Linearly Ordered Space is Compact iff Complete

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

Then $T$ is a compact space it is complete.

Necessary Condition
Let $T$ be a compact space.

Let $A \subseteq S$.

$A$ has no supremum.

Consider the sets:
 * $P_\alpha = \left\{ {x \in S: x \prec \alpha}\right\}$ for $\alpha \in A$
 * $B_\beta = \left\{ {x \in S: \beta \prec x}\right\}$ for $\beta$ an upper bound of $A$.

We have that $P_\alpha$ and $B_\beta$ cover $S$ but contain no finite subcover.

Thus $T$ is not a compact space.

Similarly if $A$ has no infimum.

It follows that $A$ has both a supremum and infimum.

As $A$ is arbitrary, it follows that $S$ is a complete ordered set.

Sufficient Condition
Let $T$ be a complete ordered set.

Let $\mathcal U$ be an open cover of $S$.

Let $a$ be the infimum of $S$.

Let $B$ be the set of the elements $y \in S$ for which $\left[{a \,.\,.\, y}\right)$ have a finite cover from elements of $\mathcal U$.

Let $\alpha$ be the supremum of $B$.

If $\alpha \in U \in \mathcal U$, then $U \subseteq B$.

So unless $\alpha$ is the supremum of $S$ itself, there exists an open interval $\left({x \,.\,.\, y}\right) \subseteq U$ such that $\alpha \in \left({x \,.\,.\, y}\right)$.

But we have that $\alpha$ be the supremum of $B$.

Hence $\left({x \,.\,.\, y}\right) = \varnothing$.

But that means $y \in B$ which is impossible.

Thus $B = S$.

That is, the set of elements of $S$ which have a finite subcover of $\mathcal U$ is $S$ itself.

Hence $T$ is a compact space by definition.