Nowhere Dense iff Complement of Closure is Everywhere Dense/Corollary

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.

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

From Closed Set equals its Closure, $H$ is closed in $T$ if and only if:

$H = H^-$

where $H^-$ is the closure of $H$.

The result then follows directly from Nowhere Dense iff Complement of Closure is Everywhere Dense.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Corollary $3.7.29$