > That means, you are willing to believe in what the Vatican says? > > [quoted text clipped - 3 lines] > Sorry, I only read it some time ago somwhere. But I think the set of > dogmas must be in the net for those who are interested. I am not. So you just state something without being able to back it up. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <0f2e464e-9791-4085-a381-0e428be84...@b9g2000yqm.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 11 Jun., 14:50, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 12 lines] > First off, the paths of the tree are not isomorphic with the real number. > There is no 1-1 mapping that is order preserving. There are two paths for every terminating rational number. The rest is a simple 1 to 1 mapping. > Moreover, the statement "the binary tree has uncountably many paths" is a > theorem, not an axiom. It can be proven. And it can be disproven. > In ZF there is no axiom that uses the word "uncountable". It is defined and > all statements using it are theorems. > > But apparently you still do not understand what an axiom is. said: I The axiom *could be*: The binary tree has uncountably many paths. > > > > I show that the end of each > > > > path p of the set P can be mapped on a node, and that all paths p of [quoted text clipped - 6 lines] > So your statment "I show that the end of each path p of the set P can be > mapped on a node" is blatant nonsense. I used the word end, that can have at least two meanings in the sense of "tail". > > But it can be shown for every node > > that it gets covered and that all nodes get covered by a countable set > > of paqths. > > You have not shown it. You only show it by assuming that there are countably > many paths. If every terminating path of the form 111...1000... is used, then every node is covered by the last 1. > > > > Therefore after having completed the > > > > covering of the whole tree, there remains no node that could be used [quoted text clipped - 15 lines] > do not satisfy the condition that from some node onwards it always goes > to the right. These path are in the tree too, after having completed the construction. > Or can you find a node that is *not* covered by a path that after some node > to the right? Can you find a node that is covered by a path that alternates > going left and right that is not covered by a path that after some node > always goes to the right? I see that every such alternating path with all its nodes is in the tree, after construction. > > > > This disproves the > > > > mentioned axiom. [quoted text clipped - 7 lines] > assume the number of paths is countable. Why are you always misreading > what people do write? Perhaps because they do not express them precisely enough? But I do not always misread. We can state: The number of nodes is countable. The number of paths required to cover all nodes is countable too. > > > > That means, you are willing to believe in what the Vatican says? > > > [quoted text clipped - 5 lines] > > So you just state something without being able to back it up. That is a ridiculous requirement in an informal discussion, in particular if not mathematics is concerned. Regards, WM In article <3f7ebb76-23f8-494a-9247-284a14720b07@3g2000yqk.googlegroups.com>, > > In article > > <0f2e464e-9791-4085-a381-0e428be84...@b9g2000yqm.googlegroups.com> WM [quoted text clipped - 25 lines] > > And it can be disproven. Until WM provides us with a reasonably complete listing of his fundamental truths, on which he claims to base all his conclusions, he cannot prove anything to anyone. > > In ZF there is no axiom that uses the word "uncountable". It is defined > > and [quoted text clipped - 4 lines] > said: I The axiom *could be*: The binary tree has uncountably many > paths. Axioms are accepted without proof. The the maximal infinite binary tree has uncountably many paths is a theorem, not an axiom, and is provable from the axioms of ZF plus relevant definitions. > > > > > I show that the end of each > > > > > path p of the set P can be mapped on a node, and that all paths p [quoted text clipped - 21 lines] > If every terminating path of the form 111...1000... is used, then > every node is covered by the last 1. But when one has the set of all nodes having only finitely many 1 branchings between them and the root node, suddenly one also has all paths with infinitely many one branchings in that same tree. > > Right. But do you use up *all* paths? *That* is what you have to prove. > > You are always going to it from the wrong way. It is right that all nodes [quoted text clipped - 15 lines] > I see that every such alternating path with all its nodes is in the > tree, after construction. Then there are uncountably many of them, as Cantor proved, and as no one, including WM, has been able to disprove. ssume it is false, it is easy to prove it is false. > > > > > > I do not assume that the number of nodes is countable, but I count [quoted text clipped - 6 lines] > Perhaps because they do not express them precisely enough? But I do > not always misread. You do it often enough, and consistently enough, to make discussions with you dificult. > We can state: The number of nodes is countable. The number of paths > required to cover all nodes is countable too. Fair enough, but that does not mean that EVERY set of paths which covers all nodes needs to be countable. The set of ALL paths, not all being required, can be, and is enough larger than any of WM's "required to cover" sets as to be uncountable. > > > > > That means, you are willing to believe in what the Vatican says? > > > > [quoted text clipped - 9 lines] > That is a ridiculous requirement in an informal discussion, in > particular if not mathematics is concerned. If that statement is so irrelevant to mathematics, why did you bring it up in a discussion of mathematics? SignatureVirgil ... > > > > > > And you are deluded. An axiom is a statement of something > > > > > > that can not be proven, neither disproven using the remainder [quoted text clipped - 13 lines] > There are two paths for every terminating rational number. The rest is > a simple 1 to 1 mapping. And so your statement that they are isomorphic is wrong. Do you agree? > > Moreover, the statement "the binary tree has uncountably many paths" is a > > theorem, not an axiom. It can be proven. > > And it can be disproven. I have not seen a disprove of that theorem. > > In ZF there is no axiom that uses the word "uncountable". It is defined > > and all statements using it are theorems. > > > > But apparently you still do not understand what an axiom is. > > said: I The axiom *could be*: The binary tree has uncountably many paths. No, that is not a valid axiom. > > > > > I show that the end of each > > > > > path p of the set P can be mapped on a node, and that all paths [quoted text clipped - 9 lines] > I used the word end, that can have at least two meanings in the sense > of "tail". So you did use consciously an ambiguous word without clarifying what you did mean? > > > But it can be shown for every node > > > that it gets covered and that all nodes get covered by a countable set [quoted text clipped - 5 lines] > If every terminating path of the form 111...1000... is used, then > every node is covered by the last 1. But that still does not show that all nodes are covered by a countable set of paths. It only shows that every node is covered by a set of paths that contains a countable subset, not that the set itself is countable. > > > > > Therefore after having completed the > > > > > covering of the whole tree, there remains no node that could be [quoted text clipped - 18 lines] > These path are in the tree too, after having completed the > construction. Yes, they are in the tree, so you have to prove that they are also used in the covering, which you have not done. You do not have to show that each node is covered, you have to show that with your covering you use each path. You did show the former. > > Or can you find a node that is *not* covered by a path that after some > > node to the right? Can you find a node that is covered by a path that [quoted text clipped - 3 lines] > I see that every such alternating path with all its nodes is in the > tree, after construction. This is not an answer to my question. The path is in the tree, but does it contain a node that is not covered by a path that after some node always goes to the right? > > > > > This disproves the > > > > > mentioned axiom. [quoted text clipped - 12 lines] > We can state: The number of nodes is countable. The number of paths > required to cover all nodes is countable too. Right. But the number of required paths is the minimal number of paths needed to cover all the nodes. That is *not* the number of all paths. Again: consider the set of paths where each path after some point always goes to the right. This set of paths covers each node. Consider the path that alternates going right and left continuously. Can you point to a node in this alternating path that is not covered by one of the paths in the earlier set? But you did agree slightly higher up that that path was in the tree. > > > > > That means, you are willing to believe in what the Vatican says? > > > > [quoted text clipped - 8 lines] > That is a ridiculous requirement in an informal discussion, in > particular if not mathematics is concerned. Oh. So in an informal discussion you can tell nonsense whatever you like and when challenged you do not need to back it up? Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > If every terminating path of the form 111...1000... is used, then > > every node is covered by the last 1. > > But that still does not show that all nodes are covered by a countable set of > paths. It only shows that every node is covered by a set of paths that > contains a countable subset, not that the set itself is countable. It is easy to construct a bijection between a countable set of paths and all nodes. These nodes can even be defined by the paths that lead to them. That shows that all nodes of the tree can be covered by a countable set of paths. > Yes, they are in the tree, so you have to prove that they are also used in the > covering, which you have not done. You do not have to show that each node > is covered, you have to show that with your covering you use each path. You > did show the former. If every node is covered, then there remains no path to be covered. > > > Or can you find a node that is *not* covered by a path that after some > > > node to the right? Can you find a node that is covered by a path that [quoted text clipped - 7 lines] > contain a node that is not covered by a path that after some node always goes > to the right? No. From that we can obtain that the number 1/3 is also in a list of terminating rationals. There does not exist a sequence 0.010101... that is longer than *every* finite sequence of that form. > > We can state: The number of nodes is countable. The number of paths > > required to cover all nodes is countable too. > > Right. But the number of required paths is the minimal number of paths needed > to cover all the nodes. That is *not* the number of all paths. If all nodes are covered then all path that can be distinguished by nodes, i.e., all reals that can be distinguished by digits, are there. > Again: > consider the set of paths where each path after some point always goes to > the right. This set of paths covers each node. Consider the path that > alternates going right and left continuously. Can you point to a node in > this alternating path that is not covered by one of the paths in the earlier > set? But you did agree slightly higher up that that path was in the tree. All parts of that path that really exist, really are in the tree. > > That is a ridiculous requirement in an informal discussion, in > > particular if not mathematics is concerned. > > Oh. So in an informal discussion you can tell nonsense whatever you like > and when challenged you do not need to back it up? I have read that there is a dogma of the catholic church stating that the existence of God can be proved by means of scientific reasoning. I do not remember the source. Does that turn this information into nonsense? (Except that its contents is nonsense.) Regards, WM In article <fd4c6395-51cf-4045-883e-f5c986b9176b@z7g2000vbh.googlegroups.com>, > > > If every terminating path of the form 111...1000... is used, then > > > every node is covered by the last 1. [quoted text clipped - 8 lines] > to them. That shows that all nodes of the tree can be covered by a > countable set of paths. Actually, WM is right for once. Every node (there are only countably many of them after all) can be "covered" by a countable set of paths. Simply pick the nth path so as to contain the nth node in some counting of the nodes. However often WM is wrong on his own, one should be careful not to assume him always wrong. He is not that consistent. SignatureVirgil In article <fd4c6395-51cf-4045-883e-f5c986b9176b@z7g2000vbh.googlegroups.com>, > > > If every terminating path of the form 111...1000... is used, then > > > every node is covered by the last 1. [quoted text clipped - 17 lines] > > If every node is covered, then there remains no path to be covered. Covering a node with a path does not automatically also cover each of the uncountably many paths through that node. > If all nodes are covered then all path that can be distinguished by > nodes, i.e., all reals that can be distinguished by digits, are there. Then WM must mean something other than the usual by "distinguished". No single digit, nor finite set of digits, distinguishes any one real from all others, just as no one node, nor even any finite set of nodes, distinguishes any one path in a maximal infinite binary tree from all others. > > Again: > > consider the set of paths where each path after some point always goes to [quoted text clipped - 5 lines] > > All parts of that path that really exist, really are in the tree. If any part of that path does not "really exist", one is no longer talking about maximal infinite binary trees at all, but just another of WM's myths. > > > That is a ridiculous requirement in an informal discussion, in > > > particular if not mathematics is concerned. [quoted text clipped - 6 lines] > do not remember the source. Does that turn this information into > nonsense? It is WM who keeps trying to insert his own dogmas in places that they do not fit. In order to even talk about infinite sets coherently, one must have at least one infinite set, any inductive set as a metaphor for the set of all naturals will do. And once one has that set, none of WM's objections obtain, since we can define a partial order on any such set which makes it into a maximal infinite binary tree having all those properties WM objects to. SignatureVirgil > > > If every terminating path of the form 111...1000... is used, then > > > every node is covered by the last 1. [quoted text clipped - 7 lines] > to them. That shows that all nodes of the tree can be covered by a > countable set of paths. Yes, they can be covered by a countable set of paths. Nothing more. It does *not* show a bijection, and not a construction of a bijection. > > Yes, they are in the tree, so you have to prove that they are also used in > > the covering, which you have not done. You do not have to show that each > > node is covered, you have to show that with your covering you use each > > path. You did show the former. > > If every node is covered, then there remains no path to be covered. What do you mean with "covering a path"? Until now you were talking about "covering nodes". And whatever it may mean, in what way does "every node is covered" show that there remains "no path to be covered"? > > > > Or can you find a node that is *not* covered by a path that after > > > > some node to the right? Can you find a node that is covered by a [quoted text clipped - 11 lines] > terminating rationals. There does not exist a sequence 0.010101... > that is longer than *every* finite sequence of that form. Ah, so now you contend that 1/3 is a terminating rational. As the sequence of terminating rationals starting with that is: 1/4, 5/16, 21/64, 84/256, ... which of those is 1/3? > > > We can state: The number of nodes is countable. The number of paths > > > required to cover all nodes is countable too. [quoted text clipped - 4 lines] > If all nodes are covered then all path that can be distinguished by > nodes, i.e., all reals that can be distinguished by digits, are there. What do you *mean* with "all path that can be distinguished by nodes"? I see two meanings: (1) All paths that can be distinguished by a node from each other path (2) All paths that can be distinguished by a node from all other paths. As the covering of all nodes obviously does not imply (1), you must mean (2). But there is no path in the set of all paths that can be distinguished by a node from all other paths. So clearly that is not "all paths". > > Again: > > consider the set of paths where each path after some point always goes to [quoted text clipped - 5 lines] > > All parts of that path that really exist, really are in the tree. That is not an answer to my question. Moreover you have not defined "really exist". > > > That is a ridiculous requirement in an informal discussion, in > > > particular if not mathematics is concerned. [quoted text clipped - 6 lines] > do not remember the source. Does that turn this information into > nonsense? (Except that its contents is nonsense.) Clearly it is not backed-up by facts, so it can be nonsense or not, I have no way of knowing. It might be true, if might be false, the author may have been wrong, you can remember wrong. But whatever, it is not an argument in a sensible discussion. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <fd4c6395-51cf-4045-883e-f5c986b91...@z7g2000vbh.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 19 Jun., 15:42, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 13 lines] > Yes, they can be covered by a countable set of paths. Nothing more. It > does *not* show a bijection, and not a construction of a bijection. I do not argue that there is a bijection with all real numbers. The reason is that real numbers with infinite bit sequences do not exist *as bit sequences". pi exists in form of many formulas, Vielta, Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist asa a subject to Cantor's proof. All existing sequences are in my tree. And obviously there are only countably many. > > > Yes, they are in the tree, so you have to prove that they are also used in > > > the covering, which you have not done. You do not have to show that each [quoted text clipped - 6 lines] > "covering nodes". And whatever it may mean, in what way does "every node > is covered" show that there remains "no path to be covered"? Try to find a path p that contains a node not yet covered by a path q used to construct the tree. In order to prove that there is such a path p you must show for every path q that there is a node not covered by q. But that is impossible. If you think, you have accomplished it at some node, there will always be another q that is identical with p to that node. That means, you cannot diagonalize the tree. It contains already all paths that can be distinguished by nodes, i.e., all reals that can be distinguished by digits. And As my construction B shows, this set is countable. > > No. From that we can obtain that the number 1/3 is also in a list of > > terminating rationals. There does not exist a sequence 0.010101... > > that is longer than *every* finite sequence of that form. > > Ah, so now you contend that 1/3 is a terminating rational. No. It is a rational that has no decimal and no binary representation. It has a ternary representation though. Regards, WM In article <eda6be39-323e-4b70-b40c-a43519b16e73@33g2000vbe.googlegroups.com>, > > In article > > <fd4c6395-51cf-4045-883e-f5c986b91...@z7g2000vbh.googlegroups.com> WM [quoted text clipped - 23 lines] > Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist > asa a subject to Cantor's proof. Cantor's proof only speaks of infinite binary sequences, not real numbers, and proves that there is no such thing as a complete listing of them. > All existing sequences are in my tree. Then your tree is incomplete, so not a maximal infinite binary tree at all. > And obviously there are only countably many. It is not obvious to anyone who can understand Cantor's proof to the contrary. > > > > Yes, they are in the tree, so you have to prove that they are also > > > > used in [quoted text clipped - 11 lines] > Try to find a path p that contains a node not yet covered by a path q > used to construct the tree. Try to discover a node of any maximal infinite binary tree not covered by uncountably many paths. > In order to prove that there is such a > path p We do not claim any such path, we only claim there will be paths not used in any construction which only uses only countably many paths, which is quite a different issue. > you must show for every path q that there is a node not covered > by q. WRONG! WE only need show a path not used in the construction. If the set of paths used in the construction is cunable, then any proof of its countability allows, by the Cantor construction, proof of existence of a path not used. > But that is impossible. No its just the Cantor proof. > If you think, you have accomplished it > at some node, there will always be another q that is identical with p > to that node. WM'S nonsense about nodes just shows how little he understands about maximal infinite binary trees. > That means, you cannot diagonalize the tree. Since the set of paths in a maximal infinite binary tree is already known to be uncountable, there is no point to even trying to "diagoanlize" it. > It contains > already all paths that can be distinguished by nodes I wonder what sort of paths WM has in mind which cannot be "distinguished by nodes"? Or at least by sets of nodes. In a real world maximal infinite binary tree, any path can be distinguished from all others by the set of all its nodes. > > > No. From that we can obtain that the number 1/3 is also in a list of > > > terminating rationals. There does not exist a sequence 0.010101... [quoted text clipped - 3 lines] > > No. It is a rational that has no decimal and no binary representation. In the larger world, outside of WM's mathUnrealism, where non-terminating decimals and binaries reside, it has both. SignatureVirgil ... > > > > But that still does not show that all nodes are covered by a > > > > countable set of paths. It only shows that every node is covered [quoted text clipped - 10 lines] > > I do not argue that there is a bijection with all real numbers. You did assert that there was a bijection with the set of all paths in the tree. You have not proven that. > The > reason is that real numbers with infinite bit sequences do not exist > *as bit sequences". Oh, well, you are apparently a non-standard form of "exist". By the axiom of infinite, the infinite set N exists, and so there does exist a function: f: N -> {0, 1} f(n) = 0 if n is odd = 1 if n is even. I would think that would be an appropriate infinite bit sequence of 1/3. > pi exists in form of many formulas, Vielta, > Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist > asa a subject to Cantor's proof. All existing sequences are in my > tree. And obviously there are only countably many. I would argue that the infinite bit sequence .0101010101 exists and is in your tree. Your bijection above does not include that bit sequence. > > > > Yes, they are in the tree, so you have to prove that they are also > > > > used in the covering, which you have not done. You do not have to [quoted text clipped - 9 lines] > Try to find a path p that contains a node not yet covered by a path q > used to construct the tree. Is *that* a definition of "covering a path"? But if your paths cover all nodes, there is no such path. > It contains > already all paths that can be distinguished by nodes, i.e., all reals > that can be distinguished by digits. And As my construction B shows, > this set is countable. And the path .0101010101... is not in that bijection, nevertheless there is no node in the tree that is *not* covered by that path. So you did show only a bijection between a subset of the paths with the nodes. Back to square 1 I would say. > > > No. From that we can obtain that the number 1/3 is also in a list of > > > terminating rationals. There does not exist a sequence 0.010101... [quoted text clipped - 4 lines] > No. It is a rational that has no decimal and no binary representation. > It has a ternary representation though. Eh? You explicitly did state (as I did quote): "From that we can obtain that the number 1/3 is also in a list of terminating rationals." I would state this to mean that 1/3 is a terminating rational. But I would also say that what I wrote higher up consists (in view of the axiom of infinity) an infinite bitstring of 1/3. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > I do not argue that there is a bijection with all real numbers. > > You did assert that there was a bijection with the set of all paths in the > tree. You have not proven that. I do not argue that there is a bijection with what you dream of as "all real numbers". I show that the number of paths that can be identified by sequences of nodes or bits is countable. > > The > > reason is that real numbers with infinite bit sequences do not exist [quoted text clipped - 6 lines] > = 1 if n is even. > I would think that would be an appropriate infinite bit sequence of 1/3. And I have shown that this sequence is covered by finite sequences to the satisfaction of everybody who is using bits to identify a number. > > pi exists in form of many formulas, Vielta, > > Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist [quoted text clipped - 3 lines] > I would argue that the infinite bit sequence .0101010101 exists and is in > your tree. Your bijection above does not include that bit sequence. The tree will not differ in any detail from the construced one, if we use all finite paths as before but add the sequence 010101... instead of 000.... > > > > > Yes, they are in the tree, so you have to prove that they are also > > > > > used in the covering, which you have not done. You do not have to [quoted text clipped - 12 lines] > Is *that* a definition of "covering a path"? But if your paths cover all > nodes, there is no such path. That' swhy you cannot find it. > > It contains > > already all paths that can be distinguished by nodes, i.e., all reals [quoted text clipped - 4 lines] > no node in the tree that is *not* covered by that path. So you did show > only a bijection between a subset of the paths with the nodes. I showed hat all paths are in the tree that can be identified by nodes. > > > > No. From that we can obtain that the number 1/3 is also in a list of > > > > terminating rationals. There does not exist a sequence 0.010101... [quoted text clipped - 9 lines] > terminating rationals." > I would state this to mean that 1/3 is a terminating rational. All that can be identified of 1/3 belongs to a terminating rational. Your "..." has no meaning in rigorous mathematics. Regards, WM > > > I do not argue that there is a bijection with all real numbers. > > [quoted text clipped - 3 lines] > I do not argue that there is a bijection with what you dream of as > "all real numbers". Sorry, no dreaming involved. You did show that you could cover each node by countably many paths. You did *not* show that the set of all paths was exhausted by that process. > I show that the number of paths that can be identified by sequences of > nodes or bits is countable. Not at all, because you did not show a bijection. > > > The > > > reason is that real numbers with infinite bit sequences do not exist [quoted text clipped - 10 lines] > And I have shown that this sequence is covered by finite sequences to > the satisfaction of everybody who is using bits to identify a number. Ah, you are easily satisfied. So 0.010101 identified 1/3? > > > pi exists in form of many formulas, Vielta, > > > Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist [quoted text clipped - 7 lines] > use all finite paths as before but add the sequence 010101... instead > of 000.... So your bijection did *not* include that path. Why do you not admit that there is a path that was not included in your bijection? And when you add the sequence 01010101... instead of 000... the bijection does not include 000... which is also a path in your tree. > > > > What do you mean with "covering a path"? Until now you were talking > > > > about "covering nodes". And whatever it may mean, in what way does [quoted text clipped - 8 lines] > > That' swhy you cannot find it. Yes, but not yet given any definition about "covering a path"... And there are paths present in the tree where all nodes in the path are also covered by other paths. Consider the tree as you presented (starting with a list of nodes, of which only finitely many are labeled with 1, and, yes, I do admit that that list is countable). You created a mapping of paths to nodes such that each node was covered by a path. Also you did admit that the path 01010101... was also in the tree, but your mapping did *not* include that path. So you did exhaust all nodes but not all paths. Now come up with a *bijection* between the set of nodes and the set of paths in your tree, and be sure that there is *no* path in the tree that is not part of that bijection. Only when that is done I will believe in your so-called bijection. We start with the following definitions: a node is a finite sequence of 0's and 1's a path is an infinite sequence of 0's and 1's If you do not agree with these definitions, provide your own definitions. > > > It contains > > > already all paths that can be distinguished by nodes, i.e., all reals [quoted text clipped - 7 lines] > I showed hat all paths are in the tree that can be identified by > nodes. Yes, so you did not show a bijection. That is more than what you do state now. > > > > > No. From that we can obtain that the number 1/3 is also in a list > > > > > of terminating rationals. There does not exist a sequence [quoted text clipped - 13 lines] > All that can be identified of 1/3 belongs to a terminating rational. > Your "..." has no meaning in rigorous mathematics. So you would like to abolish limits from rigorous mathematics? Do you have any idea about the meaning of the "..." there in rigorous mathematics? I do not think so. Moreover, one left-over question you always snip and never answer: (1) FISON{n} = {1, 2, 3, ..., n} union(k = 1...n) FISON(k) is a FISON so union(k = 1...oo) FISON(k) is a FISON (2) P(n) = 1/n! is rational sum(k = 1...n) P(k) is rational so sum(K = 1...oo) P(k) is rational can you explaing why (1) is correct but (2) is wrong (that is what you state)? Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <9dfc93db-dfea-48b1-9f2b-04f257f37...@s9g2000yqd.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 30 Jun., 16:12, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 135 lines] > dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/ Because (2) is wrong and (1) is correct below: Results 1 - 10 for musatov bijection. (0.68 seconds) 1. Answer to Dik T. Winter - sci.math | Google Groups the bijection of the rationals > 0 and the naturals I have so often shown, .... Cosmic Lensing os very handy for Musatov's proof P=NP<<>> ... May 27 by ... groups.google.com/group/sci.math/msg/2f0140fa143ab15f 2. Math Forum Discussions that for every x in X there is a bijection between ... furnishes such a bijection. If E_x is "infinite to the ... Martin Michael Musatov. 6/11/09 ... mathforum.org/kb/thread.jspa?messageID=6749040&tstart=0 3. [PDF] <a href="http://arxiv.org/abs/0904.3116v1">arXiv:0904.3116v1 [cs ... File Format: PDF/Adobe Acrobat - View as HTML Apr 20, 2009 ... performs a bijection between L ...... D. Musatov, Extractors and an effective variant of Muchnik's theorem. Diplom (Master thesis). ... arxiv.org/pdf/0904.3116 by D Musatov 4. Math Forum Discussions On May 14, 12:07 pm, Martin Musatov <marty.musa...@gmail.com> wrote: ... forums ... write bijections between any infinite set S and S \F for any ... mathforum.org/kb/thread.jspa?messageID=6714416&tstart=0 5. Integer Format Sequence for linbox-use - linbox-use | Google Groups May 3, 2009 ... From: Martin Michael Musatov <marty.musa...@gmail.com> ..... D A000045 N.S.S.Gu,N.Y.Li and T.Mansour,2-Binary trees:bijections ... groups.google.com/group/linbox-use/browse.../7bc6976f1181638c 6. Variations on a theme of Debs and Saint Raymond remarkable theorem which includes that of Ivasev-Musatov as a particular case. Theorem 8. ...... be a bijection. Set N(0) = 0, γ ... jlms.oxfordjournals.org/cgi/reprint/79/1/33.pdf by TW Korner - 2009 - Related articles - All 5 versions 7. Result for query "keyword(s)=theorem author= title=" On a theorem of Ivasev-Musatov III, ...... by Theorem; bijection, we get that the theorem holds for all $\pi(\lambda)$ with; _ \chi(G_ 1)$, see \cite KR. ... nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/.../j?... 8. [PDF] ABOUT INTERPOLATION OF SUBSPACES OF REARRANGEMENT INVARIANT SPACES ... File Format: PDF/Adobe Acrobat - View as HTML and G is bijective. Theorem 1.5. Let r.i.s.'s X ...... "Mir",. Moscow, 1965 (Russian), translated from English by O. S. Ivasev Musatov. ... www.emis.de/journals/HOA/IJMMS/Volume25_7/465.pdf by SV ASTASHKIN - Cited by 9 - Related articles 9. Proving the Parallel Postulate of Geometry from the Number Systems ... May 28, 2009 ... From: Martin Musatov <marty.musa...@gmail.com>. Date: Fri, 29 May 2009 08:46:13 -0700 (PDT) ... C (* Binary expansions as a bijection ... groups.google.com/group/sci.logic/browse.../b01f8ea2c5255dbc? lnk... + + + Martin Musatov > Sorry, no dreaming involved. You did show that you could cover each node by > countably many paths. You did *not* show that the set of all paths was > exhausted by that process. Proof by your failure to find a path that is not in the tree. The tree is constructed by terminating paths with tails 000... appended, but the same tree can also be constructed by terminating paths with tails 010101.. or pi appeded. If you, by looking at the completely constructed tree only, can find a path that is not in the tree and that was not used to construct the tree, then you are right. But you cannot. Therefore your assertion concerning missing paths is wrong. > > I show that the number of paths that can be identified by sequences of > > nodes or bits is countable. > > Not at all, because you did not show a bijection. I did use the countable set of terminating paths (extended by a special kind of tails). And you cannot identify a path that I did not use, if you don't know from what kind of paths the tree was constructed. This proves that there is nothing further in R that is related to sequences of nodes only. Regards, WM In article <8016ac03-f17e-41a9-b43b-d368288c37ae@t13g2000yqt.googlegroups.com>, > > Sorry, no dreaming involved. You did show that you could cover each node by > > countably many paths. You did *not* show that the set of all paths was > > exhausted by that process. > > Proof by your failure to find a path that is not in the tree. There are no infinite binary paths which are not in TA maximal infinite binary tree, but there are such paths which are not is any given countable set of paths, as Cantor proved. > The tree is constructed by terminating paths with tails 000... > appended, but the same tree can also be constructed by terminating [quoted text clipped - 3 lines] > But you cannot. Therefore your assertion concerning missing paths is > wrong. If all one needs is a path not used in the construction, and that construction uses only countably many paths, then it is trivial to find paths not used in that construction, even though they appear in the "constructed" tree. > > > I show that the number of paths that can be identified by sequences of > > > nodes or bits is countable. [quoted text clipped - 5 lines] > use, if you don't know from what kind of paths the tree was > constructed. But as soon as I do know which ones were used, I can find others in the tree which were not used. WM's version of the problem is like having to guess whether a tossed coin has come up heads or tails without looking at it, but the actual problem allows one to look. > This proves WM's arguments only prove that he does not understand what is going on. In WM's world, maximal infinite binary trees are prohibited from existing, so any arguments he makes about their properties is fatuous. In a ZFC world, where maximal infinite binary trees can exist, their properties derive from the ZFC axioms. SignatureVirgil > In article <db512ef4-fbff-48ee-af30-ed25dc9b7...@s16g2000vbp.googlegroups.com> George Greene <gree...