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Consider the function:

$f(x)= \begin{cases} 1/n \quad &\text{if $x= m/n$ in simplest form} \\ 0 \quad &\text{if $x \in \mathbb{R}\setminus\mathbb{Q}$} \end{cases} $

Prove that the function is continuous at every irrational point and also that the function is not continuous at every rational point. Also, we can say that the function is continuous at some point $k$ if $\displaystyle\lim_{x \to k} f(x)=f(k)$.

I was thinking of doing an epsilon delta proof backwards using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ for rational points and irrational points. Any ways on how to expand on this are welcome.

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    $\begingroup$ The way to do this is to show that in arbitrarily small neighborhoods of an irrational point $x$, rationals approximating $x$ must have large denominators in reduced form. $\endgroup$
    – Seth
    Commented Oct 4, 2012 at 12:11
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    $\begingroup$ Hint: $|\frac ab-\frac xy|=|\frac{ay-bx}{by}|\ge \frac1{by}$ if $\frac ab\ne\frac xy$ $\endgroup$ Commented Oct 4, 2012 at 12:31

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This function is known as Thomae's function and is an excellent example of a function that is continuous at uncountably infinite number of points while discontinuous at countably infinite number of points. Note that the reverse result is not possible, i.e. one cannot have a function that is discontinuous on an uncountably infinite dense set and continuous on a countably infinite dense set.

As for the proof , the discontinuity part follows from the fact that irrationals are dense in $ R $ as you rightly thought. The continuity part is a little bit more involved and it would help to employ the Archimedean principle in tandem with the $ \epsilon$-$\delta $ definition.

Crudely, look at $ x=\sqrt3 $ , see that using the non-terminating decimal expansion i.e. $ \sqrt 3=1.732148\ldots $ , one can forge a sequence of rationals $ \ \ 1 , 17/10 ,173/100 ,\ldots \ $ that converges to $ x $, while at the same time your $ f(x_n) $ is $ \ \ 1/10,1/100,1/1000,\ldots \ $ which converges to $ 0 $ which is nothing but $ f(x) $. This is merely an illustration of the continuity and not a rigorous proof.

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  • $\begingroup$ Vishesh Can you help Connection of complex $e^z$ and real Dirichlet please? $\endgroup$
    – BCLC
    Commented Aug 19, 2018 at 6:26
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    $\begingroup$ Can you tell me why the reverse result is not possible? $\endgroup$
    – Mr.xue
    Commented Aug 6, 2020 at 16:30
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    $\begingroup$ @Mr.xue Sorry for the late repsonse. That would be down to the Baire Category theorem if you are aware of that result. $\endgroup$
    – Vishesh
    Commented Aug 20, 2020 at 7:58
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You cannot do this using simple rules about limits, like de l?Hôpital's rule or the continuity of "analytic expressions", etc. You have to go back to the actual definition of continuity when solving this problem. Here are a few hints:

In the first place you want to prove different things about a given $x\in{\mathbb R}$, depending on whether $x$ is rational or irrational.

If $x$ is rational then $f(x)$ is a certain positive number. It is claimed that $f$ is not continuous at $x$. Assume to the contrary that $f$ is continuous at $x$. Then there is a neighborhood $U:=U_\delta(x)$ such that $f(t)>{f(x)\over2}$ for all $t\in U$. Hm?

If $x$ is irrational then $f(x)=0$. It is claimed that $f$ is continuous at $x$. This means that for any $\epsilon>0$ we should be able to produce a neighborhood $U:=U_\delta(x)$ such that $f(t)<\epsilon$ for all $t\in U$. Which points $t\in{\mathbb R}$ could violate the condition $f(t)<\epsilon\ $?

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  • $\begingroup$ Are they going to be the rationals? $\endgroup$ Commented Oct 4, 2012 at 13:27
  • $\begingroup$ @Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals. $\endgroup$ Commented Oct 4, 2012 at 13:31
  • $\begingroup$ @ChristianBlatter the inequality $f(t)<\epsilon$ is violated by those rational numbers $t = a/b$ in the lowest terms for which $b\leq 1/\epsilon$. Am I right? If so, what next? $\endgroup$ Commented Apr 1, 2022 at 20:45

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