Equivalence of Definitions of Separated Sets/Definition 2 implies Definition 1
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Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $A, B \subseteq S$.
Let $U, V \in \tau$ satisfy:
- $A \subset U$ and $U \cap B = \O$
- $B \subset V$ and $V \cap A = \O$
where $\O$ denotes the empty set.
Then
- $A^- \cap B = A \cap B^- = \O$
where $A^-$ denotes the closure of $A$ in $T$.
Proof
Let $U \in \tau$ be an arbitrary open set of $T$ such that $A \subseteq U$ and $U \cap B = \O$.
From Empty Intersection iff Subset of Relative Complement:
- $B \subseteq S \setminus U$
By the definition of a closed set, the relative complement $S \setminus U$ of $U$ is closed in $T$.
From Set Closure is Smallest Closed Set in Topological Space:
- $B^- \subseteq S \setminus U$
From Empty Intersection iff Subset of Relative Complement:
- $B^- \cap U = \O$
Then
\(\ds B^- \cap A\) | \(=\) | \(\ds B^- \cap \paren {U \cap A}\) | Intersection with Subset is Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {B^- \cap U} \cap A\) | Intersection is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \O \cap A\) | as $B^- \cap U = \O$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \O\) | Intersection with Empty Set |
Similarly:
- $A^- \cap B = \O$
$\blacksquare$