@email.unc.edu> writes: > > If you were posting this crap in German, you would've > > been fired long ago. > You apparently do not understand it at all. Understand WHAT?? You owe us an antecedent for "it", there. > WM has posted this stuff for > years in the German newsgroup on mathematics, in German. sh.t. There is MORE THAN ONE mathematics newsgroup in the German language. This one in America has become so degraded that what few serious scholars still read it treat it as idle amusement, not something to try to get anybody fired about. WM has NOT been posting this stuff in any forum that anybody CARES about (in German). > Moreover, he has two books about mathematics on his name, > in German. Books ABOUT mathematics are NOT likely to be respected by MATHEMATICIANS. He would NEED to write a book WITH SOME MATH in it AS OPPOSED to a book "about" mathematics, in order to be relevant. ... > > > If you were posting this crap in German, you would've > > > been fired long ago. > > > You apparently do not understand it at all. > > Understand WHAT?? You owe us an antecedent for "it", there. How the German academic works. > > WM has posted this stuff for > > years in the German newsgroup on mathematics, in German. > > sh.t. There is MORE THAN ONE mathematics newsgroup in the German > language. What is your problem? You stated that if WM has been posting this crap in German he would have been fired long ago. I state that he has been posting this crap in German and has not been fired. So you were clearly wrong in your assertion. > > Moreover, he has two books about mathematics on his name, > > in German. [quoted text clipped - 3 lines] > MATH in it AS OPPOSED to a book "about" mathematics, in order > to be relevant. You were talking about posting in German, nothing about the context. Moreover, one of his books is published as a textbook for students. It is in German, contains the same crap, and he is still not fired. Draw your conclusions. BTW, are you never able to discuss things reasonably? Using caps all over the place does not increase your credibility. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <37a98f77-6bc9-4cae-88e5-fcbadd1d0...@q2g2000vbr.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 27 Mai, 15:50, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 19 lines] > linear sets. Note that "classical logic is obtained from finite sets". > Nowhere in that quote the word linear is mentioned. Nowhere in that quote the word union in mentioned. Nevertheless the logical rules of unions are obtained from unions of finite sets. The logical rules of linear sets are obtained from finite linear sets. > > > > You said [*] is not true, but [**] is true. > > > [quoted text clipped - 10 lines] > E n A m P(m, n) <- A m E n P(m, n) > which is [***], neither [*] nor [**]. I never said so. But [*] is [**] & [***]. Therefore [*] differs from [**] only by the reverse. > > > I said that for the case involved you have to > > > *prove* that it is true, because it is not generally true. [quoted text clipped - 3 lines] > > It is not true for the infinite set of naturals. That is your claim. It is justified for potential infinity. It is wrong for complete sets. > (1) define FISON(n) be the set of naturals from 1 to n, that is: {1, ..., n}. > (2) A{m in N} E(n in N} such that FISON(m) subset FISON(n), trivial, take > n = m + 1. > (3) E{n in N} A{m in N} such that FISON(m) subset FISON(n), trivially false, > take m = n + 1. > Which part of this proof is wrong? The proof is correct for potential infinity. The proof is incorrect for actual infinity. In that latter case you need not take an n that is surpassed by m. Why don't you start with an n that has no greater m? > It clearly shows that > E n A m P(m, n) <- A m En P(m, n) [quoted text clipped - 5 lines] > implication is true, and so the implication is false (all by classical > logic). Not at all. By classical logic, a complete linear set has a last element. > > > But now you include the word "linear". Where did "Weyl" include the > > > word "linear"? [quoted text clipped - 6 lines] > But as Weyl did not include "linear" in his words, how can that quote > support your claim? There are many finite sets with many special properties that follow from classical logic. One of them is that a complete linear set has a lst element. You drop the completeness condition in certain cases but you assume it in case of Cantor's proof. That is cheating. > > > > > What is contested is that: > > > > > En Am: m =< n <== Am En m =< n [***] [quoted text clipped - 7 lines] > You just state without proof. Where in my proof above that it is false > did I go wrong? State before beginning whether the set that you assume is complete and static, i.e., every element is actually existing, or potentially infinite. > > And you > > should recognize that actually complete linear sets obey [*]. > > Strange, I give above a proof that it does not hold. I did not use > "actual infinity" nor "potentially infinity" That is the point! You use the absence of element m when you choose n = m - 1. But you use the non-absence of m when you execute Cantor's proof. Then you do not admit that for every FISON(n) there is an m = n + 1 that is not in the proof. > > Only > > potentially infinite sets do not. But you mix up things. You claim the [quoted text clipped - 3 lines] > Why is that a necessary consequence? Because you want it to be so? Can > you give a *mathematical* reason? Every finite linear set obeys [*]. That is the mathematical reason. Regards, WM In article <5df917e3-517f-4a38-b28b-3638434966c7@t21g2000yqi.googlegroups.com>, > > In article > > <37a98f77-6bc9-4cae-88e5-fcbadd1d0...@q2g2000vbr.googlegroups.com> [quoted text clipped - 24 lines] > Nowhere in that quote the word union in mentioned. Nevertheless the > logical rules of unions are obtained from unions of finite sets. While those rules may be suggested by looking at finite sets, the rules themselves are not limited by the limitations of finite sets. When one says that the union of a set of sets is the set of all members of those member sets, there is no actual or implied restriction on any of those sets to be finite. > The logical rules of linear sets are obtained from finite linear > sets. That the rules for general well ordered sets may be suggested by finite sets of finite well ordered sets, does not mean that when not finite they must behave as if they were finite. That WM is incapable of dealing honestly with infinite sets does not limit everyone to his incapacity. > > > > I said that for the case involved you have to > > > > *prove* that it is true, because it is not generally true. > > > > > > It is generally true for complete linear sets. You have to prove > > > that it is not. Complete in this case meaning finite. Since WM cannot prove that his claims hold except for finite "linear sets", which are merely finite well-ordered sets, no one has to assume that they must hold for arbitrary well-ordered sets, and there is good reason to suppose otherwise. > > It is not true for the infinite set of naturals. > > That is your claim. It is justified for potential infinity. But outside of the tiny world of WM's MathUnrealism, there are "potential infiniteness" cannot be applied to sets. > It is wrong for complete sets. By which WM means finite ones, but no one has claimed otherwise. > The proof is correct for potential infinity. But outside of the tiny world of WM's MathUnrealism, there are "potential infiniteness" cannot be applied to sets. > The proof is incorrect > for actual infinity. In that latter case you need not take an n that > is surpassed by m. Why don't you start with an n that has no greater > m? If one double quantifies with "AxEy" or " EyAx"one is not allowed to constrain x, but must consider all x, even those ones wold prefer not to consider. So that one needs to consider not only n < m but also n = m and n > m. > Not at all. By classical logic, a complete linear set has a last > element. Only if "complete" requires "finite", which,outside of the tiny world of WM's MathUnrealism, it does not. > > > > But now you include the word "linear". Where did "Weyl" > > > > include the word "linear"? [quoted text clipped - 11 lines] > from classical logic. One of them is that a complete linear set has a > lst element. What is incomplete or non-linear about the set of rational integers? > You drop the completeness condition in certain cases but you assume > it in case of Cantor's proof. That is cheating. Cantor, in his diagonal proof of the incompleteness of any list of binary sequences, does NOT assume that the given list is complete. He merely proves that any given list is incomplete. The Cantor challenge is: Provide me with a list of infinite binary sequences and I will show you how to find one not in that list. So until WM, or someone, can provide a list of infinite binary sequences for which the Cantor rule fails to provide a non-member of the list, Cantor wins. And WM loses. > > > > > > What is contested is that: > > > > > > En Am: m =< n <== Am En m =< n [***] [quoted text clipped - 11 lines] > and static, i.e., every element is actually existing, or potentially > infinite. The set of all naturals, like every set in any mathematically sane set theory, is static. It is "complete" only in the sense that it is not merely potential but is actual. Anything that will be a member of it tomorrow was already a member yesterday. It is well ordered, which means that every non-empty subset has a smallest (but need not have a largest) member. > > > And > > > you [quoted text clipped - 7 lines] > proof. Then you do not admit that for every FISON(n) there is an m = > n + 1 that is not in the proof. Nonsense. N is a well ordered set, but has no maximal member. Thus for every m in N there is an n in N such that n > m. "AmeN EneN m > n" says, in effect, that there must exist some function, say f, from N to N such that f(n) = M. In this case, f(n) = n+1 is such a function. So that if WM objects, he must find some member, say n, of N for which (n+1) is not a member of N. > > > > > > Only [quoted text clipped - 7 lines] > > Every finite linear set obeys [*]. That is the mathematical reason. But no infinite well-ordered set does, and such sets exist outside of WM's wee wee world. SignatureVirgil > > It is wrong for complete sets. > > By which WM means finite ones, but no one has claimed otherwise. Complete means finished = finite. > Cantor, in his diagonal proof of the incompleteness of any list of > binary sequences, does NOT assume that the given list is complete. [quoted text clipped - 3 lines] > The Cantor challenge is: Provide me with a list of infinite binary > sequences and I will show you how to find one not in that list. That is not difficult to satisfy : The complete list of all real numbers starts just at that position n+1 where you will cease to seek it. Regards, WM In article <aa348889-9e3b-4299-b906-e1d53ac96f85@s12g2000yqi.googlegroups.com>, > > > It is wrong for complete sets. > > [quoted text clipped - 12 lines] > numbers starts just at that position n+1 where you will cease to seek > it. If WM cannot even produce a list of binaries for Cantor's technique to work on, or a list of reals for the modified diagonal proof to work on, then WM loses by default. And whenever WM, or anyone else, does produce one of those lists, it will easily be shown to be incomplete. SignatureVirgil > And whenever WM, or anyone else, does produce one of those lists, it > will easily be shown to be incomplete. I told you alredy several times that logic fails in case of infinite sets. Do not argue that Cantor's proof is correct. Argue why my proof is incorrect. Argue facts, not opinions. See the thread concerning the reactions to the binary tree. Choose one or more of those counter arguments and try to convince people who are not trapped in matheology. That would be more commendable than parroting your stuff. Regards, WM In article <27453436-4a7f-4501-a5b6-3d36ef7a4c2e@k38g2000yqh.googlegroups.com>, > > And whenever WM, or anyone else, does produce one of those lists, it > > will easily be shown to be incomplete. > > I told you alredy several times that logic fails in case of infinite > sets. Unlike the bellman, what WM tells me three times need not be , and almost always isn't, true. > Do not argue that Cantor's proof is correct. Do not argue that it isn't until you can come up with a proof that it isn't. > Argue why my proof > is incorrect. Among other reasons, because it is in conflict with Two of Cantor's theorems, and I have a great deal more trust in Cantor than I have in WM. Argue facts, not opinions. WM rekes not his own rede, but treads the primrose path. > See the thread concerning the > reactions to the binary tree. Choose one or more of those counter > arguments and try to convince people who are not trapped in > matheology. That would be more commendable than parroting your stuff. Since WM has done nothing better that to parrot his own stuff, he is hardly in a position to criticize others for doing it. SignatureVirgil ... > > > > > > > The problem boils down to the following: > > > > > > > > > > > > > > En Am: m =< n <==> Am En m =< n [*] > > > > > > > En Am: m =< n ==> Am En m =< n [**] ... > > > > > I said: For complete linear sets [*] is true. > > > > [quoted text clipped - 9 lines] > Nowhere in that quote the word union in mentioned. Nevertheless the > logical rules of unions are obtained from unions of finite sets. What is the relevance of this? The logical rules of unions state that when you have a union of sets that union does contain an element if it is in one of the sets. There is no difference between finite unions and infinite unions. > The > logical rules of linear sets are obtained from finite linear sets. Aha, here we get to the heart of the matter. You do not believe in infinite sets (or infinite unions or infinite sets of finite linear sets). That is possible, of course, but does not rule out theories in which those things do exist. Quoting philosophers of some time ago does not make that wrong. > > > The only thing that can be stated is (symbolically): > > > E n A m P(m, n) -> A m E n P(m, n) [quoted text clipped - 8 lines] > I never said so. But [*] is [**] & [***]. Therefore [*] differs from > [**] only by the reverse. Being incomprehensible again. The reverse of *what*? The reverse of [**] is (as I wrote) [***]. You may never have said so, but it is. To recap: [*] En Am: m =< n <==> Am En m =< n [**] En Am: m =< n ==> Am En m =< n [***] En Am: m =< n <== Am En m =< n But you ask as a counterexample something where [**] is true but [*] false. But that is the wrong way around. Whenever [**] is true, [*] is also true. What is contested is that there are case where [***] is true and [*] false. And it is the latter implication that you do use. > > > > I said that for the case involved you have to > > > > *prove* that it is true, because it is not generally true. [quoted text clipped - 6 lines] > That is your claim. It is justified for potential infinity. It is > wrong for complete sets. Now you are using words that are again completely incomprehensible. You have still failed to give a definition of "potential infinity" that is valid within ZF. Moreover, you have not proven (within ZF) that the statement is wrong. > > (1) define FISON(n) be the set of naturals from 1 to n, that is: > > {1, ..., n}. [quoted text clipped - 6 lines] > The proof is correct for potential infinity. The proof is incorrect > for actual infinity. Can you provide me with definitions within ZF that shows the difference? > In that latter case you need not take an n that > is surpassed by m. Why don't you start with an n that has no greater > m? Because there is no such n. Remember: the set of natural numbers has no largest element in ZF. > > It clearly shows that > > E n A m P(m, n) <- A m En P(m, n) [quoted text clipped - 8 lines] > Not at all. By classical logic, a complete linear set has a last > element. Oh. I think that the term "classical logic" has changed a bit since the last time you looked at it. And, if you refer to Weyl's quote, he stated that classical logic was *derived* from the logic on finite sets, not that it was *identical* to logic on finite sets. > > > I did never claim that quantifier exchange is allowed in case of non- > > > linear sets, like cyclic sets as, for instance, your dice. That would [quoted text clipped - 7 lines] > from classical logic. One of them is that a complete linear set has a > lst element. Not "a complete linear set". But "a complete finite linear set". Why do you drop the word "finite" in the second sentence? To obfuscate? > You drop the completeness condition in certain cases but you assume it > in case of Cantor's proof. That is cheating. You again misunderstand the proof completely. There is an assumption that a complete list is provided and that is proven false. > > > > > > What is contested is that: > > > > > > En Am: m =< n <== Am En m =< n [***] [quoted text clipped - 11 lines] > static, i.e., every element is actually existing, or potentially > infinite. What in the world is a "potentially infinite element"? And, in ZF all sets are static, there is no place for non-static sets. I will recap my proof, I assume the axiom of infinity. That is all. > > (1) define FISON(n) be the set of naturals from 1 to n, that is: > > {1, ..., n}. > > (2) A{m in N} E(n in N} such that FISON(m) subset FISON(n), trivial, take > > n = m + 1. > > (3) E{n in N} A{m in N} such that FISON(m) subset FISON(n), trivially > > false, take m = n + 1. > > Strange, I give above a proof that it does not hold. I did not use > > "actual infinity" nor "potentially infinity" > > That is the point! You use the absence of element m when you choose n > = m - 1. Where in the proof do I use an element m when I chose n = m - 1? In the proof I chose only elements that *follow* given elements, as is assured by the axiom of infinity. Quantifier dyslexia on (3)? > But you use the non-absence of m when you execute Cantor's > proof. Then you do not admit that for every FISON(n) there is an m = n > + 1 that is not in the proof. This is really incomprehensible. > > > Only > > > potentially infinite sets do not. But you mix up things. You claim the [quoted text clipped - 5 lines] > > Every finite linear set obeys [*]. That is the mathematical reason. Clearly, the only reason is that you want it to be so. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article > <5df917e3-517f-4a38-b28b-3638434966c7@t21g2000yqi.googlegroups.com> WM > <mueckenh@rz.fh-augsburg.de> writes: > > You drop the completeness condition in certain cases but you assume it > > in case of Cantor's proof. That is cheating. > > You again misunderstand the proof completely. There is an assumption that > a complete list is provided and that is proven false. As I understand the Cantor diagonal proof, the only assumption is that whenever one is provided with a list then that list has to omit at least one sequence. I do not think it was, in its original form, an indirect proof as your statement seems to indicate. SignatureVirgil > In article <KKn5F7....@cwi.nl>, "Dik T. Winter" <Dik.Win...@cwi.nl> > wrote: [quoted text clipped - 12 lines] > one sequence. I do not think it was, in its original form, an indirect > proof as your statement seems to indicate. Cantor understood it as a proof by contradiction. "da wir sonst vor dem Widerspruch stehen würden, daß ein Ding E0 sowohl Element von M, wie auch nicht Element von M wäre." But probably you know better. The only thing that is interesting here is that the same holds for the list of all natural numbers. Give me a list of natural numbers and I will show you that it is incomplete. Of course you are not allowed to say: All n in N! That would be a contradiction, because there is no last n and consequently no chance to check whether your list would contain all n in N. (And if you did so, then one could also say: All r in R. The contracdiction would be of same size.) Regards, WM In article <d797f3cd-50a1-4403-8e1f-a04b31d252ff@s21g2000vbb.googlegroups.com>, > > In article <KKn5F7....@cwi.nl>, "Dik T. Winter" <Dik.Win...@cwi.nl> > > wrote: [quoted text clipped - 16 lines] > dem Widerspruch stehen würden, daß ein Ding E0 sowohl Element von M, > wie auch nicht Element von M wäre." But probably you know better. A quote without a source cited is not evidence. There is certainly no necessity for it to be cast as an indirect proof when the direct proof is even simpler. > The only thing that is interesting here is that the same holds for the > list of all natural numbers. Give me a list of natural numbers and I > will show you that it is incomplete. A list in the context of the Cantor diagonal argument means a mapping from the set of all naturals to the members of the listed set, so that the function, f: N --> N: x |--> x is a list of natural numbers. Now WM must shown how and why THAT listing is incomplete (in the sense of missing some member of N either as argument or as value of the function f). > Of course you are not allowed to say: All n in N! All n in N, without the factorial symbol, though, is quite legitimate. > That would be a > contradiction, because there is no last n and consequently no chance > to check whether your list would contain all n in N. Since my listing clearly includes every n in N for which n = n, WM must show that there is some n for which n = n is false to falsify my argument. SignatureVirgil > > > As I understand the Cantor diagonal proof, the only assumption is that > > > whenever one is provided with a list then that list has to omit at least [quoted text clipped - 6 lines] > > A quote without a source cited is not evidence. The quote stems from Cantor's most famous paper. That is very short and you should be able to read it. And here is even the translation from THE LOGIC MUSEUM Copyright © E.D.Buckner 2005 http://uk.geocities.com/frege@btinternet.com/index.htm From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence [Reihenform]: E1, E2, …, Ev, … otherwise we would have the contradiction, that a thing [Ding] E0 would be both an element of M, but also not an element of M. Can you see the word "contradiction" in that text? > > The only thing that is interesting here is that the same holds for the > > list of all natural numbers. Give me a list of natural numbers and I [quoted text clipped - 3 lines] > from the set of all naturals to the members of the listed set, so that > the function, f: N --> N: x |--> x is a list of natural numbers. Then define a list as a mapping from N --> R and everything is fine. > Now WM must shown how and why THAT listing is incomplete (in the sense > of missing some member of N either as argument or as value of the > function f). If you give me a number n then there are n^n^n^...^n numbers not given. Therefore your claim is nonsense. > > Of course you are not allowed to say: All n in N! > > All n in N, without the factorial symbol, though, is quite legitimate. All fractions required for the Oresme proof 1 + 1/2 + (1/3 + 1/4) + ... of the divergence of the series SUM(1/n) make up a number of 2^aleph_0, iff there are aleph_0 parentheses. > > That would be a > > contradiction, because there is no last n and consequently no chance > > to check whether your list would contain all n in N. > > Since my listing clearly includes every n in N for which n = n but not every n^n^n^...^n > WM must > show that there is some n for which n = n is false to falsify my > argument. The same could be claimed for a list of all real numbers. Here it is: r = r. Regards, WM In article <9626e9ff-11ac-481c-960c-ca6e48ff27e4@z5g2000vba.googlegroups.com>, > > > > As I understand the Cantor diagonal proof, the only assumption is that > > > > whenever one is provided with a list then that list has to omit at least [quoted text clipped - 18 lines] > > Can you see the word "contradiction" in that text? WM misunderstands what "proof by contradiction" means. Such a proof requires that one start with a contrary assumption, that the theorem in some way fails to be true and from that assumption derive a contradiction, which is not done here. The proof itself merely assumes ANY list of sequences of m's and w's and shows that there is another which is not included. The proof is quite complete before it is noted that the theorem's falsehood would produce any contradiction, so that contradiction remark is in no way essential to the proof, and the proof is not, in any sense, a proof by contradiction. WM's ignorance of formal logic again trips him up. > > > The only thing that is interesting here is that the same holds for the > > > list of all natural numbers. Give me a list of natural numbers and I [quoted text clipped - 5 lines] > > Then define a list as a mapping from N --> R and everything is fine. Since there is a unique injection f:N -> R in which each natural gets mapped to its real value, while there are reals unlisted there are no natural-valued reals unlisted. The reals have a unique multiplicative identity, which to distinguish it from the 1 in N I shall designate by one. Then the mapping from N to R defined by f)1 = one and f(sucessor(n)) = f(n) + one, where "+" is the real number addition operator, is that injection mentioned above. > > Now WM must shown how and why THAT listing is incomplete (in the sense > > of missing some member of N either as argument or as value of the > > function f). > > If you give me a number n then there are n^n^n^...^n numbers not > given. Therefore your claim is nonsense. To justify calling a claim nonsense, one must be given evidence of its falsity, which WM carefully does not attempt. For what sets does WM argue that the identity mapping from that set to itself is not a bijection? He seems now to be claiming it for N. > > > Of course you are not allowed to say: All n in N! > > [quoted text clipped - 3 lines] > + ... of the divergence of the series SUM(1/n) make up a number of > 2^aleph_0, iff there are aleph_0 parentheses. And the difference between a duck is that one leg is both the same. > > > That would be a > > > contradiction, because there is no last n and consequently no chance [quoted text clipped - 3 lines] > > but not every n^n^n^...^n Then WM is claiming that n^n^n^...^n is not equal to itself? What strange natural numbers exist in WM's world. > > WM must > > show that there is some n for which n = n is false to falsify my > > argument. > > The same could be claimed for a list of all real numbers. Here it is: > r = r. But I presented a function listing (as an image of N) all members of N, whereas surjective images of R do not count as listings until after R has been shown to be a surjective image of N, which Cantors followers have proved impossible. > Regards, WM SignatureVirgil > In article <5df917e3-517f-4a38-b28b-363843496...@t21g2000yqi.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 29 Mai, 03:58, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 19 lines] > > What is the relevance of this? The word union and the word linear are not mentioned in the quote. Nevertheless the due logica rules were obtained from unions of finite sets and linear finite sets. > > The > > logical rules of linear sets are obtained from finite linear sets. [quoted text clipped - 3 lines] > possible, of course, but does not rule out theories in which those things > do exist. But it does rule out theories which are contradicted by the fundamental logical rules. And one of these rules is that a complete linear set has a last element. > > > > The only thing that can be stated is (symbolically): > > > > E n A m P(m, n) -> A m E n P(m, n) [quoted text clipped - 19 lines] > What is contested is that there are case where [***] is true and [*] false. > And it is the latter implication that you do use. No. I use the fact that for complete linear sets always both implications are true : [**] & [***]. This means that [*] is true. If you disagree, then you should come up with a finite linear set for which only one implication is true. > > > > > I said that for the case involved you have to > > > > > *prove* that it is true, because it is not generally true. [quoted text clipped - 10 lines] > still failed to give a definition of "potential infinity" that is valid within > ZF. Moreover, you have not proven (within ZF) that the statement is wrong. ZF uses potential infinity whenever the validity of En Am: m =< n <== Am En m =< n [***] for linear sets is denied. ZF claims that this denial is correct for complete linear infinite sets, but this is a wrong claim, as we can obtain from logic. > > > (1) define FISON(n) be the set of naturals from 1 to n, that is: > > > {1, ..., n}. [quoted text clipped - 15 lines] > Because there is no such n. Remember: the set of natural numbers has no > largest element in ZF. That is the logic of potential infinity, i.e., of incomplete sets. In Cantor’s diagonal argument you can use the same logic : There is no last line, therefore there is always a line beyond the checked lines. But there you don’t. > > > It clearly shows that > > > E n A m P(m, n) <- A m En P(m, n) [quoted text clipped - 13 lines] > that classical logic was *derived* from the logic on finite sets, not that > it was *identical* to logic on finite sets. When it is derived from logic of finite sets, then it is not the reverse of the logic of finite sets. But that is claimed in ZF. > > > > I did never claim that quantifier exchange is allowed in case of non- > > > > linear sets, like cyclic sets as, for instance, your dice. That would [quoted text clipped - 10 lines] > Not "a complete linear set". But "a complete finite linear set". Why do > you drop the word "finite" in the second sentence? To obfuscate? The logic is derived from finite complete linear sets. If logic is to be applied to infinite complete linear sets, then we cannot change it to the opposite. Either those sets obey that logic or they do not exist in a science that is subject to the application of logic. In fact thoses sets contradict their own existence under the government of logic. > > You drop the completeness condition in certain cases but you assume it > > in case of Cantor's proof. That is cheating. > > You again misunderstand the proof completely. There is an assumption that > a complete list is provided and that is proven false. Small wonder. There cannot be a complete list, because the existence of a complete infinite linear set like N contradicts logic. > > > > > > > What is contested is that: > > > > > > > En Am: m =< n <== Am En m =< n [***] [quoted text clipped - 14 lines] > What in the world is a "potentially infinite element"? And, in ZF all sets > are static, there is no place for non-static sets. That is a blatant lie. Every static linear set has a last element. If you say you cannot choose the last elemement because there is none, then you apply potentially infinite sets. Then for every element that you choose there is a larger one. If it has been there all time, why the hell did you not start with this one ? > > > (1) define FISON(n) be the set of naturals from 1 to n, that is: > > > {1, ..., n}. [quoted text clipped - 12 lines] > I chose only elements that *follow* given elements, as is assured by the > axiom of infinity. That is assured by the axiom of potential infinity. If all elemeents were there, why then do you have to select one that is diminishingly small compared to most ? > > But you use the non-absence of m when you execute Cantor's > > proof. Then you do not admit that for every FISON(n) there is an m = n > > + 1 that is not in the proof. > > This is really incomprehensible. So it is, but you do it nevertheless. > > > > Only > > > > potentially infinite sets do not. But you mix up things. You claim the [quoted text clipped - 7 lines] > > Clearly, the only reason is that you want it to be so. Show me a finite linear set that does not obey [*]. Or try to explain why in your opinion this law must be reversed for infinite sets which are actually existing, i.e., every elemenmt of them can be chosen for comparison with others. Regards, WM In article <324b4dea-62b3-4a92-b1db-ccc0034649e7@j12g2000vbl.googlegroups.com>, > > In article > > <5df917e3-517f-4a38-b28b-363843496...@t21g2000yqi.googlegroups.com> WM [quoted text clipped - 28 lines] > Nevertheless the due logica rules were obtained from unions of finite > sets and linear finite sets. While they may be suggested by and be compatible with unions of finite sets, definitions of union of a set of sets in any set theory (excluding WM's) do not restricts sets, or their unions, to being finite sets. > > > The > > > logical rules of linear sets are obtained from finite linear sets. [quoted text clipped - 7 lines] > But it does rule out theories which are contradicted by the > fundamental logical rules. WM's notion of what contradicts "fundamental logical rules" conflicts with logic's notion of what contradicts logical rules. > And one of these rules is that a complete > linear set has a last element. A finite ordered set need not have a last element, so WM os wrong again. > No. I use the fact that for complete linear sets always both > implications are true : But unless "completeness" requires "finiteness" it is false, and nothing outside of WM's perverted vision of set theories requires finiteness. > If you disagree, then you should come up with a finite linear set for > which only one implication is true. (En in S) (Am in S) : m =< n <==> Am En m =< n > > > > > > I said that for the case involved you have to > > > > > > *prove* that it is true, because it is not generally true. [quoted text clipped - 43 lines] > > That is the logic of potential infinity, i.e., of incomplete sets. There is no logic of potentially infinite sets, as they are self contradictory. > In Cantor¹s diagonal argument you can use the same logic : There is no > last line, therefore there is always a line beyond the checked lines. > But there you don¹t. Then WM does not understand that argument, since Cantor does not require any constraints on a list other than that its members be binary sequences, including whether or not there is a last "line". > > Oh. I think that the term "classical logic" has changed a bit since the > > last time you looked at it. And, if you refer to Weyl's quote, he stated [quoted text clipped - 3 lines] > When it is derived from logic of finite sets, then it is not the > reverse of the logic of finite sets. But that is claimed in ZF. That may be how WM misinterpretes ZF, but no one else can be forced to accept WM's misinterpretation, and no one else does. > > > > > I did never claim that quantifier exchange is allowed in case of > > > > > non- [quoted text clipped - 16 lines] > > The logic is derived from finite complete linear sets. "Derived from" does not mean "identical to", and the set theory of ZF, among others, is compatible with those set theories of finite sets which do not impose finiteness. > If logic is to > be applied to infinite complete linear sets, then we cannot change it [quoted text clipped - 3 lines] > In fact thoses sets contradict their own existence under the > government of logic. That "government" does not rule anywhere except in WM's weird world of MathuUnrealism. If WM's arguments could be proved, he would long since have gathered supporters, but he has not. > > > You drop the completeness condition in certain cases but you assume it > > > in case of Cantor's proof. That is cheating. [quoted text clipped - 4 lines] > Small wonder. There cannot be a complete list, because the existence > of a complete infinite linear set like N contradicts logic. Actually, the Cantor argument does not assume or require a complete list, it allows ANY list, all of whose members are binary sequences and then shows that that list must be incomplete in the sense of omitting at least one potential member. So WM does not even understand the very argument he is vainly trying to oppose. > > What in the world is a "potentially infinite element"? And, in ZF all sets > > are static, there is no place for non-static sets. > " > That is a blatant lie. Every static linear set has a last element. Not necessarily in ZF. From what axiom of combination of axioms does WM claim to derive his "Every static linear set has a last element"? Unless he can do so, he lies. > If you say you cannot choose the last elemement because there is none, > then you apply potentially infinite sets. Nonsense! There is no provision for potentially infinite sets in ZF. If WM disputes this, he must show which definitions and/or axioms produce this provision. > Then for every element that > you choose there is a larger one. If it has been there all time, why > the hell did you not start with this one ? For which element does WM argue there is no larger one? Unless WM can tell us how to find "this one", we will continue to hold that his his "this one", with no larger one, does not exist. ZF says that for EVERY natural there is a successor natural. > > > > (1) define FISON(n) be the set of naturals from 1 to n, that is: > > > > {1, ..., n}. [quoted text clipped - 16 lines] > > That is assured by the axiom of potential infinity. What does "the axiom of potentail infinity" > say? If all elemeents > were there, why then do you have to select one that is diminishingly > small compared to most ? Because every one of them is "diminishingly small compared to most". WM keeps asking us to to select things which do not exist while ignoring things which do. > Show me a finite linear set that does not obey [*]. Or try to explain > why in your opinion this law must be reversed for infinite sets which > are actually existing, i.e., every elemenmt of them can be chosen for > comparison with others. No one can show anything to anyone as invincibly ignorant as WM. SignatureVirgil ... > > > > You think so, but you have to prove that it is valid for infinite > > > > complete linear sets. Note that "classical logic is obtained from [quoted text clipped - 9 lines] > Nevertheless the due