Order Isomorphism between Ordinals and Proper Class/Corollary

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Corollary to Order Isomorphism between Ordinals and Proper Class

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

$A$ is a proper class of ordinals.

It is strictly well-ordered by $\in$.

Moreover, every initial segment of $A$ is a set, since the initial segment of the ordinal is simply the ordinal itself.

Therefore, we may apply Order Isomorphism between Ordinals and Proper Class to achieve the desired isomorphism.

$\blacksquare$


Remark



This theorem shows that every proper class of ordinals can be put in a unique order-isomorphism with the set of all ordinals.


Sources