## Luminy – Coda

While at Luminy, David Asperó showed me a quick proof of a nice result on Reinhardt cardinals in ${\sf ZF}$. It complements Grigor Sargsyan’s result discussed here.

Theorem (Asperó). Work in ${\sf ZF}$. Suppose $j:V\to V$ is a nontrivial elementary embedding. Then there are a $\bar\kappa<{\rm cp}(j)$ and an ordinal $\alpha$ such that for all $\beta$ there is a $\mu$ and an elementary

$\pi:V_\mu\to V_\mu$

such that ${\rm cp}(\pi)=\bar\kappa$ and $\pi(\alpha)>\beta$.

Proof. For $\alpha$ an ordinal, set

$\kappa^\alpha=\min\{\kappa\mid\exists\mu\exists i:V_\mu\to V_\mu$ such that ${\rm cp}(i)=\kappa$ and ${\rm ot}(\{\beta<\mu\mid i(\beta)=\beta\})\ge\alpha\}$.

Note that suitable fragments of $j$ witness that $\kappa^\alpha$ is defined for all $\alpha$. Moreover, $\alpha<\beta$ implies that $\kappa^\alpha\le\kappa^\beta\le{\rm cp}(j)$, and therefore there is a $\bar\kappa\le{\rm cp}(j)$ such that $\kappa^\beta=\bar\kappa$ for all $\beta$ sufficiently large. Moreover, since it is definable, we actually have $\bar\kappa<{\rm cp}(j)$.

Let $\alpha$ be least with $\kappa^\beta=\bar\kappa$ for $\beta\ge\alpha$. We claim that $\bar\kappa$ and $\alpha$ are as wanted. For this, consider some $\beta>\alpha$, and pick $i:V_\mu\to V_\mu$ witnessing that $\bar\kappa=\kappa^\beta$. All we need to do is to check that $i(\alpha)\ge\beta$.

But note that if $\gamma\in[\alpha,\beta)$, then $V_\mu\models\kappa^\gamma=\bar\kappa$ Hence, if $i(\alpha)<\beta$, we have

$V_\mu\models \kappa^{i(\alpha)}=\bar\kappa$.

But $\kappa^{i(\alpha)}=i(\kappa^\alpha)=i(\bar\kappa)>\bar\kappa$. Contradiction. $\Box$