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: $\S 9.17$