# Countable basis problem

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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, gotta get some sleep now.

Thanks for the help!

Sweet dreams!! :tongue2:

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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.

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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.

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.

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

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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, countability is easy. The fun part is to prove that its a basis. But I dont think its to hard. The hardest part is finding what the countable basis is supposed to be...

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!

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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?

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Ah, OK, thanks a lot!

The real solutions to such problems are actually always quite simple, but require a certain amount of creativity. The proof I tried is more "definition-based", and such proofs ofteh lead to nothing. :uhh:

The hardest part was finding what the countable basis is. Proving that its a basis is indeed not so difficult. Sadly this is typical for topology, once you know what the things are supposed to be, it isn't hard to prove that they are...

Ah, OK, thanks a lot!

The real solutions to such problems are actually always quite simple, but require a certain amount of creativity. The proof I tried is more "definition-based", and such proofs ofteh lead to nothing. :uhh:

Well the problem here is that Munkres didn't define basis in a very good way. I always define basis as lemma 13.2, since that is the form one will always use when discussing a basis...

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Yes, it's interesting how definitions differ from author to author.

In a set of lecture notes on metric spaces and topology I went through earlier, the definition of a basis for a topology T is that it's a subfamily B of T such that every member of T equals a union of the members of B.

In Munkres for example, this is stated as a separate lemma.

Back in this set of lecture notes, the definition from Munkres is actually stated as a theorem.

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Of course, all these are closely related, so it's probably all a matter of personal choice and taste.

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By the way, could we define the family A as follows: (?)

Let A be the family of all such basis elements B for which there is some basis element C contained in them. Now take the family of all the basis elements C contained in some B. This family is countable.

Now, if U is any open set, and x in U, there exists some basis element B containing x. Further on, there exists some basis element C containing x and contained in B, hence this C belongs to the earlier defined countable family, which is by this argument a basis for out topology.

Frankly, I don't see a conceptual difference between this "proof" and the last one..?

I suppose that argument would work out to...

I'm still wondering if you can't prove the problem directly from the definition of a basis. I would deem it possible, but it would be more difficult.

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I suppose that argument would work out to...

I'm still wondering if you can't prove the problem directly from the definition of a basis. I would deem it possible, but it would be more difficult.

OK.

I wrote that down, I didn't give up on the other proof attempt, if I figure something out, I'll post it here.

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I may have an idea, but before I consider it, I have the following question (it may seem a bit stupid, but nevertheless):

If we have a countable collection of sets, is the number of all possible unions of these sets countable?

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Actually, I think the answer is no; I just found that the power set of the positive integers is uncountable, and my question is equivalent to asking if the power set of the positive integers is countable.

Yes, you'e correct. It isn't true