Transfinite Recursion Theorem/Formulation 1

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 1

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

By the Superinductive Class under Strictly Progressing Mapping is Proper Class, $M$ is not a set; it is a proper class.

From Minimally Superinductive Class is Well-Ordered under Subset Relation, $\struct {M, \subseteq}$ is indeed a well-ordered class.

By definition, $M$ is a $g$-tower.

Hence from $g$-Tower is Well-Ordered under Subset Relation:

$(1): \quad \O$ is the smallest element of $M$.

$(2): \quad$ The immediate successor of $x$ under the well-ordering $\subseteq$ is $\map g x$.

$(3): \quad$ For a limit element $x$ of $M$:

$x = \bigcup x^\subset$

where $\bigcup x^\subset$ denotes the union of the lower section of $x$.

From the Counting Theorem, there exists a unique order isomorphism from $\On$ to $M$.

Let $M_\alpha$ denote the element of $M$ corresponding to $\alpha$ under this order isomorphism.

We refer to $M_\alpha$ as the "$\alpha$th element of $M$".

By the definition of order isomorphism, it is immediate that $M_0$ is the smallest element of $M$ under the well-ordering induced by $\subseteq$.

Hence:

$M_0 = \O$

It is also immediate that $M_{\alpha^+}$ is the immediate successor of $M_\alpha$, that is:

$M_{\alpha^+} = \map g {M_\alpha}$

Let $\lambda$ be a limit ordinal.

Then $M_\lambda$ is a limit element of $M$:.

Hence $M_\lambda$ is the union of the set of all $M_\alpha$ such that $\alpha < \lambda$.

That is:

$\ds M_\lambda = \bigcup_{\alpha \mathop < \lambda} M_\alpha$

Hence the result.

$\blacksquare$

Proof 2

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.

$\blacksquare$