Is there an injection between $\mathbf{R}$ and $[0,1)$

Question

I want to find an injective map $f\colon\mathbb{R}\to[0,1)$ that is not a transcendental function (I prefer a rational function). Is it possible to find such a function or do I need a transcendental function like $$f(x)=\frac1{1+e^x}$$

Answer 1

How about $$ f(x) = \frac 12\left(1 + \frac{x}{|x| + 1}\right) $$ not quite a rational function, but it is continuously differentiable.

Answer 2

It is not possible for rational functions. Here is why not:

Suppose such an $f$ exists. Write $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials with no common terms. First, $q(x)$ can not have any roots. Since $f$ is continuous on $\mathbb{R}$ and $f$ is injective, it must either be increasing or decreasing. To finish it off, for any rational function $\lim_{x \to \infty} f(x) = \lim_{x \to -\infty}f(x),$ if either limit is finite. This can't happen if $f$ is increasing or decreasing.

Answer 3

A recent question asked for a proof that $f(x)=\frac{1}{x}+\frac{1}{x-1}$ defines a bijection $(0,1) \to \mathbb{R}$.

The inverse of $f$ gives a bijection $ \mathbb{R} \to (0,1)$ and so an injection $ \mathbb{R} \to [0,1)$.

This inverse function has an expression using square roots.