Bounded Rank implies Small Class

From ProofWiki
Jump to: navigation, search


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.



This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides.

If you see any proofs that link to this page, please insert this template at the top.

If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page.

Let $V$ denote the von Neumann hierarchy.


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

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