# Supercompact and Reinhardt cardinals without choice

A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:

Definition. A cardinal $$\kappa$$ is supercompact if for all ordinals $$\lambda$$, there exists a transitive inner model $$M_\lambda$$ of $$V$$ and an elementary embedding $$j_\lambda: V\rightarrow M_\lambda$$ such that $$crit(j_\lambda):=\min\lbrace\alpha\in ON: j(\alpha)\not=\alpha\rbrace=\kappa$$ and $$M_\lambda$$ is closed under $$\lambda$$-sequences (that is, $$^\lambda M_\lambda\subseteq M_\lambda$$).

Now in the definition of supercompactness, the inner models $$M_\lambda$$ are allowed to vary with $$\lambda$$. We could define a large cardinal notion, uniform supercompactness, by removing this possibility: $$\kappa$$ is uniform supercompact if there is a single $$M$$ and $$j$$ such that $$M$$ is closed under all sequences of ordinal length and $$j: V\rightarrow M$$ is an elementary embedding with critical point $$\kappa$$. But in ZFC, this is not an interesting notion. First, assuming choice, $$M$$ is closed under sequences of ordinal length if and only if $$M=V$$ (since two models with the same ordinals and sets of ordinals are equal), so $$ZFC\models$$ "$$\kappa$$ is uniformly supercompact $$\iff$$ $$\kappa$$ is Reinhardt;" second, Kunen showed that Reinhardt cardinals are inconsistent with $$ZFC$$. (EDIT: as Joel points out, this statement doesn't actually make sense, as Reinhardt-ness is not a first-order property. One needs to pass to some extension of ZFC which can talk about proper classes directly, like Morse-Kelly set theory, which is the setting Kunen used for his proof.)

However, assuming $$\neg$$Choice, both of these points seem suspect. It is still open whether Reinhardt cardinals are consistent with $$ZF$$, and it is unclear to me that $$ZF$$ alone proves that uniformly supercompact cardinals are Reinhardt. So my questions are the following:

Question 1: Does $$ZF\models$$ every uniformly supercompact cardinal is Reinhardt?

Question 2: What is the consistency strength of $$ZF+$$ there exists a uniformly supercompact cardinal?

Obviously $$ZF$$ proves that every Reinhardt cardinal is uniformly supercompact; also obviously, Question 2 is really only interesting in lieu of a positive answer to Question 1. The most basic obstruction to a positive answer to Question 1 is the fact that given an elementary embedding $$j: V\rightarrow M$$ with $$M$$ closed under ordinal-length sequences and $$crit(j)=\kappa$$, the restriction of $$j$$ to $$M$$, $$j\upharpoonright M: M\rightarrow M$$ is not obviously (to me at least) an elementary embedding.

• Like you say in the last paragraph, it is not clear that the restriction of j to M is going to be an elementary embedding into M in general. Instead you could consider the restriction of j to smaller models, like the class N consisting of all sets constructible from sequences of ordinals. Then j does restrict to an embedding N -> N but you still have might have the problem that the embedding is not amenable to the model. This seems like the bigger issue to me. – Trevor Wilson Apr 23 '12 at 4:28
• This is an interesting question. The contradiction in ZFC comes from the fact that models with the same sets of ordinals are isomorphic. In ZF it is not true. Monro proved that there is a chain of $\omega$ models, where $V_n\subseteq V_{n+1}$ and they both have the same sets-of-sets-..$n$-times of ordinals, but they are not isomorphic (Jech The Axiom of Choice, Ch. 5, Problem 17). This, however is "going up" while inner models are "going down" so there might be some difference there. – Asaf Karagila Apr 23 '12 at 6:21
• Very nice question. One quibble: it isn't really correct to say that ZFC refutes Reinhardt cardinals, since there is no first-order characterization of Reinhardt cardinals; that is, you can't say "$\kappa$ is a Reinhardt cardinal" in a first-order way. Kunen proved his inconsistency in the second-order Kelly-Morse set theory, but it is also possible to formalize the result in Goedel-Bernays set theory GBC. See jdh.hamkins.org/generalizationsofkuneninconsistency for a further discussion of this point. – Joel David Hamkins Apr 23 '12 at 12:31
• I believe (from your remark about ordinal lengths) that you intend that the assertion "for all $\lambda$..." in the definition is meant to be interpreted as, "for all ordinals $\lambda$...". The difference matters, as you likely know, because if you have $j:V\to M$ and $j"A\in M$ for all sets $A$, then it follows easily that $M=V$ even in ZF, since if $A$ is transitive, then $A$ is the Mostowski collapse of $j"A$. – Joel David Hamkins Apr 23 '12 at 14:34
• @Joel: good points. I've corrected the question. Also, building off of your first comment, is it obvious that uniform supercompactness is first-order expressible? For that matter, how is supercompactness itself first-order expressible? – Noah Schweber Apr 23 '12 at 15:44

Call a cardinal $$\lambda$$-Matryoshka if and only if for every transitive inner model $$M\prec V$$, there is a non-trivial elementary embedding $$j\colon V\rightarrow M$$ with critical point $$\kappa$$, and $$j(\kappa)>\lambda$$. (This name comes from the strong downards reflection properties exhibited by $$\kappa$$.) Partially $$\lambda$$-Matryoshka cardinals are defined in the same way, except for some instead of every transitive inner model. I claim every Reinhardt cardinal is $$\kappa$$-Matryoshka, and every $$\kappa$$-Matryoshka cardinal is uniformly supercompact, but not every uniformly supercompact cardinal is $$\kappa$$-Matryoshka.

