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.

$\blacksquare$


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