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Math Help - Least Upper Bond Proof. I have no Idea D:.

  1. #1
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    Least Upper Bound Proof. I have no Idea D:.

    Let S be a bounded set that is not empty.
    Prove that sup(S) is a boundary point of S


    can someone help me complete this proof? i've been trying for a long time and i can't come up with anything other than

    If S is a (not empty) set of real numbers and S has an upper bound y,
    then there is a least upper bound of the set S.
    That least upper bound is also called the supremum of S

    please help, i really need it
    Last edited by arturdo968; September 10th 2008 at 04:40 PM. Reason: title
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  2. #2
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    I think that you need help with the idea of boundary points.
    The point b is a boundary point of S if and only if \left( {\forall \varepsilon  > 0} \right)\left[ {\left( {\exists x \in S} \right)\left( {\exists y \notin S} \right)\left[ {\left\{ {x,y} \right\} \subseteq \left( {b - \varepsilon ,b + \varepsilon } \right)} \right]} \right].
    In other words, in every \varepsilon-\mbox{neighborhood} of b contains a point of S and a point not in S.

    Using the definition of supremum, the proof is simple.
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  3. #3
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    hmm. see i'm not very good with this. i'm just starting off, and i'm having a hard time with the writing of proofs. it's hard. thanks you though. the definition of a supremum is the LUB on a function over its domain right?kj
    Last edited by arturdo968; September 10th 2008 at 04:36 PM.
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  4. #4
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    would it help if i were to plug anything in?>
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  5. #5
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    errr, anyone?
    i hate to post in a row lke this but i really need help
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