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Math Forum / Mathematics / Undergraduate Math / February 2008Is this STRANGE topological space connected?
Consider R^2 with the standard topology, and C a subset of R^2 defined with the subspace topology (intersections of C with open sets from R^2) where: C = { (x , 0) : 0<= x <= 1} U { (1/n , y): n=1,2,3,..., 0<= y <=1} Is this "comb-like" space connected? On Thu, 7 Feb 2008 10:20:24 -0800 (PST), in alt.math.undergrad: > Consider R^2 with the standard topology, and C a subset of R^2 defined > with the subspace topology (intersections of C with open sets from > R^2) where: > C = { (x , 0) : 0<= x <= 1} U { (1/n , y): n=1,2,3,..., 0<= y <=1} > Is this "comb-like" space connected? Yes. It's even path connected. Brian > On Thu, 7 Feb 2008 10:20:24 -0800 (PST), > "David.Fail...@gmail.com" <David.Fail...@gmail.com> wrote in [quoted text clipped - 10 lines] > > Brian Proof? Presumably by contradiction, assuming it is disconnected. But would a separation on C induce one on R^2? In article <7ff5bd92-9576-4a6e-a1a8-bcb288fc1aaf@i72g2000hsd.googlegroups.com>, > > On Thu, 7 Feb 2008 10:20:24 -0800 (PST), > > "David.Fail...@gmail.com" <David.Fail...@gmail.com> wrote in [quoted text clipped - 12 lines] > > Proof? Presumably by contradiction, No, take any two points in C and join them with the obvious path. > assuming it is disconnected. But > would a separation on C induce one on R^2? On Feb 7, 6:53 pm, The World Wide Wade <aderamey.a...@comcast.net> wrote: > In article > <7ff5bd92-9576-4a6e-a1a8-bcb288fc1...@i72g2000hsd.googlegroups.com>, [quoted text clipped - 21 lines] > > assuming it is disconnected. But > > would a separation on C induce one on R^2? I'm not sure you have understood which notion of connectedness I was intending. A space, in my definition, is disconnected iff there exist two disjoint open sets U and V whose union is the entirety of C. A space is connected iff it is not disconnected. On Feb 7, 7:17 pm, "David.Fail...@gmail.com" <David.Fail...@gmail.com> wrote: > On Feb 7, 6:53 pm, The World Wide Wade <aderamey.a...@comcast.net> > wrote: [quoted text clipped - 31 lines] > > - Show quoted text - Double oops - forgot to say to include the point (0,1) as well. On Feb 7, 8:26 pm, "David.Fail...@gmail.com" <David.Fail...@gmail.com> wrote: > On Feb 7, 7:17 pm, "David.Fail...@gmail.com" <David.Fail...@gmail.com> > wrote: [quoted text clipped - 38 lines] > > - Show quoted text - Or perhaps someone could give me the hint of what open sets in the subspace topoogy on C look like? On Thu, 7 Feb 2008 18:26:00 -0800 (PST), in alt.math.undergrad: > On Feb 7, 7:17 pm, "David.Fail...@gmail.com" <David.Fail...@gmail.com> > wrote: >> On Feb 7, 6:53 pm, The World Wide Wade <aderamey.a...@comcast.net> >> wrote: >>> In article >>> <7ff5bd92-9576-4a6e-a1a8-bcb288fc1...@i72g2000hsd.googlegroups.com>, >>> "David.Fail...@gmail.com" <David.Fail...@gmail.com> wrote: >>>>> On Thu, 7 Feb 2008 10:20:24 -0800 (PST), >>>>> "David.Fail...@gmail.com" <David.Fail...@gmail.com> >>>>> wrote in >>>>> <news:dbce36a6-362a-46cc-96c7-9aea3f25df24@b2g2000hsg.googlegroups.com> >>>>> in alt.math.undergrad: >>>>>> Consider R^2 with the standard topology, and C a >>>>>> subset of R^2 defined with the subspace topology >>>>>> (intersections of C with open sets from R^2) where: >>>>>> C = { (x , 0) : 0<= x <= 1} U { (1/n , y): n=1,2,3,..., 0<= y <=1} >>>>>> Is this "comb-like" space connected? [...] > Double oops - forgot to say to include the point (0,1) as well. Now that makes a substantial difference: that space isn't path connected. Suppose that V and W are open sets in R^2 whose intersections with C are V' and W', respectively, and assume further that (0, 1) in V, V' and W' are disjoint, and C is contained in the union of V and W. Each 'tooth' {1/n} x [0, 1] is connected, so if (1/n, 1) is in V, then {1/n} x [0, 1] is a subset of V, and if (1/n, 1) is in W, then {1/n} x [0, 1] is a subset of W. Clearly the points (1/n, 1) converge to (0, 1), so (1/n, 1) is in V for some n, and hence V contains the tooth {1/n} x [0, 1]. But then V intersects [0, 1] x {0}, which is connected, so [0, 1] x {0} must lie entirely within V. Finally, this implies that each 'tooth' lies in V, and so that W' is empty. Brian In article <20467eff-e503-4841-a744-a6fb02243656@c4g2000hsg.googlegroups.com>, > On Feb 7, 7:17 pm, "David.Fail...