If $(X, \mathcal{T})$ has a countable subbasis, then it has a countable basis

Question

Given $(X, \mathcal{T})$ a topological space. Let $\mathcal{S}$ be a subbasis on $(X, \mathcal{T})$

Claim: If $\mathcal{S}$ is countable, then $\mathcal{T}$ has a countable basis $\mathcal{B}$

I am not sure how to go about approaching this quetion but here's my attempt:

I want to show that there exists a surjection $g$ from $\mathcal{S}$ to $\mathcal{B}$, and there is an injection $f$ from $\mathcal{B}$ and $\mathcal{S}$ thus $|\mathcal{S}| = \aleph_0 = |\mathcal{B}|$

Define $g$ as $$g(S_1, S_2, \ldots, S_n) = S_1 \cap S_2 \cap \ldots \cap S_n = B$$

where $S_1, \ldots, S_n \in \mathcal{S}$, and $B \in \mathcal{B}$

But how does the countability of $\mathcal{S}$ come in? I'm really lost

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Per Henno's suggestion re-attempt:

• Let $\mathcal{S}$ be a countable subbase of $(X, \mathcal{T})$. Then $\mathcal{S}$ can be listed as $\{S_1, S_2, \ldots \}, S_i \in S, i \in \mathbb{N}$

  • By definition, each basis element is the finite intersection of subbasic elements written as $\bigcap\limits_{i \in F_n} S_i$, where $F_n$ is a finite set in $\mathbb{N}$.

    • Since there exists countably many finite sets in $\mathbb{N}$, we can list all the finite sets as $\{F_1, F_2, \ldots\}$

      • Then correspondingly we can list all the basis elements as:$\{\bigcap\limits_{i \in F_1} S_i, \bigcap\limits_{i \in F_2} S_i, \ldots\}$ which is a countable set.

        • Hence $\mathcal{B}$ is countable.

Answer 1

Hint: if $A$ is a countably infinite set, then the set of all finite subsets of $A$ is also countably infinite.

And to go from a subbase to a base we take all intersections of finite subsets of the subbase. So if the subbase is countable, we can only have countably many finite subsets, so at most that many different intersections.

Answer 2

Forget about specific surjections/injections.

A countable set has countably many finite sets, so the collection of finite sets $\{S_1,...,S_n\} \subset \mathcal{B}$ is countable. Thus your basis is countable.

Answer 3

The function $g$ you describe is most naturally written in the following manner:

$$g:\bigcup_{j=0}^\infty\prod_{i=1}^j\mathcal{S}\to\mathcal{B}:(S_k)_{k=1}^n\mapsto\bigcap_{k=1}^nS_k$$

Here $\mathcal{B}$ is the induced basis from the subbasis $\mathcal{S}$, and we have the equality (by definition):

$$\mathcal{B}=\left\{\bigcap_{k=1}^nS_k\middle|n\in\mathbb{N}_0\text{ and }S_k\in\mathcal{S}\text{ for all }1\leq k\leq n\right\}$$

In this manner the proof that $g$ is a surjection is obvious. On the other hand the domain of $g$ is a countable union of finite products of countable sets, so it is countable. This implies by definition of the cardinal ordering and the assumption that $|\mathcal{S}|=\aleph_0$ that $|\mathcal B|\leq|\bigcup_{j=0}^\infty\prod_{i=1}^j\mathcal{S}|=\aleph_0$. All that remains is to show that $\mathcal{B}$ is infinite, which follows immediately by noting that $\mathcal{S}\subseteq\mathcal{B}$. Hence we have that $\mathcal{B}$ is countable as desired.