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Limit of a sequence on a metric space

  1. Dec 7, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##(X,d)## be a metric space and let ##D \subset X## a dense subset of ##X##. Suppose that given ##\{x_n\}_{n \in \mathbb N} \subset X## there is ##x \in X## such that ##\lim_{n \to \infty}d(x_n,s)=d(x,s)## for every ##s \in D##. Prove that ##\lim_{n \to \infty} x_n=x##.

    The attempt at a solution.

    What I did was:
    Let ##\epsilon>0##
    ##d(x_n,x)\leq d(x_n,s)+d(s,x)##. For each ##n \in \mathbb N## I know there is ##s \in D## such that ##d(x_n,s)<\dfrac{\epsilon}{2}##. But, this doesn't say anything, I don't know how find the ##N \in \mathbb N## such that for all ##n\geq N##, ##d(x_n,s)+d(s,x)\leq \epsilon##
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2

    PeroK

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    You know that D is dense in X. So, either:

    x [itex]\in[/itex] D (in which case the result follows)

    or

    There is a sequence in D converging to x.

    Can you take it from there?

    It's a bit tricky, this one!
     
  4. Dec 7, 2013 #3
    I'll give it another try with your suggestion:

    As ##D## is dense in ##X##, there is a sequence ##\{s_n\}_{n \in \mathbb N}## such that ##s_n \to x## when ##n \to \infty##.

    By the triangle inequality,
    ##0\leq d(x_n,x)\leq d(x_n,s_n)+d(s_n, x)##. Again I have problems, because now I know that for a given ##\epsilon##, I can pick ##n_0## : for all ##n\geq n_0 \space## ##d(s_n,x)<\dfrac{\epsilon}{2}##, but how can I control the distance ##d(x_n,s_n)##?
     
  5. Dec 7, 2013 #4

    PeroK

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    You know something about ##d(x_n,s_m)## for any m. By choosing the same subscript for the two sequences, you boxed yourself in a bit!
     
  6. Dec 7, 2013 #5
    Sure, thanks for the help.

    First, consider ##n_0## : ##d(s_n,x)<\dfrac{\epsilon}{4}## for all ##n\geq n_0##.

    Now, I know that ##d(x_n,s) \to d(x,s)## for every ##s \in D##. Then, there exists ##n_1## such that for all ##n\geq n_1##, ##|(d(x_n,s)-d(x,s)|<\dfrac{\epsilon}{2}## for all ##s \in D##. This means ##d(x_n,s)<\dfrac{\epsilon}{2}+d(x,s)## for all ##s \in D## and for all ##n\geq n_1##.

    If this inequality holds for every element in ##D##, in particular, it holds for every ##s_n \in \{s_n\}_{n \in \mathbb N}##. If I choose ##N=\max\{n_0,n_1\}##, then, for all ##n\geq N##, ##0\leq d(x_n,s)\leq d(x_n,s_n)+d(s_n,x)<2d(s_n,x)+\dfrac{\epsilon}{2}<2\dfrac{\epsilon}{4}+\dfrac{\epsilon}{2}=\epsilon##.
     
  7. Dec 7, 2013 #6

    PeroK

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    You've got the right idea, but you definitely need to use different subscripts for the two sequences. You've got a moving target for the convergence of d(x_n, s_n). Do you see the slight problem with what you've got?

    It's also a good idea in general, because when you have two sequence like this, you don't always want to compare x_n with s_n. You want to fix a given member of s: s_m, say, and then consider the sequence d(x_n, s_m) as n →∞
     
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