Suppose $\kappa$ is an inaccessible cardinal. Let $G \times H$ be $\mathrm{Col}(\omega_1,{<}\kappa) \times \mathrm{Add}(\omega,\kappa)$-generic over $V$. Let $X \subseteq \kappa$ be $\mathrm{Add}(\kappa,1)$-generic over $V[G][H]$. Since $X$ codes every bounded subset of $\kappa$ as an interval-subsequence, $V[X] \models \kappa = \omega_2 = 2^\omega$. Does there exist an inner model of $V[X]$ with the same cardinals satisfying CH?

Note: By arguments similar to those of Section 2.1 here, $G \notin V[H][X]$.

  • $\begingroup$ Mmm, $V[G]$ maybe? $\endgroup$
    – Asaf Karagila
    May 17 '20 at 10:37
  • $\begingroup$ @AsafKaragila: Probably $G \notin V[X]$. I haven't worked out a proof of this, but it is similar to the situation discussed in Section 2.1 of my paper arxiv.org/abs/1901.01160. $\endgroup$ May 17 '20 at 11:33
  • $\begingroup$ Actually I'm pretty confident it works. I will add this to the question. $\endgroup$ May 17 '20 at 11:51

The following answers the question as posed, but is a bit unsatisfactory since we will find a choiceless inner model.

In $V[X]$, let $F = \{ x \subseteq \omega_1 : \forall \alpha < \omega_1(x \cap \alpha \in V) \}$. Clearly $\mathcal P(\omega_1)^{V[G]} \subseteq F$. We claim that $\mathcal P(\omega_1)^{V[G]} = F$ using:

Lemma (Mitchell): For all $\lambda$, $\mathrm{Add}(\omega,\lambda)$ has the $\omega_1$-approximation property.

This means that any $x \subseteq \omega_1$ which is in $V[G][H] \setminus V[G]$ must have some initial segment not in $V[G]$, and thus not in $V$.

We consider the model $V(F) \subseteq V[G] \cap V[X]$. Since $\mathbb R^{V[G]} = \mathbb R^V$, $V(F)$ satisfies CH. Since it has the same subsets of $\omega_1$ as $V[G]$, it satisfies $\kappa = \omega_2$. By standard homogeneity arguments, $V(F)$ does not have a well-ordering of $F$.

At least we can say that $V[X]$ satisfies weak square $\square^*_{\omega_1}$. (The motivation for the question had to do with the tree property.)


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.