Is there a bijection from $[0,1]$ to $\mathbb{R}$?

Question

I'm looking for a bijection from the closed interval $[0,1]$ to the real line. I have already thought of $\tan(x-\frac{\pi}{2})$ and $-\cot\pi x$, but these functions aren't defined on 0 and 1.

Does anyone know how to find such a function and/or if it even exists?

Thanks in advance!

Inclusion is an injective function... There is no continuous ("nice") bijection. — Arthur, Nov 24 '16 at 17:25
$f(x)=x$ defines an injective function $[0,1]\to\mathbb R$. It is not surjective, though. Did you mean to ask for a bijection rather than an injection? — hmakholm left over Monica, Nov 24 '16 at 17:28

score 5 · Accepted Answer · edited Apr 13 '17 at 12:21

5

Use Did's method here to construct a bijection $[0,1] \to (0,1)$. Play around with $\tan$ for a bijection $(0,1) \to \mathbb{R}$

Note that any bijection cannot be continuous. This is because $[0,1]$ is compact.

edited Apr 13 '17 at 12:21

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answered Nov 24 '16 at 17:32

MathematicsStudent1122

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score 0 · Answer 2 · answered Nov 24 '16 at 17:32

A continuous bijection can't exists because $[0,1]$ is a compact set and continuous functions send compacts in compacts. You can look for a non-continuous bijection, that exists because $[0,1]$ and $\mathbb R$ have the same cardinality. It follows from Cantor-Bernstein theorem, that states "given two sets $A, B$ if exist two injective functions $f:A\rightarrow B \ g:B\rightarrow A$" than exists a bijective function $h:A\rightarrow B$".

Is there a bijection from $[0,1]$ to $\mathbb{R}$?

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