2
$\begingroup$

Let $\kappa$ be a regular, uncountable cardinal. Let $S\subseteq \kappa$ be a closed and unbounded set. Suppose that we partition $S$ into $<\kappa$ pieces. Does one of those pieces contain a closed and unbounded set?

$\endgroup$
2
  • 1
    $\begingroup$ If, instead of ZFC, you assume ZF + Axiom of Determinacy, then the answer is yes for $\kappa=\omega_1$. $\endgroup$ Commented Mar 1 at 16:52
  • $\begingroup$ @LajosSoukup That is exactly the kind of thing I was looking for. Do you have a reference? $\endgroup$ Commented Mar 1 at 17:06

1 Answer 1

8
$\begingroup$

Not necessarily. In fact, we can split $\kappa$ into $\kappa$-many disjoint stationary sets; this (and more) is due to Solovay, see "stationary splitting."

$\endgroup$
1
  • 2
    $\begingroup$ No set in such a splitting can include a club, because the other sets in the splitting are stationary. $\endgroup$ Commented Feb 29 at 16:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .