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Math Help - analysis

  1. #1
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    analysis

    Let S be a nonempty set of real numbers that is bounded above, and let Beta be the least upper bound of S. Prove that for every Epsilon greater than 0, there exists an element x such that x is greater than Beta minus Epsilon.

    Any ideas or help would be appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by eigenvector11 View Post
    Let S be a nonempty set of real numbers that is bounded above, and let Beta be the least upper bound of S. Prove that for every Epsilon greater than 0, there exists an element x such that x is greater than Beta minus Epsilon.

    Any ideas or help would be appreciated. Thanks.
    If \beta is a least upper bound then \beta - \epsilon cannot be an upper bound on the set if \epsilon > 0. Why?
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  3. #3
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    The statement that \lambda is not an upper bound of S means that \left( {\exists x \in S} \right)\left[ {\lambda  < x} \right]. That is some number in S is strictly greater that \lambda.

    The statement that \beta is the least upper bound of S means that \beta is an upper bound of S no number less than \beta is an upper bound.

    If \varepsilon  > 0 then \beta  - \varepsilon  < \beta.

    Now you put those three facts together to form a proof of the proposition.
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