# 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:

 $\displaystyle y$ $\in$ $\, \displaystyle \map {B_\epsilon} x \,$ $\displaystyle$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \map d {y, x}$ $<$ $\, \displaystyle \epsilon \,$ $\displaystyle$ Definition of Open $\epsilon$-Ball $\displaystyle \leadstoandfrom \ \$ $\displaystyle \size {y - x}$ $<$ $\, \displaystyle \epsilon \,$ $\displaystyle$ Definition of Euclidean Metric on Real Number Line $\displaystyle \leadstoandfrom \ \$ $\displaystyle -\epsilon$ $<$ $\, \displaystyle y - x \,$ $\, \displaystyle <\,$ $\displaystyle \epsilon$ Definition of Absolute Value $\displaystyle \leadstoandfrom \ \$ $\displaystyle x - \epsilon$ $<$ $\, \displaystyle y \,$ $\, \displaystyle <\,$ $\displaystyle x + \epsilon$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle y$ $\in$ $\, \displaystyle \openint {x - \epsilon} {x + \epsilon} \,$ $\displaystyle$ Definition of Open Real Interval

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.

$\blacksquare$