Nowhere Dense iff Complement of Closure is Everywhere Dense

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Theorem

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

Let $H \subseteq S$.


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


Corollary

Let $H$ be a closed set of $T$.


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


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 $T$ if and only if $\left({H^-}\right)^\circ = \varnothing$.

From the definition of interior, it follows that $\left({H^-}\right)^\circ = \varnothing$ if and only if 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.

$\blacksquare$


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