It is an immediate consequence of Cohen's forcing that if there is one countable transitive model of $\sf ZFC$, then there are many of them. Even if all of these models are of the same height, there are still many.

But the proof of existence of generic filters don't carry over for uncountable models of set theory. Not without further assumptions (like Martin's axiom) anyway.

Is it possible to have an uncountable transitive model which is the only uncountable transitive model? Since that model is going to have to be $L_\alpha$ for some $\alpha>\omega_1$, is it also consistent with $V\neq L^1$, that a unique uncountable model exists?

Can this be pushed arbitrarily high (perhaps with, or without the assistance of $\sf MA$ and the Tennenbaum-Solovay theorem) so there are many transitive models of size $<\kappa$, but only one of size $\kappa$?


  1. Clearly one can just add a Cohen real to the universe without damaging any uncountable sets in a mind-shattering way. And on the other hand, if $V\neq L$ in a strong enough way, i.e. $0^\#$ existing in the universe, then there are arbitrarily large uncountable transitive models of $\sf ZFC$. Simply $L_\kappa$ for $\kappa$ a cardinal, or a Silver indiscernible.

    So my purposely vague question about $V\neq L$ is some middle-ground between the very uninteresting $L[r]$ for some "simple" real number $r$, and the existence of large cardinals (which automatically implies there are many uncountable transitive models).

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    $\begingroup$ If $M$ is a transitive uncountable model of size $> \aleph_1,$ then take an elementary submodel $X$ of size $\aleph_1.$ It is well-founded, so isomorphic to some transitive uncountable model $M^*.$ Clearly $M$ is not equal to $M^*$, so for a positive answer to your question, the model should have size $\aleph_1.$ Also for $M$ as above, $M$ contains all countable ordinals, so $L^M$ is also uncountable, hence we must have $M=L^M.$ $\endgroup$ Dec 9, 2013 at 6:19
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    $\begingroup$ Why does knowing that $M$ has size at least $\aleph_1$ imply that $M$ has all countable ordinals? $\endgroup$ Dec 9, 2013 at 6:59
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    $\begingroup$ Let $\alpha$ denote the height of $M.$ If $\alpha<\omega_1,$ then for some $\beta<\alpha, V_{\beta}^M$ is uncountable, and as $M\models AC,$ there is an ordinal $\gamma\in M$ which is in 1-1 correspondence with $V_{\beta}^M$ and $\gamma$ is necessarily uncountable. $\endgroup$ Dec 9, 2013 at 7:17

1 Answer 1


No, this is not possible.

As Mohammad argued in the comments, such a model $M$ must contain all countable ordinals.

We may assume that $M = L^M$, otherwise $L^M$ is another transitive model of ZFC which is uncountable since $\omega_1 \subseteq L^M$. If $M = L^M$, then $M = L_\alpha$ for some $\alpha \geq \omega_1$. The cardinal $\omega_1$ must be an uncountable regular cardinal in $L$ and I will now write $\kappa$ instead of $\omega_1$ to avoid confusion with $\omega_1^L$. We may assume that $\alpha \lt (\kappa^+)^L$ (i.e. $L \vDash |M| = \kappa$). Otherwise we could apply Löwenheim-Skolem in $L$ to get $N \prec M$ with $\kappa \subseteq N$ and $|N| = \kappa$; the transitive collapse of $N$ would then be an uncountable model of ZFC different from $M$.

There are now two cases:

  • If $\alpha \gt \kappa$, then the poset $(2^{\lt\kappa})^L$ is in $M$. Working in $L$, using the fact that $2^{\lt\kappa}$ is $(\lt\kappa)$-closed and that $|M| \leq \kappa$, we can easily construct a $M$-generic $G$ for $2^{\lt\kappa}$. Then, the generic extension $M[G]$ is an uncountable transitive model of ZFC different from $M$.

  • If $\alpha = \kappa$ then every set in $M$ is countable in $V$. Therefore, every set forcing in $M$ has an $M$-generic set in $V$, which leads to a plethora of different uncountable transitive models of ZFC.

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    $\begingroup$ Very nice answer! In your first case, you get $2^{\omega_1}$ many models, arising from different choices of that diagonalization procedure. In the second case, with set forcing you'll get only continuum many models for each forcing notion, but I suppose one can look at class forcing over $M$ and then construct again $2^{\omega_1}$ many different forcing extensions. So isn't the "plethora" the same size in both cases? $\endgroup$ Dec 9, 2013 at 11:44

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