Open Ball in Real Number Line is Open Interval

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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.

$\blacksquare$


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