Indiscrete Space is Pseudometrizable

Theorem
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.

Then $T$ is pseudometrizable.

Proof
Let $d: S \times S \to \R$ be the mapping defined as:


 * $\forall x, y \in S: \map d {x, y} = 0$

Then clearly $d$ is a pseudometric.

Let $\struct {S, \tau_{\struct {S, d} } }$ be the topological space induced by $d$.

Since $\struct {S, \tau_{\struct {S, d} } }$ is a topological space, by Empty Set is Element of Topology we have that $\O$ is open.

For any $\epsilon \in \R_{>0}$ we have:


 * $\map {B_\epsilon} a := \set {x \in S: \map d {x, a} < \epsilon} = S$

where $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$.

Let $U\subset S$ be a non-empty open set.

By Open Sets in Pseudometric Space we see that for every $x \in U$, there must exist an $\epsilon > 0$ such that $\map {B_\epsilon} x \subset U$.

However, $S \subset \map {B_\epsilon} x$.

Thus $U = S$.

Hence it is seen that $\O$ and $S$ are the only open sets of this resulting pseudometric space.