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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.

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  • $\begingroup$ What's $\mathrm{rank}(\varnothing)$? And what about $\mathrm{rank}(\{\varnothing\})$? How does it change depending on whether you add $1$? $\endgroup$ Mar 27 '20 at 18:52
  • $\begingroup$ $\mathrm{rank}(\varnothing)=0$ 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? $\endgroup$
    – zz20s
    Mar 27 '20 at 18:55
  • $\begingroup$ I was only justifying why the $+1$ is needed $\endgroup$ Mar 27 '20 at 18:58
  • $\begingroup$ Oh, I see, thank you! $\endgroup$
    – zz20s
    Mar 27 '20 at 18:59
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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.

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