From Cantor's Attic:

A cardinal κ is a Berkeley cardinal, if for any transitive set $M$ with $κ∈M$ and any ordinal $α<κ$ there is an elementary embedding $j : M → M$ with $\alpha<\text{crit}(j)<\kappa$. These cardinals are defined in the context of ZF set theory without the axiom of choice.

I have seen claimed several times that the existence of a Berkeley is stronger than that of a Reinhardt cardinal (let's say in NBG without Choice, as formalizing existence of Reinhardts in ZF takes some extra consideration). I can't however find any proof of this claim.

Of course if Berkeley-ness were about elementary embeddings of arbitrary classes containing $\kappa$, then it would be obvious, but we are only talking about transitive set models here, so I don't really see what to do. Can anyone provide a proof that Berkeley are stronger than Reinhardts? What about Super Reinhardts?

  • 1
    $\begingroup$ If $\delta$ is the critical point of $j\colon V_{\kappa+1}\to V_{\kappa+1}$, set $\lambda=\sup j^n(\delta)$, and I think that $V_\lambda$ should be a model of ZF in which $\delta$ is a super-Reinhardt. $\endgroup$
    – Asaf Karagila
    Feb 17, 2018 at 21:58
  • $\begingroup$ This would be a just a model for the Wholeness Axiom. $\endgroup$
    – Master
    Aug 29, 2019 at 15:11

1 Answer 1


The following is proved by Bagaria-Koellner-Woodin:

Theorem. If $δ_0$ is the least Berkeley cardinal, then there exists $γ < δ_0$ such that $(V_γ,V_{γ+1})\models ZF_2+$“There exists a Reinhardt card., witnessed by $j$, and an $ω$-huge card. above $κ_ω(j)$”.

See Large Cardinals beyond Choice. Let me sketch the proof, skipping some details:

First, one can show that for a tail of ordinals $\beta$, if $j: V_\beta \to V_\beta$ is an elementary embedding with $crit(j)< \delta_0$, then $j(\delta_0)=\delta_0$ and $\{\alpha < \delta_0: j(\alpha)=\alpha \}$ is cofinal in $\delta_0$. Fix such a $\beta$ and $j$. Let $\kappa=crit(j), \lambda=κ_ω(j)$

Since $\delta_0$ is a Berkeley cardinal there are embeddings $j': V_\beta \to V_\beta$ with $\lambda < crit(j') < \delta_0$. Also one can show $\kappa_\omega(j')< \delta_0$ holds for any such $j'$.

So there are $\lambda'< \delta_0$ and $j': V_{\lambda'} \to V_{\lambda'}$ with $κ_ω(j')=\lambda'$ and $crit(j') > \lambda$. Let $\lambda'$ be the least ordinal with this property and let $\gamma$ be the least strongly inaccessible cardinal greater than $\lambda'$. Then $j(\lambda') = \lambda'$ and $j(\gamma) = \gamma.$ Thus, $j\restriction V_\gamma$ and $j'$ witness that $(V_γ,V_{γ+1})\models ZF_2+$“There exists a Reinhardt card., witnessed by $j$, and an $ω$-huge card. above $κ_ω(j)$”.


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