Compact Subspace of Linearly Ordered Space
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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 1, $S$ has a supremum $k$ in $Y$.
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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.
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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 $\text C$