# Complement of Closed Interval Defined by Absolute Value

## Theorem

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

Let $\delta > 0$.

Then:

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

where:

$\closedint {\xi - \delta} {\xi + \delta}$ is the closed real interval between $\xi - \delta$ and $\xi + \delta$
$\setminus$ denotes the set difference operator.

## Proof

 $\ds y$ $\in$ $\ds \set {x \in \R: \size {\xi - x} > \delta}$ $\ds \leadstoandfrom \ \$ $\ds y$ $\notin$ $\ds \set {x \in \R: \size {\xi - x} \le \delta}$ $\ds \leadstoandfrom \ \$ $\ds y$ $\notin$ $\ds \closedint {\xi - \delta} {\xi + \delta}$ Closed Interval Defined by Absolute Value $\ds \leadstoandfrom \ \$ $\ds y$ $\in$ $\ds \R \setminus \closedint {\xi - \delta} {\xi + \delta}$ Definition of Set Difference

$\blacksquare$

## Also presented as

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

which is immediate from:

$\size {x - \xi} = \size {\xi - x}$