Show that $X$ is separable.

Question

I'm working through Kunen's Set Theory and I'm not sure how to proceed on part of one exercise. Let $X$ be compact Hausdorff and $\mathbb{O}_X$ be the poset of nonempty open sets of $X$ ordered by inclusion. I want to show that all of

1. $X$ is separable.
2. $\mathbb{O}_X$ is $\sigma$-centered.
3. $\mathbb{O}_X$ is a countable union of filters.

are equivalent. I have 3 $\Rightarrow$ 2 and 1 $\Rightarrow$ 3. For 2 $\Rightarrow$ 1, we have that $\mathbb{O}_X=\bigcup_{n\in\Bbb N} C_n$ where each $C_n$ is centered, but I'm not sure how from these I should pick the points in the countable dense subset. Also, I haven't used the compact Hausdorff hypotheses yet and I am not sure how they will come into play.

Answer 1

Let $\Bbb O_X=\bigcup_{n\in\omega}\Bbb O_n$, where each $\Bbb O_n$ is centred. For $n\in\omega$ let $K_n=\bigcap\{\operatorname{cl}U:U\in\Bbb O_n\}$; use the fact that $X$ is compact to conclude that $K_n\ne\varnothing$. For $n\in\omega$ let $x_n\in K_n$, and let $D=\{x_n:n\in\omega\}$. Now let $U$ be any non-empty open set in $X$; since $X$ is compact and Hausdorff, $X$ is regular, and there is a non-empty open $V$ such that $\operatorname{cl}V\subseteq U$. There is an $n\in\omega$ such that $V\in\Bbb O_n$, so $$x_n\in D\cap K_n\subseteq D\cap\operatorname{cl}V\subseteq D\cap U\;,$$ and it follows that $D$ is dense in $X$.