Characterizing worldly cardinals as the supremum (by n) of least $\Sigma_n$-extendible cardinals A cardinal $\kappa$ is called $\Sigma_n$-extendible iff there is some $\theta$ such that $V_\kappa\prec_{n} V_\theta$. That is, they agree on the truth of $\Sigma_n$ assertions with parameters from $V_\kappa$. A cardinal $\kappa$ is called worldly iff $V_\kappa$ is a model of ZFC. Question:

Consider $\sup_n(\min\{\alpha\mid \alpha \text{ is } \Sigma_n\text{-extendible}\})$.
Is this cardinal the least worldly cardinal?

An observation: if we let $\kappa$ be the least worldly cardinal, then since $\Sigma_n$-satisfaction for $V_\kappa$ is definable, by the reflection principle we are able to find $V_\alpha\prec_n V_\kappa$ for each $n$ in $V_\kappa$. So $\sup_n(\min\{\alpha\mid \alpha \text{ is } \Sigma_n\text{-extendible}\})$ is not greater than the least worldly cardinal. Is it possible that it is smaller?
 A: The answer is negative. I claim that the supremum can see the
$\omega$-cofinal sequence $\kappa_n$, where $\kappa_n$ is the least
$\Sigma_n$-extendible cardinal.
The main point is that if $\alpha$ is $\Sigma_{n+1}$-extendible, so
$V_\alpha\prec_{\Sigma_{n+1}}V_\beta$ for some $\beta$, then
$V_\beta$ can see that $\alpha$ is $\Sigma_n$-extendible, and so
the least $\Sigma_n$-extendible is in fact less than $\alpha$. So
each next $\Sigma_{n+1}$-extendible cardinal must be beyond the $\beta$ used for
the $\Sigma_n$-extendibility of the least $\Sigma_n$-extendible cardinal. Therefore, the limit $\kappa$ of the
least $\Sigma_n$-extendibles can see that they are like that, and
so it can define the cofinal $\omega$-sequence, which means it
isn't even $\Sigma_2$-extendible.
Another argument: if $\kappa$ is the least worldly cardinal, then let $\delta<\kappa$ be $\Sigma_2$-correct in $V_\kappa$, which is possible, because ZFC proves that the $\Sigma_2$-correct cardinals form a club. But $V_\delta$ is correct about whether a cardinal is $\Sigma_n$-extendible, and whether there is one. So the whole sequence is below $\delta$.
