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Math Help - Real Analysis Question

  1. #1
    dom
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    Real Analysis Question

    Here is a question. It has multiple parts:

    Let a_n be a sequence. Denote L=lim sup a_n. Define a sequence b_m by b_m=sup{a_m, a_m+1, a_m+2,...}.

    a). Prove that b_m is nonincreasing sequence.
    b). Prove that L<= b_m for any m>= 1.(Hint: prove for m=1 and explain why true for m>=1)
    c) Prove that for any epsilon >0, there is some m such that b_m
    d). Use the above results to prove that L=lim b_m.

    I think I got part (a):

    b_m=sup{a_m, a_m+1, a_m+2,...}.=sup{a_n: n>= m}

    So, b_m+1=sup{a_n: n>= m+1}=sup{ a_m+1, a_m+2,...}
    We know that if A, B are in R, then if A contained in B means that supA<= supB, using this we have that sup { a_m+1, a_m+2,...} is contained in sup{a_m, a_m+1, a_m+2,...}, where the first one is b_m+1 and the second is b_m, so we can conclude that b_m+1<= b_m, which proves that b_m is nonincreasing.

    For part (b) I got:
    We need to prove that L<= b_m
    L<=sup{a_m,a_m+1,...}
    sup an <= sup{a_m,a_m+1,...}
    Here I get stuck.

    Or to be more precise, I get confused because a_m.....is a subsequence of a_n, and it has to be the other way around where sup a_n >= sup a_m?


    I'm not sure how to proceed.
    Thanks in advance for any help
    Last edited by dom; May 23rd 2011 at 07:41 PM.
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  2. #2
    bl2
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    What is your definition of limsup. The definition I know is exactly lim b_m, but it seems like you were given a different one to start with, and are asked to show that the two are equivalent.

    In part b), you wrote "sup an <= sup{a_m,a_m+1,...}" which is where you got stuck. You're right to get stuck there, because that's not necessarily true. But that's because L is not sup an, it's limsup an, which has a different definition
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