Rank of Set Determined by Members

Theorem
Let $S$ be a set.

Let $\map {\operatorname{rank} } S$ denote the rank of $S$.

Then:


 * $\map {\operatorname{rank} } S = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$

Proof
Let:
 * $T = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$

Let $y \in S$.

Then by Membership Rank Inequality:
 * $\map {\operatorname{rank} } x < \map {\operatorname{rank} } S$

Therefore:
 * $T \subseteq \map {\operatorname{rank} } S$

Conversely, take any $x \in T$.

By the definition of $T$:
 * $\forall y \in S: \map {\operatorname{rank} } y < x$

From Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy:
 * $\forall y \in S: y \in \map V x$

where $\map V x$ denotes the von Neumann hierarchy.

Therefore by the definition of subset:
 * $S \subseteq \map V x$

By the definition of rank.
 * $\map {\operatorname{rank} } S \le x$

Therefore:
 * $\map {\operatorname{rank} } S = T$