@gmail.com" <David.Fail...@gmail.com> > wrote: [quoted text clipped - 38 lines] > > Double oops - forgot to say to include the point (0,1) as well. Your result follows from this: Suppose X is a metric space and A is a connected subset of X. If A is a subset of B and B is a subset of the closure of A, then B is connected. On Thu, 7 Feb 2008 17:17:14 -0800 (PST), in alt.math.undergrad: > On Feb 7, 6:53 pm, The World Wide Wade <aderamey.a...@comcast.net> > wrote: >> In article >> <7ff5bd92-9576-4a6e-a1a8-bcb288fc1...@i72g2000hsd.googlegroups.com>, >> "David.Fail...@gmail.com" <David.Fail...@gmail.com> wrote: >>>> On Thu, 7 Feb 2008 10:20:24 -0800 (PST), >>>> "David.Fail...@gmail.com" <David.Fail...@gmail.com> >>>> wrote in <news:dbce36a6-362a-46cc-96c7-9aea3f25df24@b2g2000hsg.googlegroups.com> >>>> in alt.math.undergrad: >>>>> Consider R^2 with the standard topology, and C a >>>>> subset of R^2 defined with the subspace topology >>>>> (intersections of C with open sets from R^2) where: >>>>> C = { (x , 0) : 0<= x <= 1} U { (1/n , y): n=1,2,3,..., 0<= y <=1} >>>>> Is this "comb-like" space connected? >>> > Yes. It's even path connected. >>> Proof? Presumably by contradiction, >> No, take any two points in C and join them with the obvious path. >>> assuming it is disconnected. But >>> would a separation on C induce one on R^2? > I'm not sure you have understood which notion of > connectedness I was intending. A space, in my definition, > is disconnected iff there exist two disjoint open sets U > and V whose union is the entirety of C. A space is > connected iff it is not disconnected. But path connectedness implies connectedness. If you don't already know this, it's easy enough to prove. Brian >On Feb 7, 6:53 pm, The World Wide Wade <aderamey.a...@comcast.net> >wrote: [quoted text clipped - 28 lines] >two disjoint open sets U and V whose union is the entirety of C. A >space is connected iff it is not disconnected. We know that. Do you know what "path-connected" means? David C. Ullrich > On Thu, 7 Feb 2008 17:17:14 -0800 (PST), "David.Fail...@gmail.com" > [quoted text clipped - 35 lines] > > David C. Ullrich Yes, I know what path connected means, but for purposes of this problem i'm not to use it as the topology class i'm in hasn't learned it just yet On 8 Feb., 15:10, "David.Fail...@gmail.com" <David.Fail...@gmail.com> wrote: > > On Thu, 7 Feb 2008 17:17:14 -0800 (PST), "David.Fail...@gmail.com" > [quoted text clipped - 14 lines] > > >> > > > R^2) where: > > >> > > > C = { (x , 0) : 0<= x <= 1} U { (1/n , y): n=1,2,3,..., 0<= y <=1} [later corrected to C = { (x , 0) : 0<= x <= 1} u { (1/n , y): n=1,2,3,..., 0<= y <=1} u {(0,1)} ] > > >> > > > Is this "comb-like" space connected? > [quoted text clipped - 21 lines] > problem i'm not to use it as the topology class i'm in hasn't learned > it just yet OK, then assume U,V are open in R^2 and such that U /\ C and V /\ C are disjoint and C subset U u V. We have shown that C is connected if we can prove that one of A/\C, V/\C is empty. Wlog. (0,1) in U. Therefore (1/n,1) in U for n big enough. Since {(1/n,y) | 0<=y<=1} is connected, it follows that {(1/n,y) | 0<=y<=1} subset U for such n. Especially, U intersects the connected subset {(x,0) | 0<x<=1}, which must therefore be contained in U. Hence U intersects each vertical line. Again, each of these is connected and therefore contained in U. Thus C is contained in U, C/\V is empty. QED hagman > Consider R^2 with the standard topology, and C a subset of R^2 defined > with the subspace topology (intersections of C with open sets from [quoted text clipped - 3 lines] > > Is this "comb-like" space connected? C = [0,1]x{0} \/ { {1/n}x[0,1] | n in N } It's path connected. S = C \/ {0}x[0,1] is also path connected. Just go from (a,b) down to (a,0) and then over to (0,0) or (1,0) and then reverse the route to get to (r,s) in S or C. Recall lines are path connected and the union of path connected spaces that each of which intersects with the same path connected space, is path connected. The interesting comb space is S - {(0,0} for it's an example of a connected, not path connected space. |
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