Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:42:09Z http://mathoverflow.net/feeds/question/13770 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13770/model-of-zf-negc-in-which-solovays-theorem-on-stationary-sets-fails Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? Norman Lewis Perlmutter 2010-02-02T01:46:33Z 2010-02-07T03:28:53Z <p>It is a theorem of Solovay that any stationary subset of a regular cardinal, $\kappa$ can be decomposed into a disjoint union of $\kappa$ many disjoint stationary sets. As far as I know, the proof requires the axiom of choice. But is there some way to get a model, for instance a canonical inner model, in which ZF + $\neg$C holds and Solovay's Theorem fails?</p> <p>I am interested in this problem because Solovay's theorem can be used to prove the Kunen inconsistency, that is, that there is no elementary embedding j:V -->V, where j is allowed to be any class, under GBC. The Kunen inconsistency may be viewed as an upper bound on the hierarchy of large cardinals. Without choice, no one has yet proven the Kunen inconsistency (although it can be proven without choice if we restrict ourselves to definable j). So if there is hope of proving Solovay's Theorem without choice, we could use this to prove the Kunen inconsistency without choice.</p> http://mathoverflow.net/questions/13770/model-of-zf-negc-in-which-solovays-theorem-on-stationary-sets-fails/13783#13783 Answer by Justin Palumbo for Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? Justin Palumbo 2010-02-02T03:13:18Z 2010-02-02T04:21:57Z <p>The Axiom of Determinacy (AD) implies that the club filter on $\omega_1$ (the subsets of $\omega_1$ containing a club) is an ultrafilter. Certainly if that is the case then we can't even decompose $\omega_1$ into two disjoint stationary sets, because one of them would have to contain a club. Assuming sufficient large cardinal hypotheses (infinitely many Woodin cardinals and a measurable cardinal above them) one has that $L(\mathbb{R})$ satisfies AD, and hence that is a canonical inner model of the form I think you are looking for.</p> <p>What about above $\omega_1$? I believe it is a theorem of John Steel's that (again under large cardinal assumptions) in $L(\mathbb{R})$ for any regular $\kappa$ below $\Theta$, the $\omega$-club filter on $\kappa$ is an ultrafilter. (An $\omega$-club is an unbounded set closed under countable limits). So for such $\kappa$ the stationary set of ordinals of countable cofinality cannot be partitioned into two disjoint stationary sets. I don't know about getting Solovay's theorem to fail at cardinals higher than that.</p> <p>Also, that all assumes some large cardinals. I do not know if large cardinal assumptions are necessary to get the failure of Solovay's Theorem.</p> http://mathoverflow.net/questions/13770/model-of-zf-negc-in-which-solovays-theorem-on-stationary-sets-fails/14455#14455 Answer by alephomega for Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? alephomega 2010-02-07T03:28:53Z 2010-02-07T03:28:53Z <p>It seems to me that choice is used in a crucial way in Solovay's theorem. I don't know if there are any known proof without choice.</p>