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Set operation problem

  1. Sep 16, 2004 #1
    Suppose there are 3 sets: A, B, and C s.t. (U for 'union')

    (A - B) U (A - C) U (B - C)

    Now, I was wondering if there is the precedence of parentheses over set operations (union in this case).
    This is saying "every element in A but not in B" or "every element in A but not in C" or "every element in B but not in C".
    I know that nothing in C is contained in this union and that everything in A is minus those that are in both A and B and both A and C. Now, the last difference, B - C, annoys me. Would B - C override A - B and include every element in B but not in C (and not those in both A and B)?

    Would this be a case of symmetric difference? For example
    (A - B) U (B - A)

    would this include everything but the intersection (if A and B are not disjoint or empty and A does not equal B)

    (sidenote: Has there been a problem with Latex, it wouldn't show this tag:

    [tex]\left(A\setminus B\right)\cup \left(A\setminus C\right)\cup \left(B\setminus C\right)[/tex (last bracket intentionally left out so it would show text))
     
    Last edited: Sep 16, 2004
  2. jcsd
  3. Sep 16, 2004 #2

    arildno

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    "I know that nothing in C is contained in this union and that everything in A is minus those that are in both A and B and both A and C. Now, the last difference, B - C, annoys me. Would B - C override A - B and include every element in B but not in C (and not those in both A and B)?"
    This is plain wrong!
    A-B may well have elements also in C. These elements are present in the union
    There is no "overriding" present; each member of the union provides its own elements to the union; those elements are not removed from the union even if these elements are lacking from other members of the union.
     
  4. Sep 16, 2004 #3
    The definition of A - B (difference) states {(for all x)| x is an element of A and x is not an element of B).
    Therefore (A - B) does not contain elements that are in C unless they are in both A and C. But then the second difference, (A - C), removes those elements that are both in A and C. So, thus far (A - B) U (A - C) contains elements exclusive to A.
     
  5. Sep 16, 2004 #4
    I think you (Yoss) may have the wrong idea of "union". If (A-B) has elements of C, then those elements have become part of the union, regardless of what's in the set (A-C). The parentheses take precedence over the union.
     
    Last edited: Sep 16, 2004
  6. Sep 17, 2004 #5

    arildno

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    Yoss: Please read Betty-boop's reply, and the following excerpt from my own:
    "A-B may well have elements also in C. These elements are present in the union
    There is no "overriding" present; each member of the union provides its own elements to the union; those elements are not removed from the union even if these elements are lacking from other members of the union."
    You have a faulty understanding of the union concept.
     
  7. Sep 17, 2004 #6

    HallsofIvy

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    Have you tried any examples?
    Let A= {every member of the alphabet except a}, B= {every member of the alphabet ecxept b}, C= {every member of alphabet except c}.

    Then A-B= {b}, A-C= {c}, B- C= {c}

    (A-B)U(A-C)U(B-C)= {b, c}
     
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