Union of Closure with Closure of Complement is Whole Space

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

Let $H \subseteq S$ be a subset of $S$.

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

Let $S \setminus H$ denote the complement of $H$ relative to $S$.

Then:
 * $H^- \cup \left({S \setminus H}\right)^- = S$

Proof
We have that:
 * $H^- \cup \left({S \setminus H}\right)^- \subseteq S$

by definition of $S$.

From Union with Relative Complement:
 * $H \cup \left({S \setminus H}\right) = S$

From Set is Subset of its Topological Closure:

From Set Union Preserves Subsets:
 * $H \subseteq H^-, \left({S \setminus H}\right) \subseteq \left({S \setminus H}\right)^- \implies H \cup \left({S \setminus H}\right) \subseteq H^- \cup \left({S \setminus H}\right)^-$

which means:
 * $S \subseteq H^- \cup \left({S \setminus H}\right)^-$

The result follows by definition of equality of sets.