Let $f$ be a function from $\Bbb R$ to $\Bbb R$ such that $f(x)$ is rational when $x$ is irrational, and $f(x)$ is irrational when $x$ is rational. Can $f$ be continuous?
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user130550
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1No. $F$ is a non-constant function with countable range. The Intermediate Value Theorem prohibits it from being continuous. – David Mitra Feb 21 '14 at 19:30
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Yes it is exactly same question.Thanks. – user130550 Feb 21 '14 at 19:34
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Not possible.
Let $x_1,x_2$ be irrational. If $f(x_1)\neq f(x_2)$ then the range $[f(x_1),f(x_2)]$ would contain uncountably many irrationals that would have to be in the range of $f(\mathbb{Q})$, which is impossible. Therefore $f$ is constant on irrationals. By the density of irrationals $f=const$.
Michael
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This answer has the oddity of being simultaneously accepted and downvoted. Would the downvoter explain the reason for -1? – Michael Feb 21 '14 at 21:14