Order Isomorphism between Ordinals and Proper Class/Lemma

Theorem
Suppose the following conditions are met:


 * $A$ be a class, (we allow $A$ to be a proper class or a set).
 * $(A,\prec)$ is a strict well-ordering.
 * Every $\prec$-initial segment is a set, not a proper class.
 * Let $\operatorname{Im}(x)$ denote the image, or range, of $x$ (not the image of $x$ under a mapping). Set $G$ to equal the class of all ordered pairs $(x,y)$ satisfying:
 * $\displaystyle y \in ( A \setminus \operatorname{Im}(x) )$
 * The initial segment $A_y$ is a subset of $\operatorname{Im}(x)$


 * $F$ is a mapping with a domain of $\operatorname{On}$.
 * $F$ also satisfies $F(x) = G(F \restriction x)$

Then, we may conclude that:


 * $G$ is a mapping
 * $G(x) \in (A \setminus \operatorname{Im}(x)) \iff (A \setminus \operatorname{Im}(x)) \ne \varnothing$

Note that only the first three conditions need hold: we may construct classes $F$ and $G$ satisfying the other conditions using transfinite recursion.

Proof
Therefore, we may conclude, that $G$ is a single-valued relation and therefore a mapping.

For the second part:

Furthermore:


 * $\displaystyle G(x) \in ( A \setminus \operatorname{Im}(x) ) \implies (A \setminus \operatorname{Im}(x)) \ne \varnothing$ by the definition of nonempty.

Also see

 * Transfinite Recursion
 * Condition for Injective Mapping on Ordinals
 * Maximal Injective Mapping from Ordinals to a Set