Characterization of T0 Space by Closures of Singletons

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

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

where $\left\{{y}\right\}^-$ denotes the closure of $\left\{{y}\right\}$.

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

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

Aiming for a contradiction suppose that
 * $x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$

Then:
 * $\left\{{x}\right\} \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{x}\right\}^-$

By Topological Closure of Subset is Subset of Topological Closure:
 * $\left\{{x}\right\}^- \subseteq \left({\left\{{y}\right\}^-}\right)^- \land \left\{{y}\right\}^- \subseteq \left({\left\{{x}\right\}^-}\right)^-$

By Closure of Topological Closure equals Closure:
 * $\left\{{x}\right\}^- \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\}^- \subseteq \left\{{x}\right\}^-$

Then by definition of set equality:
 * $\left\{{x}\right\}^- = \left\{{y}\right\}^-$

By Characterization of $T_0$ Space by Distinct Closures of Singletons:
 * $\left\{{x}\right\}^- \ne \left\{{y}\right\}^-$

This contradicts the equality.

Thus the result by Proof by Contradiction

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

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

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

Aiming for a contradiction suppose that
 * $(2): \quad \left\{{y}\right\}^- = \left\{{x}\right\}^-$

By definition of singleton:
 * $x \in \left\{{x}\right\} \land y \in \left\{{y}\right\}$

By Set is Subset of its Topological Closure
 * $\left\{{x}\right\} \subseteq \left\{{x}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{y}\right\}^-$

Then by definition of subset and $(2)$:
 * $x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$

This contradicts the assumption $(1)$.

Thus the result by Proof by Contradiction.