logica rules were obtained from unions of finite > sets and linear finite sets. Yes, obtained from. But that does not mean they are identical. Moreover, does what Weyl wrote a long time ago still have validity now? > > > The > > > logical rules of linear sets are obtained from finite linear sets. [quoted text clipped - 7 lines] > fundamental logical rules. And one of these rules is that a complete > linear set has a last element. What fundamental rules of logic are you using? I have never seen such a logical rule, because logic does not talk about sets. > > > I never said so. But [*] is [**] & [***]. Therefore [*] differs from > > > [**] only by the reverse. [quoted text clipped - 15 lines] > implications are true : > [**] & [***]. This means that [*] is true. You just state so without proof. > If you disagree, then you should come up with a finite linear set for > which only one implication is true. Why should I show that for a *finite* linear set? Why not for an *infinite* linear set? You are *still* thinking that wat is valid for finite things is also valid for infinite things. So in your opinion: sum{i = 0..oo} 1/(i!) = lim{n -> oo} sum{i = 0..n} 1/(i!) is rational, because it is rational for each n, and so your logic requires that. Also: lim{n -> oo} 1/n > 0 because it is > 0 for each n, and so should also be > 0 in the infinite case. Now, what is it. Is it always the case that what is valid for finite things is also valid for infinite things or not? If it is sometimes true and sometimes false, I wish you would provide me with the reasons *why* it should be sometimes true and sometimes false. > > > > It is not true for the infinite set of naturals. > > > [quoted text clipped - 9 lines] > En Am: m =< n <== Am En m =< n [***] > for linear sets is denied. And you still do not answer my question. You fail to give a definition that is valid in ZF, so I have no idea what you are talking about. > ZF claims that this denial is correct for complete linear infinite > sets, but this is a wrong claim, as we can obtain from logic. What logic? Not the logic as discussed in sci.logic. So you must use your own logic and logical rules. > > > > (1) define FISON(n) be the set of naturals from 1 to n, that is: > > > > {1, ..., n}. [quoted text clipped - 8 lines] > > > > Can you provide me with definitions within ZF that shows the difference? No answer to this? > > > In that latter case you need not take an n that > > > is surpassed by m. Why don't you start with an n that has no greater [quoted text clipped - 4 lines] > > That is the logic of potential infinity, i.e., of incomplete sets. Eh? As far as I know logic is *not* about sets. It is about axioms, statements and inference rules. > In Cantor's diagonal argument you can use the same logic : There is no > last line, therefore there is always a line beyond the checked lines. > But there you don't. Because you do not check the lines in order. It is always your basic assumption that you first check the first line and after that the next line. That is wrong. You simply give a definition of a new number so it is clear that it will be different from each of the omega lines, without checking directly any of them. > > Oh. I think that the term "classical logic" has changed a bit since the > > last time you looked at it. And, if you refer to Weyl's quote, he stated [quoted text clipped - 3 lines] > When it is derived from logic of finite sets, then it is not the > reverse of the logic of finite sets. But that is claimed in ZF. It is *not* the reverse of the logic of finite sets. For finite sets it is identical. > > > There are many finite sets with many special properties that follow > > > from classical logic. One of them is that a complete linear set has a [quoted text clipped - 6 lines] > be applied to infinite complete linear sets, then we cannot change it > to the opposite. Ok. The logic of finite complete sums of rational elements gives that the sum is rational. Using your logic we get that the complete (i.e. in the limit) sum of rational elements is also rational. And so by that logic, e, pi and whatever are rational, and all numbers we do use are rational. I may note that in this sense, the limit *is* the sum of the complete set of elements. > Either those sets obey that logic or they do not > exist in a science that is subject to the application of logic. Ah, so 'e', 'pi', 'sqrt(2)' do not exist. Still I think you use at least one of them on occasion. > In fact thoses sets contradict their own existence under the > government of logic. So 'sqrt(2)' contradicts itself under the government of logic. I think you are a few thousand years behind. > > > You drop the completeness condition in certain cases but you assume it > > > in case of Cantor's proof. That is cheating. [quoted text clipped - 4 lines] > Small wonder. There cannot be a complete list, because the existence > of a complete infinite linear set like N contradicts logic. Apparently your logic. What are the rules of inference in your logic? And, also apparently, in your logic the length of the diagonal of a square with sides with size 1 does not exist. > > > State before beginning whether the set that you assume is complete and > > > static, i.e., every element is actually existing, or potentially [quoted text clipped - 4 lines] > > That is a blatant lie. Eh? What is the lie? I just state that in ZF there are no non-static sets. That is all. So you think that is false? If so, prove it, please. BTW, you are getting offensive. > Every static linear set has a last element. As that is false in ZF... > If you say you cannot choose the last elemement because there is none, > then you apply potentially infinite sets. You have still not provided a definition of "potentially infinite sets" in ZF, so I cannot comment on this. > Then for every element that > you choose there is a larger one. If it has been there all time, why > the hell did you not start with this one ? Because in ZF a set does not need to have a last element. So whatever element you chose, there is *always* a larger one. > > > > (1) define FISON(n) be the set of naturals from 1 to n, that is: > > > > {1, ..., n}. [quoted text clipped - 14 lines] > > That is assured by the axiom of potential infinity. What *is* potential infinity in ZF? Nothing in ZF talks about that. > If all elemeents > were there, why then do you have to select one that is diminishingly > small compared to most ? Because there is no largest? You know, the axiom of infinity assures that there is a set that has not a largest element. > > > But you use the non-absence of m when you execute Cantor's > > > proof. Then you do not admit that for every FISON(n) there is an m [quoted text clipped - 3 lines] > > So it is, but you do it nevertheless. I have no idea. What is the "non-absence of m when I execute Cantor's proof"? What do you mean with "there is an m == n + 1 that is not in the proof"? Your statements are simply incomprehensible. > > > > > Only > > > > > potentially infinite sets do not. But you mix up things. You [quoted text clipped - 13 lines] > are actually existing, i.e., every elemenmt of them can be chosen for > comparison with others. See my examples above about 'sqrt(2)'. There is no reversal of the law, what happens is what happens in finite cases is not necessarily valid in infinite cases. You have to prove that what is valid in finite cases is also valid in the infinite case, something you always omit. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > The word union and the word linear are not mentioned in the quote. > > Nevertheless the due logica rules were obtained from unions of finite > > sets and linear finite sets. > > Yes, obtained from. But that does not mean they are identical. Moreover, > does what Weyl wrote a long time ago still have validity now? It is 50 years younger than Canto's writings. > > But it does rule out theories which are contradicted by the > > fundamental logical rules. And one of these rules is that a complete > > linear set has a last element. > > What fundamental rules of logic are you using? I have never seen such a > logical rule, because logic does not talk about sets. Logic states that the union of a *complete* set of finite linear sets is a finite linear set > > No. I use the fact that for complete linear sets always both > > implications are true : > > [**] & [***]. This means that [*] is true. > > You just state so without proof. A proof is a derivation of theorems from axioms or basic truths by means of rules of logical inference. These rules themselves cannot be proven but can only be obtained from the behaviour of existing (i.e., finite) sets. > > If you disagree, then you should come up with a finite linear set for > > which only one implication is true. > > Why should I show that for a *finite* linear set? Why not for an *infinite* > linear set? You are *still* thinking that wat is valid for finite things > is also valid for infinite things. There is no reason to believe that always the contrary is true. > So in your opinion: > sum{i = 0..oo} 1/(i!) = lim{n -> oo} sum{i = 0..n} 1/(i!) > is rational, because it is rational for each n, and so your logic requires > that. Also: > lim{n -> oo} 1/n > 0 > because it is > 0 for each n, and so should also be > 0 in the infinite case. The limit does not belong to the series! Similarly omega is not a natural number and the number of natural numbers is not omega. > > ZF uses potential infinity whenever the validity of > > En Am: m =< n <== Am En m =< n [***] > > for linear sets is denied. > > And you still do not answer my question. You fail to give a definition that > is valid in ZF, so I have no idea what you are talking about. Potential means not complete. There always appears another number once you have found the last one. > > ZF claims that this denial is correct for complete linear infinite > > sets, but this is a wrong claim, as we can obtain from logic. > > What logic? Not the logic as discussed in sci.logic. No, I mean the correct logic. Regards, WM In article <557d4c53-cba7-4fca-aec2-aa62770788ec@l32g2000vba.googlegroups.com>, > > > The word union and the word linear are not mentioned in the quote. > > > Nevertheless the due logica rules were obtained from unions of finite [quoted text clipped - 4 lines] > > It is 50 years younger than Canto's writings. Do you mean "Cantor"? But 50 years is a mere nothing. Consider how long it took to get "Fermat's Last Theorem" settled. > > > But it does rule out theories which are contradicted by the > > > fundamental logical rules. And one of these rules is that a complete [quoted text clipped - 5 lines] > Logic states that the union of a *complete* set of finite linear sets > is a finite linear set Wherever does "logic" state that? Logic does not speak of sets or finiteness or linearity at all of itself, so there is a lot more than mere "fundamental rules of logic" involved. > > > No. I use the fact that for complete linear sets always both > > > implications are true : [quoted text clipped - 6 lines] > proven but can only be obtained from the behaviour of existing (i.e., > finite) sets. But the ones you insist on are not so derivable without the assumption that all sets are finite, for which you have nothing but your own faith to justify. > > > If you disagree, then you should come up with a finite linear set for > > > which only one implication is true. [quoted text clipped - 5 lines] > > There is no reason to believe that always the contrary is true. We are not claiming the contrary ALWAYS, merely sometimes. And if something is sometimes true then it REQUIRES proof to be able to claim tht it not true in a particular case. And WM has not provided that REQUIRED proof of his claim, so we are free to reject it. > > So in your opinion: > > sum{i = 0..oo} 1/(i!) = lim{n -> oo} sum{i = 0..n} 1/(i!) [quoted text clipped - 7 lines] > Similarly omega is not a natural number and the number of natural > numbers is not omega. The ordinal of N is omega, but depending n one's terminology one may regard aleph_0 as either the same as omega as set or different as a type of number. > > > ZF uses potential infinity whenever the validity of > > > En Am: m =< n <== Am En m =< n [***] [quoted text clipped - 6 lines] > Potential means not complete. There always appears another number once > you have found the last one. Then there is never a last one. And the requirement of being merely potential can never be met. > > > ZF claims that this denial is correct for complete linear infinite > > > sets, but this is a wrong claim, as we can obtain from logic. > > > > What logic? Not the logic as discussed in sci.logic. > > No, I mean the correct logic. To which WM has repeatedly shown he has no access. So WM is not competent to speak for "the correct logic". > Regards, WM SignatureVirgil > > > The word union and the word linear are not mentioned in the quote. > > > Nevertheless the due logica rules were obtained from unions of finite [quoted text clipped - 4 lines] > > It is 50 years younger than Canto's writings. What is the relevance? > > > But it does rule out theories which are contradicted by the > > > fundamental logical rules. And one of these rules is that a complete [quoted text clipped - 5 lines] > Logic states that the union of a *complete* set of finite linear sets > is a finite linear set As I said, logic is not talking about sets. So where in logic is such stated? > > > No. I use the fact that for complete linear sets always both > > > implications are true : [quoted text clipped - 6 lines] > proven but can only be obtained from the behaviour of existing (i.e., > finite) sets. So you are not using mathematical logic? > > > If you disagree, then you should come up with a finite linear set for > > > which only one implication is true. [quoted text clipped - 4 lines] > > There is no reason to believe that always the contrary is true. I do not believe so, so what is the relevance? > > So in your opinion: > > sum{i = 0..oo} 1/(i!) = lim{n -> oo} sum{i = 0..n} 1/(i!) [quoted text clipped - 5 lines] > > The limit does not belong to the series! According to your reasoning it does belong to the complete series. > Similarly omega is not a natural number and the number of natural > numbers is not omega. What is in this context the relevance of the second part of the statement? > > > ZF uses potential infinity whenever the validity of > > > En Am: m =< n <== Am En m =< n [***] [quoted text clipped - 5 lines] > Potential means not complete. There always appears another number once > you have found the last one. Your definition of complete is not the standard definition. In N there is always one after the other, but the complete set does exist. > > > ZF claims that this denial is correct for complete linear infinite > > > sets, but this is a wrong claim, as we can obtain from logic. > > > > What logic? Not the logic as discussed in sci.logic. > > No, I mean the correct logic. I still do not see the logic through which you obtain it. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <557d4c53-cba7-4fca-aec2-aa6277078...@l32g2000vba.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 4 Jun., 04:04, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 8 lines] > > What is the relevance? Your questuion: Does what Weyl wrote a long time ago still have validity now? > > > > But it does rule out theories which are contradicted by the > > > > fundamental logical rules. And one of these rules is that a complete [quoted text clipped - 7 lines] > > As I said, logic is not talking about sets. So where in logic is such stated? Logic is obtained from the behaviour of things. Things can be considered as sets, at least if two things are taken together. Bolzano excused himself for including 2. Later they included 1 and even 0, > > > > No. I use the fact that for complete linear sets always both > > > > implications are true : [quoted text clipped - 8 lines] > > So you are not using mathematical logic? Nobody should do so. Many "logicians" are below any level. There is one Fool Of Matheology, for instance, who thinks that the cartesian product of the set of finite alphabets is uncountable. Better stop recommending or even using that nonsense. > > > > If you disagree, then you should come up with a finite linear set for > > > > which only one implication is true. [quoted text clipped - 6 lines] > > I do not believe so, so what is the relevance? You do, because there is no further reason. > > Potential means not complete. There always appears another number once > > you have found the last one. [quoted text clipped - 10 lines] > > I still do not see the logic through which you obtain it. It is obtained from the action and reaction of physical subjects. Regards, WM In article <102f1c16-0ef5-4189-aea6-bcccf7729aed@z19g2000vbz.googlegroups.com>, > > In article > > <557d4c53-cba7-4fca-aec2-aa6277078...@l32g2000vba.googlegroups.com> WM [quoted text clipped - 15 lines] > Your questuion: Does what Weyl wrote a long time ago still have > validity now? My question: Did it ever have validity? On what evidence? The convincing authority in mathematics is not ever merely what someone says without providing supporting evidence. Citing authorities, unless they contain proofs, is not a mathematically valid form of argument. > > > > > But it does rule out theories which are contradicted by the > > > > > fundamental logical rules. And one of these rules is that a [quoted text clipped - 14 lines] > considered as sets, at least if two things are taken together. Bolzano > excused himself for including 2. Later they included 1 and even 0, So there WM is stating some of his axiom. But "Logic is obtained from the behaviour of things." os not a mathematically or logically coherent axiom. > > So you are not using mathematical logic? > > Nobody should do so. > Many "logicians" are below any level. There is > one Fool Of Matheology, for instance, who thinks that the cartesian > product of the set of finite alphabets is uncountable. What sort of "cartesian product of the set of finite alphabets" would that be? If one takes A ={"0","1"} and B as any infinite set, the the Cartesian product A^B, i,e, the set of all functions from B t A, is indeed uncountable. > Better stop > recommending or even using that nonsense. Unless you want to do actual mathematics. > > > > > If you disagree, then you should come up with a finite linear set > > for [quoted text clipped - 9 lines] > > You do, because there is no further reason. WM now adds mind reading to his list of claimed talents? > It is obtained from the action and reaction of physical subjects. Physics in not the source of logic. At least not for anyone but physicists. SignatureVirgil ... > > > > Yes, obtained from. But that does not mean they are identical. > > > > Moreover, does what Weyl wrote a long time ago still have validity [quoted text clipped - 6 lines] > Your questuion: Does what Weyl wrote a long time ago still have > validity now? My question still stands and I see no answer. > > > > > But it does rule out theories which are contradicted by the > > > > > fundamental logical rules. And one of these rules is that a [quoted text clipped - 12 lines] > considered as sets, at least if two things are taken together. Bolzano > excused himself for including 2. Later they included 1 and even 0, That is not an answer to my question. > > > > > No. I use the fact that for complete linear sets always both > > > > > implications are true : [quoted text clipped - 10 lines] > > Nobody should do so. Many "logicians" are below any level. Yes, I know you do not like logic. > There is > one Fool Of Matheology, for instance, who thinks that the cartesian > product of the set of finite alphabets is uncountable. What is "the cartesion product of the set of finite alphabets"? The only definition I know is that "the cartesion product of a set" is "the cartesian square of a set", i.e. the set of pairs (x, y) where both x and y come from the set. Apparently you mean something else. Can you properly state what you *do* mean? > Better stop > recommending or even using that nonsense. So you think I should follow nonsense where many things are not properly defined and only follow intuition? > > > > > If you disagree, then you should come up with a finite linear > > > > > set for which only one implication is true. [quoted text clipped - 8 lines] > > You do, because there is no further reason. No. There are two statements: (1) What is valid for finite things is also valid for infinite things. The contrary of that is: (2) What is not valid for infinite things is also not valid for finite things. But neither is true. But you intend the reverse: (3) What is valid for infinite things is also valid for finite things. Which is also false. So, which of (2) or (3) do I believe according to you? > > > > > ZF claims that this denial is correct for complete linear infinite > > > > > sets, but this is a wrong claim, as we can obtain from logic. [quoted text clipped - 6 lines] > > It is obtained from the action and reaction of physical subjects. What has *that* to do with logical reasoning? Logical reasoning is able to come up with algorithms like APR-CL that decide whether a number is prime or not. What is the relation with "action and reaction of physical subjects"? How does "action and reaction of physical subjects" relate to the logic that constructed algorithms (like NSF) to factorise numbers? What are the "actions and reactions of physival subjects" involved in taking the union of FISONs? Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <102f1c16-0ef5-4189-aea6-bcccf7729...@z19g2000vbz.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 11 Jun., 15:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 11 lines] > > My question still stands and I see no answer. Then take my answer: Yes. > > > > > > But it does rule out theories which are contradicted by the > > > > > > fundamental logical rules. And one of these rules is that a [quoted text clipped - 14 lines] > > That is not an answer to my question. That answer is: Logic is obtained from the behaviour of things. > > > > > > No. I use the fact that for complete linear sets always both > > > > > > implications are true : [quoted text clipped - 12 lines] > > Yes, I know you do not like logic. I like logic, but not nonsense. > > There is > > one Fool Of Matheology, for instance, who thinks that the cartesian [quoted text clipped - 4 lines] > cartesian square of a set", i.e. the set of pairs (x, y) where both x and > y come from the set. Apparently you mean something else. No. I mean exactly that: The set of finite words over a finite alphabet is countable. The set of meanings of these words, i.e., the set of languages, is countable. The set of finite alphabets is countable. The cartesian product of these, and possibly some further features, is countable. > > > I still do not see the logic through which you obtain it. > > [quoted text clipped - 7 lines] > "actions and reactions of physival subjects" involved in taking the union of > FISONs? The logic is obtained from physical objects. How else should it have come into being? Remember, even brains are physical objects. Regards, WM In article <2afbc0fd-c33b-43c1-8a0e-acaf34cc7403@x31g2000prc.googlegroups.com>, > > In article > > <102f1c16-0ef5-4189-aea6-bcccf7729...@z19g2000vbz.googlegroups.com> WM [quoted text clipped - 16 lines] > > Then take my answer: Yes. Then what Euclid wrote about the need for axioms should still be valid as well. > > > > > > > > As I said, logic is not talking about sets. So where in logic is such [quoted text clipped - 7 lines] > > That answer is: Logic is obtained from the behaviour of things. On the contrary, logic is obtained from the behavior of minds. > > > > > > > No. I use the fact that for complete linear sets always both > > > > > > > implications are true : [quoted text clipped - 14 lines] > > I like logic, but not nonsense. On the contrary, you seem enamored by nonsense. Otherwise you would not produce anywhere near as much of it. > > > There is > > > one Fool Of Matheology, for instance, who thinks that the cartesian [quoted text clipped - 10 lines] > countable. The cartesian product of these, and possibly some further > features, is countable. Since one can have a Cartesian product of any set with itself countably many times, if that set contains more than one element, one can inject the set of paths of a maximal infinite binary tree into such a product, proving the product to be uncountable. So it depends on the sort of Cartesian product one considers whether it is uncountable or not. > > > > I still do not see the logic through which you obtain it. > > > > > > It is obtained from the action and reaction of physical subjects. Mathematics is not constrained by physics, except in the minds of those physicists who are not mathematicians. > The logic is obtained from physical objects. Logic is derived from mental objects, not physical ones. > How else should it have > come into being? Remember, even brains are physical objects. Even though embedded in a physical world, minds are not. They are metaphysical. Minds are not constrained to think only of things existing in WM's physics. If they were, there would be no theist religions, no fairy tales, no novels, no art, no music. SignatureVirgil > > In article <102f1c16-0ef5-4189-aea6-bcccf7729...@z19g2000vbz.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: ... > > > > > > Yes, obtained from. But that does not mean they are identical. > > > > > > Moreover, does what Weyl wrote a long time ago still have [quoted text clipped - 10 lines] > > Then take my answer: Yes. Interesting: Q: What is the relevance? A: Yes. You state that what Weyl wrote still has validity, I doubt that. Then you reply that it is 50 years younger than Cantors writing, and I ask what the relevance is. I still have doubts about the validity of what Weyl wrote, and I still do not see the relevance of what he wrote being 50 years younger than Cantors writings. > > > > > Logic states that the union of a *complete* set of finite linear > > > > > sets is a finite linear set [quoted text clipped - 9 lines] > > That answer is: Logic is obtained from the behaviour of things. Still no answer to my question. Where in logic is it stated that the union of a *complete* set of finite linear sets is a finite linear set. > > > > > A proof is a derivation of theorems from axioms or basic truths by > > > > > means of rules of logical inference. These rules themselves cannot [quoted text clipped - 8 lines] > > I like logic, but not nonsense. Well, I see, you think mathematical logic is nonsense. > > > There is > > > one Fool Of Matheology, for instance, who thinks that the cartesian [quoted text clipped - 7 lines] > No. I mean exactly that: The set of finite words over a finite > alphabet is countable. Right. > The set of meanings of these words, i.e., the > set of languages, is countable. Is it? I would state that the set of meanings of each of those words can indeed be countable (I do not know), nothing more. > The set of finite alphabets is > countable. Is it? I would state that a finite alphabet consists of a finite number of disctinct symbols. Now you are actually stating that the number of symbols is countable. > > > > I still do not see the logic through which you obtain it. > > > [quoted text clipped - 10 lines] > The logic is obtained from physical objects. How else should it have > come into being? Remember, even brains are physical objects. Yeah, I know that you have a verr liberal view on what is part of physics and what not. As you once wrote: that we use computers to do cryptography means that we use physical objects to do cryptography, and so it is part of physics... Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > Well, I see, you think mathematical logic is nonsense. To great parts, yes. > > No. I mean exactly that: The set of finite words over a finite > > alphabet is countable. [quoted text clipped - 6 lines] > Is it? I would state that the set of meanings of each of those words can > indeed be countable (I do not know), nothing more. Every meaning of every word is defined by a language. Every language is a finite definition. The number of finite definitions is countable. > > The set of finite alphabets is > > countable. > > Is it? I would state that a finite alphabet consists of a finite number > of disctinct symbols. Now you are actually stating that the number of > symbols is countable. Every symbol is finite and is defined by a finite word. Therefore the number of symbols and the number of finite sets of symbols is counatble. As a result, the cartesian product of the set of finite alphabets and the set of finite words and the set of languages is countable. > > > > > I still do not see the logic through which you obtain it. > > > > [quoted text clipped - 15 lines] > means that we use physical objects to do cryptography, and so it is part > of physics... And that answers your question that I snipped above: Every set of linear sets in physics is finite and has a last element. Therefore logic, that is obtained from physics, requires for finite linear sets: AnEm <==> EmAn.. Regards, WM In article <9861b377-6571-4ca1-850a-2d8dd3473a20@z5g2000vba.googlegroups.com>, > > Well, I see, you think mathematical logic is nonsense. > [quoted text clipped - 12 lines] > > Every meaning of every word is defined by a language. In mathematics, not every word need have a meaning. For example, in set theory, "is a member of" need not have meaning, though it will have a grammar. WM is hung up on his ow particular model of set theory, but his is not the only possible model, and, in general, one can, and possibly should, do set theory without any particular model in mind at all. SignatureVirgil ... > > > No. I mean exactly that: The set of finite words over a finite > > > alphabet is countable. [quoted text clipped - 10 lines] > Every language is a finite definition. > The number of finite definitions is countable. For each word the meaning can indeed be countable. But that does not mean that the set of meanings for all the words in a language is countable. > > > The set of finite alphabets is > > > countable. [quoted text clipped - 6 lines] > number of symbols and the number of finite sets of symbols is > counatble. I did not know that every symbol is defined by a finite word. Can you show where I can find that result? > > > The logic is obtained from physical objects. How else should it have > > > come into being? Remember, even brains are physical objects. [quoted text clipped - 9 lines] > linear sets: > AnEm <==> EmAn.. Yeah, I know that you have a very liberal view on what is part of physics and what not. The mathematicians have a different view. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <9861b377-6571-4ca1-850a-2d8dd3473...@z5g2000vba.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 19 Jun., 16:18, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 16 lines] > For each word the meaning can indeed be countable. But that does not mean > that the set of meanings for all the words in a language is countable. No? How can that be accomplished? > > > > The set of finite alphabets is > > > > countable. [quoted text clipped - 9 lines] > I did not know that every symbol is defined by a finite word. Can you show > where I can find that result? You must scroll down quite a lot. But this margin is too small to note the address. Regards, WM In article <5c70e8d7-67e1-48e7-8dee-ec3d853bd29e@r34g2000vba.googlegroups.com>, > > > Every symbol is finite and is defined by a finite word. Therefore the > > > number of symbols and the number of finite sets of symbols is [quoted text clipped - 4 lines] > > You must scroll down quite a lot. One might argue that each symbol can be defined by a finite paragraph, but where is it written that each symbol can be defined by a single word? Such a "BIBLE" should be made public, not hidden away in secret as WM is doing. SignatureVirgil ... > > > Every meaning of every word is defined by a language. > > > Every language is a finite definition. [quoted text clipped - 4 lines] > > No? How can that be accomplished? I think by some negation of the axiom of choice. > > > Every symbol is finite and is defined by a finite word. Therefore the > > > number of symbols and the number of finite sets of symbols is [quoted text clipped - 5 lines] > You must scroll down quite a lot. But this margin is too small to note > the address. Ok, so you can not show where I can find that result. In that case I am allowed to ignore it. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <5c70e8d7-67e1-48e7-8dee-ec3d853bd...@r34g2000vba.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 22 Jun., 15:11, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 9 lines] > > I think by some negation of the axiom of choice. I did not ask how to "prove" that. I asked how to *do* it. Regards, WM > > In article <5c70e8d7-67e1-48e7-8dee-ec3d853bd...@r34g2000vba.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > > On 22 Jun., 15:11, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 13 lines] > > Regards, WM We know every language contains words, or symbolic means or presentation. Factor this into the equationa nd begin to produce a creative execution. Here is a clever take: Languages and Finite Automata is finite. Proof: We will reduce the halting problem. to this problem ... construct. YES. NO. NO. YES. halting problem decider. finite language. problem ... The language is contained and finite if to every input there is only Yes or No. The system then only ask the right question to achieve the right answer in the language. http://MeAmI.org In article <0e0f648e-4ed3-4435-92ca-72143e174731@3g2000yqk.googlegroups.com>, > > In article > > <5c70e8d7-67e1-48e7-8dee-ec3d853bd...@r34g2000vba.googlegroups.com> WM [quoted text clipped - 14 lines] > > I did not ask how to "prove" that. I asked how to *do* it. Since WM cannot "do" any proofs of his own claims, he is hardly in a position to demand 'doings" of others. SignatureVirgil > > In article <5c70e8d7-67e1-48e7-8dee-ec3d853bd...@r34g2000vba.googlegroups= > .com> WM <mueck...@rz.fh-augsburg.de> writes: [quoted text clipped - 13 lines] > > I did not ask how to "prove" that. I asked how to *do* it. Apparently you do not know how mathematical proof works, especially if the proof includes something that is in itself not provable. A: given a line and a point there is through that point exactly one line that does not meet the given line. B: eh? I do not think so, there may be more than one line. A: no? how can that be accomplished? B: I think by some negation of the parallel axiom. A: I did not ask how to *prove* that. I asked how to *do* it. B: by some negation of the parallel axiom. My answer to your question above is similar. The question whether a countable union of countably many sets is countable is implied by the axiom of countable choice, which is not provable from ZF. So negating the axiom of countable choice (which implies negating the axiom of choice) we can have situations where the conclusion is false. So it is up to *you* to prove that something is the case "the countable union of countable sets is countable", when I only state that I do not know that. