(1) Are there formulae $\varphi_D(x)$ and $\varphi_\in(x,y)$ defining an internal model $\mathcal{N}$ of $ZFC$ where $\mathcal{N}$ is not set-like and no definable, set-like, internal model $\mathcal{M}$ is elementary equivalent to $\mathcal{N}$?

(2) Are there formulae $\varphi_D(x)$ and $\varphi_\in(x,y)$ defining an internal model $\mathcal{N}$ of $ZFC$ where $\mathcal{N}$ is not set-like and no definable, set-like, $\it{well-founded}$, internal model $\mathcal{M}$ is elementary equivalent to $\mathcal{N}$?

A simple way to obtain non set-like models of $ZFC$ is to take a normal ultrafilter $U$ and take the iterated ultrapower of $V$ through length $Ord$. This gives us a non set-like model of $ZFC$, but it is elementary equivalent to $V$. (This approach uses parameters, but they are easily removed.)

In a vague nutshell, is there another way to define "long" models without some form of iteration?

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    $\begingroup$ I'm really curious to know what set-like means for a model of ZFC. Can you explain it to me? $\endgroup$
    – user141903
    Oct 31, 2020 at 1:30
  • $\begingroup$ Erin, a class relation R \subseteq V x V is set-like if for all y, {x | x R y} is a set. $\endgroup$ Oct 31, 2020 at 1:56
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    $\begingroup$ @Erin The membership relation of $\mathcal N$. $\endgroup$ Oct 31, 2020 at 2:12
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    $\begingroup$ Erin, that's right. The model $M$ is defining (a copy of) a much taller structure $N$. Precisely because it is taller, it cannot do so using the actual membership relation of $M$. But perhaps it has a definable relation that would be isomorphic to the membership relation of $N$. $\endgroup$ Oct 31, 2020 at 8:25
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    $\begingroup$ I suppose that every model thinks its own $\in$ is set-like. $\endgroup$ Oct 31, 2020 at 8:42

2 Answers 2


Let me describe another method for making long models definable.

First consider the case that $\kappa$ is inaccessible in $L$ and $\lambda$ is a worldly cardinal in $L$ above $\kappa$. Let $G$ be $L$-generic for the forcing to collapse $\lambda$ to $\kappa$. In the forcing extension, there is a set $E\subset\kappa$ that codes the structure $\langle L_\lambda,\in\rangle$. Let $L[G][H]$ be the forcing extension that codes the set $E$ into the GCH pattern up to $\kappa$. This preserves the inaccessibility of $\kappa$. Let $M=(V_\kappa)^{L[G][H]}$, which is a model of ZFC since $\kappa$ remained inaccessible. In $M$, we can define $E$ using the GCH pattern and therefore in $M$ we can define a copy of the structure $\langle L_\lambda,\in\rangle$, which is a model of ZFC. This model is actually well-founded, and so the model $M$ will also look upon it as well-founded, and it is not set-like, since it has height $\lambda$, which is taller than $\kappa$.

So this method shows how a model of ZFC can define a much taller well-founded model of ZFC.

In your question, however, you had asked for more. You wanted the defined model to have a theory not realized in any set-like interpreted model of $M$. Let me modify the construction to achieve something closer to this. I will show how to arrange that $N$ satisfies a theory not satisfied by any set structure in $M$ nor any definable well-founded set-like class structure. (Thanks, Ali, for your comments about this.)

To do it, I will suppose not that $\kappa$ is actually inaccessible in $L$, but rather merely that it is inaccessible in $L_\lambda$, which is itself a pointwise definable model. One can make this situation from the situation above simply by taking the Mostowski collapse of the definable elements of $L_\lambda$. Now, these are countable ordinals in $L$, even though $L_\lambda$ thinks $\kappa$ is inaccessible. But we can still do the forcing $G$ and $H$ and form the model $M=(V_\kappa)^{L_\lambda[G][H]}$ just as above, except that we take $G$ merely $L_\lambda$-generic and $H$ is $\langle L_\kappa,\in,E\rangle$-generic. In $M$, again we can define $L_\lambda$ as a well-founded non-set-like model of ZFC.

The point of the pointwise definability assumption is that the theory $T$ of $L_\lambda$ ensures that every model of it contains a copy of $L_\lambda$. There can be no set model of this theory in $M$, because then $M$ would be able to take the definable elements of that model and thereby produce a copy of $L_\lambda$ as a set, which is impossible in $M$ since $\kappa<\lambda$. Similarly, there can be no well-founded set-like class definable model of the theory in $M$, since any such model $N$ would have to have an ordinal with at least $\kappa$ many predecessors, and such an order relation would not be set-like in $M$.

What remains is the possibility that there could be a definable set-like class model in $M$, not necessarily well-founded, but which satisfies the theory. I don't at the moment know how to rule this out. The subtle point is that a set-like order in $M$ can contain a copy of $\lambda$ — after all, both $\kappa$ and $\lambda$ are countable ordinals and hence embed into $\mathbb{Q}$, which is set-like in $M$ — and the pointwise definability idea doesn't seem quite enough to rule out such ill-founded interpreted classs models.

Therefore, this answer doesn't quite fully answer the question.

