Math Beauties need HELP, part II

[marlen]

Homework Statement

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

Homework 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.

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

 

[HallsofIvy]

Homework Statement

Let S be the nonempty set of real numbers bounded above.

Well, that, to start with, makes no sense. I assume you mean a 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

Homework 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.

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 > ...

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?

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