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > My answer to your question above is similar. The question whether a countable > union of countably many sets is countable is implied by the axiom of countable > choice, which is not provable from ZF. So negating the axiom of countable > choice (which implies negating the axiom of choice) we can have situations > where the conclusion is false. Feferman and Levy showed that the statement that the set of all real numbers is the union of a denumerable set of denumerable sets cannot be refuted. So does it depend on the chosen axioms, whether R is countable? Regards, WM > > My answer to your question above is similar. The question whether a countable > > union of countably many sets is countable is implied by the axiom of countable [quoted text clipped - 9 lines] > > Regards, WM No, countable choice is a theorem of ZF. It is attributed to Hausdorff "countable unions of countable sets are not necessarily dissimilar from uncountable sets". However, to him is attributed a more direct statement. R can always be considered countable in for example some constructible universe containing the real numbers via Skolemization. That's where it's a countable model, even if R is still uncountable. The model is supposed to be some big ordinal, bigger than any it models, but there is no such thing in regular set theories. Of course, the universe or domain of discourse is generally assumed in naive set theories, and as well less naive set theories. Feferman considers a lot whether the universe is the constructible universe. Levy I think of more in terms of acknowledgment of Skolem in Levy collapse, in a constructivist way. Thanks, Ross F. On 2 Jul., 09:06, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > My answer to your question above is similar. The question whether a countable > > > union of countably many sets is countable is implied by the axiom of countable [quoted text clipped - 7 lines] > > > So does it depend on the chosen axioms, whether R is countable? > No, countable choice is a theorem of ZF. It is attributed to > Hausdorff "countable unions of countable sets are not necessarily [quoted text clipped - 6 lines] > supposed to be some big ordinal, bigger than any it models, but there > is no such thing in regular set theories. There is no model of ZFC. My question is: Is the set of real numbers uncountable in ZF, is it uncountale in ZFC. Is a countable union of contable sets uncountable in ZF, is it uncountable in ZFC? That boils down to the question: Does Cantor's proof show the same as Feferman and Levy showed? Or does the result depend on the chosen axioms? Regards, WM > There is no model of ZFC. You've never proven that. > My question is: Is the set of real numbers > uncountable in ZF, Yes. It's uncountable even in just Z. > is it uncountale in ZFC. Yes. > Is a countable union of > contable sets uncountable in ZF, ZF does not prove that there exists an uncountable set that is a countable union of countable sets. And ZF does not prove that there does not exist an uncountable set that is a countable union of countable sets. > is it uncountable in ZFC? No. ZFC proves that there is no uncountable set that is a countable union of countable sets. > That boils down to the question: Does Cantor's proof show the same as > Feferman and Levy showed? They're proofs of two different theorems. > Or does the result depend on the chosen > axioms? Any formal proof depends on what axioms are used (Cantor's own writeup of his proof was not formal, but we can formalize it). MoeBlee On Jul 2, 12:06 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > countable choice is a theorem of ZF. No, it's not. MoeBlee > On Jul 2, 12:06 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: [quoted text clipped - 4 lines] > > MoeBlee The definition of countable is there existing a bijective function between some set and the natural integers or here their ordinals. A choice function is representable by any such function bijecting an ordinal as a set to a set. Then, where countable choice means there are choice functions for countable sets, otherwise there are "countable" sets with no provable bijection to the naturals, where if that's undecideable then so is whether they're countable, not disproving the definition. Thanks, Ross On Jul 2, 4:12 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > On Jul 2, 12:06 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> > > wrote: > > > > countable choice is a theorem of ZF. > > > No, it's not. > The definition of countable is there existing a bijective function > between some set and the natural integers or here their ordinals. A definition of countable is ['w' stands for the set of natural numbers]: x is countable iff (there exists a bijection from x onto a natural number of there exists a bijection from x onto w) > A > choice function is representable by any such function bijecting an > ordinal as a set to a set. I have no idea what you're trying to say. ['P' stands for the power set operation:] A choice function for x is a function f whose domain is Px\{0} and such that for all y in Px\{0}, we have f(y) in y. I guess sometimes people also refer to a choice function for x as being a function whose domain is x\{0} and such that for all y in x \{0} we have f(y) in y. Context should make clear which sense is intended. > Then, where countable choice means there > are choice functions for countable sets, The axiom of countable choice is: For all x such that 0 is not in x, there exists a function f whose domain is x and such that for all y in x we have f(y) in y. Since finite sets are not at issue anyway, we can narrow: For all x such that x is denumerable, there exists a function f whose domain is x\{0} and such that for all y in x\{0} we have f(y) in y. > otherwise there are > "countable" sets with no provable bijection to the naturals, You don't know what you're talking about. We don't require any choice principles just to prove that there exists a bijection between a countable set and either a natural number or the set of natural numbers. The result follows simply by the DEFINITION of 'countable'. > where if > that's undecideable then so is whether they're countable, not > disproving the definition. Get help. Start with a book on working in the first order predicate calculus. MoeBlee > On Jul 2, 4:12 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: [quoted text clipped - 60 lines] > > MoeBlee Here a choice function is a bijective function from an ordinal (which is a collection of all lesser ordinals) to a set, or vice versa. Then for set X and ordinal O, f(x) is a well-ordering. I see that I am not using the same definition. I was presuming that a well-ordering was a choice function, which it is, not the existence of them in non-well- orderable set theories, like ZF. So, my mistake, thanks. Still, in terms of countable choice, if a set if countable but its elements are uncountable sets, in the process of building up the structures to represent that, various support structures then aren't countable. So, there are ways to distinguish between countable sets with recursively countable elements and those otherwise, in terms of ready choice functions that are easily constructed. Maybe it is something about the definition along the lines of "the choice function selects an element from each subset of a set". For something like the naturals then a choice function is always surjective, never injective, in regular set theories that would always be so. Then point here is that it is known that for there to be as well that any subset has an element selected (in terms of a course-of- passage in induction over the elements generally however they are), any choice function f is a surjection f:P(X)->X. Consider then the cartesian product P(x) x X, or P(N) x N. There is an indicator value for each p in P(N) for only one value n in N. Now, these functions aren't "choice functions" but do select each element according to at least one subset of the set. The point here is that usual operations build up constrctively from the countable see no end of selection methods. Where Feferman/Levy reiterates that a countable union of countable sets may not be, that would be a paradox with the same objects in regular set theories. Thanks, Ross On Jul 2, 6:00 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > Here a choice function is a bijective function from an ordinal (which > is a collection of all lesser ordinals) to a set, or vice versa. What does "here" refer to in your sentence? Choice functions are not necessarily bijections, nor do they necessarily involve ordinals, so I don't know what you're talking about. MoeBlee > On Jul 2, 6:00 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: [quoted text clipped - 7 lines] > > MoeBlee That's a well-ordering. (Also in a theory where the well-ordering principle is axiomatized or a theorem it's an equivalent definition under minimization of terms.) I think the definition of choice function includes that the range contains each element, so it's a surjection onto the set. Thanks, Ross On Jul 6, 10:32 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > On Jul 2, 6:00 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> > > wrote: [quoted text clipped - 5 lines] > > necessarily bijections, nor do they necessarily involve ordinals, so I > > don't know what you're talking about. > That's a well-ordering. WHAT is a well ordering? The "here" in your previous comments refers to well ordering? > (Also in a theory where the well-ordering > principle is axiomatized or a theorem it's an equivalent definition > under minimization of terms.) What is equivalent to what? What are you talking about? > I think the definition of choice > function includes that the range contains each element, What "each" element? If you mean that the definition of 'choice function on x' entails that every member of x is in the range of the choice function, then you're mistaken. > so it's a > surjection onto the set. Every function is a surjection onto its range. So what? You're a pit of wasted time. MoeBlee > On Jul 6, 10:32 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: [quoted text clipped - 36 lines] > > MoeBlee So, not every function is a surjection onto its co-image or range. That gets into definitions of "range", in classical, modern, standard, and as well modern nonstandard definitions of function. That's "roaring maw" thank you. Ross On Jul 6, 12:11 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > So, not every function is a surjection onto its co-image or range. In ordinary set theories, with the ordinary definition of 'function' as a set of ordered pairs that is single rooted, and 'co-image' as 'superset of the range', a function does not have a unique co-image, but has many co-images, but a function does have a unique range. Then evey function is a surjection onto its range; but not a surjection onto any of its co-images except the range itself. (Of course, that does not apply where, alternatively, a function is understood not just as a set of ordered pairs that is single rooted, but rather as "an operation, a domain, and a range" all together.) > That gets into definitions of "range", in classical, modern, standard, > and as well modern nonstandard definitions of function. The definition of 'the range of a set' is pretty ordinary: range(x) = {y | Ez <z y> in x}. Some authors give ranges only to relations and some, maybe, only to functions. But, as you see above, we can define 'the range' for any x whatsoever, as is common to do. MoeBlee In article <9ab23ac1-d1b9-4bb0-9946-3981b1d0b486@n16g2000yqm.googlegroups.com>, > > My answer to your question above is similar. The question whether a > > countable [quoted text clipped - 7 lines] > numbers is the union of a denumerable set of denumerable sets cannot > be refuted. Based on what set of axioms? > So does it depend on the chosen axioms, whether R is countable? Everything in formal Math depends on sets of axioms. SignatureVirgil > Apparently you do not know how mathematical proof works ... REALLY?! It must be very satisfying to discuss the validity of a alleged mathematical "proof" with a person which does not know how mathematical proof works... Man, are you sure that your behavior concerning WM is sane? (Hint: I don't think so.) Herb > > Apparently you do not know how mathematical proof works ... > [quoted text clipped - 4 lines] > Man, are you sure that your behavior concerning WM is sane? (Hint: I don't > think so.) Are you sure your behaviour concerning me is sane? Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > Dik T. Winter wrote @WM: > > [quoted text clipped - 8 lines] > > Are you sure your behaviour concerning me is sane? Are you sure you're not WM's sock puppet? Is there some other reason for keeping this spavined donkey on life support and littering up the archives all to hell? Sheesh, enough already. You're insane. Get help. -- hz ... > > > Man, are you sure that your behavior concerning WM is sane? (Hint: I don't > > > think so.) [quoted text clipped - 7 lines] > > Sheesh, enough already. You're insane. Get help. Excellent advice, you should follow it. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > "Dik T. Winter" wrote: > ... [quoted text clipped - 11 lines] > > Excellent advice, you should follow it. I will if you will. -- hz > I will if you will. Can't see any reason for that (on your side). Herb > > I will if you will. > > Can't see any reason for that (on your side). I'm willing to take one for the team. -- hz > > > Sheesh, enough already. You're insane. Get help. > > > > Excellent advice, you should follow it. > > I will if you will. Why do you think I'm insane? SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon man nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus Aatu Koskensilta says... >> > > Sheesh, enough already. You're insane. Get help. >> > [quoted text clipped - 3 lines] > >Why do you think I'm insane? I think he was speaking to Dik Winter. But to answer your question: Of course crackpots are insane. Those who argue with crackpots are insane. Those who chastise others for arguing with crackpots are insane. Those who have discussions with those who chastise others for arguing with crackpots are insane. There are a tiny number of non-insane posters, but if I mention them by name, they may become infected, as well. -- Daryl McCullough Ithaca, NY Daryl McCullough wrote: > Aatu Koskensilta says... > >> > [quoted text clipped - 7 lines] > > I think he was speaking to Dik Winter. Heh, now I get it -- clearly Aatu is suffering from paranoia. > But to answer your question: Of course crackpots are insane. > Those who argue with crackpots are insane. [quoted text clipped - 4 lines] > There are a tiny number of non-insane posters, but if I mention > them by name, they may become infected, as well. That's insane. -- hz > > Aatu Koskensilta says... > > >> > herb...@cox.net writes: [quoted text clipped - 24 lines] > -- > hz Stupid people are sane. Intelligent people, as well, the most intelligent who get to define it: psychiatrists and psychologists: sane. They are so defined in their possession of medical degrees. My field is as to them. Angry! Ross "Ross A. Finlayson" wrote: > > > Aatu Koskensilta says... > > > >> > herb...@cox.net writes: [quoted text clipped - 23 lines] > > Stupid people are sane. Aw, c'mon Ross -- that was just my feeble attempt at humor. -- hz > > > > Aatu Koskensilta says... > > > > >> > herb...@cox.net writes: [quoted text clipped - 28 lines] > -- > hz My dear sir, At least I can say it was never my attempt. Indeed, I really don't care how crazy people who read sci.math or sci.logic are. So, there are sane people all over the case, and lots of totally f.cking crazy people maintaining a veneer of sanity, and lots of people who read sci.logic are doctors in the science of mathematical logic, to an extent that there are those people, in the world. So, if you aren't here to buck up or rationally tear down the modern foundations of mathematics, then, I guess you are. Angry! Ross > Heh, now I get it -- clearly Aatu is suffering from paranoia. Even worse! I think he's an alien from outer space and part of a secret plan to conquer all earthlings!!! Herb Herbert Newman wrote: > > Heh, now I get it -- clearly Aatu is suffering from paranoia. > > > Even worse! I think he's an alien from outer space and part of a secret > plan to conquer all earthlings!!! Oh, that's old news. -- hz Dear herbzet! You wrote: > Sheesh, enough already. Agree! > You're insane. Get help. Well, that was a little bit harsh, imho. Imho it's (or might be) a case of so called co-dependence, which is NOT a form of insanity, though seeking (some form of) "help" might still be in order. Herb P.S. Are you serious? >> Man, are you sure that your behavior concerning WM is sane? (Hint: I don't >> think so.) >> > Are you sure your behaviour concerning me is sane? I don't know (and, actually, I don't care). Moreover, would this CHANGE anything? I mean, would the fact (if so) that MY behavior (concerning your person) is not sane make YOUR behavior concerning WM sane? :-o Finally, completely agree with herbzet here; one might ask: "Is there some [rational] reason for keeping this spavined donkey on life support and littering up the archives all to hell?" Imho, you are doing no good to sci.logic with "keeping this spavined donkey on life support". :-) Herb Am Thu, 2 Jul 2009 18:19:40 +0200 schrieb Herbert Newman: > [...] completely agree with herbzet here; one might ask: > [quoted text clipped - 3 lines] > Imho, you are doing no good to sci.logic with "keeping this spavined donkey > on life support". :-) BTW, if you are *so* interested in a "discussion" with WM, why don't you just use e-mail? I'm sure that most participants in this NG (with the exception of Virgil, I guess) would be glad about that. Herb > BTW, if you are *so* interested in a "discussion" with WM, why don't you > just use e-mail? I'm sure that most participants in this NG (with the > exception of Virgil, I guess) would be glad about that. I don't know about most people in this newsgroup, but your suggestion as to email definitely does not speak for me. MoeBlee Am Thu, 2 Jul 2009 12:06:34 -0700 (PDT) schrieb MoeBlee: >> BTW, if you are *so* interested in a "discussion" with WM, why don't you >> just use e-mail? I'm sure that most participants in this NG (with the >> exception of Virgil, I guess) would be glad about that. > > I don't know about most people in this newsgroup, but your suggestion > as to email definitely does not speak for me. Really? So you are interested in this rather fascinating "discussion" between WM and DTW? Is this right? You would actually MISS it, right? I guess the reason would be that there is SO MUCH (mathematical and logical) information contained in the contributions of those two guys (in the context of their "discussion" here), right? I see, this would really be A BIG LOSS, right!!! Herb P.S. Maybe you (and Virgil + DTW) could set up a WM mailing list? :-o > Am Thu, 2 Jul 2009 12:06:34 -0700 (PDT) schrieb MoeBlee: > [quoted text clipped - 11 lines] > context of their "discussion" here), right? I see, this would really be A > BIG LOSS, right!!! Whatever my thoughts about exchanges between WM and Dik T. Winter, it is not my suggestion to them that they refrain from posting therir exchanges. The purpose of my response to you was just to state for the record that my own preference is not covered by your suggestion. MoeBlee > Whatever my thoughts about exchanges between WM and Dik T. Winter, it > is not my suggestion to them that they refrain from posting their > exchanges. The purpose of my response to you was just to state for the > record that my own preference is not covered by your suggestion. I see. Right, imho it would be much better for Dik T. Winter to completely refrain from those "discussions" with WM - after all time is precious, no? Herbert > > Whatever my thoughts about exchanges between WM and Dik T. Winter, it > > is not my suggestion to them that they refrain from posting their [quoted text clipped - 3 lines] > I see. Right, imho it would be much better for Dik T. Winter to completely > refrain from those "discussions" with WM - after all time is precious, no? And you're welcome to your opinions. MoeBlee MoeBlee wrote: > > BTW, if you are *so* interested in a "discussion" with WM, why don't you > > just use e-mail? I'm sure that most participants in this NG (with the > > exception of Virgil, I guess) would be glad about that. > > I don't know about most people in this newsgroup, but your suggestion > as to email definitely does not speak for me. Really? You see some redeeming value to this ceaseless noise? -- hz > > > BTW, if you are *so* interested in a "discussion" with WM, why don't you > > > just use e-mail? I'm sure that most participants in this NG (with the [quoted text clipped - 7 lines] > -- > hz You know, I'm angry. Probably not at you, Herb, yet angry. Ross "Ross A. Finlayson" wrote: > > > > BTW, if you are *so* interested in a "discussion" with WM, why don't you > > > > just use e-mail? I'm sure that most participants in this NG (with the [quoted text clipped - 9 lines] > > You know, I'm angry. Probably not at you, Herb, yet angry. This is most regrettable. FWIW, I don't see you as being in the same category as WM. -- hz > > > > > BTW, if you are *so* interested in a "discussion" with WM, why don't you > > > > > just use e-mail? I'm sure that most participants in this NG (with the [quoted text clipped - 15 lines] > -- > hz Don't piss me off! I'm sitting here ready to quote Stephen Hawking to make MY point. Page after page after page, Hawking's description of the end of classical relativity and beginning of quantum gravity sees the passage from the standard through the strings through the polydimensional, wherein which lies trivial route solutions. Thanks, Ross > > > BTW, if you are *so* interested in a "discussion" with WM, why don't you > > > just use e-mail? I'm sure that most participants in this NG (with the [quoted text clipped - 4 lines] > > Really? You see some redeeming value to this ceaseless noise? Whether I see "redeeming value" in it or not, it is not, Newman's suggestion to Winter does not speak for me. MoeBlee >>>> ...if you are *so* interested in a "discussion" with WM, why don't you >>>> just use e-mail? I'm sure that most participants in this NG (with the [quoted text clipped - 7 lines] > Whether I see "redeeming value" in it or not, it is not, Newman's > suggestion to Winter does not speak for me. [Moe] So YOU would not be glad if Winter would stop "keeping this spavined donkey on life support" [in this NG]? That's interesting. Do you see ANY value in "keeping this spavined donkey on life support" [in this NG]? Herb > >>>> ...if you are *so* interested in a "discussion" with WM, why don't you > >>>> just use e-mail? I'm sure that most participants in this NG (with the [quoted text clipped - 10 lines] > So YOU would not be glad if Winter would stop "keeping this spavined donkey > on life support" [in this NG]? I've not said what would make me glad or not; and that subject is not of very much interest to me for me to discuss at this time. I am merely pointing out that your suggestion does not stand for me. > That's interesting. Do you see ANY value in "keeping this spavined donkey > on life support" [in this NG]? I'm not interested in commenting on that, at least not at this time. My point, which I've now repeated about three times is merely that your suggestion as to the matter does not stand for me. I don't think it will do you much good to ask me more about my feelings and judgments as to the value of the conversations with WM. That particular topic is not of much interest for me to discuss at this time. Rather, what I've already said stands: Whatever my evaluation of the conversations with WM, your suggestion about them doesn't stand for me. That's really all there is that I need to say. MoeBlee Look Moe, in my uttering >>>>>> ...if you are *so* interested in a "discussion" with WM, why don't you >>>>>> just use e-mail? I'm sure that most participants in this NG (with the >>>>>> exception of Virgil, I guess) would be glad about that. [Newman] I expressed (1.) a wish (something at least *I* would be glad about, if it happened), (2.) a believe (namely, that most participants in this NG ... would be glad about that move). Now you commented that uttering the following way: >>>>> I don't know about most people in this newsgroup, but your suggestion >>>>> as to email definitely does not speak for me. [Moe] Well, concerning my personal wish I didn't want to speak for anyone else than just me (especially not for you). It's not clear to me, why you feel a need to comment on that. So -contrary to me- you are interested in "keeping this spavined donkey on life support" in this NG, well fine. But then, why are you not engaged in this idiotic undertaking yourself? *) Herb *) Just a thought: maybe because it's idiotic? > Look Moe, > [quoted text clipped - 16 lines] > than just me (especially not for you). It's not clear to me, why you feel a > need to comment on that. You said what you think most posters feel about the matter. My comment is to put on record that I am not among that set of posters, whether it is a majority or not. I wish to put on record that I am not to be considered (whether anyone does or would consider such) as counted as one who would be glad about the event you wish for (whether I would be glad or not, I simply am putting myself on record that I do not wish to be counted as being glad if your particular wish were realized) and further that I do not wish to be considered as wishing for the event you wish for. > So -contrary to me- you are interested in "keeping this spavined donkey on > life support" in this NG, well fine. No, I just made clear that I have not commented on that matter itself. Rather, I wish to put on record that I am not to be considered as desiring the outcome you seek. Whatever my feelings are about that outcome you seek and whatever my feelings about threads with WM prominent, I simply do not wish to be counted as supporting your desired outcome (whether I even might or might not actually wish it myself). This should not require further discussion. Simply put: No matter what my thoughts and feelings about various things, I wish it to be on record that I am not to be counted as endorsing (or even necessarily OPPOSING, for that matter!) your wished for outcome. > But then, why are you not engaged in > this idiotic undertaking yourself? *) What idiotic undertaking? Posting about my response to your post about your wish? Really, I don't see what you need to pursue here. If I haven't made myself quite clear finally, even at the price of tedium, then I don't know how I can make my point any more clear. MoeBlee ... > Finally, completely agree with herbzet here; one might ask: > [quoted text clipped - 3 lines] > Imho, you are doing no good to sci.logic with "keeping this spavined donkey > on life support". :-) Well, I do not read sci.logic, but sci.math. That WM cross-posts to sci.logic is not my fault. Second, it bothers me that somebody who teaches mathematics and puts part of his nonsense in the exams to his students (the "explanation" of the binary tree), bothers me quite a lot. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ >> "Is there some [rational] reason for keeping this spavined donkey >> on life support and littering up the archives all to hell?" [quoted text clipped - 4 lines] > Well, I do not read sci.logic, but sci.math. That WM cross-posts to > sci.logic is not my fault. I see. (How about restricting your replys to sci.math?) > Second, it bothers me that somebody who teaches mathematics and puts part > of his nonsense in the exams to his students (the "explanation" of the > binary tree), bothers me quite a lot. Sure, same with me. BUT _imho_ your continued attempts to "correct" WM are counterproductive. Actually, this way WM can (rightly!) claim that his "controversial views" _are discussed_ by mathematicians, i.e in sci.math. With other words, your engagement ADDS to his "reputation"! (Note that!) Herb ... > > Well, I do not read sci.logic, but sci.math. That WM cross-posts to > > sci.logic is not my fault. > > I see. (How about restricting your replys to sci.math?) How about not reading threads started by WM? Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ >>> Well, I do not read sci.logic, but sci.math. That WM cross-posts to >>> sci.logic is not my fault. >> >> I see. (How about restricting your replys to sci.math?) >> > How about not reading threads started by WM? Would this help to keep sci.logic clean? Hey, Dik, maybe you should upgrade you logic module. ;-) C'mon, don't be an a.shole, Dik! If you are writing (just) in sci.math restrict your replies to WM to sci.math, PLEASE. This would help preventing (further) pollution of sci.logic. Herb > >>> Well, I do not read sci.logic, but sci.math. That WM cross-posts to > >>> sci.logic is not my fault. [quoted text clipped - 7 lines] > > C'mon, don't be an a.shole, Dik! I was gonna tell him to stop being such a Dik. > If you are writing (just) in sci.math > restrict your replies to WM to sci.math, PLEASE. This would help preventing > (further) pollution of sci.logic. -- hz > > >>> Well, I do not read sci.logic, but sci.math. That WM cross-posts to > > >>> sci.logic is not my fault. [quoted text clipped - 16 lines] > -- > hz Angry! Ross > > I was gonna tell him to stop being such a Dik. > Angry! Please note that I restricted my last bunch of posts to sci.logic. He probably never saw them. Nevertheless, I feel chastened -- HTH. -- hz >> C'mon, don't be an a.shole, Dik! >> > I was gonna tell him to stop being such a Dik. Note that we are not talking about Philip K. Dick. Herb > >> C'mon, don't be an a.shole, Dik! > [quoted text clipped - 3 lines] > > Herb It doesn't take a psychiatrist to know crazy. Crazy is relative. Doesn't take an idiot to know a.shole. Everybody's got one! I recommend Time Storm which within the plot is actually quite great. Similarly, Talisman for pretty much the same reason. We're talking about Dik T. Winter, not a crazy a.shole, nut gallery. What's his point? Of course, most people have read most of the works of the widest read science fiction authors, in obviously a small segment of well-informed people. Consider for example Asimov, Heinlein, Anthony, Dick and all those, Anthony, Heinlein, Asimov, Herbert, etcetera. Dik, what's your point? If it's about WM we can ignore that already. The point of interest is yours. Angry! Ross > I recommend Time Storm which within the plot is actually quite great. > Similarly, Talisman for pretty much the same reason. I'm not familiar with those, but I thought Dick's "Valis" was great. They made an opera out of it: http://en.wikipedia.org/wiki/VALIS#In_popular_culture -- hz > >> C'mon, don't be an a.shole, Dik! > [quoted text clipped - 3 lines] > > Herb It doesn't take a psychiatrist to know crazy. Crazy is relative. Compared to people who read sci.logic, most people are crazy. Doesn't take an idiot to know a.shole. Everybody's got one! Just because you have an opinion doesn't mean it's worth sh.t. I recommend Gordon Dickson's Time Storm which within the plot is actually quite great. Similarly, Talisman for pretty much the same reason. We're talking about Dik T. Winter, not a crazy a.shole, nut gallery. What's his point? Of course, most people have read most of the works of the widest read science fiction authors, in obviously a small segment of well-informed people. Consider for example Asimov, Heinlein, Anthony, Dick and all those, Anthony, Heinlein, Asimov, Herbert, etcetera. Read all that and get back to yourself. Dik, what's your point? If it's about WM we can ignore that already. The point of interest is yours. Angry! Ross > Would this help to keep sci.logic clean? Hey, Dik, maybe you should > upgrade you logic module. ;-) I suggest you upgrade your humour module. It'll be a few bucks well spent. SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon man nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus ... > > How about not reading threads started by WM? > [quoted text clipped - 4 lines] > restrict your replies to WM to sci.math, PLEASE. This would help preventing > (further) pollution of sci.logic. It is common courtesy not to restrict an answer to a selection of the newsgroups only. What could be done is set a follow-up-to to a single newsgroup, but that is defeating your purpose. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ >> C'mon, don't be an a.shole, Dik! If you are writing (just) in sci.math >> restrict your replies to WM to sci.math, PLEASE. This would help preventing >> (further) pollution of sci.logic. > > It is common courtesy not to restrict an answer to a selection of the > newsgroups only. Trying to be an a.shole? Herb > >> C'mon, don't be an a.shole, Dik! If you are writing (just) in sci.math > >> restrict your replies to WM to sci.math, PLEASE. This would help preventing [quoted text clipped - 4 lines] > > Trying to be an a.shole? You are so friendly. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ >>> C'mon, don't be an a.shole, Dik! If you are writing (just) in sci.math >>> restrict your replies to WM to sci.math, PLEASE. This would help preventing [quoted text clipped - 4 lines] > > Trying to be an a.shole? No more than you. Honestly, why should Dik's threads get under your skin? Simply filter them out or don't read them. Signature"This confused and outraged many Matrix fans, who'd already spent hours on the web explaining that man and computers could never really live in such a state of harmony and mutual benefit." -- http://www.pointlesswasteoftime.com > Well, I do not read sci.logic, but sci.math. That WM cross-posts to > sci.logic is not my fault. Second, it bothers me that somebody who > teaches mathematics and puts part of his nonsense Your mathelogical feelings let you assume paths of the binarytree that cannot be identified by nodes neither by sequences of nodes Try to explain that. Better drop it as what it is: nonsense. Regards, WM > > Finally, completely agree with herbzet here; one might ask: > > [quoted text clipped - 9 lines] > students (the "explanation" of the binary tree), bothers me quite a > lot. And what are you doing about it? Trying to convince Mueckenheim of his errors until he retires? You could look here: http://www.stmwfk.bayern.de/Start_en.aspx To wit: "The state of Bavaria, situated in the south of Germany, is proud of its universities offering both students and staff high educational standards..." Ask them how this is compatible with Mueckenheim's teaching activities at one of those universities. Contact data are here: http://www.stmwfk.bayern.de/Contact_en.aspx Ralf > > > ... this spavined donkey ... > > > >.... teaches mathematics and puts part of his nonsense in the exams to his > > students (the "explanation" of the binary tree), ... bothers me quite a > And what are you doing about it? Trying to convince Mueckenheim of his > errors until he retires? You could look here: [quoted text clipped - 5 lines] > one of those universities. Contact data are here: > http://www.stmwfk.bayern.de/Contact_en.aspx I suspect that contacting officialdom will get nowhere at all. Much more effective would be to create a critique web-site (in German of course) such that anyone contemplating going to study or sending their children to study at the Augsburg Whateverschule [capitalised because it's supposed to be a German noun] will immediately find it. It needs to be level-headed and nonabusive, but one of the wonders of the web is the absence of n-pages of A4 limits, so of course it should include a demonstration of the invalidity of some of his published "tree" nonsense. (Of course, I don't know -- perhaps there is already, but googling restricted to German doesn't seem to produce much other than book adverts [!] - http://www.google.com/search?hl=en&safe=off&client=firefox-a&rls=com.ubuntu%3Aen -US%3Aofficial&num=20&q=wolfgang+M%C3%BCckenheim+Mathematik&lr=lang_de&aq=f&oq=& aqi= Certainly it is clear that Dik's astonishing perseverance will never get anywhere. Brian Chandler On Jul 3, 11:05 pm, Brian Chandler <imaginator...@despammed.com> wrote: > > > In article <pcqum4dh0co7$.l0k3lwh9rsmi$....@40tude.net> Herbert Newman > > > <nomail@invalid> writes: ... [quoted text clipped - 31 lines] > > Brian Chandler Why does he persevere? What's his point? I don't know Dik with any novelty. Is he a modern movement of some standard thing? Angry! Ross Am Fri, 3 Jul 2009 23:05:33 -0700 (PDT) schrieb Brian Chandler: > I suspect that contacting officialdom will get nowhere at all. Much > more effective would be to create a critique web-site (in German of > course) ... "Much Ado About Nothing" - in a sense. On the other hand, it's definitely a shame that such a loon is allowed to teach mathematics. :-( Herb > > > In article <pcqum4dh0co7$.l0k3lwh9rsmi$....@40tude.net> Herbert Newman > > > <nomail@invalid> writes: ... [quoted text clipped - 16 lines] > > I suspect that contacting officialdom will get nowhere at all. Simply try it! Mr. Bader already has a good reputation as a devious informant. You could obtain similar glory, in particular you could obtain an adequate level as a disciple of your grand master Georg Cantor, who was a talented intriguer. He tried to hinder scholars in Germany to get professorships – not by officially stated and discussed arguments, but by secret action without informing the belittled scholars, just like Mr. Bader uses to do. Cantor at his times tried to prevent Darwinism (alas, always without success). But don't despair! You know, Darwins book appeared in 1859. Even now, 150 years later, there are more than enough people around who have not yet comprehended it but believe in creationism, like Cantor did. I am sure, even 50 years laer, there will be more than enough Fools Of Matheology around, discussing transfinity in earnest within their circles as if it was a grave theory of great importance and not nonsense, disconnected from any useful application. > Much > more effective would be to create a critique web-site (in German of [quoted text clipped - 5 lines] > a demonstration of the invalidity of some of his published "tree" > nonsense. Yeeeeaaaaahhh, do so, pleeeeeeeze. People must get informed. If possible, explain that in a manner that is understandable not only by "the community" but by soberly thinking intelligent beings. Perhaps you will start in this way: Once upon a time, there was a man who assumed that there are all real numbers and then proved that it is impossible that there are all real numbers. But he did not conclude that it is impossible that there are all real numbers, but he concluded that there are different finished infinities. That already must be convincing the non-community, mustn't it. If you are less unfair than Mr. Bader, you will allow me to drop only few lines on the intended sheet. Here they are: Take a countable set of infinite binary sequences as paths. Construct a complete infinite binary tree from that set of paths. Nobody will be able, when being shown the completed tree only, to define a path that is not in that tree. Therefore there are not more than countably many real umbers that can be defined by binary sequences. Will you be so glad? Regards, WM In article <7f588dc2-e7c7-408d-b4ff-6c72508e838f@a36g2000yqc.googlegroups.com>, > If you are less unfair than Mr. Bader, you will allow me to drop only > few lines on the intended sheet. Here they are: [quoted text clipped - 4 lines] > is not in that tree. Therefore there are not more than countably many > real umbers that can be defined by binary sequences. That presumes, falsely as it happens, that if a maximal infinite binary tree can be constructed at all, which requires the existence of actually infinite sets, that there exists a set whose power set injects into it. But this is false for all finite sets, so why does it suddenly become true for some actually infinite set? Outside of WM's Weird Weird World of mathUNrealism, it doesn't. SignatureVirgil > In article > <7f588dc2-e7c7-408d-b4ff-6c72508e8...