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Help, analsysis proof - inf (-A)=-sup(A) [-A= {-a : a∈A}]

  1. Sep 19, 2007 #1

    let A be a set of real numbers bounders above.
    then, inf (-A)=-sup(A) [-A= {-a : a∈A}]

    really...thanks......i have no idear..how to prove it....?
  2. jcsd
  3. Sep 19, 2007 #2
    the question didn't mention if A is a non-empty set...
    so do i need to prove A is a non-empty set?
    Or it said is real numbers, then it means is non empty set?
  4. Sep 19, 2007 #3


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    You don't have to prove A is non-empty.

    A is a set of real numbers already means there exists an element x which is a real number and belongs to A. Since there exists such an element, thus A is non-empty.
  5. Sep 20, 2007 #4


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    Notice that a "set of real numbers" is not a "subset of the set of real numbers" since the latter might be empty! It is more common for a theorem like this to start "Let A be a non-empty subset of R."

    To prove your theorem- keep multiplying by -1!

    For example, sup(A) is, by definition, an upper bound on A.

    Let b be any member of -A. Then b=-a for some a in A. We must have [itex]a\le sup(A)[/itex] so, multiplying by -1, [itex]-a= b\ge -sup(A)[/itex]. That tells you that -sup(A) is a lower bound on -A. I'll leave the rest to you.
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