Is it provable in $\sf ZFC$ that the existence of a transitive model of $\sf ZFC$ implies the existence of a sequence $(M_n)_{n \in \mathbb N}$ of transitive models of $\sf ZFC$ such that $M_m \in M_n$ for every naturals $m< n$, and such that each model proves all axioms of $\sf ZFC$ relativized to the lower models, but at the same time no model proves the lower models within it being models of $\sf ZFC$.
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$\begingroup$ Since ZFC+Con(ZFC) is not strong enough to prove the existence of a single transitive model, it will let alone not be able to prove a whole sequence of them exist. $\endgroup$– Asaf Karagila ♦Jun 17, 2022 at 11:12
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$\begingroup$ Yes! correct. I've corrected the post. $\endgroup$– ZuhairJun 17, 2022 at 11:17
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Transitive models must have the same $\omega$ as the universe, so they all must agree on the theory of True Arithmetic, which include the list of axioms of $\sf ZFC$.
Now, if you look at the second $\alpha$ such that $L_\alpha\models\sf ZFC$, then in that model there is a unique height of a transitive model of $\sf ZFC$.
answered Jun 17, 2022 at 11:32
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$\begingroup$ what happens if we remove transitivity condition? $\endgroup$– ZuhairJun 17, 2022 at 13:27
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$\begingroup$ Without transitivity yes, because the theory $\operatorname{ZFC}+\{ c_m \in c_{n} : m < n \} + \bigcup_{n} \operatorname{ZFC}^{c_n}$ is consistent if $\operatorname{ZFC}$ is consistent, by reflection and compactness. At the same time, if $\operatorname{ZFC} + \operatorname{Con}(\operatorname{ZFC})$ is inconsistent non of the $c_n$ can satisfy that $c_m$ is a model of $\operatorname{ZFC}$. $\endgroup$ Jun 17, 2022 at 15:50