Union of Interiors and Boundary Equals Whole Space
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $T$.
Then:
- $S = \Int A \cup \partial A \cup \Int {A'}$
where:
- $A' = S \setminus A$ denotes the complement of $A$ relative to $S$
- $\Int A$ denotes the interior of $A$
- $\partial A$ denotes the boundary of $A$.
Proof
\(\ds \Int A \cup \partial A \cup \Int {A'}\) | \(=\) | \(\ds \Int A \cup \Int {A'} \cup \partial A\) | Union is Associative, Union is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \Int A \cup \Int {A'} \cup \paren {\map \cl A \cap \map \cl {A'} }\) | Boundary is Intersection of Closure with Closure of Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\Int A \cup \Int {A'} \cup \map \cl A} \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }\) | Intersection Distributes over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\Int A \cup \paren {\map \cl A}' \cup \map \cl A} \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }\) | Complement of Interior equals Closure of Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\Int A \cup \paren {\paren {\map \cl A}' \cup \map \cl A} } \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }\) | Union is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\Int A \cup S} \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }\) | Union with Relative Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\Int A \cup S} \cap \paren {\Int A \cup \Int {A'} \cup \paren {\Int A}'}\) | Complement of Interior equals Closure of Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\Int A \cup S} \cap \paren {\Int A \cup \paren {\Int A' \cup \Int {A'} } }\) | Union is Associative, Union is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\Int A \cup S} \cap \paren {S \cup \Int {A'} }\) | Union with Relative Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds S \cap \paren {S \cup \Int {A'} }\) | Union with Superset is Superset | |||||||||||
\(\ds \) | \(=\) | \(\ds S \cap S\) | Union with Superset is Superset | |||||||||||
\(\ds \) | \(=\) | \(\ds S\) | Set Intersection is Idempotent |
$\blacksquare$
Sources
- Mizar article TOPGEN_1:13