If there is a "worldly ordinal," then must there be a worldly cardinal? A cardinal $\kappa$ is said to be worldly if $V_\kappa$ is a model of ZFC.  Let us (potentially) generalize this by saying an ordinal $\alpha$ is worldly if $V_\alpha$ is a model of ZFC.  The existence of a worldly ordinal implies the existence of a transitive model of ZFC, hence a countable ordinal $\alpha$ such that $L_\alpha$ is a model of ZFC.  But I'm not sure in that case if there must then be a worldly cardinal.
My questions are as follows.

*

*Does ZFC prove that, if there is a worldly ordinal, then there is a worldly cardinal?

*Does ZFC prove that every worldly ordinal is a (worldly) cardinal?

*If not, is it consistent with ZFC that every worldly ordinal is a cardinal?

Hopefully it's my last set theory question for a while.  :)
 A: Yes, every worldly ordinal is a cardinal.
Suppose $\alpha$ were a worldly ordinal that is not a cardinal, so that $\kappa := |\alpha| < \alpha$.  In particular, since $\kappa$ is a set of rank $\kappa < \alpha$, we have $\kappa \in V_\alpha$.
Let $f \in V$ be a bijection from $\alpha$ to $\kappa$.  By pushing forward the $\in$ well-ordering on $\alpha$, we get a relation $R$ on $\kappa$ which is a well-ordering of type $\alpha$.  Since $R$ is a subset of $\kappa \times \kappa$, it has rank $\kappa+3$ or something like that.  Now since $\kappa < \alpha$, and $\alpha$, being worldly, is not a successor, we also have $\kappa+3 < \alpha$.  So $R \in V_\alpha$.   But ZFC proves there exists an ordinal isomorphic to $(\kappa, R)$, so this ordinal must exist in $V_\alpha$, and it must be $\alpha$.  This is a contradiction.
A: If $V_\alpha$ is a model of ZFC, then $\alpha$ must be a cardinal, and much more. In fact it must be a strong limit cardinal, a $\beth$-fixed point, a fixed point in the enumeration of $\beth$-fixed points, and have any other strong limit property of this sort.
To see this, observe that ZFC proves that the $\beth$-hierarchy is unbounded, but also we can show that the $\beth$ hierarchy (and the Von-Neumann hierarchy) is absolute for a "full" model of the form $V_\alpha,$ since the model's power set operator is the same as the real one. So it follows that $\alpha$ is a strong limit. And the stronger properties follow from similar considerations.
In a little more detail, if $\beta <\alpha,$ then $P(\beta) \in V_\alpha$ since $P(\beta)$ is just two ranks higher than $\beta,$ and so since being a subset is absolute, $P(\beta)^{V_\alpha}=P(\beta).$ ZFC proves $2^{|\beta|}$ exists, which relativized to $V_\alpha$ is the least ordinal in $V_\alpha$ that has a bijection in $V_\alpha$ with $P(\beta).$ And being a bijection is absolute so this is a real bijection, thus there is an ordinal in $V_\alpha$ that is in one-to-one correspondence with $P(\beta),$ so $2^{|\beta|}<\alpha.$
