Isomorphic Ordinals are Equal/Proof 1
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Then $A = B$.
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 $T$
- $T$ is an initial segment of $S$.
From this contradiction it follows that $S = T$.
- 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 $1.7.10$