1. ## SupA.SupB=InfAB

Querrian ayudarme con lo siguiente:

A,B, subconjuntos de R acotados y no vacíos

a)Si SupA<0 y SupB<0
Demostrar:
Inf(AB)=SupA.SupB
gracias

2. You probably want to show:

$\inf AB \leq \sup A * \sup B$ and,
$\inf AB \geq \sup A * \sup B$.

Hmm.. Maybe I can get you started, one of the directions seems simple.

$a \leq \sup A,$ $\forall$ $a \in A$.
$b \leq \sup B,$ $\forall$ $b \in B$.

Now,

$a*b \leq \sup A * \sup B$. [MISTAKE?]

Also,

$\inf AB \leq a*b,$ $\forall$ $a \in A,$ and $b \in B$.

So,

$\inf AB \leq a*b \leq \sup A * \sup B,$ and thus

$\inf AB \leq \sup A * \sup B$

3. Originally Posted by cgiulz
$a*b \leq \sup A * \sup B$.
Hmm, I dunno bout this:

-5<-3 and -2<0 but (-5)*(-2)>(-3)*0

I don't speak the language of the original post, but I think it translates as this in English:

If A and B are non-empty subset of real numbers such that
supA< 0 and supB< 0
show that:
infAB = supA.supB

Here's what I would do:

Since supA< 0 it means there is a negative number w such that $a\leq w<0$ for all a in A. Also, there is a v<0 such that $b\leq v<0$ for all b in B.
So $ab\geq vw$ for all a in A and b in B, which we can do because of the signs of everything. This shows that vw is a lower bound for AB, i.e. supA.supB is a lower bound for AB. All that remains to show is that vw is the largest lower bound. Try to see if you can show this.

4. I think I was assuming the converse.

5. If someone could translate the original post to English it would be very helpful

6. Originally Posted by MATHFRIEND
Querrian ayudarme con lo siguiente:

A,B, subconjuntos de R acotados y no vacíos

a)Si SupA<0 y SupB<0
Demostrar:
Inf(AB)=SupA.SupB
gracias
Woud you like to help me with the following:

A,B subsets of R, bounded and non-empty.

a)If SupA<0 and SupB<0
Prove:
Inf(AB)=SupA.SupB
Thanks