$f: \mathbb{R} \to \mathbb{R}$
$f(x) \notin \mathbb{Q}$ for $ x \in \mathbb{Q}$
$f(x) \in \mathbb Q $ for $x \notin \mathbb{Q} $
I was making cases but couldnot get to anything
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For sure this function cannot be Lipschitz because of measure theory argument but there are continuous functions such that the image of a set of measure zero is nonzero. Therefore measure theory won't help us here :( – Jan Olszewski Apr 19 '23 at 23:15
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@JanOlszewski im not that advanced either. its just first course in analysis – Sophie Clad Apr 19 '23 at 23:16
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2Hint: The image of such a function would be at most countable infinite. Why is this a contradiction? (think of something similar to your previous question) – Mark Apr 19 '23 at 23:21
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@Mark why is image of such function be atmost countable infinite ? – Sophie Clad Apr 20 '23 at 00:15
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@SophieClad The image of the irrationals is contained in $\mathbb{Q}$, and so countable. The image of the rationals is surely countable, because any function would satisfy $|f(\mathbb{Q})|\leq |\mathbb{Q}|$. – Mark Apr 20 '23 at 00:17