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Math Help - Eventually bounded/bounded sequences - I can't complete the proof -_-

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    Newbie Femto's Avatar
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    Eventually bounded/bounded sequences - I can't complete the proof -_-

    Proposition

    If a sequence is eventually bounded, then it is bounded.

    Proof

    Suppose a_n \rightarrow a as n \rightarrow \infty. Then \forall \epsilon > 0\ \exists N(\epsilon) \in \mathbb{N}\ s.t.\ |a_n - a| < \epsilon \ \forall n > N. That is, the subsequence of a_n defined by a_{N + n} is bounded ( |a_{N + n} - a| < \epsilon). Since a_{N + n} is bounded \forall n>N it follows that a_n is eventually bounded.

    ------------------

    From here I have no idea where to go. If I'm right, I need to show that as a result of this sequence being eventually bounded, it must be consequently bounded. But I don't see how the first statement implies the second. It says in the margin that I should consider the idea that every set has a maximum and minimum element but at the moment I'm not sure on anything really. Any advice is welcome And if this is all completely wrong, please inform me about it.

    Also I should note that I have just started my first year at university so a lot of things are a bit alien to me at the moment. Please be patient if I don't understand something that may be above my level
    Last edited by Femto; October 21st 2012 at 05:26 AM.
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    Re: Eventually bounded/bounded sequences - I can't complete the proof -_-

    Your initial proof attempt here begins with the assumption that a_n converges. That's not justified by the assumption that a_n is eventually bounded. For instance, a_n = (-1)^n is certainly eventually bounded, but it doesn't converge.
    Try to state what "eventually bounded" means.
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    Re: Eventually bounded/bounded sequences - I can't complete the proof -_-

    Quote Originally Posted by Femto View Post
    Proposition If a sequence is eventually bounded, then it is bounded.
    Proof Suppose a_n \rightarrow a as n \rightarrow \infty.
    Bounded sequences may not converge.
    But from the given \exists B>0~\&~\exists N\in\mathbb{Z}^+ such that if n\ge N then \left|a_n\right|\le B.
    Let B' = \sum\limits_{k = 1}^{N - 1} {\left| {a_k } \right|}  + B do we have a bound for a_n~?
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    Re: Eventually bounded/bounded sequences - I can't complete the proof -_-

    Quote Originally Posted by Plato View Post
    Bounded sequences may not converge.
    But from the given \exists B>0~\&~\exists N\in\mathbb{Z}^+ such that if n\ge N then \left|a_n\right|\le B.
    Let B' = \sum\limits_{k = 1}^{N - 1} {\left| {a_k } \right|}  + B do we have a bound for a_n~?
    Thanks.

    Yes I realise that bounded sequences may not converge but I thought that all I have to prove is that a sequence is bounded because it is eventually bounded, so I could consider a case in which a sequence is eventually bounded. But yes I see what you mean - that wouldn't be correct.

    I've defined B = max(|a_1|, |a_2|,...,|a_N|, M) as my bound for a_n. I have seen what you've written before but, and do forgive me if this is a stupid question, what is B' exactly? I see you've defined B to be the bound that exists for all n \ge N but why is B' represented as a sum of bounds? If it is the bound for the whole sequence then shouldn't it be the maximum taken from a set of terms rather than a sum (sorry it just confuses me)?
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    Re: Eventually bounded/bounded sequences - I can't complete the proof -_-

    Quote Originally Posted by Femto View Post
    I've defined B = max(|a_1|, |a_2|,...,|a_N|, M) as my bound for a_n. I have seen what you've written before but, and do forgive me if this is a stupid question, what is B' exactly? I see you've defined B to be the bound that exists for all n \ge N but why is B' represented as a sum of bounds? If it is the bound for the whole sequence then shouldn't it be the maximum taken from a set of terms rather than a sum (sorry it just confuses me)?
    Both ways work. I just always use the way I did it.
    \forall n it is true that |a_n|\le B'.
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