= Solution
An <strongly inaccessible cardinal> $\kappa$ has the <Keisler extension property> when there is a proper <transitive set> $X\supsetneq V_\kappa$ such that
$$
(V_\kappa,\in)\prec(X,\in).
$$
Suppose $\kappa$ is strongly inaccessible and has this property. Because $X$ properly extends the transitive set $V_\kappa$, it contains $\kappa$. Strong inaccessibility of $\kappa$ is <downward absolute formula>[downward absolute] from the ambient universe to the transitive set $X$: any internal witness that $\kappa$ is countable, singular, or not a strong limit would also be an ambient witness. Hence
$$
X\models\text{“there exists a strongly inaccessible cardinal”},
$$
with $\kappa$ as a witness. Since $V_\kappa\prec X$, the same sentence holds in $V_\kappa$. Its witness is an ordinal $\mu<\kappa$. The set $V_\kappa$ contains $V_{\mu+1}$, so it computes all subsets of cardinals below $\mu$ correctly; strong inaccessibility of $\mu$ is therefore absolute between $V_\kappa$ and the universe. Thus there is a strongly inaccessible $\mu<\kappa$, and $\kappa$ cannot be the least strongly inaccessible cardinal.
Back to article page
