Closure Condition for Hausdorff Space

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

Then $\struct {S, \tau}$ is a Hausdorff space :


 * For all $x, y \in S$ such that $x \ne y$, there exists an open set $U$ such that $x \in U$ and $y \notin U^-$, where $U^-$ is the closure of $U$.

Necessary Condition
Let $\struct {S, \tau}$ be a Hausdorff space.

Let $x, y \in S$ with $x \ne y$.

Then by the definition of Hausdorff space there exist open sets $U, V \subseteq S$ such that:
 * $x \in U$
 * $y \in V$
 * $U \cap V = \O$

By Empty Intersection iff Subset of Complement, $U \subseteq S \setminus V$.

By definition, $S \setminus V$ is closed.

Thus by definition of closure:
 * $U^- \subseteq S \setminus V$

Since $y \in V$, by definition of relative complement it follows that $y \notin S \setminus V$.

By definition of subset, $y \notin U^-$.

Sufficient Condition
Suppose that for each $x, y \in S$ such that $x \ne y$ there exists an open set $U$ such that $x \in U$ and $y \notin U^-$.

Let $x, y \in S$ with $x \ne y$.

Let $U$ be as described.

Let $V = S \setminus U^-$.

Then $y \in V$ by the definition of set difference.

Since $U \subseteq U^-$, it follows that $U \cap V = \O$.

As such $U$ and $V$ exist for all such $x$ and $y$, it follows that $\struct {S, \tau}$ is a Hausdorff space.