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Proof about countable base of topological space 
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#1
Mar612, 02:17 AM

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1. The problem statement, all variables and given/known data
Prove that if a topological space has a countable base, then all bases contain a subset which is a countable base 2. Relevant equations A base is a subset of the topological space such that all open sets can be constructed from unions and finite intersections of open sets from the base 3. The attempt at a solution I'm only a few pages into this chapter, all I learned so far is arbitrary union and finite intersection are closed operations. I have no clue how to construct a countable subset from an arbitrary base, not to mention that it must be a base by itself. Any help is appreciated 


#2
Mar612, 05:17 AM

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#3
Mar612, 05:36 PM

P: 312

Let [itex]B=\{B_{\alpha}\alpha\in A\}[/itex] be an arbitrary base, and [itex]C=\{C_ii\in I\}[/itex] the countable base, then for each [itex]C_i[/itex], the set of sets [itex]S_i=\{B_\alpha\exists j\in I\,{\rm s.t.}\,C_j\subset B_\alpha\subset C_i\}[/itex] turn out to be countable, if for each [itex]j[/itex] so that such [itex]B_\alpha[/itex] exists, choose only one of such [itex]B_\alpha[/itex]'s. The union of the sets in [itex]S_i[/itex] is [itex]C_i[/itex](needs some constructive proof), therefore the union of countable sets [itex]S_i[/itex] for all [itex]i[/itex]'s turns out to be a countable base. 


#4
Mar712, 01:33 AM

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Proof about countable base of topological space
I wasn't able to make any progress with your problem yesterday, but I only spent 15 minutes or so on it.
Isn't the set S_i, as you have defined it, equal to ##\{B_\alpha\in BB_\alpha\subset C_i\}##? This is what I'm thinking: Let i be an arbitrary member of I. Since C_i is a member of a base, it's open. Since B is a base, this implies that there's a subset ##A_i\subset A## such that ##C_i=\bigcup_{\alpha\in A_i}B_\alpha##. The ##B_\alpha## with ##\alpha\in A_i## must be subsets of C_i, so this ensures that the set S_i is nonempty. Let ##\alpha## be an arbitrary member of ##A_i##. Since ##B_\alpha## is a member of a base, it's open. Since C is a base, this implies that there's a subset ##I_\alpha\subset I## such that ##B_\alpha=\bigcup_{j\in I_\alpha} C_j##. The C_j with ##j\in I_\alpha## must be subsets of ##B_\alpha##, so there's always a C_j (with ##j\in I_\alpha##) such that ##C_j\subset B_\alpha\subset C_i##. I don't see how to prove that each S_i are countable. That would be a surprising result actually. I would expect those sets to not be countable. But maybe you can do what you're suggesting at the end anyway: For each i in I, choose an ##\alpha\in A## such that ##B_\alpha\subset C_i##. Then maybe you can prove that the set of all ##B_\alpha## chosen this way is a countable base. (I haven't tried that myself). It's clearly countable, because of how the sets were chosen. 


#5
Mar712, 02:07 AM

P: 312

By merely reading the proof took me hours to convince myself, and it helps to try to rephrase it in my own language. 


#6
Mar712, 05:39 AM

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Let [itex]\mathcal{B}[/itex] be your arbitrary basis. The easiest proof of this is to take a countable basis [itex]\mathcal{A}[/itex] and to look at the collection
[tex]\mathcal{C}=\{(A,A^\prime)\in \mathcal{A}\times \mathcal{A}~\vert~\exists B\in \mathcal{B}:~A\subseteq B\subseteq A^\prime\}[/tex] This is countable (why?). Now, find a map [itex]\mathcal{C}\rightarrow \mathcal{B}[/itex]. 


#7
Mar712, 03:43 PM

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