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I need help solving a proof relating sup(AB) and Binf(A)

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Let A be a bounded nonempty subset of the set of all real numbers (R). B exists in R and B<0. Let BA= {Ba: a exists in A} Prove sup(BA)=Binf(A)

    2. Relevant equations

    We are able to use the ordered field axioms, Archemedian Property ect..


    3. The attempt at a solution

    I know that I need to show
    that sup(BA)<=Binf(A) and sup(BA)>=BinfA
    and If A is bounded then y<b where b is a bound for all y that exist in A
     
  2. jcsd
  3. Oct 3, 2012 #2
    What is the definition of supremum and infimum?
     
  4. Oct 3, 2012 #3
    Supremum is the lowest upper bound of the set and infimum is the highest lower bound of the set
     
  5. Oct 3, 2012 #4

    Ok, try thinking in terms of elements, that is, suppose i call β the supremum, how does that relate to elements in the set? And by supremum I'm referring to the supremum of set A.
    Think about it, we have the ordered field axioms, a negative number, and the definition of sup and inf.
     
    Last edited: Oct 3, 2012
  6. Oct 4, 2012 #5
    I still seem to be confused as to how B would relate to the elements in the set. If B is sup(A) then this is saying it is the lowest upper bound of A. If this is the lowest upper bound then B could be either less than or greater than the set of BA itself correct?
     
  7. Oct 4, 2012 #6
    Try thinking of an example. Suppose I have the set A=(1,2) and B=-1
    What is the supremum of B*A? What is B * infimum of A?

    Does this help?

    Also beta is not B, sorry i should have used a better letter. Beta is the supremum of the set.
     
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