What is the countable basis problem in topology?

[radou]

Homework Statement

So, if X has a countable basis {Bn}, then every basis C for X contains a countable basis for X.

The Attempt at a Solution

First of all, consider all the intersections of elements of C of the form Ci$$\cap$$Cj. For every x in the intersection (if it's non empty), choose a basis element Bx contained in the intersection. There are at most countable such elements. Then, for every Bx, choose an element Cx of C contained in Bx, and containing x. There are at most countable such elements, too. Now do this for all the possible intersections Ci$$\cap$$Cj.

Since X has a countable basis, every open covering for X contains a countable subcollection covering X. Apply this to C in order to obtain a countable subcollection C'. We need to show that C' is a basis for X.

First, C' obviously covers X, so there's only one more condition left - we need to show that if x is in the intersection Ci$$\cap$$Cj of elements of C', then there exists an element Ck from C' such that x$$\in$$Ck$$\subseteq$$Ci$$\cap$$Cj.

The part I'm not sure about is this: if C' is countable, then the number of intersections of two elements of C' is countable. If this is true, then for each intersection, take the countable number of sets Cx from the first paragraph, and add it to C'. A countable union of countable sets is countable, hence the new collection C'' is a countable subcollection of C which is a basis for X.

 

[micromass]

The question you ask, i.e. "If C' is countable then the set of all intersections of two elements is countable", this is true.

But I have trouble with following your first paragraph. You say

"For every x in the intersection (if it's non empty), choose a basis element Bx contained in the intersection. There are at most countable such elements."

I assume that you mean that $$\{Bx~\vert~x\in C_i\cap C_j\}$$ is countable. It was a little unclear.

"Then, for every Bx, choose an element Cx of C contained in Bx, and containing x. There are at most countable such elements, too."

This is something I don't immediately see. I agree that the number of Bx is at most countable. But if you then take $$x\in C_x\subseteq B_x$$ for every x, then there is nothing that guarantees that all the Cx are at most countable. Or am I missing something?

 

[radou]

I assume that you mean that $$\{Bx~\vert~x\in C_i\cap C_j\}$$ is countable. It was a little unclear.

Yes, this is exactly what I meant.

"Then, for every Bx, choose an element Cx of C contained in Bx, and containing x. There are at most countable such elements, too."


This is something I don't immediately see. I agree that the number of Bx is at most countable. But if you then take $$x\in C_x\subseteq B_x$$ for every x, then there is nothing that guarantees that all the Cx are at most countable. Or am I missing something?

Hm, good point. It seems this won't work after all. Any ideas on how to prove this?

 

[radou]

Yes, I see the error in my reasoning - for some reason, I intuitively concluded that every Bx contains only one element, which is absurd, it's the notation which mislead me a bit.

 

[radou]

Actually, I'll rather re-think it myself and update later. There's a hint in the book, but I didn't follow it at first.

 

[micromass]

Yes, I was mislead to the first time I read it

How to prove it then. Well, take a basis $$\mathcal{C}$$. Take a countable basis $$\mathcal{B}$$. Now consider the set

$$\mathcal{A}=\{(B,B^\prime)\in \mathcal{B}\times\mathcal{B}~\vert~\exists C\in \mathcal{C}:~B\subseteq C\subseteq B^\prime\}$$

With this set, you can find map $$\mathcal{A}\rightarrow \mathcal{C}$$. Take the image of this map as your "countable basis". You'll still need to show that it is countable and a basis...

 

[micromass]

Ah yes, you could also follow the hint in the book. That's probably easier

 

[radou]

Here's another attempt to deal with the intersection Ci$$\cap$$Cj. The concept of the proof remains the same.

For any x in Ci$$\cap$$Cj, take the element Cx containing x and contained in Ci$$\cap$$Cj. Now, Ci$$\cap$$Cj is a subspace of a second-countable space, so it is itself second-countable. Hence its open cover consisting of the Cx sets contains a countable subcollection covering Ci$$\cap$$Cj.

 

[micromass]

Yes, that is correct. But I'm not sure how you would continue. Surely there could be uncountably many $$C_i\cap C_j$$, since C is not necessairily countable??

I think that the authoer of the exercises didn't intend that you used the definition of basis. I think it would be easier if you applies lemma 13.2 of the chapter "basis for a topology". I think this will make things easier...

 

[radou]

Yes, that is correct. But I'm not sure how you would continue. Surely there could be uncountably many $$C_i\cap C_j$$, since C is not necessairily countable??

