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

Theorem

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 $k$.

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 $a$.

$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$

$\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$

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:

$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$

$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$

By its method of construction:

$\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$

$\openint {a - \epsilon} {a + \epsilon} \subseteq N$

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 $a$.

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $4 \ \text{iv)}$