Looking for an example of an increasing function $f:[a,b] \to [a,b]$ which is discontinuous at infinitely many points

Question

I am looking for an example of an increasing function $f:[a,b] \to [a,b]$ which is discontinuous at infinitely many points ; please help , thanks in advance .

Answer 1

A simple example of a continuous function with an infinite number of removable discontinuities would be $$ f(x)=\frac{1}{\frac{x}{2} - \lfloor{\frac{x}{2}}\rfloor} $$

Many more examples can be made with the floor and ceiling functions(or any other step functions, really) in the denominator, as they allow for controlled periodic zeroes(in this case, every even integer would result in a denominator of 0).

Answer 2

Classic example: define $$ I(x) = \begin{cases} 0 & x < 0\\ 1 & x \geq 0 \end{cases} $$ Take $f(x) = \sum_{j=1}^\infty 2^{-j}\cdot I(x - 1/j)$ on $[0,1]$.

If you want a function that's strictly increasing, consider $g(x) = f(x) + x$.

Answer 3

Select infinitely many points in $[a,b].$ For example, $a,a+\frac {b-a}{2},a+\frac {b-a}{2}+\frac {b-a}{2^2},\cdots.$ Define $f$ to be a line segment in each of the intervals $[a+\frac {b-a}{2}+\cdots \frac {b-a}{2^n},a+\frac {b-a}{2}+\cdots \frac {b-a}{2^n}+ \frac {b-a}{2^{n+1}})$. For example, $f(x)$ increases from $ 1$ to $1+\frac {1}{2}$ in the first interval , from $1+\frac{1}{2}+\frac{1}{2^2}$ to $1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}$ in the second, etc. Then each of the end points of these subintervals $a+\frac {b-a}{2}+\cdots \frac {b-a}{2^n}$ is a point of discontinuity