If $\kappa$ is an inaccessible cardinal and $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ is a $V$-generic filter, then in $V[G]$ the Chang model $L(\text{Ord}^\omega)$ satisfies "every set of reals has the classical regularity properties" by Solovay's theorem. (By the classical regularity properties I mean Lebesgue measurability, the Baire property, and the perfect set property.)

If instead $\kappa$ is a limit of Woodin cardinals (not necessarily inaccessible, possibly singular,) $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ is a $V$-generic filter, and we define $\mathbb{R}^*_G = \bigcup_{\alpha < \kappa} \mathbb{R}^{V[G\restriction \alpha]}$, then the model $L(\mathbb{R}^*_G)$ satisfies the Axiom of Determinacy (and hence also the weaker statement that every set of reals has the classical regularity properties) by a theorem of Woodin.

Now let's assume that $\kappa$ is inaccessible and is a limit of Woodin cardinals. Again let $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ be a $V$-generic filter. Because $\kappa$ is inaccessible we have $\mathbb{R}^*_G = \mathbb{R}^{V[G]}$. In this case it is possible to obtain a mutual strengthening of the two results mentioned above:

In $V[G]$, does the Chang model $L(\text{Ord}^\omega)$ satisfy $\mathsf{AD}$?

(I am aware that Woodin has proved that the much stronger hypothesis of the existence of a proper class of Woodin limits of Woodin cardinals implies that the Chang model of $V$ satisfies $\mathsf{AD}$.)

  • 1
    $\begingroup$ Nice question. This is probably open. Have you seen Bill's analysis of the sharp of the model? $\endgroup$ – Andrés E. Caicedo Jun 10 '15 at 2:37
  • $\begingroup$ @Andres I heard him talk about it in Toronto a few years ago; I will need to refresh my memory. $\endgroup$ – Trevor Wilson Jun 10 '15 at 2:40
  • $\begingroup$ Since every real in $V[G]$ was introduced by a bounded part of the forcing, this is quite the same idea as the usual proof that collapsing a limit of Woodin cardinals and taking a symmetric submodel (or $\rm HOD(X)$ for a suitable $\rm X$) satisfies $\sf AD$ in $L(\Bbb R)$; where do the usual arguments fail here? $\endgroup$ – Asaf Karagila Jun 10 '15 at 13:45
  • $\begingroup$ @Asaf Here is one place where things could go wrong. First, we would want to show that if a $\Sigma^2_1$ sentence $\varphi$ (such as $\neg \mathsf{AD}$) holds in the Chang model of $V[G]$, then it holds in the Chang model of $V$. Let's think about how to do this using genericity iterations. (With the stationary tower, it should be similar.) We would take a countable elementary $\pi_0 : P_0 \to V_\theta$ for some large $\theta$ and, working in $V^{\text{Col}(\omega,\mathbb{R})}$, do a genericity iteration $P_0 \to P_\omega$ to make all reals of $V$ generic over $P_\omega$ for ... $\endgroup$ – Trevor Wilson Jun 10 '15 at 22:11
  • $\begingroup$ ... the Levy collapse at its limit of Woodin cardinals, so that we can define the symmetric extension $P_\omega(\mathbb{R}^V)$ and get a set of reals witnessing $\varphi$ there. Although the $L(\mathbb{R})$ of $P_\omega(\mathbb{R}^V)$ is contained in the $L(\mathbb{R})$ of $V$, it's not clear to me that the Chang model of $P_\omega(\mathbb{R}^V)$ is contained in the Chang model of $V$. $\endgroup$ – Trevor Wilson Jun 10 '15 at 22:16

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.