Transfinite Recursion Theorem/Formulation 2

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 $F$.

This needs considerable tedious hard slog to complete it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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 $5.5$