Characteristic of Extending Operation

Theorem

Then there exists a mapping $F$ on the class of all ordinals $\On$ 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 $\alpha$.

Let $M$ be the class which is minimally superinductive under $E$.

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

where:

for an arbitrary ordinal $\alpha$, $M_\alpha$ denotes the $\alpha$th element of $M$ under the well-ordered class $\struct {M, \subseteq}$

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

We know that $M_0 = 0$, so $M_0$ is a $0$-sequence.

Let $M_\alpha$ be an arbitrary $\alpha$-sequence.

Then from $(2)$ above, $M_{\alpha^+}$ is an $\alpha^+$-sequence.

Let $\lambda$ be a limit ordinal.

Suppose that, for each $\alpha < \lambda$, $M_\alpha$ is an $\alpha$-sequence.

for every $\alpha \in \On$, $M_\alpha$ is an ordinal sequence of length $\alpha$.

Let $F = \bigcup M$ be the union of $M$.

Because each $M_\alpha \subseteq F$ and $\Dom {M_\alpha} = \alpha$, it follows that $M_\alpha = F \restriction \alpha$.

$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$

$\blacksquare$

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 $5.3$

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