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Math Help - Bounded subset of R

  1. #1
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    Bounded subset of R

    I'm having trouble with this problem. Could someone give me a hand?
    Let S be a nonempty subset of R that is bounded above. Show that if sup S is not in S, then for every \epsilon >0, the interval (sup S-\epsilon, sup S) contains infinitely many elements of S.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jackie View Post
    I'm having trouble with this problem. Could someone give me a hand?
    Let S be a nonempty subset of R that is bounded above. Show that if sup S is not in S, then for every \epsilon >0, the interval (sup S-\epsilon, sup S) contains infinitely many elements of S.
    Suppose for some \epsilon >0 that (sup S-\epsilon, sup S) contains only finitely many points of  S (that it is non-empty should be obvious).

    CB
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    Thanks a lot for your help, CB. I actually tried to assume the contrary as you suggested, but I got stuck. I have x_1, x_2, ...., x_n is in (supS- \epsilon, supS). So,  supS - \epsilon <x_1, x_2,...,x_n < sup S I think I should be able to show that this lead to sup S is in S, but I don't know how.
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  4. #4
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    Quote Originally Posted by jackie View Post
    Thanks a lot for your help, CB. I actually tried to assume the contrary as you suggested, but I got stuck. I have x_1, x_2, ...., x_n is in (supS- \epsilon, supS). So,  supS - \epsilon <x_1, x_2,...,x_n < sup S I think I should be able to show that this lead to sup S is in S, but I don't know how.
    A finite subset of the reals has a largest element, which will be an upper bound for  S and less than {\text{sup}}(S) which is a contradiction etc...
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