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Math Help - Limit points of subsequences

  1. #1
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    Limit points of subsequences

    I have to give examples of four kinds of subsequences if they exist. S∈ℝ is the limit point of a subsequence of the sequence (Xn) (n=1 -> ∞) if ∃ subsequence (Xnk) (k=1 -> ∞) : Xnk -> S.

    What could the sequences be or do they exist if

    a) limit point of subsequence = {0}
    b) limit point of subsequence = {0,1}
    c) limit point of subsequence = {infinitely many points}
    d) limit point of subsequence = Q ∩ [0,1]
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  2. #2
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    Quote Originally Posted by antero View Post
    I have to give examples of four kinds of subsequences if they exist. S∈ℝ is the limit point of a subsequence of the sequence (Xn) (n=1 -> ∞) if ∃ subsequence (Xnk) (k=1 -> ∞) : Xnk -> S.

    What could the sequences be or do they exist if

    a) limit point of subsequence = {0}
    Do you mean by this that the only subsequential is 0? Any sequence converging to 0 will do.

    b) limit point of subsequence = {0,1}
    Find a sequence, \{a_n\}, that converges to 0, a sequence, \{b_n\}, that converges to 1, and "interleave" them: a_1, b_1, a_2, b_2, ....

    c) limit point of subsequence = {infinitely many points}
    It is a property of all real numbers, that, given any \epsilon, there exist a rational number within distance \epsilon of the real number. Since the rational numbers are countable, we can order them: r_1, r_2, r_3,\cdot\cdot\cdot. That is, the set of all rational numbers form a sequence having all real numbers as subsequential limits, and there are, of course, an infinite number of them.

    d) limit point of subsequence = Q ∩ [0,1]
    Since there are an infinite number of rational numbers in 0, 1, any example for d is also an example for c. But, unfortunately, the example I gave for c does not work since its set of sequential points is too big- it includes much more than the rational numbers. I don't think you will be able to give a specific, numerical, example. You might do something like this:
    Since the set of rational numbers in [0, 1] is countable, it is possible to put them into a sequence: r_1, r_2, r_3, \cdot\cdot\cdot. It is a property of all real numbers, and in particular of rational numbers, that given any rational number, r_i, there exist a sequence of rational numbers, [tex]\{a_{ij}\}[tex] that converges to that rational r_i. Now "interleave" those: \{a_{11}, a_{12}, a_{13}, \cdot\cdot\cdot\, a_{21}, a_{22}, a_{32}, \cdot\cdot\cdot, a_{31}, a_{32}, a_{33}, \cdot\cdot\cdot\ \}.
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