# Ordinals Isomorphic to the Same Well-Ordered Set

## Theorem

Let $A$ and $B$ be ordinals.

Let $\left({\prec, S}\right)$ be a strict well-ordering.

Let $\left({\in, A}\right)$ and $\left({\prec, S}\right)$ be order isomorphic.

Let $\left({\in, B}\right)$ and $\left({\prec, S}\right)$ be order isomorphic.

Then:

$A = B$

## Proof

Let $\phi_1$ denote the mapping creating the order isomorphism between $\left({\in, A}\right)$ and $\left({\prec, S}\right)$.

Let $\phi_2$ denote the mapping creating the order isomorphism between $\left({\in, B}\right)$ and $\left({\prec, S}\right)$.

Then $\phi_2^{-1}: S \to B$ is an order isomorphism by Inverse of Order Isomorphism is Order Isomorphism.

$\phi_1 \circ \phi_2^{-1}: A \to B$ is an order isomorphism by Composite of Order Isomorphisms is Order Isomorphism.

So $\left({\in, A}\right)$ and $\left({\in, B}\right)$ are order isomorphic ordinals.

$A = B$

$\blacksquare$