Is it true that for every interval (not singleton ) $I$ in $\mathbb R$ , there exists a continuous surjection $f : I \setminus \mathbb Q \to I \cap \mathbb Q$ ?
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It can't be a singleton anyway, since if it is a singleton exactly one of $I\cap\Bbb Q$ and $I\setminus\Bbb Q$ is non-empty. – Asaf Karagila Jun 07 '16 at 07:47
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You should specify the topology you want $f$ to be continuous with respect to. – zuggg Jun 07 '16 at 13:07
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I will show it for $I=[0,1]$. Take $$ f\colon I\setminus \Bbb{Q} \to I \cap \Bbb{Q} $$ to be the constant function with value $\varphi (n)$ on $[\frac{1}{n+1},\frac{1}{n}]$ where $\varphi \colon \Bbb{N} \to \Bbb{Q}\cap I$ is any surjection.
This function is only "discontinuous" at rational points, so $f$ is continuous and by construction $f$ is surjective.
For the general case just fit $[0,1]$ into your favorite interval and extend the function by $0$ on the left and right.
user60589
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Minor pedantic point: you need to make one endpoint or the other of $[1/(n+1),1/n]$ be open so that your $f$ is not ambiguously defined at the $1/n$ points. – Ian Jun 07 '16 at 22:29
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Oh, right, yes, I guess it should be $[1/(n+1),1/n] \setminus \mathbb{Q}$ but now that's an even more pedantic point. Sorry. – Ian Jun 07 '16 at 22:32