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 a linear continuum.

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 $b$.

In fact, if $y$ is any point not in $U$, it must be an bound for $S$:

Because, if it were not bounded above by $y$, there would be some $p \in S$, such that $y < p$.

But $p \in S$ means by definition that $\left[a,p\right] \subseteq U$.

Now, $a < y < p$, so $y \in \left[a,p\right]$ and so $y \in U$, which was not the case by assumption.

So any $y \notin U$ is an upper bound for $S$, and so in particular $b$ is.

So $S$ has a supremum $m$, as a bounded above, non-empty set has a supremum. This is part of the definition of linear continuum, namely being Dedekind complete.

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 $m \in S$ implies that $[a,m] \subseteq U$, and we also have that $[m,q] \subseteq [m,p) \subseteq U$, so $[a, q] \subseteq U$.

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

Thus $m \notin S$.

So $[a,m] \nsubseteq U$ (or else $m \in S$ by definition). So there is some element $w \in X$ that is $[a,m]$ and that is not in $U$, and so lies in $V$, as these two sets cover $X$.

As already proved above, $w \notin U$ implies it is an upper bound for $S$.

But then $m$ is the minimal one upper bound by the definition of supremum, so $w < m$ is impossible.

We can only have $w = m$. So $m \notin U$, so by necessity, $m \in V$.

Then as $V$ is open, we have some $w' < m$ such that $(w',m] \subseteq V$, and again by being close packed we have some $w$ with $w' < w < m$.

This $w''$ is in $V$, so not in $U$, so is an upper bound for $S$ again, and strictly smaller than $m$. Contradiction.

This shows that the separation cannot exist.

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 lower closure of $a$
 * $b^\prec$ denotes the strict lower closure of $b$
 * $b^\succeq$ denotes the upper closure of $b$
 * $a^\succ$ denotes the 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.