5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – Asaf Karagila Aug 28 '14 at 14:16
  • $\begingroup$ Asaf, the models are all countable closed, and so they all have the sequence of alpha_n. $\endgroup$ – Joel David Hamkins Aug 28 '14 at 16:13
  • $\begingroup$ @Joel: Yes, you're right. Thanks. :) $\endgroup$ – Asaf Karagila Aug 28 '14 at 16:19

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.