Is existence of this set consistent with Zermelo set theory minus choice? Define a pre-ordinal as a transitive set of transitive sets.
Is it consistent with Zermelo set theory (without choice) to have a nonempty set $S$ such that: for every element $s \in S$ there exists a pre-ordinal $\beta$ and a set $Y$ such that $s=\langle \beta, Y \rangle $, and there exists an element $r \in S$ such that: $r= \langle \alpha, X \rangle$, where $\beta=\alpha \cup \{\alpha\}$ and $Y = \cal P$$(X)\backslash \{\emptyset\}$.
 A: First note that if $s\in S$, then there is a unique $r$ such that your condition holds. Simply because $\mathcal P(X)=\mathcal P(X')$ if and only if $X=X'$. So we can write $s^-$ for that $r$ and call it a predecessor.
Next, consider Hartogs' theorem, but without Replacement. Namely, for every set $X$ there is a smallest type of a well-order set which cannot inject into $X$. The proof is completely Replacement free, noting that we can still do transfinite recursion on a given well-ordered set, as long as all the ingredients of that recursion exist in our universe (e.g. comparability of well-orders, etc.) and so there's a meaning to the statement "Hartogs number" of a set. Let me refer to the Hartogs number of $s=\langle\alpha,Y\rangle$ to mean the Hartogs number of $Y$.
Finally, note that in the proof of Hartogs' theorem we get that if $x$ is any set, then $\mathcal{PPP}x$ will contain a well-orderable set which cannot inject into $x$.
So, fix some $s\in S$, then $s^{---}$ must have a shorter order type of its Hartogs number. But that means that $s^{-6}$ has an even smaller one, etc. And therefore this is impossible since by fixing a well-ordered witness of $s$'s Hartogs number, then $s^{-3n}$ defines a unique point within that well-order which defines an initial segment witnessing the Hartogs number of $s^{-3n}$. But now that defines a decreasing sequence, which is impossible.
