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Thread: Disjoint cartesian products

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

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

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

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

    How can this be true? For example, if $\displaystyle C_{1}(x) = (1,2) \times (5,6) $ and $\displaystyle C_{2}(x) = (3,4) \times (5,6) $, then $\displaystyle C_{1} $ and $\displaystyle C_2 $ are disjoint, because no ordered pair $\displaystyle (a,b)$ is in both $\displaystyle C_1$ and $\displaystyle C_2$. But, $\displaystyle C_{1}(x) = (5,6) $ and $\displaystyle 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 $\displaystyle (E,\mathcal{E})$ and $\displaystyle (F,\mathcal{F}) $ be two measurable spaces.

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

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

    How can this be true? For example, if $\displaystyle C_{1}(x) = (1,2) \times (5,6) $ and $\displaystyle C_{2}(x) = (3,4) \times (5,6) $, then $\displaystyle C_{1} $ and $\displaystyle C_2 $ are disjoint, because no ordered pair $\displaystyle (a,b)$ is in both $\displaystyle C_1$ and $\displaystyle C_2$. But, $\displaystyle C_{1}(x) = (5,6) $ and $\displaystyle 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 $\displaystyle C_n(x)$ are the slices of the sets $\displaystyle C_n$ obtained by fixing the first coordinate x and letting the y coordinate vary.

    Quote Originally Posted by TheProphet View Post
    For example, if $\displaystyle C_{1}(x) = (1,2) \times (5,6) $ and $\displaystyle C_{2}(x) = (3,4) \times (5,6) $, then $\displaystyle C_{1} $ and $\displaystyle C_2 $ are disjoint, because no ordered pair $\displaystyle (a,b)$ is in both $\displaystyle C_1$ and $\displaystyle C_2$. But, $\displaystyle C_{1}(x) = (5,6) $ and $\displaystyle C_2(x) = (5,6) $ ?
    In that example, you should have taken the sets to be $\displaystyle C_{1} = (1,2) \times (5,6) $ and $\displaystyle 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 $\displaystyle C_1(1.5) = (5,6)$, but the set $\displaystyle C_2(1.5)$ is empty, because $\displaystyle C_2$ 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 $\displaystyle C_1 = (1,2) \times (5,6) $ and $\displaystyle C_2 = (3,4) \times (5,6) $, this was a misprint on my part. Sorry about that.
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