# Closure Condition for Hausdorff Space

## Theorem

Let $\left({X, \tau}\right)$ be a topological space.

Then $\left({X, \tau}\right)$ is a Hausdorff space if and only if:

- 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 $U$.

## Proof

### Necessary Condition

Let $\left({X, \tau}\right)$ be a Hausdorff space.

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

Then by the definition of Hausdorff space there exist open sets $U, V \subseteq X$ such that:

- $x \in U$
- $y \in V$
- $U \cap V = \varnothing$

Then $U \subseteq X \setminus V$.

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

Thus by definition of closure:

- $U^- \subseteq X \setminus V$

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

$\Box$

### Sufficient Condition

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

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

Let $U$ be as described.

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

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

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

As such $U$ and $V$ exist for all such $x$ and $y$, it follows that $\left({X, \tau}\right)$ is a Hausdorff space.

$\blacksquare$