Separated Subsets of Linearly Ordered Space under Order Topology/Lemma

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Lemma

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 $T$.


Then:

$(1): \quad A \subseteq A^*$
$(2): \quad B \subseteq B^*$
$(3): \quad A^* \cap B^* = \O$


Proof

Let $a \in A$.

Then:

\(\ds \closedint a a\) \(=\) \(\ds \set a\)
\(\ds \leadsto \ \ \) \(\ds \closedint a a \cap B^-\) \(=\) \(\ds \O\) Definition of Separated Sets
\(\ds \leadsto \ \ \) \(\ds \closedint a a\) \(\subseteq\) \(\ds A^*\) Definition of $A^*$
\(\ds \leadsto \ \ \) \(\ds a\) \(\in\) \(\ds A^*\) Definition of $\closedint a a$
\(\ds \leadsto \ \ \) \(\ds A\) \(\subseteq\) \(\ds A^*\) Definition of Subset

Similarly, $B \subseteq B^-$.

$\Box$


Aiming for a contradiction, suppose $A^* \cap B^* \ne \O$.

Then:

$\exists p: p \in A^* \cap B^*$

Hence:

$\exists a, b \in A, c, d \in B: p \in \closedint a b \cap \closedint c d$

But because $A$ and $B$ are separated sets:

$c, d \notin A$

and:

$a, b \notin B$

and so:

$\closedint a b \cap \closedint c d = \O$

Thus $p \notin \closedint a b \cap \closedint c d$

It follows by Proof by Contradiction that $A^* \cap B^* = \O$.

$\blacksquare$


Sources

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