Bounded Set of Real Numbers
Theorem
Let $S \subseteq \R$ be a subset of the set of real numbers.
Then $S$ is bounded iff $\exists K \in \R: \forall x \in S: \left|{x}\right| \le K$.
Proof
- Suppose $\exists K \in \R: \forall x \in S: \left|{x}\right| \le K$.
Then by Negative of Absolute Value, $\forall x \in S: -K \le x \le K$.
Thus by definition, $S$ is bounded both above and below, and is therefore bounded.
- Now suppose $S$ is bounded.
Then $S$ is bounded both above and below.
As $S$ is bounded below, $\exists L \in \R: \forall x \in S: L \le x$.
As $S$ is bounded above, $\exists H \in \R: \forall x \in S: H \ge x$.
Now let $K = \max \left\{{\left|{L}\right|, \left|{H}\right|}\right\}$.
Then $K \ge \left|{L}\right|$ and $K \ge \left|{H}\right|$.
It follows from Negative of Absolute Value that $-K \le L \le K$ and $-K \le H \le K$.
In particular:
- $-K \le L$ and so $\forall x \in S: -K \le x$;
- $H \le K$ and so $\forall x \in S: x \le K$.
Thus $\forall x \in S: -K \le x \le K$ and, by Negative of Absolute Value, $\forall x \in S: \left|{x}\right| \le K$.
$\blacksquare$
