Prove that a set S of numbers has a maximum iff it is bounded above and sup S ε S

1. Apr 27, 2012

mrchris

1. The problem statement, all variables and given/known data
Sorry that should read and sup S$\in$S

2. Relevant equations
just the definitions of sup, inf, boundedness, and max

3. The attempt at a solution
I have tried a few different things, but this question is posted for advice, not just an answer. I am trying to think of a good way to start and we don't use contrapositive proofs in my class. I was thinking I could try proof by contradiction, but then I will have to do 2 cases, first assuming S is not bounded, then assuming supS is not in S. I understand the definitions of the terms, and I could explain why this is true verbally, but I am stuck trying to do it formally. Again, I am not just fishing for an answer, I really need to get these down or I wont pass anyway.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 27, 2012

kru_

This depends on your precise definition of maximum. If you assume S has a maximum, then what do you know about it?

Similarly, if S is bounded above, and the sup S, call it s0, is an element inside of S, then what do we know about s0?

3. Apr 27, 2012

4. May 4, 2012

mrchris

Case 1: S is not bounded above. So this is a direct contradiction to the definition given for a maximum in the book, "a member c of S is called the max of S provided that it is an upper bound for S". Since S is assumed to be unbounded, no number c exists s.t. c is an upper bound for S.

Case 2: Sup S is NOT in S. Then we know for every x in S, x<c, but x can never equal c, since we have assumed sup S is NOT in S. Then we take any positive number epsilon= |(c-x_0)/2|, which can never be equal to 0 because c can never equal x. If we assume some number x_0 is the max of S, then we can take (x_0 + epsilon), where x_0 < (x_0 + epsilon) < c. This is a contradiction to the choice of x_0 as the choice of a maximum. Basically, if Sup S is not in S, then no matter what number x_0 in S that you choose, there exists a number x_0 + |(c-x_0)/2|, s.t. x_0 < (x_0 + |(c-x_0)/2| ) < c. This will contradict any choice of x_0 as a maximum of S.