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Math Help - f: [0,1) -> R

  1. #1
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    Exclamation f: [0,1) -> R

    Hi there,

    I wonder if you can think of any bijective function that maps [0,1) interval into R?

    In case of (0,1) I've done the following: y=tan(pi*x - pi/2), but I have no idea how to deal with [0,1).

    Thanks in advance for any help,
    Lucas
    Last edited by srt12; October 24th 2006 at 09:29 AM.
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  2. #2
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    Quote Originally Posted by srt12 View Post
    Hi there,

    I wonder if you can think of any bijective function that maps [0,1) interval into R?...
    Hi,

    try this one: f(x)=\frac{\sqrt{x-x^2}}{x-\frac{1}{2}}

    Unfortunately this function maps [0,1]\mapsto \mathbb{R}

    But may be you find a restriction so you can use your interval.

    EB
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  3. #3
    TD!
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    Or: try mapping [0,1) to (0,1), use the composition to map [0,1) (via (0,1) to R.
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    Quote Originally Posted by earboth View Post
    Hi,

    try this one: f(x)=\frac{\sqrt{x-x^2}}{x-\frac{1}{2}}

    Unfortunately this function maps [0,1]\mapsto \mathbb{R}

    But may be you find a restriction so you can use your interval.

    EB
    Actually, the domain of f does not include x = 1/2, so this isn't quite true.

    My thought is that this mapping is impossible. The interval [0, infinity) is half closed, whereas the "interval" R is either closed or open, depending on how you look at it. I'm not positive I'm remembering this correctly, but I believe there is a topological theorem that says such functions (mapping from different types of intervals) is not possible, for a continuous function anyway.

    Naturally, someone please correct me if I'm wrong.

    -Dan
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    Quote Originally Posted by topsquark View Post

    Naturally, someone please correct me if I'm wrong.
    You are wrong topsquark.

    The idea is that [0,1) is the countinuum as does \mathbb{R}. To see a prove of this using a variation of the Diagnol argument look here.

    Now, by the countinuum hypothesis since |[0,1)|>\aleph_0 and |[0,1)|\leq \mathbb{R} we must have,
    |[0,1)|=|\mathbb{R}|.

    The countinuum hypothesis provides us that there does exists a bijection between the two sets. However, whether is is possible to explicity state one, even via Axiom of Choice might not be possible (but I think yes).
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    Quote Originally Posted by srt12 View Post
    Hi there,

    I wonder if you can think of any bijective function that maps [0,1) interval into R?

    In case of (0,1) I've done the following: y=tg(pi*x - pi/2), but I have no idea how to deal with [0,1).

    Thanks in advance for any help,
    Lucas
    \tan(\pi\ x/2)
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    \tan(\pi\ x/2)
    I'm affraid it returns only positive values for arguments from [0,1).
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    Quote Originally Posted by ThePerfectHacker View Post
    The countinuum hypothesis provides us that there does exists a bijection between the two sets. However, whether is is possible to explicity state one, even via Axiom of Choice might not be possible (but I think yes).
    Thanks a lot, hope it will do that I just show that there exist such a bijection.
    And please correct me, if I'm wrong, but continuum hypothesis cannot be neither proved nor disproved, right?
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  9. #9
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    Quote Originally Posted by srt12 View Post
    Thanks a lot, hope it will do that I just show that there exist such a bijection.
    And please correct me, if I'm wrong, but continuum hypothesis cannot be neither proved nor disproved, right?
    Yes it is completely independant from ZFC.
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  10. #10
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    Quote Originally Posted by srt12 View Post
    I'm affraid it returns only positive values for arguments from [0,1).
    Opps.

    RonL
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  11. #11
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    If we can biject (0,1] with R, then certainly we can [0,1).
    <br />
(0,1] = \bigcup\limits_{k = 1}^\infty  {\left( {\frac{1}{{k + 1}},\frac{1}{k}} \right]} <br />

    Now define a function from (0,1] to (0,1):
    0,1] \to (0,1) \\
    x \in (0,1] \Rightarrow \left[ {\exists n} \right]\left( {x \in \left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]} \right) \\
    f\left( x \right) = x + 1 - \frac{1}{n} - \frac{1}{{n + 1}} \\
    \end{array}
    " alt="
    \begin{array}{l}
    f0,1] \to (0,1) \\
    x \in (0,1] \Rightarrow \left[ {\exists n} \right]\left( {x \in \left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]} \right) \\
    f\left( x \right) = x + 1 - \frac{1}{n} - \frac{1}{{n + 1}} \\
    \end{array}
    " />

    Show that that is a bijection. Then finish the problem with you function.
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  12. #12
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    I do not know what Plato was attempting to do but when I saw that union and I had a revealation.

    If you can biject,
    f:A\to B and g:B\to C then you can biject,
    A\to C

    Thus, you divide your interval as,
    [0,1),[1,2),[2,3),...
    And you can create a bijection between any adjacent to each other.
    That means, the union of these bijective maps (there are \aleph_o many thus the set is well-founded ) will create a bijection,
    [0,1)\to \mathbb{R}^+

    I am too lazy to work on a procedure but I got the idea from seeing what Plato wrote. (I hope I did not say the same thing).
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