# Increasing

A function $f: A \to B$ where $A, B$ are totally ordered sets is said to be increasing if for every elements $x, y \in A$, $x \leq y$ implies $f(x) \leq f(y)$.

A simple example of an increasing function is the map $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x$. In fact, it's easy to see that the identity map is an increasing map on any totally ordered set.

If $x < y$ implies $f(x) < f(y)$ we say that $f$ is strictly increasing.

The notion of an increasing function is generalized to the context of partially ordered sets by order-preserving functions.

If the function $f: (a, b) \to \mathbb{R}$ is differentiable then $f$ is increasing if and only if $f'(x) \geq 0$ for all $x \in (a, b)$. If $f'(x) > 0$ for all $x \in (a, b)$ then $f$ is strictly increasing, but the converse does not hold: for example, the function $g(x) = x^3$ is strictly increasing on the interval $(-1, 1)$, but $g'(0) = 0$.

If the domain of the function $f$ is the integers (or the positive integers or nonnegative integers) then $f$ is increasing (or strictly increasing) if and only if the sequence $\ldots, f(1), f(2), f(3), \ldots$ of its values is an increasing sequence (respectively, strictly increasing sequence).