# Order Isomorphism between Ordinals and Proper Class

## Theorem

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

Let $A$ be a proper class.

Let the initial segment of $x$ be a set for every $x \in A$.

Then we may make the following definitions:

Set $G$ equal to the collection of ordered pairs $\left({x, y}\right)$ such that:

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

Use 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 $\left({A, \prec}\right)$.

### Corollary

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

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### Lemma

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 of a subclass $x$.

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

$\Box$

Assume that:

$\exists x: \left({A \setminus \operatorname{Im} \left({x}\right) = \varnothing}\right)$

Then:

$A \subseteq \operatorname{Im} \left({x}\right)$

Therefore, $A$ is a set.

This contradicts the fact that $A$ is a proper class.

Therefore by Axiom of Subsets Equivalents and De Morgan's Laws (Predicate Logic):

$\forall x: \left({A \setminus \operatorname{Im} \left({x}\right) \ne \varnothing}\right)$

Then:

 $\displaystyle$  $\displaystyle \forall x: \left({\left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing}\right)$ $(1):\quad$ $\displaystyle$ $\implies$ $\displaystyle F \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \land \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \cap A_{F \left({x}\right)} \ne \varnothing$ Lemma $\displaystyle$ $\implies$ $\displaystyle f: \operatorname{On} \rightarrowtail A$ Condition for Injective Mapping on Ordinals

where $\rightarrowtail$ denotes an injection.

Now to prove that $F$ is surjective:

 $\displaystyle y \in \operatorname{Im} \left({F}\right)$ $\implies$ $\displaystyle \exists x: y = F \left({x}\right)$ Definition of Image $\displaystyle$ $\implies$ $\displaystyle \exists x: \left({\left({A \setminus \operatorname{Im} \left({x}\right)}\right) \cap A_y = \varnothing \land y = F \left({x}\right)}\right)$ from $(1)$ $\displaystyle$ $\implies$ $\displaystyle \exists x: \forall z: \left({\left({z \prec y \land x \in A}\right) \implies z \in \operatorname{Im} \left({x}\right)}\right)$ $\displaystyle$ $\implies$ $\displaystyle \forall z: \left({\left({z \prec y \land x \in A}\right) \implies z \in \operatorname{Im} \left({F}\right)}\right)$ $\displaystyle$ $\implies$ $\displaystyle \operatorname{Im} \left({F}\right) = A \lor \exists x \in A: \operatorname{Im} \left({F}\right) = \left({A \cap A_x}\right)$ Well-Ordered Transitive Subset is Equal or Equal to Initial Segment

But if $\operatorname{Im} \left({F}\right) = \left({A \cap A_x}\right)$, then it is equal to some initial segment of $A$.

This would imply that $\operatorname{Im} \left({F}\right)$ is a set, which is a contradiction.

Therefore $A = \operatorname{Im} \left({F}\right)$ and the function is bijective.

 $\displaystyle x \in y$ $\implies$ $\displaystyle \operatorname{Im} \left({x}\right) \subseteq \operatorname{Im} \left({y}\right)$ Transitive Set is Proper Subset of Ordinal iff Element of Ordinal and the fact that the image preserves subsets $\displaystyle$ $\implies$ $\displaystyle \left({A \setminus \operatorname{Im} \left({y}\right)}\right) \subseteq \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$ $\displaystyle$ $\implies$ $\displaystyle F \left({y}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$ from $(1)$ $\displaystyle$ $\implies$ $\displaystyle F \left({y}\right) \notin A_{F \left({x}\right)}$ from $(1)$ $\displaystyle$ $\implies$ $\displaystyle F \left({x}\right) \prec F \left({y}\right) \lor F \left({x}\right) = F \left({y}\right)$ $\prec$ is a Strict Well-Ordering $\displaystyle$ $\implies$ $\displaystyle F \left({x}\right) \prec F \left({y}\right)$ $F$ is injective and $x \ne y$

Conversely, assume $F \left({x}\right) \prec F \left({y}\right)$.

Then $y \in x$ and $x = y$ lead to contradictory conclusions.

By Ordinal Membership is Trichotomy, we may conclude that $x \in y$.

$\blacksquare$