Nowhere Dense iff Complement of Closure is Everywhere Dense/Corollary
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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 $3.7.29$