# Formula to compute Von Neumann rank

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.

• What's $\mathrm{rank}(\varnothing)$? And what about $\mathrm{rank}(\{\varnothing\})$? How does it change depending on whether you add $1$? Mar 27 '20 at 18:52
• $\mathrm{rank}(\varnothing)=0$ and $\mathrm{rank}(\{\varnothing\})=1$. I see that nesting a set in another set increases the rank by $1$ (because of the powerset operation in the hierarchy?). That said, that doesn't suggest a proof to me, and how do I handle the limit case? Mar 27 '20 at 18:55
• I was only justifying why the $+1$ is needed Mar 27 '20 at 18:58
• Oh, I see, thank you! Mar 27 '20 at 18:59

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.