Math Forum / Mathematics / Algebra Help / October 2009

True. It's say that T is the collection of all unions of elements of
C(B), not the union of all B in C(B). I lack an intuitive idea of what
we're doing; I still don't see things clearly. I might have proved the
right lemma, but wrote it incorrectly and didn't notice because I have
no intuitive idea of what this is.

In fact, let me recapitulate. We have a set X. C(B) is a subcollection
of T. It doesn't need to be finite. Now, T is the collection of all
possible unions of C(B). I think I have a problem with ``union(...).''

 (*) A trivial example of some union(...)
 
 Let S = {{1,2}, {1,4,5}, {0,1,2,8}}. The axiom of unions says that
 there is a set R such that x in R if x in B subset S for some B in
 S. (If I recall correctly.) So the idea of using ``union(...)'' is
 that we specify which Bs we want, and we collect them all into one
 set. For instance, let
 
   K = union(B where B = {1,2} or B = {1,4,5})
     = {1,2,4,5}.

 We know this set K is unique. Now, if I want to express some other K
 with respect to some particular sets of a collection, I need to be
 careful about not including in the collection more sets than I want,
 or K could end up larger and give me unexpected results. For
 instance, if I had said B in S instead of explicitly have listed the
 ones I wanted, I would have ended up with 8 in K.

 Now, what are all possible unions of S? They would be {1,2},
 {1,4,5}, {0,1,2,8}, then {1,2,4,5}, {0,1,2,8}, {0,1,2,4,5,8}.

 (*) Back to a basis of a space

 A basis always has a B_x containing x. But that's not all. If x is
 shared between any two members of a basis (x in B1 intersect B2),
 then there would be another basis member B3 containing x and B3
 being no larger than the intersection.

 It is imperative here that we say something like ``no larger'' than
 the intersection; we cannot say that B3 will contain only x;
 thinking of the real numbers, it's easy to see counter-examples of
 such singleton B3 --- we can't isolate a number in an open interval.

Let's see your comments.

Sanity check. B is a collection of subsets; ``a set of subsets'' means
that. Just checking. Alright. B is a basis for a topology on X.

Alright. That's B-open.

Okay. That's what we get by following the procedure.

Okay. In other words, U is B-union if we can assemble it from a
subcollection of the basis --- assemble it through a ``union(...)''.

Right. This could be very believable now. Let me give you what I'm
seeing. You said

 (1) Temporarily define a subset U of X to be B-open if for every
 element x of U there is an element B_x of B such that x is in B_x
 and B_x is contained in U.

That is, U is B-open if we can write U in terms of a subcollection of
the basis. Every x in U must be somewhere in the basis (x in B_x); no
B_x can be larger than U. This seems to be exactly ``U is B-open if

  U = union(B_x : B_x in B and x in U and B_x subset U).''

 (*) Example. Let's take the reals R; let U = (0,1). A basis C(B) for
 R is all open intervals; C(B) = {(a,b) : a, b in R with a < b}. I
 want to express U in terms of a union of a subcollection of the
 basis. Let x in U. Since (0,1) is itself a member of the basis we're
 using, then each x in U satisfies x in B for some B in C(B).
 
 Let S = {(0,0.5), (0.4,1)}. Each member of S is in C(B) and also
 each one of them is subset U. Let's name them S1, and S2. Let K =
 union(S1, S2). This means that x in K if x in S1 or x in S2. (That
 is, K = (0,1)). And certainly U subset K. So K = U.

[...]

This will be volume one. I'll work on the rest some time later. Thanks
for your attention, Paul.

[...]

I see. The interesting part of Lemma 13.2 is that C(C) generates the
same topology T. Let me write the condition.

Lemma 13.2. Let X be a space. Suppose C(C) is a collection of open
sets of X such that for each open U subset X, and each x in U, there
is a C in C(C) such that x in C subset U. Then C(C) is a basis for
*the* topology of X.

Checking that it's basis first: as Munkres puts it, the first
condition for a basis is trivial to show because X is large enough to
have each x in it, and X is open by definition with X subset X, so
it'll be in the C(C) of Lemma 13.2. The second condition is also
trivial because an intersection between two open sets will also be
open by the definition of a space; so there'll be a C in C(C)
containing x in the intersection. So C(C) is a basis. The question now
is which topology it generates. I must show it generates *the*
topology defined for X.

Let me call them T(X) and T(C). Must show T(C) = T(X). That is, I must
show that every open set as described by C(C) is in T(X) and that each
in T(X) is in T(C).

Let U be in T(C). Then, by Lemma 13.1, U = union(C : C in C(C)). But
each C is open, and unions of open sets are open. So U is in
T(X). Hm. I'm going to have to be very careful.

Two topologies on the same set can be wildly different. I have a big
clue of what T(C) is because I have its basis. Now T(X) is simply
abstract. All I know about it is that X and empty are there, that
finite intersections of X are there, and arbitrary unions of X are
there.

How do I know that each C in C(C) is in T(X)? Because it must be an
arbitrary union of open sets of X? I guess that's it. By definition of
a basis C(C), each C in C(C) will be open by trivially applying the
procedure: for each x in C there is C in C(C) such that x in C with C
subset C. Okay, every C is open in X --- that is, C is in T(X).

Therefore, finite intersections of C will be in T(X) and also
arbitrary unions. So by Lemma 13.1, U can be written as

 U = union (C : C in C(C))

and so U is in T(X). Alright. That says that T(C) is contained in T(X).

Let V be in T(X). Must show V is in T(C). T(C) is the collection
generated by C(C). V is open; so the hypothesis of this Lemma says
that for each x in V there is a C in C(C) such that x in C subset
V. This means that there is a C in C(C) containing each element x in
V. So V will end up in T(C) because we've shown that C(C) is a basis,
and that says that T(X) is contained in T(C).

Well, perhaps this is it.

Next Lemma says that if for each x and each basis member B, we can
find a basis member B' (in another basis for X) such that x in B'
subset B, then this basis-prime is finer.

So every basis is finer than itself. That kinda sucks to say. Wouldn't
be nicer to demand ``proper subset''?