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I have sort of a silly question, since I feel confused.

Let $f(x) = \begin{cases} 1, & x\in \mathbb{Q}\\ 0, & x\in \mathbb{R} \setminus \mathbb{Q} \end{cases}$

Intuitively, this function has a discontinuity at every rational number, and this set is countable. On the other hand, I could also say that it has a discontinuity at every irrational number, and this set is uncountable.

Which one of those statements is true?

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    $\begingroup$ Here is another function that often appears in exercises: Let $f(x)=0$ when $x$ is irrational. And when $x=p/q$ where $p,q$ are co-prime integers then $f(x)=1/|q|.$ Then $ f$ is discontinuous at every non-zero rational ,but continuous at $0$ and at every irrational. $\endgroup$ – DanielWainfleet Jun 16 '17 at 18:14
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Both of your statements are true. The function is not continuous at any point, since you can always find two sequences (one consisting of rational numbers and one of irrational numbers) converging to your point, yet the limits of the function values will be different.

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  • $\begingroup$ Thus the set of discontinuities is the whole $\mathbb{R}$? $\endgroup$ – Angie Jun 16 '17 at 8:07
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    $\begingroup$ Yes - it is not continuous at any point in $\mathbb{R}$, as I indicated. $\endgroup$ – SvanN Jun 16 '17 at 8:08

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