# Is there a form of choice that can elude Kunen's inconsistency theorem?

When it is said that Kunen inconsistency theorem proves that given $$\sf ZFC$$ no elementary embedding can exist from the universe to itself. Most references quote full choice in stating that result, and contemplate a salvage by forsaking choice altogether like in Reinhardt's cardinals setting.

Is there a known weaker form of choice, like dependent or countable choice, that can evade this theorem?

• Isn't it open whether Rheinhardt cardinals are consistent with $\mathsf{ZF}$ at all? Commented Jan 12, 2023 at 19:58
• @JamesHanson, AFAIK, Yes! Commented Jan 12, 2023 at 20:01
• I'm confused as to why that doesn't answer your question. Commented Jan 12, 2023 at 20:02
• @JamesHanson, the question is about if it is an open question whether Reinhardt cardinals are consistent with ZF + some form of choice? Or is it the case that it is proved that all known forms of choice are inconsistent with having a Reinhardt's cardinal. Put it the other way, what is the minimal known form of choice needed for Kunen's inconsistency theorem to work? Commented Jan 12, 2023 at 20:06
• @JamesHanson For any large cardinal axiom A, one cannot prove con(ZF) => con(ZF + A) unless ZF is inconsistent, so that doesn't say much. Commented Jan 14, 2023 at 8:58

Work of Usuba combined with work of Woodin shows that if there is a Reinhardt cardinal $$\kappa$$ that is a limit of Lowenheim-Skolem cardinals, then there is a forcing extension in which $$\kappa$$ remains a Reinhardt cardinal but $$\text{DC}_\kappa$$ holds. This means one has $$\text{DC}_\lambda$$ where $$\lambda = \sup_{n < \omega} j^n(\kappa)$$ where $$j : V\to V$$ is elementary with critical point $$\kappa$$. In this context, $$\lambda^+$$ must be a measurable cardinal, so one really cannot have much more choice than $$\text{DC}_\lambda$$; e.g., $$\text{DC}_{\lambda^+}$$ and even $$\text{AC}_{\lambda^+}$$ fail.
The consistency proof uses Woodin's Easton iteration of collapse forcings as in SEM 1 Theorem 226 but substituting Usuba's Proposition 4.7. One only iterates up to $$\lambda$$ (so one is not forcing full choice, which of course would kill the Reinhardt). Then one has to lift $$j$$ to the forcing extension, which uses the master condition argument for rank-to-rank embeddings which can be found in Section 5 of Hamkins's "Fragile measurability" paper, in the context of $$I_1$$.
• Could we argue that $\mathsf{DC}_{\lambda^+}$ does not hold because we can carry over Woodin's proof of Kunen inconsistency with $\mathsf{DC}_{\lambda^+}$? Does the measurability of $\lambda^+$ give a stronger fact that $\mathsf{AC}_{\lambda^+}$ is incompatible with the Reinhardtness? Commented Jan 14, 2023 at 2:08
• @HanulJeon I think $\text{AC}_{\lambda^+}$ suffices to split $S^{\lambda^+}_\omega$ into $\lambda^+$ disjoint stationary sets, contrary to Woodin's proof. (The measurability of $\lambda^+$ is just a slight elaboration on Woodin's proof in the $\text{DC}_{\lambda^+}$ context; it's Woodin's proof plus Ulam splitting.) Commented Jan 14, 2023 at 2:31