Boundary of Set is Closed

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

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

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

Proof
From Boundary is Intersection of Closure with Closure of Complement:
 * $\partial H = H^- \cap \left({T \setminus H}\right)^-$

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

From Closure is Closed, both $H^-$ and $\left({T \setminus H}\right)^-$ are closed in $T$.

From Topology Defined by Closed Sets, the intersection of arbitrarily many (in particular $2$) closed sets of $T$ is a closed set of $T$.

As $\partial H$ is the intersection of $H^-$ and $\left({T \setminus H}\right)^-$ the result follows.