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Proof dealing with bounded sets

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Let A and B be nonempty bounded subsets of [itex]\mathbb{R}[/itex], and let [itex] A + B [/itex] be the set of all sums [itex]a + b[/itex] where [itex]a ∈ A[/itex] and [itex]b ∈ B[/itex].
    (a) Prove [itex] sup(A+B) = supA+supB [/itex].


    2. Relevant equations



    3. The attempt at a solution
    Let Set [itex]A=(a_1,...,a_t: a_1<...a_i<a_t)[/itex] and let set [itex]B=(b_1,...,b_s: b_1<...<b_s)[/itex] Then set [itex]A+B=(a_i+b_k: 1≤ i≤ t \text{and} 1≤ k ≤ S)[/itex] It follows that [itex] a_t= sup(A)[/itex] and [itex] b_s=sup(B) [/itex] since [itex] a_i ≤a_s [/itex] and [itex] b_s ≥ b_k[/itex] for all [itex] a_i ε A [/itex] and [itex] b_k ε B [/itex] where [itex] 1≤ k ≤s[/itex] and [itex] 1≤ k ≤s[/itex]. Now [itex] sup(A+B) = a_t+b_s[/itex] since [itex] a_t+b_s≥ a_i+b_k [/itex]
    Thus [itex] sup(A+B) = a_t+b_s= sup(A)+ sup (B) [/itex]. Now I don't know whether I should argue that [itex] a_t[/itex] and [itex] b_s [/itex] is the greatest least lower bound because I already have the elements in sets A and B are in increasing order. Other than that would my proof be correct?
     
  2. jcsd
  3. May 29, 2013 #2

    HallsofIvy

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    I don't see anything in the hypothesis that A and B must be finite sets which your proof assumes. Also you seem to be assuming that sup(A) is in A and sup(B) is in B which is true for finite sets but not for infinite sets.

    Basically, you want to prove that sup(A)+ sup(B) is the least upper bound of A+ B so you need to prove, first, that it is an upper bound on A+ B. Suppose x is in A+ B. Then, by definition of "A+ B", x= a+ b where a is in A and b is in B. [itex]a\le sup(A)[/itex] and [itex]b\le sup(A)[/itex] so that [itex]a+ b\le sup(A)+ sup(B)[/itex] showing that sup(A)+ sup(B) is an upper bound on A+ B.

    It remains to show that any other upper bound on A+ B is larger than sup(A)+ sup(B).
     
  4. May 29, 2013 #3
    Thanks. I knew something was wrong. I should break it up into two cases then one for finite sets and one for infinite sets then.
     
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