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The problem asks as, whether there exists a continuous function $f: \Bbb{R} \rightarrow \Bbb{R}$ an image of which is set:

  • $[0,1)$
  • $(0,1]\cup[2,3]$
  • $\Bbb{Q}$

So - for $[0,1)$ you can see that you can create function $f$ for which $\lim_{x\rightarrow +\infty} f(x) = 1$, and from $-\infty$ to some finite number it is constant $f(x)=0$ and than you can make a "smooth", parabolic transition between $0$ and $1$. For the second and third set you kinda see that it's not possible, as there is a gap between every rational number and between $1$ and $2$. But how to prove such things formally?

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The second and third contradict the intermediate value theorem.

For the first one, i will give you an example with image $(0,1]$: $x \mapsto \frac{1}{x^2+1}$. I think you are capable of adjusting this to $[0,1).$

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  • $\begingroup$ +1, even though one could think of topological obstructions rather than analytical ones. $\endgroup$ – Giovanni De Gaetano Feb 3 '15 at 11:18
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(topological) Hint:

$(0,1]\cup[2,3]$ and $\mathbb Q$ are both not connected.

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