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

\(\displaystyle B^- \cap A\) \(=\) \(\displaystyle B^- \cap \paren {U \cap A}\) Intersection with Subset is Subset
\(\displaystyle \) \(=\) \(\displaystyle \paren {B^- \cap U} \cap A\) Intersection is Associative
\(\displaystyle \) \(=\) \(\displaystyle \O \cap A\) as $B^- \cap U = \O$
\(\displaystyle \) \(=\) \(\displaystyle \O\) Intersection with Empty Set


Similarly:

$A^- \cap B = \O$

$\blacksquare$