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Math Help - Covering Relation

  1. #1
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    Covering Relation

    Let (S,  \preceq ) be a poset. We say that an element y  \in S covers an element x  \in S if x  \prec y and there is no element z  \in S such that  x \prec z \prec y . The set of pairs (x,y) such that y covers x is called the covering relation of (S,  \preceq )

    What is the covering relation of the partial ordering {(a,b)  \vert a divides b} on {1,2,3,4,6,12}?

    My answer so far is {(1,2), (1,3), (2,4), (2,6), (3,6), (4,12), (6,12)}

    is this right?
    Last edited by Discrete; July 10th 2007 at 02:21 PM.
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  2. #2
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    Quote Originally Posted by Discrete View Post
    Let (S,  \preceq ) be a poset. We say that an element y  \in S covers an element x  \in if x  \prec y and there is no element z  \in S such that  x \prec y \prec z . The set of pairs (x,y) such that y covers x is called the covering relation of (S,  \preceq ) What is the covering relation of the partial ordering {(a,b)  \vert a divides b} on {1,2,3,4,6,12}?
    My answer so far is {(1,2), (1,3), (2,4), (2,6), (3,6), (4,12), (6,12)}is this right?
    NO indeed!
    Look very carefully at the definition of covering elements.
    Is this true,  2 \prec 4 \prec 12 ?
    If it is true, then can 4 cover 2?
    Try again!
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  3. #3
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    Sorry I revised the question! I misstyped it. Is it correct now?
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  4. #4
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    Yes now it is correct using the corrected definition.
    The other definition give maximal elements in chains.
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