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Relations proof

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that if a set has 3 elements, then we can find 8 relations on A that all have the same symmetric closure.


    2. Relevant equations

    Symmetric closure ##R^* = R \cup R^{-1} ##


    3. The attempt at a solution

    If the symmetric closures of n relations are the same then we have,

    ## R_1 \cup R_1^{-1} = R_2 \cup R_2^{-1} = ... = R_n \cup R_n^{-1} ##

    I have to prove n=8 for |A| = 3

    Also, ##R_1##, ##R_2##,...,##R_n## can't be symmetric. A friend told me to use power sets but I don't see how that applies here.

    Do I have to write down all possible relations that can occur from A={a,b,c} or is there a better way to prove this one?

    Any help would be appreciated.
     
    Last edited: Dec 15, 2013
  2. jcsd
  3. Dec 15, 2013 #2

    Dick

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    Homework Helper

    I would imagine a symmetric relation on A and then try to picture how you could delete elements of the relation to get the 8 different ##R_i##. Suppose the relation you start with is the one where all of the elements of A are related to each other? That gives you lots of options. I think you can find more than 8.
     
    Last edited: Dec 15, 2013
  4. Dec 15, 2013 #3
    Sorry I wrote 3 relations instead of 3 elements in the question.

    This problem has appeared twice in previous years exams. My gut tells me to use permutation and combinations but I've forgot most of it.
     
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