Pseudometric Space is Metric Space iff Kolmogorov

From ProofWiki
Jump to navigation Jump to search


Let $M = \left({S, d}\right)$ be a pseudometric space.

Let $T = \left({S, \tau}\right)$ 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.


Necessary Condition

Let $M$ be a metric space.

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


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


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

it follows that $M$ is a Kolmogorov space.


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:

$\left({a \in U \land b \notin U}\right) \lor \left({b \in U \land a \notin U}\right)$

Suppose WLOG that $a \in U$ and $b \notin U$.

Then by the definition of the induced topology:

$\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subset U$

where $B_\epsilon \left({a}\right)$ denotes the open $\epsilon$-ball of $a$ in $M$.

Since $b \notin U$:

$d \left({a, b}\right) \ge \epsilon$

Since $\epsilon > 0$:

$d \left({a, b}\right) > 0$

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