Constructive bijection between [0,1] and R?

[nomadreid]

It is easy to construct a bijection between the open interval (0,1) and ℝ, and (if one isn't an intuitionist) it is easy to prove that there exists a bijection between [0,1] and ℝ, but is it possible to construct such a bijection between [0,1] and ℝ? Obviously it won't be continuous, but that's OK.

 

[pasmith]

Define $g: \mathbb{R} \to \mathbb{R}$ by $$g : x \mapsto \begin{cases} x, & x \neq 0, 1, 2, 3, \dots, \\<br /> x + 2, & x = 0, 1, 2, 3, \dots .\end{cases}$$ Note that $g$ is an injection and its image is $\mathbb{R} \setminus \{0,1\}$.

Now take any bijection $f: (0,1) \to \mathbb{R}$, and define $h: [0,1] \to \mathbb{R}$ by $$h: x \mapsto \begin{cases} (g \circ f)(x), & x \in (0,1), \\<br /> 0, & x = 0, \\<br /> 1, & x = 1.\end{cases}$$ Then $h$ is a bijection.

 

[nomadreid]

Ooooh, that's elegant. I like it! Thanks very much, pasmith.