Rank of Ordinal

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Theorem

Let $x$ be an ordinal.

Let $\map {\operatorname {rank} } x$ denote the rank of $x$.


Then:

$\map {\operatorname {rank} } x = x$


Proof

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The proof shall proceed by Transfinite Induction (Strong Induction) on $x$.

Suppose $\forall y \in x: \map {\operatorname {rank} } y = y$.


Then:

\(\ds \map {\operatorname {rank} } x\) \(=\) \(\ds \bigcap \set {z \in \On: \forall y \in x: y < z}\) Rank of Set Determined by Members
\(\ds \) \(=\) \(\ds \bigcap \set {z \in \On: x \subseteq z}\) Definition of Subset
\(\ds \) \(=\) \(\ds x\)

$\blacksquare$


Sources