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Math Help - Algebra - Please help

  1. #1
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    Algebra - Please help

    X_1, X_2 are nonempty sets. Let \mathcal{F}_1 \subset 2^{X_1} and \mathcal{F}_2 \subset 2^{X_2} be \sigma-algebras.

    Let \mathcal{R} = \{A \times B: A \in \mathcal{F}_1, B \in \mathcal{F}_2 \}

    And \mathcal{R}_1 = \left\{\bigcup_{i=1}^n R_i: n \in \mathbb{N}, \ \{R_i\}_{i=1}^n \subset \mathcal{R} \right\}

    Prove that \mathcal{R}_1 is an algebra.

    It was not hard to prove that it is colsed with respect to finite unions. However I failed to prove it is closed with respect to complement!
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  2. #2
    Forum Admin topsquark's Avatar
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    And this relates to Probability and Statistics, how?

    Thread moved.

    -Dan

    Edit: My bad. Sorry about that!
    Last edited by topsquark; June 16th 2008 at 03:18 PM.
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  3. #3
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    Quote Originally Posted by topsquark View Post
    And this relates to Probability and Statistics, how?

    Thread moved.

    -Dan
    Because axiomatic probability theory involves measures and \sigma-algebra.
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  4. #4
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    Frankly speaking I was working on some probability theory theorem when this problem occured. I'm 99% sure R_1 is algebra but the full proof seems bit too long.... If somebody knows a short proof, please post it.

    PS I think I can prove that R_1 is closed under finite intersections. Than I take complement of set that belongs to R_1 and I get finite intersection - the only problem is to find a simple proof that the "intersected" sets belong to R_1 .
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  5. #5
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    Quote Originally Posted by albi View Post
    Frankly speaking I was working on some probability theory theorem when this problem occured. I'm 99% sure R_1 is algebra but the full proof seems bit too long.... If somebody knows a short proof, please post it.

    PS I think I can prove that R_1 is closed under finite intersections. Than I take complement of set that belongs to R_1 and I get finite intersection - the only problem is to find a simple proof that the "intersected" sets belong to R_1 .
    these two trivial identities should be enough for you to complete the proof:

    1) \ \ (A_1 \times B_1) \cap (A_2 \times B_2) = (A_1 \cap A_2) \times (B_1 \cap B_2).

    2) \ \ (A \times B)^c = (A^c \times B) \cup (A \times B^c) \cup (A^c \times B^c), where Y^c is the complement of Y.
    Last edited by NonCommAlg; June 16th 2008 at 01:07 PM.
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  6. #6
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    The last one is not true. I belive it should be:
    <br />
 (A \times B)^c = (A^c \times B) \cup (A \times B^c)  \cup (A^c \times B^c),<br />

    However that will help.
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  7. #7
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    Quote Originally Posted by albi View Post
    The last one is not true. I belive it should be:
    <br />
(A \times B)^c = (A^c \times B) \cup (A \times B^c) \cup (A^c \times B^c),<br />
    of course! thanks! i fixed it. i've already done too much latex for today ... i guess i'm getting really tired!
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