Connected Subspace of Linearly Ordered Space

Theorem
Let $\left({X, \preceq}\right)$ be a totally ordered set, equipped with its order topology so that it is considered as a topological space.

Then a topological subspace $Y \subseteq X$ is connected iff all of the following hold:
 * $\left({1}\right): \quad$ $\left({Y, \preceq \restriction_{Y \times Y}}\right)$ is densely ordered, where $\restriction$ denotes restriction.
 * $\left({2}\right): \quad$ $\left({Y, \preceq \restriction_{Y \times Y}}\right)$ is Dedekind complete.
 * $\left({3}\right): \quad$ $\forall x, y \in Y: \left[{x \, . \, . \, y}\right] \subseteq Y$, where $\left[{x \, . \, . \, y}\right]$ denotes a closed interval in $X$.

Also see

 * Only Intervals are Connected
 * Compact Subspace of Totally Ordered Set