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$$f(x)=\begin{cases} x+1, & \text{if $x$ is irrational} \\ x, & \text{if $x$ is rational} \end{cases}$$

Show that $f$ is discontinuous on $\mathbb{R}$

edit: It would be preferable if this could be proved without the use of sequences or epsilon-delta

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closed as off-topic by Did, E. Joseph, Daniel W. Farlow, Namaste, Shailesh Nov 20 '16 at 1:06

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  • $\begingroup$ Any ideas? What have you tried? You tagged as epsilon-delta, have you tried applying the definition? $\endgroup$ – CompuChip Nov 19 '16 at 18:56
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    $\begingroup$ Hint: $f(x)=x+D(x)$, where $D$ is the Dirichlet function. $\endgroup$ – user378947 Nov 19 '16 at 18:57
  • $\begingroup$ The epsilon-delta definition was by mistake!Edited the question because of a mistake I made.I have tried the epsilon-delta but I'm supposed to solve it without it. $\endgroup$ – george wold Nov 19 '16 at 19:13
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Let $r$ be a rational.

we have $\;f(r)=r$

and $\;\; \forall n>0$,

$$f(r+\frac{\sqrt{2}}{n})=r+\frac{\sqrt{2}}{n}+1$$

thus

$$\lim_{n\to+\infty}f(r+\frac{\sqrt{2}}{n})=r+1\neq f(r)$$

which means that $\;f\;$ is not continuous at $x=r$.

Let $s$ be an irrational.

we have $f(s)=s+1$ and

$$s=\lim_{n\to+\infty}\frac{\lfloor 10^n s \rfloor}{10^n}$$

but

$$\lim_{n\to+\infty}f(\frac{\lfloor 10^n s \rfloor}{10^n})=\lim_{n\to+\infty}\frac{\lfloor 10^n s \rfloor}{10^n}$$

$=s\neq f(s)$ .

$f$ is not continuous at $x=s$.

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  • $\begingroup$ That's the exact proof I wanted :). Thanks a lot! $\endgroup$ – george wold Nov 19 '16 at 19:14
  • $\begingroup$ @Abdallah Hammam (Y) keep it high! $\endgroup$ – Melina Nov 19 '16 at 19:16
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Regardless of delta we can find a y within delta of x that is more than distance 1 from x (there are rationals and irrationals arbitrarily close to x). But then there exists no delta corresponding to epsilon less than 1.

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