Interval Defined by Absolute Value

Theorem
Let $$\xi, \delta \in \mathbb{R}$$ be real numbers.

Let $$\delta > 0$$.

Then $$\left\{{x \in \mathbb{R}: \left|{\xi - x}\right| < \delta}\right\} = \left({\xi - \delta \, . \, . \, \xi + \delta}\right)$$, where $$\left({\xi - \delta \, . \, . \, \xi + \delta}\right)$$ is the open real interval between $$\xi - \delta$$ and $$\xi + \delta$$.

Similarly, $$\left\{{x \in \mathbb{R}: \left|{\xi - x}\right| \le \delta}\right\} = \left[{\xi - \delta \,. \, . \, \xi + \delta}\right]$$, where $$\left[{\xi - \delta \,. \, . \, \xi + \delta}\right]$$ is the closed real interval between $$\xi - \delta$$ and $$\xi + \delta$$.

Proof
$$ $$ $$ $$

But $$\left({\xi - \delta \, . \, . \, \xi + \delta}\right) = \left\{{x \in \mathbb{R}: \xi - \delta < x < \xi + \delta}\right\}$$.

The other result follows similarly.