# Compact Subspace of Linearly Ordered Space

## Theorem

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

Let $Y \subseteq X$ be a non-empty subset of $X$.

Then $Y$ is a compact subspace of $\struct {X, \tau}$ if and only if both of the following hold:

- $(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
- $(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

## Proof

### Forward Implication

Let $S$ be a non-empty subset of $Y$.

By Compact Subspace of Linearly Ordered Space: Lemma, $S$ has a supremum $k$ in $Y$.

Aiming for a contradiction, suppose that $S$ has an upper bound $b$ in $X$ such that $b \prec k$.

Let:

- $\AA = \set {s^\preceq: s \in S} \cup \set {b^\succeq}$

where:

- $s^\preceq$ denotes the lower closure of $s$ in $S$
- $b^\succeq$ denotes the upper closure of $s$ in $S$.

Then $\AA$ is an open cover of $Y$.

But $\AA$ has no finite subcover, contradicting the fact that $Y$ is compact.

A similar argument proves the corresponding statement for infima, so $(2)$ holds.

$\Box$

### Reverse Implication

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

Let $Y \subseteq X$ be a non-empty subset of $X$.

Let the following hold:

- $(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
- $(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

Then $Y$ is a compact subspace of $\struct {X, \tau}$.

## Also see

## Sources

- 1955: John L. Kelley:
*General Topology*: $\S 5$: Problem $\text C$