\section{Closed Interval Defined by Absolute Value}
Tags: Real Intervals, Absolute Value Function

\begin{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 $\xi + \delta$.
\end{theorem}

\begin{proof}
{{begin-eqn}}
{{eqn | l = \size {\xi - x}
      | o = \le
      | r = \delta
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = -\delta
      | o = \le
      | r = \xi - x \le \delta
      | c = Negative of Absolute Value: Corollary 2
}}
{{eqn | ll= \leadstoandfrom
      | l = \delta
      | o = \le
      | r = x - \xi \le -\delta
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = \xi + \delta
      | o = \le
      | r = x \le \xi - \delta
      | c =
}}
{{end-eqn}}
But:
:$\closedint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta \le x \le \xi + \delta}$
{{qed}}
\end{proof}

