Equivalence of Definitions of Bounded Subset of Real Numbers

Theorem
The following definitions of a bounded set of real numbers are equivalent.

Definition 1 implies Definition 2
Let $S$ be a bounded set of real numbers by definition 1.

Then by definition $S$ 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:
 * $\exists H \in \R: \forall x \in S: H \ge x$

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:
 * $\forall x \in S: \left|{x}\right| \le K$

Thus $S$ is a bounded set of real numbers by definition 2.

Definition 2 implies Definition 1
Let $S$ be a bounded set of real numbers by definition 2.

Thus:
 * $\exists K \in \R: \forall x \in S: \left|{x}\right| \le K$

Then by Negative of Absolute Value:
 * $\forall x \in S: -K \le x \le K$

Thus by definition, $S$ is both bounded above (by $K$) and bounded below (by $-K$).

Thus $S$ is a bounded set of real numbers by definition 1.