# Compact Subspace of Linearly Ordered Space/Lemma 2

## Lemma

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:

- $\struct {Y, \preceq', \tau'}$ is a linearly ordered space.

## Proof

By definition of a generalized ordered space:

- $\tau'$ has a sub-basis consisting of upper and lower sections in $Y$.

To prove that $\struct {Y, \preceq', \tau'}$ is a linearly ordered space, we need to show that each element of this sub-basis is open in the $\preceq'$ order topology.

Let $U$ be a non-empty, $\tau'$-open upper section in $Y$.

If $U = \O$ or $U = Y$, then $U$ is open in the $\preceq'$-order topology by the definition of topology.

Suppose then that $\O \subsetneqq U \subsetneqq Y$.

Let $C = Y \setminus U$.

Then $C$ is non-empty and $\tau'$-closed.

$Y$ is $\tau'$-closed by Closed Set in Linearly Ordered Space.

Thus $C$ is also $\tau'$-closed.

$C$ has a supremum in $X$ by the premise.

Let $c = \sup_X C$

Since $C$ is $\tau'$-closed, Closed Set in Linearly Ordered Space shows that $c \in C$.

Since $c \in C = Y \setminus U$ and $U$ is an upper section in $Y$:

- $c \prec u$ for all $u \in U$.

Furthermore, if $y \in Y$ and $c \prec y$, then by the definition of supremum, $y \notin C$, so $y \in U$.

Thus:

- $U = {c^\succeq}_Y$

where ${c^\succeq}_Y$ denotes the upper closure of $c$ in $Y$.

So $U$ is open in the $\preceq'$-order topology for $Y$.

The same argument shows that $\tau'$-open lower sections in $Y$ are open in the $\preceq'$-order topology.

Thus $\tau'$ is the $\preceq'$-order topology.

$\blacksquare$