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

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

Let $U, V \in \tau$ satisfy:
 * $A \subset U$ and $U \cap B = \O$
 * $B \subset V$ and $V \cap A = \O$

where $\O$ denotes the empty set.

Then
 * $A^- \cap B = A \cap B^- = \O$

where $A^-$ denotes the closure of $A$ in $T$.

Proof
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

Similarly:
 * $A^- \cap B = \O$