# 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.

1. Does ZFC prove that, if there is a worldly ordinal, then there is a worldly cardinal?
2. Does ZFC prove that every worldly ordinal is a (worldly) cardinal?
3. 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. :)

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 $$2^{|\beta|}<\alpha.$$

• So $\alpha$ doesn't have to be inaccessible for the power set operator of $V_\alpha$ to be the real one? Aug 27, 2020 at 1:14
• @Jesse No, all you need for that is $\alpha$ to be a limit ordinal. Aug 27, 2020 at 1:16
• I see, so $\alpha$ being singular or regular would be irrelevant. It is still weird to me that the smallest worldly cardinal, if it exists, has cofinality $\omega$. Aug 27, 2020 at 1:28
• @JesseElliott In fact, if $\alpha$ is a limit ordinal $>\omega,$ $V_\alpha$ is already model of ZFC minus replacement. (With full power sets since the whole power set is already of rank two higher than the set itself, and any transitive model has $P(x)^M=P(x)\cap M$ since the subset relation is absolute for transitive models.) Aug 27, 2020 at 1:29
• @JesseElliott Replacement is the big issue here, since that may fail if $\alpha$ is not regular, but the key is that to violate replacement the witnessing cofinal sequence needs to be first-order-definable in the model. This will be the case if $\alpha=\omega\cdot 2,$ or $\alpha = \beth_\omega,$ or $\alpha$ is the first $\beth$-fixed point... in all these cases you can define in $V_\alpha$ sequence witnessing that $\alpha$ has cofinality $\omega.$ This is another way to see that $\alpha$ has some big-ness to it. But it doesn't need to be regular, and the smallest has cofinality $\omega.$ Aug 27, 2020 at 1:37

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.