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"
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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 $5.7$

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

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