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

Hey, this is the problem

A,B Bounded. Let

. Proof

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

Any hints?

Thank you.

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

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

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

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

I'm starting to think this statement isn't true. If we stick with the sets , then . We then have , but . If everything looks valid to you, then I think we need further assumptions on A and B to obtain the equality .

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

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

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

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.

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