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 $X$.



A property of an ultrafilter
Here's a try for $n=3$ which I think generalizes to any $n$.


Well-order $\mathcal{F}$ as $\{F\_\alpha\}$. By discarding any $F\_\alpha$ which is contained in $\bigcup\_{\beta <\alpha} F\_\beta$, we can ensure that each $F\_\alpha$ contains at least one point which is not in any previous $F\_\beta$, without affecting $U = \bigcup F\_\alpha$.


We can now inductively construct a set $V \subset U$ with the property that $|V \cap F\_\alpha|= 1$ or $2$ for all alpha. At each $\alpha$ at least one element of $F\_\alpha$ is new and we can include it or not to ensure the condition for $F\_\alpha$.


Now $U \setminus V$ has the same property, so since $U \in \mathcal{U}$, wlog we can assume $V \in \mathcal{U}$.


Let $\mathcal{E}\_1 \subseteq \mathcal{F}$ consist of those $F\_\alpha$ which intersect $V$ in exactly one point, and let $\mathcal{E}\_2$ consist of the other $F\_\alpha$, which all intersect $V$ in two points. If the union of $\mathcal{E}\_1$ belongs to $\mathcal{U}$ then we are done. Otherwise the union of $\mathcal{E}\_2$ belongs to $\mathcal{U}$ and this reduces us to the $n=2$ case.


The general reduction turns the problem for an arbitrary $n$ into the same problem for some smaller $n$.


