# A variant of Chang's model with choice

Let $M_n$, $n < \omega$, be a models of $ZFC$ with the same ordinals, closed under countable sequences. Let $\alpha_n$ be an ordinal which is a regular cardinal in $M_n$.

Question 1: Is it possible to get a model $M$ of $ZFC$ with the same ordinals such that for every $n$, $M \models \alpha_n$ is regular but $M$ is closed under countable sequences?

Note that in Chang's model, $L(On^\omega )$, each one of the ordinals $\alpha_n$ is regular, but it maybe a model of $\neg AC$. So, my question is whether there is always a model that contains the Chang's model, satisfies choice, but still believes that every one of the ordinals $\alpha_n$ is a regular cardinal.

I believe that the answer is negative and I'm interested in the consistency strength of the negative answer.

Question 2: What about two models? Assume that there are two models of ZFC $M_0, M_1$, with the same ordinals, both countably closed and $M_0 \models \alpha_0$ is regular, $M_1 \models \alpha_1$ is regular. Does there exist a countably closed model $M$ of ZFC such that its ordinals are $M_0 \cap On = M_1 \cap On$ and $M\models \alpha_0, \alpha_1$ are both regular?

• Is the sequence of the $\alpha_n$'s bounded internally? (I assume it is, and you just want to talk about inner models, just clearing that point up) – Asaf Karagila Aug 28 '14 at 14:16
• Asaf, the models are all countable closed, and so they all have the sequence of alpha_n. – Joel David Hamkins Aug 28 '14 at 16:13
• @Joel: Yes, you're right. Thanks. :) – Asaf Karagila Aug 28 '14 at 16:19