Order Isomorphism between Ordinals and Proper Class

Theorem
Let $\struct {A, \prec}$ 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 $\tuple {x, y}$ such that:


 * $y \in A \setminus \Img x$


 * $\paren {A \setminus \Img x} \cap A_y = \O$

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


 * The domain of $F$ is $\On$


 * For all ordinals $x$, $\map F x = \map G {F {\restriction_x} }$

Then $F: \On \to A$ is an order isomorphism between $\struct {\On, \in}$ and $\struct {A, \prec}$.

Lemma
Assume that:
 * $\exists x: A \setminus \Img x = \O$

Then:
 * $A \subseteq \Img x$

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):
 * $\forall x: A \setminus \Img x \ne \O$

Then:

where $\rightarrowtail$ denotes an injection.

Now to prove that $F$ is surjective:

But if $\Img F = A \cap A_x$, then it is equal to some initial segment of $A$.

This would imply that $\Img F$ is a set, which is a contradiction.

Therefore $A = \Img F$ and the function is bijective.

Conversely, assume $\map F x \prec \map F y$.

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