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Thread: Supremum

  1. #1
    mus
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    Supremum

    Hi,

    Could someone assist me in the right direction concerning the following problem:

    Deduce the existence of a supremum from the principle of nested intervals.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    I hope this will help...

    Let $\displaystyle (a_n)$ be bounded sequence. Then $\displaystyle S=limsup{a_n}$ exist and he the greater partial limit of the sequence.

    In other words: $\displaystyle limsup{a_n}=max\{L:\text{when L is partial limit}\}$



    Proof:

    Due to Bolzano-Weierstrass Theorem we may discuss on the set of partial limits, which is non-empty.

    Given $\displaystyle \epsilon>0$ there are infinite that $\displaystyle S-\epsilon<a_n$, but only a finite number of $\displaystyle S+\epsilon\leq a_n$. Therefor there is infinite $\displaystyle a_n$ in neighborhood $\displaystyle \epsilon$ of $\displaystyle S$, hence $\displaystyle S$ is partial limit.

    Now, suppose that $\displaystyle S<S'$ other partial limit then we will choose $\displaystyle \epsilon >0$ so that $\displaystyle S<S'-\eppsilon$, but we know that there is only finite number of $\displaystyle S'-\epsilon<a_n$, therefor it impossible that $\displaystyle S'$ is partial limit, thus $\displaystyle S$ is the greatest partial limit.
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  3. #3
    mus
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    Could you kindly elaborate on the proof. Its not quite clear.

    Thanks
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