[Math] Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)

axiom-of-choicelarge-cardinalslo.logicset-theory

The question is: if I assert in ZF that there exists a Reinhardt cardinal, do I really get a theory of higher consistency strength than when I assert in ZFC that there exists an I0 cardinal (the strongest large cardinal not known to be inconsistent with choice, as I understand)? This is implicit in the ordering of things on Cantor's Attic, for example, but I've been unable to find a proof (granted, I don't necessarily have the best nose for where to look!).

One thing that worries me is that when there is a ZFC analog of a ZF statement, many equivalent formulations of the ZFC statement may become inequivalent in ZFC. So we don't have much assurance that the usual definition of a Reinhardt cardinal is "correct" in the absence of choice.

I think it should be clear that Con(ZF + Reinhardt) implies Con(ZF + I0). But again, it's not clear that ZF+I0 is equiconsistent with ZFC+I0.

It's apparently not possible to formulate Reinhardt cardinals in a first-order way, so I should really talk about NBG + Reinhardt, or maybe ZF($j$) + Reinhardt, where ZF($j$) has separation and replacement for formulas involving the function symbol $j$.

EDIT

Since this question has attracted a bounty from Joseph Van Name, maybe it's appropriate to update it a bit. Now, I'm not actually a set theorist, but it's not even clear to me that Con(ZF + Reinhardt) implies Con(ZFC + an inaccessible). So perhaps the question should really be: what large cardinal strength, if any, can we extract from the theory ZF + Reinhardt?

Best Answer

The answer to your question is (almost) yes (almost is because of the addition of DC to the statement).

Recently Gabriel Goldberg has proved

''Con(NBG+DC+Reinhardt)$ \implies$ Con(ZFC+I0)''.

See the abstract of the talk by Gabriel Goldberg Choiceless cardinals and I0.

(Thanks to Rahman for pointing this to me).


Edit. The result of Goldberg is now available,where indeed something stronger is proved. See Even ordinals and the Kunen inconsistency. It is shown, assuming DC, the existence of an elementary embedding from $V_{λ+3}$ to $V_{λ+3}$ implies the consistency of ZFC + $I_0$, while by a recent result of Schlutzenberg, an elementary embedding from $V_{λ+2}$ to $V_{λ+2}$ does not suffice. The paper of Schlutzenberg is On the consistency of ZF with an elementary embedding $j : V_{λ+2} → V_{λ+2}$.

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