Isomorphism Class of Total Orderings

Theorem

Let $S$ be a finite set with $n$ elements.

There is exactly one isomorphism class containing the total orderings on $S$.

That is, every total ordering on $S$ is (order) isomorphic to every other total ordering.

Proof

This theorem requires a proof. In particular: Intuitively obvious, probably needs to be hammered out in a proof by induction. I want to move on to other stuff today. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.5$