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Math Help - Prove that an infimum equals a supremum in different sets

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    Prove that an infimum equals a supremum in different sets

    Suppose that A is contained in the set of all real numbers and is bounded above. Let U be the set of upper bounds, i.e., U={x is an element of the real numbers: x>=a for all a elements of A}. Prove that U is bounded below and that inf U=sup A. (Note: U is not the empty set by the assumption that A is bounded above.)

    Thanks for the help!
    Last edited by Plato; September 23rd 2012 at 08:03 AM.
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    Re: Prove that an infimum equals a supremum in different sets

    Quote Originally Posted by lovesmath View Post
    Suppose that A is contained in the set of all real numbers and is bounded above. Let U be the set of upper bounds, i.e., U={x is an element of the real numbers: x>=a for all a elements of A}. Prove that U is bounded below and that inf U=sup A. (Note: U is not the empty set by the assumption that A is bounded above.)
    First we need to say that A is not empty, therefore U is bounded below.
    Let \alpha=\sup(A). You need to show that \alpha is a lower bound for U.
    Then show that \alpha is the greatest lower bound of U.
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