Connected Subspace of Linearly Ordered Space

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

Let $Y \subseteq S$.

Then $Y$ is connected in $\left({S, \tau}\right)$ both of the following hold:


 * $(1): \quad Y$ is convex in $S$
 * $(2): \quad \left({Y, \preceq \restriction_Y}\right)$ is a linear continuum, where $\restriction$ denotes restriction.

Necessary Conditions
Let $Y$ be connected in $\left({S, \tau}\right)$.

$Y$ is not convex in $S$.

Then there exist $a, b, c \in S$ such that:
 * $a \prec b \prec c$
 * $a, c \in Y$ but $b \notin Y$

Recall that:
 * $b^\prec$ denotes the (strict) lower closure of $b$: $b^\prec = \left\{ {u \in S: u \prec b}\right\}$
 * $b^\succ$ denotes the (strict) upper closure of $b$: $b^\succ = \left\{ {u \in S: b \prec u}\right\}$

We have that $Y \cap b^\prec$ and $Y \cap b^\succ$ are separated in $Y$.

Hence is disconnected.

Hence by Proof by Contradiction it follows that $Y$ is convex in $S$.

$Y$ is convex in $S$, but not a linear continuum.

Then by Order Topology on Convex Subset is Subspace Topology, the subspace topology on $Y$ is the same as the order topology on $Y$.

Thus by Linearly Ordered Space is Connected iff Linear Continuum, $Y$ is disconnected.

Hence by Proof by Contradiction it follows that $Y$ is a linear continuum.

Sufficient Condition
Suppose that $Y$ is convex in $S$ and a linear continuum.

Then the result follows from:
 * Order Topology on Convex Subset is Subspace Topology
 * Linearly Ordered Space is Connected iff Linear Continuum.

Also see

 * Subset of Real Numbers is Interval iff Connected
 * Compact Subspace of Linearly Ordered Space