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 <br /> g : x \mapsto \begin{cases} x, &amp; x \neq 0, 1, 2, 3, \dots, \\<br /> x + 2, &amp; 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 <br /> h: x \mapsto \begin{cases} (g \circ f)(x), &amp; x \in (0,1), \\<br /> 0, &amp; x = 0, \\<br /> 1, &amp; x = 1.\end{cases} Then h is a bijection.

 

[nomadreid]

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