Closure of Complement of Closure is Regular Closed

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

Let $A \subseteq S$ be a subset of $T$.

Let $A^-$ denote the closure of $A$ in $T$.

Let $A^\prime$ denote the complement of $A$ in $S$: $A^\prime = S \setminus A$.

Then $A^{- \, \prime \, -}$ is regular closed.