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 $(x,y)$ such that:


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

Define $F$ by transfinite recursion to construct a mapping $F$ such that:


 * The domain of $F$ is $\operatorname{On}$.
 * For all ordinals $x$, $F(x) = G(F \restriction x)$.

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

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.