Pseudometric Space is Metric Space iff Kolmogorov

Theorem

Let $M = \struct {S, d}$ be a pseudometric space.

Let $T = \struct {S, \tau}$ be the topological space over $S$ induced by $d$.

Then $M$ is a metric space if and only if $T$ is a $T_0$ (Kolmogorov} space.

Proof

Necessary Condition

Let $M$ be a metric space.

From Metric Space is Hausdorff, $M$ is a Hausdorff space.

From:

$T_2$ (Hausdorff) Space is $T_1$ Space

and:

$T_1$ Space is $T_0$ (Kolmogorov) Space

it follows that $M$ is a Kolmogorov space.

$\Box$

Sufficient Condition

Suppose now that $M$ is a Kolmogorov space.

Let $a, b \in S$ such that $a \ne b$.

Then $\exists U \in \tau$ such that:

$\paren {a \in U \land b \notin U} \lor \paren {b \in U \land a \notin U}$

Without loss of generality, suppose that $a \in U$ and $b \notin U$.

Then by the definition of the induced topology:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subset U$

where $\map {B_\epsilon} a$ denotes the open $\epsilon$-ball of $a$ in $M$.

Since $b \notin U$:

$\map d {a, b} \ge \epsilon$

Since $\epsilon > 0$:

$\map d {a, b} > 0$

Since this holds for any pair of distinct points, $M$ is a metric space.

$\blacksquare$