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Math Help - Prove there is a subsequence of S, where L = supS, S nonempty + bdd abv.

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    Senior Member MacstersUndead's Avatar
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    Prove there is a subsequence of S, where L = supS, S nonempty + bdd abv.

    Let S be a nonempty set that is bbd abv. Let u be the supremum of S. Show that there is a sequence {x_n} in S with {x_n} approaching u.
    --

    I have two general ideas on how to solve this problem, but I don't know how to start.

    1) I use the epsilon definition of convergence, defining epsilon with relation to u.

    or

    2) Show that the limit cannot be less than supS (easy to prove, or not? if lim < sup S, lim is in S, choose an increasing sequence and {x_k} > lim), or greater than supS.

    If not(greater or less than), must be equal to.
    --

    Thanks in advance.
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    Quote Originally Posted by MacstersUndead View Post
    Let S be a nonempty set that is bbd abv. Let u be the supremum of S. Show that there is a sequence {x_n} in S with {x_n} approaching u.
    There are two cases to consider:
    • 1. u\in S
    • 2. u\notin S
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