Solution verification: Show that $a_1 <_{(-1,1)} a_2\implies f(a_1)<_{\mathbb{R}} < f(a_2)$ to prove $(-1,1)$ has the same order type as $\mathbb{R}$

Question

I want to show that $(-1,1)$ and $\mathbb{R}$ have the same order type by showing that there exists a bijective correspondence $f : (-1,1) \to \mathbb{R}$ such that $$a_1 <_{(-1,1)} a_2\implies f(a_1)<_{\mathbb{R}} < f(a_2).$$

Let $f : (-1,1) \to \mathbb{R}$ be defined by $f(x) = \dfrac{x}{1 - x^2}$. We begin by showing $f$ is injective: let $x_1, x_2 \in (-1,1)$ such that $f(x_1) = f(x_2)$. Using elementary algebra, we find that $(x_1 - x_2)(1 + x_1x_2) = 0$, which implies $x_1 = x_2$ or $1 + x_1x_2 = 0$. If the former, then we are done. If the latter, then note that $x_1 \neq x_2$, but if $x_1 = 0$ then $1 + 0x_2 = 1 \neq 0$, which is a contradiction. Therefore, $x_1 = x_2$, and so $f$ is injective. We will now show that $f$ is surjective. We believe the simplest way to do so is to show that $\exists g : \mathbb{R} \to (-1,1)$ such that $f \circ g = I_\mathbb{R}$. That is, $$\frac{g(x)}{1 - (g(x))^2} = x,$$ and using elementary algebra we find that $$g(x) = \frac{-1 -\sqrt{4x^2 + 1}}{2x} \quad\text{for }x \neq 0.$$ We then define $g(0) = 0$ and we have that $\forall x \in \mathbb{R}, f(g(x)) = x$, which implies $f$ is surjective. Therefore, $f$ is bijective.

Now we let $a_1, a_2 \in (-1,1)$ such that $a_1 <_{(-1,1)} a_2$. Since $f$ is bijective, $f(a_1) \neq f(a_2)$, so $f(a_1) < f(a_2)$ or $f(a_2) < f(a_1)$. If the former, then we are done. If the latter, then note that $f$ is strictly increasing $(-1,1)$, which is a therefore a contradiction, so we are done.

Remark: I tried not to use theorems from real analysis to prove the bijectivity of $f$, instead I used first principles as defined in Munkres, however I wasn't sure how to get a contradiction without using that $f$ is increasing.

Answer 1

There are some mistakes.

• The fact that $x_1=0$ implies $1=0$ doesn't prove that $x_1=x_2$. Instead we should observe that $-1<x_1,x_2<1$, so $1+x_1x_2>0$ and therefore $x_1=x_2$.
• The inverse function is found incorrectly. Since $g\colon \Bbb R\to (-1,1)$, we have to make sure that $|g(x)|<1$ for all $x$. Your function satisfies $|g(x)|>1$. Moreover you can't just say that we define $g(0)=0$, since it has to be shown. To find the inverse observe that $$g(x)=y \iff f(y) = x \iff xy^2+y-x=0.$$ If $x=0$ then $y=0$, so $g(0)=0$. If $x\neq 0$ then we got a quadratic equation with two solutions $$y=y_\pm:=\frac{\pm\sqrt{4x^2 + 1}-1}{2x}= \frac{2x}{1\pm\sqrt{4x^2 + 1}}. $$ Observe now that $y_+\in(-1,1)$ and $y_-\in\Bbb R\setminus [-1,1]$, so $g(x)=y_+$. The proper inverse function is therefore $$g(x)=\begin{cases}\frac{-1 +\sqrt{4x^2 + 1}}{2x};&\text{for }x \neq 0 \\0;&\text{for }x=0\end{cases} = \frac{2x}{1 +\sqrt{4x^2 + 1}}.$$
• The condition $$a_1 <_{(-1,1)} a_2\implies f(a_1)<_{\mathbb{R}} f(a_2)$$ precisely means that $f$ is increasing. You can't use the fact that $f$ is increasing to show that it's increasing. You should prove it separately. The proof is similar to the proof of injectivity of $f$, since we deal with linear (total) orders. $$f(x_1)<f(x_2) \iff (x_1-x_2)(1+x_1x_2)<0 \iff x_1<x_2,$$ since $1+x_1x_2>0$ for $x_1,x_2\in(-1,1)$.

Answer 2

$f:(-1,1)\to\Bbb{R}$ defined by $$f(x) = \dfrac{x}{1 - x^2}$$

Injective :

Let $x, y\in (-1, 1) $ and $f(x) =f(y) $

Then $(x-y) (xy-1) =0$

Hence $x=y$ as the $xy+1=0$ implies $y=-\frac{1}{x}$ and $|y|=\frac{1}{|x|}\ge 1$.Hence $y\notin (-1, 1)$

Surjective:

$\lim_{x\to -1} f(x) =-\infty$ and $\lim_{x\to 1} f(x) =\infty$

Since $f$ is continuous, it has the Darboux property (intermediate value property). Hence $f$ attained every value in between $(-\infty, \infty) $ .Hence $f$ is surjective.

Order preservation:

The bijection $f$ preserve the order if $a<b$ implies $f(a) <f(b)$ i.e $f$ is strictly increasing.

Since $f$ is injective continuous map, $f$ is strictly monotone (by Darboux property of $f$). It's increasing as $f(0) =0 $ and $f(\frac{1}{2})=\frac{2}{3}$