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Math Help - disprove completeness ; fixed point for contractions

  1. #1
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    Cool disprove completeness ; fixed point for contractions

    Space X - graph of f(x) = sin(1/x) on (0,1]
    X = {(x,y): 0<x<=1, y = sin(1/x)}

    Prove:
    a) X is not complete
    b) Every contraction h:X-->X has a fixed point

    Hints for b given by my professor:
    1.) X={(x,y) are elements in X: x = delta} U {(x,y) are elements in X: x >= delta}
    2.) If delta is sufficiently small then diam(h(x1))<2; so h(x1) is an arc
    3.) h(x2) is an arc (COMPACTNESS)
    4.) therefore h(x) bounded away from the y-axis
    5.) therefore h has a fixed point

    Any help would be useful on any parts of these problems. He said proving part a would be simple but then he said part b would be long. Again, any help would be useful. I will post what I have for part a in a little bit ... still working on it. Thank you for reading.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by derek walcott View Post
    Space X - graph of f(x) = sin(1/x) on (0,1]
    X = {(x,y): 0<x<=1, y = sin(1/x)}

    Prove:
    a) X is not complete
    b) Every contraction h:X-->X has a fixed point

    Hints for b given by my professor:
    1.) X={(x,y) are elements in X: x = delta} U {(x,y) are elements in X: x >= delta}
    2.) If delta is sufficiently small then diam(h(x1))<2; so h(x1) is an arc
    3.) h(x2) is an arc (COMPACTNESS)
    4.) therefore h(x) bounded away from the y-axis
    5.) therefore h has a fixed point

    Any help would be useful on any parts of these problems. He said proving part a would be simple but then he said part b would be long. Again, any help would be useful. I will post what I have for part a in a little bit ... still working on it. Thank you for reading.
    I'm assuming this is with the usual metric.

    What about x_n=\left(\frac{1}{2n\pi},\sin\left(\frac{1}{\frac  {1}{2n\pi}}\right)\right)? To see that it's Cauchy merely note that d(x_n,x_m)=\sqrt{\left(\frac{1}{2n\pi}-\frac{1}{2m\pi}\right)^2+\left(\sin\left(2n\pi\rig  ht)-\sin\left(2m\pi\right)\right)^2}=\frac{1}{2\pi}\le  ft|\frac{1}{n}-\frac{1}{m}\right|. Now you can easily prove that for sufficiently large m,n that last term is smaller than any given \varepsilon>0. So, x_n is cauchy but not convergent since x_n\to (0,0)\notin \Gamma_{f}
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