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 Condition
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 Condition
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:
 * $X = a^\preceq \cup b^\succeq$

and the components of this union are disjoint.

By Mind the Gap:
 * $a^\preceq = b^\prec$
 * $b^\succeq = a^\succ$

where:
 * $a^\preceq$ denotes the Definition:Lower Closure of Element|lower closure]] of $a$
 * $b^\prec$ denotes the Definition:Strict Lower Closure of Element|strict lower closure]] of $b$
 * $b^\succeq$ denotes the Definition:Upper Closure of Element|upper closure]] of $b$
 * $a^\succ$ denotes the Definition:Strict Upper Closure of Element|strict upper closure]] of $a$.

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 supremum in $X$.

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

Since $S$ has no supremum, $U$ 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:
 * $p \in s^\prec \subset X \setminus U$

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.