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

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.

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.

  • $\begingroup$ Help me understand this: So we have $\mathcal{S}$ is countabe, then $S_i \in \mathcal{S}$ is at most countable. Finite intersection of at most countable set is at most countable i.e. $S_1 \cap \ldots \cap S_n$ is at most countable. And the set of $\{S_1 \cap \ldots \cap S_n\}$ is at most countable. I can't see how I can use your hint :( $\endgroup$ – Carlos - the Mongoose - Danger Jun 12 '16 at 18:06
  • $\begingroup$ $S$ need not be countable. $\mathcal{S} = \{S_0,S_1,S_2,S_3,\ldots\}$ is countable. There are also countable many finite subsets of $\mathbb{N}$, and for a finite subset $F$ we have $\cap_{n \in F} S_n$ in the base. $\endgroup$ – Henno Brandsma Jun 12 '16 at 18:09
  • $\begingroup$ So let me try again: $\mathbb{N}$ is countable, and set of all finite set in $\mathbb{N}$ is countable. Let $F_i \subseteq \mathbb{N}$ be a finite set, then we can list all such set as $\{F_1, F_2, \ldots\}$. Then we have $\cap_{n \in F_i} S_n$, which can be listed as $\{\cap_{n \in F_1} S_n, \cap_{n \in F_2} S_n, \ldots\}$, which is a countabe set indexed by $i$. It seems I did not need to use $\mathcal{S}$ is countable to reach the conclusion there exists countably many basis elements. $\endgroup$ – Carlos - the Mongoose - Danger Jun 12 '16 at 18:23
  • $\begingroup$ You used $\mathcal{S}$ was indexed $\mathbb{N}$. $\endgroup$ – Henno Brandsma Jun 12 '16 at 18:24
  • $\begingroup$ Thanks, I think I get it now and will update with a proof $\endgroup$ – Carlos - the Mongoose - Danger Jun 12 '16 at 18:34

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.


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


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.

  • $\begingroup$ P.S. Finding a concrete surjective map from a countable set to the induced basis of $\mathcal{S}$ really is the best way to go, your intuition was good. $\endgroup$ – User12345 Jun 12 '16 at 18:06
  • $\begingroup$ The function $g$ is really complicated, I will take a close look into it $\endgroup$ – Carlos - the Mongoose - Danger Jun 12 '16 at 18:33
  • $\begingroup$ Admittedly $g$ looks funny, but it is not too bad and it does exactly what it needs to. If the notation bothers you, consider first rewriting it in whatever notation you are most comfortable with. $\endgroup$ – User12345 Jun 12 '16 at 18:40

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