sigma rings

[sto]

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") 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.

Thanks,
-sto

[Arturo Magidin]

In article <ztWdna8Ga43eGvDV...@earthlink.com>,

sto <s...@address.invalid> 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.

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.

--
======================================================================
"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

[sto]

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)?

[William Elliot]

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.

----

[Arturo Magidin]

In article <gbadnbfpfIh2APDV...@earthlink.com>,

sto <s...@address.invalid> wrote:
>Arturo Magidin wrote:
>> In article <ztWdna8Ga43eGvDV...@earthlink.com>,
>> sto <s...@address.invalid> 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...

[sto]

THanks everybody. This is very helpful. I am working through the
algebra again right now.
-sto