Let $T$ be a maximal (in the sense that either $\phi \in T$ or $\phi \not \in T$ for all $\phi \in \mathcal{L}_\in$) set of sentences consistent with $ZFC$.

Question For a countable transitive model $\mathfrak{M}$ such that $\mathfrak{M} \models T$, is $\mathfrak{M}$ the unique such model?

This is somewhat close to questions asked here:

On the number of countable models of complete theories of models of ZFC

and here:

Number of Non-isomorphic models of Set Theory

However, those answers just deal with countable models rather than countable transitive models.

It is tempting to say that $\mathfrak{M}$ should be unique; for Skolem hull $H \models T$ the collapsing isomorphism to $\mathfrak{M}$ is unique. It's then hard to see what could substantiate any difference between the result of collapsing two distinct $H_1$ and $H_2$ that both satisfy $T$.

Many thanks for any pointers!

  • 1
    $\begingroup$ Sometimes, sometimes not. If $T$ is something which includes $V=L$ and "There are no transitive models of $\sf ZFC$", then yes, but in general I think that the answer is negative. (Not to mention that assuming large cardinals there are "more" such $T$ available to us...) $\endgroup$
    – Asaf Karagila
    Sep 23, 2014 at 14:22
  • $\begingroup$ Thanks Asaf. A couple of clarifications. 1. By $T$ I meant a set that also decides statements of the form ``There exists a large cardinal of kind $x$'' (my question wasn't entirely clear in this respect). 2. Is there an easy way to see this? For example, we can easily force from $CH$ to $\neg CH$ and back again, but I'd expect the final model in the chain to satisfy some different sentences from the first... $\endgroup$ Sep 23, 2014 at 14:35
  • $\begingroup$ (1) Of course that a completion of $\sf ZFC$ decides things like there are such and such large cardinals. But the assumption that such $T$ exists requires additional assumptions (in some universes there will be no transitive models of $\sf ZFC+\exists\kappa$ inaccessible, even if there are models of that theory). So it can be a bit hard to give a complete answer, because the question is too... "independent", or too dependent on the meta-theory assumptions. (2) An easy way to see what? The case where there is a unique model which I suggested? $\endgroup$
    – Asaf Karagila
    Sep 23, 2014 at 14:45
  • $\begingroup$ Re: (1.) okay good. I'm happy with the independent assumptions. Re: (2.) I meant an easy way to see a case of non-uniqueness. The unique case you specify is AOK. $\endgroup$ Sep 23, 2014 at 14:54

2 Answers 2


Sometimes the answer is yes, for example consider $T$ which includes these two axioms:

  1. $V=L$.
  2. "There are no transitive models of $\sf ZFC$".

Because of the first statement any countable transitive model has the form $L_\alpha$ for some $\alpha<\omega_1$. But the second statement already ensures that this $\alpha$ is unique, since any larger $\alpha'$ will know $L_\alpha$ and therefore the second axiom will be false there.

Similarly we can add all sort of these statements which make the choice of $\alpha$ unique.

Sometimes the answer is no, for example if $0^\#$ exists, then there is a class of indiscernibles for $L$, in particular if $\alpha$ and $\beta$ are indiscernibles then $L_\alpha$ is an elementary submodel of $L_\beta$. Pick $\alpha$ any countable indiscernible (and there are many of them), and take $T$ as the theory of $L_\alpha$, now you have $\aleph_1$ non-isomorphic countable transitive models of $T$.

And sometimes the answer is "what are you talking about?" consider the model from the first example, and work internally to that model. Regardless to what completion we took of $\sf ZFC$ it doesn't have a transitive model, let alone a countable transitive model.

  • 2
    $\begingroup$ Great idea with $0 \sharp$. That's a very elegant and clear example. $\endgroup$ Sep 23, 2014 at 14:56

Here's a particularly simple way to show non-uniqueness for arbitrary T. Suppose there's a transitive model $M$ of ZFC containing $\mathcal P(\omega)$. Let T be the complete theory of $M$. Now, suppose there are less than $2^\omega$ ctm's (up to isomorphism) of T. Then there is some $x\subseteq \omega$ not (represented) in any of them. But the countable Skolem hull of $M$ wrt $\{x\}$ will have a transitive collapse containing $x$. Contradiction.

  • $\begingroup$ How does that not clash with my first example? $\endgroup$
    – Asaf Karagila
    Sep 23, 2014 at 15:28
  • $\begingroup$ Won't M think that there are transitive models? I mean, it will think there are well-founded models of ZFC on $\omega$. $\endgroup$
    – user104955
    Sep 23, 2014 at 15:30
  • $\begingroup$ Which $M$? I'm talking about $T$ which is $\sf ZFC+V=L+$There are no well-founded models of $\sf ZFC$. In the case of working internally to such model, note that while $\operatorname{Con}(\ldots)$ is an arithmetic statement, the existence of well-founded models is not arithmetical anymore, rather $\Sigma^1_2$, so it's not necessarily absolute to smaller models like that. $\endgroup$
    – Asaf Karagila
    Sep 23, 2014 at 15:32
  • $\begingroup$ The M I mention in my answer. It doesn't satisfy your (2) and so the T I define doesn't include (2), and so I don't see how my example clashes with your first example. What am I missing? $\endgroup$
    – user104955
    Sep 23, 2014 at 15:34
  • $\begingroup$ I didn't notice that $\mathcal P(\omega)\in M$. Sorry. $\endgroup$
    – Asaf Karagila
    Sep 23, 2014 at 15:59

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