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I was asked to find a function that is continuous on Z but discontinuous on R\Z. as I'm new to continuity I want a feedback on the function I've created, and also tips on what to look for in these types of question!

$f: \mathbb{R}\to \mathbb{R}$ such that \begin{align} f(x) = \begin{cases} 1 & \text{if $x\notin \mathbb{Z}$} \\ \lfloor x \rfloor & \text{if $x \in \mathbb{Z}$} \end{cases} \end{align}

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  • $\begingroup$ That one is continuous for $1≤x<2$, for example. $\endgroup$ – lulu Dec 28 '20 at 17:31
  • $\begingroup$ A tip for this type of question: to show continuity at a point $x\in\mathbb{R}$, you can apply the usual $\epsilon$-$\delta$ definition of continuity. Do you know it? $\endgroup$ – JWP_HTX Dec 28 '20 at 17:32
  • $\begingroup$ To get started, I suggest looking at the function defined on rationals by $f\left(\frac pq\right)=\frac pq$ and by $f(x)=0$ if $x$ is irrational. Show that $f$ is continuous only at $x=0$. Use that to build your example. $\endgroup$ – lulu Dec 28 '20 at 17:33
  • $\begingroup$ $\lfloor x \rfloor =x$ when $x \in \mathbb Z$ so you suggestion is continuous if $x=1$ or $x \in \mathbb R \backslash \mathbb Z$, but is discontinuous if $x \in \mathbb Z \backslash \{1\}$. Almost the opposite of what you are asking for $\endgroup$ – Henry Dec 28 '20 at 18:01
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Actually, the function is discontinuous at every integer other than $1$.

You can take, for instance,$$f(x)=\begin{cases}\sin(\pi x)&\text{ if }x\in\Bbb Q\\0&\text{ otherwise.}\end{cases}$$Can you check that it works?

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  • $\begingroup$ I guess any non-zero value would work in place of $\sin(\pi x)$? $\endgroup$ – Al.G. Dec 28 '20 at 18:12
  • $\begingroup$ If I put $1$ instead of $\sin(\pi x)$, then I get a function which is discontinuous everywhere. $\endgroup$ – José Carlos Santos Dec 28 '20 at 18:15
  • $\begingroup$ Oh I get it, I misunderstood the question -- I was thinking about the restriction $f|_{\mathbb Z}$ being continuous (albeit on a discrete set...). $\endgroup$ – Al.G. Dec 28 '20 at 18:24
  • $\begingroup$ Nice! I've used your suggestion, but left f(x) = sin(πx) * D(x), D(x) being the derichlet fucntion, that way it will always be 0 on intergers, but not continuous on not integers! $\endgroup$ – MathCurious Jan 4 at 7:23

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