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Math Help - Set Theory

  1. #1
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    Set Theory

    Hello!!! Please help meeeeeee!! I am just doing some homework problems for my test and I come across this one I don't understand.

    1) Is it possible to find set A and B such that both A in B and A subset of B are true? Give an example or prove this is impossible.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jenjen View Post
    Hello!!! Please help meeeeeee!! I am just doing some homework problems for my test and I come across this one I don't understand.

    1) Is it possible to find set A and B such that both A in B and A subset of B are true? Give an example or prove this is impossible.
    What about:

    A={null_set}, B={null_set, {null_set}}.

    That is A is the set whose only element is the null set,
    and B is the set with two elements, the null set, and the set
    whose only element is the null set (which is A).

    RonL
    Last edited by CaptainBlack; October 10th 2006 at 08:58 PM. Reason: spell check
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  3. #3
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    elmentary constructions on sets

    Thank you Captainblack. I hope you are still there because I just came across one more question.

    1) Let A, B, C be subsets of some fixed set S.
    Prove that (A-B)-C = (A-C) - (B-C)

    Thank you so much.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jenjen View Post
    Thank you Captainblack. I hope you are still there because I just came across one more question.

    1) Let A, B, C be subsets of some fixed set S.
    Prove that (A-B)-C = (A-C) - (B-C)

    Thank you so much.
    [x in (A-B)-C] iff [x in (A-B) and x not in C] iff [x in A and x not in B and x not in C)

    [y in (A-C)-(B-C)] iff [y in (A-C) and y not in (B-C)] iff [(y in A and y not in C) and y not in (B-C)]

    But [(y not in C) and (y not in (B-C))] iff [y not in B], so:

    [y in (A-C)-(B-C)] iff [y in A and y not in C and y not in B]

    The final steps are trivial, what we have shown is that [x in (A-B)-C] iff
    [x in (A-C)-(B-C)]

    Informally the process involved in a proof is just to expand what the two
    expressions mean, and you evantualy find that they mean the same thing.

    RonL
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