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 $j$.


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?

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    $\begingroup$ Related: mathoverflow.net/questions/87222/… $\endgroup$ – Asaf Karagila Apr 22 '16 at 12:25
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    $\begingroup$ 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). $\endgroup$ – Gro-Tsen Apr 22 '16 at 12:32
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    $\begingroup$ @Gro-Tsen: The word "Reinhardt" does not appear in either paper of the Suitable Extender Models. $\endgroup$ – Asaf Karagila Apr 22 '16 at 13:19
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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Apr 29 '16 at 2:19
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    $\begingroup$ 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. $\endgroup$ – 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).

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    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Nov 9 '16 at 6:10
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    $\begingroup$ 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. $\endgroup$ – Stefan Mesken Nov 9 '16 at 6:32
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    $\begingroup$ @Stefan For some reason, I cannot click on the link. $\endgroup$ – Victoria Gitman Nov 9 '16 at 14:57
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    $\begingroup$ @Victoria My bad. This link should work. $\endgroup$ – Stefan Mesken Nov 9 '16 at 15:04
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    $\begingroup$ 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.) $\endgroup$ – 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.


Mohammed Golshani's link doesn't work, so I have reconstructed a sketch of a proof. The key fact is this (You can find a proof in most Set Theory textbooks):

Theorem: If $DC_\omega$ holds and $D$ is an $\omega_1$-complete ultrafilter, then the ultraproduct of $N$ by $D$ is well-founded, for any inner model $N$.

Theorem: If $DC_\lambda$ holds and $\kappa$ is $I0$ (With target $\lambda$) if and only if there is a non-principal $\kappa$-complete $L(V_{\lambda+1})$-ultrafilter on $V_{\lambda+1}$.

Proof. Let $D$ be such an ultrafilter. To verify the existence of a such an ultraproduct, we can code elements of $D$ as function $F: X^{\lt\lambda}\rightarrow X'$ for every set $X'=\{\chi_x|x\in V_{\lambda+1}\}$, there is a function $f: \lambda\rightarrow X$, such that $f(\chi_x)\in\chi_x$, and so some $A\in D$, such that $\{\chi_x|x\in A\}$ admits a Choice function. Then if $M\cong Ult_D(L(V_{\lambda+1}))$ is the ultrapower $M\ni V_{\lambda+1}$. $L(V_{\lambda+1})$ inherits a well-order for each element (A well order not necessarily in $L(V_{\lambda+1})$ from $L(V_{\lambda+1})$. Then by condensation $M= L(V_{\lambda+1})$ and $\kappa$ is the critical point of the ultrapower embedding $k_D: L(V_{\lambda+1})\prec L(V_{\lambda+1})$. For the other direction, define an ultrafilter $D=\{X\subseteq V_{\lambda+1}|j\restriction V_\lambda\in j(X)\}$. This satisfies all the necessary properties.◼

Theorem: If $\kappa$ is Reinhardt, indeed even the critical point of $j: V_{\lambda+2}\prec V_{\lambda+2}$, then $\kappa$ is $I0$.

Proof. Let $\kappa$ Reinhardt as witnessed by $j: V\prec V$, and let $\lambda$ be the least fixed point above $\kappa$. Let $D=\{X\subseteq V_{\lambda+1}|j\restriction V_\lambda\in j(X)\}\cap L(V_{\lambda+1})$. Then $D$ satisfies all the necessary properties. The second case is not much trickier, as it uses the same argument.◼

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    $\begingroup$ DC is a red herring for your first theorem. For an ultrapower $M^\kappa/D$ to be elementary equivalent to $M$ you need to have $\sf AC_\kappa$ modulo $D$. Namely, for any family $\{A_\alpha\mid\alpha<\kappa\}$ of non-empty sets, there is $A\in D$ such that $\{A_\alpha\mid\alpha\in A\}$ admits a choice function. This is exactly the amount of choice necessary for Łoś to work and for you to get elementarity. Otherwise the ultrapower is not even extensional. $\endgroup$ – Asaf Karagila Aug 29 at 6:17
  • $\begingroup$ (Let me qualify my previous remark, and say that you need choice from families of sets in $M$. So if $M$ is a model of choice, e.g. HOD, you might just have this without needing more. But the point stands: DC is irrelevant here.) $\endgroup$ – Asaf Karagila Aug 29 at 19:21
  • $\begingroup$ Anything to respond to that? $\endgroup$ – Asaf Karagila Sep 4 at 6:17
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    $\begingroup$ Yes, it's true that $L(V_{\lambda+1})$ inherits a well-orderings from $V$. But this pressuposes that choice holds in $V$. If there is a Reinhardt cardinal, choice fails. As I wrote to begin with, DC is kind of a red herring here. $\endgroup$ – Asaf Karagila Sep 5 at 14:52
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    $\begingroup$ You cannot prove that $V_{\lambda+1}$ can be well-ordered just by assuming $\sf DC_\lambda$ holds. (I miscounted in a previous comment which is now deleted.) An easy example is that $\sf DC$ need not imply that the real numbers can be well-ordered. $\endgroup$ – Asaf Karagila Sep 5 at 16:58

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