The starting point of this question is the following true statement for graphs:


A simple, undirected graph $G = (V,E)$ is [bipartite](https://en.wikipedia.org/wiki/Bipartite_graph) if and only if for all $E\_0\subseteq E$ the graph $(V, E\_0)$ is bipartite.


Note that any graph is bipartite if it does not have any odd cycles. Using this, it is not hard to prove the above statement.


A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ is *bipartite* if there is $D\subseteq V$ such that whenever $e\in E$ and $|e|> 1$, we have that $D$ intersects $e$, and also $V\setminus D$ intersects $e$.


One might hope that if $H = (V, E)$ is a hypergraph such that for all finite $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ is bipartite, we get that $H$ itself is bipartite. But if $[\omega]^\omega$ is the collection of infinite subsets of $\omega$, then the hypergraph $(\omega, [\omega]^\omega)$ shows that is not true.


**Question.** Let $H = (V, E)$ be a hypergraph such that there is $n\in \omega$ with $|e|\leq n$ for all $e\in E$ and such that for all finite $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ is bipartite. Does it necessarily follow that $H$ itself is bipartite?



Hypergraphs such that all finite subhypergraphs are bipartite
Even if all edges are finite (not necessarily of uniformly bounded size), this is correct. It follows from the compactness argument: the set $2^V$ of all maps $V\to \{0,1\}$ is a Tychonoff compact set, for each $e$ the set $F(e)$ of all functions which are constant on $e$ is open. So, if such sets cover $2^V$, then finitely many of them cover $2^V$. In other words, if the hypergraph is not bipartite, a certain finite subhypergraph is not bipartite.


