Nowhere Dense iff Complement of Closure is Everywhere Dense

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

Then $H$ is nowhere dense in $T$ iff $T \setminus \operatorname{cl} \left({H}\right)$ is everywhere dense in $T$.

Corollary
A closed set $H$ of $T$ is nowhere dense in $T$ iff $T \setminus H$ is everywhere dense in $T$.

Proof
From the definition, $H$ is nowhere dense in $T$ iff $\operatorname{Int} \left({\operatorname{cl} \left({H}\right)}\right) = \varnothing$.

From the definition of interior, it follows that $\operatorname{Int} \left({\operatorname{cl} \left({H}\right)}\right) = \varnothing$ iff every open set of $T$ contains a point of $T \setminus \operatorname{cl} \left({H}\right)$.

But this is exactly the definition for $T \setminus \operatorname{cl} \left({H}\right)$ being everywhere dense.

Proof of Corollary
Follows directly from the above result and Closed Set Equals its Closure: $H$ is closed in $T$ iff $H = \operatorname{cl}\left({H}\right)$.