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Math Help - power set question

  1. #1
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    power set question

    Hey guys i have, A = {1,2,3,4} and B = {2,3,5}, list the following elements:

    1. Powerset(A) \ Powerset(B)

    Does that mean, find Powerset(A) first then Powerset(B) then if Powerset(A) has subsets the same as Powerser(B), then remove those sets. Then the remaining subsets of Powerset(A) will = Powerset(A) \ Powerset(B) ?

    thanks for the help!
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  2. #2
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    If C\in \left(\mathcal{P}(A)\cap \mathcal{P}(B)\right) then remove C from  \mathcal{P}(A).
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  3. #3
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    Quote Originally Posted by Plato View Post
    If C\in \left(\mathcal{P}(A)\cap \mathcal{P}(B)\right) then remove C from  \mathcal{P}(A).
    i dont think \mathcal{P}(A) \cap \mathcal{P}(B) is the same as  \mathcal{P}(A) \backslash \mathcal{P}(B)?
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    Quote Originally Posted by jvignacio View Post
    if Powerset(A) has subsets the same as Powerser(B), then remove those sets
    Maybe you meant, if P(A) has MEMBERS in common with P(B) then remove those common members from P(A).

    S\T = {x | x in S and x not in T}

    So:

    P(A)\P(B)
    =
    {x | x in P(A) and x not in P(B)}
    =
    {x | x subset of A and x not subset of B}

    I.e., gather the subsets of A to form P(A), then gather the subsets of B to form P(B), then remove those members of P(A) that are also members of P(B) to form P(A)\P(B).
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  5. #5
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    Quote Originally Posted by MoeBlee View Post
    So gather the subsets of A, then gather the subsets of B, then from the gathering of subsets of A remove those that are also subsets of B.
    that's what I meant remove the subsets of A that are also subsets of B
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  6. #6
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    Quote Originally Posted by jvignacio View Post
    i dont think \mathcal{P}(A) \cap \mathcal{P}(B) is the same as  \mathcal{P}(A) \backslash \mathcal{P}(B)?
    That wasn't what Plato was claiming. Imho, he was just making the suggestion that because of

    \mathcal{P}(A)\backslash \mathcal{P}(B)=\mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)

    it would by easier to determine the solution of your exercise by using the right side of this identity.
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  7. #7
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    Quote Originally Posted by Failure View Post
    That wasn't what Plato was claiming. Imho, he was just making the suggestion that because of

    \mathcal{P}(A)\backslash \mathcal{P}(B)=\mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)

    it would by easier to determine the solution of your exercise by using the right side of this identity.
    Yep I understand now. thanks!
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  8. #8
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    Quote Originally Posted by jvignacio View Post
    i dont think \mathcal{P}(A) \cap \mathcal{P}(B) is the same as  \mathcal{P}(A) \backslash \mathcal{P}(B)?
    Correct. But what the poster Plato said is correct.

    Let 'n' stand for the binary intersection operation.

    S\T = S\(SnT)

    P(A)\P(B) = P(A)\(P(A)nP(B))

    /

    Edit: I see that you got it now anyway.
    Last edited by MoeBlee; April 17th 2010 at 08:28 AM.
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  9. #9
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    Quote Originally Posted by MoeBlee View Post
    Correct. But what the poster Plato said is correct.

    Let 'n' stand for the binary intersection operation.

    S\T = S\(SnT)

    P(A)\P(B) = P(A)\(P(A)nP(B))

    /

    Edit: I see that you got it now anyway.
    In a case where it was: find elements in: A^2 \cap B^2, would it be ZERO elements? since the set A deals with 4 numbers and set B deal with 3, therefore never having any subsets in common. ?
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Failure View Post
    That wasn't what Plato was claiming. Imho, he was just making the suggestion that because of

    \mathcal{P}(A)\backslash \mathcal{P}(B)=\mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)

    it would by easier to determine the solution of your exercise by using the right side of this identity.
    I think it was more that as \mathcal{P}(A)\not\subset \mathcal{P}(B) you can't really talk about \mathcal{P}(A)\backslash \mathcal{P}(B), and instead we should talk about \mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big). For notation reasons, however, we use the former as it is "neater".
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  11. #11
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    Quote Originally Posted by Swlabr View Post
    I think it was more that as \mathcal{P}(A)\not\subset \mathcal{P}(B) you can't really talk about \color{red}\mathcal{P}(A)\backslash \mathcal{P}(B)
    Why not? Set difference is set difference.
    To say that C\in\mathcal{P}(A)\backslash \mathcal{P}(B) means C\in\mathcal{P}(A) \text{ and }C\notin \mathcal{P}(B).
    Moreover \mathcal{P}(A)\not\subset \mathcal{P}(B) has nothing to do with the general concept.
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