I am studying for a final exam and have come across the following old exam question:
Prove that any countable set of real numbers is the set of points of discontinuity of some monotone function.
The proof I gave is as follows:
Let $X$ be a countable set of real numbers.
Let $\{r_n\}_{n \in \mathbb{N}}$ be an enumeration of $X$, such that $r_n < r_{n+1}$. Define $$f(x) = \begin{cases} r_n & x \in [r_n, r_{n+1}) \\ x & (x<r_1) \text{ or } (r_n < x \forall n).\end{cases}$$ Then for all real $x \not \in X$, $f$ is continuous at $x$ and discontinuous at $r_n$. Moreover, it is clear that $f$ is monotone by our choice of enumeration of $X$. Q.E.D.
After looking online to confirm my answer, I came across two things which left me confused. First, this StackExchange post: Can I always order a countable set of numbers.
This post seems to say "I am given an enumeration, and I would like to order it" which is shown to be impossible. I believe this contrasts to my approach because I am taking an enumeration which is already ordered, instead of taking an arbitrary enumeration which is then ordered. But I could see the argument to be made that since the ordered enumeration is a permutation of any other enumeration that they are the same thing, and thus a well-ordering of $X$ is not guaranteed to exist. I think that is wrong because $X \subseteq \mathbb{R}$ so there is at least the well-ordering induced by $\mathbb{R}$. Is anyone able to confirm/deny this logic?
Second, I came across Froda's theorem. I think that the question statement is the converse of Froda's theorem, but I am unsure if it is the converse or if it is the same statement in disguise. I realize that it is kind of a dumb question to ask, but what is the relationship between Froda's theorem and this question?
Thanks for any help!
Best Answer
Froda's theorem is indeed the converse of what you are trying to do.
Froda's theorem says "Monotone function on interval" implies "set of discontinuities is countable".
Your task is "Countable set" implies "there is a monotone function whose set of discontinuities is this set".
So Froda's theorem takes a function and gives the set of discontinuities. You have the set of discontinuities, and you have to find the function itself.
Since most other comments have focused on the task of well-ordering, I will focus on tackling the main question, since it has been pointed out that the answer is wrong.
For your task, rather than going to enumerations of the countable set, we can use the fact that there are countably many elements, to create a step function with jumps at the points in the countable set.
For this, let us first consider the set itself, which we write as $\{r_n\}$ . This can be done because a countable set is in bijection with $\mathbb N$, so choosing any such bijection , we let $r_n$ be the image of $n$ under the bijection(So, no ordering is assumed). Simply define : $$ f(x) = \sum_{n : r_n < x} \frac 1{n^2} $$
That is to say, given $x \in \mathbb R$, we find all $n$ such that $r_n < x$, and take the sum of reciprocals of the squares of these $n$.
This is well defined, since $0 \leq f(x) \leq \frac{\pi^2}{6}$ for all $x$, and the set under which summation occurs is well defined. $f$ is also monotone, clearly.
Pick $r_N$, a point in the countable set. Then, for all $x > r_N$, note that $$f(x) = \sum_{n : x > r_n} \frac 1{n^2} = f(r_N) + \sum_{n : x > r_n \geq r_N} \frac 1{n^2} \geq f(r_N) + \frac 1{N^2}$$
Therefore, we have $\lim_{x \to r_N^+} f(x) \geq f(r_N) + \frac 1{N^2} > f(r_N)$, which contradicts the definition of continuity. Hence, discontinuity at these points holds.
For continuity of $f$ at points outside $\{r_n\}$, we use the fact that the tail of the series can be made as small as desired. To do this, let $s \notin \{r_n\}$, and $\epsilon > 0$. Pick $N$ large enough, so that $\sum_{n=N+1}^\infty \frac 1{n^2} < \epsilon$. Now, since $\{r_1,...,r_N\}$ is a finite set, we can find an interval around $s$, say $I=(s-\delta,s+\delta)$, which is such that $r_i \notin I$ for all $i \leq N$. This can be done because a finite set is closed, so its complement is open.
Finally, note that $f(s+\delta) - f(s-\delta) = \sum_{n : s-\delta \leq r_n < s+ \delta} \frac 1{N^2} < \epsilon$ by the choice of $\delta$. By the monotonicity of $f$, we conclude that $f(x)-f(y) < \epsilon$ for all $x,y \in (s-\delta,s+\delta)$. You can use this to see that continuity holds at $s$.
Thus, without enumeration, it is possible to construct such a function, by inducing jumps at the right places.
EDIT : The above works for $\{r_n\}$ being "at most countable" i.e. either finite or countable.