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

Question

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.

Answer 1

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 $V_\kappa.$

Answer 2

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.