Proof dealing with bounded sets

1. May 29, 2013

bonfire09

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

2. Relevant equations

3. The attempt at a solution
Let Set $A=(a_1,...,a_t: a_1<...a_i<a_t)$ and let set $B=(b_1,...,b_s: b_1<...<b_s)$ Then set $A+B=(a_i+b_k: 1≤ i≤ t \text{and} 1≤ k ≤ S)$ It follows that $a_t= sup(A)$ and $b_s=sup(B)$ since $a_i ≤a_s$ and $b_s ≥ b_k$ for all $a_i ε A$ and $b_k ε B$ where $1≤ k ≤s$ and $1≤ k ≤s$. Now $sup(A+B) = a_t+b_s$ since $a_t+b_s≥ a_i+b_k$
Thus $sup(A+B) = a_t+b_s= sup(A)+ sup (B)$. Now I don't know whether I should argue that $a_t$ and $b_s$ 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. May 29, 2013

HallsofIvy

Staff Emeritus
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. $a\le sup(A)$ and $b\le sup(A)$ so that $a+ b\le sup(A)+ sup(B)$ 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).

3. May 29, 2013

bonfire09

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.