Can exist a continuous function $f:[0, 1]\rightarrow \mathbb{R}$ so that $f(x)\in\mathbb{Q}$ if $x\in\mathbb{I\cap [0, 1]}$ and $f(x)\in\mathbb{I}$ if $x\in\mathbb{Q\cap [0, 1]}$? Why yes? why not?
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Are you thinking of this question? – David Mitra May 09 '13 at 00:00
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See this excellent answer. – Cameron Buie May 09 '13 at 00:03
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Minor variants of this question have been asked several times before, see the two questions linked above, or this one: http://math.stackexchange.com/questions/251683/f-mathbbr-setminus-mathbbq-subseteq-mathbbq-and-f-mathbbq-sub?lq=1 All of these should provide you with many ways to tackle the problem! – Tom Oldfield May 09 '13 at 00:44
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HINT: I assume that $\Bbb I$ stands for the set of irrational numbers. (In my experience $\Bbb P$ is more common.) For each $q\in\Bbb Q$ let $F(q)=\{x\in[0,1]:f(x)=q\}$; then $\Bbb I\cap[0,1]=\bigcup_{q\in\Bbb Q}F(q)$. Apply the Baire category theorem to conclude that there are $a,b\in\Bbb I\cap[0,1]$ and a $q\in\Bbb Q$ such that $a<b$ and $\Bbb I\cap[a,b]\subseteq F(q)$. Then use continuity to conclude that $f$ is constant on $[a,b]$ and so get a contradiction.
Brian M. Scott
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