finite fields of sets
Note that, if is a partition of and , we have
so is a field of sets.
Now assume that is a field of subsets of a finite set . Let us define the set of “prime elements” of as follows:
The choice of terminology “prime element” is meant to be a suggestive mnemonic of how the only divisors of a prime number are 1 and the number itself.
We claim that is a partition. To justify this claim, we need to show that elements of are pairwise disjoint and that .
Suppose that and are prime elements. Since, by definition, and and is a field of sets, . Since , we must either have or . In the former case, and are disjoint, whilst in the latter case .
Suppose that is any element of . Then we claim that the set defined as
is a prime element of . To begin, note that, since is finite, a forteriori any subset of is finite and, since fields of sets are assumed to be closed under intersection, it follows that the intersection of a susbet of is an element of , in particular .
Suppose that and . If , then . Since is a field of sets, . Hence, by the construction of , it is the case that , hence . Together with , this implies . If , then, by construction, , which implies .
Thus, we see that is a prime set. Since was arbitrarily chosen, this means that every element of is contained in a prime element of , so the union of all prime elements is itself. Together with the previously shown fact that prime elements are pairwise disjoint, this shows that the prime elements for a partition of .
Let be an arbitrary element of . Since , it is the case that . Since is a partition of ,
so every element of can be expressed as a union of elements of .
|Title||finite fields of sets|
|Date of creation||2013-03-22 15:47:49|
|Last modified on||2013-03-22 15:47:49|
|Last modified by||rspuzio (6075)|