Equivalence of Definitions of Separated Sets/Definition 1 implies Definition 2

Theorem

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

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.

Then there exist $U,V \in \tau$ with:

$A \subset U$ and $U \cap B = \O$

$B \subset V$ and $V \cap A = \O$

Proof

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

$V \cap A = \O$

$\blacksquare$