Separated Subsets of Linearly Ordered Space under Order Topology

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 $T$.

Then $A^*$ and $B^*$ are themselves separated sets of $T$.

Proof
From the lemma:


 * $A \subseteq A^*$


 * $B \subseteq B^*$


 * $A^* \cap B^* = \varnothing$