It is known that the consistency strength of $\sf ZFC+\rm Con(\sf ZFC)$ is greater than that of $\sf ZFC$ itself, but still weaker than asserting that $\sf ZFC$ has a transitive model. Let us denote the axiom "There is a transitive model of $T$" by $\rm St(\sf ZFC)$.

If $M$ is a transitive model of $\sf ZFC$ of size $\kappa$ then we can easily generate transitive models of any smaller [infinite] cardinal by using the downward Löwenheim–Skolem theorem and the Mostowski collapse. Note that we can use the latter because the former guarantees that the model uses the real $\in$ relation, so it is well-founded.

But the upward Löwenheim–Skolem makes use of compactness, which can easily generate models which are not well-founded, and therefore cannot be collapsed to a transitive model.

So while $\rm Con(\sf ZFC)$ can prove that there is a model of $\sf ZFC$ of any cardinality, can $\rm St(\sf ZFC)$ do the same, or do we have a refinement of the consistency axioms in the form of bounding the cardinality of transitive models?

It is tempting to take a countable transitive model and apply forcing and add more and more sets, but forcing doesn't add ordinals and the result is that we cannot increase the size of the model without bound.


2 Answers 2


No. Because the construction of the inner model $L$ is absolute for transitive models of ZFC the ordinal heights of transitive models of ZFC are precisely the ordinals $\alpha$ such that $L_\alpha$ is a model of ZFC. If $\alpha_0,\alpha_1$ are the first two ordinals such that $L_{\alpha_0}$ and $L_{\alpha_1}$ are models of ZFC, then $L_{\alpha_1}$ is a model of ZFC in which there is a transitive model of ZFC, namely $L_{\alpha_0}$, and every transitive model of ZFC has rank $\alpha_0$ and is therefore bounded in size by $|V_{\alpha_0}|$.

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    $\begingroup$ (Third time is the charm, let's hope. I keep leaving typos in the comment.) Well, it is even stronger: $L_{\alpha_0}$ is countable in $L_{\alpha_1}$, so all transitive models of $\mathsf{ZF}$ (not even choice is needed) are countable in this example. $\endgroup$ May 5, 2013 at 14:45
  • $\begingroup$ Andres, awesome comment. Can a similar approach show that we can bound the size of transitive models by every cardinal? $\endgroup$
    – Asaf Karagila
    May 5, 2013 at 17:27
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    $\begingroup$ @AndrésE.Caicedo I'm not sure I buy that for $\mathsf{ZF}$, since $\mathsf{ZF}$-models can have more elements than ordinals (and in particular that seems to contradict a result of Friedman, see here). Am I missing something? $\endgroup$ May 19, 2020 at 2:09
  • $\begingroup$ @Noah You are correct, my comment that choice is not needed is wrong, and Harvey's result is a counterexample. $\endgroup$ May 19, 2020 at 3:12

In a comment to François's answer, I point out that the least $\alpha$ such that $L_\alpha$ is a model of $\mathsf{ZFC}$ is countable. In what follows, "model" means "transitive model of $\mathsf{ZFC}$."

If $M$ is transitive, then $L^M=L_\beta\models\mathsf{ZFC}$ for $\beta=\mathsf{ORD}\cap M$, so the least height of a model is countable. Moreover, any model $M$ proves that there is a bijection between each level of its cumulative hierarchy, and one of its ordinals, so if $M$ has height $\kappa$, so is $\kappa$ its size. This proves that the existence of a model does not imply the existence of an uncountable one: If $M$ has least height, let $\alpha_0$ be least such that $L_{\alpha_0}$ is a model, and $\alpha_0$ is larger than the height of $M$ ($\alpha_0$ could be $\mathsf{ORD}$). We see that in $L_{\alpha_0}$ the height of $M$ is countable and there are no models of height larger than the height of $M$.

Asaf asked whether this generalizes, that is, whether for each $\kappa$ the existence of models of size $\kappa$ does not imply the existence of models of larger size. That this is indeed the case follows from extending the argument from the previous paragraph: Let $L_\alpha$ be a model of height (and size) $\kappa$, and let $\alpha_0$ be such that $\alpha<\alpha_0$ and $L_{\alpha_0}$ is a model. We may as well assume that $\alpha_0$ exists (that is, it is an ordinal), or else there is nothing to prove. Now let $X$ be an elementary substructure of $L_{\alpha_0}$ containing both $L_\alpha\cup\{L_\alpha\}$ (as a subset) and a bijection between $\alpha$ and $|\alpha|^{L_{\alpha_0}}$, and of size $\kappa$, which exists by a standard application of the Lowenheim-Skolem argument. The transitive collapse of $X$ is $L_\beta$ for some $\beta$, and has size $\kappa$. This means that if $\alpha_0$ is least (so the collapse of $X$ is again $L_{\alpha_0}$), then $L_{\alpha_0}$ is a model of $\mathsf{ZFC}$ plus the assertion that there are no set models of height (and therefore size) larger than $|\alpha|=\kappa$.

Without choice, I do not know whether models of $\mathsf{ZF}$ of height $\kappa$ must have size $\kappa$.

