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.
By Isomorphic Ordinals are Equal:
- $A = B$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 7.39$