# Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?

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?

• – Asaf Karagila Apr 22 '16 at 12:25
• I think the proof is supposed to be somewhere inside Woodin's supercompact set of papers with titles like "suitable extender sequences" or "suitable extender models". At least that's what I remember Woodin stating in one of his survey papers (but I already found the surveys a bit hard to digest so I won't try to go beyond that, and I certainly can't give a precise reference). – Gro-Tsen Apr 22 '16 at 12:32
• @Gro-Tsen: The word "Reinhardt" does not appear in either paper of the Suitable Extender Models. – Asaf Karagila Apr 22 '16 at 13:19
• The ordering of cardinals on Cantor's attic was based on the ZFC version of Reinhardt cardinals, which are inconsistent. Reinhardt had originally proposed his cardinals in ZFC. The vestige of his idea lives on ZF. – Joel David Hamkins Apr 29 '16 at 2:19
• The "I" stands for "inconsistent", because it was initially thought that those embeddings might be inconsistent. Richard Laver disliked the notation, first because of that, and second, because the numbers $I_0$, $I_1$, $I_2$ and $I_3$ are in the wrong order. – Joel David Hamkins May 16 '16 at 11:26

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).

• Uhh, I don't see where in the linked abstract it says that. I says "come very close to improving this lower bound to Con(ZFC+I0)". It doesn't say that any such implication was actually proved. – Asaf Karagila Nov 9 '16 at 6:10
• I took notes during Gabe's talk yesterday and he indeed outlined of proof of $\operatorname{NBG} + \operatorname{DC} + \text{Reinhardt} \vdash \operatorname{Con}(\operatorname{ZFC} + \operatorname{I_0})$. The use of $\operatorname{DC}$ - according to Gabe - is unfortunate, but it's unknown how to avoid it for now. If Gabe is fine with it (I'll ask him later today), I'd be happy to share my notes in case anyone is interested in them. – Stefan Mesken Nov 9 '16 at 6:32
• @Stefan Thanks Stefan, I edited the answer to make the statement precise. Surely, I'm interested to see the notes – Mohammad Golshani Nov 9 '16 at 9:25
• @Victoria My bad. This link should work. – Stefan Mesken Nov 9 '16 at 15:04
• With Gabe's permission, I uploaded my notes here. He is also finalizing a paper on the subject, so keep an eye out for that. (Sorry about the bad handwriting - it was freezing cold inside the lecture hall.) – Stefan Mesken Nov 9 '16 at 15:06

Regarding the edit, one can easily show some simple lower bounds for a Reinhardt cardinal that are far stronger than an inaccessible cardinal. For example, if $\kappa$ is a Reinhardt cardinal, assuming ZF only, then it is clear that $\kappa$ is inaccessible and weakly compact and much more in $L$, because it is the critical point of an elementary embedding $j:V\to V$, which therefore gives rise to an elementary embedding $j\upharpoonright L:L\to L$, and any such $\kappa$ must be inaccessible in $L$ and weakly compact in $L$ and much more. Indeed, one easily gets the consistency of a measurable cardinal, since if $\mu$ is the measure on $\kappa$ induced by the original embedding $j:V\to V$, then $L[\mu]$ will be the canonical inner model in which $\kappa$ is measurable.

It seems to me that one will be able to carry this argument completely through the standard inner model of large cardinals. Thus, from a Reinhardt cardinal in ZF set theory, I expect that the critical point $\kappa$ of the corresponding embedding $j:V\to V$ will be very large in the corresponding core models.

What is less clear to me is the extent to which one gets models of ZFC plus $\kappa$ has large cardinal properties that are not witnessed by the standard inner model theory, and this is how one should interpret the question.