Boundary of Boundary is Contained in Boundary

Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.

Then:

$\map \partial {\partial H} \subseteq \partial H$

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:

$\partial B = B^- \cap \paren {T \setminus B}^-$

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:

$\map \partial {\partial H} \subseteq \partial H$

$\blacksquare$

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