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Real Analysis Problem

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data
    (a) Suppose that A and B are nonempty subsets of R. Define subsets -A={-x: x[itex]\in[/itex]A} and A+B={x+y: x[itex]\in[/itex]A and y[itex]\in[/itex]B}. Show that if A and B are bounded above, then the greatest lower bound of -A = - least upper bound of A and the least upper bound of (A+B) = the least upper bound of A plus the least upper bound of B.

    (b) Use part (a) to prove the Greatest Lower Bound Property: Any nonempty shubset of R that is bounded below has a greatest lower bound.

    2. Relevant equations
    If 0<a and 0<b, then there is a positive integer n such that b<a+a+...+a (n summands).
    If A is any nonempty subset of R that is bounded above, then there is a least upper bound for A.


    3. The attempt at a solution
    I've proven that first part of (a), that the greatest lower bound of -A = -least upper bound of A, but I can't figure out why the least upper bound of (A+B) would = the least upper bound of A plus the least upper bound of B. [or sup(A+B) = sup(A)+sup(B)]
     
    Last edited: Sep 8, 2011
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  3. Sep 8, 2011 #2

    micromass

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    Hi major_maths! :smile:

    So you need to prove that

    [tex]\sup(A+B)=\sup A+\sup B[/tex]

    Can you first prove that [itex]\sup A+\sup B[/itex] is an upper bound of A+B??

    That is, take an arbitrary element z in A+B, can you prove that [itex]z\leq \sup A+\sup B[/itex]??
     
  4. Sep 8, 2011 #3
    Thanks! I got through part (a) by proving (A+B) must be nonempty and then proving that there was an upper bound in (A+B) since both A and B had upper bounds, using the Least Upper Bound Property to prove that there must be a least upper bound since there was an upper bound to begin with.

    I'm stuck again on part (b) though. I know that since inf(-A) exists, -sup(A) must exist as well. I don't know how to go about proving the Greatest Lower Bound Property from there though. I was thinking about using the Least Upper Bound Property somehow.
     
  5. Sep 8, 2011 #4

    micromass

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    Yes, you proved that A+B has a least upper bound. But did you prove that sup(A)+sup(B) is that exact upper bound??

    You do need to prove the least upper bound property!! Just transform the inf into a sup and use the least upper bound property.
     
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