I am arguing with someone that the following function have or does not have any jump discontinuities: $$f = \begin{cases} 1, & x \in \mathbb{Q}\\ 0, & x \notin \mathbb{Q} \\ \end{cases}$$

Can someone verify?

  • $\begingroup$ Do you believe that it does or doesn't have jump discontinuities? $\endgroup$ – Ethan Alwaise Feb 22 '16 at 2:58

I would define a jump discontinuity of $f(x)$ as a point $x_0$ where the one-sided limits $\lim_{x\to x_0^+}f(x)$ and $\lim_{x\to x_0^-}f(x)$ exist but are not equal.

The function you've defined does not have one-sided limits at any $x_0$, so doesn't have any jump discontinuities.


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