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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.

Community
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user42912
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    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 $A\subset (0,1)$?

     – JP McCarthy Jan 15 '13 at 14:32
  • @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. – user42912 Jan 16 '13 at 10:38

2 Answers2

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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.

André Nicolas
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  • there is a problem, which element goes to 1/2? – user42912 Jan 15 '13 at 14:40
  • For the inverse of $H$? $3/4$. – André Nicolas Jan 15 '13 at 14:49
  • @user42912: Well that depends on which direction you mean. $H(0)=\frac12$, but $h(\frac34)=\frac12$. – Cameron Buie Jan 15 '13 at 14:50
  • @CameronBuie but the map has to be bijective. – user42912 Jan 15 '13 at 15:09
  • @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? – Cameron Buie Jan 15 '13 at 16:09
  • @CameronBuie the problem is the injectivity, because as you said: $H(0)=\frac12$ and $h(\frac34)=\frac12$. – user42912 Jan 15 '13 at 16:21
  • Note that $H$ is being used for the mapping from $[0,1)$ to $(0,1)$, and $h$ forits inverse. – André Nicolas Jan 15 '13 at 16:23
  • @user42912: You seem to be confusing $H$ with $h$. If we can prove that $H:[0,1)\to(0,1)$ is a bijection, then $h=H^{-1}:(0,1)\to[0,1)$ is also a bijection. Let $A=\left{1-\frac1{2^n}:n\text{ a nonnegative integer}\right}$, a subset of $[0,1)$. Now $H:[0,1)\to(0,1)$ is given by $$H(x)=\begin{cases}g(x) & x\in A\f(x) & x\notin A.\end{cases}$$ Since $f,g$ are injective, then $H$ fails to be injective only if $\exists x_1,x_2\in[0,1)$ with $x_1\in A,x_2\notin A$ and $g(x_1)=f(x_2):=x_2$. But if $x_1\in A$, then $g(x_1)\in A$ (check that), so no such $x_1,x_2$ exist. Thus, $H$ is injective. – Cameron Buie Jan 15 '13 at 18:47
  • @CameronBuie thank you for your comments, but how can you find this $A$? – user42912 Jan 16 '13 at 02:29
  • @user42912: If you're looking for a way to find it yourself, see Henning's post--specifically, the paragraph starting with "What happens next...." He describes there explicitly how it can be found. (Note that since $f=f^{-1}$, his definition of $H$ and $h$ are precisely the same as the one given in this post, though it may be a little clearer from Henning's definition how Cantor-Bernstein is coming into play.) – Cameron Buie Jan 16 '13 at 05:29
  • @CameronBuie 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? – user42912 Jan 16 '13 at 10:29
  • @user42912: Note that the map you propose fails to be onto, since nothing maps to $0$. You've got your cases backward. – Cameron Buie Jan 16 '13 at 14:38
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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 $h$.

hmakholm left over Monica
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