Sunday, 23 June 2013

real analysis - Proving a function is discontinuous by constructing a sequence

Let f:[0,1]R be a function that maps rationals to irrationals and irrationals to rationals. I could show that any such f can't be continuous by showing that its range is at most countable and going from there. We also know that for a function g to be continuous, for all sequences xn that converge to x, g(xn) has to converge to g(x). So since f is discontinuous, there exists a sequence for which this does not hold. Is it possible to construct it without assuming anything further about f? In other words, is it possible to prove that f is discontinuous by constructing such a sequence?

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