@a36g2000yqc.googlegroups.com>, [quoted text clipped - 18 lines] > -- > Virgil Mueckenheim raises a point of interest: application. Transfinite cardinals and the construction of standard measure theory about them is a reflexive shoehorn of real analysis and particularly that of the unit interval towards making suitably rigorous the infinitesimal analysis. What it doesn't do is satisfy all intuitions, reasonable intuitions, about what the infinitesimal differential is. These are available from geometry and ancient methods of exhaustion as rigorous. Transfinite cardinals aren't directly used or necessary in real-world applications. Consider the integer lattice. These are all the points of the product Z x Z. (Z is for Zahlen, the integers). There are countably many of those points. Then, start building out the paths of the infinite binary tree in, say, a quadrant. Do so leaving each node of the tree with neighbor points not in the tree, having each level of the tree in particular bands of distance from the origin. Then consider a ray for each path that is defined iteratively (in the way as an expansion defines a real of the unit interval) by the nodes of that path. Then, it seems that there don't exist any integer points between those lines, each and all defined by integers points with integer points between them. Otherwise there are. Ross MeAmI.org wrote: > > In article > > <7f588dc2-e7c7-408d-b4ff-6c72508e8...@a36g2000yqc.googlegroups.com>, [quoted text clipped - 43 lines] > > Ross The name is "Musatov" not "Muckenheim". Dik, how might this thread mathematically be different if your name contained a "c" variable, instead of sharing similarity to the popular childrens programming company? As to the transfinite cardinals we are onto something. Mark the binary character "/<...>\" as both a division and continuation in logic. See: Cantor was able to show the first transfinite cardinal number is the cardinality of <...>/K^\ is the first cardinal after all Np where P precedes A. Thus, \\ http://books.google.com/books?isbn=019875244X \\ \\ TRANSFINITE NUMBERS AND NATIVE AMERICAN COSMOLOGY: A COURSE ABOUT /... \A topic like Cantor's theory of transfinite cardinals is nor- /.....\ about P vs NP. Are there patterns in 7f? Do numbers exist? What is intuitionism? \\ http://www.informaworld.com/index/781284139.pdf \\ \\ Consider the Arithmetic of Cardinal Numbers. The sum of two transfinite cardinal numbers is equal to the larger of the two. /...\ (mn)p) = mnp mn+p = mnmp. Below are proofs that these properties apply to \\ http://www.applet-magic.com/cardinals.htm \\ \\ Logic, logic, and logic. Along with the more familiar transfinite cardinals NQ , the cardinals 3Q are defined in set /...\ (Since Dp < P < Np < Dp, Dp = Np.) I want a number that is \\ http://books.google.com/books?isbn=067453767X \\ \\ Characteristics of discrete transfinite time Turing machine models. Reference the work of Deolalikar, V., Hamkins, J.D. and Schindler, R- D., P¿NP¿CO-NP for infinite time Turing /...\ \\ Drake, F.R., Set Theory: An introduction to large cardinals. \\ ÷÷ http://portal.acm.org/citation.cfm%3Fid%3D1498066 ÷÷ "Uncountable" transfinite graphs and electrical networksp-node np as the set. 8 t - l. UJYt-'. The single element of /.....\ the cardinal number of a denumerably infinite set. Consequently, the cardinal \\ http://doi.wiley.com/10.1002/cta.4490220507 \\ Is between logic and intuition: per the essays in honor of Charles Parsons Along with the more familiar transfinite cardinals Na, (Salt) the cardinals 3a are /...\ 3p < p < Np < 3p, 3p = Np.) I want a number that is problematically big even \\ http://books.google.com/books?isbn=0521650763 \\ \\ The chromatic index of an infinite complete hypergraph: A /...\ Let p , m and n be three cardinals so that. L. i p < m < n and n p = n . /....\ Our transfinite induction is complete, and so the family (Ak)k< /...\ \\ http://www.springerlink.com/index/9552220056h0x861.pdf \\ \g\ Per the UCLA Department of Mathematics, numbers, cardinality, axiom of choice, transfinite numbers. P/NP or letter grading. transfinite recursion, and cardinal and ordinal arithmetic. \\ http://www.math.ucla.edu/ugrad/courses/math114S/index.shtml \\ Falls within. \\ Set theory and logic. The natural ordering and the cardinal ordering agree on N- to say, for all natural numbers p and 7, q <np iff g <cp. Proof. \\ -- Musatov http://MeAmI.org "Better than Google. I mean it." On Jul 5, 12:59 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > In article > > <7f588dc2-e7c7-408d-b4ff-6c72508e8...@a36g2000yqc.googlegroups.com>, [quoted text clipped - 41 lines] > lines, each and all defined by integers points with integer points > between them. Otherwise there are. See, there are ways to construct an infinite balanced binary tree, leaving rays from the origin through integer points between each pair of paths of the tree, which is constructed to have its nodes rest on integer points, where rays sweep through in lexicographic order the paths. That is where there are only countably many integer points, so only countably many rays. The rays through integer points are dense in the paths. That is where given any two rays through the origin there is a ray through an integer point between them, and the non- crossing paths of the tree approximate a ray. Consider for example a layout of a binary tree basically in having it with the levels in one direction and the nodes lexicographically in one direction on a level. Make it instead a 5-ary tree, using the 1- and 3- values for 0- and 1- of the binary tree, with the other nodes as integer points serving to indicate rays from the origin. There are only countably many of those rays from the origin through integer points, where the edges of the channels bound the paths. The integer lattice extended through a countable well-founded infinite ordinal is countable. In that sense the paths are still defined piecewise by ordinal points, their placement admits as well structures dense among them that are countable. Ross On Jul 6, 10:04 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > On Jul 5, 12:59 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: [quoted text clipped - 67 lines] > piecewise by ordinal points, their placement admits as well structures > dense among them that are countable. That rays through ordinal points are dense among the paths where the paths are infinite would be similar as to how rationals are dense in the reals, but there is an extra feature afforded by the serial "continuum" of ordinal points out among the paths that given an ordinal point through which there is a path, the following points ascend in a well-ordering, and, as there are rays through the origin dense in the paths, then they alternate, rays refined by paths and rays through ordinal points at that level. Ross On Jul 7, 9:48 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > On Jul 6, 10:04 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: [quoted text clipped - 79 lines] > dense in the paths, then they alternate, rays refined by paths and > rays through ordinal points at that level. The binary tree then seems to offer another example of a case where each of a countable and uncountable set are dense within their union and disjoint yet not sharing the same property of being able to map one onto the other. In the consideration of the rationals and irrationals and how they are each dense in the reals and disjoint, where their union is the reals, each also everywhere not gapless, a consequence is that there is no well ordering of an infinite set of numbers within a non-zero and finite interval that has the well-ordering matching the normal ordering. There are infinite intervals having subsets with those properties, like the natural integers, and "zero" intervals, like one minus the negative powers of two. Then a notion is how to approach such a thing, towards tractability in terms of the most simply defined bounds and shortest description. One consideration is to start with halving, then putting into thirds, quarters, etcetera, the unit interval. In that way a series of numbers is modeled as (0/d, 1/ d, ..., n/d), for n from 1 to d, and then d from 1 towards infinity. This series has that it is well-ordered as are the naturals, and infinite as are the naturals, yet its existence would be a uniform partition of the unit interval into infinitesimals. Now by no means is that a novel consideration, indeed it is rather that the conclusion of the modern rigorization of analysis has that particular thing not existing. It's ancient taboo, yet for some still used in practice, in acknowledgment of its intuitive place in infinitesimal analysis, where the partitions and subsequent summation of the partitions are so modeled as above in the standard. Then it's re-emplacement into the modern, standard, Cauchy, to be post-modern, non-standard, non-Cauchy, is to so offer mathematical tools unavailable to the standard, for application. Ross In article <7f588dc2-e7c7-408d-b4ff-6c72508e838f@a36g2000yqc.googlegroups.com>, > > > > In article <pcqum4dh0co7$.l0k3lwh9rsmi$....@40tude.net> Herbert Newman > > > > <nomail@invalid> writes: ... [quoted text clipped - 68 lines] > is not in that tree. Therefore there are not more than countably many > real umbers that can be defined by binary sequences. So that WM claims that there exists a set which surjects onto the set of all its subsets. That is not even true for finite sets, so why does WM claim it to be true for infinite sets? SignatureVirgil > In article > <7f588dc2-e7c7-408d-b4ff-6c72508e8...@a36g2000yqc.googlegroups.com>, [quoted text clipped - 75 lines] > all its subsets. That is not even true for finite sets, so why does WM > claim it to be true for infinite sets? No, actuaylly you have not yet understood it. I did not claim but prove, namely by means of your incapability to distinguish the elements, that there is no set of all subsets (because even the complete set is not complete). If you insist, for instance, that the set N actually exists and all its subsets actually are elements of P(N), then you get the contradiction that it is impossible to distinguish them. Got it? Regards, WM In article <9258d2e8-a1c1-464e-aa5c-d28dbfd5756d@c1g2000yqi.googlegroups.com>, > > In article > > <7f588dc2-e7c7-408d-b4ff-6c72508e8...@a36g2000yqc.googlegroups.com>, [quoted text clipped - 82 lines] > > No, actuaylly you have not yet understood it. WM's position is not capable of being understood by anyone whose understanding is constrained by logic. > I did not claim but prove, namely by means of your incapability to > distinguish the elements, that there is no set of all subsets (because > even the complete set is not complete). For any finite set there is a set of all its subsets so what impediment does WM find to having the same for infinite sets? WM is so certain that anything that holds for finite sets should hold for all sets that this reversal of his philosophy is unprecedented. > If you insist, for instance, that the set N actually exists and all > its subsets actually are elements of P(N), then you get the > contradiction that it is impossible to distinguish them. Got it? There is nothing there to be "got"! What WM seems to object to is that for any infinite set it is impossible to LIST its subsets, but that is just what Cantor concluded. > Regards, WM SignatureVirgil > For any finite set there is a set of all its subsets so what impediment > does WM find to having the same for infinite sets? The infinity of the set. > WM is so certain that anything that holds for finite sets should hold > for all sets that this reversal of his philosophy is unprecedented. Facts show that there are only finite sets. So there is no switch in philosophy. > > If you insist, for instance, that the set N actually exists and all > > its subsets actually are elements of P(N), then you get the [quoted text clipped - 4 lines] > What WM seems to object to is that for any infinite set it is impossible > to LIST its subsets, but that is just what Cantor concluded. It is not only impossible to "list", it is also impossible to distinguish. It is impossible to exist. You know: Once upon a time, there was a man who assumed that there are all real numbers and then proved that it is impossible that there are all real numbers. But he did not conclude that it is impossible that there are all real numbers, but he concluded that there are different finished infinities. Regards, WM > You know: > [quoted text clipped - 3 lines] > all real numbers, but he concluded that there are different finished > infinities. This parable is about as interesting and profound as the ah-so-funny story about a Buddhist monk and the hod-dog: A Buddhist monk, visiting New York City for the first time in twenty years, walked up to a hot dog vendor, handed him a twenty dollar bill, and said, "Make me one with everything." The vendor pocketed the money, and handed the Buddhist monk his hot dog. The monk, after waiting for a moment, asked for his change. The vendor looked at him and said, "Change comes from within." With a wistful smile, the monk walked away. Make the effort --- spice it up a bit, captivate your readers with a riveting yarn! SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon mann nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus In article <a6310c36-3184-4b30-946b-7d245f3e627b@q11g2000yqi.googlegroups.com>, > > For any finite set there is a set of all its subsets so what impediment > > does WM find to having the same for infinite sets? [quoted text clipped - 6 lines] > Facts show that there are only finite sets. So there is no switch in > philosophy. WM's notion of set theory requires his own form of particular physical implimentation of sets, i.e., a particular model, but nothing in set theory requires that model, or even requires set theory to be comapatible with that model. > > > If you insist, for instance, that the set N actually exists and all > > > its subsets actually are elements of P(N), then you get the [quoted text clipped - 7 lines] > It is not only impossible to "list", it is also impossible to > distinguish. I have no idea what that means. If it means that there are sets which are different but cannot be shown to be different, then it requires more proof than WM has produced. If it means anything else, it is, as so often with WM's claims, mere nonsense. SignatureVirgil Just a short (additional) comment: You wrote (@WM): > > > Apparently you do not know how mathematical proof works ... Look, Dik, WM does not just APPARENTLY not know how mathematical proof works: he DOES NOT know how mathematical proof works --- and YOU KNOW that (that he doesn't know how etc.). So why this _never-ending_ pseudo discussion?! *sigh* Herb > > Apparently you do not know how mathematical proof works ... > [quoted text clipped - 6 lines] > > Herb That brings up the questin of whether WM himself is sane, which is at least equally dubious. SignatureVirgil > > That is the logic of potential infinity, i.e., of incomplete sets. > > Eh? As far as I know logic is *not* about sets. It is about axioms, > statements and inference rules. Logic has been obtained from the bahviour of parts of reality that can be called sets. > > In Cantor's diagonal argument you can use the same logic : There is no > > last line, therefore there is always a line beyond the checked lines. [quoted text clipped - 3 lines] > assumption that you first check the first line and after that the next line. > That is wrong. That is necessary because you cannot find the n-th line unless you know the line number n - 1 or some equivalent mark. > You simply give a definition of a new number so it is clear > that it will be different from each of the omega lines, without checking > directly any of them. And in my binary tree I give a definition of an end of a path that contains aleph_0 nodes. Every end of a path contains aleph_0 nodes. These nodes can be mapped on that path. Every node will eventually be mapped on one path. Therefore all nodes are used up for constructing a countable set of paths. > > When it is derived from logic of finite sets, then it is not the > > reverse of the logic of finite sets. But that is claimed in ZF. > > It is *not* the reverse of the logic of finite sets. For finite sets it is > identical. For infinite sets it is not identical. And the reverse of being identical is being not identical. > Ok. The logic of finite complete sums of rational elements gives that the > sum is rational. Using your logic we get that the complete (i.e. in the > limit) sum of rational elements is also rational. And so by that logic, e, > pi and whatever are rational, and all numbers we do use are rational. In fact there are no binary expansions of irrational numbers. > > Either those sets obey that logic or they do not > > exist in a science that is subject to the application of logic. > > Ah, so 'e', 'pi', 'sqrt(2)' do not exist. Still I think you use at least > one of them on occasion. They exist as ideas but not as sums of series. > > In fact thoses sets contradict their own existence under the > > government of logic. > > So 'sqrt(2)' contradicts itself under the government of logic. I think you > are a few thousand years behind. I am only a few years ahead. > > Small wonder. There cannot be a complete list, because the existence > > of a complete infinite linear set like N contradicts logic. > > Apparently your logic. What are the rules of inference in your logic? The union of a complete set of linear sets is one of the sets. > And, also apparently, in your logic the length of the diagonal of a square > with sides with size 1 does not exist. It exists, but not as binary or decimal expansion. . > > > > That is a blatant lie. > > Eh? What is the lie? I just state that in ZF there are no non-static sets. Then every element should be accessible. Why do you start always with a diminishingly small one? > > Every static linear set has a last element. > > As that is false in ZF... This shows that ZF deals with non-static sets if necessary. > > If you say you cannot choose the last elemement because there is none, > > then you apply potentially infinite sets. > > You have still not provided a definition of "potentially infinite sets" in > ZF, so I cannot comment on this. A potentially infinite set is a set that has no last element. > > Then for every element that > > you choose there is a larger one. If it has been there all time, why > > the hell did you not start with this one ? > > Because in ZF a set does not need to have a last element. So whatever > element you chose, there is *always* a larger one. That is potential infinity. Regards, WM In article <8d9c63c5-b605-4a98-81f2-815875aaea4f@21g2000vbk.googlegroups.com>, > > > That is the logic of potential infinity, i.e., of incomplete sets. > > [quoted text clipped - 3 lines] > Logic has been obtained from the bahviour of parts of reality that can > be called sets. That is not the parts of reality from which Logic was derived. Logic preceded any notion of sets. The roots of logic are at least as old as Aristoteles, so precede sets by a couple of millenia. > > > In Cantor's diagonal argument you can use the same logic : There is no > > > last line, therefore there is always a line beyond the checked lines. [quoted text clipped - 6 lines] > That is necessary because you cannot find the n-th line unless you > know the line number n - 1 or some equivalent mark. Unless one can check ALL lines simultaneoulsly by a gerneral rule, which is the case. > > You simply give a definition of a new number so it is clear > > that it will be different from each of the omega lines, without checking > > directly any of them. > > And in my binary tree I give a definition of an end of a path that > contains aleph_0 nodes. Then the whole path is its own end. > Every end of a path contains aleph_0 nodes. If there can be a set of aleph_0 nodes then there can be MORE THAN aleph_0 subsets of that set, each of which is a path. > These nodes can be mapped on that path. Every node will eventually be > mapped on one path. Therefore all nodes are used up for constructing a > countable set of paths. But that need not, and does not, exhaust the paths, since through each node in a maximal infinite binary tree pass uncountably many paths. > > > When it is derived from logic of finite sets, then it is not the > > > reverse of the logic of finite sets. But that is claimed in ZF. [quoted text clipped - 4 lines] > For infinite sets it is not identical. And the reverse of being > identical is being not identical. AS there are, in general, myriads of ways of being not identical with a given finite-set rule, but only one of them can be the reverse of that finite-set rule, WM is claiming that many_more_than_one = one. > > Ok. The logic of finite complete sums of rational elements gives that the > > sum is rational. Using your logic we get that the complete (i.e. in the > > limit) sum of rational elements is also rational. And so by that logic, e, > > pi and whatever are rational, and all numbers we do use are rational. > > In fact there are no binary expansions of irrational numbers. There are binary expansions of some of them which are as well defined as those binary expansions which eventually become repeating. > > > Either those sets obey that logic or they do not > > > exist in a science that is subject to the application of logic. Mathematics is not a science, however often non-mathematicians may mistakenly claim it to be. > > Ah, so 'e', 'pi', 'sqrt(2)' do not exist. Still I think you use at least > > one of them on occasion. > > They exist as ideas but not as sums of series. No number exists other than as an idea, so that WM's sums of series cannot be used as a test. > > > In fact thoses sets contradict their own existence under the > > > government of logic. [quoted text clipped - 3 lines] > > I am only a few years ahead. Those like WM who are hopelessly behind often delude themselves that way. > > > Small wonder. There cannot be a complete list, because the existence > > > of a complete infinite linear set like N contradicts logic. > > > > Apparently your logic. What are the rules of inference in your logic? > > The union of a complete set of linear sets is one of the sets. If that is the only rule, then WM is hopelessly out of it. And absent a clear definition of "complete set" is meaningless. > > And, also apparently, in your logic the length of the diagonal of a square > > with sides with size 1 does not exist. > > It exists, but not as binary or decimal expansion. So that in WM's world some numbers are "more equal" than others? Shades of "Animal Farm". > . > > > [quoted text clipped - 4 lines] > Then every element should be accessible. Why do you start always with > a diminishingly small one? Because they are all diminishingly small. Or does WM claim to be able to name one that isn't? > > > Every static linear set has a last element. > > > > As that is false in ZF... > > This shows that ZF deals with non-static sets if necessary. Not unless "static" means "finite" in WM-ese. > > > If you say you cannot choose the last elemement because there is none, > > > then you apply potentially infinite sets. [quoted text clipped - 3 lines] > > A potentially infinite set is a set that has no last element. Then {} is a potentially infinite set in WM's matheology. > > > Then for every element that > > > you choose there is a larger one. If it has been there all time, why [quoted text clipped - 4 lines] > > That is potential infinity. Not in ZF, whatever it may look like in WM's unrealistic matheology. SignatureVirgil > A potentially infinite set is a set that has no last element. Where "last" is regarded in terms of WHAT ordering on the set? Or do you mean: S is potentially infinite <-> there exists an ordering R on X such that there is no R-maximal member of X MoeBlee > Or do you mean: > > S is potentially infinite <-> there exists an ordering R on X such > that there is no R-maximal member of X Correction (and incorporating Virgil's remark about the empty set): S is potentially infinite <-> (S is non-empty & there exists a linear ordering R on S such that there is no R-maximal member of S). But, isn't that equivalent with 'infinite' (i.e. 'not finite', i.e., not equinumerous with any natural number') anyway? MoeBlee In article <13ee1342-16b3-47ed-8cb4-48900dab1398@o36g2000vbi.googlegroups.com>, > > Or do you mean: > > [quoted text clipped - 10 lines] > > MoeBlee It is! So if that is WM's definition, it is the same as at least one definition of actual infiniteness. SignatureVirgil > In article > <13ee1342-16b3-47ed-8cb4-48900dab1...@o36g2000vbi.googlegroups.com>, [quoted text clipped - 14 lines] > It is! So if that is WM's definition, it is the same as at least one > definition of actual infiniteness. My definition is: The union of finite segments is a finite segment. There is always a last segment. But it can be surpassed. The idea that the infinite union of finite segments1 1,2 1,2,3 ... results in an infinite segment is false. You can verify this by forming an infinite union of finite segments 1 1 1 ... If the infinite union of finite segments always yields an infinte segment, why then is the result of this infinite union not an infinite segment? Briefly: The set of all natural numbers that exists has always a last element, but potentially infinite sets are not static. They can grow (and they can shrink). Regards, WM > Briefly: The set of all natural numbers that exists has always a last > element, but potentially infinite sets are not static. They can grow > (and they can shrink). Do you have an opinion on the (intuitionistic) theory of choice sequences? Admittedly they do not shrink but perhaps otherwise they're to your liking. SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon mann nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus > > Briefly: The set of all natural numbers that exists has always a last > > element, but potentially infinite sets are not static. They can grow [quoted text clipped - 3 lines] > sequences? Admittedly they do not shrink but perhaps otherwise they're > to your liking. Most (of the few) mathematicians who share my standpoint do not talk about vanishing information. But consider any lawless choice sequence that has been fixed to a certain n (by writing it in a memory). If this memory is destroyed, the due information has disappeared. Regards, WM > Most (of the few) mathematicians who share my standpoint do not talk > about vanishing information. Who are these few mathematicians who share your standpoint? > But consider any lawless choice sequence that has been fixed to a > certain n (by writing it in a memory). If this memory is destroyed, > the due information has disappeared. On your conception what becomes of such basic principles as the density axiom Every finite sequence <a1, ..., an> is the initial segment of a lawless choice sequences. the principle of open data If a property P holds of a lawless choice sequence alpha, there is an initial segment <a1, ..., an> of alpha such that P holds of all lawless choice sequences with <a1, ..., an> as an initial segment. and so on? SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon mann nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus > > Most (of the few) mathematicians who share my standpoint do not talk > > about vanishing information. > > Who are these few mathematicians who share your standpoint? Here you will find some of them, if you click through till the end of that lesson. http://www.hs-augsburg.de/~mueckenh/GU/GU11.PPT#385,50,Folie 50 > > But consider any lawless choice sequence that has been fixed to a > > certain n (by writing it in a memory). If this memory is destroyed, [quoted text clipped - 13 lines] > > and so on? I need no axioms. 5. A constructive compromise. [...] We accept the hierarchy of types; but we assume only one category of primary objects, the numbers; and one basic binary relation between numbers, namely "x is followed by y." All other relations of the various types are explicitly constructed, the quantifiers (x) and (x) being applied only to numbers and not to arguments of higher type. No axioms are postulated. (H. Weyl) You can find it on p. 274 in his collected works: http://books.google.de/books?id=lNPriL6kG7AC&pg=PA272&lpg=PA272&dq=%22transcende ntal+world+taxes+the+strength&source=bl&ots=9HDX06hGG-&sig=njIZrXtyKYjBPDmjqDg_s ftAEzo&hl=de&ei=FAEwSvCaDsuFsAajkfi6CQ&sa=X&oi=book_result&ct=result&resnum=4#PP A272,M1 Regards, WM In article <1348b29b-3f39-40a7-be0e-5bd24a4e3cb0@s21g2000vbb.googlegroups.com>, > I need no axioms. Then there is absolutely nothing that you take to be true? There are absolutely no truths from which you can deduce other things? That makes for a very simple mathematics in which nothing need be true. Consider axioms like in the American Declaration of Independence: "We hold these truths to be self evident..." Unless WM can list the things he "holds to be self evident", he does not have anything to build on. So, WM, list them or shut up. SignatureVirgil > In article > <1348b29b-3f39-40a7-be0e-5bd24a4e3...@s21g2000vbb.googlegroups.com>, [quoted text clipped - 4 lines] > > There are absolutely no truths from which you can deduce other things? There are truths, like 1 + 1 = 2, or, for linear sets, ExAy <==> AyEx, but they cannot be chosen by free will. Incorrect are such lies as the axiom of infinity or the axiom of choice. > That makes for a very simple mathematics in which nothing need be true. > [quoted text clipped - 4 lines] > Unless WM can list the things he "holds to be self evident", he does not > have anything to build on. There is no necessity to list them in mathematics. Every sober mind knows them. Regards, WM In article <552bb80e-30b7-447e-a386-693d8fc0c6ca@m19g2000yqk.googlegroups.com>, > > In article > > <1348b29b-3f39-40a7-be0e-5bd24a4e3...@s21g2000vbb.googlegroups.com>, [quoted text clipped - 7 lines] > There are truths, like 1 + 1 = 2, or, for linear sets, ExAy <==> AyEx, > but they cannot be chosen by free will. Then they cannot be chosen at all. > Incorrect are such lies as the axiom of infinity or the axiom of > choice. Incorrect is WM. > > That makes for a very simple mathematics in which nothing need be true. > > [quoted text clipped - 7 lines] > There is no necessity to list them in mathematics. Every sober mind > knows them. There is every need to list them as without such a list, there can be no agreement on what they are. Not everything that seems "self evident" to one person is necessarily so to all, as the warlike history of mankind evidences. I find that WM's claims of self-evidentness to be quite false, as do many, if not all, of those who still bother to read his tirades. SignatureVirgil > > Unless WM can list the things he "holds to be self evident", he does not > > have anything to build on. > > There is no necessity to list them in mathematics. Every sober mind > knows them. They are known to all Scotsmen. That is, they are known to all *true* Scotsmen. -- hz "Mathematics is best done without axioms." -- Wolfgang Mückenheim corrects Euclid -- > > > Unless WM can list the things he "holds to be self evident", he does not > > > have anything to build on. [quoted text clipped - 5 lines] > > That is, they are known to all *true* Scotsmen. Aye, *true* Scotsmen, there's the rub. And, of course, to WM, who is hardly a true anything. > -- > hz > "Mathematics is best done without axioms." > -- Wolfgang M?ckenheim corrects Euclid -- SignatureVirgil >> In article >> <1348b29b-3f39-40a7-be0e-5bd24a4e3...@s21g2000vbb.googlegroups.com>, [quoted text clipped - 6 lines] > > There are truths, like 1 + 1 = 2, or, for linear sets, ExAy <==> AyEx These "truths" are the axioms of the theory you're using, or are implied by them. > Incorrect are such lies as the axiom of infinity or the axiom of > choice. The mathematics community mostly does not pursue a universally true set of axioms. Instead, mathematics takes sets of axioms and attempts to demonstrate that they lead to interesting results. The results derived from a given theory (a set of axioms) are only valid within that theory. There are informal "families" of theories, within which the informal definitions of various terms are portable (for example, there is a family of "set theories," each of which has a loosely similar definition of what a "set" is), but informal similarity does not mean that proofs from theories in the same "family" can necessarily be used together. ZF set theory is no more "true" or "false" than it is "blue" or "grape-flavoured" — it is only a set of axioms, from which results that are, most of the time, consistent with intuition can be derived. It also implies some results that are less intuitive, like the existence of more than one size of infinite set. There are other theories which imply that there are no infinite sets at all. Both are useful; neither is more right than the other, and the existence of infinite sets in ZF is not "inconsistent" (in the technical sense of the word) with the non-existence of infinite sets in another theory. The pursuit of a single "true" set of axioms is at best an exercise in philosophy, not mathematics, and at worst a grave misunderstanding of how math works. Fortunately, you are not alone in this search; there are others posting to sci.math who believe that there is a "true mathematics." >> That makes for a very simple mathematics in which nothing need be true. >> [quoted text clipped - 7 lines] > There is no necessity to list them in mathematics. Every sober mind > knows them. If that were true, this thread would've ended years ago. You really do have to get your head around the idea that the people you're talking to aren't idiots, don't have any personal animosity towards you or your ideas, aren't yanking your chain (much), and quite honestly don't share your intuition. The repeated requests by some posters that you break your claims down into very small, easily digested pieces are not attempts to waste your time, but rather (mostly) honest attempts to understand the rules of the intellectual framework you're using to make your pronouncements. Mathematicians, by long and painful experience, generally distrust "intuitive" truths. The parallel postulate is intuitively true, but its contradiction is not only mathematically consistent but extremely useful: geometries without the parallel postulate are used to model the way light and matter interact. Similarly, it was once intuitively true that all numbers could be expressed as the ratio between two whole numbers, but the contradiction of that axiom is (again) both mathematically consistent and useful. So it is with the idea that all sets are finite. The ZF axiom of infinity is only the contradiction of the axiom "there are no infinite sets," and the resulting theory is both consistent and useful. There are also consistent, useful theories that do not prove or require the existence infinite sets. If you would only name the theory you're using, or sketch out its fundamental truths (that is, its axioms), you'd probably find people a lot more receptive to your ideas. -o > >> In article > >> <1348b29b-3f39-40a7-be0e-5bd24a4e3...@s21g2000vbb.googlegroups.com>, [quoted text clipped - 81 lines] > fundamental truths (that is, its axioms), you'd probably find people a > lot more receptive to your ideas. WM seems to want us to take HIM purely on faith, but refuses to reciprocate. And then gets up tight when we don't take him on faith. SignatureVirgil > >> In article > >> <1348b29b-3f39-40a7-be0e-5bd24a4e3...