Working in a suitable extension of $\sf Z$ like $\sf ZfC + wholeness \ axiom$, or $\sf ZFj + Reinhardt \ axiom$.


Can we have a sequence $(j\_n)\_{n \in \mathbb N} $ of nontrivial elementary embeddings from $V$ to $V$, such that for some $V\_\lambda$ we have:


$$\exists (\alpha\_n)\_{n \in \mathbb N} : \forall n \in \mathbb N \ \big {(} \,\alpha\_n < \alpha\_{n+1} < \lambda, \, \\ j\_n(V\_{\alpha\_n})= V\_{\lambda +1} \big{)} \land \\\bigcup^{n\in \mathbb N} V\_{\alpha\_n} = V\_\lambda \,$$


In English, there exists a strictly increasing sequence $(\alpha\_n)\_{n \in \mathbb N}$ of ordinals $< \lambda$ such that $V\_{\lambda+1}$ is the image of each embedding $j\_n$ from $V\_{\alpha\_n}$, and such that the union of all $V\_{\alpha\_n}$sets is $V\_\lambda$



Is this sequence of embeddings possible?
Work in ZF+AC$\_\omega$. Suppose $V\_\theta$ is inaccessible. Suppose $j,k:V\_{\theta+1}\to V\_{\theta+1}$ are elementary and $\mathrm{crit}(j)=\mathrm{crit}(k)=\kappa$ and $\kappa\_\omega(j)<k(\kappa)$. Here $\kappa\_n(j)$ for $n\leq\omega$ is the critical sequence of $j$; that is, $\kappa\_0(j)=\mathrm{crit}(j)$ and $\kappa\_{n+1}(j)=j(\kappa\_n(j))$ and $\kappa\_\omega(j)=\sup\_{n<\omega}\kappa\_n(j)$.


(Of course, this is stronger than Reinhardt, but after all you did say "like".)


Recall that we can define $k(j):V\_{\theta+1}\to V\_{\theta+1}$ in a natural way, and it is fully elementary. (In fact this holds for any two elementary $\ell\_1,\ell\_2:V\_{\theta+1}\to V\_{\theta+1}$).


We will find a sequence as desired with $\lambda=k(\kappa\_\omega(j))=\kappa\_\omega(k(j))$. By AC$\_\omega$ in $V\_{\theta+1}$, it suffices to show that for each $n<\omega$, there is an elementary $\ell:V\_\theta\to V\_\theta$ and $\lambda'$ such that $\kappa\_n(k(j))<\lambda'<\kappa\_{n+1}(k(j))$ and $\ell(\lambda')=\lambda$. (Then using AC$\_\omega$, choose a sequence $\left<\ell\_n,\lambda'\_n\right>$ witnessing this and set $\alpha\_n=\lambda'\_n+1$, and note that the model $(V\_\theta,\left<\ell\_n\right>\_{n<\omega})$ has the desired properties.)


So, by composing maps, we get an elementary $h:V\_{\theta+1}\to V\_{\theta+1}$ such that $\mathrm{crit}(h)=\kappa$ and $h(\kappa)=\kappa\_{n+1}(k(j))=\kappa^\*$ and $h(\kappa\_\omega(j))=\lambda$. (I.e. first apply $k$, then $k(j)$, then $k(j(j))$, etc, until getting $h$.) Because $h\upharpoonright V\_\theta\in V\_{\theta+1}$, for each $\alpha<\kappa$, $V\_{\theta+1}\models$"There is an elementary $h':V\_\theta\to V\_\theta$ and $\lambda'$ such that $\alpha<\kappa'=\mathrm{crit(h')}<\lambda'<\kappa^\*$ and $h'((\kappa',\lambda'))=(\kappa^\*,\lambda)$". So this statement pulls back under $h$, so that $V\_{\theta+1}$ models the same regarding $(\kappa,\kappa\_\omega(j))$ instead of $(\kappa^\*,\lambda)$ (and $\alpha$ the same). (Note it only says $h':V\_\theta\to V\_\theta$ is elementary, not its extension to $V\_{\theta+1}$.) Since this holds for all $\alpha<\kappa$, the all-quantified statement lifts back again with $h$ to $(\kappa^\*,\lambda)$, and this yields what we want.


I haven't tried to reduce the large cardinal assumption.


