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Math Help - Help! Intro to Real Analysis problem

  1. #1
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    Help! Intro to Real Analysis problem

    I'm having problems with some of the proofs of the supremum and infimum properties in my text. here are two that I'm stuck on:

    a) Let S be a non-empty subset of R that is bounded below. Prove that
    inf S = -sup{-s: s in S}

    b) Let A and B be bounded non-empty subsets of R, and let
    A+B := {a+b: a in A, b in B}.
    Prove that sup(A+B) = supA + supB and inf(A+B) = infA + infB

    Any comments and/or help would be much appreciated, it's getting near the end of the semester, and I'm really struggling with these concepts!

    -Robitar
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  2. #2
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    Quote Originally Posted by Robitar View Post
    I'm having problems with some of the proofs of the supremum and infimum properties in my text. here are two that I'm stuck on:

    a) Let S be a non-empty subset of R that is bounded below. Prove that
    inf S = -sup{-s: s in S}
    "inf S" is defined as the largest of all lower bounds on S. Let a= inf S. Then a is a lower bound on S so if x is any member of S, then a< x. Can you use that to show that a is an upper bound on {-s: s in S}? Try multiplying by -1. If y is any other lower bound on S, then a< y. Prove that -y is also an upper bound on {-s: s in S} and -a> -y.

    b) Let A and B be bounded non-empty subsets of R, and let
    A+B := {a+b: a in A, b in B}.
    Prove that sup(A+B) = supA + supB and inf(A+B) = infA + infB
    If x is in A+ B, then there exist a in A and b in B such that x= a+ b.
    Show that a< sup(A) and b< sup(B) so a+b< sup(A)+ sup(B) so sup(A)+ sup(B) is an upperbound on A+ B. What does that prove about sup(A+ B), the least upperbound on A+ B?
    Any comments and/or help would be much appreciated, it's getting near the end of the semester, and I'm really struggling with these concepts!

    -Robitar[/QUOTE]
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Robitar View Post

    a) Let S be a non-empty subset of R that is bounded below. Prove that
    inf S = -sup{-s: s in S}
    Define A=\left(a_1,a_2,\cdots,a_n\right) and -A=\left(-a_1,-a_2,\cdots,-a_n\right)=B. So your question is equivalent to showing that \inf\left(A\right)=-\sup\left(B\right).

    1. Since A\subset\mathbb{R}, it is nonempty, and bounded we have that \inf\left(A\right) exists since \mathbb{R} has the greatest lower bound property. Furthermore since \mathbb{R} is an ordered field we have that a_1\leqslant{a_2}\leqslant\cdots\leqslant{a_n}. So we know that \forall{\gamma}\in{A}~a_1\leq{\gamma}, where it follows that a_1 is a lower bound of A. Now it is pretty obvious that if a_1<\gamma then \gamma is not a lower bound of A. Therefore we conclude that a_1 is a greatest lower bound of A, or \inf\left(A\right)=a_1.

    2. Now consider once again a_1\leqslant{a_2}\leqslant\cdots\leqslant{a_n}, now by the ordering of \mathbb{R} we would have that -a_1\geqslant{-a_2}\geqslant\cdots\geqslant{-a_n}, so we can see that since B=\left(-a_1,-a_2,\cdots,-a_n\right) that \forall{\gamma}\in{B}~\gamma\leqslant{-a_1}, so B is bounded above by -a_1. Now noting that B=-A gives us two things, B\subset\mathbb{R} and B\ne\varnothing, so once again because \mathbb{R} has the leas upper bound property we can conclude that \sup\left(B\right) exists. Now as noted before -a_1 is an upper bound of B, and it can be seen that if \gamma<-a_1 then \gamma is not an upper bound of B. So we can see that -a_1 is a least upper bound of B or in other words \sup\left(B\right)=-a_1

    3. So conclude that since \inf\left(A\right)=a_1 and \sup\left(B\right)=-a_1 that \inf\left(A\right)=-\sup\left(B\right)~\blacksquare
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