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

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $x$ be an ordinal.

Let $S$ be a small class.

Let $\map V x$ denote the von Neumann hierarchy on the ordinal $x$.


Then $x$ is a subset of the rank of $S$ if and only if $S \notin \map V x$.


Proof

NotZFC.jpg

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.


Necessary Condition

Let $x \subseteq \map {\operatorname{rank} } S$.

Then by Von Neumann Hierarchy Comparison:

$S \in \map V x \implies S \in \map V {\map {\operatorname{rank} } S}$

But by Ordinal Equal to Rank:

$S \notin \map V {\map {\operatorname{rank} } S}$

By contraposition:

$S \notin V \left({x}\right)$

$\Box$


Sufficient Condition

Let $S \notin \map V x$.

Then:

\(\ds S\) \(\in\) \(\ds \map V {\map {\operatorname{rank} } S + 1}\) Ordinal Equal to Rank
\(\ds \leadsto \ \ \) \(\ds \map V {\map {\operatorname{rank} } S + 1}\) \(\nsubseteq\) \(\ds \map V x\) Rule of Transposition
\(\ds \leadsto \ \ \) \(\ds \map {\operatorname{rank} } S + 1\) \(\nsubseteq\) \(\ds x\) Von Neumann Hierarchy Comparison
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \map {\operatorname{rank} } S + 1\) Transitive Set is Proper Subset of Ordinal iff Element of Ordinal and Ordinal Membership is Trichotomy
\(\ds \leadsto \ \ \) \(\ds x\) \(\subseteq\) \(\ds \map {\operatorname{rank} } S\) Definition of Successor Set

$\blacksquare$


Sources