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 $Y\subseteq [0,1]$).


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.