[Edit: In response to the last paragraph above, Joel and Asaf pointed out some results. I am including them here, to increase visibility.]


Ali Enayat. Models of set theory with definable ordinals, Arch. Math. Logic 44 (3), (2005), 363–385. MR2140616 (2005m:03098),

Ali discusses some results about transitive models of $\mathsf{ZF}$ that show that the situation is much more subtle than in the presence of choice. One of the most incredible results is due to Friedman, in

Harvey Friedman. Large models of countable height, Trans. Amer. Math. Soc. 201 (1975), 227–239. MR0416903 (54 #4966).

Let me quote from Harvey's paper:

The first examples of transitive models of $\mathsf{ZF}$ of power $\omega_1$ with countably many ordinals were constructed by Cohen. Later Easton, Solovay, and Sacks showed that every countable transitive model of $\mathsf{ZF}$ has an ordinal-preserving extension satisfying $\mathsf{ZF}$, of power $2^{\omega}$. We prove here that every countable transitive model $M$ of $\mathsf{ZF}$ has an ordinal preserving extension satisfying $\mathsf{ZF}$, of power $\beth_{M\cap\mathsf{ORD}}$.

Harvey's argument uses forcing. For his first result, given $M$ a countable transitive model of $\mathsf{ZF}$, he says that $x\subset\omega^\omega$ is $M$-generic iff any finite sequence of distinct elements of $x$ is $M$-generic (for the product of the appropriate number of copies of Cohen forcing), $x$ is infinite, and dense. The models he builds are of the form $M(x)$ so, in particular, they are transitive. He shows that there are $M$-generics $x$ of size continuum with $M(x)$ a model of $\mathsf{ZF}$ (and $M(x)$ has the same height as $M$). He then builds on the machinery introduced here, and proves that, starting with an $M$-generic $x$, a family of sets $C_\alpha$, $\alpha\lt M\cap\mathsf{ORD}$, can be found with $|C_\alpha|\ge\beth_\alpha$, and such that $M[(C_\alpha)_\alpha]$, properly defined, is a model of $\mathsf{ZF}$ of the claimed size.

Ali builds on this results to produce Paris models of $\mathsf{ZF}$, that is, models $M$ all of whose ordinals are first order definable in $M$. In prior work, he had shown that from the assumption that $L$ satisfies that there are uncountable transitive models of $\mathsf{ZFC}$, it follows that there are unboundedly many $\alpha<\omega_1^L$ such that $L_\alpha$ is Paris. He shows now that from the same assumption, we have that for every infinite $\kappa$ there are Paris models of $\mathsf{ZF}$ of size $\kappa$; this uses Harvey's result, since generic (or simply, ordinal preserving) extensions of $L_\alpha$ are Paris if $L_\alpha$ itself is Paris. It follows that there is a complete extension of $\mathsf{ZF}$ admiting in $L$ Paris models of size $\beth_\alpha$ for each countable $\alpha$. The theory, including the requirement that its models are Paris, can be described in $L_{\omega_1\omega}$. Since the Hanf number of this logic is $\beth_{\omega_1}$, the result follows.

This produces large transitive models indeed, in view of a result of Paris: If a completion $T$ of $\mathsf{ZF}$ has a well-founded model, then every Paris model of $T$ is well-founded.

As pointed out by Mohammad here, Solovay's construction referenced in Friedman's answer (and a clear influence in Friedman's argument) can be found in

Ulrich Felgner. Choice functions on sets and classes. In Sets and classes (on the work by Paul Bernays), pp. 217–255. Studies in Logic and the Foundations of Math., Vol. 84, North-Holland, Amsterdam, 1976. MR0424566 (54 #12525).

  • $\begingroup$ With regard to your last comment, Ali Enayat has some very interesting models of ZF that are enormous in size, but have only countably many ordinals. He even does it with Paris models, where every ordinal is definable without parameters. $\endgroup$ May 5, 2013 at 20:15
  • $\begingroup$ Joel, that sounds crazy. Do you mean this paper? dx.doi.org/10.1007/s00153-004-0256-9 $\endgroup$
    – Asaf Karagila
    May 5, 2013 at 20:21
  • $\begingroup$ Thanks, Joel. I figured Ali should have something on this question, and was planning to track down a reference. $\endgroup$ May 5, 2013 at 20:30
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    $\begingroup$ Asaf, yes, that paper has some of the theorems I had in mind. Item 3 implies large models with only countably many ordinals (since they are definable), but when the models are too large, they cannot be well-founded. But I think we gets up to size $\beth_{\omega_1}$ with transitive models, in the proof. I think that a result of Harvey Friedman comes into it... $\endgroup$ May 5, 2013 at 21:07
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    $\begingroup$ Yes, there was a theorem in the paper (I think 2.18 or something around that number) which stated that if $M$ is an uncountable Paris model of $\sf ZF$ and it is well-founded then it has some element which is not the iterated well-ordered union of well-ordered sets. The quoted result of Friendman is unbelievable. "Every countable transitive model $M$ of $\sf ZF$ has a generic extension of cardinality $\beth_\alpha$ where $\alpha={\sf Ord}^M$". $\endgroup$
    – Asaf Karagila
    May 5, 2013 at 21:23

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