By using the Löwenheim–Skolem theorem & Mostowski collapse, in every model $V$ of $ZF+Con(ZF)$ there is a countable transitive set $M$ such that $(M,\in_M) \models ZF$. Is the following "converse" true?

In every model $V$ of $ZF$ and every transitive set $M \in V$ such that $(M,\in_M) \models ZF$, there exists a transitive set $N \in V$ such that:

  1. $M \in N$

  2. $(N,\in_N) \models ZF$

  3. $M$ is countable inside $N$

  • $\begingroup$ If by "in every model $V$ there is $M$" you mean that $M$ is an element of $V$, then your claim about LS is not correct. $V$ might be a transitive model of "ZF + (ZF is inconsistent)", in which case there is no such $M$. $\endgroup$ – Goldstern Mar 10 '15 at 9:26
  • $\begingroup$ You demand "$M\subseteq N$" in your first condition. I assume you mean "moreover $M\in N$", as indicated by condition 3. $\endgroup$ – Goldstern Mar 10 '15 at 9:28
  • $\begingroup$ You're right. I corrected the question. $\endgroup$ – Andrei Sipoș Mar 10 '15 at 9:43
  • 1
    $\begingroup$ Your claim about LS is still not correct. Con(ZF) will give your a countable model, but not necessarily a transitive $\in$-model. Assuming that there is such a model, there will be some countable ordinal $\delta$ such that $L_\delta$ is a model of ZF+Con(ZF), but $L_\delta$ thinks that there is no such mode. $\endgroup$ – Goldstern May 11 '15 at 10:04

The answer to your question is no. For example, suppose $\kappa$ is inaccessible and $M = V_{\kappa}$. Then $M$ is a model of $\sf ZF$ but $M$ cannot be countable in $N$ as it is not countable in $V$.

Of course there will always be a set generic extension of $V$ containing such an $N$ (just generically add a bijection from $M$ to omega)

  • $\begingroup$ Thanks! I guess I have to learn more about how to work with $V[G]$ instead of $M[G]$. (The question I originally wanted to ask used the multiverse, but I didn't know how to phrase it rigorously.) $\endgroup$ – Andrei Sipoș Mar 10 '15 at 9:46

Let me point out that a close variant of the question has a positive answer. Namely,

Theorem. Every model of set theory $M\models\text{ZFC}$ has an elementary extension $M\prec M^*$ such that there is a model $N\models\text{ZFC}$ inside of which $\langle M^*,\in^{M^*}\rangle$ is a countable transitive model.

Proof. Fix any model $\langle M,\in^M\rangle\models\text{ZFC}$. First, I claim that there is a model $W\models\text{ZFC}$ with a cardinal $\delta$ such that $$M\prec V_\delta^W\prec W.$$ To see this, consider the theory $T$, in the language of set theory augmented with a constant symbol $\delta$, which asserts the elementary diagram of $M$, together with the assertions that $\delta$ is an ordinal and every element of $M$ is in $V_\delta$ and also the scheme $V_\delta\prec V$, asserting that $\delta$ is a correct cardinal. The reflection theorem shows that every finite subset of this theory is consistent, and so $T$ has a model $W$, which means that $M\prec V_\delta^W\prec W$, as desired.

Second, we will go to a further extension in which (the image of) $V_\delta^W$ becomes countable. Specifically, inside $W$ consider the forcing that makes $\delta$ countable. Let $U\in W$ be an ultrafilter on the corresponding complete Boolean algebra $\mathbb{B}$, and let $j:W\to \check W_U\subset W^{\mathbb{B}}/U$ be the Boolean ultrapower embedding. This embedding and these models exist inside $W$, without any actual forcing; I am just taking a quotient by an ultrafilter in $V$. You can find further explanation in my paper, Well-founded Boolean ultrapowers as large cardinal embeddings.

Note that $j:V_\delta^W\to j(V_\delta^W)=[{\check V}_\delta^W]_U$ is an elementary embedding, and furthermore, $j(V_\delta^W)$ is a countable transitive set inside $W^{\mathbb{B}}/U$. So let $M^*=j(V_\delta^W)$ and $N=W^{\mathbb{B}}/U$, and we have achieved the statement of the theorem. QED

The theorem applies to the case $M=V_\kappa$ in Nate's answer as follows: the model $M^*$ will be $\omega$-nonstandard (as in Asaf's answer), with $\kappa$ many natural numbers, and so the former contradiction of Nate's answer does not engage. The uncountable model $M$ is thought countable inside $N$, since $N$ has a lot of natural numbers.

One can actually improve the theorem to have $N\models\text{ZFC}+V=L$, since every countable model is a transitive model inside a model of $V=L$. This and many other similar results are proved and discussed in my paper, A multiverse perspective on the axiom of constructibility.


While the original question has been answered, remember that Skolem's paradox says that if $M$ is a countable model of set theory, then it knows only about countably many real numbers.

The reverse Skolem paradox, if so, is the following situation that can occur:

There $(M,E)\models\sf ZF$ such that $\{x \mid M\models x\text{ is an integer}\}$ is uncountable. (Assuming that there is a model of $\sf ZF$ to begin with, but let's dispense of that issue for now.)

The point is that $M$ thinks that an uncountable set is in fact countable. This $M$ cannot be transitive (or well-founded), and obtaining it is a simple exercise in compactness or ultrapowers. This shows that not only that "uncountable" is relative and internal to the model, but also "countable", and in fact in this model $M$, even "finite" is relative.

Since $M$ will think about certain infinite set that they are finite (e.g. various bounded sets of "natural numbers" in $M$ will contain infinitely many objects, but $M$ will still think they are finite because $M$ has a twisted view as for what is finite and countable).


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