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So I've been studying Stone duality and the following question arises naturally:

Let $X$ be a Stone space (compact, zero-dimensional Hausdorff) and $B$ the associated Boolean algebra of clopen sets. What condition on $B$ is equivalent to $X$ being separable? First countable?

I know that $X$ is second-countable iff $B$ is countable. I don't really know if there is any possible condition depending only on the order of $B$, since separability and first countability are "point" properties.

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A Stone space (zero-dimensional compact Hausdorff space) $X$ is first countable iff every point (singleton set, really) is the intersection of a sequence of clopen sets. So using points = ultrafilters we get that

$\mathrm{st}(B)$ is first countable, iff for every ultrafilter $\mathcal{F}$ on $B$, there is a countable subset $B(\mathcal{F})$ of $\mathcal{F}$ such that $\mathcal{F}$ is the unique ultrafilter that contains $B(\mathcal{F})$, so all ultrafilters are "uniquely countably determined" to coin a phrase.

Separability I'll have to think about; at least it implies ccc and this corresponds to the fact that in $B$ all antichains (all $b \neq b'$ in an antichain have infimum $0$) are at most countable. But it's stronger than that.

[Added] I think the straight-forward translation is: there is a countable set in $\mathrm{st}(B)$ that intersects every non-empty clopen set.

So this happens iff $B$ has a countable set of ultrafilters $\{\mathcal{F}_n: n \in \omega\}$ such that for every $b \neq 0$ in $B$ there is some $n$ such that $b \in \mathcal{F}_n$, or

$$B\setminus \{0\} = \bigcup_n \mathcal{F}_n$$

or $B$ "has a countable cover by ultrafilters", to coin another phrase. Note that this indeed implies the antichain property, as an uncountable antichain would have (more than) two elements in one $\mathcal{F}_n$ by the pigeon hole principle, which cannot be, as filter elements have a non-$0$ infimum.

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