# Completeness axiom/theorem and supremum

1. Dec 11, 2005

"Let A and B be bounded nontempty sets of real numbers. Let C={ab:a in A, b in B}. Prove that sup(C)=sup(A)sup(B)."

Here's what I've done so far:

By the completeness axiom/theorem A and B have suprema. Let sup(A)=z and sup(B)=y. For all e>0, there exists a in A and b in B such that z-e<a and y-e<b. Multiplying the two inequalities and we have

zy-ze-ye+e^2 < ab

I'm stuck here.

2. Dec 11, 2005

### HallsofIvy

Staff Emeritus
It would be a good idea to show first that zy is an upper bound on C!
You are using the fact that since z is the least upper bound on A there must be an a between z-e and z and since y is the least upper bound on B there must be a b between z-e and z. What does that tell you about there being an ab between zy- e and z?

3. Dec 11, 2005

### matt grime

step 1 show it's an upper bound,

step 2 show it is a least upper bound which can be messy.

it's easier to use the definition of sup as the maximum of the accumulation points.

You need to show that give d>0 you can find e greater than zero such that 0<ey+ez-e^2<d, so do that.

4. Dec 11, 2005

I don't know, but I want to arrive at the conclusion that zy-x<ab for all x>0. If I could prove that zy-(???)<ab where (????) is positive, then I'm done, since I can let x=(????).

5. Dec 11, 2005

### matt grime

we may suppose e<z and e<y, can you see how that might help?

6. Dec 11, 2005

e>0. z and y could be negative.

7. Dec 12, 2005

### matt grime

How can the supremum of a set of positive numbers be negative?

8. Dec 12, 2005

A and B aren't sets of strictly positive numbers.

9. Dec 12, 2005

### matt grime

Then the proposition is trivially false.

10. Dec 12, 2005