Rank of Set Determined by Members

Theorem
Let $S$ be a set.

Let $\operatorname{rank} \left({ S }\right)$ denote the rank of $S$.

Then:


 * $\operatorname{rank} \left({ S }\right) = \bigcap \left\{ x \in \operatorname{On} : \forall y \in S: \operatorname{rank} \left({ y }\right) < x \right\}$

Proof
Set $T = \bigcap \left\{ x \in \operatorname{On} : \forall y \in S: \operatorname{rank} \left({ y }\right) < x \right\}$

Suppose $y \in S$.

Then $\operatorname{rank} \left({ x }\right) < \operatorname{rank} \left({ S }\right)$ by Membership Rank Inequality.

Therefore, $T \subseteq \operatorname{rank} \left({ S }\right)$.

Conversely, take any $x \in T$.


 * $\forall y \in S: \operatorname{rank} \left({ y }\right) < x$ by the definition of $T$.

It follows, by Ordinal Subset of Rank, that $\forall y \in S: y \in V \left({ x }\right)$ where $V\left({x}\right)$ denotes the von Neumann hierarchy.

Therefore, $S \subseteq V\left({x}\right)$ by the definition of subset.


 * $\operatorname{rank} \left({ S }\right) \le x$ by the definition of rank.

Therefore, $\operatorname{rank} \left({ S }\right) = T$