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
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 9.18$