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Constructive bijection between [0,1] and R?

  1. Apr 30, 2015 #1
    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.
     
  2. jcsd
  3. Apr 30, 2015 #2

    pasmith

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

    Now take any bijection [itex]f: (0,1) \to \mathbb{R}[/itex], and define [itex]h: [0,1] \to \mathbb{R}[/itex] by [tex]
    h: x \mapsto \begin{cases} (g \circ f)(x), & x \in (0,1), \\
    0, & x = 0, \\
    1, & x = 1.\end{cases}[/tex] Then [itex]h[/itex] is a bijection.
     
  4. Apr 30, 2015 #3
    Ooooh, that's elegant. I like it! Thanks very much, pasmith. :smile:
     
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