Equivalence of Definitions of Separated Sets
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B \subseteq S$.
The following definitions of the concept of Separated Sets are equivalent:
Definition 1
$A$ and $B$ are separated (in $T$) if and only if:
- $A^- \cap B = A \cap B^- = \O$
where:
Definition 2
$A$ and $B$ are separated (in $T$) if and only if there exist $U,V\in\tau$ with:
- $A \subset U$ and $U \cap B = \O$
- $B \subset V$ and $V \cap A = \O$
where $\O$ denotes the empty set.
Proof
Definition 1 implies Definition 2
Let $A, B \subseteq S$ satisfy:
- $A^- \cap B = A \cap B^- = \O$
where $A^-$ denotes the closure of $A$ in $T$, and $\O$ denotes the empty set.
From Topological Closure is Closed, $B^-$ is closed in $T$.
Let $U = S \setminus B^-$ be the relative complement of $B^-$.
By the definition of a closed set, $U$ is open in $T$.
From Empty Intersection iff Subset of Relative Complement:
- $A \subseteq S \setminus B^- = U$
From Relative Complement of Relative Complement:
- $S \setminus U = B^-$
By the definition of the closure of a subset:
- $B \subseteq B^- = S \setminus U$
From Empty Intersection iff Subset of Relative Complement:
- $U \cap B = \O$
Similarly, let $V = S \setminus A^-$ then $V \in \tau$ with:
- $B \subset V$
and
- $V \cap A = \O$
$\Box$
Definition 2 implies Definition 1
Let $A, B \subseteq S$.
Let $U, V \in \tau$ satisfy:
- $A \subset U$ and $U \cap B = \O$
- $B \subset V$ and $V \cap A = \O$
Let $U \in \tau$ be an arbitrary open set of $T$ such that $A \subseteq U$ and $U \cap B = \O$.
From Empty Intersection iff Subset of Relative Complement:
- $B \subseteq S \setminus U$
By the definition of a closed set, the relative complement $S \setminus U$ of $U$ is closed in $T$.
From Set Closure is Smallest Closed Set in Topological Space:
- $B^- \subseteq S \setminus U$
From Empty Intersection iff Subset of Relative Complement:
- $B^- \cap U = \O$
Then
| \(\ds B^- \cap A\) | \(=\) | \(\ds B^- \cap \paren {U \cap A}\) | Intersection with Subset is Subset | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {B^- \cap U} \cap A\) | Intersection is Associative | |||||||||||
| \(\ds \) | \(=\) | \(\ds \O \cap A\) | as $B^- \cap U = \O$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \O\) | Intersection with Empty Set |
Similarly:
- $A^- \cap B = \O$
$\blacksquare$