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In general, for a set $X$, why is it true that $\mathrm{rank}(X)=\sup\{\mathrm{rank}(y)+1\mid y\in X\}$?

The definition of rank I have is that $\text{rank}(x)=\text{ the least }\alpha\text{ such that }x\in V_{\alpha+1}$. I suppose these must be the same thing?

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It is the definition of the rank. – Asaf Karagila Dec 12 '11 at 21:25
As @AsafKaragila said, this is generally the definition. If, by chance, it's not the definition you've been given, but stated somewhere as a consequence, you should provide the definition you have. Otherwise, the question would be better as something like "why does this definition make sense?". – Quinn Culver Dec 12 '11 at 21:36
up vote 8 down vote accepted

This is the definition of the rank of the set $X$. $\newcommand{rank}{\operatorname{rank}}$

We use the fact that $\in$ is well-founded, and that the ordinals are well-ordered. These properties allow us to define recursively this function. $$\rank:V\to\operatorname{Ord}$$

And allows us to talk about the sets $V_\alpha=\{x\in V\mid \rank (x)<\alpha\}$, for an ordinal $\alpha$. This is known as the von Neumann hierarchy.

Given the definition of $\rank$ as the least $\alpha$ such that $x\subseteq V_\alpha$, we can show the equivalence by induction.

Suppose that for an ordinal $\alpha$ we have that if $\rank(y)<\alpha$ then the equivalence holds.

If $\sup\{\rank(y)+1\mid y\in x\}=\beta<\alpha$ then we have that $y\in x$ implies that $y\in V_\beta$ and therefore $x\in V_{\beta+1}\subseteq V_\alpha$ and thus $\rank(x)<\alpha$, and the equivalence holds.

Otherwise, $\sup\{\rank(y)+1\mid y\in x\}=\alpha$, and given $\beta<\alpha$ we can find $y\in x$ such that $y\notin V_\beta$. In particular this means that $x\nsubseteq V_\beta$ therefore $x\notin V_{\beta+1}$. And indeed $\alpha$ is the minimal ordinal such that $x\in V_{\alpha+1}$, as we wanted.

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Thanks for adding that part. – rankled Dec 13 '11 at 0:12
No problem, it is my pleasure! :-) – Asaf Karagila Dec 13 '11 at 0:16
In the first half of your answer: don't we need that the ordinals are a strict well-order? – Rudy the Reindeer Mar 6 '13 at 7:58
If the set that constitutes the range of the function is not a strict linear order the $\sup$-definition goes broken because the resulting function is not well-defined. – Rudy the Reindeer Mar 6 '13 at 7:59
@Matt, well obviously. :-) But if the order wasn't strict there was some $\alpha\lea\alpha$, which is $\alpha\in\alpha$ which means $\{\alpha\}$ has no $\in$-minimal element. – Asaf Karagila Mar 6 '13 at 8:04

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