User:Dfeuer/Compact Subspace of Linearly Ordered Space strengthened
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Theorem
Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space.
Let $Y \subseteq X$ be a nonempty subset of $X$.
Then if $Y$ is a compact subspace of $\left({X, \tau}\right)$ then both of the following hold:
- $(1): \quad$ Every non-empty subset of $Y$ has a supremum and infimum in $X$.
- $(2): \quad$ For each non-empty $S \subseteq Y$, $\sup S \in Y$ and $\inf S \in Y$.
Proof
$(1)$ holds by Compact Subspace of Linearly Ordered Space.
Since a linearly ordered space is Hausdorff, and a compact subset of a Hausdorff space is closed, $Y$ is closed in $X$.
Suppose that for some non-empty $S \subseteq Y$, $\sup S \notin Y$.
Then for every $x < \sup S$, $\left({{x}\,.\,.\,{\sup S}}\right] \cap Y \ne \varnothing$.
Thus $\sup S$ is a limit point of $Y$ not contained in it, contradicting the fact that $Y$ is closed.