Monotone increasing bijection from $\mathbb{R}$ to $(0,1)$. Give an example of a monotone increasing function $f: \mathbb{R} \to (0,1)$ such that $f$ is a bijection.
I have an example in mind that follows $$g(x)=\frac{1}{1+e^x} ,x \in \mathbb{R}.$$ Then $g$ is a monotone decreasing bijection from $\mathbb{R}$ to $(0,1)$.  
Then consider the function $f: \mathbb{R} \to (0,1)$ as $$f(x)=1-g(x).$$ Then is it the required example?
Is my answer is correct that, above define $f$ is a monotone increasing bijection from $\mathbb{R}$ to $(0,1)$?
If this example is wrong, please suggest me an appropriate example.
Thanks.
 A: Yes, that works.
If you are ever in doubt in a situation like this, try to convince yourself by proof.
In particular, you need only prove that


*

*strictly increasing, 

*an injection, and

*a surjection.


(1) can be established by looking at the derivative:
$$
f^{\prime}(x)=\frac{\exp x}{\left(1+\exp x\right)^{2}}>0.
$$
(2) is implied by (1).
Indeed, if $u \neq v$, then either $u < v$ (and hence $f(u) < f(v)$) or $v < u$ (and hence $f(v) < f(u)$).
(3) follows from the fact that if $0 < y < 1$ then
$$
y=\frac{1}{1+\exp x}\iff x=\ln\left(\frac{1}{y}-1\right).
$$
Therefore, given any $0 < y < 1$, we can find a real $x$ such that $f(x)=y$.
A: Your example is correct.
You may take other bijective function as $$f:(-\infty,\infty) \rightarrow (0,1),
f(x)=\frac{1}{2}(1+\mbox{erf}(x)), \mbox{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} dt,$$ then $$f'(x)=\frac{2}{\sqrt{\pi}} e^{-x^2}>0, f(\pm \infty)=\pm 1.$$
Another example is $$f:(-\infty,\infty) \rightarrow (0,1), f(x)=\frac{1}{2} \left (1+\frac{2}{\pi} \tan^{-1} x\right), f'(x)=\frac{1}{\pi (1+x^2)}>0.$$
Yet another example is 
$$f(x)= e^x,~ \mbox{if}~x<0;~ f(x)=2-e^{-x},~\mbox{if}~ x>0.$$
