Suppose $ZFC$ is consistent. Is there a model $\mathcal{M}$ of $ZFC$ and formulae $\varphi_D(x)$ and $\varphi_\in(x,y)$ that define (in $\mathcal{M}$) the domain and membership relation of a model $\mathcal{N}$ of $ZFC$ such that we have $\mathcal{N}\cong\mathcal{M}$ externally (i.e., in $V$) and yet $\mathcal{M}$ thinks $\mathcal{N}$ is well-founded but not set-like?

I think it's clear that if this were true $\mathcal{M}$ would need to be ill-founded.

I suspect the Ehrenfeucht-Mostowski theorem, which gives automorphisms for such models, can be adapted to obtain an embedding from some $\mathcal{N}$ onto an initial segment of itself, say $\mathcal{M}$, but where the respective spines of indiscernibles used to generate these models are (externally) isomorphic. If that works, I think we'd have $\mathcal{N}\cong\mathcal{M}$ but - among other things - I can't see that (an isomorphic copy of) $\mathcal{N}$ would be definable in $\mathcal{M}$.

If the issue were that assuming $ZFC$ was consistent was not sufficient, I'd still be interested in the assumption that did suffice.


1 Answer 1


The answer is yes.

Let $M$ be a countable computably saturated model of ZFC with a measurable cardinal $\kappa$. Let $N$ be the Ord-length iterated ultrapower of a measure on $\kappa$. The model $M$ thinks $N$ is a definable well-founded class structure, which is strictly taller than $M$. But $M$ and $N$ are two countable models of ZFC with the same theory, same standard system, and computably saturated, and this ensures that they are isomorphic (externally). The model $N$ is computably saturated, since any definable class in a computably saturated structure is also computably saturated.

Here is my previous answer to your question. Here is an example of the dual situation, where a model of set theory can define a model that it thinks is strictly shorter, but to which it is actually isomorphic. I am posting this because I find it interesting and related, even though it doesn't answer your question.

Theorem. Every countable computably saturated model of ZFC is isomorphic to a rank-initial segment of itself.

Proof. Suppose that $M$ is a countable computably saturated model of ZFC. It follows that the theory of $M$ is in the standard system of $M$. That is, there is a natural number $t$ in $M$ that codes a sequence of natural numbers, such that the standard part of that sequence is exactly the Gödel codes of the sentences true in $M$. By reflection, increasing large standard fragments of this theory are true in the rank-initial segments $V_\alpha^M$, and so by overspill, there must be some $N=(V_\alpha)^M$ that satisfies a nonstandard fragment of the theory coded by $t$. Thus, $N$ and $M$ have the same theory, and the same standard system. From this, it follows that they are isomorphic. $\Box$

The model $N$ is definable in $M$ from parameters, since $M$ thinks $N$ is just $V_\alpha$.

You can see various versions of this argument in my paper:

Here is another variation on the theme:

Theorem. There is a countable model $M$ of ZFC that is isomorphic to a forcing extension $M[c]$ of itself.

This is a little closer to what you asked for, since $M[c]$ can define $M$, and it is isomorphic to $M$. But this model is still set-like, so ultimately it doesn't fulfill your requirement.

Proof. Let $M_0$ be a countable computably saturated model of ZFC, and let $M=M_0[d]$ be the model obtained by forcing to add a Cohen real $d$. Let $c$ be $M$-generic, and consider $M[c]$. Since both are obtained by forcing over $M_0$ to add a Cohen real, they have the same theory, and they are both computably saturated and have the same standard system, hence isomorphic. $\Box$

  • $\begingroup$ Could something like this work? Let M be a computably saturated model of ZFC with a measurable cardinal (an extra assumption, I know). Let U be a normal ultrafilter witnessing this. Take the iterated ultrapower of M via U and its images of length Ord as defined in M (but don't try to collapse it). Call this N. I think N and M would have the same standard system, but is N computably saturated? $\endgroup$ Oct 30, 2020 at 18:41
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    $\begingroup$ Yes, I think this works, and I have posted an edit to my answer. $\endgroup$ Oct 30, 2020 at 19:03
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    $\begingroup$ But is this really longer? In what sense does $\Bbb R$ longer than $(0,1)$? It's just a scaling issue. On the other hand, $\omega_1$ is certainly longer than $\omega$. $\endgroup$
    – Asaf Karagila
    Oct 30, 2020 at 20:39
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    $\begingroup$ It's definitely longer. M has a definable map from it's ordinals to an initial segment of the ordinals of N. Namely, the iteration takes $\kappa$ to Ord itself, and the ordinals above $\kappa$ to "ordinals" above Ord. There is no definable converse map, even if there is one externally. $\endgroup$ Oct 30, 2020 at 20:50
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    $\begingroup$ Joel, it might be worth adding that if we take just a single ultrapower and collapse it in our computably saturated M, then we get an inner model which is isomorphic to M. Then shorter, wider, taller and thinner are all covered. $\endgroup$ Oct 30, 2020 at 23:51

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