$\newcommand{\rank}{\operatorname{rank}}$
$\newcommand{\ord}{\text{Ord}}$
(This question, in a sense, is a follow up to ℋolo's answer.)
Let $V_\alpha$ be the members of the von Neumann Hierarchy: $$V_0 = \emptyset, V_{\alpha + 1 } = \mathcal{P} (V_\alpha), V_\lambda = \bigcup_{\alpha < \lambda} V_\alpha.$$
Define $$\rank(x)=\sup\{\rank(y)+1\mid y\in x\}.$$
I'm interested in the implication $\rank(x) < \kappa \implies x\in V_\kappa$ (maybe this is only true for regular cardinals $\kappa$, I'm not sure). I believe it's indirectly proved in Kunen's book, but he uses different definitions and many minor results that are spread over the previous chapters. I was wondering how to prove the mentioned implication given the definitions that I have?
Also I was wondering how the above implication related to the definition $$\rank_V(x)=\min\{\alpha\in \ord\mid x\in V_{\alpha+1}\}$$
Does the implication that I'm interested in essentially prove the equivalence of the two definitions of rank (or is it rather the opposite — the equivalence of the definitions can be used to prove the implication I'm interested in)?
Best Answer
The implication is true for all ordinals, not just regular cardinals. The converse implication holds as well, i.e. $V_\alpha$ is precisely the set of all sets with rank less than $\alpha$. This immediately implies that for any set $x,$ $\operatorname{rank}(x)$ is the least ordinal $\alpha$ such that $x\in V_{\alpha+1}$.
We can prove the equivalence by induction on $\alpha$. Let $\operatorname{rank}(x)<\alpha.$ If $y\in x$ then since $\operatorname{rank}(y)<\operatorname{rank}(x)$, we have $y\in V_{\operatorname{rank}(x)}$ by the induction hypothesis. Therefore, $x\subseteq V_{\operatorname{rank}(x)}$, and so we have $x\in V_{\operatorname{rank}(x)+1}\subseteq V_\alpha$.
Conversely, let $x\in V_\alpha$. Then, $x\in V_{\beta+1}$ for some $\beta<\alpha.$ So if $y\in x,$ then $y\in V_\beta$, so by the induction hypothesis, $\operatorname{rank}(y)<\beta.$ Therefore, $\operatorname{rank}(y)+1\le \beta$ for all $y\in x,$ so $\operatorname{rank}(x)\le \beta < \alpha.$