Transfinite Recursion Theorem/Formulation 2/Lemma
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Lemma for Transfinite Recursion Theorem: Formulation $2$
Let $\On$ denote the class of all ordinals.
Let $S$ denote the class of all ordinal sequences.
Let $g$ be a mapping such that $S \subseteq \Dom g$.
There exists an extending operation $E$ such that for every mapping $F$ on $\On$:
- $\forall \alpha \in \On: \map {\paren {\map E {F \restriction \alpha} } } \alpha = \map g {F \restriction \alpha}$
Proof
Let $\theta_\alpha$ be an arbitrary ordinal sequence of length $\alpha$.
Let $\map E {\theta_\alpha}$ be the extending operation defined as:
- $\map E {\theta_\alpha} := \theta_\alpha \cup \set {\tuple {\alpha, \map g {\theta_\alpha} } }$
Thus $\map E {\theta_\alpha}$ is an ordinal sequence of length $\alpha^+$.
Also, for $\beta \in \On$ such that $\beta < \alpha$:
- $\map {\paren {\map E {\theta_\alpha} } } \beta = \map {\theta_\alpha} \beta$
but:
- $\map {\paren {\map E {\theta_\alpha} } } \alpha = \map g {\theta_\alpha}$
Then for any mapping $F$ on $\On$:
- $F \restriction \alpha$ is an ordinal sequence of length $\alpha$.
Hence:
- $\map {\paren {\map E {F \restriction \alpha} } } \alpha = \map g {F \restriction \alpha}$
$\blacksquare$
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: Lemma $5.4$