Show that any open interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to $\mathbb{R}$ Show that any open interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to $\mathbb{R}$. I've seen similar questions on here, but they all use the fact that $(a,b)$ is homeomorphic to $(c,d)$, but can we prove this with without this fact, or without using the fact that the composition of homeomorphisms is homeomorphic?
 A: Well in fact the classical examples are all mapping $\mathbb R$ to either $(0,1)$ or $(-1,1)$.
For instance considering $\mathbb R\mapsto(-1,1)$ there is $f(x)=\begin{cases}\tanh(x)\\\frac 2{\pi}\arctan(x)\\\frac x{|x|+1}\end{cases}\quad$ to cite the most commonly used, and of course their reciprocals since all are continuous bijections $f^{-1}(x)=\begin{cases}\operatorname{argth}(x)\\\tan(\frac{\pi x}2)\\\frac x{1-|x|}\end{cases}$
A simple affine function $A(x)=\frac{b-a}2\,x+\frac{a+b}2$ and $A^{-1}(x)=\frac 2{b-a}\,x-\frac{a+b}{b-a}$ will scale and translate any $(a,b)$ to $(-1,1)$.
This is why composition of homeomorphisms is almost invariably invoked since $A(f(x))$ and $f^{-1}(A^{-1}(x))$ will do the job perfectly in mapping $\mathbb R$ and $(a,b)$. You can devise a direct mapping of course, but I guess the final form will resemble very much to this method.

Edit: 03/2022
To devise a mapping between $(a,b)$ and $\mathbb R$ you need a strictly monotonic function (let say increasing), and also since it should be continuous on $(a,b)$ then the limits in $a,b$ must be infinite (to cover all real values).
This is therefore calling for vertical asymptotes in $x=a$ and $x=b$.
You proposal $f(x)=\frac{x(a+b-2x)}{(a-x)(b-x)}$ verifies these two conditions as long as $a$ and $b$ are the same sign (else it doesn't work, not monotone), but the asymptotic behaviour is respected since $a,b$ are roots of the denominator.
In fact the additional $x$ on the numerator is the one causing problems, $f(x)=\frac{(a+b-2x)}{(a-x)(b-x)}$ work fine, this is because the root of numerator $x=\frac {a+b}2$ is between $a$ and $b$ and this make the discriminant of the numerator of the derivative negative, thus $f'$ of constant sign.
