Is this sequence of embeddings possible?

Question

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 $V_\lambda$

Answer 1

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.