No Order Isomophism Between Distinct Initial Segments of Woset

Theorem

Let $E$ be a well-ordered set.

Let $S_\alpha, S_\beta$ be initial segments of $E$ that are order isomorphic.

Then $S_\alpha = S_\beta$.

Proof

Aiming for a contradiction, suppose $S_\alpha \ne S_\beta$.

Then $\alpha \ne \beta$.

By the trichotomy law, $\alpha \prec \beta$ or $\beta \prec \alpha$.

Without loss of generality assume $\alpha \prec \beta$.

Then $S_\alpha \subsetneqq S_\beta$.

That is, $S_\alpha$ is an initial segment of $S_\beta$.

We have by hypothesis that $S_\alpha$ and $S_\beta$ are order isomorphic.

Thus there is an order isomorphism between $S_\beta$ and an initial segment of $S_\beta$.

This contradicts Well-Ordered Class is not Isomorphic to Initial Segment.

Thus $S_\alpha = S_\beta$.

$\blacksquare$

Sources

• 2000: James R. Munkres: Topology (2nd ed.): Supplementary Exercise $1.2$