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Jan 3, 2019 at 21:45 vote accept Zuhair Al-Johar
Jan 2, 2019 at 19:25 comment added Wojowu 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.
Jan 2, 2019 at 18:55 answer added Noah Schweber timeline score: 5
Jan 2, 2019 at 18:43 history asked Zuhair Al-Johar CC BY-SA 4.0