"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.

If x is in A+ B, then there exist a in A and b in B such that x= a+ b.b) Let A and B be bounded non-empty subsets ofR, 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

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), theleastupperbound 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]