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Math Help - Disjoint cartesian products

  1. #1
    Junior Member TheProphet's Avatar
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    Disjoint cartesian products

    Let (E,\mathcal{E}) and  (F,\mathcal{F}) be two measurable spaces.

    Let  \mathcal{E} \otimes \mathcal{F} = \sigma(\mathcal{E} \times \mathcal{F}), where  \mathcal{E} \times \mathcal{F} = \{A \subset E \times F : A = \Lambda \times \Gamma, \Lambda \in \mathcal{E}, \Gamma \in \mathcal{F}\} .

    For  C \in \mathcal{E} \otimes \mathcal{F} , let  C(x) = \{y : (x,y) \in C \} . If we now let  C_{n} be a pairwise disjoint sequence in  \mathcal{E} \otimes \mathcal{F} , then the  C_{n}(x) are also pairwise disjoint. <- HERE IS MY QUESTION.

    How can this be true? For example, if  C_{1}(x) = (1,2) \times (5,6) and  C_{2}(x) = (3,4) \times (5,6) , then  C_{1} and C_2 are disjoint, because no ordered pair (a,b) is in both C_1 and C_2. But, C_{1}(x) = (5,6) and C_2(x) = (5,6) ?
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  2. #2
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    Opalg's Avatar
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    Re: Disjoint cartesian products

    Quote Originally Posted by TheProphet View Post
    Let (E,\mathcal{E}) and  (F,\mathcal{F}) be two measurable spaces.

    Let  \mathcal{E} \otimes \mathcal{F} = \sigma(\mathcal{E} \times \mathcal{F}), where  \mathcal{E} \times \mathcal{F} = \{A \subset E \times F : A = \Lambda \times \Gamma, \Lambda \in \mathcal{E}, \Gamma \in \mathcal{F}\} .

    For  C \in \mathcal{E} \otimes \mathcal{F} , let  C(x) = \{y : (x,y) \in C \} . If we now let  C_{n} be a pairwise disjoint sequence in  \mathcal{E} \otimes \mathcal{F} , then the  C_{n}(x) are also pairwise disjoint. <- HERE IS MY QUESTION.

    How can this be true? For example, if  C_{1}(x) = (1,2) \times (5,6) and  C_{2}(x) = (3,4) \times (5,6) , then  C_{1} and C_2 are disjoint, because no ordered pair (a,b) is in both C_1 and C_2. But, C_{1}(x) = (5,6) and C_2(x) = (5,6) ?
    Your example is obviously correct. Given disjoint sets in a product space, it certainly does not follow that their projections onto one of the coordinate spaces are disjoint.
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Disjoint cartesian products

    Ok. The reason why I asked is because they make this claim in a proof in a book I´m reading. This is a big error, or it is something else that I am missing.
    Thank you anyway.
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  4. #4
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    Re: Disjoint cartesian products

    Oops, I was wrong first time, and the book is right. The fact that x appears in the notation C(x) implies that x is meant to be fixed. So the sets C_n(x) are the slices of the sets C_n obtained by fixing the first coordinate x and letting the y coordinate vary.

    Quote Originally Posted by TheProphet View Post
    For example, if  C_{1}(x) = (1,2) \times (5,6) and  C_{2}(x) = (3,4) \times (5,6) , then  C_{1} and C_2 are disjoint, because no ordered pair (a,b) is in both C_1 and C_2. But, C_{1}(x) = (5,6) and C_2(x) = (5,6) ?
    In that example, you should have taken the sets to be  C_{1} = (1,2) \times (5,6) and  C_{2} = (3,4) \times (5,6) (in other words, the parameter x does not appear in the names of the product sets). If you now choose x = 1.5, for example, then C_1(1.5) = (5,6), but the set C_2(1.5) is empty, because C_2 does not contain any points with x-coordinate 1.5.
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  5. #5
    Junior Member TheProphet's Avatar
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    Re: Disjoint cartesian products

    Ah, thank you. I actually meant C_1 = (1,2) \times (5,6) and  C_2 = (3,4) \times (5,6) , this was a misprint on my part. Sorry about that.
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