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For sets $A, B$ we write $B^A$ for the set of all functions $f:A\to B$.

Let $H = (V,E)$ be a hypergraph such that $V,E\neq\varnothing$ and $|e| \geq 2$ for all $e\in E$. Let $\kappa>1$ be a cardinal, finite or infinite, and let $c:V\to \kappa$ be a map. By $\text{Mono}(H,c)$ we denote the set of mono-chromatic edges with respect to $c$, that is

$$\text{Mono}(H,c)=\{e\in E: c\restriction_e \text{ is constant}\}.$$

Let ${\cal M}(H, \kappa) = \{\text{Mono}(H,c): c\in \kappa^V\}$, ordered by set inclusion.

Question. Do we necessarily have $\text{Min}({\cal M}(H, \kappa)) \neq \varnothing$?

Notes.

  1. If $(P,\leq)$ is a partially ordered set, by $\text{Min}(P)$ we denote the set of minimal elements of $P$.
  2. We have that $\varnothing$ is an element (the only element) of $\text{Min}({\cal M}(H, \kappa))$ if and only if $\kappa$ is at least the hypergraph chromatic number of $H=(V,E)$.
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  • $\begingroup$ So just to be clear the hypergraph $H$ itself may be infinite? Does that mean both infinite number of vertices, infinite number of edges, and possibly the edges contain infinitely many vertices? $\endgroup$ Commented Jan 27, 2021 at 13:35
  • $\begingroup$ That's right @SamHopkins - sorry for not specifying this in the post. For finite $H$, the poset ${\cal M}(H,\kappa)$ is finite as well, so it contains minimal elements, i.e. $\text{Min}({\cal M}(H,\kappa))\neq \varnothing$ . $\endgroup$ Commented Jan 27, 2021 at 21:07
  • $\begingroup$ Oh, right -- thanks @bof! $\endgroup$ Commented Jan 28, 2021 at 10:58

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Let $H=(V,E)$ be a hypergraph, $\kappa$ a cardinal.

Observation 1. $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$ if and only if $H$ has a maximal $\kappa$-colorable subhypergraph.

Proof. If $F\subseteq E$, then $(V,F)$ is a maximal $\kappa$-colorable subhypergraph of $H$ if and only if $E\setminus F$ is a minimal element of $\mathcal M(H,\kappa)$.

Observation 2. If $\kappa$ is infinite, then $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$ if and only iff $H$ is $\kappa$-colorable. (Hence we get a simple example where $\operatorname{Min}(\mathcal M(H,\kappa))=\emptyset$ by taking $\kappa=\aleph_0$ and $H=K_{\aleph_1}$, the complete graph on $\aleph_1$ vertices.)

Proof. For the nontrivial direction, since $\kappa+1=\kappa$, we can always make another hyperedge non-monochromatic by creating a new color.

Observation 3. If $\kappa$ is finite and all hyperedges (elements of $E$) are finite, then $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$.

Proof. By Observation 1, we have to prove that $H$ has a maximal $\kappa$-colorable subhypergraph. This follows from Zorn's lemma and the hypergraph version of the De Bruijn–Erdős theorem, which says that for finite $\kappa$ a hypergraph with finite hyperedges is $\kappa$-colorable iff all of its finite subhypergraphs are $\kappa$-colorable.

Observation 4. If $1\lt\kappa\lt\aleph_0$ and if $E$ is a nonprincipal ultrafilter on $V$, then $\operatorname{Min}(\mathcal M(H,\kappa))=\emptyset$.

Proof. Consider any coloring $c:V\to\kappa$. Since $E$ is an ultrafilter and $\kappa$ is finite, there is some $i\in\kappa$ such that $V_i=\{v\in V:c(v)=i\}\in E$, whence $V_i\in\operatorname{Mono}(H,c)$. Now we can make $V_i$ non-monochromatic by changing the color of one vertex in $V_i$; and this will not result in any previously non-monochromatic hyperedge becoming monochromatic, because each hyperedge has infinite intersection with $V_i$. Therefore $\operatorname{Mono}(H,c)$ is not a minimal element of $\mathcal M(H,\kappa)$.

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  • $\begingroup$ Sorry if I repeat myself, but thank you @bof for another amazing answer! $\endgroup$ Commented Jan 31, 2021 at 20:23

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