Membership Rank Inequality

Theorem

Let $S$ and $T$ be sets.

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

Then:

$S \in T \implies \map {\operatorname{rank} } S < \map {\operatorname{rank} } T$

Proof

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By Ordinal Equal to Rank:

$T \in \map V {\map {\operatorname{rank} } T + 1}$

By the definition of rank:

$T \subseteq \map V {\map {\operatorname{rank} } T}$

Since $S \in T$:

$S \in \map V {\map {\operatorname{rank} } T}$

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

$\map {\operatorname{rank} } T \nsubseteq \map {\operatorname{rank} } S$

Therefore by Ordinal Membership is Trichotomy and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:

$\map {\operatorname{rank} } S < \map {\operatorname{rank} } T$

$\blacksquare$

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 9.16$