Theorem: If $$j:V\rightarrow M$$ witness $$\kappa$$ is partially $$\kappa$$-Matryoshka, then there is some $$V[G]$$ such that $$\kappa$$ is uniformly supercompact in $$V[G]$$, but $$M\nprec V[G]$$.

Proof. Force via $$P$$ being the set of all partial injection $$p\colon \mathrm{Ord}^V\rightarrow M$$, and let $$G$$ be a filter on $$P$$. Then $$M^{V[G]}\vDash \mathrm{NBG}$$, then if we can prove $$V[G]\nvDash \mathrm{NBG}$$ and $$j'(V[G])\prec M$$ for some $$j'$$, we will have completed the proof. Suppose $$\exists x(\phi(x,x_0,\dots,x_n))$$ is $$\Sigma_1$$. Then $$V\vDash\exists x(\phi(x,x_0,\dots,x_n))$$ if and only if $$V\vDash\exists \alpha(\exists z\in V_\alpha(V_\alpha\vDash \phi(z,x_0,\dots,x_n)))$$ if and only if $$V[G]\vDash\exists \alpha(\exists z\in V_\alpha(V_\alpha\vDash \phi(z,x_0,\dots,x_n)))$$ if and only if $$V[G]\vDash\exists x(\phi(x,x_0,\dots,x_n))$$. Therefore $$V\prec_{\Sigma_1}V[G]$$, and as global choice is $$\Sigma_1$$, then $$V[G]\nvDash \mathrm{NBG}$$. Finally, let $$j'=\{(x,y)\in j\mid y\in M\}$$. Then $$j'(V)=j(V)\cap M$$ and so $$j'(V)\prec M$$. ■

Theorem: If $$\kappa$$ is $$\kappa$$-Matryoshka in $$V$$, then there is some $$V'$$ such that $$\kappa$$ is uniformly supercompact in $$V'$$, but not $$\kappa$$-Matryoshka.

Proof. We add to the first theorem that $$G$$ adds no new Matryoshka embeddings. Suppose $$j\colon V[G]\rightarrow M$$ is elementary. Then we simply need to show $$j\restriction V\colon V\rightarrow M$$ is $$\Sigma_1$$ elementary. But if $$V[G]\vDash j(V[G])\prec_{\Sigma_1}M$$, as that statement is $$\Pi_2$$, then $$V\vDash j(V)\prec_{\Sigma_1}M$$ and so $$j(V)\prec_{\Sigma_1}M$$. Then let $$V^0=V$$ and $$V^{\alpha+1}=V^\alpha[G]$$, continuous at limits, where $$G$$ is the forcing to remove a Matryoshka embedding. Then, if $$\lambda=|\{j\mid j\text{ is a Matryoshka embedding}\}|$$, take $$V'=V^{\lambda}$$. ■

Now, to show the important part. That a given cardinal $$\kappa$$ is $$\kappa$$-Matryoshka if and only if $$\kappa$$ is Reinhardt.

Theorem: $$\kappa$$ is $$\kappa$$-Matryoshka if and only if $$\kappa$$ is Reinhardt.

Proof. The forward direction is simple. For the other direction let $$j'=\{(x,y)\in j\mid y\in M\}$$. Then $$j'(V)=j(V)\cap M$$ and so $$j'(V)\prec M$$. ■

For the second question, I believe you are asking for a lower bound on the consistency strength of uniformly supercompact cardinals, not an upper bound (As they are bellow Reinhardt cardinals). Obviously every uniformly supercompact cardinal is I2. Beyond that I don’t know much.

• I fixed it. It is every. Partially $\lambda-$Matryoshka cardinals are defined in the same way, except for some instead of every transitive inner model. – Master Jul 2 '19 at 3:45
• What is a uniformly measurable? Measurable cardinals carry measures, they are not critical points if you don't assume choice. – Asaf Karagila Jul 2 '19 at 6:12
• (Also, several tips from a former type-setter: use \text for text inside math, e.g. set definitions; use \mid for $\mid$ separator in set definitions; $-$ (- in math mode) is a minus not a hyphen, so $\kappa$-Matryoshka rather than $\kappa-$Matryoshka.) – Asaf Karagila Jul 2 '19 at 8:35
• An inner model that is an elementary substructure of $V$ is equal to $V$. – Gabe Goldberg Jul 3 '19 at 7:25
• For all ordinals $\alpha$, $(V_\alpha)^M = V_\alpha$ by elementarity, so $V_\alpha\subseteq M$ by transitivity. – Gabe Goldberg Jul 3 '19 at 20:21