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Compactness and continuity.

  1. Feb 8, 2008 #1
    I need to prove that for every continuous function f:X->X of a metric and compact space X, which satisfy for each two different x and y in X p(f(x),f(y))<p(x,y) where p is the metric on X, there's a fixed point, i.e there exist x0 s.t f(x0)=x0.

    obviously i thought assuming there isnt such a point i.e that for every x in X f(x)!=x
    now because X is compact and it's a metric space it's equivalent to sequence compactness, i.e that for every sequence of X there exist a subsequence of it that converges to x0.

    now [tex]p(f(x_{n_k}),x_{n_k})[/tex], because they are not equal then there exist e0 such that: [tex]p(f(x_{n_k}),x_{n_k})>=e0[/tex]
    now if x_n_k=f(y_n_k) y_n_k!=x_n_k, we can write it as:
    now if y_n_k were converging to x0, it will be easier, not sure how to procceed...

    what do you think?
  2. jcsd
  3. Feb 8, 2008 #2


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    This is a special case of the Banach contraction mapping theorem. A proof would go as follows: Let x0 be any point in X, and let xn=f(xn-1) for n>1. Claim: {x_n} converges.

    Post back if you need more hints.
  4. Feb 9, 2008 #3
    well cauchy sequnce obviously will do here.
    now if x_{n_k-n} converges to x_0 (which can be assumed cause it's compact and metric), it will be easy to prove your claim, cause then for every e>0 s.t k is big enough:
    p(x_0,x_{n_k-1})<e/2 and also p(x0,x_{n_k-n})<e/2
    so p(x_n-1,x0)<=p(f(x_n),f(x_n_k))+p(f(x_n_k),x0)<...<e.

    is this wrong? I have a sneaky suspicion that yes.
  5. Feb 9, 2008 #4


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    If n > n_k, then [itex]x_{n_k - n}[/itex] doesn't make sense. Also I don't see how you can assume that "x_{n_k-n} converges to x_0", because it isn't true.

    You started out with the right idea. Let n>m, and consider p(xn, xm). Show that we can make this arbitrarily small. This would imply that {xn} is Cauchy and hence convergent (because X is compact). Then we can use the continuity of f to conclude that f has a fixed point (how?).
  6. Feb 11, 2008 #5
    well, p(xn,xm)<p(xn-1,xm-1)<....<p(xn-m,x0)
    now how do i procceed from here?
    I mean if I assume n-m is big enough, s.t x_n-m->x0 then that will do, not sure that this is correct...
  7. Feb 12, 2008 #6
    I have another two questions, I need to answer if the next spaces satisfy S2 or Sep, the spaces are with they metrics affiliated with them, in here:
    in questions 4,5 (disregard the herbew words near them) there listed the spaces.

    well what i think is that because if a space is metric and it satisfies S2 then it also satisifes S2, and always when S2 is satisifes then also Sep is satisifed, then it's easy to check fo Sep, i think it follows that for the first the space follows both of them, while in the second it doesnt satisify either of them.

    not sure how argue that?
    I mean can I find a countable basis for the C^k[0,1]?
    or a countable dense set in it?
    what do you think?
  8. Feb 16, 2008 #7


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    To finish off, you can use the triangle inequality
    [tex]\rho(x_{n-m},x_0) < \rho(x_{n-m},x_{n-m-1}) + \rho(x_{n-m-1}, x_{n-m-2}) + \cdots + \rho(x_1, x_0)[/tex]
    coupled with the observation that
    [tex]\rho(x_n, x_{n-1}) < \rho(x_1, x_0)[/tex].
  9. Feb 16, 2008 #8


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    As for your other question: I'm guessing S_2 means second countable (has a countable basis) and Sep means separable (has a countable dense subset), right?

    And you have the right idea: a metric space is separable iff it's second countable. I would use separability here. For C^k[0,1], try to see if Weierstrauss's theorem is helpful. For l_2, I would think about the subspace consisting of sequences with only finitely many terms. This is certainly dense in l_2, but is it countable? No. So how about we restrict these sequences to those with rational terms?
  10. Feb 16, 2008 #9
    it seems eventually that munkres has a similar questions with hints which were helpful.
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