Transfinite Recursion Theorem/Formulation 2
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Theorem
Let $\On$ denote the class of all ordinals.
Let $S$ denote the class of all ordinal sequences.
Let $g$ be a mapping such that $S \subseteq \Dom g$.
Then there exists a unique mapping $F$ on $\On$ such that:
- $\forall \alpha \in \On: \map F \alpha = \map g {F \restriction \alpha}$
where $F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$.
Proof
First we establish the following
Lemma
There exists an extending operation $E$ such that for every mapping $F$ on $\On$:
- $\forall \alpha \in \On: \map {\paren {\map E {F \restriction \alpha} } } \alpha = \map g {F \restriction \alpha}$
$\Box$
By Characteristic of Extending Operation, there exists a mapping $F$ on $\On$ such that:
- $\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
where:
- $F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$
- $\alpha^+$ denotes the successor ordinal of $\alpha$.
Then:
\(\ds \forall \alpha \in \On: \, \) | \(\ds \map F \alpha\) | \(=\) | \(\ds \map {\paren {F \restriction \alpha^+} } \alpha\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\map E {F \restriction \alpha} } } \alpha\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map g {F \restriction \alpha}\) | by the lemma |
It remains to demonstrate uniqueness of $F$.
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Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.5$