Open Set Disjoint from Set is Disjoint from Closure

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $A, B \in \tau$ such that $A \cap B = \varnothing$.

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

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

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

Thus: