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Math Help - 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

    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.
    Last edited by gordo151091; August 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

    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.
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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 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).
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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; August 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

    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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