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

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?

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).
• It's also worth pointing out that the smallest $\kappa$ such that $V_\kappa\models\mathsf{ZFC}$ - if such a $\kappa$ exists at all, of course - is not inaccessible. (Also, the same line of attack applies to well-founded models in general: $\mathsf{ZFC}+Con(\mathsf{ZFC})$ does not prove that $\mathsf{ZFC}$ has a well-founded model.) May 17, 2020 at 6:10
• @NoahSchweber That's an excellent point. In my mind, the collapsing argument is more elementary, no pun intended, than $\omega$-models, but that's of course rather subjective. I've never seen a source that collects nice facts about $\omega$-models, the very little I know of them comes from off-hand remarks I've read here and there. If you know of a more comprehensive reference, I'd love to hear about it! May 17, 2020 at 6:21
• The collapsing argument actually uses $\omega$-model-ness, though, when fully unwound. You have to argue that $V_\kappa$ detects models of $\mathsf{ZFC}$ correctly to go from $M\models\mathsf{ZFC}$ to $V_\kappa\models Con(\mathsf{ZFC})$ by way of $V_\kappa\models(M\models\mathsf{ZFC})$. And this isn't trivial: consider for example the fact that ever model of $\mathsf{ZFC}$ contains a model of $\mathsf{ZFC}$, in light of the second incompleteness theorem. May 17, 2020 at 6:23
• The ingredient that makes it work is exactly that $V_\kappa$ is an $\omega$-model: $V_\kappa$'s version of the $\mathsf{ZFC}$ axioms is just the $\mathsf{ZFC}$ axioms themselves (as opposed to the usual $\mathsf{ZFC}$ axioms + some nonstandard sentences). So the same key idea is needed in either approach. May 17, 2020 at 6:29
• "Do you know of a reference for a proof of arithmetical statements being absolute for $\omega$-models?" It's basically immediate from the definition. In set theory, an arithmetical statement is just a sentence with all quantifiers relativized to (the canonical formula defining) $V_\omega$. (Relativizing to (the canonical formula defining) $\omega$ is going too far: $(\omega,\in)$ is a very weak structure.) Consequently, two models with isomorphic $V_\omega$s satisfy the same arithmetical sentences. And again by definition $\omega$-models are just those $M$s with $(V_\omega)^M\cong V_\omega$. May 17, 2020 at 19:57