@s21g2000vbb.googlegroups.com>, [quoted text clipped - 9 lines] > These "truths" are the axioms of the theory you're using, or are > implied by them. I do not deny that. But I deny that axioms can be chosen. > > Incorrect are such lies as the axiom of infinity or the axiom of > > choice. > > The mathematics community mostly does not pursue a universally true set > of axioms. But mathematics does. > Instead, mathematics takes sets of axioms and attempts to > demonstrate that they lead to interesting results. Here is an interesting quote by Weyl (1946): "We accept the hierarchy of types; but we assume only one category of primary objects, the numbers; and one basic binary relation between numbers, namely "x is followed by y." All other relations of the various types are explicitly constructed, the quantifiers (Ex) and (Ax) being applied only to numbers and not to arguments of higher type. No axioms are postulated." Does he no longer belong to the math. community? > The results derived > from a given theory (a set of axioms) are only valid within that > theory. Mathematics (if done correctly) is true without any theory. > There are informal "families" of theories, within which the > informal definitions of various terms are portable (for example, there > is a family of "set theories," each of which has a loosely similar > definition of what a "set" is), but informal similarity does not mean > that proofs from theories in the same "family" can necessarily be used > together. All transfinite set theories are nonsense, formal or informal. > ZF set theory is no more "true" or "false" It is false, no doubt. A man who believes in or even attempts to prove something by means of "higher cardinals" is a fool. > So it is with the idea that all sets are finite. The ZF axiom of > infinity is only the contradiction of the axiom "there are no infinite > sets," and the resulting theory is both consistent and useful. That is wrong. Not a single useful bit has ever been obtained from finished infinity. > There > are also consistent, useful theories that do not prove or require the > existence infinite sets. This is a prerequisite of a consisten theory like the convergence of a sequence to zero, shall the due series converge. > If you would only name the theory you're using, or sketch out its > fundamental truths (that is, its axioms), you'd probably find people a > lot more receptive to your ideas. I am doing simply correct mathematics, at least I am striving for that goal. Regards, WM In article <6ae5a704-1bba-4cab-8a18-c77ded4167d0@p6g2000pre.googlegroups.com>, > > >> In article > > >> <1348b29b-3f39-40a7-be0e-5bd24a4e3...@s21g2000vbb.googlegroups.com>, [quoted text clipped - 19 lines] > > But mathematics does. Then how can mathematics deal with both Euclidean and non-Euclidean geometries? By WM's standards, only one of them can be true, so that mathematics, by WM's standards, should be forbidden from dealing with all but the true one. Which one would that be, WM? You claim to know so much about what is really true, so tell us which geometry is the true one. And then tell us how you know. > > Instead, mathematics takes sets of axioms and attempts to > > demonstrate that they lead to interesting results. > > Here is an interesting quote by Weyl (1946): Mathematics does not accept truth-by-authority. > > The results derived > > from a given theory (a set of axioms) are only valid within that > > theory. > > Mathematics (if done correctly) is true without any theory. Where do the first "truths" come from, that everything else derives from? They cannot come from the world of physics because we have no direct perception of that world. > > There are informal "families" of theories, within which the > > informal definitions of various terms are portable (for example, there [quoted text clipped - 4 lines] > > All transfinite set theories are nonsense, formal or informal. Then all of WM's theories are informal nonsense, too. And I prefer the coherent nonsense of mathematics to the inchoate idiotic nonsense of WM. > > ZF set theory is no more "true" or "false" > > It is false, no doubt. To claim that without logical proof, as WM repeatedly does, is to proclaim oneself illogical. > A man who believes in or even attempts to prove > something by means of "higher cardinals" is a fool. The creative foolishness of mathematics can be, and historically has been, highly productive and useful, but WM's stupidity will never be. > > So it is with the idea that all sets are finite. The ZF axiom of > > infinity is only the contradiction of the axiom "there are no infinite > > sets," and the resulting theory is both consistent and useful. > > That is wrong. The creative foolishness of mathematics can be, and historically has been, highly productive and useful, but WM's stupidity will never be. > > There > > are also consistent, useful theories that do not prove or require the > > existence infinite sets. > > This is a prerequisite of a consisten theory like the convergence of a > sequence to zero, shall the due series converge. In a set theory without infinite sets, and therefore also without infinite sequences, there can be no convergence because every sequence must have a last term. > > If you would only name the theory you're using, or sketch out its > > fundamental truths (that is, its axioms), you'd probably find people a > > lot more receptive to your ideas. > > I am doing simply correct mathematics, at least I am striving for that > goal. And failing. Partly because you have not, or cannot, sketch out what you consider to be the fundamental truths on which you base your theory. Without being able to derive things from only fundamental truths and formal definitions, there is no way to check the validity of anything. So if WM feels compelled to keep his list of fundamental truths secret, he cannot publicly validate any of his arguments. SignatureVirgil > Consider axioms like in the American Declaration of Independence: > > "We hold these truths to be self evident..." > > Unless WM can list the things he "holds to be self evident", he does not > have anything to build on. No one can "list the things he holds to be self evident" in the sense of the American Declaration of Independence. Such things are also not axioms in any usual logical or mathematical sense -- they are more like the axioms of Objectivism, in that they have an indefinite range of possible applications, consequences, meaning. SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon mann nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus > > Consider axioms like in the American Declaration of Independence: > > [quoted text clipped - 8 lines] > the axioms of Objectivism, in that they have an indefinite range of > possible applications, consequences, meaning. In that case WM is SOL in trying to build a coherent system. SignatureVirgil > In that case WM is SOL in trying to build a coherent system. Mückenheim is short on luck in trying to build a coherent system because things like "all men are created equal" are not axioms in any usual logical or mathematical sense? SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon mann nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus We agree that in Cantor's diagonal argument, applied to real numbers, the numbers are represented and identified solely by their digits. No further information is available. We assume that a real number p can be distinguished from a set Q of real numbers q by general considerations, for instance, if p is a transcendental number and Q consists of rational numbers q only. Of course it would be impossible to distinguish p from all q, because for every digit d_n of p, there is a number q that shares all digits up to d_n with p. It is possible to construct the complete binary tree from the numbers in Q. That means, it is possible to construct all possible digit sequences from the numbers in Q. It is also possible to construct the complete binary tree from the set Tp that consists of all termintaing rationals which are appended by a tail consisting of p. Therefore it is impossible to distinguish p from the binary tree. But the binary tree has been errected by using a countable set of paths only. Therefore it is impossible to distinguish, by means of digits, any real number from that countable set of paths which has been used to construct the tree. So it is impossible to distinguish pi from all terminating reals. Therefore there is no pi. Regards, WM In article <8f32eea2-9d46-4033-8a78-d3a05e88fe32@h2g2000yqg.googlegroups.com>, > We agree that in Cantor's diagonal argument, applied to real numbers, > the numbers are represented and identified solely by their digits. No [quoted text clipped - 7 lines] > for every digit d_n of p, there is a number q that shares all digits > up to d_n with p. It is possible to distinguish any real from any finite set of reals which does not contain it, and from many infinite sets of reals which do not contain it. The only ones which cannot be so distinguished are those sets containing a sequence converging to the real in question, which begs the question. > It is possible to construct the complete binary tree from the numbers > in Q. But it is not possible to complete it without automatically creating uncountably many binary sequences. > That means, it is possible to construct all possible digit > sequences from the numbers in Q. All rational numbers have a binary expansion which is eventually periodic. How does one construct from such eventually periodic binary expansions a binary expansion that is not eventually periodic? One can find a sequence of such eventually periodic binaries converging to a non-perodic seqeunce, but it never actually arrives. So one cannot construct them exactly. One gets them purely as a bonus for completing the tree, not by any explicit construction of more than a dense "subset" o them.. SignatureVirgil > In article > <8f32eea2-9d46-4033-8a78-d3a05e88f...@h2g2000yqg.googlegroups.com>, [quoted text clipped - 12 lines] > > It is possible but not sufficient for the following > to distinguish any real from any finite set of reals > which does not contain it, and from many infinite sets of reals which do > not contain it. > > The only ones which cannot be so distinguished are those sets containing > a sequence converging to the real in question, which begs the question. Every set of all terminating paths appended by arbitrary tails is countable and covers all real numbers of the unit interval. > > It is possible to construct the complete binary tree from the numbers > > in Q. > > But it is not possible to complete it without automatically creating > uncountably many binary sequences. That is the result of superstition. There is a countable set at the start and that set has not changed at the end. > > That means, it is possible to construct all possible digit > > sequences from the numbers in Q. [quoted text clipped - 4 lines] > How does one construct from such eventually periodic binary expansions > a binary expansion that is not eventually periodic? That is not the right question to be put here. The right question is: If all periodic expansions cover all expansions of real numbers: can such expansions exist that are not eventually periodic? And the answer is a resounding "no". > One can find a sequence of such eventually periodic binaries converging > to a non-perodic seqeunce, but it never actually arrives. So it is. Therefore Cantor's diagonal never actually arrives. It is never completed, it is not a number. > So one cannot construct them exactly. Yes, just this is impossible. One can construct this diagonal to any desired finite epsilon, but not better. Regards, WM In article <3310a94f-4ec5-4c03-bb4e-d6c884fb2cdb@37g2000yqp.googlegroups.com>, > Every set of all terminating paths appended by arbitrary tails is > countable and covers all real numbers of the unit interval. If those "tails" are sufficiently arbitrary as to "cover" by inclusion all reals, then there are uncountably many of them. > > > It is possible to construct the complete binary tree from the numbers > > > in Q. [quoted text clipped - 4 lines] > That is the result of superstition. There is a countable set at the > start and that set has not changed at the end. Then, as Cantor has proved, there are some missing. > > > That means, it is possible to construct all possible digit > > > sequences from the numbers in Q. [quoted text clipped - 6 lines] > > That is not the right question to be put here. That is precisely the right question here. WM only objects to it because he cannot answer it. > The right question is: > If all periodic expansions cover all expansions of real numbers: can > such expansions exist that are not eventually periodic? And the answer > is a resounding "no". The those eventually periodic numbers do not cover all reals, as it is quite easy to construct reals whose representations are not eventually periodic. > > One can find a sequence of such eventually periodic binaries converging > > to a non-perodic seqeunce, but it never actually arrives. > > So it is. Therefore Cantor's diagonal never actually arrives. It is > never completed, it is not a number. The various "Cantor diagonals" for decimals are RULES for constructing a decimal number from given a sequence of such numbers, so are not themselves a numbers. Each may be thought of as a function whose arguments are lists of decimal numbers and whose value is a decimal number. WM is confusing the value of a function with the function itself. An associated number only "arises" when someone provides a list of numbers. > > So one cannot construct them exactly. > > Yes, just this is impossible. One can construct this diagonal to any > desired finite epsilon, but not better. Zero is a finite epsilon. And whichever particular Cantor rule one invokes for a particular list, its value differs from any particular member of that list by some positive epsilon. > Regards, WM SignatureVirgil > Every set of all terminating paths appended by arbitrary tails is > countable NO, it isn't. Even the set of ONE terminating path "appended by arbitrary tails" is NOT countable because there are an UNcountable number of "arbitrary" tails. Of course, the fact that you don't know how to say what you mean here EITHER in English OR with quantifiers is, obviously, not helping the situation. > In article <87r5xofrgi....@alatheia.truth.invalid>, > [quoted text clipped - 12 lines] > > In that case WM is SOL in trying to build a coherent system. Apropos: "I am convinced that the platonism which underlies Cantorian set theory is utterly unsatisfactory as a philosophy of our subject [...] platonism is the medieval metaphysics of mathematics; surely we can do better" [Soloman Feferman: "Infinity in Mathematics: Is Cantor Necessary?"] Furthermore, a case can be made that higher set theory is dispensable in scientifically applicable mathematics, that is, in that part of everyday mathematics which finds its applications in the other sciences. Put in other terms: the actual infinite is not required for the mathematics of the physical world. The reasons for this depend on other recent developments in mathematical logic, the description of which is the final aim of this essay [now chapter 12; cf. also chapter 14]. In order to explain the objections to Cantor's ideas in mathematics ... [Solomon Feferman: IN THE LIGHT OF LOGIC, p. 30] IN THE LIGHT OF LOGIC From the table of contents: V COUNTABLY REDUCIBLE MATHEMATICS 12 Infinity in mathematics: Is Cantor necessary? (Conclusion) 229 13 Weyl vindicated: The Kontinuum 70 years later 249 14 Why a little bit goes a long way: Logical foundations of scientifically applicable mathematics 284 From the cover: At least to that extent the question "Is Cantor necessary?" is answered with a resounding "no" Regards, WM In article <cef492a8-b817-4312-a6b5-78e91180b390@t21g2000yqi.googlegroups.com>, > > In article <87r5xofrgi....@alatheia.truth.invalid>, > > [quoted text clipped - 19 lines] > better" [Soloman Feferman: "Infinity in Mathematics: Is Cantor > Necessary?"] WM again argues from authority. Mathematics does not accept such arguments unless accompanied by proofs. > Furthermore, a case can be made that higher set theory is dispensable > in scientifically applicable mathematics, that is, in that part of > everyday mathematics which finds its applications in the other > sciences. WM is certainly not able to make any such case, and what is not mathematics, but only application, is irrelevant to what mathematics is. > Put in other terms: the actual infinite is not required for > the mathematics of the physical world. That physicists do not need, or use, all of mathematics is their problem, not that of mathematicians. There is no need in mathematics for any "applications" to exist. In fact, G.H. Hardy regarded such mathematics as could be applies to be unworthy of study. SignatureVirgil > In fact, G.H. Hardy regarded such mathematics as could be applies to > be unworthy of study. A fine illustration of the well-known phenomenon that renowned mathematicians are just as capable of mouthing inanities as anyone. SignatureAatu Koskensilta (aatu.koskensilta@uta.fi) "Wovon mann nicht sprechen kann, darüber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus > > In fact, G.H. Hardy regarded such mathematics as could be applies to > > be unworthy of study. > > A fine illustration of the well-known phenomenon that renowned > mathematicians are just as capable of mouthing inanities as anyone. And hardhy was qquite wrong about what could be applied, as a considerable part of his work was in the bits of number theory that are the basis of current security techniques in todays internet transactions. SignatureVirgil Aatu Koskensilta wrote: > > In fact, G.H. Hardy regarded such mathematics as could be applies to > > be unworthy of study. > > A fine illustration of the well-known phenomenon that renowned > mathematicians are just as capable of mouthing inanities as anyone. They're no logicians. -- hz >> In fact, G.H. Hardy regarded such mathematics as could be applies to >> be unworthy of study. > > A fine illustration of the well-known phenomenon that renowned > mathematicians are just as capable of mouthing inanities as anyone. It's foremost a fine example of selective reading, or at least of anachronistic reading, because Hardy also wrote: "But here I must deal with a misconception. It is sometimes suggested that pure mathematicians glory in the uselessness of their work, and make it a boast that it has no practical applica- tions." [...] "I have been accused of taking this view myself." Apparently Hardy considered this a -misrepresentation- of his views! Some other telling quotations: "... there is no mathematician so pure that he feels no interest at all in the physical world" (Is this still true today? Hardy shows no disapproval of the situation as it was in his time.) "... there is probably less difference between the positions of a mathematician and of a physicist than is generally supposed" "I count Maxwell and Einstein, Eddington and Dirac, among 'real' mathematicians." "I hope that I need not say that I am trying to decry mathemati- cal physics, a splendid subject with tremendous problems where the finest imaginations have run riot." And one more noteworthy passage that starts with: "My eyes were first opened by Professor Love [...] primarily an applied mathematician." We shouldn't forget that in Hardy's days, physics occupied a much more important place in mathematical work than today. People like Turing were half engineers, iirc. Could it perhaps be that Hardy would have defended a -larger- role for physics in todays mathematics had he known the current state of affairs? That, had he lived today, his opinions had been much closer to that of our own HdB or David Petry? SignatureCheers, Herman Jurjus > >> In fact, G.H. Hardy regarded such mathematics as could be applies to > >> be unworthy of study. [quoted text clipped - 4 lines] > It's foremost a fine example of selective reading, or at least of > anachronistic reading, because Hardy also wrote: I think that Aatu alluded to the fine example of appeal to authority which Virgil just had critizised. Regards, WM In article <a2466748-54c4-4317-89d6-af8d54450c5d@c36g2000yqn.googlegroups.com>, > > >> In fact, G.H. Hardy regarded such mathematics as could be applies to > > >> be unworthy of study. [quoted text clipped - 9 lines] > > Regards, WM As I was just showing how irrelevant appeals to authority are in mathematics, any such criticism of my own example merely supports my case. SignatureVirgil > In article > <a2466748-54c4-4317-89d6-af8d54450...@c36g2000yqn.googlegroups.com>, [quoted text clipped - 10 lines] > > I think that Aatu alluded to the fine example of appeal to authority > > which Virgil just had critizised. > As I was just showing how irrelevant appeals to authority are in > mathematics, any such criticism of my own example merely supports my > case. In particular if it stems from an authority (like Aatu, for instance). Regards, WM In article <b4aab16a-9706-4a79-86f1-dc84945385c9@37g2000yqp.googlegroups.com>, > > In article > > <a2466748-54c4-4317-89d6-af8d54450...@c36g2000yqn.googlegroups.com>, [quoted text clipped - 16 lines] > > In particular if it stems from an authority (like Aatu, for instance). I do not know whether Aatu qualifies as an authority on infinite binary trees, but the evidence clearly shows that WM does not. SignatureVirgil > In article > <b4aab16a-9706-4a79-86f1-dc8494538...@37g2000yqp.googlegroups.com>, [quoted text clipped - 22 lines] > I do not know whether Aatu qualifies as an authority on infinite binary > trees, He is certainly not an authority on trees, but on criticism of appeal to authority. Regards, WM In article <e35a9421-14ea-4443-9791-00509f6ceb7f@f19g2000yqh.googlegroups.com>, > > In article > > <b4aab16a-9706-4a79-86f1-dc8494538...@37g2000yqp.googlegroups.com>, [quoted text clipped - 27 lines] > He is certainly not an authority on trees, but on criticism of appeal > to authority. Since WM is definitely not an authority on trees, except possibly in his own mind, he is in no position to judge Aatu's qualifications as an authority on trees. SignatureVirgil Herman Jurjus wrote: > Could it perhaps be that Hardy would have defended a -larger- role for > physics in todays mathematics had he known the current state of affairs? > That, had he lived today, his opinions had been much closer to that of > our own HdB or David Petry? Ack!! -- hz > >> In fact, G.H. Hardy regarded such mathematics as could be applies to > >> be unworthy of study. [quoted text clipped - 45 lines] > That, had he lived today, his opinions had been much closer to that of > our own HdB or David Petry? Quite possibly, but what do you suppose Hardy's opinion of WM would be? SignatureVirgil > In article <4a36423f$0$1642$703f8...@textnews.kpn.nl>, > [quoted text clipped - 49 lines] > > Quite possibly, but what do you suppose Hardy's opinion of WM would be? Appeal to assumed opinion of authority? Regards, WM In article <6c5557b1-4a3f-4aa1-8f09-e1139389c9a7@j32g2000yqh.googlegroups.com>, > > In article <4a36423f$0$1642$703f8...@textnews.kpn.nl>, > > [quoted text clipped - 51 lines] > > Appeal to assumed opinion of authority? hardy had a fairly wry sense of humor, so his opinion of someone like WM might have been quite amusing. Though possibly less so than that of his frequent collaborator, Littlewood. SignatureVirgil In article <069a1acd-d0f6-42f3-a8b5-d6f9c27e942e@s16g2000vbp.googlegroups.com>, > > > Briefly: The set of all natural numbers that exists has always a last > > > element, but potentially infinite sets are not static. They can grow [quoted text clipped - 6 lines] > Most (of the few) mathematicians who share my standpoint do not talk > about vanishing information. We have yet to see any evidence that any mathematician, indeed anyone other than your students, shares your "standpoint". Even those who insist on only finite sets, do it quite in a different fashion quite incompatible with WM's nonsense. SignatureVirgil A nice pink elephant <snip> WM: The idea that the infinite union of finite segments WM: results in an infinite segment is false. WM now gives a "counterexample", finding an infinte union of finite segments that does not result in an infinite segment. Look! Over There! A Pink Elephant! WM: If the infinite union of finite segments *always* WM: yields an infinte segment ... - William Hughes > WM: The idea that the infinite union of finite segments > WM: results in an infinite segment is false. > > WM now gives a "counterexample", finding an infinte union > of finite segments that does not result in > an infinite segment. Nobody has any evidence of an example supporting the idea that the infinite union of finite segments results in an infinite segment. No form of logic does support it. I gave a counter example. The only pro-argument is that set theorists feel obliged to believe in that idea. But that is not enough. Regards, WM > I gave a counter example. Nope, you tried to distact people with a Pink Elephant while you sliped in the word "always". - William Hughes In article <e2d06e23-d946-42c9-a58b-e4303afd6b0a@l28g2000vba.googlegroups.com>, > > WM: The idea that the infinite union of finite segments > > WM: results in an infinite segment is false. [quoted text clipped - 6 lines] > infinite union of finite segments > results in an infinite segment. We have plenty of examples showing that the infinite union of a sequence of sets each a proper subset of the next, cannot be a finite set. And there is no consistent set theory extant in which there is any such thing as a potentially infinite set. > No form of logic does support it. > I gave a counter example. Which has itself been countered, so is invalid. > The only pro-argument is that set theorists feel obliged to believe in > that idea. But that is not enough. They feel obliged to follow the logic which says that if there can be an infinite sequence of finite sets, each a proper subset of its successor, which existence WM accepts, then there must be a union of all those sets, i.e., a set of which all those nested sets are proper subsets, and any such a union must be itself infinite. SignatureVirgil In article <4c99a86a-1dd0-44da-b717-a691722a7e81@r3g2000vbp.googlegroups.com>, > A nice pink elephant > [quoted text clipped - 8 lines] > > Look! Over There! A Pink Elephant! Actually, WM's supposed example was not an infinite union of DISTINCT sets but a union of just one set infinitely repeated. Which doesn't count as an infinite union. > WM: If the infinite union of finite segments *always* > WM: yields an infinte segment ... > > - William Hughes SignatureVirgil In article <67c38928-9314-44de-bce5-e884f1fe8cb4@l28g2000vba.googlegroups.com>, > > In article > > <13ee1342-16b3-47ed-8cb4-48900dab1...@o36g2000vbi.googlegroups.com>, [quoted text clipped - 17 lines] > My definition is: The union of finite segments is a finite segment. > There is always a last segment. But it can be surpassed. That does not appear to be a definition of anything, but might possibly be an axiom. > The idea that the infinite union of finite segments1 > 1,2 > 1,2,3 > ... > results in an infinite segment is false. The idea that an INFINITE union of finite "segments", no two of which are identical, must result in a finite segment is not even false. It is ridiculous. > You can verify this by forming an infinite union of finite segments > 1 [quoted text clipped - 4 lines] > If the infinite union of finite segments always yields an infinte > segment Wm seems to think the the union of {1} and {1} must be a set containing more than one element. In unions, no greater-than-one number of repetitions of any one element counts more than a single instance of that element counts. In a unions, one only considers distinct objects, so WM's alleged union is of only one object. > why then is the result of this infinite union not an infinite > segment? Because that is not how unions work, at least in any sensible form of mathematics. What goes on in WM's peculiar non-mathematical world is only his own problem, not ours. > Briefly: The set of all natural numbers that exists has always a last > element, but potentially infinite sets are not static. They can grow > (and they can shrink). Not in mathematics. At least not until WM, or someone, comes up with an axiom system in which those properties WM p[osttulates are less self-contradictory than they are at present. SignatureVirgil > > The idea that the infinite union of finite segments1 > > 1,2 [quoted text clipped - 5 lines] > are identical, must result in a finite segment is not even false. It is > ridiculous. Nearly correct. Absolutely correct is this statement: The idea of an INFINITE union of finite "segments", no two of which are identical, is ridiculous. > At least not until WM, or someone, comes up with an axiom system in > which those properties WM posttulates are less self-contradictory than > they are at present. There is no axiom system required. We accept the hierarchy of types; but we assume only one category of primary objects, the numbers; and one basic binary relation between numbers, namely "x is followed by y." All other relations of the various types are explicitly constructed, the quantifiers (Ex) and (Ax) being applied only to numbers and not to arguments of higher type. No axioms are postulated. Regards, WM In article <d02b20eb-7587-4584-9cdd-619735a34878@j18g2000yql.googlegroups.com>, > > > The idea that the infinite union of finite segments1 > > > 1,2 [quoted text clipped - 9 lines] > The idea of an INFINITE union of finite "segments", no two of which > are identical, is ridiculous. Not outside of WM's deceitful world of MathUnealism. > > At least not until WM, or someone, comes up with an axiom system in > > which those properties WM posttulates are less self-contradictory than > > they are at present. > > There is no axiom system required. An axiom system need be no more than a list of the truths that one holds to be self evident. If WM holds nothing to be self evident, then why is he arguing? > We accept the hierarchy of types; Which "hierarchy of types" would that be? Google has over 5 million hits on that subject. > but we assume only one category of primary objects, the numbers; and > one basic binary relation between numbers, namely "x is followed by > y." All other relations of the various types are explicitly > constructed, the quantifiers (Ex) and (Ax) being applied only to > numbers and not to arguments of higher type. No axioms are postulated. If you accept some hierarchy of types, a category of primary objects , etc. You are proposing a set of axioms. But if the above are not axioms, then we shall ignore them. SignatureVirgil > > > That is the logic of potential infinity, i.e., of incomplete sets. > > [quoted text clipped - 3 lines] > Logic has been obtained from the bahviour of parts of reality that can > be called sets. O. Still, as far as I know logic is *not* about sets. > > > In Cantor's diagonal argument you can use the same logic : There is no > > > last line, therefore there is always a line beyond the checked lines, [quoted text clipped - 6 lines] > That is necessary because you cannot find the n-th line unless you > know the line number n - 1 or some equivalent mark. You are wrong. A list is a mapping from N to the elements of the list. Through that list, given a number n, you find the n-th element of the list without referring to any previous elements of the list. To give an example, let's have a list of positive rational numbers through the mapping given a lnong time ago by David Tribble. To get the 51st element of the list, we calculate the mapping: write 51 in binary, create from it a continued fraction as described on <http://homepages.cwi.nl/~dik/english/mathematics/mueck/mapping.html> which is [0, 1, 2, 3], calculate it and we find 7/9. So we know the 51st element without knowing the 50th element. Where did I come at the 47th rational in that list (which is 7/11)? > > You simply give a definition of a new number so it is clear > > that it will be different from each of the omega lines, without checking > > directly any of them. > > And in my binary tree I give a definition of an end of a path that > contains aleph_0 nodes. I have not seen such a definition. What is the definition of "end of a path"? > Every end of a path contains aleph_0 nodes. > These nodes can be mapped on that path. Every node will eventually be > mapped on one path. Therefore all nodes are used up for constructing a > countable set of paths. Perhaps right, depends on how you actually do define things. But are there nodes mapped to all paths? That is what you assert. > > > When it is derived from logic of finite sets, then it is not the > > > reverse of the logic of finite sets. But that is claimed in ZF. [quoted text clipped - 3 lines] > > For infinite sets it is not identical. No, of course not, because it is originally available for finite sets only, it can not be identical. > And the reverse of being > identical is being not identical. Not in this case bacause you apply the words to different things. *Unless* you assume that what is valid in finite cases also is valid in infinite cases. But let's see: sum{i = 0 .. n} 1/(i!) is a rational number. So according to your logic: e = sum{i = 0 .. n} 1/(i!) is also a rational number. > > Ok. The logic of finite complete sums of rational elements gives that > > the sum is rational. Using your logic we get that the complete (i.e. in [quoted text clipped - 3 lines] > > In fact there are no binary expansions of irrational numbers. What is the relevance? Where am I talking about binary expansions? > > > Either those sets obey that logic > > > or they do not exist in a science that is subject to the application > > > of logic. > > > > Ah, so 'e', 'pi', 'sqrt(2)' do not exist. Still I think you use at least > They exist as ideas but not as sums of series. So the equation e = sum{i = 0 .. oo} 1/(i!) is invalid in your logic? > > > Small wonder. There cannot be a complete list, because the existence > > > of a complete infinite linear set like N contradicts logic. > > > > Apparently your logic. What are the rules of inference in your logic? > > The union of a complete set of linear sets is one of the sets. Eh? That is not a rule of inference. > > And, also apparently, in your logic the length of the diagonal of a square > > with sides with size 1 does not exist. > > It exists, but not as binary or decimal expansion. Eh? Now suddenly sqrt(2) does exist. > > > That is a blatant lie. > > [quoted text clipped - 3 lines] > Then every element should be accessible. Why do you start always with > a diminishingly small one? Every element is accessible. And you start always at the beginning because there is no last one. > > > Every static linear set has a last element. > > > > As that is false in ZF... > > This shows that ZF deals with non-static sets if necessary. No. It just shows that your idea of sets is not the idea ZF has about sets. > > > If you say you cannot choose the last elemement because there is none, > > > then you apply potentially infinite sets. [quoted text clipped - 3 lines] > > A potentially infinite set is a set that has no last element. Ok. Than N is a potentially infinite set. > > > Then for every element that > > > you choose there is a larger one. If it has been there all time, why [quoted text clipped - 4 lines] > > That is potential infinity. Ok. Than N is a potentially infinite set that actually does exist in ZF. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <8d9c63c5-b605-4a98-81f2-815875aae...@21g2000vbk.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 4 Jun., 04:04, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 22 lines] > Through that list, given a number n, you find the n-th element of the list > without referring to any previous elements of the list. But you cannot find number n without referring to the numbers less than n. > To give an example, > let's have a list of positive rational numbers through the mapping given a [quoted text clipped - 5 lines] > element without knowing the 50th element. Where did I come at the 47th > rational in that list (which is 7/11)? How can you write 51 without knowing what it is? Of course you must count. In unary this is more difficult than in decimal or binary, but the principle is the same. How do you obtain this number 1111111111111111111111111111111111111 unless you count the digits? > > > You simply give a definition of a new number so it is clear > > > that it will be different from each of the omega lines, without checking [quoted text clipped - 5 lines] > I have not seen such a definition. What is the definition of "end of a > path"? Virgil calls it tail. That is a good name. I begin the construction with one path p_0. Then I construct another path p_1. All nodes that p_1 does not have in common with p_0, is the tail of p_1. All nodes of the tail are mapped on p_1. Then p_2 is constructed. All nodes of p_2 that differ from p_0 and p_1 are mapped on p_2. And so on. You see the ratio of paths and nodes per path at any stage during the construction is 0. http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#382,45,Folie 45 > > Every end of a path contains aleph_0 nodes. > > These nodes can be mapped on that path. Every node will eventually be [quoted text clipped - 3 lines] > Perhaps right, depends on how you actually do define things. But are there > nodes mapped to all paths? That is what you assert. Unless there is at least one node occupied by the path p_n that is not occupied by the paths p_0 to p_n-1, p_n is not a new path. > > And the reverse of being > > identical is being not identical. [quoted text clipped - 6 lines] > e = sum{i = 0 .. n} 1/(i!) > is also a rational number. I don't see a difference. But I can assuer you, there is no decimal expansion for irrational numbers. > > > Ok. The logic of finite complete sums of rational elements gives that > > > the sum is rational. Using your logic we get that the complete (i.e. in [quoted text clipped - 5 lines] > > What is the relevance? Where am I talking about binary expansions? Binary, decimal, whatever. It does not exist. > > > > Either those sets obey that logic > > > > or they do not exist in a science that is subject to the application [quoted text clipped - 4 lines] > > So the equation e = sum{i = 0 .. oo} 1/(i!) is invalid in your logic? sum{i = 0 .. oo} 1/(i!) is a finite word with about 12 symbols. I suffices to communicate what you mean. But what you think to have been abbreviating is not abbreviated by this. > > > > Small wonder. There cannot be a complete list, because the existence > > > > of a complete infinite linear set like N contradicts logic. [quoted text clipped - 4 lines] > > Eh? That is not a rule of inference. But it is true. > > > And, also apparently, in your logic the length of the diagonal of a square > > > with sides with size 1 does not exist. > > > > It exists, but not as binary or decimal expansion. > > Eh? Now suddenly sqrt(2) does exist. Not suddenly. It accept that it exists. But it has no binary or decimal expansion. > > > > That is a blatant lie. > > > [quoted text clipped - 6 lines] > Every element is accessible. And you start always at the beginning because > there is no last one. But you claim that there is always a larger one than you atrt with. Why that? > > > > A potentially infinite set is a set that has no last element. > > Ok. Than N is a potentially infinite set. That is true. But as such a set is never exhausted, Cantor waits and waits and waits ... > > > > Then for every element that > > > > you choose there is a larger one. If it has been there all time, why [quoted text clipped - 6 lines] > > Ok. Than N is a potentially infinite set that actually does exist in ZF. If one rejects actual infinity, then there is not "a continuous function nowhere differentiable". The actual infinity is a necessary condition of Cantor's diagonal proof of the power-set theorem. If one rejects actual infinity, then the theorem becomes unprovable and therefore there are not any cardinals greater than aleph_0. [Prof. Alexander A. Zenkin, Doctor of Physical and Mathematical Sciences, Leading Research Scientist of the Computer Center of the Russian Academy of Sciences.1937 - 2006] http://www.math.rutgers.edu/~zeilberg/fb68.html Regards, WM WM a écrit : > Virgil calls it tail. That is a good name. > I begin the construction with one path p_0. Then I construct another [quoted text clipped - 4 lines] > any stage during the construction is 0. > http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#382,45,Folie 45 Cranky stuff hosted on an academic server. Let's make info@hs-augsburg.de know about it. In article <bfcd6301-e7f0-4ced-906d-c111d378429e@j32g2000yqh.googlegroups.com>, > > In article > > <8d9c63c5-b605-4a98-81f2-815875aae...