# Inaccessible cardinal in standard transitive models of ZFC

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 ?

• 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. Commented May 16 at 22:55
• @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. Commented May 16 at 22:57
• @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 $<\kappa$. Commented May 16 at 23:03
• @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. Commented May 16 at 23:14
• @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.) Commented May 16 at 23:41