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Math Help - Equivalence relation over set A={1,2,3,4}

  1. #1
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    Equivalence relation over set A={1,2,3,4}

    Hello all, first post here :-)

    I'm not sure what all the proper English names are for this subject, so bear with me, please.

    Given a set A = {1,2,3,4}
    And set K = set of all equivalence relations of A.

    Is {(1,1)(2,2)(3,3)(4,4)(1,2)(2,1)} an equivalence relation over A, and as such a member of K?

    The question I need help with is : given these sets, if Ia is removed - what is the minimal member of K? Give two examples and prove that they are minimal.

    I'm just starting this University math stuff and feeling very lost.

    Thanks,
    Emil.
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  2. #2
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    Re: Equivalence relation over set A={1,2,3,4}

    Quote Originally Posted by emil1369 View Post
    Given a set A = {1,2,3,4}
    And set K = set of all equivalence relations of A.

    Is {(1,1)(2,2)(3,3)(4,4)(1,2)(2,1)} an equivalence relation over A, and as such a member of K?
    The question I need help with is : given these sets, if Ia is removed - what is the minimal member of K? Give two examples and prove that they are minimal.
    I'm just starting this University math stuff and feeling very lost.
    I can tell you that \{(1,1)(2,2)(3,3)(4,4)(1,2)(2,1)\} is an equivalence relation.

    Also I think that by Ia you mean the set \{(1,1)(2,2)(3,3)(4,4)\}.
    We call that the diagonal and name it \Delta_A.

    Beyond that, I don't really understand what you have written.
    Maybe you can explain more?
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  3. #3
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    Re: Equivalence relation over set A={1,2,3,4}

    Thanks alot, that helps already!

    The question has two parts.
    In the first part, A is just an abstract set and K is the set of all its equivalence relations (of A);
    The question is to prove that there is in fact a "smallest" member and a "largest" member in K. I gave the answer that the diagonal is the smallest because it is the smallest equivalence relation, and that AxA is the largest.
    Hope that's true.
    In the second part they say that A equals {1,2,3,4} and that K is again the set of equivalence relations, only i'm asked to remove the smallest and largest members that were found in part 1.

    So the diagonal isn't a member of K now.

    The question is to give an example of a "minimal" member and prove that it is in fact minimal.

    Again, maybe you guys don't call it minimal and smallest, but I hope you can understand what I'm asking.

    Thanks!
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  4. #4
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    Re: Equivalence relation over set A={1,2,3,4}

    Before you can talk about "minimum" or "smallest" in a set, you have to have an order relation. I don't believe there is a "natural" ordering of equivalence relations.
    Thanks from topsquark
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    Re: Equivalence relation over set A={1,2,3,4}

    Quote Originally Posted by emil1369 View Post
    In the first part, A is just an abstract set and K is the set of all its equivalence relations (of A);
    The question is to prove that there is in fact a "smallest" member and a "largest" member in K. I gave the answer that the diagonal is the smallest because it is the smallest equivalence relation, and that AxA is the largest.
    Hope that's true.
    In the second part they say that A equals {1,2,3,4} and that K is again the set of equivalence relations, only i'm asked to remove the smallest and largest members that were found in part 1.

    I will guess that your set is partially ordered by subset inclusion.

    Let \math{K'} be the set \math{K} with both \Delta_A~\&~A\times A removed.

    Now \{(1,1),(2,2).(3,3),(4,4),(1,2),(2,1)\}\in \math{K'} and no proper subset of that set is in \math{K'}.
    That means it is minimal.

    BUT \{(1,1),(2,2),(3,3),(4,4),(1,3),(3,1)\}\in \math{K'} and is also minimal.
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  6. #6
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    Re: Equivalence relation over set A={1,2,3,4}

    Y, I forgot to mention the order...because I don't really understand this yet.
    Thanks so much for your comprehensive answers, this is great!
    I can now finish this exercise, finally.

    Thanks again.
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