Isomorphic Ordinals are Equal/Proof 1

From ProofWiki
Jump to navigation Jump to search

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 No Isomorphism from Woset 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