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?

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.


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.

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.

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.

  • $\begingroup$ I guess you can show that large cardinals are required if Solovay's theorem fails in the extreme way that there were no disjoint stationary subsets of a cardinal kappa at all, because this is equivalent to asserting that the club filter is an ultrafilter, which implies that kappa is measurable. Perhaps you can show that if Solovay's theorem on omega_1 fails in the stated way, then omega_1 must be measurable? And I would guess that you don't need the full strength of AD to make omega_1 measurable in a ZF model. What happens if you Levy collapse a measurable kappa and build a symmetric model? $\endgroup$ – Joel David Hamkins Feb 2 '10 at 4:30
  • 4
    $\begingroup$ Indeed if you Levy collapse a measurable to become $\omega_1$ there's a symmetric submodel in which it is still measurable, and still $\omega_1$. Ditto for weakly compact. math.uni-bonn.de/people/dimitri/Symmetric_models_basics.pdf gives a nice exposition of these facts. $\endgroup$ – Justin Palumbo Feb 2 '10 at 5:43
  • $\begingroup$ Thanks -- this fully answers my question. What does $\Theta$ refer to in the second paragraph? $\endgroup$ – Norman Lewis Perlmutter Feb 3 '10 at 16:42
  • $\begingroup$ It's the least ordinal that the reals cannot be surjected onto. $\endgroup$ – Justin Palumbo Feb 3 '10 at 17:03
  • 4
    $\begingroup$ I think, under DC, the club filter on $\omega_1$ being an ultrafilter is equiconsistent with the existence of some $\kappa$ which is $\kappa^+$-Mitchell (Mitchell). Without DC, the consistency does not require large cardinals (Spector, JSL, 1981). $\endgroup$ – Péter Komjáth Jun 14 '10 at 5:01

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.

  • 1
    $\begingroup$ There is certainly not a known proof without choice -- otherwise the open question as to whether the Kunen inconsistency can be proven in ZF would be solved. $\endgroup$ – Norman Lewis Perlmutter Feb 10 '10 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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