Order Isomorphism between Ordinals and Proper Class/Corollary

Corollary to Order Isomorphism between Ordinals and Proper Class
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 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.