Disjoint Open Sets remain Disjoint with one Closure

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

Let $A, B$ be disjoint open sets of $T$.

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

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

Proof
Since $B \in \tau$, it follows by definition that $S \setminus B$ is closed.

Thus: