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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?

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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).

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    $\begingroup$ 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.) $\endgroup$ May 17, 2020 at 6:10
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    $\begingroup$ @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! $\endgroup$
    – Reveillark
    May 17, 2020 at 6:21
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    $\begingroup$ 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. $\endgroup$ May 17, 2020 at 6:23
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    $\begingroup$ 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. $\endgroup$ May 17, 2020 at 6:29
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    $\begingroup$ "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 $(V_\omega)^M\cong V_\omega$. $\endgroup$ May 17, 2020 at 19:57

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