Ranks agree and hierarchy membership is absolute

Ordinals and omega in transitive models: In ambient ZF, ordinalhood is absolute between transitive membership domains containing the parameter. A transitive set model of ZF contains precisely the real finite ordinals as its natural numbers and has ${\omega }^M=\omega$. Its ordinals form an initial segment of the actual ordinals.

Membership rank under Foundation: Now assume ZF, including Foundation. Membership on the universe is well-founded and setlike, so its ordinal rank is defined for every set. Write

$\mathrm{rank}(x)=\sup \{\mathrm{rank}(y)+1:y\in x\}\mathrm{.}$

The empty supremum is $0$, so $\mathrm{rank}(\mathrm{\varnothing })=0$. If $y\in x$, then $\mathrm{rank}(y)<\mathrm{rank}(x)$. This is the Foundation-dependent special case of relation rank. The earlier construction of $V_{\alpha }$ did not require Foundation.

Conventions and prerequisites: def-rank-of-a-well-founded-relation, thm-foundation-equivalent-to-hierarchy-exhaustion.

Absolute basic set operations and relations: The graphs of empty set, subset, unordered pair, singleton, union, intersection (with $\bigcap \mathrm{\varnothing }=\mathrm{\varnothing }$), difference, Kuratowski ordered pair, Cartesian product, relation domain/range, functionhood, evaluation and injection have ${\mathrm{\Delta }}_0$ definitions. Thus their values agree between transitive membership structures whenever the input and output sets are in the smaller domain. This is graph agreement, not an assertion that an arbitrary transitive domain is closed under these operations.

Rank characterizes hierarchy membership: In ZF, for every set $x$ and ordinal $\alpha$,

$x\in V_{\alpha }⟺\mathrm{rank}(x)<\alpha ,x\subseteq V_{\alpha }⟺\mathrm{rank}(x)\le \alpha \mathrm{.}$

Thus $\mathrm{rank}(x)$ is the least $\alpha$ with $x\subseteq V_{\alpha }$, and $\mathrm{rank}(x)=\alpha$ iff $x\in V_{\alpha +1}\setminus V_{\alpha }$.