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

  • Thread starter mrchris
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  • #1
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Homework Statement


Sorry that should read and sup S[itex]\in[/itex]S


Homework Equations


just the definitions of sup, inf, boundedness, and max



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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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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
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I understand the definitions of the terms, and I could explain why this is true verbally[/b]

Then that's how you should start.

Write down explicitly exactly what the definitions are; what you are supposed to assume; and what it is you are required to prove.

Then write down your intuitive verbal explanation.

When you're stuck those are two things to do to get started.
 
  • #4
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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.
 

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