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Math Help - Equinumerous

  1. #1
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    Equinumerous

    Can anyone help me solve this?
    Show that the interval (0,1] is equinumerous to the interval (0,1) by giving an example of a bijection from (0,1] to (0,1).
    Thanks.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by nbluo View Post
    Can anyone help me solve this?
    Show that the interval (0,1] is equinumerous to the interval (0,1) by giving an example of a bijection from (0,1] to (0,1).
    Thanks.
    Let A=(0,1]\subseteq\mathbb{R} and B=(0,1)\subseteq\mathbb{R} and define f:A\to B as follows:

    f(x)=\begin{cases}\frac{1}{2^{n+1}}\quad \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n}\\ x\quad\ \ \not \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n} \end{cases}

    This is very similar to the question in this thread.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nbluo View Post
    Can anyone help me solve this?
    Show that the interval (0,1] is equinumerous to the interval (0,1) by giving an example of a bijection from (0,1] to (0,1).
    Thanks.
    I don't understand the point of having to exhibit the bijection. Clearly one exists since the countable union of countable sets is countable, and using this (decompose your set into the disjoint union of a countable set and an uncountable set) one can prove that if E is uncountable then E\cup N\simeq E for any N\simeq\mathbb{N}
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  4. #4
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    Thank you undefined, it rings the bell. Then I find there are many ways to show such bijection.

    Quote Originally Posted by undefined View Post
    Let A=(0,1]\subseteq\mathbb{R} and B=(0,1)\subseteq\mathbb{R} and define f:A\to B as follows:

    f(x)=\begin{cases}\frac{1}{2^{n+1}}\quad \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n}\\ x\quad\ \ \not \exists\ n \in \mathbb{Z}: x=\frac{1}{2^n} \end{cases}

    This is very similar to the question in this thread.
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