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Let S be a subset of the reals, and suppose that S is bounded above. Let B be the set of upper bounds of S and suppose that B has no lower bound. What do you conclude about S?

I know that it probably has something to do with the Completeness Axiom. I guessed that S is empty, because B basically "pushed" the upper bound of S lower and lower (because B is not bounded below). I tried to do this more formally: If S has an upper bound x where x ≥ y for all y in S, but there is a set B of zs where z gets smaller and smaller, there will always be a z in B such that z < x. So the set S is empty.

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Every $s\in S$ serves as lower bound for $B$. So if $B$ has no lower bounds then $S$ must be empty. – drhab Apr 3 '14 at 21:24
If you have found one of out answers satisfying you should accept it or at least vote it up. – Vanio Begic Apr 4 '14 at 11:50

If B has no lower bound then S does not have lowest upper bound,meaning that it does not have a supremum(least upper bound) thus it violates the Completeness axiom(if set has upper bound then it has lowest upper bound) thus the set S would be unbound above which is a contradiction meaning that set B must have an infinum or...

In another case if set S is an empty set then every number is a upper bound for it.Thus B would contain all real numbers,and would be equal to R and thusly B would not have a smallest member.

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Every $s\in S$ has the property that $s\le b$ for every $b\in B$. In other words: every $s\in S$ serves as a lower bound for $B$. If $B$ has no lower bounds then it follows immediately that $S=\emptyset$.

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