Complete Linearly Ordered Space is Compact

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Theorem

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

Let $\left({X, \preceq}\right)$ be a complete lattice.


Then $\left({X, \tau}\right)$ is compact.


Proof

By Compactness from Basis, it is sufficient to prove that an open cover of $X$ consisting of open intervals and rays has a finite subcover.

Let $\mathcal A$ be an open cover of $X$ consisting of open rays and open intervals.

Let $m = \inf X$. This infimum exists because $\left({X, \preceq}\right)$ is complete.

Let $C$ be the set of all $x \in X$ such that a finite subset of $\mathcal A$ covers $\left[{m \,.\,.\, x}\right]$.

$C$ is non-empty because $m \in C$.

Let $s = \sup C$.

Since $\mathcal A$ covers $X$, there is a $U \in \mathcal A$ such that $s \in U$.

Then we must have $U = \left({a \,.\,.\, b}\right)$, $U = {\dot\uparrow} a$, or $U = {\dot\downarrow} b$.

Suppose that $U = \left({a \,.\,.\, b}\right)$.

Let $V \in \mathcal U$ contain $b$.

Then by the definition of supremum, there is an $x \succ a$ such that there is a finite $\mathcal F \subseteq \mathcal A$ that covers $\left[{m \,.\,.\, x}\right]$.

Then $\mathcal F \cup \left\{{U, V}\right\}$ covers $\left[{m \,.\,.\, b}\right]$, contradicting the fact that $s$ is an upper bound of $C$.


Suppose next that $U = \dot\downarrow b$.

Then for some $V \in \mathcal A$, $b \in V$.

Then $\left[{m \,.\,.\, b}\right]$ is covered by $\left\{{U, V}\right\}$, contradicting the fact that $s$ is the supremum of $C$.


Thus $U = \dot\uparrow a$.

By the definition of supremum, $a$ is not an upper bound of $C$.

So there is an $x \succ a$ such that there is a finite subset $\mathcal F$ of $\mathcal A$ that covers $\left[{m \,.\,.\, x}\right]$.

Thus $\mathcal F \cup \left\{{U}\right\}$ is a finite subcover of $A$.

$\blacksquare$


Sources

1955: John L. Kelley: General Topology: $\S 1$: Problem $\text I$, $\S 5$: Problem $\text C$