Bounded Rank implies Small Class

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Theorem

Let $S$ be a class.

Suppose the rank, denoted $\operatorname{rank} \left({x}\right)$, 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:

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


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

$\blacksquare$


Sources