I have a question about filters which I suspect has a very simple answer (hence my asking it here as opposed to MO):
Let $F$ be a filter on an infinite set $X$. Then $F$ is "countably closed" if for any sequence $(A_i)_{i\in\omega}$ of elements of $F$, the intersection $\bigcap_{i\in\omega}A_i$ is in $F$.
If we demand that $F$ be an ultrafilter as well, then this is a very strong condition: either $F$ is principal (i.e., generated by a singleton), or $\vert X\vert$ is a measurable cardinal. However, there are plenty of examples of countably closed filters which are not ultrafilters: the filters of cocountable, comeager, and measure 1 sets of real numbers all have this property, assuming (I think?) countable choice.
Now to my question. Consider the following property of a filter $F$: for any sequence $(A_i)_{i\in\omega}\in F$, there is an infinite $S\subseteq\omega$ such that $\bigcap_{j\in S}A_j\in F$. Call such a filter "countably thick."
My question is the following. Is there a countably thick filter which is not countably closed? I am also interested in whether there is a countably thick ultrafilter which is not countably closed.
This question, so far as I know, has no greater mathematical significance - I just ran across it while thinking about infinitary combinatorics.
The answer to the question is no.
Suppose that the filter $\mathscr{F}$ is not countably closed, and fix a sequence $\langle F_n:n\in\omega\rangle$ in $\mathscr{F}$ such that $\bigcap\limits_{n\in\omega}F_k\notin\mathscr{F}$. For $n\in\omega$ let
$$H_n=\bigcap_{k\le n}F_k\in\mathscr{F}\;.$$
The sets $H_n$ are nested, so for any infinite $S\subseteq\omega$ we have
$$\bigcap_{n\in S}H_n=\bigcap_{n\in\omega}H_n=\bigcap_{n\in\omega}F_n\notin\mathscr{F}\;,$$
and therefore $\mathscr{F}$ is not countably thick.