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:

\(\ds H\)

\(\subseteq\)

\(\ds H^-\)

\(\ds \left({S \setminus H}\right)\)

\(\subseteq\)

\(\ds \left({S \setminus H}\right)^-\)

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.

$\blacksquare$