Nowhere Dense iff Complement of Closure is Everywhere Dense/Corollary

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

Let $H \subseteq S$. Let $H$ be a closed set of $T$.

Then $H$ is nowhere dense in $T$ $S \setminus H$ is everywhere dense in $T$.

Proof
From Closed Set equals its Closure, $H$ is closed in $T$ :
 * $H = H^-$

where $H^-$ is the closure of $H$.

The result then follows directly from Nowhere Dense iff Complement of Closure is Everywhere Dense.