Inaccessible cardinal and well founded set

How can I prove the following fact? Let be a regular cardinal $$\kappa$$ such that $$V_\kappa$$ is a model of ZFC. Then $$\kappa$$ is inaccessible.

• Note that the proof that $\kappa$ is a strong limit doesn’t use the regularity of $\kappa$. So a more informative wording would be “if $V_{\kappa}$ is a model of ZFC then $\kappa$ is a strong limit.” (So of course if $\kappa$ also happens to be regular, then it is inaccessible.) – spaceisdarkgreen Sep 23 '19 at 16:40

Since we assume $$\kappa$$ to be regular, we only need to prove that for any $$\lambda < \kappa$$ we have $$2^\lambda < \kappa$$. Note that for every ordinal $$\alpha$$ we have $$\operatorname{rank}(\alpha) = \alpha$$ (see, e.g. this answer). In particular, that means that the collection of ordinals in $$V_\kappa$$ is just $$\kappa$$. So for $$\lambda < \kappa$$, we have $$\lambda \in V_\kappa$$ and all of its subsets belong to $$V_\kappa$$. Then since $$V_\kappa$$ is a model of ZFC, the powerset $$\mathcal{P}(\lambda)$$ must belong to $$V_\kappa$$ again. Thus $$2^\lambda$$ (as an ordinal) is also in $$V_\kappa$$, and so we see $$2^\lambda < \kappa$$.

• The question asks about regular $\kappa$ and a regular wordly cardinal is inaccessible – Alessandro Codenotti Sep 23 '19 at 11:27
• @AlessandroCodenotti Thanks for pointing that out, I did not read properly. Hopefully my answer now makes sense and actually answers the question. – Mark Kamsma Sep 23 '19 at 12:05
• @Mark Kamsma Thank you. However, here I’m assuming that k is regular. So I’m not considering the case in which k is the least worldly cardinal. So the statement should be correct but I’ve no idea how to prove that. – SeTh Sep 23 '19 at 12:22
• @SeTh Refresh the page ;) Alessandro pointed this out as well, and I edited. Hopefully it now actually answers your question. – Mark Kamsma Sep 23 '19 at 12:23
• Thank you so much! Very elegant proof – SeTh Sep 23 '19 at 12:57