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Math Help - sequences and continuity

  1. #1
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    sequences and continuity

    Here are a few analysis problems I am having trouble with:

    1. Let {a_n} be a sequence of real numbers and let r be a real number that satisfies 0 < r < 1. Suppose that | a_n+1 - a_n | <= r*| a_n - a_n-1| for n > 1. Prove that {a_n} is a Cauchy sequence and hence a sequence that converges to a limit.

    2. Let {a_n} be a sequence of real numbers. Prove that if {a_n} has two subsequences that converge to different limits, then the sequence {a_n} does not converge.

    3. Prove the following theorem: Let {a_n} be a bounded sequence of real numbers and let S be the set of sequential limits of {a_n}. Then the set S contains its greatest lower bound and its least upper bound and
    lim inf a_n = inf S & lim sup a_n = sup S.

    4. Let S be a bounded nonempty set of real numbers and suppose that supS is not an element of S. Prove that there is a nondecreasing sequence {s_n} of elements of S such that the limit as n approaches infinity of s_n is supS.

    Thanks for the help.
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  2. #2
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    Quote Originally Posted by eigenvector11 View Post
    1. Let {a_n} be a sequence of real numbers and let r be a real number that satisfies 0 < r < 1. Suppose that | a_n+1 - a_n | <= r*| a_n - a_n-1| for n > 1. Prove that {a_n} is a Cauchy sequence and hence a sequence that converges to a limit.
    Proof

    2. Let {a_n} be a sequence of real numbers. Prove that if {a_n} has two subsequences that converge to different limits, then the sequence {a_n} does not converge.
    If a_n convergese to a then it means |a_n-a| < \epsilon for n>N. If a_{n_k} is a subsequence then n_k \geq n > N thus |a_{n_k}-a|<\epsilon. This shows a convergent sequence has all its sequences converge to the same limit.
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  3. #3
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    Quote Originally Posted by eigenvector11 View Post
    3. Prove the following theorem: Let {a_n} be a bounded sequence of real numbers and let S be the set of sequential limits of {a_n}. Then the set S contains its greatest lower bound and its least upper bound and
    lim inf a_n = inf S & lim sup a_n = sup S.
    Let a_{n_k} be a subsequence having limit L. Then \liminf a_{n_k} = \limsup a_{n_k} = L but since \{ a_{n_k}| k\geq N\}\subseteq \{a_n| n\geq N\} it means \liminf a_n \leq \liminf a_{n_k} = L = \limsup a_{n_k} \leq \limsup a_n. Thus \limsup a_n is upper bound on S and \liminf a_n is lower bound on S this means \liminf a_n \leq \inf S \leq \sup S \leq \limsup s_n. Now there is a theorem that says that for any bounded sequence we can find two subsequences one subsequence convergening to its limsup and the other one converging to its liminf. This means the limsup and liminf belong to the set. Thus \liminf a_n = \inf S and \limsup a_n = \sup S.
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    Quote Originally Posted by eigenvector11 View Post
    4. Let S be a bounded nonempty set of real numbers and suppose that supS is not an element of S. Prove that there is a nondecreasing sequence {s_n} of elements of S such that the limit as n approaches infinity of s_n is supS.
    Let S be a bounded non-empty set of real numbers. Then \sup S = a is a real numbers. For \epsilon = 1 we have that a - 1 is not an upper bound (why is that?) thus there exists s_1 so that s_1 \geq a-1. Likewise s_2\geq a - 1/2, and in general s_n \geq a - 1/n. Thus we have s_n - a \geq - 1/n \implies 0\leq a - s_n\leq 1/n and thus |s_n - a| \leq 1/n. This means \lim s_n = a. Now this constructed sequence is not necessary nondecreasing, try to correct this proof so that it is nondecreasing.
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  5. #5
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    Thanks for the help on the first three. I'm not 100% sure how to connect the last proof to a nondecreasing sequence though. I know this means a_n+1 >= a_n, but I don't see how to tie this into the proof.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by eigenvector11 View Post
    Thanks for the help on the first three. I'm not 100% sure how to connect the last proof to a nondecreasing sequence though. I know this means a_n+1 >= a_n, but I don't see how to tie this into the proof.
    you said it, make sure that s_{n + 1} \ge s_n
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