Nowhere Dense iff Complement of Closure is Everywhere Dense

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Theorem

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

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

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

$\blacksquare$


Warning

Note that in the result:

$H$ is nowhere dense in $T$ if and only if the relative complement of its closure is everywhere dense in $T$

must be applied to the closure of $H$.


Otherwise, consider the real numbers $\R$ and the rational numbers $\Q$.

We have that:

$\R \setminus \Q$ is the set of irrational numbers.

We have that Irrationals are Everywhere Dense in Reals.

But we also have from Rationals are Everywhere Dense in Reals:

$\map \cl \Q = \R$

and so:

$\paren {\map \cl \Q}^\circ = \R$

So it is not the case that $\R \setminus \Q$ is nowhere dense in $\R$.

However:

$\R \setminus \map \cl \Q = \O$

which is indeed nowhere dense in $\R$.


Sources