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

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


But:

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

$\blacksquare$


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