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

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 $x = 0$?

• Where exactly is the jump? At some specific $x\neq 0$ or at every $x\neq 0$? – MPW Feb 28 '14 at 2:50
• formatted now, I meant for every x value. – Shuri Feb 28 '14 at 2:52
• If I am not mistaken, I am pretty sure you cannot have a function with jump discontinuity everywhere. – Pratyush Sarkar Feb 28 '14 at 2:55
• To have a jump, the left and right limits must both exist but be different. This can't happen everywhere. – MPW Feb 28 '14 at 3:03
• @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. – 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 $0$.
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 $x=0$...
• OP asks for jump discontinuity for all $x \neq 0$ in the comments. – Pratyush Sarkar Feb 28 '14 at 3:03