Transfinite Recursion Theorem/Formulation 5
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Theorem
Let $\On$ denote the class of all ordinals.
Let $h$ be a mapping.
Then there exists a unique mapping $F$ such that:
- $\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$
Proof
This is a special case of the Transfinite Recursion Theorem: Formulation $2$.
Recall:
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 $\alpha$.
$\Box$
Consider $\map g x$.
If $x$ is a mapping, take $\map g x = \map h {\Img x}$.
If $x$ is not a mapping, take $\map g x = x$.
The result follows.
$\blacksquare$
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 $5.8$