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Math Help - Subsequential limit problem

  1. #1
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    Subsequential limit problem

    If {an} is bounded sequence with exactly one subsequential limit a, the {an} converges to a.

    Could you help me understand what bounded sequence with one subsequential limit is? I don't understand the question. Thanks very much in advance
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  2. #2
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    A sequence is "bounded", if there is a constant C>0 , such that
    |a_n|< C for all n.
    A "subsequential limit" of a sequence is a limit of a subsequence a_n_k of a_n.
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  3. #3
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    so by definition lim {a_n_k} = a so for every e > 0 there exists n* in N such that for all n_k>= n* we have with any arbitrary a_n_k being subsequence of a_n:

    |a_n_1 - a| < e/n for all n >= n1*
    |a_n_2 - a| < e/n for all n >= n2*
    |a_n_3 - a| < e/n for all n >= n3*
    .
    .
    |a_n_k - a| < e/n for all n_k >= nk*
    .
    .
    .
    .
    we need to prove that |a_n - a| < e

    we know what a_n is bounded so by definition there exist some real number M such that |a_n| < M

    since a_n_k is a subsequence of a_n so we have: n_1 < n_2 < n_3,...with i <= n_i , i in N.

    Pick n* = min (n_1, n_2, n_3, ....) so for all n >= n* we have
    |an_1 + an_2 + an_3 +...+ an_k+.....-n.a| <e

    I started getting confused from this. Please help *_*
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  4. #4
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    Quote Originally Posted by tommy11 View Post
    If {an} is bounded sequence with exactly one subsequential limit a, the {an} converges to a. Could you help me understand what bounded sequence with one subsequential limit is? I don't understand the question. Thanks very much in advance
    Because every sequence by definition is a subsquence of itself, there really nothing to prove.
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  5. #5
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    Quote Originally Posted by tommy11 View Post
    so by definition lim {a_n_k} = a so for every e > 0 there exists n* in N such that for all n_k>= n* we have with any arbitrary a_n_k being subsequence of a_n:

    |a_n_1 - a| < e/n for all n >= n1*
    |a_n_2 - a| < e/n for all n >= n2*
    |a_n_3 - a| < e/n for all n >= n3*
    .
    .
    |a_n_k - a| < e/n for all n_k >= nk*
    .
    .
    .
    .
    we need to prove that |a_n - a| < e

    we know what a_n is bounded so by definition there exist some real number M such that |a_n| < M

    since a_n_k is a subsequence of a_n so we have: n_1 < n_2 < n_3,...with i <= n_i , i in N.

    Pick n* = min (n_1, n_2, n_3, ....) so for all n >= n* we have
    |an_1 + an_2 + an_3 +...+ an_k+.....-n.a| <e

    I started getting confused from this. Please help *_*
    By assumption we know that every convergent subsequence of (a_n) converges against the same number a.
    Furthermore |a_n|<C for all n.
    We need to show that a_n also converges.
    Assume that this is not the case. In particular a_n does not converge against a. Then there is an e>0 , so that there is a subsequence a_n_j of a_n with |a_n_j-a|>=e for all j.
    But a_n_j is also a bounded sequence and any bounded sequence has a convergent subsequence. So a_n_j has a subsequence which converges, but it can't converge against a, because its distance from a is always greater or equal than e.
    This subsequence is also a convergent subsequence of a_n. => Contradiction to our assumption.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Because every sequence by definition is a subsquence of itself, there really nothing to prove.
    But not every subsequence needs to converge.
    The assumption is only that every convergent subsequence of (a_n) has the same limit.
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  7. #7
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    Quote Originally Posted by Iondor View Post
    But not every subsequence needs to converge.
    The assumption is only that every convergent subsequence of (a_n) has the same limit.
    And you proof shows that every subsequence does in fact converge.
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