# Compact Subspace of Linearly Ordered Space/Lemma 1

## Theorem

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

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

Let $Y$ be a compact subspace of $\struct {X, \tau}$.

Then $\struct {Y, \preceq \restriction_Y}$ 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 $U = \set {b \in Y: b \text { is an upper bound of } S}$.

Note that $U$ may be empty.

Let:

- $\UU = \set {u^\succeq: u \in U}$
- $\LL = \set {s^\preceq: s \in S}$.

where:

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

Then $\UU \cup \LL$ covers $Y$:

Let $y \in Y$.

Suppose $y$ is an upper bound for $S$.

Then because $S$ has no least upper bound in $Y$, there exists a $y' \in U$ such that $y' \prec y$.

Thus $y \in y'^\succeq$, so, in the case where $y$ is an upper bound for $S$:

- $\ds y \in \bigcup \UU$

Otherwise, that is if $y$ is *not* an upper bound for $S$, there exists an $s \in S$ such that $y \prec s$.

Hence in this case:

- $\ds y \in \bigcup \LL$

$\UU \cup \LL$ has no finite subcover:

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

Define:

- $\FF_U = \set {u^\succeq: u \in F_S}$
- $\FF_L = \set {s^\preceq: s \in F_L}$

We will show that $\FF_U \cup \FF_L$ does not cover $Y$.

Suppose to the contrary that $\FF_U \cup \FF_L$ 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 $\ds s \notin \bigcup \FF_L$.

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

So $\ds m \notin \bigcup \FF_U$.

Thus $\FF_L \cup \FF_U$ does not cover $Y$.

This is a contradiction of our supposition that it does.

Hence it cannot be the case that $\FF_U \cup \FF_L$ is a cover for $Y$.

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 $\ds u \notin \bigcup \FF_U$.

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

So $\ds u \notin \bigcup \FF_L$.

Thus $\FF_L \cup \FF_U$ does not cover $Y$, a contradiction.

So $\UU \cup \LL$ 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 $\struct {Y, \preceq \restriction_Y}$ is a complete lattice.

$\blacksquare$