Separated Sets are Disjoint

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:

\(\ds A^- \cap B\)

\(=\)

\(\ds \O\)

Definition of Separated Sets: $A^-$ is the closure of $A$

\(\ds \leadsto \ \ \)

\(\ds \paren {A \cup A'} \cap B\)

\(=\)

\(\ds \O\)

Definition of Set Closure: $A'$ is the derived set of $A$

\(\ds \leadsto \ \ \)

\(\ds \paren {A \cap B} \cup \paren {A' \cap B}\)

\(=\)

\(\ds \O\)

Intersection Distributes over Union

\(\ds \leadsto \ \ \)

\(\ds A \cap B\)

\(=\)

\(\ds \O\)

Union is Empty iff Sets are Empty

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): separated sets