Compact Subspace of Linearly Ordered Space/Lemma 1

Theorem
Let $\left({X, \preceq}\right)$ be a totally ordered set, equipped with its order topology so that it is considered as a topological space.

Then a topological subspace $Y \subseteq X$ is compact iff all of the following hold:
 * $\left({1}\right): \quad$ $Y$ has a greatest element and a smallest element.
 * $\left({2}\right): \quad$ $\left({Y, \preceq \restriction_{Y \times Y}}\right)$ is Dedekind complete, where $\restriction$ denotes restriction.
 * $\left({3}\right): \quad$ $Y$ is closed in $X$.

Also see

 * Heine-Borel Theorem (Special Case)
 * Connected Subspace of Totally Ordered Set