# Is this infinite sequence of transitive models of ZFC equiconsistent with Con(ZFC)?

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 $$\sf ZFC$$.

• 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. Jun 17, 2022 at 11:12
• Yes! correct. I've corrected the post. Jun 17, 2022 at 11:17

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 $$\sf ZFC$$.
• 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 $\operatorname{ZFC}$. Jun 17, 2022 at 15:50