Order Isomorphism between Ordinals and Proper Class/Corollary

Theorem
Let $A$ be a proper class of ordinals.

We will take ordering on $A$ to be $\in$.

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


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


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

Define $F$ by 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 $\left({A, \in}\right)$.

Proof
$A$ is a proper class of ordinals.

It is strictly well-ordered by $\in$.

Moreover, every initial segment of $A$ is a set, since the initial segment of the ordinal is simply the ordinal itself.

Therefore, we may apply Order Isomorphism between Ordinals and Proper Class/Theorem to achieve the desired isomorphism.

Remark
This theorem shows that every proper class of ordinals can be put in a unique order-isomorphism with the set of all ordinals.