Boundary is Intersection of Closure with Closure of Complement
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Theorem
Let $T$ be a topological space.
Let $X \subseteq T$.
Let $\partial X$ denote the boundary of $X$, defined as:
- $\partial X = X^- \setminus X^\circ$
Let $\overline X = T \setminus X$ denote the complement of $X$ in $T$.
Let $X^-$ denote the closure of $X$.
Then:
- $\partial X = X^- \cap \paren {\overline X}^-$
Proof
\(\ds \partial X\) | \(=\) | \(\ds X^- \setminus X^\circ\) | Definition 1 of Boundary (Topology): $X^\circ$ is the interior of $X$ | |||||||||||
\(\ds \) | \(=\) | \(\ds X^- \cap \overline {\paren {X^\circ} }\) | Set Difference as Intersection with Relative Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds X^- \cap \paren {\overline X}^-\) | Complement of Interior equals Closure of Complement |
$\blacksquare$
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