# 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 $\struct {A, \prec}$ be a strict well-ordering.

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

Let $\Img x$ denote the image of a subclass $x$.

Let $G$ equal the class of all ordered pairs $\tuple {x, y}$ satisfying:

$y \in A \setminus \Img x$
The initial segment $A_y$ of $\struct {A, \prec}$ is a subset of $\Img x$

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

Let $F$ also satisfy:

$\map F x = \map G {F \restriction x}$

Then:

$G$ is a mapping
$\map G x \in A \setminus \Img x \iff A \setminus \Img x \ne \O$

Note that only the first four conditions need hold: we may construct classes $F$ and $G$ satisfying the other conditions using the First Principle of Transfinite Recursion.

## Proof

 $\ds \tuple {x, y}$ $\in$ $\ds G$ $\, \ds \land \,$ $\ds \tuple {x, z}$ $\in$ $\ds G$ $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds A \setminus \Img x$ Definition of $G$ $\, \ds \land \,$ $\ds z$ $\in$ $\ds A \setminus \Img x$ $\ds \leadsto \ \$ $\ds y$ $\notin$ $\ds A_z$ $A_y$ is disjoint with $A \setminus \Img x$. Same with $A_z$. $\, \ds \land \,$ $\ds y$ $\notin$ $\ds A_y$ $\ds \leadsto \ \$ $\ds y$ $\nprec$ $\ds z$ Definition of Initial Segment $\, \ds \land \,$ $\ds z$ $\nprec$ $\ds y$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds z$ $\prec$ is a strict well-ordering

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

For the second part:

 $\ds A \setminus \Img x$ $\ne$ $\ds \O$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds \exists y \in A \setminus \Img x: \,$ $\ds \paren {A \cap A_y} \setminus \Img x$ $=$ $\ds \O$ Proper Well-Ordering Determines Smallest Elements $\ds \leadsto \ \$ $\ds \map G x$ $=$ $\ds y$ Conditions are satisfied for $\tuple {x, y} \in G$. Follows from first part. $\ds \leadsto \ \$ $\ds \map G x$ $\in$ $\ds A \setminus \Img x$ equation $(1)$, $y \in A \setminus \Img x$

Furthermore:

$\map G x \in A \setminus \Img x \implies A \setminus \Img x \ne \O$ by the definition of non-empty.

$\blacksquare$