William Elliot wrote:
> On Thu, 3 Jul 2008, sto wrote:
>
>> Let X be a set, and define a (Boolean) sigma-ring _S_ as a non-empty
>> class of subsets of X that is closed under the formation of differences
>> and countable unions.
>>
>> Let _E_ be any class of subsets of X, and denote by _S_(_E_) the
>> smallest sigma-ring containing _E_.
>>
> Pardon me while I make your notation manageable.
>
> S_E = sigma ring generated by E
>
>> Let A be a subset of X, and denote by E the generic element of the
class
>> _E_. Denote by intersection(_E_,A) the class of sets
>> {intersection(E,A): E in _E_}. Denote by _S_(intersection(_E_,A)) the
>> smallest sigma-ring containing the class of sets intersection(_E_,A).
>>
> E*A = { U /\ A | U in E }
>
>> Based on these definitions, create a class of sets _C_ to be {union(B,
>> diff(E,A)): B in _S_(intersection(_E_,A)), E in _S_(_E_)}. In other
>> words, each element of the class _C_ is the union of an element B of
>> _S_(intersection(_E_,A)) with the difference of an element E of
_S_(_E_)
>> and the set A.
>>
> C = { B \/ U\A | B in S_E*A, U in S_E }
>
> Let B \/ U\A and D \/ V\A be two elements of C
> (B \/ U\A) - (D \/ V\A) = (B \/ U\A) /\ (X\D /\ (X - V\A))
> . . = (B \/ U\A) /\ X\D /\ (X\V \/ A)
> . . = (B \/ U\A) /\ ((X - D\/V) \/ A\D)
>
> . . = X\D /\ ((B\V \/ U\(A \/ V) \/ (B /\ A) \/ (U /\ A))
> . . = B\(V \/ D) \/ U\(A \/ D \/ V) \/ ((B /\ A)\D) \/ ((U /\ A)\D)
> . . = B\(V \/ D) \/ U\(A \/ V) \/ (B\D \/ ((U /\ A)\D)
> . . = (B\D - V) \/ (U\V - A) \/ K
>
> Where K = (B\D \/ ((U /\ A)\D) in S_E*A
> . . B\D - V in S_E*A
>
> which upon putting it together, shows C closed under set difference.
>
> Since S_E*A and S_E are sigma rings, it's easy to see
> that C includes countable unions of elements of C.
>
>> How do you prove that the class _C_ is a sigma-ring? (this is supposed
>> to be "easy")
>
> It isn't when coming to showing closure under set difference.
> That's a set algebra grind.
>
> ----
THanks everybody. This is very helpful. I am working through the
algebra again right now.
-sto
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