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Let's define cardinal $\kappa$ as hyper-Berkeley if for any transitive set $M$ such that $\kappa\in M$ there exists an elementary embedding $j: M\prec M$ with fixed point $\lambda$ and $\text{crit}j\lt\lambda<\kappa$.

  1. Are hyper Berkeley cardinals equiconsistent with $\sf ZF$+"club Berkeley cardinal"?
  2. Are hyper Berkeley cardinals equiconsistent with $\sf ZF+BC$ (Berkeley cardinal)?
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  • $\begingroup$ Asking equiconsistency in the "above ZFC" portion of large cardinals is hard, since we don't actually have a lot of techniques there. Specifically, we have "more or less" direct implications (e.g. Berkeley cardinal implies there is a set model with a Reinhardt; or super-Reinhardt is Reinhardt), and forcing (e.g. start with a Berkeley cardinal of cofinality $\omega_1$, and collapse $\omega_1$ to be countable, then all the embeddings lift). And even then, the forcing results are still quite rudimentary and insufficient to provide "simple exercise" answers in most cases. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 26, 2020 at 12:25
  • $\begingroup$ (Also, try to give your questions a descriptive title.) $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 26, 2020 at 12:27
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    $\begingroup$ Every Berkeley cardinal is hyper-Berkeley. Clearly Berkeley cardinals of uncountable cofinality are hyper-Berkeley. Suppose $\kappa$ is hyper-Berkeley of countable cofinality and $s\subseteq \kappa$ is cofinal and ordertype $\omega$. Fix a transitive set $M$ containing $\kappa$. Bagaria-Koellner-Woodin tricks yield a $j : M\to M$ such that $j(\alpha) = \alpha$ for all $\alpha\in s$ and $\text{crit}(j) < \kappa$. The first fixed point of $j$ above $\text{crit}(j)$ is strictly below $\kappa$ since $j$ fixes every ordinal in the cofinal set $s$. $\endgroup$
    – Gabe Goldberg
    Commented Aug 26, 2020 at 18:29

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