Proof about countable base of topological space

In summary: 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##.
  • #1
sunjin09
312
0

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
 
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  • #2
sunjin09 said:

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.

sunjin09 said:
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.
 
  • #3
Fredrik said:
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 [itex]B=\{B_{\alpha}|\alpha\in A\}[/itex] be an arbitrary base, and [itex]C=\{C_i|i\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
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.
 
  • #5
Fredrik said:
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.
 
  • #6
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
micromass said:
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].

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.
 

1. What is a countable base of a topological space?

A countable base of a topological space is a collection of open sets that can be used to generate all other open sets in the space. In other words, every open set in the topological space can be written as a union of sets from the countable base.

2. Why is it important for a topological space to have a countable base?

A countable base is important because it allows us to define the topology of a space in a simpler way. Instead of having to define all the open sets individually, we can just specify a countable base and use it to generate the rest of the open sets. This makes it easier to study and understand the properties of a topological space.

3. How do we prove that a topological space has a countable base?

To prove that a topological space has a countable base, we need to show that there exists a collection of open sets that satisfies two conditions: (1) every open set in the space can be written as a union of sets from this collection, and (2) the collection itself is countable. This can be done by constructing a specific countable base or by showing that the space is second-countable, meaning it has a countable basis.

4. Can a topological space have more than one countable base?

Yes, a topological space can have multiple countable bases. This is because a countable base is not unique - there can be different sets of open sets that still satisfy the two conditions mentioned above. However, all countable bases for a given topological space will generate the same topology.

5. Is a countable base necessary for all topological spaces?

No, a countable base is not necessary for all topological spaces. In fact, there are topological spaces that do not have a countable base, such as the uncountable long line. However, many commonly studied topological spaces, such as Euclidean spaces and metric spaces, do have a countable base.

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