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Math Help - Sub-sequential limit point

  1. #1
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    Sub-sequential limit point

    1) Is it Possible to find a sequence for which the set of sub-sequential limit points is [0,1]?

    2)Is any closed set in R,the set of sub-sequential limit points of a sequence?
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  2. #2
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    Quote Originally Posted by math.dj View Post
    1) Is it Possible to find a sequence for which the set of sub-sequential limit points is [0,1]?

    2)Is any closed set in R,the set of sub-sequential limit points of a sequence?
    Yes (to both questions). Use the fact that R is separable. A closed subset of R has a countable dense subset. Construct a sequence which visits each element of that subset infinitely often.
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  3. #3
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    In other words, for (a) construct a sequence that converges to 0 ({1/n} will do), a sequence that converges to 1 ({(n-1)/n} will do) and "blend" the two: a_n= \frac{1}{n} if n is odd, a_n= \frac{n-1}{n} if n is even.
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  4. #4
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    a) Consider the sequence {x_n} such that there is a one to one correspondence between the elements of the sequence and the rational numbers in [0,1]. Then the sequence has sub-sequential limits [0,1].

    b) Yes. Any finite set is closed. So try to find a sequence with finite no. of limit points.
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  5. #5
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    are you trying to say there is a bijection between N and Q intersection [0,1]?
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  6. #6
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    Yes. Since the set of rational nos. Q are countable, we can arrange the rational nos Q∩[0,1] as the sequence {x1,x2,x3,.....}. Then for any no. α in [0,1] there is a sub-sequence from {xn} converging to α.
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  7. #7
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    There exist a bijection between any two countable sets:

    If A is countable then, by definition, there exist a bijection f:N-> A.

    If B is countable then, by definition, there exist a bijection g:N-> B.

    The function A->B, g\circ f^{-1} is a bijection from A to B.
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