Well, the collection of all such intersections is uncountable, but since C is an open cover for a second-countable space X, it contains a countable sub-collection which also covers X, and hence we're only interested in the intersections of the elements from this countable sub-cover. It's what I proposed in the original post.

 

[micromass]

Yes, bu who says that this countable subcollection is a basis?

For example the collection $$\{B(x,\epsilon)~\vert~x\in \mathbb{R}, \epsilon>0\}\cup \{\mathbb{R}\}$$ is a basis for $$\mathbb{R}$$. The subcollection $$\mathbb{R}$$ covers the space and is countable. But I fail to see how this would be of any interessest.

 

[radou]

Yes, bu who says that this countable subcollection is a basis?

Well, that's what the intersection issues were all about.

This countable collection:

a) covers X
b) if x is in Ci$$\cap$$Cj, where Ci and Cj are elements of our countable subcollection, there exists Ck such that it contains x and is contained in Ci$$\cap$$Cj

So it is a basis for X.

I'll try to sum it all up once again.

Assumption: out space is second countable.

Let C be a basis for X. Since C is an open cover, it contains a countable subcollection which is an open cover (only a cover, for now!) for X, call it C'.

Now, for any intersection in C (not yet C' !), by what we just showed, we can find another countable subcollection of C, call it C'' such that for any x in Ci$$\cap$$Cj we have some Ck which contains x and is contained in the intersection.

For now, the number of these countable collections C'' is generally uncountable.

But since we're only interested in the intersections from the countable collection C', their number is countable. So, our basis would be the union of C' and the countable family of the sets C'', and this whole family is now countable.

 

[ColdCoffee]

Which book is this coming from? If I have it I will take a look at the chapter tonight. This problem is interesting.

 

[radou]

Munkres, Topology, 2nd edition, Chapter 30, Exercise 2.

 

[micromass]

Ah yes, you are correct. You have indeed found a basis for A topology. But who says that it is a basis for the topology you want. I.e. does this basis indeed generate the right topology.

Example: if you take $$\{B(x,\epsilon)~\vert~x\in \mathbb{R},\epsilon >0\}$$ as a basis for $$\mathbb{R}$$. Then your construction would end up with $$\mathbb{R}$$, which is indeed a basis, but not for the Euclidean topology! And we do want a basis for the Euclidean topology...

 

[radou]

Ah yes, you are correct. You have indeed found a basis for A topology. But who says that it is a basis for the topology you want. I.e. does this basis indeed generate the right topology.

Example: if you take $$\{B(x,\epsilon)~\vert~x\in \mathbb{R},\epsilon >0\}$$ as a basis for $$\mathbb{R}$$. Then your construction would end up with $$\mathbb{R}$$, which is indeed a basis, but not for the Euclidean topology! And we do want a basis for the Euclidean topology...

Hmm, good point. For example, there is no way we could write the open set B(x, ε/2) as a union of elements of type B(x, ε).

Wow, these examples are really useful. I'll have to rethink this whole thing again, look at your hints in the posts above, and try to solve it from a "different angle". And I thought I was done

 

[radou]

Actually, just to point it out, the collection I've found is only a basis which generates a topology which is coarser than the original one, right? (i.e. the original topology is finer) All open sets in the topology generated by this basis are containd in the oroginal topology, but the original topology may contain sets which are not in this topology, right?

 

[micromass]

Well, we could write $$B(x,\epsilon/2)$$ as a countable union of sets of the form $$B(x,\epsilon)$$. But it's hard to explicitly do so. In fact, we can write every open set as a countable union of balls. So that's not incorrect...

 

[micromass]

To be honest, this is one of the harder topology problems. So it's no shame if you get stuck It would be impossible without hints...

 

[radou]

Well, we could write $$B(x,\epsilon/2)$$ as a countable union of sets of the form $$B(x,\epsilon)$$. But it's hard to explicitly do so. In fact, we can write every open set as a countable union of balls. So that's not incorrect...

What do you mean? If what you wrote is true, than B(x, ε) is a basis for the standard topology, not? Since any interval <a, b> can (as you claim) be written as a union of such elements.

 

[radou]

To be honest, this is one of the harder topology problems. So it's no shame if you get stuck It would be impossible without hints...

Well, that's a relief. Frankly, I thought it was easy.

 

[micromass]

Actually, just to point it out, the collection I've found is only a basis which generates a topology which is coarser than the original one, right? (i.e. the original topology is finer) All open sets in the topology generated by this basis are containd in the oroginal topology, but the original topology may contain sets which are not in this topology, right?

