Closure Condition for Hausdorff Space

Theorem
Let $(X, \tau)$ be a topological space.

Then $(X, \tau)$ is a Hausdorff space iff:


 * For all $x, y \in X$ 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 $(X, \tau)$ be a Hausdorff space.

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

Then by the definition of Hausdorff space there are open sets $U$ and $V$ in $X$ such that $x \in U$, $y \in V$, and $U \cap V = \varnothing$.

Then $U \subseteq X \setminus V$.

$X \setminus V$ is closed.

Thus $U^- \subseteq X \setminus V$ by the definition of closure.

Since $y \in V$, $y \notin X \setminus V$, so $y \notin U^-$.

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

Let $x, y \in X$ with $x ≠ y$.

Let $U$ be as described.

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

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

Since $U \subseteq U^-$, $U \cap V = \varnothing$.

As such $U$ and $V$ exist for all such $x$ and $y$, $(X, \tau)$ is a Hausdorff space.