# Show that $X$ is separable.

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.

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what does $\sigma$-centered mean? – Stefan Hamcke Aug 18 '13 at 1:27
@Stefan: A poset $\langle P,\le\rangle$ is centred if for each finite $F\subseteq P$ there is a $p\in P$ such that $p\le q$ for each $q\in F$. A poset is $\sigma$-centred if it is the union of countably many centred posets. – Brian M. Scott Aug 18 '13 at 1:31
I don't think the topology is a countable union of filters since filters are closed under supersets and a topology is not. Do you mean filter bases? – Stefan Hamcke Aug 18 '13 at 1:36
@Stefan: Filters in the partial order $\langle\Bbb O_x,\subseteq\rangle$, not in $\langle\wp(X),\subseteq\rangle$. – Brian M. Scott Aug 18 '13 at 1:37
This seems as good a place as any to mention two terminological points. First, one speaks (or at least one should speak) of filters in posets and filters on or over sets --- exactly as Stefan did in his comment here. Correct use of the prepositions here can save a lot of confusion in those situations where one actually needs a filter on a set that happens to also have a partial ordering. Second, "centered" and "centred" are (fortunately) the same; the former is American and the latter English (or should I say British?). – Andreas Blass Aug 18 '13 at 2:10

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 $X$.

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Do you mean "compact" instead of "Hausdorff" in the second line? – Stefan Hamcke Aug 18 '13 at 1:58
@Stefan: I sure do; thanks! – Brian M. Scott Aug 18 '13 at 1:59