Closed Intervals 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:
- $\mathcal B_a := \left\{ {\left[{a - \epsilon \,.\,.\, a + \epsilon}\right]: \epsilon \in \R_{>0} }\right\}$
that is, the set of all closed intervals of $\R$ with $a$ as a midpoint.
Then the $\mathcal B_a$ 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:
- $B_\epsilon' \left({a}\right) = \left({a - \epsilon \,.\,.\, a + \epsilon}\right)$
Let $\epsilon \in \R_{>0}: \epsilon < \epsilon'$.
Then by definition of closed interval and open interval:
- $\left[{a - \epsilon \,.\,.\, a + \epsilon}\right] \subseteq \left({a - \epsilon' \,.\,.\, a + \epsilon'}\right)$
From Subset Relation is Transitive:
- $\left[{a - \epsilon \,.\,.\, a + \epsilon}\right] \subseteq N$
Also by definition of closed interval and open interval:
- $\left({a - \epsilon \,.\,.\, a + \epsilon}\right) \subseteq \left[{a - \epsilon \,.\,.\, a + \epsilon}\right]$
From Open Real Interval is Open Ball, $\left({a - \epsilon \,.\,.\, a + \epsilon}\right)$ is the open $\epsilon$-ball at $a$.
Thus by definition, $\left[{a - \epsilon \,.\,.\, a + \epsilon}\right]$ 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 $a$.
$\blacksquare$
Sources
- 1962: Bert Mendelson: Introduction to Topology ... (previous) ... (next): $\S 2.4$: Open Balls and Neighborhoods: Exercise $4 \ \text{i)}$