# $B$ countable boolean algebra then St(B) separable.

I want to prove that :
if $B$ is a countable boolean algebra then $St(B)$ ( = Stone space ) is separable.
Assuming $BPI(B)$ i.e. the boolean prime ideal princible for the boolean algebra $B$, there is a theorem stating that $St(B)$ is a compact, Housdorff, zero-dimensional topological space and $B \cong CLOP(St(B))$ where $CLOP(St(B))$ is the set of the Clopen sets of $St(B)$.
I have the homomorfism $N : B \to P(St(B))$ defined by $N(b) = \{U \in St(B) | b \in U\}$ and $\{N(b) | b \in B \}$ is a base of clopen set.
Since $B$ is countable i can enumerate the clopen sets : $\{N(b_i) | b_i \in B \land i \in \omega\}$. For every $i \in \omega$ we have that $\bigcap N(b_i)$ is non-empty (because every $U \in N(b_i)$ contains $b_i$) and it is a basis for a filter on $B$ that can be extended to an ultrafilter $U_i$ by $BPI(B)$. I'm pretty sure that the set $V = \bigcup_{i \in \omega} \{U_i\}$ is the countable dense set that witnesses the separability of St(B) but I can't prove that its closure is St(B). Any advice? Any misteake? Any element $N(b_i)$ of the base of clopen sets has non-empty intersection with $V$; is that enough ? if yes, why?

• Are you trying to avoid the axiom of choice other than assuming $BPI(B)$? (Note that you don't actually need to assume $BPI(B)$ as a form of choice; you can prove $BPI$ for any well-orderable Boolean algebra in $ZF$.) – Eric Wofsey Jan 18 '18 at 20:34
• Well, I ask because in choosing your sequence of ultrafilters $U_i$, you are seemingly using the axiom of choice. (This use can be avoided again though because $B$ is well-orderable.) – Eric Wofsey Jan 18 '18 at 20:46
• $B$ is well-orderable (because it's countable) then there's a choice function on it! right? – M.B. Jan 18 '18 at 20:51

A set $D\subseteq X$ is dense in $X$ iff every non-empty (basic) open set of $X$ intersects $D$.
• That's the idea, but how can I prove that $V$ is dense in St(B)? – M.B. Jan 18 '18 at 20:27