Closed Interval Defined by Absolute Value

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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:

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

$\blacksquare$