Rank of Set Determined by Members

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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

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Let:

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


Let $y \in S$.

Then by Membership Rank Inequality:

$\operatorname{rank} \left({ x }\right) < \operatorname{rank} \left({ S }\right)$



Therefore:

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


Conversely, take any $x \in T$.

By the definition of $T$:

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

From Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy:

$\forall y \in S: y \in V \left({ x }\right)$

where $V \left({x}\right)$ denotes the von Neumann hierarchy.

Therefore by the definition of subset:

$S \subseteq V \left({x}\right)$

By the definition of rank.

$\operatorname{rank} \left({ S }\right) \le x$


Therefore:

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

$\blacksquare$


Sources