Friday, January 19, 2018

real analysis - Bijection from (0,1) to [0,1)




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.



Answer



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 $h$.


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