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.

$\ds y$

$\in$

$\ds \set {x \in \R: \size {\xi - x} > \delta}$

$\ds \leadstoandfrom \ \ $

$\ds y$

$\notin$

$\ds \set {x \in \R: \size {\xi - x} \le \delta}$

$\ds \leadstoandfrom \ \ $

$\ds y$

$\notin$

$\ds \closedint {\xi - \delta} {\xi + \delta}$

$\ds \leadstoandfrom \ \ $

$\ds y$

$\in$

$\ds \R \setminus \closedint {\xi - \delta} {\xi + \delta}$

Definition of Set Difference

$\blacksquare$

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

which is immediate from:

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