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If there is a nontrivial elementary embedding $j:V \to V$, then there is a universe which contains all the large cardinals.

Is there such a universe? Does this imply there is one universe from which we force as many extensions as possible?

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closed as unclear what you're asking by Andres Caicedo, Bjørn Kjos-Hanssen, Neil Strickland, j.c., Stefan Kohl Feb 13 at 10:22

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

I'm not quite sure I understand the question. Could you elaborate? –  Noah S Feb 13 at 3:07
Erin, I don't really understand the question. Can you explain it more precisely? –  Joel David Hamkins Feb 13 at 3:58
It seems to somehow relate to the HOD conjecture and Woodin's Ultimate L, perhaps? –  Asaf Karagila Feb 13 at 5:05
There are no Reinhardt cardinals, so the question as written makes no sense. If the question is over $\mathsf{ZF}$, it is expected that the hierarchy of large cardinals can be continued further past this assumption, so the answer is no. If you mean something else, such as the existence of a non-amenable embedding (as in Corazza's wholeness axiom), the answer is again no. I suspect that either there is some confusion on your part, or else the question needs serious editing to turn into what you are really after. –  Andres Caicedo Feb 13 at 6:23
Ah, I think the question might be the following: is there a (consistent) strongest large cardinal notion, one which when realized in a model causes all the others also to be realized there? I would argue that the answer is no, since for any large cardinal property $P$, we can form the axiom "$P$ holds in some rank initial segment $V_\kappa$," which will be a strictly stronger large cardinal property, of a commonly considered and acceptable type. (Of course, one could also just say $\text{Con}(\text{ZFC}+P)$, which is also stronger.) In this sense, the large cardinal hierarchy has no top. –  Joel David Hamkins Feb 13 at 13:35