# A question on existence of ultrafilters and cardinality.

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 $$ , is it true that there exists an ultrafilter $$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 $$ finite sets, so has cardinal $$. 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.

• 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 $\cap A_n= \varnothing$. Jun 2 at 15:29
• And how can it be proved if the cardinality of $P$ is less than the cofinality of $k$? Any references? Jun 2 at 15:42
• I don't understand "If all the subsets of $P$ are cofinal" but maybe you meant "If all the elements of $P$ are cofinite". Jun 2 at 16:13
• @AndreasBlass thanks, you are right, sorry. I corrected the question. Jun 2 at 16:53

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.