Boundary of Union is Subset of Union of Boundaries

Theorem

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

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

Then:

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

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

Proof

By Intersection is Subset:

$\relcomp S A \cap \relcomp S B \subseteq \relcomp S A \land \relcomp S A \cap \relcomp S B \subseteq \relcomp S B$

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

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

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

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

Thus

\(\ds \map \partial {A \cup B}\)

\(=\)

\(\ds \paren {\relcomp S {A \cup B} }^- \cap \paren {A \cup B}^-\)

Boundary is Intersection of Closure with Closure of Complement

\(\ds \)

\(=\)

\(\ds \paren {\relcomp S A \cap \relcomp S B}^- \cap \paren {A \cup B}^-\)

Complement of Union

\(\ds \)

\(=\)

\(\ds \paren {\relcomp S A \cap \relcomp S B}^- \cap \paren {A^- \cup B^-}\)

Closure of Finite Union equals Union of Closures

\(\ds \)

\(=\)

\(\ds \paren {\relcomp S A \cap \relcomp S B}^- \cap A^- \cup \paren {\relcomp S A \cap \relcomp S B}^- \cap B^-\)

Intersection Distributes over Union

\(\ds \)

\(\subseteq\)

\(\ds \partial A \cup \partial B\)

Set Union Preserves Subsets

$\blacksquare$

Sources

• Mizar article TOPS_1:33