Set is Open iff Disjoint from Boundary

Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.

Then $H$ is open in $T$ if and only if:

$\partial H \cap H = \O$

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

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 $H$.

Hence from Intersection is Subset we have that:

$\partial H \subseteq \paren {T \setminus H}^-$

But from Closed Set Equals its Closure, $\paren {T \setminus H}^- = T \setminus H$ if and only if $T \setminus H$ is closed in $T$.

That is, if and only if $H$ is open in $T$.

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

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

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors