Characterization of T0 Space by Distinct Closures of Singletons

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

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

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
 * $x \ne y$

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

, suppose
 * $\exists U \in \tau: x \in U \land y \notin U$

By definition of singleton:
 * $x \in \set x$

By Set is Subset of its Topological Closure:
 * $\set x \subseteq \set x^-$

Then by definition of subset:
 * $x \in \set x^-$

Hence by definition of intersection:
 * $\set x^- \cap U \ne \O$

By definition of singleton:
 * $\forall z: z \in \set y \implies z = y$

Then by definition of intersection:
 * $\set y \cap U = \O$

Hence by Disjoint Open Sets remain Disjoint with one Closure
 * $\set y^- \cap U = \O$

Thus:
 * $\set x^- \ne \set y^-$

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

By Characterization of $T_0$ Space by Closed Sets it suffices to prove that
 * $\leftparen {\exists F \subseteq S: F}$ is closed $\rightparen {\land x \in F \land y \notin F}$

or
 * $\leftparen {\exists F \subseteq S: F}$ is closed $\rightparen {\land x \notin F \land y \in F}$

Then assume:
 * $(2): \quad \forall F \subseteq S: F$ is closed $\land x \in F \implies y \in F$

Define $F := \set y^-$

By Topological Closure is Closed:
 * $F$ is closed.

We will prove now that
 * $x \notin F$


 * $x \in F$

Then:
 * $\set x \subseteq F$

By Topological Closure of Subset is Subset of Topological Closure:
 * $\set x^- \subseteq F^-$

By Set is Closed iff Equals Topological Closure:
 * $(3): \quad \set x^- \subseteq F$

By definition of singleton:
 * $x \in \set x$

By Set is Subset of its Topological Closure:
 * $\set x \subseteq \set x^-$

Then by definition of subset:
 * $x \in \set x^-$

Because $\set x^-$ is closed therefore by $(2)$:
 * $y \in \set x^-$

Then:
 * $\set y \subseteq \set x^-$

By Topological Closure of Subset is Subset of Topological Closure:
 * $F \subseteq \paren {set x^-}^-$

Then by Closure of Topological Closure equals Closure:
 * $F \subseteq \set x^-$

Hence by $(3)$ and definition of set equality:
 * $F = \set x^-$

This contradicts the assumption $(1)$.

Thus $x \notin F$.

By definition of singleton:
 * $y \in \set y$

By Set is Subset of its Topological Closure:
 * $\set y \subseteq \set y^- = F$

Thus by definition of subset:
 * $y \in F$

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