Let $f : [0, 1] \to \mathbb{R}$ 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 $x_n$ that converge to $x$, $g(x_n)$ 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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