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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 $\alpha<\text{crit}(j)<\kappa$.

My question is about the restrictions involved in that definition of $\kappa \in M$? and of $M$ being transitive? and of $M$ being a set? why not the following?

....if for any class $M$ with $κ \subseteq M$ and any ordinal $α<κ$ there is an elementary embedding $j : M → M$ with $\alpha<\text{crit}(j)<\kappa$.

Where the underlying theory is $MK$ like, with the axiom of limitation of size replaced by an axiom of limitation of size on sets, more specifically any class that is subnumerous to a set, is a set. And of course without $AC$.

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    $\begingroup$ I think assuming the condition for every class doesn't immediately lead to a contradiction, but it might increase the strength of the notion - suppose there is an inaccessible $\lambda$ above a Berkeley $\kappa$. Consider $V_\lambda$ - every class in it is a set in $V$, and an elementary embedding from it to itself is again a class in $V_\lambda$, so $V_\lambda$ satisfies the Berkeley condition for all classes. One reason for restricting the definition to sets is that it turns the notion into something expressible in ZF - you can't quantify, not even explicitly talk about, classes there. $\endgroup$
    – Wojowu
    Jan 2, 2019 at 19:25

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Dropping transitivity doesn't actually add anything: given $M\supseteq\kappa$, just consider $\hat{M}=$ the Mostowski collapse of $M$. Given $\alpha<\kappa$ and $j:\hat{M}\rightarrow \hat{M}$ nontrivial elementary with $\alpha<crit(j)<\kappa$, $j$ lifts to a nontrivial elementary embedding $\hat{j}$ of $M$ into itself with $\alpha<crit(\hat{j})<\kappa$. So if $\kappa$ is Berkeley in the usual sense, it's Berkeley in the nontransitive sense too.

Finally, if you drop transitivity and replace "$\kappa\subseteq M$" by the original "$\kappa\in M$," then the argument above breaks down but the notion becomes inconsistent: taking $M=\{\kappa\}$ doesn't leave room for any nontrivial elementary embeddings, for example, let alone ones with critical points (and we can cook up less-silly examples too).

I had a stupid moment earlier, where I mixed up ZF and ZFC - it's not immediately obvious to me now that replacing "set" with "class" results in inconsistency. I suspect it does however.

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    $\begingroup$ The question is definitely in the context of ZF, not ZFC. In ZFC we already have a problem by taking $M=V_{\kappa+2}$. $\endgroup$
    – Wojowu
    Jan 2, 2019 at 19:09
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    $\begingroup$ You quoted Kunen's inconsistency theorem. $\endgroup$
    – Wojowu
    Jan 2, 2019 at 19:09
  • $\begingroup$ @Wojowu Yeah I noticed that just as I finished typing :P. Fixed! $\endgroup$
    – Noah Schweber
    Jan 2, 2019 at 19:10
  • $\begingroup$ it is much shorter to say $\kappa \subseteq M$ than saying $M$ is transitive and $\kappa \in M$. What remain is the class condition? $\endgroup$
    – Zuhair Al-Johar
    Jan 2, 2019 at 19:36
  • $\begingroup$ @ZuhairAl-Johar Yes, but I don't think efficiency is always the highest goal. Focusing explicitly on transitive sets makes the picture easier to think about, in my opinion; we don't have to pay attention to Mostowski collapses. It also plays more nicely with other notions where transitivity/$\ni\kappa$ is more important. Re: the class condition, like I said (in my edit) I don't know. $\endgroup$
    – Noah Schweber
    Jan 2, 2019 at 19:38

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