Nowhere Dense iff Complement of Closure is Everywhere Dense

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

Let $H \subseteq S$.

Then $H$ is nowhere dense in $T$ $S \setminus H^-$ is everywhere dense in $T$, where $H^-$ denotes the closure of $H$.

Proof
Let:
 * $H^\circ$ denote the interior of any $H \subseteq S$
 * $H^-$ denote the closure of any $H \subseteq S$.

From the definition, $H$ is nowhere dense in $T$ $\left({H^-}\right)^\circ = \varnothing$.

From the definition of interior, it follows that $\left({H^-}\right)^\circ = \varnothing$ every open set of $T$ contains a point of $S \setminus \left({H^-}\right)$.

Thus $S \setminus \left({H^-}\right)$ is everywhere dense by definition.