# What is wrong with this proof there is no $\omega$-th worldly cardinal

Call a cardinal $$\kappa$$ worldly iff $$V_\kappa\vDash ZFC$$. Let $$\kappa_\alpha$$ be the $$\alpha$$th worldly cardinal, i.e. the least worldly cardinal such that $$\{\beta\lt\kappa_\alpha|\beta\text{ is worldly}\}$$ has order-type $$\alpha$$. Since the class of worldly cardinals are closed, then $$\kappa_\omega=\sup\kappa_n$$.

Therefore, $$V_{\kappa_\omega}\vDash\forall\alpha(\exists\delta\gt\alpha(V_\delta\vDash ZFC))$$. Now define $$f: \omega\rightarrow\kappa$$ by $$f(n)=\kappa_n$$. We have that $$f(n)$$ is definable and absoloute to $$V_{\kappa_\omega}$$. Therefore, $$f"\omega\in V_{\kappa_\omega}$$ by the axiom of replacement. But this is impossible.

The worldly cardinals are not closed, and you have proved this. You are probably remembering the similar-sounding fact that $$\{\alpha: V_\alpha\prec V_\kappa\}$$ is a club in $$\kappa$$ for worldly $$\kappa$$ of uncountable cofinality. (And note that the argument for closure relies on an elementary chain argument.) Picking out this cofinal sequence requires $$V_\kappa$$ know what is elementary to it, which it doesn't. And nor does $$V_\alpha$$ for the $$\omega$$-th such $$\alpha$$ know which worldly cardinals beneath it are elementary submodels of $$V_\kappa.$$

The claim $$V_{\kappa_{\omega}} \models \forall\alpha(\exists\delta > \alpha)(V_{\delta} \models ZFC))$$ presupposes that not only is $$\kappa_{\omega}$$ the $$\omega$$th worldly cardinal, it is in fact the limit of the $$\kappa_n$$ for $$n < \omega$$. Your argument in fact handily shows that the limit of the first $$\omega$$ worldly cardinals is not worldly, so the $$\omega$$th worldly cardinal will be much larger.

For a similar, sillier example of what's going on, consider the following "proof":

• Let $$\lambda_{\alpha}$$ denote the $$\alpha$$th infinite regular cardinal.
• Consider $$\lambda_{\omega}$$.
• The sequence $$\langle\lambda_n\mid n < \omega\rangle$$ is a countable sequence cofinal in $$\lambda_{\omega}$$.
• But $$\lambda_1 = \omega_1$$ is uncountable, so $$\lambda_{\omega}$$ is also uncountable.
• Thus $$\lambda_{\omega}$$ is an uncountable cardinal with a countable cofinal sequence, so $$\lambda_{\omega}$$ is not regular. Contradiction.

This argument is essentially the same as the one you've proposed, but talking about regularity rather than worldliness hopefully makes the flaw (step 3) clearer.