- #31
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Well, it follows immediately that A is countable, not? Since both B and B' are countable basis?
What is the countable basis problem in topology?
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The discussion focuses on the countable basis problem in topology, specifically addressing the implications of having a countable basis for a topological space X. It is established that if X has a countable basis, any basis C for X must also contain a countable subcollection that serves as a basis. Participants explore the intersections of basis elements and the need for these intersections to yield a countable collection that still covers X. The conversation highlights the importance of ensuring that the resulting collection is indeed a basis for the topology generated by C, noting that it may end up being coarser than the original topology. The discussion concludes with a consensus on the necessity of using specific lemmas to validate the construction of a valid basis.
- #32
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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?
- #33
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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...
- #34
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radou said:
Yes, I think you've got it!
- #35
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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?
- #36
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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.
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.
- #37
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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...
- #38
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radou said: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.
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...
- #39
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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.
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.
- #40
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Of course, all these are closely related, so it's probably all a matter of personal choice and taste.
- #41
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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..?
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..?
- #42
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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.
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.
- #43
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micromass said:
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.
- #44
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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?
If we have a countable collection of sets, is the number of all possible unions of these sets countable?
- #45
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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.
- #46
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Yes, you'e correct. It isn't true 
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