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We have a function $f: \mathbb R \to \mathbb R$, $x \mapsto \frac{1}{n}$ when $x \in \mathbb Q, x = \frac{z}{n}, z \in \mathbb Z, n \in \mathbb N, z \text{ and } n$ are coprime, $x \mapsto 0$ when $x \notin \mathbb Q$.

We need to show that $f$ is continuos in any $x \notin \mathbb Q$.

I could not find a formal argument to prove the claim. I guess the general idea here is that, any sequence of rationals approaching an irrational has denominators going to infinity.

zesy
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1 Answers1

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It's more a counting argument.

It is sufficient to show this for $x\in [0,1]$. Let $\varepsilon >0 $ be given. Then the set $A_n$ of integers $n$ such that $\frac{1}{n}\ge\varepsilon$ is finite, so this is also true for the set $B_n$ of pairwise different rationals with denominator $\le n$ (in $[0,1]$). Hence for each $z\in [0,1]\backslash B_n$ you have $f(z) < \varepsilon$ Now choose any irrational number. It's distance to $B_n$ is positive, allowing you to choose $\delta$ for the usual continuity relation.

Thomas
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  • above you meant numerator not denominator, right? – zesy Aug 24 '16 at 12:55
  • @SergeyZykov no. you have to count the number of the form $\frac{i}{1},\frac{i}{2}, \frac{i}{3}, \dots, \frac{i}{n}$. The number of pairwise different rationals of this form in $[0,1]$ is quite obviously bounded by $n+ (n-1) + (n-2) + \dots +1$. So you look at rationals in $[0,1]$ with denominator $\le n$. The numerator is then automatically bounded. If you bound the numerator you have infinitely many rationals. – Thomas Aug 24 '16 at 15:23
  • @ OK, I have got how you mean it. Thanks for the explanation. – zesy Aug 25 '16 at 12:32