# Is there a bijection from $[0,1)$ to $\mathbb R ?$ [duplicate]

I know a continuous bijection from $[0,1)$ to $\mathbb R$ cannot exists but what happens if we lift the restriction of continuous $?$

Can there exists a bijection , not necessarily continuous from $[0,1)$ to $\mathbb R ?$

$(0,1)$ is bijective with $\mathbb R.$ Although I doubt that would be any useful here.

## marked as duplicate by user228113, Vlad, Watson, zz20s, JMPApr 27 '16 at 13:18

• $(0,1)$ being bijective to $\Bbb R$ is very useful if you know how to construct a bijection from $[0, 1)$ to $(0,1)$. – Arthur Apr 27 '16 at 8:30
• @Arthur : I don't. help please? – user118494 Apr 27 '16 at 8:32
• You can construct a bijection between $[0,1)$ and $(0,1)$ in the same way you construct a bijection between $\Bbb N$ and $\Bbb N\setminus\{0\}$. – user228113 Apr 27 '16 at 8:33
• Possible duplicate of How to define a bijection between $(0,1)$ and $(0,1]$?. And this, and this. – user228113 Apr 27 '16 at 8:35
• If you instead of writing the formula for that bijection, think about what it does: It takes $0$, and moves it to $1$. To make "room" for that, it has to take $1$ and move it to $2$. And so on. In the same way, for the bijection from $[0,1)$ to $(0,1)$ you take $0$ and move it somewhere, say $1/2$. To make "room" for that, you take $1/2$ and move it somewhere, say $1/3$. And so on. All other numbers you leave untouched. – Arthur Apr 27 '16 at 8:38

Consider the function

$f:[0,1) \to (0,1)$

$f(x):= \begin{cases} \frac{1}{2} & \text{ if } x = 0 \\ \frac{1}{2^{n+1}} & \text{ if }x=\frac{1}{2^n} \text{ for some } n\in \mathbb N \\ x & \text{ else} \end{cases}$

This is obviously a bijection.

Then consider $g(x) = \tan(\pi (x-\frac{1}{2}))$ where $g:(0,1) \to \mathbb R$. Then $g$ is obviously a bijection too. Then

$$h(x) := (g \circ f)(x)$$ is your desired bijection between $[0,1)$ and $\mathbb R$.

• Damn dollar signs always intruding. – B. Pasternak Apr 27 '16 at 8:43
• @B.Pasternak Haha Americanization everywhere=) Lets petition a change from $ to €. – flawr Apr 27 '16 at 8:44 • Count me in. Europe might be done soon anyways! – B. Pasternak Apr 27 '16 at 8:45 here is a way to do it : Step one : take the set$\{0;1/2;3/4;7/8;15/16;...;1-1/2^n;...\}$and map it (in increasing order, for example) to$\{0;1;2;3;4;...\}$Step two : make a bijection between the remaining sets, that is make a bijection between$(0;1/2)$and$(-\infty;0)$, a bijection between$(1/2;3/4)$and$(0;1)$, a bijection between$(3/4;7/8)$and$(1;2)\$, and so on.

And it's done.