Compact Subspace of Linearly Ordered Space

Theorem
Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space.

Let $Y \subseteq X$ be a non-empty subset of $X$.

Then $Y$ is a compact subspace of $\left({X, \tau}\right)$ both of the following hold:
 * $(1):\quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
 * $(2):\quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

Forward Implication
Let $S$ be a non-empty subset of $Y$.

By Compact Subspace of Linearly Ordered Space: Lemma, $S$ has a supremum $k$ in $Y$.

Aiming for a contradiction, suppose that $S$ has an upper bound $b$ in $X$ such that $b \prec k$.

Let:
 * $\mathcal A = \left\{{s^\preceq: s \in S}\right\} \cup \left\{{b^\succeq}\right\}$

where:
 * $s^\preceq$ denotes the lower closure of $s$ in $S$
 * $b^\succeq$ denotes the upper closure of $s$ in $S$.

Then $\mathcal A$ is an open cover of $Y$.

But $\mathcal A$ has no finite subcover, contradicting the fact that $Y$ is compact.

A similar argument proves the corresponding statement for infima, so $(2)$ holds.

Also see

 * Heine–Borel Theorem: Dedekind-Complete Space
 * Connected Subspace of Linearly Ordered Space