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Math Help - Real analysis: Limit superior proof

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    Real analysis: Limit superior proof




    [note: also under discussion in s.o.s. math board]
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post



    [note: also under discussion in s.o.s. math board]
    Since we have that \lim\text{ }b_n=a we have that for every \varepsilon>0 there exists an N\in\mathbb{N} such that N\leqslant n\implies \left|b_n-a\right|<\varepsilon\implies -\varepsilon<b_n-a<\varepsilon\implies b_n-\varepsilon<b_n<a+\varepsilon more pertinently we have that b_N<a+\varepsilon. But, b_N=\sup\text{ }\left\{a_n:N\leqslant n\right\} so that N\leqslant n\implies a_n\leqslant b_N<a+\varepsilon
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    Quote Originally Posted by Drexel28 View Post
    Since we have that \lim\text{ }b_n=a we have that for every \varepsilon>0 there exists an N\in\mathbb{N} such that N\leqslant n\implies \left|b_n-a\right|<\varepsilon\implies -\varepsilon<b_n-a<\varepsilon\implies b_n-\varepsilon<b_n<a+\varepsilon more pertinently we have that b_N<a+\varepsilon. But, b_N=\sup\text{ }\left\{a_n:N\leqslant n\right\} so that N\leqslant n\implies a_n\leqslant b_N<a+\varepsilon
    Sorry, I don't understand your meaning of b_n. (otherwise the proof makes sense to me)

    Let (a_n) be a sequence of real numbers.
    We have sup{a_n: n≥k} = sup{a_k,a_k+1,a_k+2,...} = b_k
    and
    lim sup{a_n: n≥k} = lim sup a_n
    k->infinity

    So you mean b_n = sup{a_n: n≥n} ??
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    Sorry, I don't understand your meaning of b_n. (otherwise the proof makes sense to me)

    Let (a_n) be a sequence of real numbers.
    We have sup{a_n: n≥k} = sup{a_k,a_k+1,a_k+2,...} = b_k

    So b_n = sup{a_n: n≥n} ??
    I'm sorry. I guess I'm just so used to using n as a subscript. Change it to k.
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    Quote Originally Posted by Drexel28 View Post
    I'm sorry. I guess I'm just so used to using n as a subscript. Change it to k.
    OK, then I think I understand your proof. Thank you!
    But now I have a general question about the way lim sup is usually defined.

    Let (a_n) be a sequence of real numbers. Then we define lim sup to be
    lim [sup{a_n: n≥k}] = lim sup a_n = b_k
    k->∞
    Here, my understanding is that the indices n and k are independent.

    But I have seen some textbooks doing the following:
    Let (a_n) be a sequence of real numbers. Then they define lim sup to be
    lim [sup{a_m: m≥n}] = lim sup a_n = b_n
    n->∞
    i.e. they are using the same subscript "n", but "n" is the subscript in the original sequence (a_n), so they can't be independent.
    Is it correct to do this?

    If we use the latter definition, would this ruin our proof? (becuase at the end, I don't think we'll get the same conclusion...)
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    OK, then I think I understand your proof. Thank you!
    But now I have a general question about the way lim sup is usually defined.

    Let (a_n) be a sequence of real numbers. Then we define lim sup to be
    lim [sup{a_n: n≥k}] = lim sup a_n = b_k
    k->∞
    Here, my understanding is that the indices n and k are independent.

    But I have seen some textbooks doing the following:
    Let (a_n) be a sequence of real numbers. Then they define lim sup to be
    lim [sup{a_m: m≥n}] = lim sup a_n = b_n
    n->∞
    i.e. they are using the same subscript "n", but "n" is the subscript in the original sequence (a_n), so they can't be independent.
    Is it correct to do this?

    If we use the latter definition, would this ruin our proof? (becuase at the end, I don't think we'll get the same conclusion...)
    That would make no sense if I understand what you're saying.

    Supremum Limit -- from Wolfram MathWorld

    This is the exact way you defined it, except they have n and k switched.
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