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Partition sets

  1. Feb 5, 2010 #1

    Pengwuino

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    I was given a problem where I was to find two disjoint partitions, [tex]S_1[/tex] and [tex]S_2[/tex] and a set A such that |A| = 4 and [tex]|S_1| = 3[/tex] and [tex]|S_2| = 3[/tex].

    Now the set I was using and the book eventually used was A = {1,2,3,4} and [tex]S_1 = [/tex]{{1},{2},{3,4}} and [tex]S_2 = [/tex]{{1,2},{3},{4}}.

    The question I have is probably a few definition questions that the book just doesn't seem to be clear about. Do the S's have to be a collection of sets and not simply a set of numbers? For example, is [tex]S_1 =[/tex] {1,2,3} not a correct partition?

    Also, the text asks for "disjoint" partitions, which I assume means [tex]S_1[/tex] and [tex]S_2[/tex] don't share any elements. However, isn't this part of the definition of a partition? That is, any two sets don't share any elements?
     
  2. jcsd
  3. Feb 5, 2010 #2
    A partition of a set A is a collection that simply separates every element of the set A into disjoint non-empty subsets. Every element of the set A must appear once and only once in one of the subsets of the partition. [tex]S_1[/tex] = {{1,2,3}} is not a partition of the set A since 4 does not appear is any of the subsets. However, [tex]S_1[/tex] = {{1,2,3}, {4}} would be a partition, however its cardinality would be 2.

    It is true that a partition is always a collection of disjoint subsets, however it is possible to have two partitions that are not disjoint, but distinct. For example [tex]S_1[/tex]={{1,2}, {3}, {4}} and [tex]S_2[/tex]={{1,3}, {2}, {4}} would be two distinct partitions of A, both with cardinality 3, however they are not disjoint because they share the element (of a partition which is a subset) {4}.
     
    Last edited: Feb 5, 2010
  4. Feb 5, 2010 #3

    Pengwuino

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    Ok, I think my text is a bit confusing as it made it seem like S1 and S2 were required to make the set A.

    For example, would S = {{1,2,3,4}} be a partition of A? And would S={1,2,3,4} be a partition of A?
     
  5. Feb 5, 2010 #4
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