Closed Interval Defined by Absolute Value

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

Let $\delta > 0$.

Then:
 * $\set {x \in \R: \size {\xi - x} \le \delta} = \closedint {\xi - \delta} {\xi + \delta}$

where $\closedint {\xi - \delta} {\xi + \delta}$ is the closed real interval between $\xi - \delta$ and $\xi + \delta$.

Proof
But:
 * $\closedint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta \le x \le \xi + \delta}$

Also presented as

 * $\set {x \in \R: \size {x - \xi} \le \delta} = \closedint {\xi - \delta} {\xi + \delta}$

which is immediate from:
 * $\size {x - \xi} = \size {\xi - x}$

Also see

 * Open Interval Defined by Absolute Value


 * Complement of Closed Interval Defined by Absolute Value
 * Complement of Open Interval Defined by Absolute Value