@21g2000vbk.googlegroups.com> WM [quoted text clipped - 29 lines] > But you cannot find number n without referring to the numbers less > than n. Nor without equally implicitly referring to the numbers greater than it. > > To give an example, > > let's have a list of positive rational numbers through the mapping given a [quoted text clipped - 9 lines] > How can you write 51 without knowing what it is? Of course you must > count. I can write lots of numbers by suitably randomized processes without even knowing the number I write, much less what numbers might precede it. > > > And in my binary tree I give a definition of an end of a path that > > > contains aleph_0 nodes. [quoted text clipped - 9 lines] > on p_2. And so on. You see the ratio of paths and nodes per path at > any stage during the construction is 0. Actually, WM is still suffering from a misunderstanding of the definition of a path. That definition does snot require any sort of "construction" such as Wm envisions. The "parent of" relation between nodes generates an "ancestor of" partial order on the nodes of a tree in the obvious way. A subset of the set of nodes of that tree is a path if and only if (1) it is totally ordered by the ancestor relation., so that for any two of its nodes one must be an ancestor of the other, and (2) it is maximal among such totally ordered sets, so that there is no node that can be to it added without destroying the total property. Every set of nodes in every tree satisfying both properties is a path, and every set of nodes failing either is not a path. Those properties, and only those, confer pathhood. For WM to claim that any set of nodes in a tree which satisfies those properties is not a path is both false and stupid, but he persists in doing so anyway. > I don't see a difference. But I can assuer you, there is no decimal > expansion for irrational numbers. WM's assurances have to often proved false to be trusted. Besides which, who says that the existence of a decimal expansion is required for a number to exist? For a particular form of numeral, but there is no requirement that any number have any particular form of name before it can exist as as number. At least not in serious mathematics. > If one rejects actual infinity, then there is not "a continuous > function nowhere differentiable". Then there are also no continuous functions at all, and nothing to replace them with. SignatureVirgil > > In article <8d9c63c5-b605-4a98-81f2-815875aae...@21g2000vbk.googlegroups.= > com> WM <mueck...@rz.fh-augsburg.de> writes: ... > > > > Because you do not check the lines in order. It is always your > > > > basic assumption that you first check the first line and after > > > > that the next line. That is wrong. > > > > > > That is necessary because you cannot find the n-th line unless you > > > know the line number n - 1 or some equivalent mark. But why is getting the number n in any way related to the checking of the previous lines? > > You are wrong. A list is a mapping from N to the elements of the list. > > Through that list, given a number n, you find the n-th element of the list > > without referring to any previous elements of the list. > > But you cannot find number n without referring to the numbers less > than n. That does not mean that you check the line with number n. So you assertion that you need to check all lines in order is false. > > To give an example, > > let's have a list of positive rational numbers through the mapping given [quoted text clipped - 11 lines] > 1111111111111111111111111111111111111 > unless you count the digits? So you agree now that you can check the n-th element of the list before checking any previous elements of the list? (How you get at n is irrelevant here, because getting at n does not involve checking lines preceding the n-th line.) Pray use tail in the future rather than end, to avoid confusion. > > > Every end of a path contains aleph_0 nodes. > > > These nodes can be mapped on that path. Every node will eventually be [quoted text clipped - 6 lines] > Unless there is at least one node occupied by the path p_n that is not > occupied by the paths p_0 to p_n-1, p_n is not a new path. Assuming countability again. Suppose we have the set of paths where each path goes to the right after some specific (for that path) node. Is there a node in your tree that is not occupied by any of the paths in that set? Are there possible other paths that are *not* in that set of paths (like a path that alternates going left and right)? > > Not in this case bacause you apply the words to different things. > > *Unless* you assume that what is valid in finite cases also is valid [quoted text clipped - 6 lines] > I don't see a difference. But I can assuer you, there is no decimal > expansion for irrational numbers. Where am I talking about decimal expansions? According to you the in finite sum above (which equals e by definition) is rational. Is that true? > > > > Ok. The logic of finite complete sums of rational elements gives > > > > that the sum is rational. Using your logic we get that the complete [quoted text clipped - 7 lines] > > Binary, decimal, whatever. It does not exist. I am not talking about whatever expansion. Pray keep to the question. According to your logic the complete (i.e. in the limit) sum of rational elements is also rational. And so by that logic, e, pi and whatever are rational and all numbers we do use are rational. > > > > Ah, so 'e', 'pi', 'sqrt(2)' do not exist. Still I think you use > > > > at least [quoted text clipped - 5 lines] > suffices to communicate what you mean. But what you think to have been > abbreviating is not abbreviated by this. I do not think that you know what that abbreviation stands for. According to a statement you have posted quite some time ago it means (according to you) the infinite sum of the elements 1/(i!). But now you state it does not exist. Are you able to explain? > > > > Apparently your logic. What are the rules of inference in your > > > > logic? [quoted text clipped - 4 lines] > > But it is true. I asked for the rules of inference, so apparently you are not able to supply the rules of inference in your logic. Ny question again, what are the rules of inference in your logic? If you are not willing to supply them it is impossible to follow your logic (apparently it consists of some ad-hoc statements that must be true). > > > > And, also apparently, in your logic the length of the diagonal of > > > > a square with sides with size 1 does not exist. [quoted text clipped - 5 lines] > Not suddenly. It accept that it exists. But it has no binary or > decimal expansion. What is the relevance to have binary or decimal expansion? 1/3 does not have a binary and decimal expansion either (in your sense of not having it). > > > > > That is a blatant lie. > > > > [quoted text clipped - 9 lines] > But you claim that there is always a larger one than you atrt with. > Why that? Because that is the definition of N. > > > A potentially infinite set is a set that has no last element. > > > > Ok. Than N is a potentially infinite set. > > That is true. But as such a set is never exhausted, Cantor waits and > waits and waits ... I wonder at the relevance. > > > > Because in ZF a set does not need to have a last element. So > > > > whatever element you chose, there is *always* a larger one. [quoted text clipped - 9 lines] > theorem becomes unprovable and therefore there are not any cardinals > greater than aleph_0. Great, you finally admit it. You just reject the axiom of infinity. So ZF is not inconsistent, it is inconsistent with your rejection of the axiom of infinity. That is normal: a theory is inconsistent with the rejection of one of its axioms. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <bfcd6301-e7f0-4ced-906d-c111d3784...@j32g2000yqh.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 11 Jun., 15:46, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 9 lines] > But why is getting the number n in any way related to the checking of the > previous lines? Because by blind choice you cannot be sure to hit what you want. > > > You are wrong. A list is a mapping from N to the elements of the list. > > > Through that list, given a number n, you find the n-th element of the list [quoted text clipped - 5 lines] > That does not mean that you check the line with number n. So you assertion > that you need to check all lines in order is false. In unary or in von Neumann's representation there is the whole counting process incorporated in each number. In decimal, you need only count the powers of 10. Anyhow addressing a number means counting. > > > To give an example, > > > let's have a list of positive rational numbers through the mapping given [quoted text clipped - 16 lines] > here, because getting at n does not involve checking lines preceding the > n-th line.) Of course you can check the n-th element before checking any other. But in order to find the n-th element, you have to count from 1 to n, either in unary or at least in the powers of 10. Therefore it is not correct to talk about simultaneity. > Pray use tail in the future rather than end, to avoid confusion. So Ido. > > > > Every end of a path contains aleph_0 nodes. > > > > These nodes can be mapped on that path. Every node will eventually be [quoted text clipped - 8 lines] > > Assuming countability again. Countability of nodes only. I am only interested to cover every node, i.e., to exhaust the tree. I am a follower of Eudoxos. > Suppose we have the set of paths where each > path goes to the right after some specific (for that path) node. That would imply all paths with tails 111... No problem. > Is there > a node in your tree that is not occupied by any of the paths in that set? No, it isn't. Every path node is occupied by at least the head of a path. > Are there possible other paths that are *not* in that set of paths (like > a path that alternates going left and right)? It is said so. But there is no possibility, after having completed the construction, to introduce such a path. They have sneaked in. > > > Not in this case bacause you apply the words to different things. > > > *Unless* you assume that what is valid in finite cases also is valid [quoted text clipped - 10 lines] > in finite sum above (which equals e by definition) is rational. Is that > true? True is: There are only rational sums of decimal or binary expansions. > > > > > Ok. The logic of finite complete sums of rational elements gives > > > > > that the sum is rational. Using your logic we get that the complete [quoted text clipped - 12 lines] > elements is also rational. And so by that logic, e, pi and whatever are > rational and all numbers we do use are rational. No. The limit is not rational. But the limit cannot be represented by a sequence. Therefore the limit is not in any Cantor-list. That is the big error of set theory: to believe that irrational numbers have rational representations. Every binary or decimal sequence is a rational representation. I skip the rest because the main point now has become fairly clear: The paths in the binary tree. We should concentrate on that problem. Regards, WM In article <55abc2e2-0ae1-4e8d-b05d-79761a6de727@m19g2000yqk.googlegroups.com>, > In unary or in von Neumann's representation there is the whole > counting process incorporated in each number. In decimal, you need > only count the powers of 10. Anyhow addressing a number means > counting. I can address sqrt(2) other than by counting. > Of course you can check the n-th element before checking any other. > But in order to find the n-th element, you have to count from 1 to n, > either in unary or at least in the powers of 10. Therefore it is not > correct to talk about simultaneity. The problem of accessing a number is quite different from the problem of its existence. There are even numbers which exist in certain contexts none of which are individually accessible. > > Assuming countability again. > > Countability of nodes only. I am only interested to cover every node, > i.e., to exhaust the tree. I am a follower of Eudoxos. Countability of members of set of nodes does not imply countability of subsets of that set. > > Suppose we have the set of paths where each > > path goes to the right after some specific (for that path) node. [quoted text clipped - 12 lines] > It is said so. But there is no possibility, after having completed the > construction, to introduce such a path. They have sneaked in. However sneaky, they are there, in the sense that there are sets of nodes that satisfy the definition of being paths that were not built in by WM's construction but which appeared despite he attempts to eliminate them. So to keep those unwanted paths out, WM will have to redefine paths so that only sets of nodes explicitly constructed can be counted as paths. > True is: There are only rational sums of decimal or binary expansions. True is: any sequence of decimal digits following "0." which can be unambiguously described defines a real number in [0,1]. And that includes lots of irrationals. > No. The limit is not rational. But the limit cannot be represented by > a sequence. Therefore the limit is not in any Cantor-list. That is the > big error of set theory: to believe that irrational numbers have > rational representations. Every binary or decimal sequence is a > rational representation. There are real numbers which have precise definitions but no known decimal expansions. WM conflates a particular form of number name with the number itself. A name is not the thing named. And there are, in general, lots of different names for any number, any one of which suffices to establish its existence. > I skip the rest because the main point now has become fairly clear: > The paths in the binary tree. We should concentrate on that problem. As soon as one has the set of all nodes of members of an infinite sequence of nested finite binary trees, one has a countable, partially ordered by the transitive closure of the 'parent of' relation set of nodes in which every maximal totally ordered subset is, by definition, a path. So there are uncountably many such paths in that tree. SignatureVirgil > There are real numbers which have precise definitions but no known > decimal expansions. May be. But those numbers are not able to participate in Cantor's list, neither as entries nor as anti diagonal- > WM conflates a particular form of number name with the number itself. > A name is not the thing named. Cantor's list requires decimal representations. > And there are, in general, lots of different names for any number, any > one of which suffices to establish its existence. But not sufficient to apply diagonalization. > > I skip the rest because the main point now has become fairly clear: > > The paths in the binary tree. We should concentrate on that problem. [quoted text clipped - 6 lines] > > So there are uncountably many such paths in that tree. non sequitur. Regards, WM In article <7d2446c5-4d49-454b-9af5-43881655b98e@h2g2000yqg.googlegroups.com>, > > There are real numbers which have precise definitions but no known > > decimal expansions. > > May be. But those numbers are not able to participate in Cantor's > list, neither as entries nor as anti diagonal- Cantor's remarks on lists of binary sequences is not about decimal expansions of real numbers. > > WM conflates a particular form of number name with the number itself. > > A name is not the thing named. > > Cantor's list requires decimal representations. Caantor's remarks about lists refers only to liss of binary sequences. It was not Cantor who revised those comments to apply to decimal expansions > > And there are, in general, lots of different names for any number, any > > one of which suffices to establish its existence. > > But not sufficient to apply diagonalization. Which is irrelevant here. Diagonalization only show that decimals aren't enough. The existence of other numbers which cannot be decimalized is just a bonus. > > > I skip the rest because the main point now has become fairly clear: > > > The paths in the binary tree. We should concentrate on that problem. [quoted text clipped - 8 lines] > > non sequitur. Follows from Cantor's anti-diagonal binary proof. SignatureVirgil ... > > > > > > Because you do not check the lines in order. It is always > > > > > > your basic assumption that you first check the first line [quoted text clipped - 7 lines] > > Because by blind choice you cannot be sure to hit what you want. Why does it imply checking the previous lines? Why do you not answer that question? > > > But you cannot find number n without referring to the numbers less > > > than n. [quoted text clipped - 6 lines] > only count the powers of 10. Anyhow addressing a number means > counting. And that *still* does not imply checking. So you assertion that you need to check all lines in order is false... > > So you agree now that you can check the n-th element of the list before > > checking any previous elements of the list? (How you get at n is irrel= [quoted text clipped - 6 lines] > either in unary or at least in the powers of 10. Therefore it is not > correct to talk about simultaneity. But the whole point is that in Cantor's argument, whatever n we choose, it is shown that the line numbered with that n is not in the list. This is true for an arbitrary n. There is no specific line that is checked. And how we come to the n-th line is also irrelevant. For any n, the n-th line is not equal to the diagonal. > > > > Perhaps right, depends on how you actually do define things. But > > > > are there nodes mapped to all paths? That is what you assert. [quoted text clipped - 6 lines] > Countability of nodes only. I am only interested to cover every node, > i.e., to exhaust the tree. I am a follower of Eudoxos. You have to exhaust the paths, not the nodes. If you think that exhausting the nodes implies exhausting the paths you are assuming that the set of paths is countable. > > Suppose we have the set of paths where each > > path goes to the right after some specific (for that path) node. [quoted text clipped - 12 lines] > It is said so. But there is no possibility, after having completed the > construction, to introduce such a path. They have sneaked in. And so paths have sneaked in that were not introduced by the construction, and so when you use the paths given in the construction to map to nodes you do not show that the set of paths is countable, only that the subset of paths that you used in the construction is countable. There are other paths, that have sneaked in, that are not mapped to a node, and can not be mapped to a node to which no path has yet been mapped. So you have not proven the the set of paths is countable. > > > > Not in this case bacause you apply the words to different things. > > > > *Unless* you assume that what is valid in finite cases also is valid [quoted text clipped - 12 lines] > > True is: There are only rational sums of decimal or binary expansions. That is (again) not an answer to my question. By your logic e is a rational number. Because that is what your logic dictates. That is, every sum of a finite initial set of the summands is rational, and so according to your logic also the complete sum is rational. Do you indeed think e is rational? > > > > What is the relevance? Where am I talking about binary expansions? > > > [quoted text clipped - 7 lines] > No. The limit is not rational. But the limit cannot be represented by > a sequence. We have: sum{i = 0 .. n} 1/(i!) is rational for each n. By *your* reasoning so also sum{i = 0 .. oo} 1/(i!) is rational, and so e is rational. You use that reasoning when you state union{i = 0 .. n} FISON(i) is a FISON and so union{i = 0 .. oo} FISON(i) os a FISON. I see no reason why you get at the conclusion in the second case but not in the first case. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <55abc2e2-0ae1-4e8d-b05d-79761a6de...@m19g2000yqk.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 12 Jun., 04:54, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 13 lines] > Why does it imply checking the previous lines? Why do you not answer that > question? You cannot check line n without knowing line n-1. > > > > But you cannot find number n without referring to the numbers less > > > > than n. [quoted text clipped - 8 lines] > > And that *still* does not imply checking. Counting is checking: how many. > > Of course you can check the n-th element before checking any other. > > But in order to find the n-th element, you have to count from 1 to n, [quoted text clipped - 3 lines] > But the whole point is that in Cantor's argument, whatever n we choose, it > is a last and largest member of a finite set. > shown that the line numbered with that n is not in the list. This is > true for an arbitrary n that is a last and largest member of a finite set. > There is no specific line that is checked. And > how we come to the n-th line is also irrelevant. For any n, the n-th line > is not equal to the diagonal. And for any n, without counting, the next line is equal to the diagonal in this example: 0.0 0.1 0.11 0.111 ... Why do you think that this is not contradicting Cantor? > > > > > Perhaps right, depends on how you actually do define things. But > > > > > are there nodes mapped to all paths? That is what you assert. [quoted text clipped - 10 lines] > the nodes implies exhausting the paths you are assuming that the set of paths > is countable. No, I am showing that the set of paths is counatble. > > It is said so. But there is no possibility, after having completed the > > construction, to introduce such a path. They have sneaked in. [quoted text clipped - 6 lines] > to a node to which no path has yet been mapped. So you have not proven the > the set of paths is countable. If you believe in sneeking (Virgil claimed that they say write it with double e) then you believe in ghosts. I do not. Constructing the tree from the list is nothing but (or at least can be done by) writing some digits or bits in a more economical way. This does not increase the number of sequences or paths. > > > Where am I talking about decimal expansions? According to you the > > > in finite sum above (which equals e by definition) is rational. Is [quoted text clipped - 4 lines] > That is (again) not an answer to my question. By your logic e is a rational > number. No, the contrary can be proven. e is not rational, but there is no binary sequence for e. There are only approximations. But they are not suitable to occur in Cantor's list. > Because that is what your logic dictates. That is, every sum of a > finite initial set of the summands is rational, and so according to your logic > also the complete sum is rational. Do you indeed think e is rational? No. See above. Do you indeed think that you can communicate e to some person by means of a bit sequence? Regards, WM In article <046c5a1d-9623-406f-88f4-b79b4447dcfc@r34g2000vba.googlegroups.com>, > > But the whole point is that in Cantor's argument, whatever n we choose, it > > is > > a last and largest member of a finite set. Also the first and smallest member of an infinite set. > > shown that the line numbered with that n is not in the list. This is > > true for an arbitrary n > > that is a last and largest member of a finite set. Also the first and smallest member of an infinite set. So what? > > There is no specific line that is checked. And > > how we come to the n-th line is also irrelevant. For any n, the n-th line [quoted text clipped - 10 lines] > > Why do you think that this is not contradicting Cantor? Because Cantor's diagonal proof only refers to infinite binary sequences, which what WM listed are NOT, and any infinite binary sequence is automatically different from every finite sequence such as the ones WM has listed. > No, I am showing that the set of paths is counatble. Not in this world! Your set of FINITE paths is countable, but its countability is irrelevant to the countability of a set of paths that it is disjoint from. > > > It is said so. But there is no possibility, after having completed the > > > construction, to introduce such a path. They have sneaked in. [quoted text clipped - 10 lines] > If you believe in sneeking (Virgil claimed that they say write it with > double e) then you believe in ghosts. I do not. How is it creating a ghost to note that an infinite binary string with infinitely 1's is not in any set of infinite binary strings each having at most finitely many 1's? Wm is becoming delusional again. Or is it "still"? > Constructing the tree from the list is nothing but (or at least can be > done by) writing some digits or bits in a more economical way. This > does not increase the number of sequences or paths. Constructing a tree requires constructing its node set together with the appropriate "left child", "right child" and "parent of" relations. Paths are merely a consequence of that construction. For all finite complete binary trees, any set of paths whose union is the nodes set of that tree must have a path for each leaf node. Since the set of paths bijects in an obvious way with the set of leaf nodes, one has to have every path in order to get every node. In maximal infinite binary trees, there are no leaf nodes, and there is no bijection between the set of all paths and any set of nodes. WM does not understand this difference. SignatureVirgil ... > > > > > > > > Because you do not check the lines in order. It is always > > > > > > > > your basic assumption that you first check the first line [quoted text clipped - 12 lines] > > You cannot check line n without knowing line n-1. But why does that imply *checking* line n-1? Why do you not answer that question? > > > > > But you cannot find number n without referring to the numbers > > > > > less than n. [quoted text clipped - 10 lines] > > Counting is checking: how many. Aha, now we come closer. So when you state that checking line n in the list implies checking previous lines, you are using the word "checking" with two different meanings. In the first instance you are meaning that it is checked whether it is different from the diagonal. In the second instance you are meaning that you count to that line. Why do you use the word "checking" the second time with a meaning that is not related to the standard meaning of checking? > > > Of course you can check the n-th element before checking any other. > > > But in order to find the n-th element, you have to count from 1 to n, [quoted text clipped - 5 lines] > > a last and largest member of a finite set. Irrelevant. > > shown that the line numbered with that n is not in the list. This is > > true for an arbitrary n > > that is a last and largest member of a finite set. Irrelevant. > > There is no specific line that is checked. And > > how we come to the n-th line is also irrelevant. For any n, the n-th line [quoted text clipped - 8 lines] > 0.111 > ... Eh? The diagonal is 0.111111111... I see no line that is equal to that diagonal. > Why do you think that this is not contradicting Cantor? Because it is nonsense. Or are you actually stating that when we check 0.1, 0.11 is equal to 0.111111111...? > > > > > > Perhaps right, depends on how you actually do define things. > > > > > > But are there nodes mapped to all paths? That is what you [quoted text clipped - 13 lines] > > No, I am showing that the set of paths is counatble. No, you are asuming that you can exhaust the paths by exhausting the nodes, this implies that you are assuming that the set of paths is countable, because the two statements are equivalent. > > > It is said so. But there is no possibility, after having completed the > > > construction, to introduce such a path. They have sneaked in. [quoted text clipped - 9 lines] > If you believe in sneeking (Virgil claimed that they say write it with > double e) then you believe in ghosts. I do not. It was *you* who wrote about that. See above, quoted here again: > > > It is said so. But there is no possibility, after having completed the > > > construction, to introduce such a path. They have sneaked in. so by your own statement you are believing in ghosts. > Constructing the tree from the list is nothing but (or at least can be > done by) writing some digits or bits in a more economical way. This > does not increase the number of sequences or paths. In that case 1/3 is not in your tree. And your tree is not a set of nodes, but a set of paths. That are two different things. > > > > Where am I talking about decimal expansions? According to you the > > > > in finite sum above (which equals e by definition) is rational. Is [quoted text clipped - 4 lines] > > That is (again) not an answer to my question. By your logic e is a > > rational number. > No, the contrary can be proven. e is not rational, but > there is no binary sequence for e. There are only approximations. But > they are not suitable to occur in Cantor's list. What are you rambling about? I am not talking about binary expansion. I am talking about a special series of sums. > > Because that is what your logic dictates. That is, every sum of a > > finite initial set of the summands is rational, and so according to your [quoted text clipped - 3 lines] > No. See above. Do you indeed think that you can communicate e to some > person by means of a bit sequence? I am not talking about a bit sequence, I am talking about an infinite sum. To recap: *You* state: union{i = 0..n} FISON(i) is a FISON and so: union{i = 0..oo} FISON(i) is a FISON With exactly the same reasoning I get: sum{i = 0..n} 1/(i!) is a rational number and so: sum{i = 0..oo} 1/(i!) is a rational number. So why is the reasoning correct in the first case and wrong in the second case (according to you)? A mathematician will tell you that the reasoning is wrong in both cases. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <046c5a1d-9623-406f-88f4-b79b4447d...@r34g2000vba.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 19 Jun., 16:35, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 18 lines] > But why does that imply *checking* line n-1? Why do you not answer that > question? It implies knowing line n-1. It implies counting till that number. Regards, WM In article <8b2855cd-18d0-44b1-be59-8dd3d8dabbb7@v2g2000vbb.googlegroups.com>, > > In article > > <046c5a1d-9623-406f-88f4-b79b4447d...@r34g2000vba.googlegroups.com> WM [quoted text clipped - 26 lines] > > It implies knowing line n-1. It implies counting till that number. It only implies knowing that there is a line n-1 for all but one n. SignatureVirgil ... > > > > > > > > > > Because you do not check the lines in order. > > > > > > > > > > It is always your basic assumption that you [quoted text clipped - 4 lines] > > > > > > > > > line unless you know the line number n - 1 or some > > > > > > > > > equivalent mark. ... > > > > Why does it imply checking the previous lines? Why do you not answer > > > > that question? [quoted text clipped - 5 lines] > > It implies knowing line n-1. It implies counting till that number. And still no answer to my question, and what you write here is wrong. In the bijection of the rationals > 0 and the naturals I have so often shown, to know the rational maped to 66 you need only know the rationals mapped to lines 1, 2, 4, 8, 16, 32 and 33. So I need not to know the rational mapped to 65. But my question was: "why is knowing a line the same as checking a line"? Because you originally asserted: "when checking line n of Cantor's list you need to check all previous lines". What is that? Nonsense or not? Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > > > It implies knowing line n-1. It implies counting till that number. [quoted text clipped - 4 lines] > to lines 1, 2, 4, 8, 16, 32 and 33. So I need not to know the rational > mapped to 65. How did you show that? However you agree to a process of counting. > But my question was: "why is knowing a line the same as checking a line"? It is by no means the same. But knowing a line is prerequisite to checking it. > Because you originally asserted: "when checking line n of Cantor's list > you need to check all previous lines". Here "checking" means checking its position, not checking ts contents. Regards, WM > > > It implies knowing line n-1. It implies counting till that number. > > [quoted text clipped - 5 lines] > > How did you show that? However you agree to a process of counting. How I did show that? Did you ever study the mapping I provided? I think not. > > But my question was: "why is knowing a line the same as checking a line"? > [quoted text clipped - 5 lines] > > Here "checking" means checking its position, not checking ts contents. Apparently you are deliberately obtuse, using words with two different meanings in the same sentence. Because that "checking" implied the comparing to the diagonal, not just checking its position. Or do you consider checking positions of lines sufficient to decide that the line is not equal to the diagonal? Try to free yourself from your obsession to use common words with different meanings in a single paragraph, or even sentence. