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Let $M$ denote a set-sized model of $\mathrm{ZFC},$ not necessarily well-founded. Is it true that for any $x \in M,$ there is an outer model $O$ of $M$ such that $(O \models x \mbox{ is countable})$?

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  • $\begingroup$ How precisely do you define outer models of $M$? Do you mean with the same ordinals, and an end-extension of $M$? (End-extension: If $M'$ is the outer model, $y\in M$, and $M'\models x\in y$, then $x\in M$.) In that case, the answer is negative in general, if by "there is" you mean "in $V$". If you mean instead whether there is such a potential outer model (in a Boolean valued extension of the universe, for instance), then yes. $\endgroup$ – Andrés E. Caicedo Jun 2 '14 at 13:47
  • $\begingroup$ @AndresCaicedo, you write: "Do you mean with the same ordinal, and an end-extension of $M$"? I think so. I mean that $O$ believes that $M$ is an inner model; so in particular, $O$ believes that $M$ is a transitive class-model with the same ordinals. And yes, I mean in $V$ rather than in some outer model of $V.$ Does that answer the question? $\endgroup$ – goblin Jun 2 '14 at 14:01
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(Following the clarification in the comments,) I am interpreting an outer model of $M$ as a model $M'$ with the same ordinals that is an end-extension of M. (End-extension: If $M′$ is the outer model, $y\in M$, and $M′\models x\in y$, then $x\in M$.) In that case, the answer is negative in general, if by "there is" one means "in $V$".

For instance: We could have an $\alpha$ such that $V_\alpha$ is a model of $\mathsf{ZFC}$, and a set $x$ that is uncountable there (and therefore in $V$). If $M'$ thinks that there is a bijection $f$ between $\omega$ and $x$, this gives us a bijection (in $V$) between $\omega$ and $x$, simply setting $\hat f=\{(n,t)\mid M'\models f(n)=t\}$.

Of course, the answer could be yes in some cases: It could be that $M$ is countable, in which case there are (in $V$) models that are (isomorphic to) generic extensions of $M$ where the set in question is countable. Or it could even be that $M$ is uncountable, but "thin". For instance, if $0^\sharp$ exists, then there are many sets that are uncountable in $L$ but countable in $V$.

If one means instead whether there is such a potential outer model (in a Boolean valued extension of the universe, for instance), then yes: One can simply pass to a forcing extension $V'$ of $V$ where (the transitive closure of) $M$ is countable, and then in $V'$ we have models $M'$ that are (isomorphic to) generic extensions of $M$ where the relevant sets are now countable.

Let me add a short remark, since the result seems interesting.

Suppose that $M$ is a transitive model of $\mathsf{ZFC}$. Then for every $x\in M$ there is an outer model $M'$ where $x$ is countable iff $\mathsf{ORD}\cap M\le\omega_1$.

(In the statement above, we also allow the possibility that $M$ is a proper class.) In one direction, suppose that $\mathsf{ORD}\cap M>\omega_1$. We then have that $\omega_1^V\in M$ and there is no outer model where $\omega_1^V$ is seen countable. This is as in the second paragraph above.

Suppose now that $\mathsf{ORD}\cap M\le \omega_1$. Since choice holds in $M$, for every $\alpha<\mathsf{ORD}\cap M$ we have that $V_\alpha^M$ is in bijection (in $M$, and therefore in $V$) with some ordinal of $M$, so each $V_\alpha^M$ is countable, and so is $\mathcal P(V_\alpha^M)^M$. It follows that generics for any poset $\mathbb P\in M$ exist in $V$ (because $\mathbb P\in V_\alpha^M$ for some $M$, and the number of dense subsets of $\mathbb P$ that belong to $M$ is countable in $V$). In particular, given any $x\in M$, this is true for $\mathbb P=\mathrm{Col}(\omega,x)$, the poset that adds a surjection from $\omega$ onto $x$. It follows that if $G\in V$ is $\mathbb P$-generic over $M$, then $M[G]\models x$ is countable (and $M[G]$ is an outer model of $M$).

The fact that choice holds in $M$ is essential here. As explained here, it is consistent to have transitive models $M$ of $\mathsf{ZF}$ of countable height but uncountable size. This implies that for some $\alpha<\mathsf{ORD}\cap M$, we have that $V_\alpha^M$ is uncountable (in $V$ and thus also in $M$), and therefore there is no outer model where it would be seen to be countable.

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  • $\begingroup$ How can we be sure that an outer model $V'$ exists in which the transitive closure of $M$ is countable, given that the first part of your answer appears to explain why in fact we cannot, in general, do just that? $\endgroup$ – goblin Jun 2 '14 at 14:17
  • $\begingroup$ This uses forcing: We can take as $V'$ the Boolean valued model $V^{\mathbb B}$, where $\mathbb B$ is the Boolean completion of the poset $\mathrm{Col}(\omega,\kappa)$, where $\kappa$ is the cardinality of the transitive closure of $M$. The effect of going to this extension is precisely to add a surjection $f:\omega\to\kappa$. In $V'$, of course, $\kappa$ is no longer a cardinal. The theory of Boolean-valued models explains how to formalize this. This is discussed (in great detail) in Bell's book. $\endgroup$ – Andrés E. Caicedo Jun 2 '14 at 14:23
  • $\begingroup$ Thanks for the reference. Why cannot we apply the same reasoning to $M$, though? $\endgroup$ – goblin Jun 2 '14 at 14:24
  • $\begingroup$ The Boolean valued model is not a $2$-valued (classical) structure, which seems to be a requirement of your notion of outer model. If you are fine with Boolean-valued structures, then certainly we can do this with $M$. The point of passing to $V^{\mathbb B}$ was to ensure an environment where we could realize generic extensions of $M$ as actual (classical) models. $\endgroup$ – Andrés E. Caicedo Jun 2 '14 at 14:32
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    $\begingroup$ Not quite. What you write is correct, except for the "if $M$ is uncountable, then there is necessarily an $x\in M$ such that ..." For example, if $0^\sharp$ exists (an overkill), $L_{\aleph_1^V}$ is an uncountable model of $\mathsf{ZFC}$ such that any $x$ in there belongs (in $V$) to an outer model where $x$ is countable. To summarize: If $M$ is countable, there are always such outer models. If $M$ is uncountable, sometimes there are and some times there are no such outer models. And yes, whether $M$ is well-counded, or even an $\omega$-model is irrelevant here. $\endgroup$ – Andrés E. Caicedo Jun 2 '14 at 14:55

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