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Suppose there exists a transitive model of $\sf ZFC$. Is it the case that every consistent theory that extends $\sf ZF$ must have a model that is an element of the minimal transitive model of $\sf ZFC$?

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The answer is no, because by the Gödel-Rosser theorem, there are continuum many consistent completions of ZF, but the minimal transitive model of ZFC is countable, and so has only countably many theories. So some of the consistent extensions are not realized in that model.

Another argument is simply this: the theory of the minimal model itself is a consistent extension of ZF, but this theory cannot be an element of the minimal transitive model (that is, it cannot have its own theory as an element), because from this theory you can reconstruct the model itself---some of the sentences of the theory assert that there is an object satisfying a certain definition, and all that one would want to know about those objects is asserted as part of the theory. So if the theory of the minimal model were inside that model, then that model would be able to construct a copy of itself as an element, which is impossible by minimality.

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  • $\begingroup$ I don’t follow your second argument. How can you reconstruct the model from its theory? This surely isn’t possible in general: a complete consistent extension of ZF isn’t necessarily categorical. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jan 3, 2023 at 10:19
  • $\begingroup$ @PeterLeFanuLumsdaine In general, you cannot construct a model from its theory. But with a pointwise definable model, you can, because all the objects are definable, and how they relate to one another and so on. So from the theory you can extract everything you need to know to build the model. $\endgroup$
    – Joel David Hamkins
    Commented Jan 3, 2023 at 12:53
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    $\begingroup$ Thankyou; I hadn’t known that the minimal model was pointwise definable. Searching it, I see you gave a nice explanation very recently, which I link here for the convenience of my fellow ignorami! $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jan 3, 2023 at 23:35
  • $\begingroup$ Ah, sorry, I should have mentioned it again---I was thinking of the two questions as part of a series asked by the same OP. $\endgroup$
    – Joel David Hamkins
    Commented Jan 3, 2023 at 23:40
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As Joel said, most of the consistent extensions $T$ of ZFC are not in the minimal transitive model $M$ of ZFC and therefore don't have models in $M$. It seems worth noting that this is the only reason the answer to the question is negative. If a consistent extension $T$ of ZFC is an element of $M$, then it has a model in $M$. The proof of this just consists of the known facts that $M$ satisfies "$T$ is consistent" (i.e., consistency is absolute for $M$) and Gödel's completeness theorem is provable in ZFC and therefore holds in $M$ (and additional absoluteness: Whatever $M$ considers to be a model of $T$ really is a model of $T$).

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