# Isomorphic Ordinals are Equal/Proof 1

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## Theorem

Let $A$ and $B$ be ordinals that are order isomorphic.

Then $A = B$.

## Proof

Let $S \cong T$.

Aiming for a contradiction, suppose that $S \ne T$.

Then from the corollary to Relation between Two Ordinals, either:

- $S$ is an initial segment of $T$

or:

- $T$ is an initial segment of $S$.

But as $S \cong T$, from Well-Ordered Class is not Isomorphic to Initial Segment, neither $S$ nor $T$ can be an initial segment of the other.

From this contradiction it follows that $S = T$.

$\blacksquare$

## Sources

- 1993: Keith Devlin:
*The Joy of Sets: Fundamentals of Contemporary Set Theory*(2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.10$