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

Question

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?

Answer 1

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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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 $j : V_{λ+2} → V_{λ+2}$.

Answer 2

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.

Answer 3

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.◼