Open Reciprocal-N Balls form 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 $\BB_a$ be defined as:
- $\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N} }$
that is, the set of all open $\epsilon$-balls of $a$ for $\epsilon$ which are reciprocals of integers.
Then $\BB_a$ 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 $\epsilon \le \epsilon'' < \epsilon'$
and so:
- $\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
From Open Real Interval is Open Ball
- $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$
is the open $\epsilon$-ball at $a$.
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 there exists a neighborhood $\openint {a - \epsilon} {a + \epsilon}$ of $a$ which is a subset of $N$.
Hence the result by definition of basis for the neighborhood system of $a$.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $4 \ \text{iii)}$