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


Sources

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