Open Reciprocal-N Balls form Neighborhood Basis in Real Number Line

Theorem
Let $\R$ be the real number line considered as a metric space under the usual metric.

Let $a \in \R$ be a point in $\R$.

Let $\mathcal B_a$ be defined as:
 * $\mathcal B_a := \left\{ {B_\epsilon \left({a}\right): \epsilon \in \left\{ {\dfrac 1 n: n \in \N}\right\} }\right\}$

that is, the set of all open $\epsilon$-balls of $a$ for $\epsilon$ which are reciprocals of integers.

Then $\mathcal B_a$ is a basis for the neighborhood system of $a$.

Proof
Let $N$ be a neighborhood of $a$ in $M$.

Then by definition:
 * $\exists \epsilon' \in \R_{>0}: B_\epsilon' \left({a}\right) \subseteq N$

where $B_\epsilon' \left({a}\right)$ is the open $\epsilon'$-ball at $a$.

From Open Ball in Real Number Line is Open Interval:
 * $B_\epsilon' \left({a}\right) = \left({a - \epsilon' \,.\,.\, a + \epsilon'}\right)$

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

Let $\epsilon''$ be expressed in canonical form:


 * $\epsilon'' = \dfrac p q$

Let $\epsilon = \dfrac 1 q$

Then $\epsilon \le \epsilon'' < \epsilon'$

and so:
 * $\left({a - \epsilon \,.\,.\, a + \epsilon}\right) \subseteq \left({a - \epsilon' \,.\,.\, a + \epsilon'}\right)$

From Open Real Interval is Open Ball
 * $B_\epsilon \left({a}\right) = \left({a - \epsilon \,.\,.\, a + \epsilon}\right)$

is the open $\epsilon$-ball at $a$.

From Subset Relation is Transitive:
 * $\left({a - \epsilon \,.\,.\, a + \epsilon}\right) \subseteq N$

From Open Ball is Neighborhood of all Points Inside, $\left({a - \epsilon \,.\,.\, a + \epsilon}\right)$ is a neighborhood of $a$ in $M$.

Hence there exists a neighborhood $\left({a - \epsilon \,.\,.\, a + \epsilon}\right)$ of $a$ which is a subset of $N$.

Hence the result by definition of basis for the neighborhood system of $a$.