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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $3$