Compact Subspace of Linearly Ordered Space/Lemma

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Theorem

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

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

Let $Y$ be a compact subspace of $\left({X, \tau}\right)$.


Then $\left({Y, \preceq \restriction_Y}\right)$ is a complete lattice, where $\restriction$ denotes restriction.


Proof

Aiming for a contradiction, suppose $Y$ is not a complete lattice.

Then there is an $S \subseteq Y$ with no least upper bound in $Y$.

Let $U = \left\{{b \in Y: b \text{ is an upper bound of } S}\right\}$.

Note that $U$ may be empty.

Let:

$\mathcal U = \left\{ {u^\succeq: u \in U}\right\}$
$\mathcal L = \left\{ {s^\preceq: s \in S}\right\}$.

where:

$u^\succeq$ is the upper closure of $u$ in $U$
$s^\preceq$ is the lower closure of $s$ in $S$.

Then $\mathcal U \cup \mathcal L$ covers $Y$:

Let $y \in Y$.

If $y$ is an upper bound for $S$, then since $S$ has no least upper bound in $Y$, there is a $y' \in U$ such that $y' \prec y$.

Thus $y \in y'^\succeq$, so $\displaystyle y \in \bigcup \mathcal U$.

Otherwise, there is an $s \in S$ such that $y \prec s$, so $\displaystyle y \in \bigcup \mathcal L$.


$\mathcal U \cup \mathcal L$ has no finite subcover:

Let $F_U$ and $F_L$ be finite subsets of $U$ and $S$, respectively.

Define:

$\mathcal F_U = \left\{ {u^\succeq: u \in F_S}\right\}$
$\mathcal F_L = \left\{ {s^\preceq: s \in F_L}\right\}$

We will show that $\mathcal F_U \cup \mathcal F_L$ does not cover $Y$.


Suppose to the contrary that $\mathcal F_U \cup \mathcal F_L$ is a cover for $Y$.

Since $Y$ is non-empty, $F_L$ or $F_U$ is non-empty.

If $F_L$ is non-empty, it has a maximum $m$.

Since $S$ has no least upper bound, it has no maximum

Therefore there is an $s \in S$ such that $m \prec s$.

Thus $\displaystyle s \notin \bigcup \mathcal F_L$.

By the definition of upper bound, no upper bound of $S$ precedes $m$.

So $\displaystyle m \notin \bigcup \mathcal F_U$.

Thus $\mathcal F_L \cup \mathcal F_U$ does not cover $Y$, a contradiction.


If $F_U$ is non-empty, it has a minimum $m$.

Then since $S$ has no least upper bound in $Y$, there is a $u \in U$ such that $u \prec m$.

Thus $\displaystyle u \notin \bigcup \mathcal F_U$.

But since $u$ is an upper bound of $S$, it is not succeeded by any element of $S$.

So $\displaystyle u \notin \bigcup \mathcal F_L$.

Thus $\mathcal F_L \cup \mathcal F_U$ does not cover $Y$, a contradiction.


So $\mathcal U \cup \mathcal L$ is an open cover of $Y$ with no finite subcover

So $Y$ is not compact, which contradicts the supposition.

By Proof by Contradiction, it follows that $\left({Y, \preceq \restriction_Y}\right)$ is a complete lattice.

$\blacksquare$