The De Bruijn–Erdős theorem states that when all finite subgraphs of a graph $G$ can be colored with $n$ colors, the same is true for the whole graph.

There is a natural notion of coloring for hypergraphs which is as follows. Let $H= (V, E)$ be a hypergraph, and let $\kappa\neq \emptyset$ be a cardinal. Then a map $c:V\to\kappa$ is said to be a (hypergraph) coloring if the restriction $c\restriction_e : e \to \kappa$ is non-constant whenever $e$ has more than $1$ element.

Is the following statement true?

Let $n>1$ be an integer, and let $H=(V,E)$ be a hypergraph such that for all finite $E_0\subseteq E$, the hypergraph $(V,E_0)$ can be colored with $n$ colors. Then $H$ can be colored with $n$ colors.


1 Answer 1


The space $X=\{1,\dots,n\}^V$ of all colorings (proper or not) of $H=(V,E)$ with $n$ colors is compact in the product topology. Given a finite set $F \subset E$, the set $K_{F}$ of proper colorings of $(V,F)$ is a closed set in $X$.

For any finite collection $F_1,\dots,F_k$ of finite subsets of $E$, the intersection $$\cap_{j=1}^k K_{F_j}=K_{\cup_{j=1}^k F_j}$$ is nonempty by the given hypothesis. Therefore, by compactness of $X$, the intersection $$\bigcap \{K_F : \, F \; \text{finite}, \; \, F \subset E \} $$ is nonempty, and any coloring in this intersection is a proper coloring of $H=(V,E)$.

  • $\begingroup$ Very cool - I would never have thought of applying compactness and Tikhonov here! $\endgroup$ Commented Oct 17, 2022 at 8:08

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