Example of discontinuous function?

(a) Let f(x) = x, where x = 1/n for any n in Z\{0}, and 0 otherwise.
Then f(0) = 0 and f(x) is continuous at 0 since for any ε > 0, if |x| < ε then |f(x)| < ε.
However, in any neighbourhood of 0, there are countably infinitely many points of the form 1/n for n in Z\{0}, and at each of these points f is discontinuous [f(1/n) = 1/n; let ε = 1/2n, then for any δ > 0 we can find an irrational z with |z - 1/n| < δ. But f(z) is 0, so |f(z) - f(1/n)| = 1/n > ε]. So f is discontinuous at countably infinitely many points in any neighbourhood of 0.

For (b), consider the function
f(x) = { x sin (π/x), x rational and not 0,
-x sin (π/x), x irrational,
0, x = 0

f is continuous at x = 0 because f(0) = 0 and |f(x)| < |x|, as in part (a).

For convenience, let's write g(x) = x sin (π/x).

Note that at any point other than 0, g(x) and -g(x) are both continuous functions and will be equal iff sin (π/x) = 0, i.e. x = 1/n for some integer n, in which case they are both 0. So f is continuous at these points (let x0 = 1/n; for any ε > 0, choose δ > 0 such that |x - x0| < δ => |g(x)| < ε; the same δ will work for -g(x), and hence |x - x0| < δ => |f(x) - f(x0)| = |f(x)| < ε).

If x0 is not one of these points, let ε = |g(x0)| > 0. Since g is continuous, choose δ' such that |x - x0| < δ' => |g(x) - g(x0)| < ε. For any δ > 0, let δ" = min{δ, δ'}.

If x0 is rational, choose x to be an irrational point in (x0 - δ", x0 + δ"). Then f(x0) = g(x0) and f(x) = -g(x). But we know |g(x) - g(x0)| < ε = |g(x0)|, so g(x) is strictly between 0 and 2g(x0). Hence f(x) is strictly between 0 and -2g(x0), and f(x) - f(x0) = f(x) - g(x0) is strictly between -g(x0) and -3g(x0). So |f(x) - f(x0)| > |g(x0)| = ε. Hence any neighbourhood of x0 contains a point x with |f(x) - f(x0)| > ε and so f is not continuous at x0.

If x0 is irrational, choose x to be an irrational point in (x0 - δ", x0 + δ"). Then we have f(x0) = -g(x0) and f(x) = g(x), and by a similar argument to the above we get that f(x) - f(x0) is strictly between g(x0) and 3g(x0), so |f(x) - f(x0)| > |g(x0)| = ε and again f is discontinuous at x0.

So f(x) is continuous at 0 and at any point x = 1/n for nonzero integers n, but discontinuous everywhere else. Any neighbourhood of 0 contains countably infinitely many points of the form 1/n for n in Z\{0} so we have proven what we needed to.

For (b) the Thomae function is continuous at the irrationals and discontinuous at the rationals
see https://www.math.washington.edu/~morrow/334_11/thomae.pdf

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