Separated Subsets of Linearly Ordered Space under Order Topology

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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 $T$.


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:

\(\ds \paren {A^*}^-\) \(\subseteq\) \(\ds \paren {A^* \cup A^-} \cap B^*\)
\(\ds \) \(=\) \(\ds \paren {A^* \cap B^*} \cup \paren {A^- \cap B^*}\)
\(\ds \) \(=\) \(\ds \O\)

Hence the result.


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$\blacksquare$


Sources

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