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 $$H = \operatorname{cl}\left({H}\right)$$.