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?


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  • $\begingroup$ Does the interval have to be closed? $\endgroup$ – icurays1 Dec 4 '12 at 19:04

There are many bijections from an open interval $(a, b)\to R$, e.g.

$g(x) = \cot(\frac{\pi}{2}x)$ 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.

  • $\begingroup$ How would you go about composing g(f(x))? $\endgroup$ – user319635 Mar 24 '16 at 20:50
  • $\begingroup$ The real task is to prove that $f$ is a bijection, and to show how and why it works. $\endgroup$ – chharvey Oct 28 '16 at 19:46

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.


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