Math Beauties need HELP, part II

1. Oct 6, 2007

marlen

1. The problem statement, all variables and given/known data

Let S be the nonempty set of real numbers bounded above. Prove that S^3 = {x^3 : x $$\in$$ S} is bounded above and sup S^3 = (sup S)^3

2. Relevant equations

given S^3 = { y $$\in$$ R : $$\exists$$ x, x $$\in$$ S and y = x^3}

and

for all $$\epsilon$$ > 0, there is y $$\in$$ S^3 such that $$\alpha$$^3 < y $$\leq$$ $$\alpha$$^3.

3. The attempt at a solution

This is what we attempted, but were told we are wrong:

Let S= { s1, s2, s3,....} s.t. s1 > s2 > s3 > ...
Then s1 $$\geq$$ sn, for all sn $$\in$$ S.
This implies S is bounded above by S1 and so supS = s1

Now:
(supS)^3 = (s1)^3

if s1 is negative, then (s1)^3 = (-s1)(-s1)(-s1) = -s1^3
if s1 is positive, then (s1)^3 = (s1)(s1)(s1) = s1^3
which implies that (supS)^3 = s1^3

For S^3 = {s1^3, s2^3, s3^3,...} and s1^3 > s2^3 > s3^3 >...
Then s1^3 $$\geq$$ sn^3, for all sn^3 $$\in$$ S^3.
This implies S^3 is bounded above by S1^3 and so supS^3 = s1^3

therefore, supS^3 = (supS)^3 = s1^3

2. Oct 6, 2007

HallsofIvy

Staff Emeritus
Well, that, to start with, makes no sense. I assume you mean a nonempty set of real numbers bounded above

Are you assuming the set is countable? That's the only way you could write them like this. And sets of real numbers, in general, are not countable.[/quote]
Then s1 $$\geq$$ sn, for all sn $$\in$$ S.
This implies S is bounded above by S1 and so supS = s1[/quote]
But you are also assuming that s1 is IN S and you were not told that sup(S) was in S.
"sup(S)" is the least upper bound of S. Assuming x is in S3, then x= s3 for some s in S and so $s\le sup(S)$. Can you then prove that $x= s^3\le (sup(S))^3$ (so that (sup(S))3 is an upper bound on S3)? Can you now prove that (sup(S))3 is the LEAST upper bound?