Compact Subspace of Linearly Ordered Space/Lemma 1

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

Let $Y \subseteq X$ be a nonempty subset of $X$.

Then $Y$ is a compact subspace of $\left({X, \tau}\right)$ iff both of the following hold:
 * $(1): \quad Y$ is closed in $\left({X, \tau}\right)$.
 * $(2): \quad \left({Y, \preceq \restriction_Y}\right)$ is a complete lattice, where $\restriction$ denotes restriction.

Sufficient Conditions
Suppose $Y$ is not a complete lattice.

Then there is an $S \subseteq Y$ with no least upper bound in $Y$.

Let $U = \{ b \in Y: b \text{ is an upper bound of } S \}$.

Note that $U$ may be empty.

Let $\mathcal U = \{ {\uparrow}u: u \in U \}$

Let $\mathcal L = \{ {\downarrow}s: s \in S \}$.

Then $\mathcal U \cup \mathcal L$ covers $Y$:

Let $y \in Y$.

If $y$ is an upper bound for $S$, then since $S$ has no least upper bound in $Y$, there is a $y' \in U$ such that $y' \prec y$.

Thus $y \in {\uparrow y'}$, so $y \in \bigcup \mathcal U$.

Otherwise, there is an element of $s \in S$ such that $y \prec s$, so $y \in \bigcup \mathcal L$.

$\mathcal U \cup \mathcal L$ has no finite subcover:

Let $F_U$ and $F_L$ be finite subsets of $U$ and $S$, respectively.

Define:
 * $\mathcal F_U = \{ {\uparrow}u: u \in F_S \}$
 * $\mathcal F_L = \{ {\downarrow}s: s \in F_L \}$

We will show that $\mathcal F_U \cup \mathcal F_L$ does not cover $Y$.

Suppose to the contrary that it does. Since $Y$ is non-empty, $F_L$ or $F_U$ is non-empty.

If $F_L$ is non-empty, it has a maximum $m$.

Since $S$ has no least upper bound, it has no maximum, so there is an $s \in S$ such that $m \prec s$.

Thus $s \notin \bigcup \mathcal F_L$.

By the definition of upper bound, no upper bound of $S$ precedes $m$, so $m \notin \bigcup \mathcal F_U$.

Thus $\mathcal F_L \cup \mathcal F_U$ does not cover $Y$, a contradiction.

If $F_U$ is non-empty, it has a minimum $m$.

Then since $S$ has no least upper bound in $Y$, there is a $u \in U$ such that $u \prec m$.

Thus $u \notin \bigcup \mathcal F_U$.

But since $u$ is an upper bound of $S$, it is not succeeded by any element of $S$, so $u \notin \bigcup \mathcal F_L$.

Thus $\mathcal \F_L \cup \mathcal F_U$ does not cover $Y$, a contradiction.

So $\mathcal U \cup \mathcal L$ is an open cover of $Y$ with no finite subcover, so $Y$ is not compact.

Suppose instead that $Y$ is not closed in $X$.

Since $X$ is a linearly ordered space, it is a Hausdorff space.

Compact Subspace of Hausdorff Space is Closed, so $Y$ is not compact.

Also see

 * Heine–Borel Theorem: Special Case
 * Connected Subspace of Linearly Ordered Space