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.

  • 1
    $\begingroup$ 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

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 $x \in B_x \subseteq B_1 \cap B_2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.