Separated Subsets of Linearly Ordered Space under Order Topology

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

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

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

Proof
From the lemma:


 * $A \subseteq A^*$


 * $B \subseteq B^*$


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

Let $p \notin A^* \cup A^-$.

Thus $p \notin A^*$ and $p \notin A^-$.

Then there exists an open interval $\openint s t$ which is disjoint from $A$ such that $p \in \openint s t$.

Now $\openint s t$ can only intersect $A^*$ only if it intersects some $\closedint a b \subseteq A^*$ where $a, b \in A$.

But we have:
 * $\openint s t \cap A = \O$

and as $a, b \in A$ it follows that:
 * $\openint s t \subseteq \openint a b$

That means $p \in A^*$.

But we have $p \notin A^*$.

Therefore:
 * $\openint s t \cap A^* = \O$

Thus:
 * $p \notin \paren {A^*}^-$

Hence:

Hence the result.