Let $X$ be a set and ${\cal P}(X)$ its powerset. We say that ${\cal F} \subseteq {\cal P}(X)$ has the splitting property (SP) if there is $A\in {\cal P}(X)$ such that for all $F\in {\cal F}$ we have $$F \cap A \neq \emptyset \neq F\cap (X\setminus A).$$

Let $\text{SP}(X)$ denote the collection of all subsets of ${\cal P}(X)$ with (SP), and we order it with $\subseteq$.

If $X$ is infinite, and ${\cal F}\in \text{SP}(X)$, is there ${\cal M}\in\text{SP}(X)$ such that ${\cal M}$ is maximal in $(\text{SP}(X),\subseteq)$ and ${\cal F}\subseteq {\cal M}$?

(The obvious tool to try to use, Zorn's Lemma seems to be of no help in this, but I might be wrong.)

  • $\begingroup$ You could just say that $\mathcal F$ is $2$-colorable, i.e., the hypergraph $(X,\mathcal F)$ has chromatic number $\le2.$ $\endgroup$
    – bof
    Aug 13, 2018 at 11:46

1 Answer 1


Yes, any family of subsets of $X$ with the splitting property can be extended to a maximal such family. The axiom of choice is not needed for this.

Suppose $\mathcal F\in \operatorname{SP}(X),$ and let $A\in\mathcal P(X)$ be such that for all $F\in\mathcal F$ we have $$F\cap A\ne\emptyset\ne F\cap(X\setminus A).$$

We may assume that $\mathcal F\ne\emptyset,$ so that $\emptyset\ne A\ne X.$ Now it is easy to see that the set $$\mathcal M=\{F\in\mathcal P(X):F\cap A\ne\emptyset\ne F\cap(X\setminus A)\}$$ has the desired properties.


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