Solovay's Theorem states

If $\kappa$ is regular uncountable, then any stationary set in $\kappa$ can be partitioned into $\kappa$-many pairwise disjoint stationary sets.

For regular uncountable successor $\kappa$, this follows from a result by Ulam [Kunen, p.79].

My questions are:

1) Can I apply the theorem for $\kappa=\mathfrak c =2^\omega$, when CH fails ($\mathfrak c >\omega_1$)? (is $\mathfrak c$ necessarily regular?).

2) Might I be able to prove the special case in (1) using methods less sophisticated than those found in Solovay's proof? Something like Ulam's matrix?


  • 3
    $\begingroup$ The only requirement in $\sf ZFC$ that $\frak c$ must satisfy is that $\operatorname{cf}(\mathfrak c)>\aleph_0$. In particular it is consistent that $\mathfrak c=\aleph_{\omega_1}$, in which case it is singular, and the theorems no longer apply. $\endgroup$ – Asaf Karagila Jul 7 '13 at 22:41
  • 4
    $\begingroup$ Pretty much the proof will come down to whether $\mathfrak{c}$ is a successor cardinal or simply just regular. If it is successor than you can use an Ulam matrix, whereas if it is just regular I believe you will have to use Solovay's approach. And as Asaf points out it is consisten that $\mathfrak{c}$ is singular in which case neither method work. $\endgroup$ – Ryan Sullivant Jul 7 '13 at 22:46
  • 3
    $\begingroup$ Solovay's approach is really not that complicated, in hindsight. (The presentation in Introduction to Cardinal Arithmetic, by M. Holz, K. Steffens, and E. Weitz is particularly accessible.) I do not know of a proof specific to the continuum, or of any "easy" approaches for (small?) weakly inaccessibles. $\endgroup$ – Andrés E. Caicedo Jul 7 '13 at 22:50
  • 2
    $\begingroup$ There is no forcing anywhere in the proof. $\endgroup$ – Andrés E. Caicedo Jul 7 '13 at 23:09
  • 2
    $\begingroup$ See theorem 3 here. $\endgroup$ – Andrés E. Caicedo Jul 7 '13 at 23:12

Your Answer

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

Browse other questions tagged or ask your own question.