Give an example of function that satisfies this theorem?

Question

Give an example of function that satisfies this theorem ?

Theorem :The set of points of discontinuity of a monotonic function $f :\mathbb{R} \rightarrow \mathbb{R}$ is at most countable

My attempt : i take $f(x) = \begin{cases} 1 \ \text{ if x} \in \mathbb{Q} \\ 0 \ \text{if x} \in \mathbb{Q^c} \end{cases}$

Is its true ?

If it is a theorem then wouldn't any monotonic function work? — John Douma, Apr 03 '19 at 13:05
To find an example seems too trivial. You could take $f(x)=x$, or $g(x)=e^x$. Pretty much any continuous monotonic function satisfies this theorem — NazimJ, Apr 03 '19 at 13:05
@NazimJ $ f(x) = x$ has no discontniuity. How can u take that — jasmine, Apr 03 '19 at 13:07
Because the set of points of discontinuity of a continuous function is the empty set. Which has size 0. This satisfies the condition "at most countable" — NazimJ, Apr 03 '19 at 13:09
I think an interesting example would be to find a monotone function with countably many discontinuities. See https://math.stackexchange.com/q/69317/72031 — Paramanand Singh, Apr 03 '19 at 17:20

score 3 · Accepted Answer · answered Apr 03 '19 at 13:03

3

No, this is not correct. The function $f$ is discontinuous at every point of $\mathbb R$, besides, of course, not being monotonic.

An example would be $x\mapsto\lfloor x\rfloor$.

answered Apr 03 '19 at 13:03

1 Answers1