$$f: \displaystyle{R\to[0,1]}$$
I can't come up with one, can anyone else ? :D
How about : $$f(x)=\begin{cases} x, x \in Q \\ 0, x\in I \end {cases}$$ ??
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A monotonous function? – Barry Jun 04 '15 at 19:00
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What about
$$ f(x)=\sum_{y\in \Bbb{Q}, y\leq x} v_y, $$
where the $v_y$ are positive with $\sum_{y\in \Bbb{Q}} v_y<\infty$?
PhoemueX
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Concretely, you can take $q_n$ to be an enumeration of the rationals and $f(x)=\sum_{n : x \leq q_n} 2^{-n}$. – Ian Jun 04 '15 at 19:14
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Could you guys explain this more concretely? Im having trouble grasping it? What is $q_n$ more detailedly ? – Bozo Vulicevic Jun 04 '15 at 19:22
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@Jim Lowe: The set of rationals $\Bbb{Q}$ is countably infinite, so there is a bijective $p :\Bbb{N}\to \Bbb{Q}$. Now set $q_n =p(n)$. It is also easy to see that every number in $[0,1]$ is a jump discontinuity of $f$. One could also show that no other discontinuities exist, but you do not need that for your claim. – PhoemueX Jun 04 '15 at 19:27
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Can you concretely define this bijective or they theory that it exist would be sufficient you would say if you were a professor? – Bozo Vulicevic Jun 04 '15 at 19:42
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@Jim Lowe: If at some point you already heard that the rationals are countable, then just stating that there is a bijective will be sufficient. For a few constructions (though mostly only f injections/surjections), see this post: http://math.stackexchange.com/questions/659302/how-to-prove-that-mathbbq-the-rationals-is-a-countable-set – PhoemueX Jun 05 '15 at 01:05