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, $S$ 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