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Showing Convergent Subsequence Exists

  1. Feb 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider the space ##([0, 1], d_1)## where ##d_1(x, y) = |x-y|##. Show that there exists a sequence ##(x_n)## in ##X## such that for every ##x \epsilon [0, 1]## there exists a subsequence ##(x_{n_k})## such that ##\lim{k\to\infty}\space x_{n_k} = x##.

    2. Relevant equations
    N/A

    3. The attempt at a solution
    We had no idea where to even begin with this one. It doesn't seem to make sense in the first place when you take the theorem:

    Let ##(x_n)## be sequence, and let ##x \epsilon X##. Prove that if ##\lim{n\to\infty}\space x_n = x##, then for every subsequence ##(x_{n_k})## we have ##\lim{k\to\infty} \space x_{n_k} = x##

    With the problem saying that there is specific sequence such that for any limit point we can find a subsequence, this implies there a multiple subsequences with different limits, which contradicts what this previous theorem says.

    Any suggestions on where to begin with this one?
     
  2. jcsd
  3. Feb 5, 2017 #2

    Ray Vickson

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    Homework Helper

    There are no contradictions in the problem statement. The main sequence ##\{x_n\}## does not have a limit; it has numerous limit points, meaning that it has numerous subsequences converging to their own, individual limits. The problem asserts that there is a sequence such that any ##x \in [0,1]## is a limit of some subsequence. Believe it or not, there is a very easy solution.
     
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