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
- $A \cap B \subseteq A \land A \cap B \subseteq B$
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
- Mizar article TOPS_1:32