Boundary of Boundary is Contained in Boundary
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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