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In $V$, let me call a set theory structure A is a $\omega_1$ model if the $\omega_1$ of $A$ is the same as the $\omega_1$ in $V$ (up to isomorphism). The question I would like to ask is the following: Given any transitive structure $(B,\in)$ containing $\omega_1$ in $V$, is it always possible to find an $\omega_1$ model $A$ such that $B\prec A$ and $B\cap H(\omega_2)$ is in the well-founded part of $A$.

Some remark: Lowenheim-Skolem or Compactness theorem can be employed to construct an extension. But in general this extension is not an $\omega_1$ model. To get one $\omega_1$ model, one can either use some large cardinal or force a generic ultrafilter to construct an ultrapower which is well founded below $\omega_1$. But I am not sure how to get this using ZFC alone. On the other hand, it is known that the existence of $\omega$-model is correct between transitive ZFC models and thus can be constructed under ZFC.

I know very little about model theory of set theory. So it would be very helpful if anyone can provide some reference about this topic.

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  • $\begingroup$ Sorry, I do mean $H(\omega_2)^B$ is in the well-founded part of $H(\omega_2)^A$. It would be better $B$ itself is an element in the well-founded part of $A$. In this case, the $\omega_2$ of $A$ and $B$ are not correct. $\endgroup$ – user47286 Feb 22 '14 at 15:24
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In this generality, this is impossible, if you want $B\prec A$ to be a proper extension.

For a counterexample, let $B=V_{\omega_2}$, the rank initial segment of the universe of height $\omega_2$. I claim that there can be no proper extension $\langle B,\in\rangle\prec\langle A,\hat\in\rangle$, where the well-founded part of $A$ includes $H_{\omega_2}$. Suppose that we have such an $A$, and let $\beta$ be any of the new ordinals in $A$, that are not in $B$. (There must be such a new ordinal, since $B$ thinks every set has an ordinal rank, and for the ranks up to $\omega_2$, the original structure $B$ already had all the sets of that rank.) So $\beta$ is larger than every ordinal in $B$. But since $B$ thinks every ordinal has size $\omega_1$, it follows that $A$ thinks it has a bijection of $\beta$ with $\omega_1$. But it is correct about $\omega_1$, and so we can form an actual bijection of the objects that $A$ thinks are below $\beta$ with the actual $\omega_1$. This is impossible, since the former set includes all of $\omega_2$. Contradiction.

In the comments, you inquire whether one can find a counterexample $B$ of size $\omega_1$, and the answer is yes. Let $B=L_{\omega_1+1}$, which is a transitive set that contains $\omega_1$ as an element (I mean to use true $\omega_1$ here, which might be larger than $\omega_1^L$), and indeed $\omega_1$ is the largest ordinal of $B$. Suppose that $\langle B,\in\rangle\prec \langle A,\hat\in\rangle$ is a nontrivial extension, where $A$ is an $\omega_1$-model and $B\cap H_{\omega_2}=B$ is in the well-founded part of $A$. It follows that $A$ is well-founded up to and including $\omega_1$, and by elementarity, since $\omega_1$ is the largest ordinal of $B$, it must also be the largest ordinal of $A$. So the ordinals of $A$ are precisely $\omega_1+1$. Since also $A\models V=L$, it follows by elementarity that $A=L_{\omega_1+1}$, which is to say, that $A=B$, and so this isn't a nontrivial extension after all. Contradiction. Perhaps this example shows that you haven't quite asked the question you intended to ask.

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  • $\begingroup$ Thanks. It's great. And does your answer work for $\omega_1$-size $B$? $\endgroup$ – user47286 Feb 22 '14 at 15:36
  • $\begingroup$ This argument does not seem to work with $B$ of size $\omega_1$, since it used the fact that all ordinals up to $\omega_2$ were in $B$. $\endgroup$ – Joel David Hamkins Feb 22 '14 at 17:18
  • $\begingroup$ But meanwhile, I've found a different counterexample of size $\omega_1$. $\endgroup$ – Joel David Hamkins Feb 23 '14 at 0:31
  • $\begingroup$ I see. There is no such $A$ unless $B$ is closed enough. $\endgroup$ – user47286 Feb 23 '14 at 3:15

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