Set is Open iff Disjoint from Boundary

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

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

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

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

where $\operatorname{cl} \left({H}\right)$ is the closure of $H$.

Hence from Intersection Subset we have that:
 * $\partial H \subseteq \operatorname{cl} \left({T \setminus H}\right)$

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

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

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

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