I'm trying to solve the following question:

Let $f:(0,1)\to [0,1)$ and $g:[0,1)\to (0,1)$ be maps defined as

$f(x)=x$ and $g(x)=\frac{x+1}{2}$. Use these maps to build a bijection $h:(0,1)\to [0,1)$

I've already proved that these maps are injectives, and following the others questions on the site such as

Continuous bijection from $(0,1)$ to $[0,1]$

How to define a bijection between $(0,1)$ and $(0,1]$?

I think I can found such $h$, but the problem is that we have to use only $f$ and $g$ to build $h$.

I need help.

Thanks a lot.

  • 1
    $\begingroup$ How about something like $\displaystyle h(x):=\left\{\begin{array}{cc}f(x) & \text{ for }x\in A\\ g(x) & \text{ for }x\in (0,1)\backslash A\end{array}\right.$, for some $A\subset (0,1)$? $\endgroup$ – JP McCarthy Jan 15 '13 at 14:32
  • $\begingroup$ @JpMcCarthy yes I've just found a map like this, with A equals to $1-2^{-n}$. Am I correct? thank you for your comment. $\endgroup$ – user42912 Jan 16 '13 at 10:38

The slick answer is

Since each of $f$ and $g$ are clearly injective, apply the Cantor-Bernstein theorem to $f$ and $g$. This gives a bijection $(0,1)\to[0,1)$.

It may be implicit in the question that you're supposed to "crank the handle" on the proof of Cantor-Bernstein in order to find an explicit description of the particular bijection it produces. In that case you'd start by finding the ranges (not merely the domain) of $f$ and $g$, namely $(0,1)$ and $[\frac12,1)$.

What happens next depends on the details of the proof of Cantor-Bernstein you're working with, but in one that is easiest to apply here, we look for the the part of $[0,1)$ that is not in the range of $f$, which is the singleton $\{0\}$. If we iterate $f\circ g$ on this, we get $A=\{1-2^{-n}\mid n\in \mathbb N\}$. Then the bijection $H:[0,1)\to(0,1)$ is $$ H(x) = \begin{cases} g(x) & x\in A \\ f^{-1}(x) & x\notin A \end{cases} $$ Now unfold this definition and invert it to get $h$.

  • $\begingroup$ Thanks for your answer, I'm seeing right now the Cantor-Bernstein theorem, how can you get this $A$? $\endgroup$ – user42912 Jan 16 '13 at 1:26
  • $\begingroup$ Can you add a little bit of information how you get this A, please? $\endgroup$ – user42912 Jan 16 '13 at 5:36
  • $\begingroup$ I found this function $F:(0,1)\to[0,1)$ where $F(x)=f(x)$, if $x$ is in $A$ and $F(x)=g^{-1}(x)$, if x isn't in A. Am I correct? $\endgroup$ – user42912 Jan 16 '13 at 10:34
  • $\begingroup$ @user42912: A is just $\{0, f(g(0)), f(g(f(g(0))), f(g(f(g(f(g(0)))))),\ldots\}$. Your proposed definition of $F$ can't be right, because $1/3$ is not in A, but $g^{-1}(1/3)$ is not even defined. $\endgroup$ – Henning Makholm Jan 16 '13 at 14:41
  • $\begingroup$ which proof of Cantor-Bernstein are you using? thank you $\endgroup$ – user42912 Jan 17 '13 at 1:59

If $x$ has shape $1-\frac{1}{2^n}$, where $n$ is a non-negative integer, let $H(x)=g(x)$. Otherwise, let $H(x)=f(x)$.

This gives a bijection in the "wrong" direction. Take the inverse.

  • $\begingroup$ there is a problem, which element goes to 1/2? $\endgroup$ – user42912 Jan 15 '13 at 14:40
  • $\begingroup$ For the inverse of $H$? $3/4$. $\endgroup$ – André Nicolas Jan 15 '13 at 14:49
  • $\begingroup$ @user42912: Well that depends on which direction you mean. $H(0)=\frac12$, but $h(\frac34)=\frac12$. $\endgroup$ – Cameron Buie Jan 15 '13 at 14:50
  • $\begingroup$ @CameronBuie but the map has to be bijective. $\endgroup$ – user42912 Jan 15 '13 at 15:09
  • $\begingroup$ @user42912: Ah, but both $H:[0,1)\to(0,1)$ and $h=H^{-1}:(0,1)\to[0,1)$ are bijections. Are you having trouble seeing injectivity or seeing surjectivity? $\endgroup$ – Cameron Buie Jan 15 '13 at 16:09

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