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作者  fongya (......) 站內  97RA_I
標題  CH.2 Exercise 1
時間  2008/11/15 Sat 19:00:03

972201011

Given a collection of sets F_1 , F_2,...,F_n,

                                *     *         *
construct another collection F_1 , F_2 ,..., F_N ,

                            n         N     *
with N = 2^n -1 , so that  ∪  F_k = ∪  F_j   ;
                           k=1       j=1

                   *
the collection {F_j }  is disjoint ;

                           *
also  F_k =        ∪   F_j                 , for every k.
               *
            F_j  is contained in F_k


p.f.
                                                                     c
    Consider Γ= {F_1'∩ F_2'∩ ...∩ F_n'| F_k' is either F_k or F_k  }

                      n     c           n      c     n      c
                 - { ∩  F_k  }  (  ∵ ∩   F_k  = (∪  F_k)   )
                     k=1               k=1          k=1

                  n
    =>  # (Γ) = 2 -1

                 *         *
    Let Γ = {F_1 , ... , F        }
                           2^n -1


   (1) Γ is disjoint :

              *      *
       Let F_i  , F_j  be belonged to Γ with i≠j


       Then there exists 1 ≦ k ≦ n

                     *                           *                  c
            s.t.  F_i  contained in F_k  and  F_j  contained in  F_k

             *      *
       => F_i ∩ F_j  =  empty set

                      n
         n           2 -1      *
   (2)  ∪  F_k  =    ∪    F_j  :
        k=1           j=1


       " > "           其中 > 表示"包含"
                            n
             For all 1≦j≦2 -1 ,

                     *
                  F_j  which is contained in  F    , for some 1≦ k_j ≦ n
                                               k_j

                    *                   n
             =>  F_j  is contained in  ∪  F_k
                                       k=1

       " < "           其中 < 表示"包含於"


             For all 1 ≦ k ≦ n ,

                                                                            c
let S= {F_1'∩...∩ F_(k-1)'∩ F_(k+1)'∩...∩F_n'| F_i'is either F_i or F_i }


         Then

              F_k =      ∪         (F_k ∩ F)
                    F belongs to S
                                                n
                                               2 -1    *
              and  F_k ∩ F is contained in    ∪   F_j
                                               j=1
                                    n
                                   2 -1    *
         =>  F_k is contained in   ∪   F_j
                                   j=1


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□ Modify: 2008/11/15 Sat 19:04:04  59-115-78-118.dynamic.hinet.net 修改

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