Question

AB is a proper subset of R^+. Show that
sup(AB)=supAsupB
Please help

Answer 1

@SithsAndGiggles

Answer 2

whats sup here

Answer 3

My attempt:
Let \(U_A\) = sup A
\(U_B\) = sup B
Let z \(\in \) (AB), then \(x\in A\) and \(y\in B\)
\(x\leq U_A\)
\(y\leq U_B\)
then \(xy\leq U_AU_B\)

Answer 4

sup is supremum

Answer 5

Now, show \(supA *supB= sup (AB)\)

Answer 6

I have to show \(xy > U_A*U_B -\varepsilon\)

Answer 7

A more beautiful way to do it :
a,b>0 a in sup(A) and b in sup(B)

sup(AB) ≥ ab <=> 1/a sup(AB) ≥ b
So B is bounded above by 1/a sup(AB) ie.
1/a sup(AB) ≥ sup(B) <=> 1/sup(B) sup(AB) ≥ a
So A is bounded above by 1/sup(B) sup(AB) meaning:
1/sup(B) sup(AB) ≥ sup(A) <=> sup(AB) ≥ sup(A)sup(B)

Answer 8

Thanks a lot. It is helpful. :)

Answer 9

Welcome =)