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Let $M=\{\frac{1}{n} : n\in\mathbb{N}\}$.

I am trying to find a simple surjective and injective function $[0,1]\to [0,1]\setminus M$.

let define that $[0,1]\setminus M = Y$.

I can't understand how to handle with questions like this. I understand that the set $Y$ has all the elements as $[0,1]$ without elements like $1,\frac{1}{2},\frac{1}{3},...\frac{1}{n}$ but for example if I want to send $1$, what the value of function $f(1)$ will be if all other elements are already taken and this function should be injective (because $Y\subseteq [0,1]$).

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$$f(x) = \begin{cases} x & \nexists k \in \Bbb N: x = \frac 2k \\ \frac 2{2k-1} & \exists k \in \Bbb N: x = \frac 2k \end{cases}$$

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  • $\begingroup$ Can you explain the process of thinking of how did you get to this function? $\endgroup$ – kickstart Dec 30 '17 at 0:25
  • $\begingroup$ @kickstart added picture $\endgroup$ – Kenny Lau Dec 30 '17 at 0:27
  • $\begingroup$ You should have an explicit $f(0)=0$ case. $\endgroup$ – Henning Makholm Dec 30 '17 at 1:02
  • $\begingroup$ @HenningMakholm why? $\endgroup$ – Kenny Lau Dec 30 '17 at 1:02
  • $\begingroup$ Because $\frac{1}{2x}$ is not defined (and $x=\frac2k$ is false) for $x=0$. $\endgroup$ – Henning Makholm Dec 30 '17 at 1:04

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