How to prove sup AC = supA supC Proving sup AC = supA supC

[kingwinner]

Fact: If A and C are subsets of R, let AC={ac: a E A, c E C}. If A and C are bounded and A and C consist of strictly positive elements only, then sup AC = supA supC.

I am trying to understand and prove this fact, and this is what I've got so far...
A, C bounded => supA, supC exist (by least upper bound axiom)
0<a≤supA for all a E A
0<c≤supC for all c E C
=> 0<ac≤supA supC for all ac E AC
=> supA supC is an upper bound of AC
=> supAC≤supA supC

But how can we prove the other direction? I think we somehow have to use the fact "for all ε>0, there exists a E A such that sup A - ε < a."
At the end, we have to show that for all ε>0, supA supC - ε is NOT an upper bound for AC. But I'm not sure how it is going to work out here in our case.

Does anyone have any idea?
Thanks for any help!

 

[Gib Z]

From what you have you know already that for all ε > 0,

$$(\sup A - \epsilon)(\sup C - \epsilon) < ac \leq \sup AC \leq \sup A \sup C$$, where in this case, a and c are some specific elements of A and C, not general ones.

Can you see a sandwich argument approaching here?

 

[kingwinner]

I don't think (supA-ε)(supC-ε)<ac is true for ALL ε>0, e.g. if supA-ε<0, then the inequality is reversed...

Also, at the end, we want to show that for all ε>0, supA supC - ε is NOT an upper bound for AC.
But expanding the LHS of your inequality, we have
supA supC -ε(supA+supC) + ε², instead of supA supC - ε, how can we fix this?

Thanks!

 

[Gib Z]

I don't think (supA-ε)(supC-ε)<ac is true for ALL ε>0, e.g. if supA-ε<0, then the inequality is reversed...

Sorry, my last post should have said “for ε sufficiently small” instead of “all”.

Also, at the end, we want to show that for all ε>0, supA supC - ε is NOT an upper bound for AC.
But expanding the LHS of your inequality, we have
supA supC -ε(supA+supC) + ε², instead of supA supC - ε, how can we fix this?

Those things are pretty much exactly the same. What is epsilon? A constant greater than zero that we can choose to be arbitrarily small. Realizing this, we can see that
supA supC -ε(supA+supC) + ε² = supA supC -ε(supA+supC - ε) = supA supC – ε(Some Bounded, Positive Terms)
yields the required result.