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:

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 $M_\alpha = F \restriction \alpha$.

Hence from Union of Nest of Ordinal Sequences which is Proper Class:
 * $\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$