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Let $S$ be an infinite set with cardinality $k$. Let $U$ be an ultrafilter containing the filter of cofinite sets.

Then, for any set $P \subset U$ such that the cardinal of $P$ is $<k$ , is it true that there exists an ultrafilter $V$, $V\ne U$, such that $P\subset V$?

If all the subsets of $P$ are cofinite, I can show that the intersection of all of them is non-empty, as its complementary is the union of all the complements of subsets in $P$, which are finite, and therefore is the union of $<k$ finite sets, so has cardinal $<k$. Hence there is one element $s\in S$ contained in all of them. Then the principal ultrafilter $V$ associated to $s$ does the job.

So my guess is that the same should be true in general, so that the hypothesis imply the intersection of all subsets in $P$ is non-empty.

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    $\begingroup$ I think you need that $\vert P \vert \lt \operatorname{cf} k$. So I think your conjecture may be false for singular $k$. The hypothesis that the intersection of $P$ is nonempty definitely can fail for singular cardinals. There is an ultrafilter containing the cofinite subsets of $\aleph_{\omega}$ and also $A_n=\aleph_{\omega} \setminus \aleph_n$ for each $n \in \Bbb N$, but $\cap A_n= \varnothing$. $\endgroup$
    – Robert Shore
    Jun 2 at 15:29
  • $\begingroup$ And how can it be proved if the cardinality of $P$ is less than the cofinality of $k$? Any references? $\endgroup$
    – Nulhomologous
    Jun 2 at 15:42
  • $\begingroup$ I don't understand "If all the subsets of $P$ are cofinal" but maybe you meant "If all the elements of $P$ are cofinite". $\endgroup$
    – Andreas Blass
    Jun 2 at 16:13
  • $\begingroup$ @AndreasBlass thanks, you are right, sorry. I corrected the question. $\endgroup$
    – Nulhomologous
    Jun 2 at 16:53

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No, this is not at all true if all we require is that $U$ is free.

Let $\kappa=(2^{\aleph_0})^+$ and let $U_0$ be any free ultrafilter on $\omega$. We define $U=\{A\subseteq\kappa\mid A\cap\omega\in U_0\}$, and we claim that $U$ is a free ultrafilter on $\kappa$.

  • Clearly, $U$ is free, since otherwise there is some $\{x\}\in U$, in which case $\{x\}\cap\omega\in U_0$, but either $\{x\}\cap\omega=\varnothing$, in which case $\{x\}\notin U$ or $x\in\omega$ and $\{x\}\cap\omega=\{x\}$, but $U_0$ is free as well, so $\{x\}\notin U_0$ and therefore not in $U$ either.

  • $U$ is a filter, that much is easy to check.

  • If $A\notin U$, then $A\cap\omega\notin U_0$, which means that $\omega\setminus A\in U_0$, since $\omega\setminus A\subseteq\kappa\setminus A$, we get that $U$ is an ultrafilter indeed.

Since $U_0$ is an ultrafilter on $\omega$, $|U_0|=2^{\aleph_0}<\kappa$, but it is very easy to show that $U$ is the unique ultrafilter on $\kappa$ generated by $U_0$.

The correct generalisation of "free" here is not "containing the cofinite sets", but rather the "small sets". Namely, requiring that $U$ is uniform, or that it avoids $[S]^{<\kappa}$, rather than $[S]^{<\omega}$.

In this case, it is indeed true that any ultrafilter base has size of at least $\kappa^+$, or in other words, given any $A\subseteq U$ such that $|A|\leq\kappa$, the filter that $A$ generates is not $U$ itself, and therefore it can be extended to at least two different ultrafilters.

You can find a short proof in the following paper as Claim 1.2.

Garti, S.; Shelah, S., The ultrafilter number for singular cardinals, Acta Math. Hung. 137, No. 4, 296-301 (2012). ZBL1289.03030.

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