Real Analysis Question: continuity of $f(x) = \sum_{q_n<x} 1/n^2$

Sorry, I was having trouble giving the question because I can't figure out how to type in mathematical symbols. Let me try again:

$\{q\}$ is an enumeration of the rational numbers, and

$$f(x)=\sum_{q_n< x}\frac1{n^2}$$

for $x\in\Bbb R$.

Or $f(x)=\sum(1/n^2)$, the index of summation $q_n< x$, and $q$ is an enumeration of the rational numbers.

The goal is to prove that $f$ is continuous at each irrational and discontinuous at each irrational.

I'm having trouble visualizing this series so I'm not sure why it is supposed to be different at irrational as opposed to a rational $x$ ...

I think I need to use the epislon-delta definition of continuity but I'm not sure what epsilon to use.

Solution 1:

For each $x\in\Bbb R$ let $Q_x=\{n\in\Bbb Z^+:q_n<x\}$; then

$$f(x)=\sum_{q_n<x}\frac1{n^2}=\sum_{n\in Q_x}\frac1{n^2}\;.$$

Suppose that $x$ is rational; then $x=q_m$ for some $m\in\Bbb Z^+$. Note that $m\notin Q_x$. However, for any real number $y>x$, $x=q_m<y$, and therefore $m\in Q_y$. It’s also clear that $Q_x\subseteq Q_y$: any rational that’s less than $x$ is certainly also less than $y$. Thus,

$$f(y)=\sum_{n\in Q_y}\frac1{n^2}\ge\sum_{n\in Q_x\cup\{m\}}\frac1{n^2}=f(x)+\frac1{m^2}\;.$$

Since $f(y)\ge f(x)+\dfrac1{m^2}$ for every $y>x$, we have

$$\lim_{y\to x^+}f(y)\ge f(x)+\frac1{m^2}>f(x)\;,\tag{1}$$

implying that $f$ is not continuous at $x$. (The limit in $(1)$ exists because the function $f$ is clearly monotone increasing, and $\{f(y):y>x\}$ is bounded below.) The function $f$ has a jump discontinuity at $x=q_m$: it jumps by $\frac1{m^2}$.

Now suppose that $x$ is irrational, and try to use some of these ideas to show that

$$\lim_{y\to x^-}f(y)=f(x)=\lim_{y\to x^-}f(y)\;.$$

If you get completely stuck, leave a comment, and I’ll see if I can unstick you.