Separated Sets are Disjoint

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $A, B \subseteq S$ such that $A$ and $B$ are separated in $T$.


Then $A$ and $B$ are disjoint:

$A \cap B = \O$


Proof

Let $A$ and $B$ be separated in $T$.


Then:

\(\displaystyle A^- \cap B\) \(=\) \(\displaystyle \O\) Definition of Separated Sets: $A^-$ is the closure of $A$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {A \cup A'} \cap B\) \(=\) \(\displaystyle \O\) Definition of Set Closure: $A'$ is the derived set of $A$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {A \cap B} \cup \paren {A' \cap B}\) \(=\) \(\displaystyle \O\) Intersection Distributes over Union
\(\displaystyle \leadsto \ \ \) \(\displaystyle A \cap B\) \(=\) \(\displaystyle \O\) Union is Empty iff Sets are Empty

$\blacksquare$


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