Prove $\sup(AB) = \sup(A)\sup(B)$

Question

I know $\sup(AB) = \sup(A)\sup(B)$ if $A$ and $B$ are nonnegative, but what if the assumption is dropped that $A$ and $B$ are nonnegative. Does this change the answer?

score 5 · Answer 1 · answered Sep 18 '13 at 17:38

5

Consider $A=[0,1]$, $B=\{-1\}$, $Sup(AB)=0$, $Sup(A)Sup(B)=-1$

answered Sep 18 '13 at 17:38

helmonio

725

score 2 · Answer 2 · answered Sep 18 '13 at 17:37

2

Yes it does. Consider the set $$A=B=\{-1/n : n\in \mathbb{N}\}.$$ Then $\sup(AB)=sup\{ab: a\in A, b\in B\}=1$ but $sup(A)=sup(B)=0$

answered Sep 18 '13 at 17:37

Quickbeam2k1

2,139

Prove $\sup(AB) = \sup(A)\sup(B)$

2 Answers2