Prove sup(ab)=sup(a)sup(b)

**gordo151091** · August 5th 2012, 03:40 PM

Hey, this is the problem

$A,B \subset \mathbb{R}$ A,B Bounded. Let
$AB :=\left \{ ab:a \in A, b \in B \right \}$ . Proof $sup(A)sup(B)=sup(AB)$

I proofed that $xy \leq sup(A)sup(B) \forall x \in A, y \in B$ ( this is false as pointed out by GJA). So by definition $sup(A)sup(B)$ bounds $AB$ . I want to show that indeed $sup(A)sup(B)=sup(AB)$ by showing that any there's no element in $AB$ lower than $sup(a)sup(b)$ than bounds the set. I trying to make the proof by contradiction, by contradicting the fact that theres a element in etheir set $A$ or $B$ that bounds it and is lower than its sup.

Any hints?

Thank you.

**GJA** · August 5th 2012, 04:05 PM

Hi, gordo151091. Do we know if these sets contain only nonnegative numbers? Here's why I ask:

Originally Posted by gordo151091

$A,B \subset \mathbb{R}$ A,B Bounded. Let
$AB :=\left \{ ab:a \in A, b \in B \right \}$ . Proof $sup(A)sup(B)=sup(AB)$

I proofed that $xy \leq sup(A)sup(B) \forall x \in A, y \in B$ .

What if we take $A=[-2,-1]$ and $B=[-2,-1]$ . Then $\sup A=\sup B = -1$ . Now take $x=y=-2$ . Then $xy=4>1=\sup A\sup B$ . Am I missing something? Let me know if I'm in error here.

**gordo151091** · August 5th 2012, 04:12 PM

You are not missing anything. I'm wrong here. Any hints on how to start again?

**GJA** · August 5th 2012, 04:20 PM

I'm starting to think this statement isn't true. If we stick with the sets $A=B=[-2,-1]$ , then $AB=[1,4]$ . We then have $\sup(AB)=4$ , but $\sup(A)\sup(B)=(-1)(-1)=1$ . If everything looks valid to you, then I think we need further assumptions on A and B to obtain the equality $\sup(AB)=\sup(A)\sup(B)$ .

**gordo151091** · August 5th 2012, 04:29 PM

I think the assumption we have to make is that they are subset of the posite real numbers.

**Deveno** · August 5th 2012, 06:18 PM

because of difficulties like these, in defining the real numbers, it is often preferable to define the set of positive real numbers, first. in fact, defining a negative real number is a bit subtle, if one is proceeding by using dedekind cuts.

**gordo151091** · August 8th 2012, 04:46 PM

Originally Posted by gordo151091

Hey, this is the problem

$A,B \subset \mathbb{R}$ A,B Bounded. Let
$AB :=\left \{ ab:a \in A, b \in B \right \}$ . Proof $sup(A)sup(B)=sup(AB)$

I proofed that $xy \leq sup(A)sup(B) \forall x \in A, y \in B$ ( this is false as pointed out by GJA). So by definition $sup(A)sup(B)$ bounds $AB$ . I want to show that indeed $sup(A)sup(B)=sup(AB)$ by showing that any there's no element in $AB$ lower than $sup(a)sup(b)$ than bounds the set. I trying to make the proof by contradiction, by contradicting the fact that theres a element in etheir set $A$ or $B$ that bounds it and is lower than its sup.

Any hints?

Thank you.

We do have to assume $A,B \subset \mathbb{R}_{+}$ . This is the proof:

Since $A,B \subset \mathbb{R}_{+}$ we have $ab \leq sup(A)sup(B) \forall a \in A, b \in B$ . So $Sup(AB)\leq Sup(A)Sup(B)$ . For the other inequality we have that $ab\leq Sup(AB)$ , it follows that $a \leq Sup(AB)b^{-1}$ . So $Sup(A)b \leq Sup(AB)$ . We repeat with $b$ . $b \leq Sup(AB)Sup(A)^{-1}$ , which implies $Sup(B)Sup(A) \leq Sup(AB)$ . Q.E.D

I hope this helps someone

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