Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set

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

Let $A$ and $B$ be separated sets of $T$.

Let $A^*$ and $B^*$ be defined as:
 * $A^* := \displaystyle \bigcup \left\{ {\left[{a \,.\,.\, b}\right]: a, b \in A, \left[{a \,.\,.\, b}\right] \cap B^- = \varnothing}\right\}$
 * $B^* := \displaystyle \bigcup \left\{ {\left[{a \,.\,.\, b}\right]: a, b \in B, \left[{a \,.\,.\, b}\right] \cap A^- = \varnothing}\right\}$

where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.

Let $A^*$, $B^*$ and $\complement_S \left({A^* \cup B^*}\right)$ be expressed as the union of convex components of $S$:
 * $\displaystyle A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \complement_S \left({A^* \cup B^*}\right) = \bigcup C_\gamma$

where $\complement_S \left({X}\right)$ denotes the complement of $X$ with respect to $S$.

Then the set $M = \left\{ {A_\alpha, B_\beta, C_\gamma}\right\}$ inherits a linear ordering from $S$, and so is a linearly ordered set.