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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.

Martin Sleziak
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Kathy
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    I think there is something missing from the definition of $f$... – David Mitra Dec 08 '12 at 03:02
  • I don't understand: what's the relation between a general $,x\in\Bbb R,$ and that sum?? – DonAntonio Dec 08 '12 at 03:02
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    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]. – davidlowryduda Dec 08 '12 at 03:06
  • 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 09 '12 at 00:49
  • For the rational points, consider $f(q_n)$ for some $n$ and then $f(q_n+\delta)$ for any small $\delta$. – Francis Adams Dec 09 '12 at 01:11
  • See also: http://math.stackexchange.com/questions/421750/a-problem-on-continuity-of-a-function-on-irrationals – Martin Sleziak Jan 30 '14 at 08:19

1 Answers1

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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.

Brian M. Scott
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  • Just a tiny bit stuck...do I need two separate proofs for if y is rational or irrational or can I use one proof for either y? It seems obvious that if x and y are irrational, then as y-->x, f(y)-->f(x) but I'm not sure if it is evident if x is irrational but y is rational. Do I need the delta-epsilon def if y is rational? – Kathy Dec 11 '12 at 01:58
  • Also, thanks for your help! The explanation really made it clearer, at first I wasn't sure why it was continuous at irrationals not rationals, but now I can completely understand the concept. – Kathy Dec 11 '12 at 01:59
  • I didn't quite understand the justification that it is continuous nis irrational, it seems to me to say the obvious, how did you formalize it? – Pierre Jun 30 '22 at 08:22
  • @PedroCosta: Here’s a sketch of one way. If $f$ is not continuous at $x$, there is a sequence $\langle x_n:n\in\Bbb N\rangle$ converging to $x$ such that $\langle f(x_n):n\in\Bbb N\rangle$ does not converge to $f(x)$. Show that without loss of generality we may assume that this sequence is either strictly increasing or strictly decreasing. If it’s strictly increasing, $Q_x=\bigcup_{n\in\Bbb N}Q_{x_n}$, if it’s strictly decreasing, $Q_x=\bigcap_{n\in\Bbb N}Q_{x_n}$, and in either case it’s easy to check that $f(x)=\lim_{n\to\infty}f(x_n)$, contradicting the discontinuity of $f$ at $x$. – Brian M. Scott Jul 01 '22 at 04:21