Set is Closed iff it Contains its Boundary

Theorem

Let $T$ be a topological space, and let $H \subseteq T$.


Then $H$ is closed in $T$ if and only if:

$\partial H \subseteq H$

where $\partial H$ is the boundary of $H$.


Proof

From Boundary is Intersection of Closure with Closure of Complement:

$\partial H = H^- \cap \paren {T \setminus H}^-$

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

Hence from Intersection is Subset we have that:

$\partial H \subseteq H^-$

Then from Closed Set Equals its Closure, $H$ is closed in $T$ if and only if $H = H^-$.

Hence the result.

$\blacksquare$


Sources

  • 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 30 \ \text {(i)}$
  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
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