Recall the so-called Von Neumann Universe of sets: $V_0=\emptyset$, $V_{\beta+1}=\mathcal{P}(V_\beta)$, $V_\lambda = \bigcup_{\beta < \lambda}V_\beta$, where $\lambda$ is a nonzero limit ordinal,
On the same page, there is the following definition:
$$\text{rank}(S)=\text{the least $\alpha$ such that $S\subseteq V_\alpha$}$$
and the following formula: $$\operatorname{rank} (S) = \bigcup \{ \operatorname{rank} (z) + 1 \mid z \in S \}$$
Does anyone have a proof of this fact or a reference to one? Intuitively it makes sense, the rank can be computed by recursively computing the ranks of the elements of $S$ and "combining" them via union, but I don't quite see why the $+1$ is neccessary.
Best Answer
To see the need for the $+1$, consider $S=\{0\}$: $S$ is not a subset of $\varnothing=V_0$, so $\operatorname{rank}(S)>0=\operatorname{rank}(\varnothing)$.
Suppose that $\operatorname{rank}(S)=\alpha$; then $S\subseteq V_\alpha$, so $x\in V_\alpha$ for each $x\in S$, and there is therefore a $\beta<\alpha$ such that $x\subseteq V_\beta$ and hence $\operatorname{rank}(x)\le\beta$. In particular, $\operatorname{rank}(x)<\alpha$ for each $x\in S$, so $\operatorname{rank}(x)+1\le\alpha$ for each $x\in S$, and hence $\alpha\ge\sup\{\operatorname{rank}(x)+1:x\in S\}$.
Now let $\beta=\sup\{\operatorname{rank}(x)+1:x\in S\}$. Let $x\in S$; then $x\subseteq V_{\operatorname{rank}(x)}$, so $x\in V_{\operatorname{rank}(x)+1}\subseteq V_\beta$. But then $S\subseteq V_\beta$, and it follows from the minimality of $\alpha$ that $\alpha\le\beta$. Combining inequalities, we see that $\alpha=\beta$, as desired.