Yes, you've created a coarses topology, so you'll need more open sets.
Still, I highly recommend to use lemma 13.2 of page 80. The usual definition of basis is just for being a basis for a topology. But lemma 13.2 gives a criterium for being a basis for the right topology. This is why this lemma is useful.

 

[micromass]

What do you mean? If what you wrote is true, than B(x, ε) is a basis for the standard topology, not? Since any interval <a, b> can (as you claim) be written as a union of such elements.

But $$\{B(x,\epsilon)~\vert~x\in \mathbb{R},\epsilon >0\}$$ does form a basis for the topology. But you probably mean something else...

 

[radou]

Yes, you've created a coarses topology, so you'll need more open sets.
Still, I highly recommend to use lemma 13.2 of page 80. The usual definition of basis is just for being a basis for a topology. But lemma 13.2 gives a criterium for being a basis for the right topology. This is why this lemma is useful.

Yes, I have this Lemma in mind now from one of your previous posts, I'll think about it.

Example: if you take $$\{B(x,\epsilon)~\vert~x\in \mathbb{R},\epsilon >0\}$$ as a basis for $$\mathbb{R}$$. Then your construction would end up with $$\mathbb{R}$$, which is indeed a basis, but not for the Euclidean topology! And we do want a basis for the Euclidean topology...

There's probably misunderstanding here. Do you, by "euclidean topology" on R, mean the collection generated by basis elements of the form <a, b>? Clearly, this topology is induced by the euclidean metric d(x, y) = |x - y|, i.e. its open balls Bd(x, ε), right?

From post #18 I concluded that any element of our topology can be written as a union of balls B(x, ε)? (probably wrong, it doesn't look right, anyway?)

 

[micromass]

Not sure what you mean here. But it is indeed true that any open set in R can be written as union of open balls. This is essentially lemma 13.1 of page 80... It can even be shown that this union is countable.

 

[radou]

Not sure what you mean here. But it is indeed true that any open set in R can be written as union of open balls. This is essentially lemma 13.1 of page 80... It can even be shown that this union is countable.

OK, actually it's not important, let's ignore it.

So, the point is - from my original basis C, I ended up with its countable subcollection C'', which is a basis for a topology coarser than the one generated by C. Actually this is quite obvious, since C'' is obtained by removing sets from C.

I'll try to solve this tomorrow, got to get some sleep now.

Thanks for the help!

 

[micromass]

Sweet dreams! :tongue2:

 

[radou]

When considering your map from post #6, one may conclude that it's possible, for any x in X, first to choose a basis element B containing x, then a basis element C containing x and contained in B, and then a basis element B' containing x and contained in C. But I don't really see how to proceed now. Somehow I need to pull out a countable collection from C.

The lemma we considered in the previous posts stated that, given a topology on X, if, for every open set U in X and for every element x from U there is an element from a collection C of open sets which contains x and is contained in U, then C is a basis for this very topology.

 

[radou]

By the way, that's bugging me a bit, is it possible to "extend" my previous proof attempt somehow, or should I just forget it? I somehow have a feeling I reached a dead end up there, since I only found a basis for a coarser topology.

 

[micromass]

Well, at the moment I see no way to save your proof attempt. I think you should forget it for now, and possibly return to it if you've found the proof.

As for my hint. So I said to consider

$$\mathcal{A}=\{(B,B^\prime)\in \mathcal{B}\times\mathcal{B}~\vert~\exists C\in \mathcal{C}:~B\subseteq C\subseteq B^\prime\}$$

First you need to check that $$\mathcal{A}$$ is countable.
Then you can choose for every $$(B,B^\prime)$$ a C between B and B'. The collection of these sets C will be the desired countable basis.

 

[radou]

Well, it follows immediately that A is countable, not? Since both B and B' are countable basis?

 

[radou]

Hence, the collestion C is countable. Now, if we apply our lemma here, let U be any open set and x some element in U, then ma post #28 implies that there is a C from the collection A such that x is in C and C is in U, hence C is a countable basis?

 

[micromass]

Yes, countability is easy. The fun part is to prove that its a basis. But I don't think its to hard. The hardest part is finding what the countable basis is supposed to be...

 

[micromass]

Hence, the collestion C is countable. Now, if we apply our lemma here, let U be any open set and x some element in U, then ma post #28 implies that there is a C from the collection A such that x is in C and C is in U, hence C is a countable basis?

Yes, I think you've got it!

 

[radou]

This seems too simple...The only thing we need to apply is the lemma (see my post #32) and we know that the countable collection C is a basis?