Countably closed ultrafilters on incomplete Boolean algebras

Suppose that $$B$$ is a Boolean algebra. Say that an ultrafilter, $$U$$, on $$B$$ is countably closed iff whenever $$X\subseteq U$$ is countable and the meet $$\bigwedge X$$ exists, $$\bigwedge X\in U$$.

I understand that when $$B=P(X)$$ for some $$X$$, the existence of a countably closed non-principal ultrafilter implies $$X$$ is really big (is of measurable cardinality). But I would like to know whether countably closed non-principal ultrafilters can exist on other, possibly incomplete, Boolean algebras consistently with ZFC or with $$B$$ being comparatively small.

1. Is there a small (e.g. whose existence is consistent with ZFC) Boolean algebra admitting a countably closed non-principal ultrafilter?
2. Is there a Boolean algebra with the property that every filter extends to a countably closed ultrafilter?

Relating to the second question, are there any general theorems of the form: every filter on $$B$$ (with maybe some additional properties) extends to a countably closed ultrafilter that I should know about?

• For question 1, start with your favorite uncountable set $X$, and take the Boolean algebra consisting of the countable subsets of $X$ and their complements. The complements constitute a nonprincipal, countably complete ultrafilter. – Andreas Blass Nov 28 '18 at 21:20

Let $$X$$ be an uncountable set and let $$B$$ be the algebra of subsets of $$X$$ that are either countable or cocountable. Then the set of cocountable subsets of $$X$$ is a countably closed nonprincipal ultrafilter on $$B$$. More generally, if $$F$$ is any countably closed filter on $$B$$, then $$F$$ extends to a countably closed ultrafilter (either every element of $$F$$ is cocountable so it extends to the cocountable filter, or else some element of $$F$$ is countable and then it follows easily that $$F$$ is principal so it extends to a principal ultrafilter generated by some singleton).
There are also more degenerate examples: there are Boolean algebras $$B$$ in which no nontrivial countable meets exist (i.e., if $$S\subseteq B$$ is countable and $$\bigwedge S$$ exists then there is a finite subset $$S_0\subseteq S$$ such that $$\bigwedge S_0=\bigwedge S$$), and so trivially every filter is countably closed. For instance, this is true of any finite Boolean algebra. Less obviously, it is true of $$\mathcal{P}(\mathbb{N})/\mathrm{fin}$$, where $$\mathrm{fin}$$ is the ideal of finite sets (this algebra is atomless and so all its ultrafilters are nonprincipal).
A more robust definition of a "countably closed filter" $$F$$ would be that if $$S\subseteq F$$ is a countable subset, then there exists $$a\in F$$ such that $$a\leq b$$ for all $$b\in S$$ (even if $$\bigwedge S$$ does not exist). With this definition, the first example above is still an example of a nonprincipal countably closed ultrafilter, but there is no infinite Boolean algebra in which every filter (not necessarily countably closed) extends to a countably closed ultrafilter. This follows immediately from the fact that every infinite Boolean algebra $$B$$ has an ultrafilter which is not countably closed (and therefore trivially cannot be extended to a countably closed ultrafilter).
As a proof of this last fact, let $$X$$ be the Stone space of $$B$$. Then $$X$$ is an infinite, so we can pick a sequence $$(x_n)$$ of distinct points in $$X$$. Since $$X$$ is compact, these points $$(x_n)$$ accumulate at some point $$x\in X$$ which we may assume is not equal to $$x_n$$ for any $$n$$. Now for each $$n$$ we can pick some $$a_n\in B$$ such that $$a_n\in x$$ but $$a_n\not\in x_n$$. If the ultrafilter $$x$$ were countably closed, then there would exist $$a\in x$$ such that $$a\leq a_n$$ for all $$n$$, and so $$a\not\in x_n$$ for all $$n$$. But then $$a$$ would correspond to an open neighborhood of $$x$$ in $$X$$, and so it would have to contain some $$x_n$$ since $$(x_n)$$ accumulates at $$x$$. This iis a contradiction.