Give an example of a function f:(0,1)-->Reals which is continuous at exactly the irrational points in (0,1). I think the function f such that f(x)=1/n if x is rational in (0,1) (x=m/n for some n not 0) and f(x)= 0 if x is irrational in (0,1) should work. I get the reason why f is continuous at the irrationals, but what would be a convincing argument to show that f is not continuous at the rationals? I mean, there should be an e>0 s.t. for every d>0, we have |x-xo|<d but |f(x)-f(xo)|> or eq. to e. (for every rational xo).