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
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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:
- $\map {\operatorname{rank} } x < \map {\operatorname{rank} } S$
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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:
- $\map {\operatorname{rank} } S = T$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 9.17$