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. :)


2 Answers 2


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

  • $\begingroup$ So $\alpha$ doesn't have to be inaccessible for the power set operator of $V_\alpha$ to be the real one? $\endgroup$ Aug 27, 2020 at 1:14
  • $\begingroup$ @Jesse No, all you need for that is $\alpha$ to be a limit ordinal. $\endgroup$ Aug 27, 2020 at 1:16
  • $\begingroup$ 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 $\omega$. $\endgroup$ Aug 27, 2020 at 1:28
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    $\begingroup$ @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.) $\endgroup$ Aug 27, 2020 at 1:29
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    $\begingroup$ @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 $\omega.$ $\endgroup$ 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.


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