Order Isomorphism between Ordinals and Proper Class

Theorem
Let $\left({A, \prec}\right)$ be a strict well-ordering.

Let $A$ be a proper class.

Let the initial segment of $x$ be a set for every $x \in A$.

Then we may make the following definitions:

Set $G$ equal to the collection of ordered pairs $\left({x, y}\right)$ such that:


 * $y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$


 * $\left({A \setminus \operatorname{Im}\left({x}\right)}\right) \cap A_y = \varnothing$

Use transfinite recursion to construct a mapping $F$ such that:


 * The domain of $F$ is $\operatorname{On}$


 * For all ordinals $x$, $F \left({x}\right) = G \left({F {\restriction_x} }\right)$

Then $F: \operatorname{On} \to A$ is an order isomorphism between $\left({\operatorname{On}, \in}\right)$ and $\left({A, \prec}\right)$.

Lemma
Assume that:
 * $\exists x: \left({A \setminus \operatorname{Im} \left({x}\right) = \varnothing}\right)$

Then:
 * $A \subseteq \operatorname{Im} \left({x}\right)$

Therefore, $A$ is a set.

This contradicts the fact that $A$ is a proper class.

Therefore by Axiom of Subsets Equivalents and De Morgan's Laws (Predicate Logic):
 * $\forall x: \left({A \setminus \operatorname{Im} \left({x}\right) \ne \varnothing}\right)$

Then:

where $\rightarrowtail$ denotes an injection.

Now to prove that $F$ is surjective:

But if $\operatorname{Im} \left({F}\right) = \left({A \cap A_x}\right)$, then it is equal to some initial segment of $A$.

This would imply that $\operatorname{Im} \left({F}\right)$ is a set, which is a contradiction.

Therefore $A = \operatorname{Im} \left({F}\right)$ and the function is bijective.

Conversely, assume $F \left({x}\right) \prec F \left({y}\right)$.

Then $y \in x$ and $x = y$ lead to contradictory conclusions.

By Ordinal Membership is Trichotomy, we may conclude that $x \in y$.

Also see

 * Transfinite Recursion
 * Condition for Injective Mapping on Ordinals
 * Maximal Injective Mapping from Ordinals to a Set