Compact Subspace of Linearly Ordered Space

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Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space.

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

Then $Y$ is a compact subspace of $\left({X, \tau}\right)$ 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 $S \subseteq Y$: $\sup S, \inf S \in Y$.


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 $b \prec k$.


$\mathcal A = \left\{{s^\preceq: s \in S}\right\} \cup \left\{{b^\succeq}\right\}$


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

Then $\mathcal A$ is an open cover of $Y$.

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

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


Reverse Implication

Let $\mathcal F$ be an ultrafilter on $Y$.

For $S \in \mathcal F$, let $f(S) = \inf S$.

Let $p = \sup f(\mathcal F)$.

Then $\mathcal F$ converges to $p$:

Upward rays

Let $a \in X$ with $a \prec p$.

Since $\mathcal F$ is an ultrafilter, either $Y \cap {\uparrow}a \in \mathcal F$ or $Y \cap {\bar\downarrow}a \in \mathcal F$.

Suppose for the sake of contradiction that $Y \cap {\bar\downarrow}a \in \mathcal F$.

For each $S \in \mathcal F$:

$S \cap {\bar\downarrow}a \in \mathcal F$ because an ultrafilter is a filter.
$S \cap {\bar\downarrow}a \ne \varnothing$ because a filter on a set is proper.

By applying the definition of supremum to $p$, it follows that there exists an $S \in \mathcal F$ such that $a \prec \inf S$.

By the definition of infimum, $S \cap {\bar\downarrow}a = \varnothing$, a contradiction.

Thus $Y \cap {\uparrow}a \in \mathcal F$.

Downward rays

Let $b \in X$ with $p \prec b$.

Either $Y \cap {\downarrow} b \in \mathcal F$ or $Y \cap {\bar \uparrow} b \in \mathcal F$.

Suppose, for the sake of contradiction, that $Y \cap {\bar \uparrow} b \in \mathcal F$.

Let $b' = \inf \left({ Y \cap {\bar \uparrow} b}\right)$.

We have that $b$ is a lower bound of $Y \cap {\bar \uparrow} b$

So by the definition of infimum:

$b \preceq b'$

Since $p \prec b$ and $b \preceq b'$, $p \prec b'$ by Extended Transitivity.

By the definition of $b'$ and the definition of $f$:

$b' \in f \left({\mathcal F}\right)$

But this contradicts the fact that $p$ is the supremum, and hence an upper bound, of $f \left({\mathcal F}\right)$.


By the definition of the order topology, the upward and downward rays containing each point form a neighborhood sub-basis for that point.

Thus by the Neighborhood Sub-Basis Criterion for Filter Convergence, $\mathcal F$ converges.

Since every ultrafilter on $Y$ converges, $Y$ is compact by Equivalent Definitions of Compactness.

Also see