# User:Dfeuer/Compact Subspace of Linearly Ordered Space strengthened

## 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.