Interval Defined by Absolute Value

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

Let $\delta > 0$.

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

where $\left({\xi - \delta \,.\,.\, \xi + \delta}\right)$ is the open real interval between $\xi - \delta$ and $\xi + \delta$.

Similarly:
 * $\left\{{x \in \R: \left\vert{\xi - x}\right\vert \le \delta}\right\} = \left[{\xi - \delta \,.\,.\, \xi + \delta}\right]$

where $\left[{\xi - \delta \,.\,.\, \xi + \delta}\right]$ is the closed real interval between $\xi - \delta$ and $\xi + \delta$.

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

The other result follows similarly.