Characterization of T0 Space by Closures of Singletons

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

Then
 * $T$ is a $T_0$ space :
 * $\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$

where $\set y^-$ denotes the closure of $\set y$.

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

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


 * $x \in \set y^- \land y \in \set x^-$

Then:

By Characterization of $T_0$ Space by Distinct Closures of Singletons:
 * $\set x^- \ne \set y^-$

This contradicts the equality.

Thus the result by Proof by Contradiction

Necessary Condition
Assume that:
 * $(1): \quad \forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$

By Characterization of $T_0$ Space by Distinct Closures of Singletons it suffices to prove
 * $\forall x, y \in S: x \ne y \implies \set x^- \ne \set y^-$

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


 * $(2): \quad \set x^- = \set y^-$

Then:

This contradicts the assumption $(1)$.

Thus the result by Proof by Contradiction.