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Math Help - Discrete - closures

  1. #1
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    Discrete - closures

    Find the symmetric closure of the relation R = {(a,b)|a>b} on the set of real numbers.
    I dont know how to slow this closure..please help
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  2. #2
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    Hi

    Condition a>b So
    Assume a={1,2,3,4,5}

    Then Relation will be
    R = {(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4)}

    It is not Symmetric.
    For Symmetric (2,1) is in R but (1,2) is not. Simmilar (3,1) is in R, but (1,3) not.
    Formula: If (x,y) in R tthen (y,x) must be in R.
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  3. #3
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    yes, it is not symmetric, but what is the symmetric closure?
    Thanks.
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  4. #4
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    The symmetric closure of a binary relation R on X is the smallest relation S on X such that S is symmetric and contains R as a subrelation. By 'smallest' we mean with respect to inclusion. Put another way, S is defined to have the following properties:

    1) it is a symmetric relation
    2) if xRy (read "x is R-related to y"), then also xSy. This is the same as saying that R, thought of as a collection of ordered pairs from X, is a subset of S
    3) if T is any other symmetric relation on X containing R, then T also contains S (this is what we mean by 'smallest').

    One can show that the intersection of any collection of symmetric relations is again symmetric. This is the essential fact that proves that symmetric closures actually and always exist. With this in mind, one can show:

    The symmetric closure of R is precisely the intersection of all symmetric relations containing R.

    But this so far has been non-constructive. Given R, how do we actually construct its symmetric closure? Easy. One can finally show that the symmetric closure of R is exactly the following set:

    R union {(x,y) such that (y,x) is a member of R}

    Put more simply, if R contains (x,y), and doesn't contain (y,x), then just throw it in. Once you do this for every pair of elements in R, you have the symmetric closure.

    I think this should be enough to do the actual exercise.
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