I stumbled an a past test question where the student was asked to provide any example of a graph that has those two particular properties.

I now there should be a piece-wise function, an maybe a case where I could set x to equal a random y value, but I haven't gotten much further than that.

How can a graph have a jump discontinuty at $x \not= 0$ and removable discontinuity at $x = 0$?

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    $\begingroup$ Where exactly is the jump? At some specific $x\neq 0$ or at every $x\neq 0$? $\endgroup$ – MPW Feb 28 '14 at 2:50
  • $\begingroup$ formatted now, I meant for every x value. $\endgroup$ – Shuri Feb 28 '14 at 2:52
  • $\begingroup$ If I am not mistaken, I am pretty sure you cannot have a function with jump discontinuity everywhere. $\endgroup$ – Pratyush Sarkar Feb 28 '14 at 2:55
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    $\begingroup$ To have a jump, the left and right limits must both exist but be different. This can't happen everywhere. $\endgroup$ – MPW Feb 28 '14 at 3:03
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    $\begingroup$ @MPW: In fact, for every function $f:{\mathbb R} \rightarrow {\mathbb R},$ the set of points $x \in \mathbb R$ where the left subsequent limits of $f$ at $x$ differs from the set of right subsequent limits of $f$ at $x$ is countable. This was proved by William H. Young sometime around 1908. $\endgroup$ – Dave L. Renfro Feb 28 '14 at 15:38

$f(x) = \chi_{[1,\infty)}(x) + \chi_{\{0\}}(x)$ has a jump at $1$ and a removable discontinuity at $0$.

EDIT: Withdrawn, since you have clarified your requirement in a comment.

Perhaps you are thinking of something like $g(0) = 1$ and $g(x) = x\cdot(\chi_{\mathbb{Q}}(x)-\chi_{\mathbb{R\setminus Q}}(x))$ for $x\neq 0$? But I don't think this has jumps since the one-sided limits don't exist away from $x=0$...

  • $\begingroup$ OP asks for jump discontinuity for all $x \neq 0$ in the comments. $\endgroup$ – Pratyush Sarkar Feb 28 '14 at 3:03
  • $\begingroup$ @PratyushSarkar: Yes, I saw that was added after I posted answer. He didn't answer my original query quickly enough. And I agree with your comment--I think this is impossible. $\endgroup$ – MPW Feb 28 '14 at 3:06

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