# Order Isomorphism between Ordinals and Proper Class

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## Theorem

Let $\struct {A, \prec}$ 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 $\tuple {x, y}$ such that:

- $y \in A \setminus \Img x$

- $\paren {A \setminus \Img x} \cap A_y = \O$

Use the First Principle of Transfinite Recursion to construct a mapping $F$ such that:

- The domain of $F$ is the class of all ordinals $\On$

- For all ordinals $x$, $\map F x = \map G {F {\restriction_x} }$

Then $F: \On \to A$ is an order isomorphism between $\struct {\On, \in}$ and $\struct {A, \prec}$.

### 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 $\tuple {x, y}$ such that:

- $y \in A \setminus \Img x$

- $\paren {A \setminus \Img x} \cap A_y = \O$

Define $F$ by the First Principle of Transfinite Recursion to construct a mapping $F$ such that:

- The domain of $F$ is $\On$.

- For all ordinals $x$, $\map F x = \map G {F \restriction x}$.

Then $F: \On \to A$ is an order isomorphism between $\struct {\On, \in}$ and $\struct {A, \in}$.

## 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 $\struct {A, \prec}$ 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
- $\map G x \in A \setminus \Img x \iff A \setminus \Img x \ne \O$

$\Box$

Assume that:

- $\exists x: A \setminus \Img x = \O$

Then:

- $A \subseteq \Img x$

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):

- $\forall x: A \setminus \Img x \ne \O$

Then:

\(\ds \forall x: \, \) | \(\ds A \setminus \Img x\) | \(\ne\) | \(\ds \O\) | |||||||||||

\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map F x\) | \(\in\) | \(\ds A \setminus \Img x\) | Lemma | |||||||||

\(\, \ds \land \, \) | \(\ds \paren {A \setminus \Img x} \cap A_{\map F x}\) | \(\ne\) | \(\ds \O\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds f: \On\) | \(\rightarrowtail\) | \(\ds A\) | Condition for Injective Mapping on Ordinals |

where $\rightarrowtail$ denotes an injection.

Now to prove that $F$ is surjective:

\(\ds y\) | \(\in\) | \(\ds \Img f\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \exists x: \, \) | \(\ds y\) | \(=\) | \(\ds \map F x\) | Definition of Image of Mapping | |||||||||

\(\ds \leadsto \ \ \) | \(\ds \exists x: \, \) | \(\ds \paren {A \setminus \Img x} \cap A_y\) | \(=\) | \(\ds \O\) | from $(1)$ | |||||||||

\(\, \ds \land \, \) | \(\ds y\) | \(=\) | \(\ds \map F x\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \exists x: \forall z: \, \) | \(\ds \paren {z \prec y \land x \in A}\) | \(\implies\) | \(\ds z \in \Img x\) | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \forall z: \, \) | \(\ds \paren {z \prec y \land x \in A}\) | \(\implies\) | \(\ds z \in \Img f\) | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \Img F = A\) | \(=\) | \(\ds \) | Well-Ordered Transitive Subset is Equal or Equal to Initial Segment | ||||||||||

\(\ds \exists x \in A: \, \) | \(\, \ds \lor \, \) | \(\ds \Img F\) | \(=\) | \(\ds A \cap A_x\) |

But if $\Img F = A \cap A_x$, then it is equal to some initial segment of $A$.

This would imply that $\Img F$ is a set, which is a contradiction.

Therefore $A = \Img F$ and the function is bijective.

\(\ds x\) | \(\in\) | \(\ds y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \Img x\) | \(\subseteq\) | \(\ds \Img y\) | Transitive Set is Proper Subset of Ordinal iff Element of Ordinal and the fact that the image preserves subsets | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds A \setminus \Img y\) | \(\subseteq\) | \(\ds A \setminus \Img x\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map F y\) | \(\in\) | \(\ds A \setminus \Img x\) | from $(1)$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map F y\) | \(\notin\) | \(\ds A_{\map F x}\) | from $(1)$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map F x\) | \(\prec\) | \(\ds \map F y\) | $\prec$ is a Strict Well-Ordering | ||||||||||

\(\, \ds \lor \, \) | \(\ds \map F x\) | \(=\) | \(\ds \map F y\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map F x\) | \(\prec\) | \(\ds \map F y\) | $F$ is injective and $x \ne y$ |

Conversely, assume $\map F x \prec \map F y$.

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

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

$\blacksquare$

## Also see

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

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 7.49$