Open Ball in Real Number Line is Open Interval

Theorem
Let $\struct {\R, d}$ denote the real number line $\R$ with the usual (Euclidean) metric $d$.

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

Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball at $x$.

Then $\map {B_\epsilon} x$ is the open interval $\openint {x - \epsilon} {x + \epsilon}$.

Proof
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$.

Let $y \in \map {B_\epsilon} x$.

Then:

As the implications go both ways:
 * $\map {B_\epsilon} x \subseteq \openint {x - \epsilon} {x + \epsilon}$

and
 * $\map {B_\epsilon} x \supseteq \openint {x - \epsilon} {x + \epsilon}$

By definition of set equality:
 * $\map {B_\epsilon} x = \openint {x - \epsilon} {x + \epsilon}$

Hence the result.

Also see

 * Open Real Interval is Open Ball