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 $\left({x, y}\right)$ 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

$A$ 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