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Math Help - Help understanding limsup definition

  1. #1
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    Help understanding limsup definition

    I'm trying to show that limsup a_n=inf{ sup S_n|n \in N} where S_n={ a_k|k \geq n} and a_n is a bounded sequence using only the fact that limsup a_n=sup(T), where T is the set of all subsequential limits of a_n.

    I realize that what I'm asking is a "definition," but I'm just trying to find a way to derive it, basically. Any help on this?
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  2. #2
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    Quote Originally Posted by paupsers View Post
    I'm trying to show that limsup a_n=inf{ sup S_n|n \in N} where S_n={ a_k|k \geq n} and a_n is a bounded sequence using only the fact that limsup a_n=sup(T), where T is the set of all subsequential limits of a_n.

    I realize that what I'm asking is a "definition," but I'm just trying to find a way to derive it, basically. Any help on this?

    I think you say it all and in the clearest possible way: it is the supremum of all the partial limits of the sequence, or: if the sequence has finitely many partial limits, then the lim sup is the greatest one.

    Tonio
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  3. #3
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    Can you show a chain of equalities that links them? I still don't get how they are related...
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  4. #4
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    Quote Originally Posted by paupsers View Post
    Can you show a chain of equalities that links them? I still don't get how they are related...

    Chain of equalities? Links "them"? I don't understand.

    Tonio
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  5. #5
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    Can you show that the definition I gave is equivalent to the definition I'm trying to understand?
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  6. #6
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    Here's as far as I can get...

    I know that S_n=\{a_k|k \geq n\}=\{a_n, a_{n+1}, a_{n+2}...\}

    Then, how do I define \sup S_n=? since S_n has an infinite number of terms?

    Moreover, I don't understand the convention of writing \limsup a_n=\inf\{\sup S_n\} because isn't \sup S_n a single value, so the \inf\{\sup S_n\} is the same value?

    The definition my book gives for \limsup is: if T is the set of all subsequential limits of \{a_n\}, then \sup T is \limsup \{a_n\}. I don't understand how this definition is equivalent to \limsup a_n=\inf\{\sup S_n\}.

    Does anyone understand what I'm asking? Lol. I'm just looking for a way to show that the two definitions are equivalent.
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  7. #7
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    Quote Originally Posted by paupsers View Post
    Here's as far as I can get...

    I know that S_n=\{a_k|k \geq n\}=\{a_n, a_{n+1}, a_{n+2}...\}

    Then, how do I define \sup S_n=? since S_n has an infinite number of terms?


    So?? The definition of supremum is interesting ONLY if there's an infinite number of elements, otherwise we can always choose the largest one by observation which then would be the maximum

    Moreover, I don't understand the convention of writing \limsup a_n=\inf\{\sup S_n\} because isn't \sup S_n a single value, so the \inf\{\sup S_n\} is the same value?

    Yes, \sup S_n is a single value FOR EVERY SINGLE n, and you take the infimum over all the possible n's!

    The definition my book gives for \limsup is: if T is the set of all subsequential limits of \{a_n\}, then \sup T is \limsup \{a_n\}. I don't understand how this definition is equivalent to \limsup a_n=\inf\{\sup S_n\}.


    If t is a partial limit then there's an infinite subsequent which converges to t and, thus there's an infinite number of elements of the original sequence which converge to t.
    don't you try some examples? Take for example \{a_n\}=\left\{-1,1,1,-1,1,\frac{1}{2},-1,1,\frac{1}{3},...., -1,1,\frac{1}{n},....\right\}

    In this sequence, we'll get for infinite n's that \sup S_n = 1\Longrightarrow \inf \sup S_n=1\,,\,\,and\,\,\inf S_n=-1\Longrightarrow \sup \inf S_n=-1, so the lim sup is 1 and the lim inf is -1.

    Tonio

    Does anyone understand what I'm asking? Lol. I'm just looking for a way to show that the two definitions are equivalent.
    .
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