Characterization of T0 Space by Closed Sets

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Then
 * $T$ is a $T_0$ space
 * $\forall x, y \in S: x \ne y \implies$
 * $\left({\exists F \subseteq S: F}\right.$ is closed $\left. {\land\, x \in F \land y \notin F}\right)$
 * $\lor$
 * $\left({\exists F \subseteq S: F}\right.$ is closed $\left. {\land\, x \notin F \land y \in F}\right)$

Sufficient Condition
Let $T$ be a $T_0$ space.

Let $x, y \in S$ such that
 * $x \ne y$

By definition of $T_0$ space:
 * $\left({\exists U \in \tau: x \in U \land y \notin U}\right) \lor \left({\exists U \in \tau: x \notin U \land y \in U}\right)$

WLOG: Suppose:
 * $\exists U \in \tau: x \in U \land y \notin U$

By definition:
 * $\complement_S\left({U}\right)$ is closed

where $\complement_S\left({U}\right)$ denotes the relative complement of $U$ in $S$.

By definition of relative complement:
 * $x \notin \complement_S\left({U}\right) \land y \in \complement_S\left({U}\right)$

Thus:
 * $\exists F \subseteq S: F$ is closed $\land\, x \notin F \land y \in F$

Necessary Condition
This statement follows mutatis mutandis.