Ordinal Equal to Rank

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 $x$ equals the rank of $S$ if and only if $S \in \map V {x + 1} \land S \notin \map V x$.

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

Then $S \in \map V {x + 1}$ by the definition of rank.

Suppose $S \in \map V x$.

Then $x \ne 0$, so $x = y^+$ for some ordinal $y$ or $x$ is a limit ordinal.

Suppose $x = y^+$.

Then, $S \in \map V {y + 1}$.

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, $S \in \map V y$ for some $y$, so $S \in \map V {y + 1}$ 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, $S \notin \map V x$.

$\Box$

Suppose $S \notin \map V x$ and $S \in \map V {x + 1}$.

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

If $x < y$, then $S \in \map V y$.

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

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

Therefore, $x = \map {\operatorname{rank} } S$.

$\blacksquare$