Proof about countable base of topological space

[sunjin09]

Homework Statement

Prove that if a topological space has a countable base, then all bases contain a subset which is a countable base

Homework 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

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

 

[Fredrik]

Homework 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

By my definitions, that's a sub-base. A base is a collection of open sets such that every open set is a union of members of that collection. Check your book to make sure that it uses this definition.

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.

You don't have to specify what sets are members of this countable subset. You just have to show that such a set exists. I don't immediately see the solution to the problem. I will take a few minutes to think about it.

 

[sunjin09]

By my definitions, that's a sub-base. A base is a collection of open sets such that every open set is a union of members of that collection. Check your book to make sure that it uses this definition.


You don't have to specify what sets are members of this countable subset. You just have to show that such a set exists. I don't immediately see the solution to the problem. I will take a few minutes to think about it.

Thank you for correcting the mistake in the definition about the base, it is indeed only unions are allowed. As I researched on the internet, such a space is called second countable space, and the property I'm going to prove is well known, although the proof is not easy to understand. I'll outline below:
Let B=\{B_{\alpha}|\alpha\in A\} be an arbitrary base, and C=\{C_i|i\in I\} the countable base, then for each C_i, the set of sets S_i=\{B_\alpha|\exists j\in I\,{\rm s.t.}\,C_j\subset B_\alpha\subset C_i\} turn out to be countable, if for each j so that such B_\alpha exists, choose only one of such B_\alpha's. The union of the sets in S_i is C_i(needs some constructive proof), therefore the union of countable sets S_i for all i's turns out to be a countable base.

 

[Fredrik]

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 $\{B_\alpha\in B|B_\alpha\subset C_i\}$? 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 non-empty.

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.

 

[sunjin09]

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 $\{B_\alpha\in B|B_\alpha\subset C_i\}$? 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 non-empty.

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.

It is indeed how to construct S_i that is the trickiest part, for each x in C_i, there exist a B_α containing x that is a subset of C_i; since B_α is also the union of C's, there is a C_j containing x that is a subset of B_α, keep this B_α, and change to another x and do the same thing, if the resultant C_j is different, keep the B_α, otherwise do not keep the B_α; after exhausting all the x's in C_i, one obtains at most countably many B_α's corresponding to the C_j's, and these B_α's contain all the points x in C_i, and they are all subsets of C_i as well, therefore their union is C_i.

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

 

[micromass]

Let \mathcal{B} be your arbitrary basis. The easiest proof of this is to take a countable basis \mathcal{A} and to look at the collection

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

This is countable (why?).
Now, find a map \mathcal{C}\rightarrow \mathcal{B}.

 

[sunjin09]

Let \mathcal{B} be your arbitrary basis. The easiest proof of this is to take a countable basis \mathcal{A} and to look at the collection

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

This is countable (why?).
Now, find a map \mathcal{C}\rightarrow \mathcal{B}.

This is certainly a much clearer way of thinking, C is countable because it's a subset of the Cartesian product of two countable sets. The mapping f: C → B is obvious by definition, and is non-unique. Any one of such mappings f should have that Im(f) is a countable base, since for any fixed A' and all pairs (A'',A') where A'' is a subset of A', the union of A''s is A' since A is a base; and f(A'',A') exists, since B is a base; the union of f(A'',A') is also A', since it includes A' and is simultaneously included in A'. Therefore Im(f) is a base, since any A' is a union of Im(f) elements.