# 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.

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I think there is something missing from the definition of $f$... –  David Mitra Dec 8 '12 at 3:02
I don't understand: what's the relation between a general $\,x\in\Bbb R\,$ and that sum?? –  DonAntonio Dec 8 '12 at 3:02
Welcome to MSE! I see that you are a new user, and I wanted to let you know a few things. We try to keep our questions self-contained. We don't know what Rudin 7.11 is, for example. Nor do we know how the enumeration of the rational numbers has to do with your function, nor how your function depends on $x$. Also, I encourage you to share your motivation and work so far - this will likely improve the odds that people will help you. For more information, please see the faq. –  mixedmath Dec 8 '12 at 3:06
I suspect that you want $\Bbb Q=\{q_n:n\in\Bbb N\}$ to be an enumeration of the rationals, and you want to define $f:\Bbb R\to\Bbb R$ by $$f(x)=\sum_{q_n<x}\frac1{n^2}\;;$$ is that right? I’m going to edit your question based on that assumption; let me know if it comes out wrong. –  Brian M. Scott Dec 9 '12 at 0:49
For the rational points, consider $f(q_n)$ for some $n$ and then $f(q_n+\delta)$ for any small $\delta$. –  Francis Adams Dec 9 '12 at 1:11

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

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