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

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

$x \ne y$

By definition of $T_0$ space:

$\paren {\exists U \in \tau: x \in U \land y \notin U} \lor \paren {\exists U \in \tau: x \notin U \land y \in U}$

Without loss of generality, suppose

$\exists U \in \tau: x \in U \land y \notin U$

By definition of singleton:

$x \in \set x$
$\set x \subseteq \set x^-$

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:

$\set y \cap U = \O$
$\set y^- \cap U = \O$

Thus:

$\set x^- \ne \set y^-$

$\Box$

### 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 $F := \set y^-$

$F$ is closed.

We will prove now that

$x \notin F$

$x \in F$

Then:

$\set x \subseteq F$
$\set x^- \subseteq F^-$
$(3): \quad \set x^- \subseteq F$

By definition of singleton:

$x \in \set x$
$\set x \subseteq \set x^-$

Then by definition of subset:

$x \in \set x^-$

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

$y \in \set x^-$

Then:

$\set y \subseteq \set x^-$
$F \subseteq \paren {set x^-}^-$
$F \subseteq \set x^-$

Hence by $(3)$ and definition of set equality:

$F = \set x^-$

This contradicts the assumption $(1)$.

Thus $x \notin F$.

By definition of singleton:

$y \in \set y$
$\set y \subseteq \set y^- = F$

Thus by definition of subset:

$y \in F$

Hence

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

$\blacksquare$