Partition Space is Pseudometrizable

Theorem
Let $T = \left({S, \vartheta}\right)$ be a partition space.

Then $T$ is pseudometrizable.

Proof
Let $\mathcal P$ be the partition which is the basis for $T$.

Let us define $d: S^2 \to \R$ by:
 * $\forall x, y \in S: d \left({x, y}\right) = \begin{cases}

0 & : \exists U \in \mathcal P: x, y \in U \\ 1 & : \text{otherwise} \end{cases}$

That is, $d \left({x, y}\right) = 0$ when $x$ and $y$ are in the same set in the partition $\mathcal P$, and $1$ otherwise.

This is easily established as being a pseudometric on $S$.

Hence the result.