Derive the reverse triangle inequality from the triangle inequality.
Let \(E\subseteq \RR\). Define \(\sup E\) and \(\inf E\).
What is the Archimedean property?
Solution
- In a normed space, \(\big||x| - |y|\big| \le |x - y|\).
1. \(|x| - |y| \le |x - y|\).
The triangle inequality applied to \(x = (x - y) + y\) gives \(|x| \le |x - y| + |y|\).
2. \(|y| - |x| \le |x - y|\).
The triangle inequality applied to \(y = (y - x) + x\) gives \(|y| \le |y - x| + |x|\), and \(|y - x| = |x - y|\).
3. Q.E.D.
Steps 1 and 2 bound both \(|x| - |y|\) and its negative by \(|x - y|\).
Let \(E \subseteq \RR\) be nonempty. If \(E\) is bounded above, \(\sup E\) is the real number \(s\) such that \(x \le s\) for all \(x \in E\), and \(s \le t\) for every \(t\) with \(x \le t\) for all \(x \in E\); equivalently, \(s\) is an upper bound and for every \(\eps > 0\) there is \(x \in E\) with \(x > s - \eps\). By completeness of \(\RR\) it exists. If \(E\) is not bounded above, \(\sup E = +\infty\). Similarly \(\inf E\) is the greatest lower bound of \(E\), or \(-\infty\) if \(E\) is not bounded below, and \(\inf E = -\sup(-E)\).
The Archimedean property: for every \(x \in \RR\) there is \(n \in \NN\) with \(n > x\). Equivalently, for every \(\eps > 0\) there is \(n \in \NN\) with \(1/n < \eps\).
Suppose no \(n \in \NN\) exceeds \(x\). Then \(\NN\) is bounded above, so by completeness it has a least upper bound \(s\). Since \(s - 1\) is not an upper bound, some \(n \in \NN\) has \(n > s - 1\), so \(n + 1 > s\) with \(n + 1 \in \NN\), contradicting that \(s\) is an upper bound.