ZF is sufficient to construct the von Neumann hierarchy, and prove that every set appears at some stage $V\_\alpha$. This is the basis for Scott's trick, for instance. But how much of ZF is needed? Is bounded Zermelo/Mac Lane set theory enough, no Choice assumed? I know Foundation is necessary, and I'm not getting rid of that. I've seen something called the "rank axiom" in discussion of second-order version of original Zermelo ([these notes](http://math.boisestate.edu/%7Eholmes/hierarchy.pdf)), but I'm sure people have finely calibrated what precisely is needed.


To be honest, all I really want is an ordinal-valued rank function such that sets of rank at most $\alpha$ form a set, for all $\alpha$, and all sets have a rank. So if the von Neumann hierarchy doesn't work, I'm happy to work with something else (and for 'ordinals', I don't *need* von Neumann ordinals).



Does bounded Zermelo construct any cumulative hierarchy?
KP (Kripke-Platek set theory) is the most well-known fragment of $\sf{ZF}$ which suffices for the development of the rank function, thus $\sf{KPR}$ = $\sf{KP}$ + "for all ordinals $\alpha$, $V(\alpha)$ exists" is the usual minimal theory in which one can be assured of the stratification of the universe into $V(\alpha)$s.


On the other hand, as observed by Mathias, $\sf{KPR}$ proves that Zermelo set theory $\sf{Z}$ has a transitive model, so in particular, $\sf{KPR}$ proves that $\sf{Z}$ is consistent; see Lemma 6.31 of [this preprint](https://www.dpmms.cam.ac.uk/%7Eardm/maclane.pdf), which was later published in APAL (2001).


Therefore, by the second incompleteness theorem, even Zermelo set theory (let alone bounded Zermelo set theory) cannot interpret $\sf{KPR}$.


Finally, I will add that $\sf{KPR}$ is provable in the well-known extension $\sf{KP}^{\cal{P}}$ of $\sf{KP}$, which is also studied in Mathias' paper.