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > On 22 Jun., 15:36, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > ... [quoted text clipped - 27 lines] > Because you originally asserted: "when checking line n of Cantor's list > you need to check all previous lines". What is that? Nonsense or not? Whether or not it is nonsense, it is irrelevant, since the Cantor argument applies to all lines simultaneoulsy with a single rule. SignatureVirgil 1 2 3 4 5 6 7 8 9 10 Next > > > > > > > > > > > > Because you do not check the lines in order. > > > > > > > > > > > > It is always your basic assumption that you [quoted text clipped - 33 lines] > > - Show quoted text - If it is known that if any NP-complete language is sparse (contains no more than a polynomial number of strings of length n), then P = NP. [BH08] improved this result, showing that if any language in NP has an NP-hard set of subexponential density, then coNP is contained in NP/ poly and thus, by [Yap82], PH collapses to the third level. Conceptually, a decision problem is a problem that takes as input some string, and outputs "yes" or "no". If there is an algorithm (say a Turing machine, or a computer program with unbounded memory) which is able to produce the correct answer for any input string of length Failed to parse (<math_output_error>): n in at most c \cdot n^k steps, where k and Failed to parse (<math_output_error>): c are constants independent of the input string, then we say that the problem can be solved in polynomial time and we place it in the class P. Formally, P is defined as the set of all languages which can be decided by a deterministic polynomial-time Turing machine. That is, P = {L:L = L(M) for some deterministic polynomial-time Turing machine M} where L(M) = \{ w\in\Sigma^{*}: M \text{ accepts } w \} and a deterministic polynomial-time Turing machine is a deterministic Turing machine M which satisfies the following two conditions: 1. M halts on all input w; and 2. there exists k \in N such that T_{M}(n)\in\; O(nk), where T_{M}(n) = \max\{ t_{M}(w) : w\in\Sigma^{*}, \left|w \right| = n \} and tM(w) = number of steps M takes to halt on input w. NP can be defined similarly using nondeterministic Turing machines (the traditional way). However, a modern approach to define NP is to use the concept of certificate and verifier. Formally, NP is defined as the set of languages over a finite alphabet that have a verifier that runs in polynomial time, where the notion of "verifier" is defined as follows. Let Failed to parse (<math_output_error>): L be a language over a finite alphabet, Σ. L\in\mathbf{NP} if, and only if, there exists a binary relation R \subset\Sigma^{*}\times\Sigma^{*} and a positive integer k such that the following two conditions are satisfied: 1. For all x\in\Sigma^{*}, x\in L \Leftrightarrow\exists y\in\Sigma^ {*} such that (x,y)\in R\; and \left|y\right|\in\;O(\left|x\right|^ {k}); and 2. the language L_{R} = \{ x\# y:(x,y)\in R\} over \Sigma\cup\{\#\} is decidable by a Turing machine in polynomial time. A Turing machine that decides LR is called a verifier for L and a y such that (x,y)\in R is called a certificate of membership of x in L. In general, a verifier does not have to be polynomial-time. However, for L to be in NP, there must be a verifier that runs in polynomial time. --MartinMichaelMusatov 07:18, 24 February 2009 (UTC) NPC: NP-Complete The class of decision problems such that (1) they're in NP and (2) every problem in NP is reducible to them (under some notion of reduction). In other words, the hardest problems in NP. Two notions of reduction from problem A to problem B are usually considered: 1. Karp or many-one reductions. Here a polynomial-time algorithm is given as input an instance of problem A, and must produce as output an instance of problem B. 2. Turing reductions, in this context also called Cook reductions. Here the algorithm for problem B can make arbitrarily many calls to an oracle for problem A. Some examples of NP-complete problems are discussed under the entry for NP. The classic reference on NPC is [GJ79]. Unless P = NP, NPC does not contain any sparse problems: that is, problems such that the number of 'yes' instances of size n is upper- bounded by a polynomial in n [Mah82]. A famous conjecture [BH77] asserts that all NP-complete problems are polynomial-time isomorphic -- i.e. between any two problems, there is a one-to-one and onto Karp reduction. If that's true, the NP-complete problems could be interpreted as mere "relabelings" of one another. NP-complete problems are p-superterse unless P = NP [BKS95]. This means that, given k Boolean formulas F1,...,Fk, if you can rule out even one of the 2k possibilities in polynomial time (e.g., "if F1,...,Fk-1 are all unsatisfiable then Fk is satisfiable"), then P = NP. ------------------------------------------------------------------------------------------------------------ > In article <KM22v0...@cwi.nl>, "Dik T. Winter" <Dik.Win...@cwi.nl> > wrote: > > > In article <8b2855cd-18d0-44b1-be59-8dd3d8dab...@v2g2000vbb.googlegroups.com> > > > On 22 Jun., 15:36, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > ... Results 1 - 10 for cantor p=np. (0.25 seconds) Cantor's Non-Diagonal Proof « Gödel's Lost Letter and P=NP“No one” referred to Leopold Kronecker who thought Cantor a “corrupter of youth”, and to Henri Poincaré who thought Cantor's set theory a malady that would ... rjlipton.wordpress.com/2009/04/18/cantors-non-diagonal-proof/ Re: Cantor's argument is erroneous Jun 14, 2009 ... Gravitational Cosmic Lensing os very handy for Musatov's proof P=NP<<» ... May 27 by Martin Musatov - 9 messages - 5 authors. Cantor's ... sci.tech-archive.net/Archive/sci.math/2009-06/msg01643.html Musatov "null zero rrror!" sets dimensional time block, to ...P=NP For example, one of his results is the famous proof, in 1969, . <...> http://groups.google.com/group/comp.theory?lnk=rgr. Cantor's argument is ... Cantor's, Godel's, Tarski's, and Turing's Apr 16, 2009 ... to revise current interpretations of Cantor's, Godel's, .... 4.4 P = NP under a constructive interpretation of Peano. Arithmetic ... Cantor's argument is erroneous - sci.math | Google Groups Jun 13, 2009 ... Message from discussion Cantor's argument is erroneous ... So it seems likely enough that you, the guy flogging the P=NP ... If not empty, NP−P is topologically large Zim and proves that, in a combination of Cantor and superset topologies, the set NP−P, if not empty, is of the second (Baire) category, while NP-complete ... P, NP, Co-NP and weak systems of arithmetic - Elsevier(b) For some 1 e N, PA f U( ,, k 1 Q (m, k) p PA NP = CoNP. ...... function (x)Z defined in result 3.5, not the inverse functions of the Cantor pairing). ... An algebraic version of Cantor-Bendixson analysisdescribe the Cantor- Bendixson pair (i.e. the derivative and nucleus) ... point p of S let p = p- n p where p = A{X c CS : p E X} , pO = {U 0S : p E U}. ... Cook's Class Contains Pi « Gödel's Lost Letter and P=NPClearly, this can be computed in polynomial time to {p} bits of precision in space at most .... Cantor's Non-Diagonal Proof « Gödel's Lost Letter and P=NP ... The Cantor-Lebesgue and Denjoy-Luzin properties for double systems ...Cantor--Lebesgue and Denjoy--Luzin properties. 299. We denote by .... lim rain (mp, np) = poo and. < 1. (3.2) flo,, d(x)-_ -7 (p = 1,2 .... ). ... > There is no difference between finite unions and infinite > unions. There should not be, but there is --- at least if ZF is taken to be true. The union of FISONs (of natural numbers or of other finite linear sets) is the last FISON. Regards, WM In article <5b9b7d5b-11d3-4125-8052-ad45e6e7a2ec@x5g2000yqk.googlegroups.com>, > > There is no difference between finite unions and infinite > > unions. [quoted text clipped - 3 lines] > The union of FISONs (of natural numbers or of other finite linear > sets) is the last FISON. If WM claims the existence of a last FISON, let him present it to us. But outside of WM's U.N. Realistic world, every FISON is a proper subset of another FISON, so that the union of all of them cannot itself be one of them. Or does WM's set theory require existence of sets which are proper subsets of themselves? SignatureVirgil > > There is no difference between finite unions and infinite > > unions. > > There should not be, but there is --- at least if ZF is taken to be > true. You are again wrongly interpreting things. There is no question about ZF being true or not. ZF is only one of the many possible theories. Mathematics is not a science that tries to find truth, it is a science where it is determined what a given set of axioms (i.e. presupposed valid statements) leads to. > The union of FISONs (of natural numbers or of other finite linear > sets) is the last FISON. ZF does agree, when there is a last FISON. Your problem is that the axiom of infinity states that here is no last FISON of all FISONs. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > In article <5b9b7d5b-11d3-4125-8052-ad45e6e7a...@x5g2000yqk.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 3 Jun., 04:25, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: [quoted text clipped - 6 lines] > You are again wrongly interpreting things. There is no question about ZF > being true or not. ZF is only one of the many possible theories. It is an impossible theory because finished infinity is impossible. > Mathematics is not a science that tries to find truth, There is a mathematics, namely basic arithmetic of integers, that is right. And there are some logical rules that are right. And from these foundations some theorems can be obtained that are right. There is no need and no space for any axiom at all. There is no need , in particular, for "a set N" with some "binary relations". It is impossible, to change that stuff. And Hilberts famous model including 1 + 1 = 0 does nothing but express the display of the last wheel of a binary calculator (or Leibniz's calculator that all time long had problems with the digit transfer). > it is a science where > it is determined what a given set of axioms (i.e. presupposed valid statements) > leads to. A given set of axioms is sometimes impossible. This is so with the axiom of infinity that Zermelo created, misunderstanding Dedekind's idea of potential infinity. > > The union of FISONs (of natural numbers or of other finite linear > > sets) is the last FISON. > > ZF does agree, when there is a last FISON. Your problem is that the axiom > of infinity states that here is no last FISON of all FISONs. That is not my problem. For potentially infinite sets it is not a problem at all. For a completely existing set ZF is wrong. But that is usually veiled by applying potential infinity whenever the problem occurs. Regards, WM In article <278180e4-8675-40bc-871c-a84af38f6635@k20g2000vbp.googlegroups.com>, > > In article > > <5b9b7d5b-11d3-4125-8052-ad45e6e7a...@x5g2000yqk.googlegroups.com> WM [quoted text clipped - 10 lines] > > It is an impossible theory because finished infinity is impossible. What WM decalred to be impossible in WM' world of mathUnrealism is not binding anywhere else. > > Mathematics is not a science that tries to find truth, > > There is a mathematics, namely basic arithmetic of integers, that is > right. If so, the WM has no access to it. >And there are some logical rules that are right. If so, the WM has no access to them. > And from these > foundations some theorems can be obtained that are right. Without a comprehensive description of what one is required to accept as true, and reject as false, there is no way to tell which statements are theorems and which are not in that system. > There is no > need and no space for any axiom at all. If there are any statements which must be accepted in WM's "system" then those statements are 'axioms' of that system whether WM likes it or not. > There is no need , in > particular, for "a set N" with some "binary relations". It is > impossible, to change that stuff. And Hilberts famous model including > 1 + 1 = 0 does nothing but express the display of the last wheel of a > binary calculator (or Leibniz's calculator that all time long had > problems with the digit transfer). Without some explicit method for determining any naturals that are not explicitly listed, there will be a last natural beyond which not even WM's potential infinity can proceed, and that last one will be too small to allow much practical mathematics. > > it is a science where > > it is determined what a given set of axioms (i.e. presupposed valid > > statements) > > leads to. > > A given set of axioms is sometimes impossible. Then one ow does one decide which statements are theorems and which are not? > This is so with the > axiom of infinity that Zermelo created, misunderstanding Dedekind's > idea of potential infinity. Only in WM's world. > > > The union of FISONs (of natural numbers or of other finite linear > > > sets) is the last FISON. [quoted text clipped - 3 lines] > > That is not my problem. It is if you want to do actual mathematics instead of your own personal psuedomath. > For potentially infinite sets it is not a > problem at all. There is no self-consistent logical system which Wm has ever shown to allow such abominations. > For a completely existing set ZF is wrong. But that is > usually veiled by applying potential infinity whenever the problem > occurs. The thing is that ZF's alleged wrongness produces nothing that conflict with its axioms whereas WM's system's alleged rightness produces too many things that conflict with his own assumptions. SignatureVirgil > > In article <5b9b7d5b-11d3-4125-8052-ad45e6e7a...@x5g2000yqk.googlegroups.= > com> WM <mueck...@rz.fh-augsburg.de> writes: [quoted text clipped - 9 lines] > > It is an impossible theory because finished infinity is impossible. Again giving opinion only. > > Mathematics is not a science that tries to find truth, > > There is a mathematics, namely basic arithmetic of integers, that is > right. And there are some logical rules that are right. And from these > foundations some theorems can be obtained that are right. There is no > need and no space for any axiom at all. Oh. I would have thought that in that model the basic arithmetic of the integers forms the axioms... > There is no need , in > particular, for "a set N" with some "binary relations". It is > impossible, to change that stuff. And Hilberts famous model including > 1 + 1 = 0 does nothing but express the display of the last wheel of a > binary calculator (or Leibniz's calculator that all time long had > problems with the digit transfer). I think you are confusing a few things here. In your model you may not need a set N with binary relations, but mathematics tries to find the minimum that is required to get something done. From that follows fields like group theory, etc. > > it is a science where > > it is determined what a given set of axioms (i.e. presupposed valid [quoted text clipped - 3 lines] > axiom of infinity that Zermelo created, misunderstanding Dedekind's > idea of potential infinity. There is no misunderstanding. And that it is impossible is just opinion, nothing more. Like earlier people thought that non-Euclidean space was impossible. And some people thought that irrational numbers were impossible, or negative numbers, or imaginary numbers. > > > The union of FISONs (of natural numbers or of other finite linear > > > sets) is the last FISON. [quoted text clipped - 4 lines] > That is not my problem. For potentially infinite sets it is not a > problem at all. ZF does not know about the impossible set. > For a completely existing set ZF is wrong. Again, just opinion. > But that is > usually veiled by applying potential infinity whenever the problem > occurs. ZF does not apply something that is not defined in it. Signaturedik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > What is the relevance of this? The logical rules of unions state that when > you have a union of sets that union does contain an element if it is in one > of the sets. DAMN. This is the only reason I am still here (because you and Virgil are). YOU ARE TOO STUPID to be doing this! THERE ARE NO SUCH THINGS as "the logical rules of unions"! THERE IS AN *AXIOM* of union in ZFC! This axiom IS NOT a logical truth, or it would NOT NEED to exist at all! To the extent that "rules of unions" exist at all, THEY ARE *NOT* logical and MUST not be logical! IN ADDITION to the axiom of union, there is also, USUALLY, an axiom of pairing! THERE IS *NO LOGICAL* connection WHATEVER between the two! Either could get along FINE WITHOUT the other! But the axiom of pairing suffices to create ALL THE FINITE unions! YOU ONLY NEED the axiom of union FOR INFINITE unions! > There is no difference between finite unions and infinite > unions. THERE IS *SO*, TOO!! Finite unions can be built up ONE ELEMENT AT A TIME, THROUGH PAIRING! INFINITE UNIONS *CANNOT*!! THAT IS PRECISELY WHY WM IS WRONG to be generalizing from what happens with EVERY finite case TO the infinite case! a > But frequently I made use of what you call quatifier exchange and what > is allowed in case of complete linear sets. Nope. It is easy to show quantifier exchange is allowed for linear sets if and only if there is a largest element. Since complete linear sets without largest element exist, quantifier exchange is not valid. - William Hughes > > But frequently I made use of what you call quatifier exchange and what > > is allowed in case of complete linear sets. [quoted text clipped - 5 lines] > quantifier exchange is allowed for linear sets > if and only if there is a largest element. Rules of logic were obtained from finite sets. Rules of logic for linear sets were obtained from linear finite sets. Everything else is matheology of Wolkenkuckucksheim. Proof: A complete linear set without a last element is self contradictory, as can be proven by the only logical rules that must be accepted, namely those obtained from finite sets. Regards, WM In article <82f69cbc-0f68-4c74-a78a-c57fbf5d13fe@h18g2000yqj.googlegroups.com>, > > > But frequently I made use of what you call quatifier exchange and what > > > is allowed in case of complete linear sets. [quoted text clipped - 8 lines] > Rules of logic were obtained from finite sets. Rules of logic for > linear sets were obtained from linear finite sets. But we have linear (meaning well ordered) non-finite sets outside of WM's wee wee word of MathUnrealism. > Everything else is matheology of Wolkenkuckucksheim. > > Proof: A complete linear set without a last element is self > contradictory, as can be proven by the only logical rules that must be > accepted, namely those obtained from finite sets. If only finite sets can exist then there are only finitely many of them , and the union of all finitely many of them must itself be a finite set: a finite universe outside of which there can be nothing to add to it when the things inside are used up. But if there are only finitely many elements in that universal set, then WM's potentially infinite sets, being of necessity subsets of a finite universal set, can not be potentially infinite, as they must eventually exhaust their finite universal set. So WM's own rules kill off his idiotic notion of potential infiniteness. SignatureVirgil > But if there are only finitely many elements in that universal set, then > WM's potentially infinite sets, being of necessity subsets of a finite > universal set, can not be potentially infinite, as they must eventually > exhaust their finite universal set. Your universe of numbers is all numbers that you can construct. If you increase your capabilities, your universe grows. That's why it is infinite. Regards, WM In article <eebb8310-e58d-4f85-9eda-37af9b74b6c0@c19g2000yqc.googlegroups.com>, > > But if there are only finitely many elements in that universal set, then > > WM's potentially infinite sets, being of necessity subsets of a finite [quoted text clipped - 4 lines] > increase your capabilities, your universe grows. That's why it is > infinite. If all sets are finite, but ever changing as WM insists, then one can never have a set theory at all, as sets, including universal sets if any, do not change. Whatever WM is going on about, it is not sets. So in WM's world there can never be any sets. SignatureVirgil > In article > <eebb8310-e58d-4f85-9eda-37af9b74b...@c19g2000yqc.googlegroups.com>, [quoted text clipped - 11 lines] > never have a set theory at all, as sets, including universal sets if > any, do not change. This is stubborn nonsense. Of course sets are conceivable that can change. The complete sets like N have only been introduced because Cantor had a strong religious affinity. Regards, WM In article <c024871b-c949-4f54-afd4-7666b01189b8@y7g2000yqa.googlegroups.com>, > > In article > > <eebb8310-e58d-4f85-9eda-37af9b74b...@c19g2000yqc.googlegroups.com>, [quoted text clipped - 14 lines] > This is stubborn nonsense. Of course sets are conceivable that can > change. They have yet to be conceived in any mathematical set theory that I am acquainted with. The only possible construction that might serve would be set-valued functions of time, but even there, for any fixed time, there would have to be a fixed set. > The complete sets like N have only been introduced because > Cantor had a strong religious affinity. Wm must have a strong religious affinity of his own,believing God speaks directly to him, to hold that N cannot be a set. SignatureVirgil Virgil God proclaims: > In article > <c024871b-c949-4f54-afd4-7666b01189b8@y7g2000yqa.googlegroups.com>, [quoted text clipped - 31 lines] > -- > Virgil 1. When is truth vacuous? Is infinity a bunch of nothing?So to say that we can map the set N onto a unit length line would be .... that the statistical truth is truth, since we cannot examine every element of N? ...www.geocities.com/n_fold/vactruth.html - 32k - Cached - Similar pages - 2. Truth and Paradox: Solving the Riddles - Google Books Resultby Tim Maudlin - 2006 - Philosophy - 209 pagesBut if the rules of 2* are truth-preserving, this cannot occur. ... T(x)) DT(y)) is not true, so no set of truth-preserving rules can be used to prove it. ...books.google.com/books?isbn=0199203911... - 3. Truth and Beauty Bombs :: View topic - Can someone please explain ...4 posts - 2 authors - Last post: 6 FebNP is the set of problems where given a solution, you can check if the solution is valid ... for a given algorithm and input of length n, you can find how long the algorithm ... You cannot delete your posts in this forum ...www.truthandbeautybombs.com/bb/viewtopic.php?t=18587 - 27k - Cached - Similar pages - 4. Gödel's Theorems and Truth21 Apr 1998 ... Using improved symbolic logic, he and Alfred North Whitehead set out to do just that .... Gödel's theorem means that the universe cannot be a vast ... Spiritual truth, we are taught, can be apprehended only by the spirit ...www.rae.org/godel.html - 29k - Cached - Similar pages - 5. Degrees of Truth, Degrees of FalsityLet us use the symbol 'n' for this truth value and continue to use '1' and ... valued 0 or n — valid arguments cannot take us from something with truth to .... As can be seen, this set of truth values takes us beyond those present in ...www.amirrorclear.net/academic/ideas/degrees/ - 34k - Cached - Similar pages - 6. Fire Fighters For 9-11 Truth » Petition1 Sep 2008 ... The truth will set us free! The answer to 1984 is 1776!" .... "Firefighters should know more than anyone than fire can not collapse .... Patricia A Gala N/A. Patrick Holman, "The truth cannot be hacked or compromised. ...firefightersfor911truth.org/?page_id=469 - 59k - Cached - Similar pages - 7. Tractatus Logico-Philosophicus - Wikipedia, the free encyclopediaIt must set limits to what cannot be thought by working outwards through what can be ... Wittgenstein is to be credited with the invention of truth tables (4.31) and ... Wittgenstein's N-operator is however an infinitary analogue of the ... What Wittgenstein then goes on to show that this operator can cope with the ...en.wikipedia.org/wiki/Tractatus_Logico-Philosophicus - 63k - Cached - Similar pages - 8. Philosophy of mathematics: an anthology - Google Books Resultby Dale Jacquette - 2002 - Philosophy - 428 pagesI now wish to give the truth conditions for the language of set theory in a ... However, I cannot do this. The standard model of set theory is much more ...books.google.com/books?isbn=063121870X... - 9. Functional Completeness and Non-tukasiewiczian Truth Functionsto {Ί, D} results in a set that is functionally complete [2]. The question arises ... Let / be any pure three-valued truth function of degree n, and consider an ... We can write a representative formula R^ for row / where Ri has the value β on ... These results cannot be generalized to the ^-valued systems £ n ...projecteuclid.org/DPubS/Repository/1.0/ Disseminate?handle=euclid.ndjfl/1093883177&view=body&content- type=pdf_1 - Similar pages - by F T T T T - 1980 - Related articles - All 2 versions10. Math Forum - Ask Dr. Math Archives: College Logic/ Set Theory... [10/28/2002]: I cannot find any unsolved proofs that I can just solve for fun. ... Basic Truth Tables and Equivalents in Logic [05/23/2000]: What are the truth ... Can Rewriting P -> Q as ~Q - > ~P Lead to a False Conclusion? ... (c) The set of all subsets of N of size n, for some fixed (but arbitrary) n E N; ...mathforum.org/ library/drmath/sets/college_logic.html - 22k - Cached - Similar pages - Searches related to: N cannot can be a set truthbertrand russell truthself evident truthlogic and truth -- Virgil In article <351d0bd7-fe7d-4005-a10a-42b63a9037e5@q16g2000yqg.googlegroups.com>, > On 26 Mai, 04:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > In article [quoted text clipped - 16 lines] > Therefore I asked you: Show me a linear complete finite set, that > makes your claim [**] right and my claim [*] wrong. What Weyl may have said in no way limits logic, classical or otherwise, to finite sets. Or does WM claim that Weyl was speaking ex cathedra? > This is so simple that it should be understandable even for someone > who is not "very deep in logic". [quoted text clipped - 6 lines] > course the equivalence includes the implication [***] as well as the > implication [**] Then WM is not operating in the same world as everyone else. > > There is a trivial finite > > counter-example. Take three dice where on each of the sides one of the [quoted text clipped - 13 lines] > Of course there is no best die. Therefore this set is not linear. > Every finite set of natural numbers has a "best" number. It may have a largest, but without a definition of being "better" that is transitive, it need not have a "best". > Why do you bother with such nonsense examples? Why does WM bother with his much more nonsensical examples? I doubt that even WM knows. > But the answer is easy: Because you have no other examples. We have them, but WM has steadfastly refused to understand them so we are trying to simplify things to a point which WM can understand them. So far we have not been able to get down to his level. > > Doesn't it bother you that he gets letters from > > other mathematicians in Germany complaining [quoted text clipped - 11 lines] > But you think it would increase this value if I taught, as you > propose, that the sum of all natural numbers can be zero? Considering some of the things you do claim to teach, it would hardly change things. > Or if I > taught, contrary to fact, Cantor's claim that a real number is the > limit of its finite initial segments but all real numbers are not the > limits of all their finite initial segments? That may be WM's claim, but it is certainly not Cantor's. Cantor may have viewed reals as equivalence classes of Cauchy seequnces, or, more likely, as Dedekind cuts, but where did WM get his delusion that Cantor ever thought simultaneously that a real was something and at the same time was not that thing? SignatureVirgil > > > > The problem boils down to the following: > [quoted text clipped - 3 lines] > > > > You know: Classical logic was obtained from finite sets ... > > > > Show me a finite set that obeys [**] but not [*]. > > I said: For complete linear sets [*] is true. > > You said [*] is not true, but [**] is true. [quoted text clipped - 4 lines] > What Weyl may have said in no way limits logic, classical or otherwise, > to finite sets. Or does WM claim that Weyl was speaking ex cathedra? Weyl has recognized, by the end of his life (when obtaining his doctorate under Hilbert he was preoccupied with Hilbert's opinion), that logic is not included into the 10 commandments, but has to be constructed or obtained as an abstraction from the reality. Therefore we have no guarantee that it is suitable for sets that cannot belong to reality. And I added: We cannot use logical theorems that are counterfactual, i.e., contrary to what we obtained from reality. There is no model of a complete linear set in reality that makes [*] false and [**] true. But there are many models showing that [*] is true whenever [**] is true. > > No. I do not use the implication only, I use the full equivalence. Of > > course the equivalence includes the implication [***] as well as the > > implication [**] > > Then WM is not operating in the same world as everyone else. Most of your ilk intermingle the worlds in which they work. They work in potential infinity if the question is put whether there is a largest natural number. But they work in actual infinity if the Cantor- diagonal or any other irrational number is concerned. Only the possibility to complete it by a limit process, leads to the erroneous assumption of uncountability. This had already been recognized by the late Alexander Zenkin, one of the brave scientists who dared to condemn this hypocritical behaviour: Cantor's 'paradise' as well as all modern axiomatic set theory is based on the (self-contradictory) concept of actual infinity. Cantor emphasized plainly and constantly that all transfinite objects of his set theory are based on the actual infinity. Modern AST-people try to persuade us to believe that the AST does not use actual infinity. It is an intentional and blatant lie, since if infinite sets, X and N, are potential, then the uncountability of the continuum becomes unprovable, but without the notorious uncountablity of continuum the modern AST as a whole transforms into a long twaddle about nothing. > > Or if I > > taught, contrary to fact, Cantor's claim that a real number is the [quoted text clipped - 4 lines] > Cantor may have viewed reals as equivalence classes of Cauchy sequences, > or, more likely, as Dedekind cuts, but where did WM get his delusion. Cantor did not hold Dedekind's work in high esteem. Here is one example: Die Schrift von Dedekind „Was sind und was sollen die Zahlen" ist, wenn auch ihrer Tendenz nach, die Arithmetik rein logisch zu begründen, lobenswerth, nicht nach meinem Geschmack. ... Das künstliche System der 172 sich nur um das Elementarste und zum Theil Trivialste drehenden Dedekindschen Sätze scheint mir mehr geeignet, die Natur der Zahlen zu verdunkeln als sie aufzuhellen. [Letter from Cantor to Vivanti, 1888, April 2] Cantor used his Fundamentalreihen: Bemerkungen mit Bezug auf den Aufsatz: Zur Weierstraß-Cantorschen Theorie der Irrationalzahlen. [Math. Annalen Bd. 33, S. 476 (1889).] Es möge mir gestattet sein, nur ganz kurz auf die Bedenken zu antworten, welche Herr Illigens in bezug auf meine Theorie der Irrationalzahlen ausgesprochen hat. ... sqrt(3) ist also nur ein Zeichen für eine Zahl, welche erst noch gefunden werden soll, nicht aber deren Definition. Letztere wird jedoch in meiner Weise etwa durch (1,7, 1,73, 1,732, ...) befriedigend gegeben. Cantor talks about "his" theory of irrational numbers, neither Cauchy's nor Dedekind's. Regards, WM In article <2056d314-cabe-420c-a859-1b0d95e69fe6@i6g2000yqj.googlegroups.com>, > > > > > The problem boils down to the following: > > [quoted text clipped - 20 lines] > to reality. And I added: We cannot use logical theorems that are > counterfactual, i.e., contrary to what we obtained from reality. WM presumes that everything true can be, and must be, obtained only from reality, but the laws of logic are derived from thoughts of an ideal world and such thoughts are not obtained solely reality, but from largely from an unreal vision of the ideal. There > is no model of a complete linear set in reality that makes [*] false > and [**] true. But there are many models showing that [*] is true > whenever [**] is true. And many models for which "ExAy P(x,y) ==> AyEx P(x,y)" is false. > > > No. I do not use the implication only, I use the full equivalence. Of > > > course the equivalence includes the implication [***] as well as the [quoted text clipped - 8 lines] > possibility to complete it by a limit process, leads to the erroneous > assumption of uncountability. Then WM better confine his attentions to areas in which no limiting processes are wanted and no infinite sets are wanted, which excludes him from all calculus. > This had already been recognized by the late Alexander Zenkin, one of > the brave scientists who dared to condemn this hypocritical behaviour Scientists may mess with physics to their hearts content, but as scientists, have no business messing with pure mathematics. So if anyone introduced any "false logic", it would be physicists rather than mathematicians. SignatureVirgil > but the laws of logic are derived from thoughts of an ideal > world and such thoughts are not obtained solely reality, but from > largely from an unreal vision of the ideal. The result is in due shape. > There > [quoted text clipped - 3 lines] > > And many models for which "ExAy P(x,y) ==> AyEx P(x,y)" is false. Not linear models. And no others! > Then WM better confine his attentions to areas in which no limiting > processes are wanted and no infinite sets are wanted, which excludes him > from all calculus. Either Cantor’s diagonal proof shows that the limit of all omega indices can be reached by defining b_n =/= a_n for every n. Then my binary tree proof shows that the limit of all omega levels can be reached by showing that the number of distinct lines is countable at every level n. Then all reals are countable. (Only a very confused mind could consider the possibility that paths of the infinite tree and decimal expansions of real numbers might represent different mathematical objects). Or the limit of the paths in the binary tree does not yield all real numbers. Then Cantors diagonal proof does not establish complete real number either, and the proof is void. > > This had already been recognized by the late Alexander Zenkin, one of > > the brave scientists who dared to condemn this hypocritical behaviour > > Scientists may mess with physics to their hearts content, but as > scientists, have no business messing with pure mathematics. Mathematics is science done by scientists. Matheology is what you may have in mind. And in fact : That has as much to do with science as has astrology to with astronomy. > So if anyone introduced any "false logic", it would be physicists rather > than mathematicians. I should refrain from calling what you and your ilk do : matheology. It could be understood by lurkers as if your hobby was based upon logic. Regards, WM In article <503d7653-f1a6-4d24-b9eb-b39b34a78b5e@i6g2000yqj.googlegroups.com>, > > but the laws of logic are derived from thoughts of an ideal > > world and such thoughts are not obtained solely reality, but from [quoted text clipped - 11 lines] > > Not linear models. The naturals and the integers, and the set of unit fractions, and a lot of other models. > > Then WM better confine his attentions to areas in which no limiting > > processes are wanted and no infinite sets are wanted, which excludes him > > from all calculus. > > Either Cantor¹s diagonal proof shows that the limit of all omega > indices can be reached by defining b_n =/= a_n for every n. Or what? > Then my binary tree proof shows that the limit of all omega levels can > be reached by showing that the number of distinct lines is countable > at every level n. Such a "proof" is no proof, as it requires assumptions contrary to fact. Then all reals are countable. In WM's personal world of MathUnrealism, he may, if he chooses , declare that 2 = 1, but he has no power to declare anything new outside that world without better "proofs" than he has yet been able to create. > (Only a very confused > mind could consider the possibility that paths of the infinite tree > and decimal expansions of real numbers might represent different > mathematical objects). The WM must be saying that every path in a complete infinite binary tree "is identical to" the decimal expansion of a real number AND every decimal expansion of a real number "is identical to" a path in such a binary tree. Nonsense. For one thing, 0.1(0) and 0.0(1), where (x) indicates an infinite sequence of repetitions of x, represent the same real but different paths. > Or the limit of the paths in the binary tree does not yield all real > numbers. They can be made to produce reals, but to get a bijection between strings and reals is not anywhere as trivial as WM thinks it is, and there is no single way to do it that is more natural and obvious that anall others > Then Cantors diagonal proof does not establish complete real number > either, and the proof is void. Cantor's diagonal proof does not refer to real numbers at all. Suitable modifications of it done later by others also shows that no list of reals can exhaust all reals. But Cantor had already proved that separately before coming up with his diagonal argument. > > > This had already been recognized by the late Alexander Zenkin, one of > > > the brave scientists who dared to condemn this hypocritical behaviour [quoted text clipped - 3 lines] > > Mathematics is science done by scientists. That may be the egotist scientist's pint of view, but it is a false view. > Matheology is what you may > have in mind. And in fact : That has as much to do with science as has > astrology to with astronomy. What goes on in WM's wee weird world of MathUnrealism is not mathematics, so much as bad engineering. > > So if anyone introduced any "false logic", it would be physicists rather > > than mathematicians. > > I should refrain from calling what you and your ilk do : matheology. > It could be understood by lurkers as if your hobby was based upon > logic. I have all sorts of hobbies. One of them is revealing the foolishness of egotistists like WM. SignatureVirgil In article <503d7653-f1a6-4d24-b9eb-b39b34a78b5e@i6g2000yqj.googlegroups.com>, > > but the laws of logic are derived from thoughts of an ideal > > world and such thoughts are not obtained solely reality, but from [quoted text clipped - 11 lines] > > Not linear models. N is such a model, as are the set of all predecessors of any limit ordinal. > And no others! Actually as many other as there are limit ordinals, which is more than WM can count. > > Then WM better confine his attentions to areas in which no limiting > > processes are wanted and no infinite sets are wanted, which excludes him > > from all calculus. > > Either Cantor¹s diagonal proof shows that the limit of all omega > indices can be reached by defining b_n =/= a_n for every n. Or WM is again dead wrong and an a.s. SignatureVirgil |
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