Show: sup AC = supA supC

  • Thread starter kingwinner
  • Start date
  • #1
1,270
0
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!
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
5
From what you have you know already that for all ε > 0,

[tex](\sup A - \epsilon)(\sup C - \epsilon) < ac \leq \sup AC \leq \sup A \sup C[/tex], 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?
 
  • #3
1,270
0
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!
 
Last edited:
  • #4
Gib Z
Homework Helper
3,346
5
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.
 
Last edited:

Related Threads on Show: sup AC = supA supC

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
11
Views
3K
Replies
10
Views
10K
Replies
1
Views
2K
Replies
1
Views
3K
Replies
6
Views
3K
Replies
11
Views
3K
Replies
4
Views
4K
Replies
4
Views
2K
Top