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Sequences, subsequences and limits

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Let (rn) be an enumeration of the set Q of all rational numbers. Show that there exists a subsequence (rnk) such that limk[tex]\rightarrow\infty[/tex] rnk = +[tex]\infty[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Im not sure how to even attack this
     
  2. jcsd
  3. Nov 12, 2009 #2
    If you can make your sequence strictly increasing and unbounded, then it works. Remember that there are only finitely many possible rational numbers before the nth rational number, and yet there are infinitely many rational numbers bigger than whatever the nth rational number is.
     
  4. Nov 12, 2009 #3
    So how about this: Let (rnk) be a sequence of real numbers such that:

    1. rn1 > a, a[tex]\epsilon[/tex] Q
    2. (rnk) is strictly increasing
    ie. rn1 < rn2 ... < rnk >

    Then the lim[tex]_{k \rightarrow \infty}[/tex] r[tex]_{n}_{k}[/tex] = [tex]\infty[/tex]
     
  5. Nov 12, 2009 #4
    If your 'a' is simply any element of the rational numbers, then your r_n_1 is going to be the supremum of the rationals, which is not in the rationals.

    Also, after you show that something's strictly increasing, you have to show that it's also unbounded, meaning that for any M, you can pick r_n_k such that r_n_k > M. After all, the partial sums of sigma from i=1 to infinity of (1/2)^i are constantly increasing, but always less than 2.
     
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