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Let $(x)$ be the fractional part of a real number $x$. Prove that the function on $\mathbb{R}$ $$f(x)=\sum_{n=1}^{\infty}\frac{(nx)}{n^2}$$ is discontinuous at a countable set of points that is dense in $\mathbb{R}$.

The solution to this part is that $f$ is discontinuous on $\mathbb{Q}$. It is true when when the domain of $f$ is restricted to $\mathbb{I}$, it would be continuous. However, if we consider the domain $\mathbb{R}$, discontinuity of $f$ on the rational numbers implies discontinuity everywhere since $\mathbb{Q}$ is dense in $\mathbb{R}$, so technically speaking, $f$ is discontinuous everywhere.

Am I correct to say this? If yes, should things be stated more clearly, or are they supposed to be understood as an implicit implication?

Jia Cheng Sun
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    No, a function can be discontinuous at all rational points and continuous at the irrational points. A classical example is $f(x)=0$ for $x$ irrational and $f(x)=1/q$, when $x$ is rational and $x=p/q$ its representation as a reduced fraction of integers. –  Mar 25 '20 at 14:28
  • Yeah I believe there was an exercise like that in chapter 3. Thanks for reminding me of this. – Jia Cheng Sun Mar 25 '20 at 14:35
  • Some previous discussion here: https://math.stackexchange.com/questions/1652576/prove-that-sum-n-1-infty-fracnxn2-is-discontinuous-at-x-in-m?rq=1 – Sean Eberhard Mar 25 '20 at 14:42
  • and here: https://math.stackexchange.com/questions/136684/series-with-fractional-part-of-nx?rq=1 – Sean Eberhard Mar 25 '20 at 14:44

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No, your claim after the word "technically" is wrong. A function can be discontinuous at every rational and continuous elsewhere. A well-known example is the function $f$ defined by

$$ f(x) = \begin{cases} 1/q &: x = p/q, ~ \gcd(p, q) = 1, \\ 0 &: x ~\text{irrational}. \end{cases}. $$

Clearly this function is discontinuous at rationals, because the irrationals are dense. On the other hand, if $x$ is rational then there are only finitely many points rationals $y = p /q$ with $|x - y| \leq 1$ and with $q$ less than some bound, so $f$ is continuous at $x$.

Your function $f$ has the same behaviour: it is discontinous at rationals but continuous at irrationals. Suppose $x$ is irrational. Let $\epsilon>0$. Choose $N$ large enough that $\sum_{n > N} 1/n^2 < \epsilon/3$. The function $\sum_{n \leq N} (ny) / n^2$ is continuous at $x$, since $(nx) \neq 0$ for all $n \leq N$, so there is a $\delta>0$ such that, for $|x-y| < \delta$, $$\left|\sum_{n\leq N} (nx)/n^2 - \sum_{n\leq N} (ny)/n^2\right| < \epsilon/3.$$ It follows that $|f(x) - f(y)| \leq \epsilon$.

Sean Eberhard
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