# 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

 $\displaystyle \size {\xi - x}$ $\le$ $\displaystyle \delta$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle -\delta$ $\le$ $\displaystyle \xi - x \le \delta$ Negative of Absolute Value: Corollary 2 $\displaystyle \leadstoandfrom \ \$ $\displaystyle \delta$ $\le$ $\displaystyle x - \xi \le -\delta$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \xi + \delta$ $\le$ $\displaystyle x \le \xi - \delta$

But:

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

$\blacksquare$