Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). We say that $C\subseteq V$ is a *choice set* if $|C\cap e| = 1$ for all $e\in E$.


**Question.** Let $H=(V,E)$ be a hypergraph with $e$ finite for all $e\in E$, and suppose that for all finite sets $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ has a choice set. Does $H$ itself necessarily have a choice set?



Choice sets in hypergraphs with finite edges
For finite edges this follows indeed from compactness. Consider the Cartesian product $K=\prod e$ of all edges in Tychonoff topology. This $K$ may be understood as the set of simultaneous choices $v\_e\in e$ for all edges $e$. For every pair $e\ne f$ of edges consider the open set $U(e,f)\subset K$ determined as follows: a vertex chosen in $e$ belongs to $f$. If these open sets do not cover $K$, then a choice set exists. If they do cover $K$, then finitely many of them cover $K$, that means that the choice set does not exist for finitely many edges.


