# Complete Linearly Ordered Space is Compact

## Theorem

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.

Let $\struct {X, \preceq}$ be a complete lattice.

Then $\struct {X, \tau}$ is compact.

## Proof

By Space is Compact iff exists Basis such that Every Cover has Finite Subcover, it is sufficient to prove that an open cover of $X$ consisting of open intervals and rays has a finite subcover.

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

Let $m = \inf X$.

This infimum exists because $\struct {X, \preceq}$ is complete.

Let $C$ be the set of all $x \in X$ such that a finite subset of $\AA$ covers $\closedint m x$.

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

Let $s = \sup C$.

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

Then we must have $U = \openint a b$, $U = {\dot\uparrow} a$, or $U = {\dot\downarrow} b$.

Suppose that $U = \openint a b$.

Let $V \in \UU$ contain $b$.

Then by the definition of supremum, there is an $x \succ a$ such that there is a finite $\FF \subseteq \AA$ that covers $\closedint m x$.

Then $\FF \cup \set {U, V}$ covers $\closedint m b$, contradicting the fact that $s$ is an upper bound of $C$.

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

Then for some $V \in \AA$, $b \in V$.

Then $\closedint m b$ is covered by $\set {U, V}$, 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 $\FF$ of $\AA$ that covers $\closedint m x$.

Thus $\FF \cup \set U$ is a finite subcover of $A$.

$\blacksquare$