For a standard model of ZFC $\langle N,\in \rangle$ (where $N$ is a transitive set), if $\kappa\in N$ is an inaccessible cardinal in the model, does it imply that it is an inaccessible cardinal (in general, in the universe) ?

Id est, if for $\kappa\in N$, we have $\langle N,\in\rangle\models SI(\kappa)$, then $\kappa$ is an inaccessible cardinal. (Where SI(x) is the formula saying that x is an inaccessible cardinal). Is it true ? If yes, how to show it ?


1 Answer 1


No. If there is any transitive model containing inaccessibles, there will be (by using Lowenheim-Skolem to get a countable elementary submodel and then Mostowski collapsing it) a countable transitive model that is elementarily equivalent, hence contains ordinals that it believes are inaccessible cardinals. And since it has a countable height, none of the ordinals in that model are actually inaccessible. Note this works with literally any (first-order expressible) large cardinal property.

  • $\begingroup$ Thanks a lot ! But then, how would you show that the consistency of $ZFC$ implies the consistency of $ZFC + \neg \exists \kappa SI(\kappa)$ ? I am trying to prove it, I thought you suppose the existence of a model of $ZFC$ $\langle N,\in \rangle$ that satisfies $\exists \kappa SI(\kappa)$ wlog, you take the smallest such $\kappa\in N$, and then you show that $V_{\kappa}\models \neg \exists \kappa SI(\kappa)$. But then it doesn't make really sense if $\kappa$ is not a "real" inaccessible cardinal. $\endgroup$ Commented May 16 at 22:55
  • $\begingroup$ @MamounAich Models of the form $V_\alpha$ have some rather special properties not shared by all transitive models. An inaccessible cardinal in such a model is really inaccessible. Try to prove it directly. $\endgroup$ Commented May 16 at 22:57
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    $\begingroup$ @MamounAich That's the right idea, but you consider $V_\kappa^N$, the submodel of $N$ consisting of those elements which $N$ thinks have rank $<\kappa$. $\endgroup$ Commented May 16 at 23:03
  • $\begingroup$ @MamounAich Sorry, I didn't read your comment closely and mistook your problem for a different one... what Alex said... then use the fact that I mentioned, relativized to $N$ to show that $V_\kappa\cap N$ must not contain anything it thinks is an inaccessible cardinal, since it doesn't contain any inaccessible cardinals (all relative to $N$). Also, though it might reasonably be considered implied, I should have mentioned above that $V_\alpha$ needs to be a model of ZFC for that to be true in general. $\endgroup$ Commented May 16 at 23:14
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    $\begingroup$ @MamounAich FWIW, the way I usually think about it is that if ZFC proved there were inaccessibles, then every $V_\kappa$ for $\kappa$ inaccessible would have to contain inaccessibles (since they're all models of ZFC and innaccesibles are the same in a $V_\kappa$ and the universe)... but clearly that's not the case for the least inaccessible $\kappa$. (Or you can just lean on the incompleteness theorem.) $\endgroup$ Commented May 16 at 23:41

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