Nowhere Dense iff Complement of Closure is Everywhere Dense

Theorem
Let $T = \struct {S, \tau}$ 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 $T$ $\paren {H^-}^\circ = \O$.

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

Thus $S \setminus \paren {H^-}$ is everywhere dense by definition.