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 $U \cap B = \empty$
 * $B \subset V$ and $V \cap A = \empty$

where $\empty$ denotes the empty set.

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

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

Proof
From Empty Intersection iff Subset of Relative Complement, $B \subseteq S \setminus U$.

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

From Set Closure is Smallest Closed Set, $B^- \subseteq S \setminus U$.

From Empty Intersection iff Subset of Relative Complement, $B^- \cap U = \empty$.

Then

Similarly, $A^- \cap B = \empty$.