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

  1. #1
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    Prove sup(ab)=sup(a)sup(b)

    Hey, this is the problem

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

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

    Any hints?

    Thank you.
    Last edited by gordo151091; Aug 5th 2012 at 04:13 PM.
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  2. #2
    GJA
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    Re: Prove sup(ab)=sup(a)sup(b)

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

    Quote Originally Posted by gordo151091 View Post

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

    I proofed that $\displaystyle xy \leq sup(A)sup(B) \forall x \in A, y \in B$.
    What if we take $\displaystyle A=[-2,-1]$ and $\displaystyle B=[-2,-1]$. Then $\displaystyle \sup A=\sup B = -1$. Now take $\displaystyle x=y=-2$. Then $\displaystyle xy=4>1=\sup A\sup B$. Am I missing something? Let me know if I'm in error here.
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  3. #3
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    Re: Prove sup(ab)=sup(a)sup(b)

    You are not missing anything. I'm wrong here. Any hints on how to start again?
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  4. #4
    GJA
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    Re: Prove sup(ab)=sup(a)sup(b)

    I'm starting to think this statement isn't true. If we stick with the sets $\displaystyle A=B=[-2,-1]$, then $\displaystyle AB=[1,4]$. We then have $\displaystyle \sup(AB)=4$, but $\displaystyle \sup(A)\sup(B)=(-1)(-1)=1$. If everything looks valid to you, then I think we need further assumptions on A and B to obtain the equality $\displaystyle \sup(AB)=\sup(A)\sup(B)$.
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  5. #5
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    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.
    Last edited by gordo151091; Aug 5th 2012 at 05:23 PM.
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  6. #6
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    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.
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  7. #7
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    Re: Prove sup(ab)=sup(a)sup(b)

    Quote Originally Posted by gordo151091 View Post
    Hey, this is the problem

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

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

    Any hints?

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

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

    I hope this helps someone
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