Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) such that $\emptyset\notin E$. We say that $C\subseteq V$ is a *(vertex) cover* if for all $e \in E$ we have $C\cap e\neq \emptyset$. The minimum size that a cover can have is denoted by $\nu(H)$.


A set $M\subseteq E$ is said to be a *matching* if $M$ consists of pairwise disjoint members of $E$. Clearly, for any matching $M\subseteq E$ we have $|M|\leq \nu(H)$. We say that $H$ is *balanced* if there is a matching $M$ with $|M| = \nu(H)$.


**Question.** Let $H=(V,E)$ be a hypergraph such that for all finite $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ is balanced. Does this imply that $H$ itself is balanced?



Hypergraphs with finite matching / covering balance
If infinite edges are allowed, there are trivial counterexamples. Let $H=(V,E)$ where $V$ is an infinite set and $E$ is the set of all cofinite subsets of $V$; then $\nu(H)=\aleph\_0$ while $\nu((V,E\_0))=1$ for any nonempty finite subset $E\_0$ of $E$.


If the edges of the hypergraph $H=(V,E)$ are nonempty **finite** sets, then the answer to your question is yes. Let $M\subseteq E$ be a maximal matching, so that $\bigcup M$ is a vertex cover and $|M|\le\nu(H)\le|\bigcup M|$. If $M$ is infinite then $\nu(H)=|M|$ and we're done, so suppose $M$ is finite. Then $\nu(H)$ is finite, and a simple compactness argument shows that there must be a finite set $E\_0\subseteq E$ with $\nu((V,E\_0))=\nu(H)$. Since $(V,E\_0)$ is balanced, there is a matching $M\_0\subseteq E\_0$ with $|M\_0|=\nu(H)$.


