Open Ball in Real Number Line is Open Interval

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

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

Let $B_\epsilon \left({x}\right)$ be the open $\epsilon$-ball at $x$.

Then $B_\epsilon \left({x}\right)$ is the open interval $\left ({x - \epsilon \,.\,.\, x + \epsilon} \right)$.

Proof
Let $S = B_\epsilon \left({x}\right)$ be an open $\epsilon$-ball at $x$.

Let $y \in B_\epsilon \left({x}\right)$.

Then:

{{eqn | ll= \iff | l = d \left({y, x}\right) | o = < | m = \epsilon | c = {Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}} }}

As the implications go both ways:
 * $B_\epsilon \left({x}\right) \subseteq \left ({x - \epsilon \,.\,.\, x + \epsilon} \right)$

and
 * $B_\epsilon \left({x}\right) \supseteq \left ({x - \epsilon \,.\,.\, x + \epsilon} \right)$

By definition of set equality:
 * $B_\epsilon \left({x}\right) = \left ({x - \epsilon \,.\,.\, x + \epsilon} \right)$

Hence the result.

Also see

 * Open Real Interval is Open Ball