# Hypergraphs with finite matching / covering balance

Let $$H=(V,E)$$ be a 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?

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 $$|M_0|=\nu(H)$$.