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Question On Partitions

  1. Jun 10, 2014 #1
    1. The problem statement, all variables and given/known data
    "A family of sets is called pairwise disjoint if any two distinct sets in the family are disjoint".
    so if ANY of the two sets are disjoint with each other then the whole family can be called pairwise disjoint..

    "If A is a nonempty set, a family P of subsets of A is called a partition of A (and the sets in P are called the cells of the partition) if
    1) No cells are empty
    2) The cells are pairwise disjoint
    3) Every element of A belongs to some cell.

    "If P is a partition of A, (2) and (3) clearly imply that each element of A lies in exactly one cell of P."

    Say A = {1,2,3,4,5,6} P= {{1,2},{3,4},{5,6,1}}, This partition is pairwise disjoint as {3,4} have no intersection with the {1,2} (as well as {5,6,1} for that matter). And even though there is intersection of the 1 between {1,2} and {5,6,1} it only takes one disjoint subset to be considered pairwise disjoint. I feel like my example did not violate (1) (2) or (3). What am I missing here?





    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jun 10, 2014
  2. jcsd
  3. Jun 10, 2014 #2

    Fredrik

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    In your example, each element of A is an element of exactly one element of P.
    ##1\in\{1,2\},\ 1\notin\{3,4\},\ 1\notin\{5,6\}##
    ##2\in\{1,2\},\ 2\notin\{3,4\},\ 2\notin\{5,6\}##
    ##3\notin\{1,2\},\ 3\in\{3,4\},\ 3\notin\{5,6\}##
    ...
    You clearly need the statements 1-3 to prove that every family of subsets of A that satisfies 1-3 is such that every element of A is in exactly one cell. You don't need them when you're dealing with a specific example of a partition.
     
  4. Jun 10, 2014 #3

    jbunniii

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    I think you are misunderstanding the use of the word "any" in the definition:
    Although it's common to use the word "any" in this way, what is really meant is "every." In other words, all pairs of distinct sets ##A## and ##B## have the property that ##A## and ##B## are disjoint.
     
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