Complete Linearly Ordered Space is Compact
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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 $A$.
$\blacksquare$
Sources
- 1955: John L. Kelley: General Topology: $\S 1$: Problem $\text I$, $\S 5$: Problem $\text C$