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: $\S 7.49$