# Are these large cardinals properties equivalent?

Consider the three following large cardinal axioms:

1. there exists a nontrivial elementary embedding $j:V\to V$.
2. there exists a n.e.e. $j:V\to M$ such that $M^{j^\omega(crit(j))}\subseteq M$.
3. there exists a n.e.e. $j:V_{\lambda+2}\to V_{\lambda+2}$ for some $\lambda$.

Kunen's inconsistency theorem states that, in ZFC (with replacement extended to formulas containing $j$), these three statements are false. However there is no such (known) result within ZF (with replacement sufficiently extended) alone.

Clearly (1) implies (2). I remember seeing somewhere that (2) implies (1), but I don't remember where, so I might be mistaken. Does anyone knows how these three statements relates to each other?

EDIT: this question already deals with the case (1)=>(3)

• I find statement (1) ambiguous without further elaboration about how it is to be formalized. $\exists j$ is a second-order quantifier, so I guess we understand this in GBC. In any case, one intends $\text{ZFC}(j)$, for otherwise it is weak. To express "$j$ is elementary" seems to require a truth predicate, whose existence is not provable in GBC. Perhaps one means "$\Sigma_0$-elementary and cofinal," which is an expressible approximation to full elementarity, for it implies elementarity for standard formulas as a scheme, but this is weaker than elementarity with respect to a truth predicate. – Joel David Hamkins Dec 2 '17 at 14:01
• In the case of the Kunen inconsistency, proved in ZFC, it was fine to be a little sloppy, because the Kunen argument in effect ruled out the first-order assertion of statement (3), which is a clear consequence of whatever one might have meant in (1). But now that set theorists are considering the specific statement (1) seriously, I think it is time for set theorists to be a little more careful in formulating the statement precisely. – Joel David Hamkins Dec 2 '17 at 20:43

No.

This is a very recent work in progress of myself with Juan Aguilera.

Definition. We say that $\kappa$ is a Kunen cardinal if there is a nontrivial elementary embedding $j\colon V_{\lambda+2}\to V_{\lambda+2}$ with $\lambda=\sup j^n(\kappa)$.

(Sometimes it is easier to talk about the critical points of the sequence, and sometimes on $\lambda$, like in the case of $I0$ and such.)

Theorem. If $\kappa$ is a Reinhardt cardinal, i.e. the critical point of an elementary embedding $j\colon V\to V$, then $\kappa$ is the limit of Kunen cardinals.

Proof. Note that being a Kunen cardinal is a first-order property in the language of set theory. And note that a Reinhardt cardinal itself is Kunen by a fairly easy verification.

Let $j\colon V\to V$ be a nontrivial elementary embedding, and $\kappa$ its critical point. Now, it is easy to see that the set $A\subseteq\kappa$ defined as $\{\mu<\kappa\mid\mu\text{ is a Kunen cardinal}\}$ satisfies $\kappa\in j(A)$. Therefore it follows that $A$ is in the normal measure derived from $j$, and therefore it is stationary.

In particular, a Reinhardt cardinal is the limit of Kunen cardinals. $\square$

• We thought about the obvious $I(-1)$ name, but decided it's a silly name and it should be given a proper name. Kunen cardinals seemed appropriate. – Asaf Karagila Nov 2 '17 at 13:43
• What do you know about the relation between (2) and (3)? – Yair Hayut Nov 2 '17 at 13:46
• Well. I have no idea. I'm inclined to say that just ordinal-sequences are not enough here to ensure that (2) implies (3); and it seems to me that cutting the universe at the least critical cardinal above the least Kunen would mean that (3) does not imply (2) either. But again, I don't think there is too much known about how closure under ordinal-sequences works in ZF. – Asaf Karagila Nov 2 '17 at 13:50
• Since the existence of a Reinhardt cardinal implies every set has a sharp, a Kunen cardinal is also strictly weaker than a Reinhardt cardinal in consistency strength. – Gabe Goldberg Nov 3 '17 at 23:27