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 $AC$.

  • 2
    $\begingroup$ 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 '19 at 19:25

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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 '19 at 19:09
  • 1
    $\begingroup$ You quoted Kunen's inconsistency theorem. $\endgroup$ – Wojowu Jan 2 '19 at 19:09
  • $\begingroup$ @Wojowu Yeah I noticed that just as I finished typing :P. Fixed! $\endgroup$ – Noah Schweber Jan 2 '19 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 '19 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 '19 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.