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

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## Theorem

Let $T = \struct{S, \tau}$ be a topological space.

Let $A, B \subseteq S$ satisfy:

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

where $A^-$ denotes the closure of $A$ in $T$, and $\O$ denotes the empty set.

Then there exist $U,V \in \tau$ with:

- $A \subset U$ and $U \cap B = \O$
- $B \subset V$ and $V \cap A = \O$

## Proof

From Topological Closure is Closed, $B^-$ is closed in $T$.

Let $U = S \setminus B^-$ be the relative complement of $B^-$.

By the definition of a closed set, $U$ is open in $T$.

From Empty Intersection iff Subset of Relative Complement:

- $A \subseteq S \setminus B^- = U$

From Relative Complement of Relative Complement:

- $S \setminus U = B^-$

By the definition of the closure of a subset:

- $B \subseteq B^- = S \setminus U$

From Empty Intersection iff Subset of Relative Complement:

- $U \cap B = \O$

Similarly, let $V = S \setminus A^-$ then $V \in \tau$ with:

- $B \subset V$

and

- $V \cap A = \O$

$\blacksquare$