Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.