Boundary of Intersection is Subset of Union of Boundaries

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $A, B$ be subsets of $S$.

Then:

$\map \partial {A \cap B} \subseteq \partial A \cup \partial B$

where $\partial A$ denotes the boundary of $A$.


Proof

By Intersection is Subset:

$A \cap B \subseteq A \land A \cap B \subseteq B$

Then by Topological Closure of Subset is Subset of Topological Closure:

$\paren {A \cap B}^- \subseteq A^- \land \paren {A \cap B}^- \subseteq B^-$

Hence by Boundary is Intersection of Closure with Closure of Complement:

$\paren {A \cap B}^- \cap \paren {\relcomp S A}^- \subseteq \partial A \land \paren {A \cap B}^- \cap \paren {\relcomp S B}^- \subseteq \partial B$


Thus

\(\ds \map \partial {A \cap B}\) \(=\) \(\ds \paren {A \cap B}^- \cap \paren {\relcomp S {A \cap B} }^-\) Boundary is Intersection of Closure with Closure of Complement
\(\ds \) \(=\) \(\ds \paren {A \cap B}^- \cap \paren {\relcomp S A \cup \relcomp S B}^-\) Complement of Intersection
\(\ds \) \(=\) \(\ds \paren {A \cap B}^- \cap \paren {\paren {\relcomp S A}^- \cup \paren {\relcomp S B}^-}\) Closure of Finite Union equals Union of Closures
\(\ds \) \(=\) \(\ds \paren {A \cap B}^- \cap \paren {\relcomp S A}^- \cup \paren {A \cap B}^- \cap \paren {\relcomp S B}^-\) Intersection Distributes over Union
\(\ds \) \(\subseteq\) \(\ds \partial A \cup \partial B\) Set Union Preserves Subsets

$\blacksquare$


Sources