# Equivalence of Definitions of Separated Sets/Definition 2 implies Definition 1

## 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 $U \cap B = \empty$
- $B \subset V$ and $V \cap A = \empty$

where $\empty$ denotes the empty set.

Then

- $A^- \cap B = A \cap B^- = \empty$

where $A^-$ denotes the closure of $A$ in $T$.

## Proof

From Empty Intersection iff Subset of Relative Complement, $B \subseteq S \setminus U$.

By the definition of a closed set, the relative complement of $S \setminus U$ is closed in $T$.

From Set Closure is Smallest Closed Set, $B^- \subseteq S \setminus U$.

From Empty Intersection iff Subset of Relative Complement, $B^- \cap U = \empty$.

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 | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \empty \cap A\) | As $B^- \cap U = \empty$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \empty\) | Intersection with Empty Set |

Similarly, $A^- \cap B = \empty$.

$\blacksquare$