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

$\blacksquare$


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