A function $f:[0,1]\to[0,1]$ which is strictly increasing and discontinuous.

Question

Can there be any function $f:[0,1]\to [0,1]$ which is strictly increasing and discontinuous?

Again is a strictly increasing function $f:[0,1]\to[0,1]$ onto?

Any help is appreciated. Thank you.

Answer 1

Let $$f(x)=\begin{cases} \frac{x}{2} & \mathrm{if\ } x<\frac{1}{2}, \\ \frac{x+1}{2} & \mathrm{if\ } x\geq \frac{1}{2}.\end{cases}$$

It is simply the function $g(x)=\frac{x}{2}$, but shifted up $\frac{1}{2}$ if $x\geq \frac{1}{2}$. It is thus strictly increasing, discontinuous, and not onto.

Answer 2

$$ f(x) = \begin{cases} x/3 & \text{if } 0\le x\le 1/2, \\ (x+2)/3 & \text{if } 1/2<x\le 1. \end{cases} $$ This is a strictly increasing function and it has a discontinuity at $x=1/2.$

Can a strictly increasing function from $[0,1]$ to $[0,1]$ be discontinuous everywhere?

No. But it can have a set of points of discontinuity that is dense in its domain, i.e. between any two numbers in $[0,1],$ no matter how close together there is one point at which it is discontinuous.

All discontinuities of an increasing function $f$ are jumps, i.e. points $a$ at which $\lim\limits_{x\,\downarrow\, a} f(x) > \lim\limits_{x\,\uparrow\, a} f(x).$ The difference between these two one-sided limits is the size of the jump. So let us put a jump at every rational number between $0$ and $1$ with the following sizes: $$ \begin{array}{c|c|c} \text{denominator} & \text{point of discontinuity} & \text{size of jump} \\ \hline 2 & 1/2 & 1/2 \\ 3 & 1/3,\,2/3 & 1/8 \text{ at each point, so the sum is } 1/4 \\ 4 & 1/4,\,3/4 & 1/16 \text{ at each point, so the sum is } 1/8 \\ 5 & 1/5,\,2/5,\,3/5,\,4/5 & 1/64 \text{ at each point, so the sum is } 1/16 \\ 6 & 1/6,\,5/6 & 1/16 \text{ at each point, so the sum is } 1/32 \\ \vdots & \vdots & \vdots \end{array} $$ For each denominator, we exclude the fractions that are not in lowest terms, and make the jump sizes of the included fractions equal to each other, having such values that the sum is always half what it was for the previous denominator. That way the sum of all the jump sizes is $1.$ Now let $$ f(x) = \sum_{\text{rational } u \, \le \, x} (\text{jump size at } u). $$ This function is strictly increasing on the domain $[0,1],$ and in every open interval within the domain $[0,1],$ there are infinitely many points at which $f$ is discontinuous.

(I leave it as an exercise to show that $f$ is continuous at every irrational point in its domain.)