Closure of Complement of Closure is Regular Closed

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

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

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

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

Then $A^{-'-}$ is regular closed.

Proof
Let $A^\circ$ denote the interior of $A$.

From Set is Closed iff Equals Topological Closure, $A^-$ is closed in $T$.

Since $A^-$ is a closed set, $A^{-'}$ is open.

Therefore:


 * $\forall x \in A^{-'}: \exists \epsilon \in \R_{>0}$

Additionally, because $A^{-'}$ is the complement of $A^-$ relative to $S$, we have:


 * $\exists \epsilon \in \R_{>0}: \map {V_\epsilon} x \cap A^- = \O \iff x \in A^{-'}$

Therefore, for $y: y \in A^{-'-}$ and $ y \notin A^{- '}$ we know that:
 * $\forall \epsilon \in \R_{>0}: \map {V_\epsilon} y \cap A^- \ne \O$

meaning that $y$ is a limit point of $A^-$.

Therefore:
 * $y \notin A^{-'-\circ}$

Now:
 * $x \in A^{-'-\circ} \iff x \in A^{-'}$

In conclusion, this means that:
 * $A^{-'-\circ-} = A^{-'-},$ verifying the definition of $A^{-'-}$ being regular closed.