Equivalence of Definitions of Bounded Subset 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, $$\forall x \in S: \left|{x}\right| \le K$$.