# 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:

$A^-$ denotes the closure of $A$ in $T$
$\O$ denotes the empty set.

### Definition 2

$A$ and $B$ are separated (in $T$) if and only if there exist $U,V\in\tau$ with:

$A\subset U$ and $U\cap B = \varnothing$
$B\subset V$ and $V\cap A = \varnothing$

where $\varnothing$ 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 $T$.

$A \subseteq S \setminus B^- = U$
$S \setminus U = B^-$

By the definition of the closure of a subset:

$B \subseteq B^- = S \setminus U$
$U \cap B = \O$

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 $U \cap B = \O$.

$B \subseteq S \setminus U$

By the definition of a closed set, the relative complement $S \setminus U$ of $U$ is closed in $T$.

$B^- \subseteq S \setminus U$
$B^- \cap U = \O$

Then

 $\displaystyle B^- \cap A$ $=$ $\displaystyle B^- \cap \paren {U \cap A}$ Intersection with Subset is Subset $\displaystyle$ $=$ $\displaystyle \paren {B^- \cap U} \cap A$ Intersection is Associative $\displaystyle$ $=$ $\displaystyle \O \cap A$ as $B^- \cap U = \O$ $\displaystyle$ $=$ $\displaystyle \O$ Intersection with Empty Set

Similarly:

$A^- \cap B = \O$

$\blacksquare$