We have known that in general the smallest field containing k subsets has, in the general case, size 2 2 k .However, in this question, follows the above result, we will have 2 2 3 = 256 elements in the smallest field F . How can we represent it?

Question

We have known that in general the smallest field containing   k subsets has, in the general case, size       2                  2        k            .However, in this question, follows the above result, we will have       2                  2        3              =  256 elements in the smallest field       F  . How can we represent it?

Kaylie Mcdonald · Accepted Answer

Hint: Given   A  ,  B  ,  C consider   A  ∩  B  ∩  C  ,      A          c        ∩  B  ∩  C  ,  A  ∩      B          c        ∩  C  ,  A  ∩  B  ∩      C          c        ,  A  ∩      B          c        ∩      C          c        ,      A          c        ∩  B  ∩      C          c        ,      A          c        ∩      B          c        ∩  C and       A          c        ∩      B          c        ∩      C          c      . (Some of these may be empty). The field generated by   A  ,  B  ,  C is the collection of all possible unions of these sets. In our case you get   (  0  ,      1    4    ]  ,  (      1    4    ,      1    2    ]  ,  (      1    2    ,      3    5    ]  (      3    5    ,      2    3    ] and   (      2    3    ,  1  ]. The field generated consists of all possible unions of these (including the empty set). There are       2          5       sets in it.

We have known that in general the smallest field containing k subsets has, in the general case

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