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Math Help - Bijection between Sets(again)

  1. #1
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    Bijection between Sets(again)

    Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the sets in each pair.

    S = [0,1] and T = [0,1)
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    Quote Originally Posted by p00ndawg View Post
    Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the sets in each pair.

    S = [0,1] and T = [0,1)

    How about this f: S -> T
    for x<1/2 f(x)=x
    for x>=1/2 f(x) = 3/2-x

    I guess it works

    Sorry - This seems wrong !!
    Last edited by aman_cc; October 20th 2009 at 09:08 AM. Reason: wrong post
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  3. #3
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    Quote Originally Posted by p00ndawg View Post
    Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the sets in each pair.

    S = [0,1] and T = [0,1)
    Try f(x) = \left\{ {\begin{array}{rl}<br />
   {x,} & {~x \ne \frac{1}<br />
{n},\,\left( {\forall n \in \mathbb{Z}^ +  } \right)}  \\<br />
   {\frac{1}<br />
{{n + 1}},} & {x = \frac{1}<br />
{n},\,\left( {\exists n \in \mathbb{Z}^ +  } \right)}  \\<br /> <br />
 \end{array} } \right.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Try f(x) = \left\{ {\begin{array}{rl}<br />
   {x,} & {~x \ne \frac{1}<br />
{n},\,\left( {\forall n \in \mathbb{Z}^ +  } \right)}  \\<br />
   {\frac{1}<br />
{{n + 1}},} & {x = \frac{1}<br />
{n},\,\left( {\exists n \in \mathbb{Z}^ +  } \right)}  \\<br /> <br />
 \end{array} } \right.
    Plato - Thanks.

    Please validate my understanding. You have used a technique similar to the argument used in Hotel Infinity Paradox - To make room available ask every occupant to shift to the next room.

    Consider F: [0,1]->[0,1]

    If I remove finite points from the range - I can use your argument to construct a bijection. Correct?
    Say if range is (0,1), I will map 1/n -> 1/(n+2) and 0 -> 1/2

    Correct plz?
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    Quote Originally Posted by Plato View Post
    Try f(x) = \left\{ {\begin{array}{rl}<br />
   {x,} & {~x \ne \frac{1}<br />
{n},\,\left( {\forall n \in \mathbb{Z}^ +  } \right)}  \\<br />
   {\frac{1}<br />
{{n + 1}},} & {x = \frac{1}<br />
{n},\,\left( {\exists n \in \mathbb{Z}^ +  } \right)}  \\<br /> <br />
 \end{array} } \right.
    hmm, i dont understand this,im not seeing the bijection.

    why the 1/(n+1)?
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  6. #6
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    Quote Originally Posted by p00ndawg View Post
    hmm, i dont understand this,im not seeing the bijection.
    You can prove that f is an injection and a surjection for [0,1]\mapsto [0,1).

    Have you tried?
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    what do you mean by when x != (1/n)?

    im working on the injection on surjection right now, but when x != (1/n) is that for everything in S = [0,1] ?
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  8. #8
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    Quote Originally Posted by p00ndawg View Post
    what do you mean by when x != (1/n)?

    im working on the injection on surjection right now, but when x != (1/n) is that for everything in S = [0,1] ?
    Here are some examples,
    f(1)=\frac{1}{2},~f(\frac{1}{2})=\frac{1}{3},~f(\f  rac{2}{3})=\frac{2}{3},~f(\frac{\pi}{4})=\frac{\pi  }{4}
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  9. #9
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    Quote Originally Posted by Plato View Post
    Here are some examples,
    f(1)=\frac{1}{2},~f(\frac{1}{2})=\frac{1}{3},~f(\f  rac{2}{3})=\frac{2}{3},~f(\frac{\pi}{4})=\frac{\pi  }{4}
    Okay! thank you! I think im finally getting it.

    I just have one more question, when im looking at these two sets how would I know to
    separate it into a function like that?

    is it just practice? or is there anything from that set that is setting off some alarms in your head?

    thanks a bunch!
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