Boundary of Boundary is Contained in Boundary
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Theorem
Let $T$ be a topological space, and let $H \subseteq T$.
Then:
- $\partial \left({\partial H}\right) \subseteq \partial H$
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:
- $\partial B = B^- \cap \left({T \setminus B}\right)^-$
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:
- $\partial \left({\partial H}\right) \subseteq \partial H$
$\blacksquare$
Sources
- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology ... (previous) ... (next): $\text{I}: \ \S 1$: Closures and Interiors