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?


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 '20 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 '20 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 '20 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 '20 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 '20 at 19:57

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