Open Ball in Real Number Line is Open Interval

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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.

$\blacksquare$


Also see


Sources