Rank of a set as the supremum of ranks its elements. 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?
 A: 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.
