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    Yep, I need help again...

    Let (bn) be a sequence of real numbers.
    Suppose that each subsequence (bnr) has a subsequence (bnrs) such that (bnrs) converges to 0.
    Prove that (bn) tends to zero as n tends to infinity.

    So far I've assumed that (bn) does not tend to zero and tryed to get a contradiction but haven't really got there (but apparently there is a solution that argues by contradiction).
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    (Original post by flyinghorse)
    Yep, I need help again...

    Let (bn) be a sequence of real numbers.
    Suppose that each subsequence (bnr) has a subsequence (bnrs) such that (bnrs) converges to 0.
    Prove that (bn) tends to zero as n tends to infinity.

    So far I've assumed that (bn) does not tend to zero and tryed to get a contradiction but haven't really got there (but apparently there is a solution that argues by contradiction).
    If (bn) doesn't tend to 0 then for you can find arbitrarily large nr such that |bnr| >1 (this comes from taking epsilon=1). This sequence |bnr| has no subsequence tending to 0.
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    Suppose (b_n) doesn't tend to 0. Then there is an epsilon > 0 such that the following statement is false:

    |b_n| < epsilon, for all sufficiently large n.

    So

    |b_n| >= epsilon, for infinitely many n.

    So there are positive integers n(1) < n(2) < n(3) < ... such that

    |b_{n(r)}| >= epsilon, for all r.

    The subsequence (b_{n(r)}) of the original sequence does not have a subsubsequence that tends to zero. Contradiction.
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    thanks, rep is on its way when i can give/be bothered to give
 
 
 
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Updated: November 14, 2004

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