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

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