Boundary of Union of Separated Sets equals Union of Boundaries
Theorem
Let $T$ be a topological space.
Let $A, B$ be subsets of $T$.
Let $A$ and $B$ are separated.
Then:
- $\map \partial {A \cup B} = \partial A \cup \partial B$
where:
Proof
By definition of separated sets:
- $(1): \quad A^- \cap B = A \cap B^- = \O$
By Separated Sets are Disjoint:
- $A \cap B = \O$
\(\ds \partial A \cup \partial B\) | \(=\) | \(\ds \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}\) | Union of Boundaries | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \partial {A \cup B}\cup \O \cup \paren {\partial A \cap \partial B}\) | Boundary of Empty Set is Empty | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \partial {A \cup B} \cup \paren {\partial A \cap \partial B}\) | Union with Empty Set |
We will prove that:
- $\partial A \cap \partial B \subseteq \map \partial {A \cup B}$
Let $x \in \partial A \cap \partial B$.
Then by definition of intersection:
- $x \in \partial A \land x \in \partial B$
Hence by Boundary is Intersection of Closure with Closure of Complement:
- $x \in A^- \cap \paren {\relcomp T A}^- \land x \in B^- \cap \paren {\relcomp T B}^-$
where:
- $A^-$ denotes the closure of $A$
- $\relcomp T A = T \setminus A$ denotes the relative complement of $A$ in $T$.
Then by definition of intersection:
- $x \in A^- \land x \in B^-$
Therefore by definition of empty set, intersection, and $(1)$:
- $x \notin B \land x \notin A$
Then by definition of union:
- $x \notin A \cup B$
By definition of relative complement:
- $x \in \relcomp T {A \cup B}$
By definition of closure:
- $\relcomp T {A \cup B} \subseteq \paren {\relcomp T {A \cup B} }^-$
Then by definition of subset:
- $x \in \paren {\relcomp T {A \cup B} }^-$
By Closure of Finite Union equals Union of Closures:
- $A^- \cup B^- = \paren {A \cup B}^-$
By definition of union:
- $x \in \paren {A \cup B}^-$
Then by definition of intersection:
- $x \in \paren {A \cup B}^- \cap \paren {\relcomp T {A \cup B} }^-$
Thus by Boundary is Intersection of Closure with Closure of Complement:
- $x \in \map \partial {A \cup B}$
This ends the proof of inclusion.
Thus by Union with Superset is Superset the result:
- $\partial A \cup \partial B = \map \partial {A \cup B}$
$\blacksquare$
Sources
- Mizar article TOPGEN_4:8