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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?

Anonymous
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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).

Reveillark
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    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.) – Noah Schweber May 17 '20 at 06:10
  • Your proof that $V_\kappa\models Con(\mathsf{ZFC})$, while correct, is massive overkill: we have the assumption that $\mathsf{ZFC}$ has a model in the first place, and so any $\omega$-model satisfies $Con(\mathsf{ZFC})$ since $Con(\mathsf{ZFC})$ is a true arithmetic sentence. – Noah Schweber May 17 '20 at 06:13
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    @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! – Reveillark May 17 '20 at 06:21
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    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. – Noah Schweber May 17 '20 at 06:23
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    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. – Noah Schweber May 17 '20 at 06:29
  • @NoahSchweber Fair enough. I omitted mentioning the absoluteness of the satisfaction relation in my answer, which I will proceed to amend now. I usually think of $\models$ being absolute for transitive models due to it's recursive definition using absolute notions. Or it being $\Delta_1^{KP}$ if more precision is required. Do you know of a reference for a proof of arithmetical statements being absolute for $\omega$-models? – Reveillark May 17 '20 at 06:38
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    "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 $(V_\omega)^M\cong V_\omega$. – Noah Schweber May 17 '20 at 19:57
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    (Incidentally, note that models with isomorphic $\omega$s can have nonisomorphic $V_\omega$s - the recursive construction of $V_\alpha$ from $\alpha$ is not first order in $\alpha$.) Meanwhile, the issue of the satisfaction relation is the following. The statement "$\models$ is absolute for transitive models" is really an instance of a more general phenomenon - that $\models$ is absolute from a model to an end extensions (transitive models have joint end extensions). This is orthogonal to the $\omega$-model issue here since a non-$\omega$-model is not an end-extension of an $\omega$-model. – Noah Schweber May 17 '20 at 20:04
  • @NoahSchweber Thanks a lot! – Reveillark May 17 '20 at 20:10