Hint: Every interval of the real line contains infinite rationals and infinite irrationals.
f:[0,2pi] -> R with f(x)= 0 for x in Q and f(x)=sin(x) for x in R\Q.
At which points c in [0,2pi] is f continuos?
I'm thinking it's discontinuous everywhere except 0,pi and 2pi.
I've come up with this to show discontinuity at the rational points in [0,2pi]
Consider a sequence with for every natural n. Then does not converge so discontinuous for every s.t c is rational.
But I'm not sure about how to prove the discontinuity/continuity at other points?