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Let $B$ a Boolean Algebra and $St(B)$ its stone space. I want to prove that $B$ is countable if and only if $St(B)$ is separable (i.e. there is a countable dense subset).

For the first implication: the base of colpen sets of the form $N(b)=\{U \in St(B): b\in U\}$ is countable. By AC I can choose $U_b \in N(b)$ and the set $\{U_b : b \in B\}$ is dense and countable. Is there a way not involving AC?

For the second implication, I have no idea.

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  • $\begingroup$ $B$ is countable iff the stone space is second countable. $\endgroup$ – Henno Brandsma Jan 19 at 22:34
  • $\begingroup$ @HennoBrandsma ok, but how does this imply the equivalence with separability? Second countable implies separable, but isn't the converse true only in metric spaces? Sorry but I'm not very confident with this notions $\endgroup$ – MarcoM Jan 19 at 22:54
  • $\begingroup$ it does not. There are separable Stone spaces that are not second countable. So your idea cannot be shown. $\endgroup$ – Henno Brandsma Jan 19 at 22:56
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You cannot show what you want: let $X$ be a separable compact zero-dimensional space that is not second countable (the double arrow will do, or $\{0,1\}^{\omega_1}$ e.g.), then its clopen algebra is a counterexample to your claim.

In fact $B$ is countable iff $\textrm{St}(B)$ is second countable, where the reverse is shown by refining the standard clopen base $N(b), b \in B$ to a countable one.

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