Ordinal Equal to Rank

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $x$ be an ordinal.

Let $S$ be a small class.

Let $V \left({ x }\right)$ denote the von Neumann hierarchy on the ordinal $x$.


Then $x$ equals the rank of $S$ iff $S \in V \left({x+1}\right) \land S \notin V \left({x}\right)$.


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.


Sufficient Condition

Suppose that $x$ is equal to the rank of $S$.

Then $S \in V \left({x+1}\right)$ by the definition of rank.


Suppose $S \in V \left({x}\right)$. Then $x \ne 0$, so $x = y^+$ for some ordinal $y$ or $x$ is a limit ordinal.


Suppose $x = y^+$.

Then, $S \in V \left({y+1}\right)$.

This contradicts the fact that the rank is the smallest set $x$ such that $S \in V\left({x+1}\right)$.


Suppose $x$ is a limit ordinal.

Then, $S \in V \left({y}\right)$ for some $y$, so $S \in V \left({y+1}\right)$ and $y+1 < x$.

This contradicts the fact that the rank is the smallest set $x$ such that $S \in V \left({x+1}\right)$.


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

$\Box$


Necessary Condition

Suppose $S \notin V \left({x}\right)$ and $S \in V \left({x+1}\right)$.

Then, if $y < x$, then $S \notin V \left({y+1}\right)$ by Von Neumann Hierarchy Comparison.

If $x < y$, then $S \in V\left({y}\right)$.


Therefore, $x$ is the unique ordinal that satisfies $S \notin V \left({x}\right)$ and $S \in V \left({x+1}\right)$.

Moreover, the rank of $S$ also satisfies $S \notin V \left({x}\right)$ and $S \in V \left({x+1}\right)$.

Therefore, $x = \operatorname{rank} \left({S}\right)$.

$\blacksquare$


Sources