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

Necessary Conditions
Suppose $Y$ is not convex in $X$. Then there are $a, b, c \in S$ such that:
 * $a \prec b \prec c$
 * $a,c \in Y$ but $b \notin Y$

Then $Y$ is separated by $Y \cap {\downarrow}b$ and $Y \cap {\uparrow}b$, and hence is disconnected.

Suppose $Y$ is convex 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.

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

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

Also see

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