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.

  • $\begingroup$ Well, you indeed beat me to it. I deleted my answer since it was not relevant at all to the question. $\endgroup$ – Asaf Karagila Dec 11 '11 at 21:08
  • $\begingroup$ Yup, that was straightforward. This has not been my finest moment . . . : ) Thanks! $\endgroup$ – user13568 Dec 14 '11 at 22:35

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