New Home Forums Math Olympiad - IOQM Combinatorics Equivalence Relation to Partition

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  • #24686

    Does every equivalence relation defined on a set, automatically partitions the set?

    If yes, give a rigorous proof.

    If no, give an example.

    #24687
    Pinaki Biswas
    Participant

    Proof:1. If there is x in a set, then it will have to be in a partition.(symmetric)

    2.If x ~y, they are in the same partition.

    =>y~x(reflexive)

    3.If x~y,y~z,

    Then x~z as all are in the same partition.(transitive)

     

    #24689

    We can partition the points/elements according to their equivalence like - suppose we have a set S which has 6 points/elements- {A1,A2,A3,A4,A5,A6}

    • SYMMETRIC-A1~A3 then we can arrange them like {A1,A3}
    • REFLEXIVE - A2~A2 then we can arrange it like {A2}
    • TRANSITIVE- A4~A5 , A5~A6 and A4 ~A6 then we can arrange them like {A4,A5,A6}

    So,  we can get partitions from equivalence relations

    PROVED

    #24690
    Harshit Shah
    Participant

     

    Equivalence relation -> Partition

    Constrtuct partitions from the relation as follows,

    1) pick an element from set S, take all y in S such that x~y and put them all in a new partition

    2) repeat step 1 until all elements of S are exhausted

    step 1 guarantees disjointedness and step 2 guarantees exhaustiveness

     

    Partition -> Equivalence relation

    Declare all elements in a single partition to be equivalent,

    1) reflexivity: trivial (x is in the same partition as x, so x~x)

    2) symmetry: if x is in the same partition as y, so y is also in the same partition. Therefore, x~y=>y~x

    3) transitivity: if x is in the same partition as y, y is in the same partition as z, then x is in the same partition as z. Therefore, x~y,y~z => x~z

     

     

     

     

    #24691
    Writaban Sarkar
    Participant

    Yes, since making equivalence relations means that the elements are in the same set, we can say that stating elements equivalent means to break the set and form some new set (partition).

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