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Math Help - Limit Inferior Proof

  1. #1
    Junior Member utopiaNow's Avatar
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    Limit Inferior Proof

    Let \{s_n\}_{n \in \mathbb{Z}^+} be a sequence of real numbers that is bounded below. Prove that \displaystyle\liminf_{n\rightarrow\infty} s_n = \sup_m \left(\inf_{n > m} s_n\right).

    I honestly can't even understand the question. I understand that if we let E be the set of numbers x (in the extended real number system) such that s_{nk} \rightarrow x for some subsequence \{s_{nk}\} then \liminf_{n\rightarrow\infty} s_n = \inf{E} refers to the lower limit.

    But I'm not really understanding the term on the RHS, much less how to go about proving the equality.

    Any hints or suggestions would be greatly appreciated.
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    Quote Originally Posted by utopiaNow View Post
    Let \{s_n\}_{n \in \mathbb{Z}^+} be a sequence of real numbers that is bounded below. Prove that \displaystyle\liminf_{n\rightarrow\infty} s_n = \sup_m \left(\inf_{n > m} s_n\right).

    I honestly can't even understand the question. I understand that if we let E be the set of numbers x (in the extended real number system) such that s_{nk} \rightarrow x for some subsequence \{s_{nk}\} then \liminf_{n\rightarrow\infty} s_n = \inf{E} refers to the lower limit.

    But I'm not really understanding the term on the RHS, much less how to go about proving the equality.

    Any hints or suggestions would be greatly appreciated.
    Take a sequence a_n and define a new sequence b_n as b_n = inf{a_n, a_(n+1), a_(n+2), ...}. Then lim inf a_n is defined as lim b_n - that's the mathematical meaning of the term "lower limit." You're being asked to prove lim b_n = sup b_n. This can be done by proving b_n is nondecreasing.
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  3. #3
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by rn443 View Post
    Take a sequence a_n and define a new sequence b_n as b_n = inf{a_n, a_(n+1), a_(n+2), ...}. Then lim inf a_n is defined as lim b_n - that's the mathematical meaning of the term "lower limit." You're being asked to prove lim b_n = sup b_n. This can be done by proving b_n is nondecreasing.
    Sorry if this is a naive question, but how does that define a sequence? Shouldn't the inf{a_n, a_(n+1), a_(n+2), ...} be just a single value?

    Secondly won't showing that b_n is nondecreasing just show that lim b_n exists, won't I need to do extra work to show lim b_n = sup b_n?
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    Quote Originally Posted by utopiaNow View Post
    Sorry if this is a naive question, but how does that define a sequence? Shouldn't the inf{a_n, a_(n+1), a_(n+2), ...} be just a single value?
    It is a single value: the greatest lower bound of all the numbers a_n, a_(n+1), a_(n+2), ...

    Secondly won't showing that b_n is nondecreasing just show that lim b_n exists, won't I need to do extra work to show lim b_n = sup b_n?
    The limit of a nondecreasing sequence, if it exists, is always its supremum. If we're dealing with the extended reals, we can drop the "if it exists" proviso.
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  5. #5
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by rn443 View Post
    It is a single value: the greatest lower bound of all the numbers a_n, a_(n+1), a_(n+2), ...
    If it is a single value, doesn't that mean it's just a constant sequence and then won't proving that it's nondecreasing be unnecessary?
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  6. #6
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    Quote Originally Posted by utopiaNow View Post
    If it is a single value, doesn't that mean it's just a constant sequence and then won't proving that it's nondecreasing be unnecessary?
    No. As you remove more and more terms, the greatest lower bound may change. For example, let a_n = n/(n+1). Then b_n = inf{n/(n+1), (n+1)/(n+2), (n+2)/(n+3), ...} = n/(n+1), which clearly changes each time you increase n.
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