1
$\begingroup$

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 ${\cal S}$

?

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 8, 2017 at 7:28
  • $\begingroup$ Thanks...! Should I delete the question or do you want to post this as an answer? $\endgroup$ Nov 8, 2017 at 8:06
  • 1
    $\begingroup$ @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? $\endgroup$
    – Asaf Karagila
    Nov 9, 2017 at 4:25

1 Answer 1

1
$\begingroup$

Let $\kappa$ be any infinite cardinal.

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

Then all the given conditions are satisfied.

$\endgroup$
1
  • $\begingroup$ (This solution was inspired by @AsafKaraglia's comment from November 9.) $\endgroup$ Nov 25, 2017 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.