Equivalence of Definitions of Bounded Real-Valued Function

Proof
Let $S$ be a set.

Let $f: S \to \R$ be a real-valued function.

Definition 1 implies Definition 2
Let $f$ be bounded according to definition 1:
 * bounded above in $S$ by $H \in \R$
 * bounded below in $S$ by $L \in \R$

Thus by definition:
 * $\forall x \in S: L \le \map f x$
 * $\forall x \in S: \map f x \le H$

By Equivalence of Definitions of Bounded Subset of Real Numbers:
 * $\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$

So $f$ is bounded according to definition 2.

Definition 2 implies Definition 1
Let $f$ be bounded according to definition 2:
 * $\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$

Then by Equivalence of Definitions of Bounded Subset of Real Numbers:
 * $\exists K \in \R_{\le 0}: \forall x \in S: -K \le \map f x$

and so $f$ is bounded below in $S$ by $-K \in \R$

and
 * $\exists K \in \R_{\le 0}: \forall x \in S: \map f x \le K$

and so $f$ is bounded above in $S$ by $K \in \R$.

So $f$ is bounded according to definition 2.

Definition 1 implies Definition 3
Let $f$ be bounded according to definition 1:
 * bounded above in $S$ by $H \in \R$
 * bounded below in $S$ by $L \in \R$

Thus by definition:
 * $\forall x \in S: L \le \map f x$
 * $\forall x \in S: \map f x \le H$

That is:
 * $\map f x \in \closedint L H$

Thus $f$ is bounded according to definition 3.

Definition 3 implies Definition 1
Let $f$ be bounded according to definition 3:
 * $\exists a, b \in \R_{\le 0}: \forall x \in S: \map f x \in \closedint a b$

Thus by definition:
 * $\forall x \in S: a \le \map f x$

and so $f$ is bounded below in $S$ by $a \in \R$


 * $\forall x \in S: \map f x \le b$

and so $f$ is bounded above in $S$ by $b \in \R$.

So $f$ is bounded according to definition 1.