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.


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*Is there a small (e.g. whose existence is consistent with ZFC) Boolean algebra admitting a countably closed non-principal ultrafilter?

*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?
 A: 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.
