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Possible Duplicate:
Construct a monotone function which has countably many discontinuities

How would I construct a monotone function on $[0,1]$ that is discontinuous at each rational point.

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    One rational at a time?2011-10-31
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    I think titi is looking for a monotone function $f$ such that the set of points of continuity of $f$ is $[0,1] \setminus\mathbb{Q}$2011-10-31
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    @mixedmath: I don't understand what you mean by one rational at a time.2011-10-31
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    @leo: That's exactly what I'm looking for.2011-10-31
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    Take $r:\mathbb{N}\to\mathbb{Q}$ an enumeration of the rational numbers in $[0,1]$. I think the function $$f(x)=\sum_{r(k)\lt x}2^{-k}$$ is a example. This function was discussed here before but I can't find the post.2011-10-31
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    Just clarifying, I mean $$f(x)=\sum_{k:r(k)\lt x} 2^{-k}$$2011-10-31
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    Finally found, actually exact duplicate: http://math.stackexchange.com/q/69317/82712011-10-31
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    @leo: Thanks for the link. I'd appreciate it though if you or anyone could explain to me why the function is not right continuous.2011-10-31
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    ok, consider $x$ rational in $[0,1]$. Then $x+1/n\to x$ and note that $$f(x+1/n)\to \sum_{k:r(k)\leq x}2^{-k}\neq f(x)$$2011-10-31
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    @leo: Thanks....2011-10-31

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