Order Isomorphism between Ordinals and Proper Class

Theorem
Suppose $\left(A,\prec\right)$ is a strict well-ordering

Suppose $A$ is a proper class

Suppose, for every $x \in A$, that the initial segment of $x$ is a set.

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 $\left( A , \prec \right)$.

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

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