- #1

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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).