# Partitionability and colorability of hypergraphs

Motivation. If $$\kappa\neq\emptyset$$ is a cardinal, then a simple, undirected graph $$G=(V,E)$$ is $$\kappa$$-colorable if and only if there is a partition of $$V$$ into at most $$\kappa$$ blocks such that every edge $$e\in E$$ intersects $$2$$ blocks.

$$\kappa$$-colorability of a hypergraph. Let $$H=(V,E)$$ be a hypergraph, that is $$V$$ is a set and $$E\subseteq {\cal P}(V)$$. If $$\kappa \neq \emptyset$$ is a cardinal, we say that a map $$c:V\to \kappa$$ is a (hypergraph) coloring if for every $$e\in E$$ with $$|e|>1$$ the restriction $$c|_e$$ is non-constant.

$$\kappa$$-partitionability of a hypergraph. If $$H=(V,E)$$ is a hypergraph and $$\kappa\neq\emptyset$$ is a cardinal, we say that $$H$$ is $$\kappa$$-partitionable if there is a partition of $$V$$ into at most $$\kappa$$ blocks such that for every $$e\in E$$ with $$|e|>1$$ we have that $$e$$ is not a subset of any block of the partition.

It is easy to see that if $$H$$ is $$\kappa$$-partitionable, then it is $$\kappa$$-colorable (just give every block a different color). Does the converse hold?

• If the answer is "Yes", then this could be the motivation for the fact that the requirement that a coloring be non-constant on non-singleton edges (instead of injective). – Dominic van der Zypen Feb 3 at 7:26
• How is a partition functionally different from a coloring? Gerhard "If You Label, It Isn't" Paseman, 2020.02.03. – Gerhard Paseman Feb 3 at 8:01
• Oh right, via pre-images, I suppose... – Dominic van der Zypen Feb 3 at 8:46