Metric Space is First-Countable

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

Then $M$ is first-countable.

Proof
Let $x \in A$.

Let:
 * $\BB = \set {\map {B_{1/n} } x: n \in \N_{>0} }$

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

By Surjection from Natural Numbers iff Countable, we have that $\BB$ is countable.

By the definition of a first-countable space, it suffices to show that $\BB$ is a local basis at $x$.

By Open Ball of Pseudometric Space is Open Set, every element of $\BB$ is an open neighborhood of $x$.

Let $U$ be an open neighborhood of $x$.

By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$.

By the Archimedean Principle, there exists a natural number $n > \dfrac 1 \epsilon$.

That is, $\dfrac 1 n < \epsilon$, and so:
 * $\map {B_{1/n} } x \subseteq \map {B_\epsilon} x \subseteq U$

The result follows from Subset Relation is Transitive.