Open Interval Defined by Absolute Value

Theorem

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

Let $\delta > 0$.

Then:

$\set {x \in \R: \size {\xi - x} < \delta} = \openint {\xi - \delta} {\xi + \delta}$

where $\openint {\xi - \delta} {\xi + \delta}$ is the open real interval between $\xi - \delta$ and $\xi + \delta$.

Proof

 $\displaystyle \size {\xi - x}$ $<$ $\displaystyle \delta$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle -\delta$ $<$ $\displaystyle \xi - x < \delta$ Negative of Absolute Value: Corollary 1 $\displaystyle \leadstoandfrom \ \$ $\displaystyle \delta$ $>$ $\displaystyle x - \xi > -\delta$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \xi + \delta$ $>$ $\displaystyle x > \xi - \delta$

But:

$\openint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta < x < \xi + \delta}$

$\blacksquare$