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Let $(X, \leq)$ be a linearly ordered set and let $\mathcal{T}$ denote the order topology on $X$. Prove that $(X, \mathcal{T})$ is compact if and only if every nonempty set of $X$ has a greatest lower bound and a least upper bound.

I don't know how to apply the definition of compactness in this topology. Can anyone give an idea?

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For one direction, suppose that $A\subseteq X$ is a non-empty set with no least upper bound. Let $U$ be the set of upper bounds for $A$; show that $U$ is open. ($U$ may of course be empty.) Then show that

$$\{U\}\cup\{(\leftarrow,a):a\in A\}$$

is an open cover of $X$ with no finite subcover. You can use much the same idea if you have a non-empty set with no greatest lower bound.

For the other direction suppose that every non-empty subset of $X$ has a greatest lower bound and a least upper bound. In particular, $X$ has a least element $a$ and a greatest element $b$. Now imitate the proof that $[0,1]$ is compact to show that $X$ is compact.

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