# Must there be a proper class of Reinhardt cardinals if there is a Reinhardt cardinal?

A cardinal is Reinhardt if $$\kappa$$ is the critical point of a nontrivial elementary embedding of $$V$$ to itself, where $$V$$ is the class of all sets. As Reinhardt cardinals are inconsistent with $$\mathrm{ZFC}$$, work in $$\mathrm{ZF}j$$, which is $$\mathrm{ZF}$$ with replacement and separation for formulas including the function symbol $$j$$, and an axiom schema stating $$j$$ is a nontrivial elementary embedding $$V\to V$$.

Let $$(V_\alpha)_{\alpha\in\mathrm{Ord}}$$ be the usual von Neumann hierarchy whose union is $$V$$. For two elementary embeddings $$j,k:V\to V$$, define the embedding $$j\cdot k$$, the application of $$j$$ to $$k$$, to be $$\bigcup_{\alpha\in\mathrm{Ord}}j(k\cap V_\alpha)$$. The application operation is familiar from the study of rank-into-rank axioms like $$\mathrm I3$$, a usual reference is Laver's "On the Algebra of Elementary Embeddings of a Rank Into Itself". It is a usual exercise to verify that if $$j$$ and $$k$$ are elementary, then $$j\cdot k$$ is elementary.

If there exists a nontrivial elementary embedding $$j:V\to V$$ along with a Reinhardt cardinal $$\kappa$$, then $$j\cdot j$$ is also elementary, and $$j(\kappa)$$ is also Reinhardt as it is the critical point of $$j\cdot j$$. Similarly, $$j\cdot (j\cdot j)$$ is elementary, and $$j(j(\kappa))$$ is Reinhardt. More generally, define the critical sequence of $$j$$ as usual, by $$\kappa_0=\kappa$$, $$\kappa_{n+1}=j(\kappa_n)$$, and $$\lambda=\mathrm{sup}\{\kappa_n\mid n<\omega\}$$. Additionally define $$j^0=j$$ and $$j^{n+1}=j\cdot j^n$$. Each $$\kappa_n$$ in the critical sequence is Reinhardt, as witnessed by the elementary embedding $$j^n$$. So if there exists a Reinhardt cardinal, there must exist countably many.

Assume there exists a Reinhardt cardinal, and let $$\alpha$$ be an arbitrary von Neumann ordinal. Does it follow that there is a Reinhardt cardinal $$>\alpha$$?

Some possible direction: In section 6 of "I0 and rank-into-rank axioms", Dimonte considers direct systems $$(M_\beta,j_{\beta,\gamma})_{\beta<\gamma<\alpha}$$ of elementary embeddings (although these embeddings are only from $$V_\lambda$$ to $$V_\lambda$$), taking direct limits at limit ordinal steps, but due to the different definition $$j^{\alpha+1}=j^\alpha\cdot j^\alpha$$, if I am correct the critical point of $$j^2$$ is no longer $$j(j(\kappa))$$, but only $$\kappa$$, so no larger Reinhardt cardinals are produced.

• By the way I don’t think you’re correct about the infinite iterations. The reason they don’t produce Reinhardt cardinals is that e.g. $j_{\omega,\omega+1}$ is not an embedding from $V$ to itself, but rather from $M_\omega$ to itself. The fact that each finite iterate $M_n$ is $V$ does not imply the same for $M_\omega$. See Schlutzenberg’s paper on iteratations of embeddings from $V$ to $V$ for more information. Apr 25 at 22:12
• arxiv.org/abs/2002.01215 Apr 25 at 22:17
• @GabeGoldberg Good point, I now think you are correct. I will also make a small change about what I originally intended with the remark about the embeddings in Dimonte's paper.
– C7X
Apr 26 at 3:59
• I guess the point I was making was unclear. When you iterate an embedding (as in Dimonte's paper), the critical points do increase. In fact, $j^\alpha(j^\alpha) = j^0(j^\alpha)$. The same construction can be done for $j : V\to V$ as well as $j:V_\lambda\to V_\lambda$. Also I think it is bad notation to put the $\alpha$ in the superscript since it makes it look like we are composing $j$ with itself. Apr 26 at 15:56

No, if the existence of a Reinhardt is consistent, then it is consistent with a Reinhardt cardinal that the class of inaccessible cardinals is bounded in the ordinals. Indeed, if $$j : V\to V$$ is a nontrivial elementary embedding and $$\lambda$$ is the least fixed point of $$j$$ above its critical point, then let $$\delta$$ be the least inaccessible above $$\lambda$$ if there is one -- if not, we're done. Note that $$j(\delta) = \delta$$, so $$(V_\delta,V_{\delta+1})$$ is a model of full second-order ZF + there is a Reinhardt cardinal + every inaccessible cardinal is less than $$\lambda$$.
One interesting question here is whether it is consistent that there are exactly $$\omega$$-many Reinhardt cardinals (in ordertype). It seems like one might have a chance of showing this is true in the model of NBG where all classes are definable from a single elementary $$j : V\to V$$.