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Fixed points on compact spaces

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Let X be a compact metric space. if f:X-->X is continous and d(f(x),f(y))<d(x,y) for all x,y in X, prove f has a fixed point.


    2. Relevant equations



    3. The attempt at a solution
    Assume f does not have a fixed point. By I problem I proved before if f is continous with no fixed point then there exists an epsilon>0 st d(f(x),x)>=epsilon for all x in X. Using this I wanted to get a contradiction. i wanted to prove d(f(x),f(y))>=d(x,y) which leads to a contradiction of
    d(f(x),f(y))<d(x,y) but I got stuck.
     
  2. jcsd
  3. Jul 18, 2011 #2

    micromass

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    Hi l888l888l888! :smile:

    Try to use that the function

    [tex]\Phi:X\rightarrow \mathbb{R}:x\rightarrow d(x,f(x))[/tex]

    is continuous. What can you infer about this function using that X is compact?
     
  4. Jul 18, 2011 #3
    you can infer that it attains its bounds on X. ?
     
  5. Jul 18, 2011 #4

    micromass

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    Indeed, so you can deduce that there is an a such that d(f(a),a) is the minimum value. Either this minimum value is 0, and then we got a fixpoint. If the minimum value is not 0, then we must obtain a contradiction, can you find one?? (Use that [itex]d(f(x),f(y))<d(x,y)[/itex])
     
  6. Jul 18, 2011 #5
    ummmmm nothin comes to mind at this time...??? if o is not the min value then d(f(a),a)>0 and we must come up with a contradiction that contradicts d(f(x),f(a))<d(x,a)??????
     
  7. Jul 18, 2011 #6

    micromass

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    What is d(f(f(a)),f(a))?
     
  8. Jul 18, 2011 #7
    d(f(f(a)),f(a))<d(f(a),a). but then this contradicts the fact that d(f(a),a) is a min value?
     
  9. Jul 18, 2011 #8

    micromass

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    Yes!
     
  10. Jul 18, 2011 #9
    Great. thanks!!!!
     
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