Bijection from finite (closed) segment of real line to whole real line Is there a bijection from a finite (closed) segment of the real line to $\mathbb{R}$? For example, is there a bijection from $[0,1]$ to $\Bbb{R}$?
If so, is there a straightforward example? If not, why?
 A: Yes. There is such function, but it is less straightforward than one would think.
The reason is that "straightforward" functions are usually continuous, and a continuous function from $[0,1]$ would either have values of $\pm\infty$ or will have a range of a closed interval $[a,b]$ and not the entire real line.
However there are relatively simple ways of removing the two endpoints and then you can write a bijection from $(0,1)$ to $\mathbb R$ simply by $\frac{1-2x}{2x(x-1)}$ or some other function which you can find.
Composing these two bijections will give you a bijection between $[0,1]$ and $\mathbb R$. Examples for both bijections have been given plenty of times on this site before.
A: There are many bijections from an open interval $(a, b)\to R$, e.g.
$g(x) = \cot(\pi x)$ is a bijection $g: (0, 1)\to \mathbb{R} $.
Now, we need to find a bijection from the closed interval $[a, b]\to R$, by showing that there exists a bijection from the closed interval $[a, b]$ to the open interval $(a, b)$.
Taking the interval: $[0,1]$. Define $f(x)$ as following:
$$f(x) = \left\{
 \begin{array}{1 1}
  \frac{1}{2}  & \mbox{if } x = 0\\
  \frac{1}{2^{n+2}} & \mbox{if } x = \frac{1}{2^n}\\
                x & \mbox{otherwise}
 \end{array}
\right.$$
Then $f: [0, 1] \to (0, 1)$ is a bijection.
Now, compose: $g(f(x)): [1, 0] \to \mathbb{R}$, and you have your bijection.
