It is known that the consistency strength of $\sf ZFC+\rm Con(\sf ZFC)$ is greater than that of $\sf ZFC$ itself, but still weaker than asserting that $\sf ZFC$ has a transitive model. Let us denote the axiom "There is a transitive model of $T$" by $\rm St(\sf ZFC)$.


If $M$ is a transitive model of $\sf ZFC$ of size $\kappa$ then we can easily generate transitive models of any smaller [infinite] cardinal by using the downward Löwenheim–Skolem theorem and the Mostowski collapse. Note that we can use the latter because the former guarantees that the model uses the real $\in$ relation, so it is well-founded.


But the upward Löwenheim–Skolem makes use of compactness, which can easily generate models which are not well-founded, and therefore cannot be collapsed to a transitive model.


So while $\rm Con(\sf ZFC)$ can prove that there is a model of $\sf ZFC$ of any cardinality, can $\rm St(\sf ZFC)$ do the same, or do we have a refinement of the consistency axioms in the form of bounding the cardinality of transitive models?


It is tempting to take a countable transitive model and apply forcing and add more and more sets, but forcing doesn't add ordinals and the result is that we cannot increase the size of the model without bound.



If ZFC has a transitive model, does it have one of arbitrary size?
No. Because the construction of the inner model $L$ is absolute for transitive models of ZFC the ordinal heights of transitive models of ZFC are precisely the ordinals $\alpha$ such that $L\_\alpha$ is a model of ZFC. If $\alpha\_0,\alpha\_1$ are the first two ordinals such that $L\_{\alpha\_0}$ and $L\_{\alpha\_1}$ are models of ZFC, then $L\_{\alpha\_1}$ is a model of ZFC in which there is a transitive model of ZFC, namely $L\_{\alpha\_0}$, and every transitive model of ZFC has rank $\alpha\_0$ and is therefore bounded in size by $|V\_{\alpha\_0}|$.


