# Set is Closed iff it Contains its Boundary

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## 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:

- $\partial H = H^- \cap \left({T \setminus H}\right)^-$

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

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 1$: Closures and Interiors