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

Theorem
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.

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

Let $A^*$ and $B^*$ be defined as:


 * $A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$


 * $B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$

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

Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$:
 * $\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$

where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.

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