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Let $M$ be some non-well-founded model of $\sf ZF$, can we have a sequence $(S_n)_{n \in \mathbb N}$ of nonempty sets in $M$, where each $S_n \subset \mathcal P(S_{n+1})$; and such that there exists a sequence of bijective functions $(f_n)_{n \in \mathbb N}: S_{n+1} \to S_n$, having : $f_n(x) = f_{n+1}[x]$?

This can work in $\sf ZF-Reg.$ like singleton maps "$x \mapsto \{x\}$" between sets of iterated singletons. But here $M$ satisfies Foundation.

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    $\begingroup$ Do you want the sequences themselves to be in $M$? Otherwise, it seems that this should be easy to construct via compactness using your example. $\endgroup$
    – tomasz
    Commented Oct 5, 2022 at 9:19
  • $\begingroup$ @tomasz, No. The sequences themselves need not be elements in $M$, though they need to be subsets of $M$. $\endgroup$
    – Zuhair Al-Johar
    Commented Oct 5, 2022 at 10:04

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(Note that I am assuming the axiom of choice in the metatheory. If you want to ask if this holds without choice, clarify and add the appropriate tag. I suspect that this particular compactness instance should be true even without choice, but I am not an expert, nor did I think much about it.)

This is true by a standard compactness argument (assuming ZF is consistent). You can even demand that each $S_n$ be singleton.

Namely, fixing an arbitrary model $M$ of ZF, the set $\{x_n=\{x_{n+1}\}\mid n\in \mathbf N\}^\dagger$ is a type in $M$ in variables $(x_n)_{n\in\mathbf N}$ (clearly, every finite part is realised in $M$), so there is an elementary extension $N\succeq M$ in which it is realised. In particular, $N$ is a model of ZF (in fact, of the theory of $M$) with the sequences you wanted.

$\dagger$: naturally, here, $x_n=\{x_{n+1}\}$ is short for $x_{n+1}\in x_n\land\forall y(y\in x_n\rightarrow y=x_{n+1})$

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