1
$\begingroup$

I have to Show $f:(-1,1) \to \mathbb{R} ;\quad f(x)=\frac{x}{1-|x|}$ is a homeomorphism from (-1,1) to $\mathbb{R}$

I am considering a function to be a homeomorphism if it is continuous, has an inverse and this inverse is also continuous.

So firstly I argued $f$ is continuous at $(-1,1)$ since both $(x) \quad \& \quad \ (1-|x|)$ are continuous there. Also, I should add that $(1-|x|)$ is never 0 at $(-1,1)$.

So here comes my first problem: I wanted to show that $f$ is an injection because than I could use a theorem which states:

if $f$ is a continuous injection from any interval to $\mathbb{R}$, then its inverse is also continuous.

The thing is, I can't manage to prove $f$ is an injection. I've tried to show that $\frac{x}{1-|x|}=\frac{y}{1-|y|} \Rightarrow x=y $ but I couldn't isolate things properly, I guess.

Any help would be appreciated. Also, this was the only way I could think to solve the problem. If you have any other method, feel free to share :)

$\endgroup$
3
  • 1
    $\begingroup$ Another one: math.stackexchange.com/q/1138344/42969 – both found with Approach0 $\endgroup$
    – Martin R
    Oct 19, 2019 at 17:29
  • $\begingroup$ Multiplying your equation by $(1-|x|)(1- |y|)$ results in $x(1-|y|) = y(1-|x|)$. $\endgroup$
    – amsmath
    Oct 19, 2019 at 17:31
  • $\begingroup$ you can split the function to two intervals according to its behaviour, then differentiate to see its strictly increasing on each interval and hence a bijection. $\endgroup$
    – omer
    Oct 19, 2019 at 18:23

3 Answers 3

2
$\begingroup$

You want to show that the function $f : (-1,1)\to\mathbb R$ is bijective.

First we prove injectivity, as you suggested. For this, let $f(x) = f(y)$, that is, $\frac x{1-|x|}=\frac y{1-|y|}$. This can only be true if $x$ and $y$ have the same signs (or one of them is zero, but then the other one also must be zero). Multiplying by $(1-|x|)(1-|y|)$ gives $x(1-|y|) = y(1-|x|)$, i.e., $x-x|y| = y-y|x|$. But as $x$ and $y$ have the same sign, we have $x|y| = y|x|$. So, it follows that $x=y$.

Now, surjectivity. For this, let $z\in\mathbb R$ be arbitrary. We have to find some $x$ such that $f(x) = z$, i.e., $\frac x{1-|x|} = z$. This is equivalent to $x+z|x| = z$. If $z\ge 0$, then we try $x\ge 0$ for which the equation is $(1+z)x = z$ and so $x = \frac z{1+z}$, which is in $[0,1)$ and therefore indeed a solution. Let $z<0$. Here, we try $x<0$, for which the equation is $(1-z)x = z$ with solution $x = \frac z{1-z}$. This is indeed $<0$ and $>-1$ (i.e., $x\in (-1,0)$). Hence, for each $z\in\mathbb R$ we have found a solution of $f(x) = z$.

So, $f:(-1,1)\to\mathbb R$ is bijective and thus has an inverse function. As you said already, this inverse is then continuous and we're done.

$\endgroup$
1
$\begingroup$

$$f(0<x<1)=\frac{x}{1-x}, f(-1 < x<0)=\frac{x}{1+x} \implies f(0)=0, f'(0 <x<1)=\frac{1}{(1-x)^2}>0, f'(-1<x<0)=\frac{1}{(1+x)^2}>0.$$ The function is increasing on $x \in (1,1)$ So the function $f(x):(-1,1) \rightarrow R$ is bijection and $f^{-1}(x)$ exists and it is given as $$f^{-1}(x<0)=\frac{x}{1-x},~ f^{-1}(x>0)=\frac{x}{1+x},~f^{-1}(0)=0.$$ See the fig. belpw $f(x)$ is blu and $f^{-1}(x)$ is the red one.

enter image description here

$\endgroup$
0
$\begingroup$

Hint:

You can check that the map $\;\left\{\begin{aligned}g:\mathbf R&\longrightarrow (-1,1),\\ x&\longmapsto \dfrac x{1+|x|}\end{aligned}\right.$ is continuous and satisfies $$f\circ g=\operatorname{id}_{\mathbf R},\qquad g\circ f=\operatorname{id}_{(-1,1)}.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .