Let $M$ be a transitive model of ZFC.

From my understanding, if $x \in M$ then what $M$ believes to be its power set $\mathcal{P}(x)^M$ does not necessarily agree with the external power set $\mathcal{P}(x)$ (i.e. $\mathcal{P}(x)^M \neq \mathcal{P}(x)$), because $M$ might not contain all subsets of $x$.

Here is where my confusion begins: Let $\varphi(x,p) = \forall y (y \in p \leftrightarrow y \subseteq x)$ be the formula saying that $p$ is the power set of $x$. As $M$ is a model of ZFC we have $\varphi^M (x ,p) \leftrightarrow \varphi(x, p)$ for any $x,p \in M$. But $\varphi^M (x , \mathcal{P}(x)^M)$ holds, which implies that $\mathcal{P}(x)^M = \mathcal{P}(x)$ and that $M$ is closed under subsets by transitivity. This doesn't agree with my understanding of transitive models mentioned above.

I should note, that I don't have much background in model theory and it is very likely that I'm missing something obvious.


Transitive models are closed under elementhood, not under subsets. In other words, $M$ is transitive if $x\in y\in M$ implies $x\in M$, and not as you suggest, $x\subseteq y\in M$ implies $x\in M$.

You are correct that if $M$ is transitive, and $x,y\in M$, then $M\models x\subseteq y$ if and only if $x\subseteq y$ (in $V$, that is). The only problem is that perhaps $x\notin M$.

But what you can say is that if $M$ is transitive, then $\mathcal P(x)^M=\mathcal P(x)\cap M$.

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  • $\begingroup$ I am aware that transitive models are not necessarily closed under subsets, but I cannot see why my reasoning is faulty. Am I wrong in my assumption, that if $M$ is a model, then for any formula $\varphi$ with free variables among $x_1, \ldots ,x_n$ we have $\varphi^M(x_1,\ldots , x_n) \leftrightarrow \varphi(x_1,\ldots ,x_n)$ for all $x_1,\ldots,x_n\in M$? $\endgroup$ – Alexander Constantin Jun 6 at 15:47
  • $\begingroup$ @AlexanderConstantin That is indeed wrong. And it's wrong exactly because transitive models need not be closed under subsets. $\endgroup$ – Noah Schweber Jun 6 at 15:48
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    $\begingroup$ No, this is only true for $\Delta_1$ formulas, and the parameters need to be in $M$. So you can only conclude that sets that are in $M$ are in $M$. $\endgroup$ – Asaf Karagila Jun 6 at 15:48
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    $\begingroup$ @AlexanderConstantin Maybe you're mixing up transitive submodels and elementary submodels? Indeed, what the above shows is that no countable elementary submodel of (a large enough part of) $V$ can be transitive. $\endgroup$ – Noah Schweber Jun 6 at 15:50
  • $\begingroup$ That's exactly what happened and now I feel a bit embarrassed. Thank you both for your help. $\endgroup$ – Alexander Constantin Jun 6 at 15:57

Asaf Karagila's answer covers the properties of transitive models, and where the argument in the OP breaks down. Let me add to that answer by saying a bit about the situation vis-a-vis a very different type of submodel: elementary submodels. Early on it's quite easy to mix these up, so this seems worthwhile to write out.

Ignore class/set issues for the moment (or replace $V$ with some big enough transitive set like $V_{somethingreallybig}$). By downward Lowenheim-Skolem we can find some countable $M\prec V$. Now we do have the bi-implication $\varphi^M\leftrightarrow\varphi$ in general, and the argument of the OP is now worrying again:

  • Since $M\models\mathsf{ZFC}$ we have some $a\in M$ such that $M\models$ "$a$ is uncountable."

  • By elementarity, $a$ is in fact uncountable.

  • But $M$ is countable. What gives?

The resolution of the above issue is that $a\not\subseteq M$ - which is to say, $M$ must not be transitive. So the argument in the OP really reveals a tension between two conflicting "niceness" notions, namely transitivity and elementarity, and shows that while each individually is compatible with countability we can't have a countable submodel which is both transitive and elementary.

(Note that the argument above is closely related to Skolem's paradox, which was the original appearance of the downward Lowenheim-Skolem theorem in the first place.)

Both transitive and elementary countable submodels of $V$ play important roles in set theory; the above shows that they're really fundamentally different types of objects. That said:

  • We can always turn an elementary submodel into a transitive submodel via the Mostowski collapse (note that this kills elementarity in general, of course).

  • That said, transitive elementary submodels of $V$ do exist (under mild hypotheses) - they just have to be really really really really big. In particular, if $M$ is an elementary submodel of $V$ then any ordinal definable in $V$ has to be in $M$, and then by transitivity we get all the smaller ordinals too. And there could be really really big definable ordinals: maybe $\mathsf{GCH}$ fails somewhere but the first point of failure (which is definable) is really big, or your favorite large cardinal property actually shows up (at which point its least instance, which has to be really big, is definable), or so forth. So transitive elementary submodels of $V$ only show up rarely.

  • More technically, countable elementary submodels do have some weak transitivity properties - in particular, if $M\prec V$ then $\omega_1\cap M$ is closed downwards. This winds up being extremely useful down the road.

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