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# subsequences... of subsequences(?!) watch

1. 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).
2. (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.
3. 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.
4. thanks, rep is on its way when i can give/be bothered to give

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