Can there exist such a sequence of elementary embeddings of the universe to itself?

Question

Working in ZfC + Wholeness:

Can we have a countable sequence of non-trivial elementary embeddings of the universe to itself, such that the range of each embedding is a subclass of the range of its successor, and such that the union of all the ranges of those embeddings is the universe itself?

Formally can we have a proper class sequence $j`\mathbb N$ such that:

$\forall n \in \mathbb N : j_n: V \prec V \land rng(j_{n+1}) \supset rng(j_n) \\ \forall x \exists n \in \mathbb N : x \in rng(j_n)$

Where $\mathbb N$ is the set of all finite von Neumann ordinals.

Note: ZfC means ZFC but with replacement restricted to the pure language of set theory, i.e. it doesn't use the symbol $j$; however, $j$ is allowed in all instances of separation.

Answer 1

It was pointed out by Monroe that if $\lambda$ is a limit and $j:V_\lambda\to V_\lambda$ is elementary and $j_n$ is the $n$th iterate, i.e. $j_0=j$ and $j_{n+1}=j_n(j_n)$, and $\kappa_n=\mathrm{crit}(j_n)$, then since $\lim_{n<\omega}\kappa_n=\lambda$, we get every $x\in V_\lambda$ is in $\mathrm{rng}(j_n)$ for some $n<\omega$.

Given this, we can arrange the increasing range condition as follows. Choose a strictly increasing sequence $\left<n_i\right>_{i<\omega}$ such that for each $\alpha<\lambda$ there is $i<\omega$ such that $j_{n_{i-1}}\circ j_{n_{i-2}}\circ\ldots\circ j_{n_0}(\alpha)<\mathrm{crit}(j_{n_i})$. (Just construct the sequence recursively on $i$, and at stage $i$, make it work for $\alpha=\kappa_i$.) Now define $k_i:V_\lambda\to V_\lambda$ as the direct limit \begin{equation}\ldots \circ j_{n_{i+2}}\circ j_{n_{i+1}}\circ j_{n_i},\end{equation} i.e. for each $x\in V_\lambda$ set $k_i(x)=$ the common value of \begin{equation}j_{n_{i+\ell}}\circ j_{n_{i+\ell-1}}\circ\ldots\circ j_{n_i}(x)\end{equation} for large $\ell<\omega$, noting that by the choice of the $n_i$'s, for every $x\in V_\lambda$, $k_i(x)\in V_\lambda$ is well-defined; and $k_i$ is easily elementary. Note that $k_{i}=k_{i+1}\circ j_{n_i}$, and therefore $\mathrm{rng}(k_i)\subseteq\mathrm{rng}(k_{i+1})$. And note $\lim_{i<\omega}\mathrm{crit}(k_i)=\lambda$. So we get the desired conditions.