Point in Metric Space has Countable Neighborhood Basis

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

Let $a \in A$.

Then there exists a basis for the neighborhood system of $a$ which is countable.

Proof
Consider the countable set:
 * $\BB_a := \set {\map {B_\epsilon} a: \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}$

That is, let $\BB_a$ be the set of all open $\epsilon$-balls of $a$ such that $\epsilon$ is of the form $\dfrac 1 n$ for (strictly) positive integral $n$.

Let $N$ be a neighborhood of $a$.

Then by definition of neighborhood:
 * $\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$

From Between two Real Numbers exists Rational Number:
 * $\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$

Let $\epsilon''$ be expressed in canonical form as:
 * $\epsilon'' = \dfrac p q$

where $p$ and $q$ are coprime integers and $q > 0$.

Then:
 * $\epsilon := \dfrac 1 q \le \dfrac p q$

Thus $0 < \epsilon < \epsilon'$.

So:

From Open Ball is Neighborhood of all Points Inside, $\map {B_\epsilon} a$ is a neighborhood of $a$.

But as $\epsilon = \dfrac 1 q$ where $q \in \Z_{>0}$ it follows that:
 * $\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}$

Hence the result.