Does there exist a bijection from $[0,1]$ to $\mathbb R$?

Question

We can find a bijection from $(0,1)$ to $\mathbb R$. For example, we can use $f(x)=\frac{2x-1}{1+|2x-1|}$ composed of parts of two hyperbolas, see the graph here. Or we could appropriately scale the tangent function to get $g(x)=\tan\pi\left(x-\frac12\right)$, see the graph here. Several such bijections are suggested in the answers to this post: Is there a bijective map from $(0,1)$ to $\mathbb{R}$?

But does there exist a bijection from $[0,1]$ to $\mathbb R$? If yes, then what is it?

Answer 1

Let’s fix $f:(0,1)\to\mathbb{R}$.

Define $g:[0,1]\to\mathbb{R}$ as follows:

• $g(0) = -1$
• $g(1) = 1$

and for $0<x<1$,

• if $f(x)\in\mathbb{N}^*$, then $g(x) = f(x)+1$
• if $-f(x)\in\mathbb{N}^*$, then $g(x) = f(x)-1$
• otherwise, $g(x) = f(x)$

Then, if $f$ is a bijection, so is $g$.