Transfinite Recursion Theorem/Formulation 4

Theorem

Let $\On$ denote the class of all ordinals.

Let $g$ be a mapping defined for all sets.

Let $c$ be a set.

Then there exists a unique $\On$-sequence $S_0, S_1, \dots, S_\alpha, \dots$ such that:

\((1)\)	$:$					\(\ds S_0 \)	\(\ds = \)	\(\ds c \)
\((2)\)	$:$			\(\ds \forall \alpha \in \On:\)		\(\ds S_{\alpha + 1} \)	\(\ds = \)	\(\ds \map g {S_\alpha} \)
\((3)\)	$:$			\(\ds \forall \lambda \in K_{II}:\)		\(\ds S_\lambda \)	\(\ds = \)	\(\ds \bigcup_{\alpha \mathop < \lambda} S_\alpha \)

where $K_{II}$ denotes the class of all limit ordinals.

Proof

This is a special case of the Transfinite Recursion Theorem: Formulation $3$.

Recall:

Let $\On$ denote the class of all ordinals.

Let $h$ and $f$ be mappings defined for all sets.

Let $c$ be a set.

Then there exists a unique mapping $F$ on $\On$ such that:

\((1)\)	$:$					\(\ds \map F 0 \)	\(\ds = \)	\(\ds c \)
\((2)\)	$:$			\(\ds \forall \alpha \in \On:\)		\(\ds \map F {\alpha^+} \)	\(\ds = \)	\(\ds \map h {\map F \alpha} \)
\((3)\)	$:$			\(\ds \forall \lambda \in K_{II}:\)		\(\ds \map F \lambda \)	\(\ds = \)	\(\ds \map f {F \sqbrk \lambda} \)

where $K_{II}$ denotes the class of all limit ordinals.

$\Box$

Set $\map f x = \bigcup x$, so that:

$\map f {F \sqbrk \lambda}$

is then:

$\ds \bigcup_{\alpha \mathop < \lambda} \map F \alpha$

Let $S_\alpha$ be the set $\map F \alpha$

and the given form of the Transfinite Recursion Theorem follows.

It remains to demonstrate uniqueness of $S_0, S_1, \dots, S_\alpha, \dots$.

This needs considerable tedious hard slog to complete it. In particular: "an obvious transfinite argument" 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 $5.7$