Bounded Rank implies Small Class

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Theorem

Let $S$ be a class.

Suppose the rank, denoted $\map {\operatorname{rank} } x$, of each $x \in S$ is bounded above by some ordinal $y$.




Then $S$ is a small class.


Proof

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Let $V$ denote the von Neumann hierarchy.

Then:

\(\ds \forall x \in S: \, \) \(\ds \map {\operatorname{rank} } x\) \(\le\) \(\ds y\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \map V {y + 1}\) Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy and Ordinal Equal to Rank
\(\ds \leadsto \ \ \) \(\ds S\) \(\subseteq\) \(\ds \map V {y + 1}\) Definition of Subset


Therefore, by Axiom of Subsets Equivalents, it follows that $S$ is a small class.

$\blacksquare$


Sources