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, 2020 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, 2020 at 11:33
  • $\begingroup$ Actually I'm pretty confident it works. I will add this to the question. $\endgroup$ May 17, 2020 at 11:51

1 Answer 1


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.)


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