# 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:

 $\displaystyle y$ $\in$ $\, \displaystyle B_\epsilon \left({x}\right) \,$ $\displaystyle$ $\displaystyle \iff \ \$ $\displaystyle d \left({y, x}\right)$ $<$ $\, \displaystyle \epsilon \,$ $\displaystyle$ Definition of Open $\epsilon$-Ball $\displaystyle \iff \ \$ $\displaystyle \left\vert{y - x}\right\vert$ $<$ $\, \displaystyle \epsilon \,$ $\displaystyle$ Definition of Euclidean Metric on Real Number Line $\displaystyle \iff \ \$ $\displaystyle -\epsilon$ $<$ $\, \displaystyle y - x \,$ $\, \displaystyle <\,$ $\displaystyle \epsilon$ Definition of Absolute Value $\displaystyle \iff \ \$ $\displaystyle x - \epsilon$ $<$ $\, \displaystyle y \,$ $\, \displaystyle <\,$ $\displaystyle x + \epsilon$ $\displaystyle \iff \ \$ $\displaystyle y$ $\in$ $\, \displaystyle \left ({x - \epsilon \,.\,.\, x + \epsilon} \right) \,$ $\displaystyle$ Definition of Open Real Interval

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.

$\blacksquare$