# 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 if and only if:
$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$

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

## Proof

### 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:

 $\ds x$ $\in$ $\ds \set y^-$ $\, \ds \land \,$ $\ds \set y$ $\subseteq$ $\ds \set x^-$ $\ds \leadsto \ \$ $\ds \set x$ $\subseteq$ $\ds \set y^-$ Definition of Singleton $\, \ds \land \,$ $\ds \set y$ $\subseteq$ $\ds \set x^-$ $\ds \leadsto \ \$ $\ds \set x^-$ $\subseteq$ $\ds \paren {\set y^-}^-$ Topological Closure of Subset is Subset of Topological Closure $\, \ds \land \,$ $\ds \set y^-$ $\subseteq$ $\ds \paren {\set x^-}^-$ $\ds \leadsto \ \$ $\ds \set x^-$ $\subseteq$ $\ds \set y^-$ Closure of Topological Closure equals Closure $\, \ds \land \,$ $\ds \set y^-$ $\subseteq$ $\ds \set x^-$ $\ds \leadsto \ \$ $\ds \set x^-$ $=$ $\ds \set y^-$ Definition 2 of Set Equality
$\set x^- \ne \set y^-$

Thus the result by Proof by Contradiction

$\Box$

### 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:

 $\ds x$ $\in$ $\ds \set x$ Definition of Singleton $\, \ds \land \,$ $\ds y$ $\in$ $\ds \set y$ $\ds \set x$ $\subseteq$ $\ds \set x^-$ Set is Subset of its Topological Closure $\, \ds \land \,$ $\ds \set y$ $\subseteq$ $\ds \set y^-$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \set x^-$ Definition of Subset $\, \ds \land \,$ $\ds y$ $\in$ $\ds \set y^-$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \set y^-$ by hypothesis: $(2)$ $\, \ds \land \,$ $\ds y$ $\in$ $\ds \set x^-$

This contradicts the assumption $(1)$.

Thus the result by Proof by Contradiction.

$\blacksquare$