Transfinite Recursion Theorem/Formulation 1/Proof 1

Proof
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.