If there is a Reinhardt cardinal, then there is one universe? [closed]

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 KohlFeb 13 at 10:22

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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