I have been told that there exist free ultrafilters $ \mathcal U$ such that for any $ S \subset \mathcal U$ countable, we have $ \bigcap_{U \in S} U \neq \emptyset $. However, since free ultrafilters can't be constructed, I don't know how to show this.

I have thought a little about insisting that the sets in the ultrafilter are cocountable or cofinite, but don't know enough about ultrafilters to say that this is possible, or show that it would give the result that I want.

Grateful for any help.


This is not entirely correct.

Using the axioms of $\sf ZF$ we cannot even prove there exists free ultrafilters, so there cannot be an explicit definition of a free ultrafilter, on any set.

Even if we assume the axiom of choice, and can therefore prove the existence of a free ultrafilter, we cannot prove the existence of a countably complete free ultrafilter.

If $\kappa$ is the least cardinality of a set $X$ on which there is a countably complete free ultrafilter, then in fact there is a $\kappa$-complete ultrafilter on $X$, namely the intersection of less than $\kappa$ large sets is large. Such $\kappa$ is called a measurable cardinal.

We can prove now that $\kappa$ is a strongly inaccessible cardinal, and that this proves the consistency of $\sf ZFC$ as a whole. This means we cannot prove from $\sf ZFC$ the existence of such ultrafilter.

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  • $\begingroup$ Thanks for your answer. I don't follow why an ultrafilter satisying the hypothesis given would have to be countably complete. Theintersection has to be nonempty but doesn't have to be in the ultrafilter. $\endgroup$ – Joe L Oct 28 '16 at 16:30
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    $\begingroup$ If it is non-empty, but not in the intersection, remove the intersection from each of the sets, these modified sets must still be in the ultrafilter. What is the new intersection? $\endgroup$ – Asaf Karagila Oct 28 '16 at 17:31

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