Boundary of Set is Closed

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Theorem

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

Let $\partial H$ denote the boundary of $H$.


Then $\partial H$ is closed in $T$.


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 $H$

From Topological Closure is Closed, both $H^-$ and $\paren {T \setminus H}^-$ are closed in $T$.

From Topology Defined by Closed Sets, the intersection of arbitrarily many (in particular two) closed sets of $T$ is a closed set of $T$.

As $\partial H$ is the intersection of $H^-$ and $\paren {T \setminus H}^-$ the result follows.

$\blacksquare$


Sources