Does the Consistency of ZFC imply $V_\kappa$?

Question

It is well known that the existence of an inaccessible cardinal implies the consistency of ZFC. However, I am curious about the converse. Does the consistency of ZFC (plus ZFC) imply the existence of an inaccessible cardinal? If we suppose ZFC is consistent, then it has a model. Could that model be used to construct an inaccessible cardinal? Could perhaps the cardinality of that model itself be an inaccessible cardinal?

Answer 1

The answer is no.

Suppose there is an inaccessible cardinal. We will construct a model of ZFC + Con(ZFC) + "There is no inaccessible cardinal".

Let $\kappa$ be the smallest inaccessible cardinal. It is routine to verify that $V_\kappa\models ZFC + $ "There is no inaccessible cardinal".

Let $X\preceq V_\kappa$ be countable. Take the Mostowski collapse $M\cong X$. As $M$ is countable and transitive, $M\in V_\kappa$. By elementarity, $X\models ZFC$, so $M\models ZFC$, so $V_\kappa\models (M\models ZFC)$, so $V_\kappa \models $ Con (ZFC).