Suppose that $A$ is a free $\sigma$-algebra on a countable set of generators. Does $A$ have a $\sigma$-ultrafilter?
I have no difficulty in proving that $A$ contains at least an ultrafilter. For example, consider the set $F$ of all non-zero elements of $A$. Then
(1) $1$ (the top element of $A$) is non-empty, so $1\in F$.
(2) Suppose that $a\in F$ and that there exists $b\in A$ such that $a\leq b$. Then $b$ is also non-empty because it contains $a$, so $b\in F$.
(3) If $a,b\in F$, then $a\wedge b$ is also non-empty, so $a\wedge b\in F$.
This proves that $F$ is a filter.
(4) Suppose now that both $a,-a\in F$. Then so does their meet, which is a contradiction because $a\wedge -a$ is empty. Hence, $F$ is an ultrafilter.
Consider now a sequence $a_1,a_2,...$ of non-empty elements of $A$. Does $a_1\wedge a_2 \wedge ...\in F$?
As for me, I don't see why it should not be the case. After all, the only one way for $a_1\wedge a_2 \wedge ...$ to be empty is if one of $a_1,a_2,...$ is, but all of them are by definition non-empty, so it must be the case that $a_1\wedge a_2 \wedge ...$ is in $F$. Yet I learnt to be cautious with free $\sigma$-algebras. Could someone please confirm (or repudiate) that $F$ is a $\sigma$-ultrafilter of $A$?