Boundary of Boundary is Contained in Boundary

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

Then:
 * $\partial \left({\partial H}\right) \subseteq \partial H$

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

That is, the boundary of the boundary of $H$ is contained in the boundary of $H$.

Proof
Let $B = \partial H$.

From Boundary of Set is Closed we have that $B$ is closed in $T$.

Let $B^-$ denote the closure of $B$.

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

and so from Intersection is Subset:
 * $\partial B \subseteq B^-$

But from Closed Set Equals its Closure:
 * $B = B^-$

and so:
 * $\partial B \subseteq B$

That is:
 * $\partial \left({\partial H}\right) \subseteq \partial H$