TITLE: A property of an ultrafilter
QUESTION [9 upvotes]: Let $\mathcal U$ be a free ultrafilter on a set $X$. For $n\in\mathbb N$ let $\mathcal F$ be a family of $n$-element subsets of $X$ such that $\bigcup\mathcal F\in\mathcal U$.

Question. Is there a set $U\in\mathcal U$ and a subfamily $\mathcal E\subset\mathcal F$ such that $\bigcup\mathcal E\in\mathcal U$ and $|U\cap E|=1$ for every $E\in\mathcal E$?

I do not know the answer even for $n=2$ and countable set $X$.

REPLY [2 votes]: Consider this with assuming that $\mathcal{F}$ consists of finite subsets of bounded cardinal; let $n$ be the max of these cardinals and let us argue by induction on $n$. The goal is to find $\mathcal{E}\subset\mathcal{F}$ and $U\in\mathcal{U}$ such that $U\subset\bigcup\mathcal{E}$ and and $|E\cap U|\le 1$ for every $E\in\mathcal{E}$.
First, consider a maximal subset $\mathcal{E}$ of the cover such that no element is in the union of others (we call this ``minimal"). So $V:=\bigcup\mathcal{F}=\bigcup\mathcal{E}$. Write $V=V_1\cup V_2$, where $V_2$ is the set of elements of $V$ that belong to at least two elements of $\mathcal{E}$. The minimality of $\mathcal{E}$ implies that no $E\in\mathcal{E}$ is contained in $V_2$ (i.e., has nonempty intersection with $V_1$).
Since the intersection of elements of $\mathcal{E}$ with $V_1$ are pairwise equal or disjoint, we can partition $V_1$ into $n$ subsets each intersecting, each element of $\mathcal{E}$ in at most a singleton. Hence we can conclude if $V_1\in\mathcal{U}$ (with $U$ being one of those $n$ subsets of $V_1$).
Otherwise, $V_2\in\mathcal{U}$. Since $|E\cap V_2|\le n-1$ for all $E\in\mathcal{E}$, we can conclude by induction (and find $U\subset V_2$).
PS Nik Weaver posted an answer while I was writing this one, but I still post, although I guess it's the same idea.