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I am trying to prove this theorem:

Let X be a compact Hausdorff space, such that $\mathbf{C}\left(X\right)$ is separable, then X is second-countable.

I found a sketch of the proof here, but I am not able to show that the following set is truly a basis for the topology: $$\mathcal{B} = \{f_n^{-1}\left(\left(p,q\right)\right)\mid p,q\in\mathbb{Q}, n\in\mathbb{N}\}$$

It is easily shown that this is a countable collection of open sets in $X$, and I could also show that given an open set $U\subseteq X$, for each $x\in U$, there is $B\in \mathcal{B}$ such that $x\in B\subseteq U$. Therefore, if this is truly a basis for a topology, then it generates the same topology on $X$.

To show that it is a basis, I need to show that it is a cover (done), and also the following, which I could not prove:

$$\forall x\in X; B_1, B_2 \in \mathcal{B}: x\in B_1 \cap B_2 \Longrightarrow \exists B_3\in\mathcal{B} \text{ s.t. } x\in B_3 \subseteq B_1 \cap B_2.$$

If $x\in f_n^{-1}\left(\left(p_1,q_1\right)\right) \cap f_m^{-1}\left(\left(p_2,q_2\right)\right)$, then $f_n(x)\in\left(p_1,q_1\right)$ and $f_m(x)\in\left(p_2,q_2\right)$. I need to find a function $f_k$ and a domain $\left(p,q\right)$, such that if $f_k(x)\in\left(p,q\right)$, then this implies the two former belongings.

That is, a function that can decide where two other functions map $x$ to. This does not sound right. Any ideas?

Edit: I could also show that every open set is a union of such sets from $\mathcal{B}$. Perhaps this is enough? I don't see it.

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    $\begingroup$ You wrote "I could also show that given an open set $U\subseteq X$, for each $x\in U$, there is $B\in \mathcal{B}$ such that $x\in B \subseteq U$". That is what you need. Take $U = B_1 \cap B_2$. $\endgroup$ – Daniel Fischer Mar 9 '18 at 14:28
  • $\begingroup$ @DanielFischer You're right, since $B_1\cap B_2$ is open in X, then what I showed applies to it, as well. Thank you! $\endgroup$ – AAN4EVA Mar 9 '18 at 16:21
  • $\begingroup$ See lemma 2.3 p 81 in Munkres 2nd ed. $\endgroup$ – Henno Brandsma Mar 10 '18 at 8:33
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There is no need any more to check whether it obeys the pre-conditions for a base, it’s automatic:

Suppose $(X, \mathcal{T})$ is a space and $\mathcal{B}$ is a collection of open sets such that

$$\forall O \in \mathcal{T}: \forall x \in O: \exists B_x \in \mathcal{B}: x \in B_x \subseteq O (\ast)$$

Then $\mathcal{B}$ fulfills the two conditions for being a base for a topology. That $\bigcup \mathcal{B} = X$ is clear, we apply $(\ast)$ to $O=X$ and every $x$. And then $X = \bigcup\{B_x: x \in X\} \subseteq \bigcup \mathcal{B} \subseteq X$.

If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1\cap B_2$, then we apply $(\ast)$ to $O=B_1 \cap B_2$, and we get $B_x \in \mathcal{B}$ with $x \in B_x \subseteq B_1 \cap B_2$.

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