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Math Help - Bijection

  1. #1
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    Bijection

    .to construct a bijection

    between [0,1] and [0,1)......................


    I just think it is impossible , cuz the second one is one element less than the first....
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  2. #2
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    Well by that logic it would then be impossible to construct a bijection between \mathbb{Z} and \mathbb{Q}.
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  3. #3
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    Quote Originally Posted by felixmcgrady View Post
    .to construct a bijection

    between [0,1] and [0,1)......................


    I just think it is impossible , cuz the second one is one element less than the first....
    Hi,
    It must be possible.

    If you know Cantor-Bernstein Theorem, it is easy, because there obviously exist an injective map from [0,1) to [0,1] (the identity map) and an injective map from [0,1] to [0,1) (x -> x/2).


    If you don't know the theorem, consider set A=\{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4  },\frac{1}{5},\ldots\} \subseteq [0,1] and set B=\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5  },\ldots\} \subseteq [0,1). You can construct bijection g:A\rightarrow B by setting g\left(\frac{1}{i}\right)=\frac{1}{i+1} (verify it is a bijection).

    Next, we see that [0,1]\smallsetminus A = [0,1)\smallsetminus B, so the identity on [0,1]\smallsetminus A is a bijection from [0,1]\smallsetminus A to [0,1)\smallsetminus B.

    Now we're ready to construct a bijection f:[0,1]\rightarrow [0,1) as follows:

    f(x) = \begin{cases} g(x), & \mbox{if } x \in A \\ x, & \mbox{if } x\in [0,1]\smallsetminus A \end{cases}
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