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    $\begingroup$ @JoelDavidHamkins Joel, in the first sentence of the last paragraph of your answer you wrote "You wanted the defined model to have a theory not realized in any model of $M$". But Toby's question seems to require the defined taller model to have a theory not realized in any set-like model of $M$, can your solution be modified to take care of this stronger demand? $\endgroup$
    – Ali Enayat
    Oct 31, 2020 at 20:36
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    $\begingroup$ Ah, @AliEnayat, you are right. However, my argument does handle this version, since when you have a pointwise definable model, any model of the theory will be at least as tall as it (since the definable ordinals will have that order type as a part of the theory), and so there can be no set-like model of this theory in $M$, even as a class model, as the order type will be at least $\lambda$, which is taller than $\kappa$. $\endgroup$ Oct 31, 2020 at 21:13
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    $\begingroup$ @JoelDavidHamkins, Thanks for the answer Joel. There is one remaining concern: your argument (to rule out set-like taller models) equates the notion of "internal model"' in Toby's question with "inner model". In my mind, inner models are always well-founded relative to the ambient model, but "internal models" need not be. Perhaps Toby mean "inner model" when he wrote "internal model". $\endgroup$
    – Ali Enayat
    Oct 31, 2020 at 22:36
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    $\begingroup$ @AliEnayat You are right. This is a very subtle issue. My argument rules out set models of the theory in M, and also rules out class-sized set-like well-founded models of the theory in M. But it doesn't quite seem to work with definable set-like ill-founded models. I have edited my answer to clarify. $\endgroup$ Nov 1, 2020 at 10:42
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    $\begingroup$ Ah, I think he is concerned about the ill-founded definable set-like models. I hadn't realized at first they pose a special problem for my argument. $\endgroup$ Nov 1, 2020 at 19:44

To complement Joel's answer that a positive outcome is consistently possible, let me show that the negative outcome is also possible, at least in the case of boldface definability.

For our background universe $V$, suppose that we have some second-order resources available to us, namely a global well-order and truth predicates for all class-sized structures. (I'll comment below on what this assumption entails.) This allows us to then carry out the standard argument for downward Löwenheim–Skolem. If $\mathcal N$ is a class-sized structure then we have Skolem functions for it and so can get a set-sized elementary submodel $\mathcal M$ of $\mathcal N$. In particular, $\mathcal M$ will be set-like and definable (with parameters), entailing a negative answer to your question.

Note, however, that this approach doesn't give you a negative answer for lightface definability, i.e. without parameters. Consider the case of $\mathcal N = (V,\in)$ and suppose we could find $\mathcal M$ a set-sized elementary submodel of $\mathcal N$ which is definable without parameters. But then we could define without parameters the set of true sentences in $(V,\in)$, contradicting Tarski's theorem on the undefinability of truth.

On the other hand, we can get lightface definability if we allow class quantifiers. If your global well-order is definable then, because the truth predicate is definable with class quantifiers, you get definable Skolem functions and so $\mathcal M$ is definable with class quantifiers. It's $\Delta^1_1$-definable, to be precise. Also, $\mathcal M$ is implicitly definable (without parameters), in the sense of Hamkins and Leahy - Algebraicity and implicit definability in set theory.

Let me now address what this assumption on $V$ entails, and why it doesn't apply to Joel's positive case. Having a global well-order is cheap—you can always add a generic one by class forcing, without adding any new sets. The easiest way to do this: just add a Cohen-generic subclass to $\mathrm{Ord}$. But having truth predicates comes with a cost. There's the consistency strength cost, of course, since having a truth predicate for $(V,\in)$ let's you see that ZFC is consistent. But we can say more.

Proposition (Essentially Krajewski): If the structure $(V,\in,\mathrm{Tr})$ satisfies ZF in the expanded language plus the assertion that $\mathrm{Tr}$ satisfies the Tarskian recursion for truth of $(V,\in)$, then $V$ contains a club of ordinals so that $V_\alpha$ is elementary in $V$.

Proof sketch: The point is, you can do the usual reflection argument but using $\mathrm{Tr}$ as a parameter to get the desired $V_\alpha$s, since you can express "$\mathrm{Tr}$ satisfies Tarski" as a first-order formula in the expanded language. There's a small bit more to be said about the $\omega$-nonstandard case, but it's not hard.

In particular, having truth predicates implies you have lots and lots of undefinable ordinals. So any Paris model—one whose ordinals are all definable without parameters—cannot admit a truth predicate. (Or rather—since of course every structure externally admits a truth predicate—if you add a truth predicate you destroy Replacement in the expanded language.) This includes Joel's pointwise definable $L_\lambda$.

Finally, let me note that since the $\mathcal M$ produced is a set, all we need for the counterexample is that $V$ can be extended to have the necessary second-order resources. So, for example, if your $V$ is $V_\kappa$ for some inaccessible $\kappa$, where you only look at definable classes, then you still get the counterexample. For you could expand your classes to the full powerset of $V$, apply the argument there to your definable class-sized structure $\mathcal N$, and thereby get $\mathcal M$ in $V$. But if it's in $V$, then we didn't need the extra classes to define it.


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