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