Product of Positive Strictly Increasing Mappings is Strictly Increasing

Theorem
Let $A$ be an ordered set.

Let $B$ be an ordered field.

Let $f, g: A \to B$ be strictly increasing mappings with positive values.

Let $h \colon A \to B$ be defined by $h \left({x}\right) = f \left({x}\right) g \left({x}\right)$.

Then $h$ is strictly increasing.

Proof
Let $x, y \in A$ such that $x < y$.

If $h \left({x}\right) = 0$.

By the definition of strictly increasing:
 * $f \left({y}\right) > f \left({x}\right) \geq 0$

and:
 * $g \left({y}\right) > g \left({x}\right) \geq 0$

So:
 * $h \left({y}\right) > 0 = h \left({x}\right)$

If $h \left({x}\right) \neq 0$, then $h \left({x}\right) > 0$ so:
 * $f \left({x}\right) > 0$

and:
 * $g \left({x}\right) > 0$

Also:
 * $f \left({x}\right) < f \left({y}\right)$

and:
 * $g \left({x}\right) < g \left({y}\right)$

by the definition of strictly increasing.

Because $g \left({x}\right) > 0$:
 * $f \left({x}\right) g \left({x}\right) < f \left({y}\right) g \left({x}\right)$

Because $f \left({y}\right) > 0$:
 * $f \left({y}\right) g \left({x}\right) < f \left({y}\right) g \left({y}\right)$

By transitivity:


 * $h \left({x}\right) = f \left({x}\right) g \left({x}\right) < f \left({y}\right) g \left({y}\right) = h \left({y}\right)$