Order Isomorphism between Ordinals and Proper Class/Lemma

Lemma for Order Isomorphism between Ordinals and Proper Class
Suppose the following conditions are met:

Let $A$ be a class.

We allow $A$ to be a proper class or a set.

Let $\left({A, \prec}\right)$ be a strict well-ordering.

Let every $\prec$-initial segment be a set, not a proper class.

Let $\operatorname{Im} \left({x}\right)$ denote the image, or range, of $x$ (not the image of $x$ under a mapping).

Let $G$ equal the class of all ordered pairs $\left({x, y}\right)$ satisfying:
 * $y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$
 * The initial segment $A_y$ of $\left({A, \prec}\right)$ is a subset of $\operatorname{Im} \left({x}\right)$

Let $F$ be a mapping with a domain of $\operatorname{On}$.

Let $F$ also satisfy:
 * $F \left({x}\right) = G \left({F \restriction x}\right)$

Then:


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

Note that only the first four 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:


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

Also see

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