Set is Open iff Disjoint from Boundary

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

Then $H$ is open in $T$ :
 * $\partial H \cap H = \varnothing$

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 Subset we have that:
 * $\partial H \subseteq \left({T \setminus H}\right)^-$

But from Closed Set Equals its Closure, $\left({T \setminus H}\right)^- = T \setminus H$ $T \setminus H$ is closed in $T$.

That is, iff $H$ is open in $T$.

So $\partial H \subseteq T \setminus H$ $H$ is open in $T$.

From Intersection with Complement is Empty iff Subset it follows that $\partial H \cap H = \varnothing$ $H$ is open in $T$.