Arturo Magidin wrote:
> In article <ztWdna8Ga43eGvDVnZ2dnUVZ_gGdnZ2d@[EMAIL PROTECTED]
>,
> sto <sto@[EMAIL PROTECTED]
> 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_.
>>
>> 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).
>>
>>
>> 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.
>>
>>
>> How do you prove that the class _C_ is a sigma-ring? (this is supposed
>> to be "easy")
>
> You prove that it is closed under differences and under countable
> unions, of course.
>
>> I managed to prove that the simpler class {diff(E,A):E in
>> _S_(_E_)} is a sigma-ring, but can't find any way to prove that _C_
>> itself is a sigma-ring.
>
> An arbitrary element of _C_ is, as you note, the union of B_1 in
> _S_(int(_E_,A)) and (E_1-A) for some E_1 in _S_(_E_).
>
> So to show _C_ is closed under differences, you consider
>
> (B_1 \/ (E_1-A)) - (B_2 \/ (E_2-A)
>
> for some B_1, B_2 in _S_(int(_E_,A)), and some E_1,E_2 in
> _S_(_E_). Try to express it as the union of something in
> _S_(in(_E_,A)) and some (E'-A) for E' in _S_(_E_). You'll want to use
> the fact that the elements are in specific sigma rings.
>
This is exactly the approach I took originally, but I keep running into
the problem that
(B1 \/ E1 - A) - (B2 \/ E2 - A)
reduces to
(B1 - B2) - (E2 - A) \/ [(E1 - E2) - A] - B2
Of course the B1 - B2, E1 - E2, and even [(E1 - E2) - A] terms belong to
their respective sigma-fields, but in the end I don't see that the
expression reduces to the form B \/ E - A for some B in _S_(int(_E_,A))
and E in _S_(_E_). I've been checking my algebra all day. Maybe I've
just been without sleep too long, but I can't see how to prove it this
way.
I wonder whether there isn't some deeper significance to the fact that
E = E /\ A \/ E - A
and the fact that each element of _C_ is the union of one element from
_S_(int(_E_,A)) and one from _S_(_E_-A)?
> The closure under countable unions is simpler, since if you have a
> family {B_i \/ (E_i-A)} i=1,2,3,... then the union of the family is
> just (\/ B_i) \/ (\/E_i - A), and now you can use the fact that the
> B_i live in a sigma ring and the E_i live in a sigma ring to deduce
> this is of the desired form.
>
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