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  1. #1
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    one prove question

    Let A and C be subset of $\displaystyle R^n$ with boundaries B(A), B(C) respectively. Prove or disprove:
    (1) B(AUC) = B(A)UB(C)
    (2) B(A$\displaystyle \cap$C) = B(A)$\displaystyle \cap$B(C)
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  2. #2
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    Quote Originally Posted by wopashui View Post
    Let A and C be subset of $\displaystyle R^n$ with boundaries B(A), B(C) respectively. Prove or disprove:
    (1) B(AUC) = B(A)UB(C)
    (2) B(A$\displaystyle \cap$C) = B(A)$\displaystyle \cap$B(C)
    I haven't studied this formally per se, but both (1) and (2) seem to fail when we consider the simple case of two intersecting spheres in R^3, or even just intersecting discs in R^2.

    Edit: Actually easier to write in $\displaystyle \,\mathbb{R}$,

    let $\displaystyle A = [0,3], C = [1,2]$,

    so

    $\displaystyle B(A) = \{0, 3\}$,

    $\displaystyle B(C) = \{1, 2\}$,

    $\displaystyle B(A \cup C) = \{0, 3\}$,

    $\displaystyle B(A) \cup B(C) = \{0,1,2,3\}$,

    $\displaystyle B(A \cap C) = \{1,2\}$,

    $\displaystyle B(A) \cap B(C) = \emptyset$
    Last edited by undefined; Sep 14th 2010 at 10:13 PM.
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  3. #3
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    Quote Originally Posted by undefined View Post
    I haven't studied this formally per se, but both (1) and (2) seem to fail when we consider the simple case of two intersecting spheres in R^3, or even just intersecting discs in R^2.

    Edit: Actually easier to write in $\displaystyle \,\mathbb{R}$,

    let $\displaystyle A = [0,3], C = [1,2]$,

    so

    $\displaystyle B(A) = \{0, 3\}$,

    $\displaystyle B(C) = \{1, 2\}$,

    $\displaystyle B(A \cup C) = \{0, 3\}$,

    $\displaystyle B(A) \cup B(C) = \{0,1,2,3\}$,

    $\displaystyle B(A \cap C) = \{1,2\}$,

    $\displaystyle B(A) \cap B(C) = \emptyset$
    is this situation only in $\displaystyle R^1$ or applied in $\displaystyle R^n$
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by wopashui View Post
    is this situation only in $\displaystyle R^1$ or applied in $\displaystyle R^n$
    The edit was written just for $\displaystyle \mathbb{R}$, but in order to disprove a statement you only need one counterexample.

    Like I said though I haven't studied this formally but it seems all right.
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