# Can we recover an inner model of CH after forgetting some generic information?

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, $$G \notin V[H][X]$$.

• Mmm, $V[G]$ maybe? May 17 '20 at 10:37
• @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. May 17 '20 at 11:33
• Actually I'm pretty confident it works. I will add this to the question. May 17 '20 at 11:51

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.)