Transfinite Recursion Theorem/Formulation 1/Proof 2

Theorem

Let $M$ be a class.

Let $g$ be a strictly progressing mapping on $M$.

Let $M$ be minimally superinductive under $g$.

For an arbitrary ordinal $\alpha$, let $M_\alpha$ be the $\alpha$th element of $M$ under the well-ordered class $\struct {M, \subseteq}$.

Then:

\((1)\)	$:$		Zeroth Ordinal:			\(\ds M_0 \)	\(\ds = \)	\(\ds \O \)
\((2)\)	$:$		Successor Ordinal:	\(\ds \forall \alpha \in \On:\)		\(\ds M_{\alpha^+} \)	\(\ds = \)	\(\ds \map g {M_\alpha} \)
\((3)\)	$:$		Limit Ordinal:	\(\ds \forall \lambda \in K_{II}:\)		\(\ds M_\lambda \)	\(\ds = \)	\(\ds \bigcup_{\alpha \mathop < \lambda} M_\alpha \)

where:

$\On$ denotes the class of all ordinals

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

Proof

This is a special case of Transfinite Recursion Theorem: Formulation $4$.

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.

$\Box$

Take $c$ to be $0$, that is $\O$.

Let us take the special case where $g$ is a strictly progressing mapping.

The result follows directly.

$\blacksquare$

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