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Homework Help: Continuity on metric spaces.

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data

    T is a compact metric space with metric d. f:T->T is continous and for every x in T f(x)=x. Need to show g:T->R is continous, g(x)=d(f(x),x).

    2. Relevant equations



    3. The attempt at a solution
    f is continous for all a in T if given any epsilon>0 there is a delta>0 st d(x,a)<delta implies d(f(x),f(a))<epsilon. Need to show there is a delta st d(x,a)< delta implies |g(x)-g(a)|<epsilon.
    by a previous problem i did... |g(x)-g(a)|=|d(f(x),x)-d(f(a),a)|<= d(f(x),f(a))+d(x,a)<d(f(x),f(a)) + delta. This is where I got stuck. from the assumption that f is continous we got from there that d(f(x),f(a))<epsilon. but if i say that then i would have...|g(x)-g(a)|<epsilon + delta and inorder for |g(x)-g(a)|to be <epsilon we would have to choose delta to be 0 which it cant be because it has to be greater than 0. any suggestions on what i should do?
     
  2. jcsd
  3. Jul 9, 2011 #2
    Hi l888l888l888! :smile:

    This must be true for every epsilon, so also for [itex]\varepsilon/2[/itex]. So, what happens to the following if you take

    [tex]|g(x)-g(a)|<\varepsilon/2[/tex]

     
  4. Jul 9, 2011 #3
    hello micromass!!! im not really understanding your question. can you clarify? did you mean d(f(x),f(a))<epsilon/2?
     
  5. Jul 9, 2011 #4
    I'm so sorry!! What was I thinking...

    But yes, that is what I meant. You can take [itex]d(f(x),f(a))<\varepsilon/2[/itex]
     
  6. Jul 9, 2011 #5
    Actually, I think I have it now. correct me if im wrong...
    f is continous for all a in T if given any epsilon>0 there is a delta>0 st d(x,a)<delta implies d(f(x),f(a))<epsilon/2. Need to show there is a delta st d(x,a)< delta implies |g(x)-g(a)|<epsilon.
    |g(x)-g(a)|=|d(f(x),x)-d(f(a),a)|<= d(f(x),f(a))+d(x,a)<epsilon/2 + delta. choose delta to be epsilon/2. so therefore |g(x)-g(a)|=|d(f(x),x)-d(f(a),a)|<= d(f(x),f(a))+d(x,a)<epsilon/2 + epsilon/2=epsilon, as desired...
     
  7. Jul 9, 2011 #6
    Seems good! :smile:
     
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