# Isomorphic Ordinals are Equal

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## Theorem

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

Then $A = B$.

## Proof 1

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 $S = T$.

$\blacksquare$

## Proof 2

From Well-Ordered Class is not Isomorphic to Initial Segment, neither $A$ nor $B$ can be an initial segment of the other.

By definition, every element of an ordinal is an initial segment of it.

Hence, neither $A$ nor $B$ can be an element of the other.

By Ordinal Membership is Trichotomy, it follows that $A = B$.

$\blacksquare$