Subset of Open Reciprocal-N Balls forms Neighborhood Basis in Real Number Line

Theorem
Let $\R$ denote the real number line with the usual (Euclidean) metric.

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

Let $k \in \Z$ be some fixed integer.

Let $\BB_a$ be defined as:
 * $\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$

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

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

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

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

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

From Open Ball in Real Number Line is Open Interval:
 * $\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$

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

If $q \le k$, let $\epsilon = \dfrac 1 {k + 1}$

Otherwise, let $\epsilon = \epsilon'''$.

Then:
 * $\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$

From Open Real Interval is Open Ball
 * $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$

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

By its method of construction:
 * $\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$

From Subset Relation is Transitive:
 * $\openint {a - \epsilon} {a + \epsilon} \subseteq N$

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

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