Linearly Ordered Space is Connected iff Linear Continuum

Theorem
Let $T = \left({S, \preceq, \tau}\right)$ be a linearly ordered space.

Then $S$ is a connected space it is a linear continuum.

Necessary
Suppose that $X$ is disconnected and close packed.

Then there are non-empty open sets $U$, $V$ that separate $X$.

Let $a \in U$ and $b \in V$.

, suppose that $a \prec b$.

Let $S = \left\{{p \in X: \left[{a \,.\,.\, p}\right] \subseteq U}\right\}$

$S$ contains $a$, so it is non-empty, and it is bounded above by $b$.

Suppose for the sake of contradiction that it has a supremum $m$.

First suppose that $m \in S$.

Then $m \in U$.

Since $U$ is open, there exists a $p \in X$ such that $\left[{m \,.\,.\, p}\right) \subseteq U$.

Since $X$ is close packed, it has an element $q$ strictly between $m$ and $p$.

Then $q \in S$, contradicting the fact that $m$ is an upper bound for $S$.

Thus $m \notin S$.

Then:
 * $\exists: w \in V: w \in \left[{a \,.\,.\, m}\right]$

But then $w$ is an upper bound for $S$, contradicting the minimality of $m$.

Thus $S$ is non-empty and bounded above, but has no supremum.

Therefore $X$ is not Dedekind complete.

Sufficient
Suppose that $X$ is not a linear continuum.

Then $X$ is not close packed or $X$ is not Dedekind complete.

Suppose first that $X$ is not close packed.

Then there are points $a,b \in X$ such that $a \prec b$ and no point lies strictly between $a$ and $b$.

Thus $X = {\bar\downarrow}a \cup {\bar\uparrow}b$, and the components of this union are disjoint.

By Mind the Gap:
 * ${\bar\downarrow}a = {\downarrow}b$
 * ${\bar\uparrow}b = {\uparrow}a$

Thus these two sets are open sets that separate $X$.

Therefore $X$ is disconnected.

Suppose next that $X$ is not Dedekind complete.

Then there is a non-empty set $S \subset X$ which is bounded above in $X$ but has no least upper bound in $X$.

Let $U$ be the set of upper bounds of $S$ (non-empty by assumption)

Since $S$ has no least upper bound, $U$ is open.

Let $A = X \setminus U$ be the set of points that are not upper bounds for $S$.

Let $p \in X \setminus U$.

Then there is an element of $s \in S$ such that $p \prec s$.

Then $p \in {\downarrow}s \subset X \setminus U$

Thus $X \setminus U$ is also open. It is non-empty because it contains all elements of $S$.

So $U$ and $X \setminus U$ are open sets separating $X$, so $X$ is disconnected.