# Choice sets in “thick” sets of sets

Let $X\neq \emptyset$ be a set and let ${\cal S} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ be a collection of non-empty subsets of $X$. We say $C\subseteq X$ is a choice set for ${\cal S}$ if for all $s\in S$ we have $|s\cap C| = 1$. As Bjørn Kjos-Hanssen pointed out, choice sets do not always exist.

Is there an infinite cardinal $\kappa$ and ${\cal S} \subseteq {\cal P}(\kappa)\setminus\{\emptyset\}$ such that

1. for all $x\in\kappa$ we have $|\{s\in {\cal S}: x\in s\}| = \kappa$, every member of ${\cal S}$ has cardinality $\kappa$, and
2. there is a choice set $C\subseteq \kappa$ for ${\cal S}$

?

• Are you sure that this is what you want? Seems easy: one has to construct functions $f_\alpha:\kappa\to\kappa$ ($\alpha<\kappa$) such that each $f_\alpha$ attains 0 at exactly one point and for any pair $(\xi,\eta)\in \kappa\times \kappa$ there are $\kappa$ many $f_\alpha$ such that $f_\alpha(\xi)=\eta$. This can be done by bookkeeping. – Péter Komjáth Nov 8 '17 at 7:28
• Thanks...! Should I delete the question or do you want to post this as an answer? – Dominic van der Zypen Nov 8 '17 at 8:06
• @Péter: Isn't it easier to consider the tail segments of the ordinal $\kappa+1$, with the exception of $\{\kappa\}$, with the choice set being just $\{\kappa\}$ itself? – Asaf Karagila Nov 9 '17 at 4:25

Let $\kappa$ be any infinite cardinal.
Let $\cal S=\{\kappa\setminus\{x\}:0<x<\kappa\}$ and $C=\{0\}$.