# Characteristic of Extending Operation

## Theorem

Let $E$ be an extending operation.

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 $\alpha$.

## Proof

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

Recall the Transfinite Recursion Theorem:

 $(1)$ $:$ 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.

Then by Length of Union of Chain of Ordinal Sequences, $M_\lambda$ is a $\lambda$-sequence.

Hence by the Second Principle of Transfinite Induction:

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

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

Then by Length of Union of Chain of Ordinal Sequences again, $F$ is a mapping on $\On$.

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

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

$\blacksquare$