Date: Mon, 10 Nov 2003 12:41:41 -0600 (CST)
   From: Mark Jacobson <jacobson@math-cs.cns.uni.edu>
     To: 810-080-01@uni.edu, 810-080-03@uni.edu
Subject: Partitions of any set with cardinality 4

Suppose the set is {a, b, c, d}

How many ways can it be partitioned up into 4 sets?  4 blocks?  Only one way.

                 {{a}, {b), {c}, {d}}

How many ways can it be partitioned up into 3 disjoint subsets?  3 blocks?

                 {{a, b}, {c}, {d}} is one of the ways.

    A related question is what is the value of C(4, 2)?  The answer
    to that is the answer to how many ways any 4 element set can
    be partitioned into 3 disjoint subsets!

How many ways can it be partioned into 2 disjoint subsets?  2 blocks?

    {{a, b}, {c, d}} is one of the ways.

    A related question is what is the value of C(4, 2)?  The answer,
    however, is NOT the same as the value of C(4, 2)?  Why is that?

How many ways can it be partioned into 1 disjoint subsets, or just
    one block.

        {{a, b, c, d}} is the only partition possible, where
                       every element is in the same subset.

This should help you to understand partitioning of a set into
a set of disjoint subsets.

If the cardinality of the original set is n, then
   there is one partition into n different subsets, and

   there is one partition into 1 single subset, and

   there are partitions into 2 disjoint subsets, and
                        into 3 disjoint subsets, and
                        into 4 disjoint subsets, and
                        ...
                        into n-1 disjoint subjsets

  From today's class.  Injective = one to one = 1-1  I. = one = 1
                                                     V. = 5
                                                     II. = 2 in Roman
                                                                Numerals
                       injective = one to one    i. = 1 = one
                       -

                       Surjective = onto

    If A is the domain and B is the codomain,

      when cardinality(A) < cardinality(B) there are no functions
                                           from A to B which are
                                           surjective (onto).

     when cardinality(A) > cardinality(B)  there are no functions
                                           from A to B which are
                                           injective (one-to-one).

     when cardinality(A) = cardinality(B)  it is possible for there
                                           to be a bijective function
                                           from A to B.

For Wednesday's class:  Think about this question:
           How many bijective functions are there from A to B
           when we assume the cardinality(A) = cardinality(B) = n?

           Give an exact answer in terms of n.

     Mark