V is a set

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement.

In ZF every set $x$ belongs to $V_{\mathrm{rank}(x)+1}$. Consequently $V={\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }$ in the class sense: every set lies in a stage. This is not a union indexed by a set of all ordinals. (The universe is the class union of its stages)

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. (Membership rank under Foundation)