In article <gbadnbfpfIh2APDVnZ2dnUVZ_rrinZ2d@[EMAIL PROTECTED]
>,
sto <sto@[EMAIL PROTECTED]
> wrote:
>Arturo Magidin wrote:
>> In article <ztWdna8Ga43eGvDVnZ2dnUVZ_gGdnZ2d@[EMAIL PROTECTED]
>,
>> sto <sto@[EMAIL PROTECTED]
> wrote:
[...]
>>> 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
You'll obviously want to rewrite it in some way. Just doing the
opeartions will not be good enough. You can try decomposing some of
these terms further, naturally.
>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.
There seems to be another reply that shows how to decompose this.
>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)?
Quite possibly...
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
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