- #1
Treadstone 71
- 275
- 0
"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.
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.