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From: tmoon on 25 Apr 2010 01:47
I am looking at Sterling Berberian's "Measure and Integration" and he
has a definition on rings and sigma rings that I find confusing. In
this book, a ring is defined such that it is "closed under the
formation of set theoretic differences and <i>finite</i> unions" (p.
3). Then a sigma ring is defined to be "closed under the formation of
set theoretic differences and <i>countable</i> unions" (p. 4).
Finally, he says "it is clear that every sigma ring is a ring".

The part I don't get is how a number of operations that are defined to
be "finite" can be greater than operations that are "countable" (and
hence sigma rings are a subset of rings). Isn't it the opposite? Isn't
the number of operations implied by "finite unions" less than that
implied by "countable unions"?
From: Virgil on 25 Apr 2010 02:09
In article
<04b2236e-8c85-497a-b3bf-49effe06d5de(a)b23g2000yqn.googlegroups.com>,
tmoon <tsunmoon(a)gmail.com> wrote:

> I am looking at Sterling Berberian's "Measure and Integration" and he
> has a definition on rings and sigma rings that I find confusing. In
> this book, a ring is defined such that it is "closed under the
> formation of set theoretic differences and <i>finite</i> unions" (p.
> 3). Then a sigma ring is defined to be "closed under the formation of
> set theoretic differences and <i>countable</i> unions" (p. 4).
> Finally, he says "it is clear that every sigma ring is a ring".
>
> The part I don't get is how a number of operations that are defined to
> be "finite" can be greater than operations that are "countable" (and
> hence sigma rings are a subset of rings). Isn't it the opposite? Isn't
> the number of operations implied by "finite unions" less than that
> implied by "countable unions"?

Every finite union is also a countable union, so sets of set that are
closed under countable unions are also closed under finite unions.
From: Ray Vickson on 25 Apr 2010 03:11
On Apr 24, 10:47 pm, tmoon <tsunm...(a)gmail.com> wrote:
> I am looking at Sterling Berberian's "Measure and Integration" and he
> has a definition on rings and sigma rings that I find confusing. In
> this book, a ring is defined such that it is "closed under the
> formation of set theoretic differences and <i>finite</i> unions" (p.
> 3). Then a sigma ring is defined to be "closed under the formation of
> set theoretic differences and <i>countable</i> unions" (p. 4).
> Finally, he says "it is clear that every sigma ring is a ring".
>
> The part I don't get is how a number of operations that are defined to
> be "finite" can be greater than operations that are "countable" (and
> hence sigma rings are a subset of rings). Isn't it the opposite? Isn't
> the number of operations implied by "finite unions" less than that
> implied by "countable unions"?

Being closed under countable unions is more restrictive than just
being closed under finite unions. Therefore, some collections of sets
may have the finite-union closure property but not the countably-
infinite union closure property.

R.G. Vickson
From: William Elliot on 25 Apr 2010 04:16

On Sat, 24 Apr 2010, tmoon wrote:

> I am looking at Sterling Berberian's "Measure and Integration" and he
> has a definition on rings and sigma rings that I find confusing. In
> this book, a ring is defined such that it is "closed under the
> formation of set theoretic differences and i>finite/i> unions" (p.
> 3). Then a sigma ring is defined to be "closed under the formation of
> set theoretic differences and i>countable/i> unions" (p. 4).
> Finally, he says "it is clear that every sigma ring is a ring".
>
> The part I don't get is how a number of operations that are defined to
> be "finite" can be greater than operations that are "countable" (and
> hence sigma rings are a subset of rings). Isn't it the opposite?

> Isn't the number of operations implied by "finite unions" less than that
> implied by "countable unions"?
>
Yes. That's why every sigma ring is a ring.

If A1,.. Aj are j subsets of a sigma ring R, then
A1 \/ A2 \/..\/ Aj = A1 \/ A2 \/..\/ Aj \/ Aj \/ Aj \/ ...
is in R. Thus R is a ring

Exercise. Construct a ring that's not a sigma ring.

As you see above, html or whatever, doesn't appear correctly
at "finite", which can be emphasized as _finite_.
From: David C. Ullrich on 25 Apr 2010 08:36
On Sat, 24 Apr 2010 22:47:43 -0700 (PDT), tmoon <tsunmoon(a)gmail.com>
wrote:

>I am looking at Sterling Berberian's "Measure and Integration" and he
>has a definition on rings and sigma rings that I find confusing. In
>this book, a ring is defined such that it is "closed under the
>formation of set theoretic differences and <i>finite</i> unions" (p.
>3). Then a sigma ring is defined to be "closed under the formation of
>set theoretic differences and <i>countable</i> unions" (p. 4).
>Finally, he says "it is clear that every sigma ring is a ring".
>
>The part I don't get is how a number of operations that are defined to
>be "finite" can be greater than operations that are "countable" (and
>hence sigma rings are a subset of rings). Isn't it the opposite? Isn't
>the number of operations implied by "finite unions" less than that
>implied by "countable unions"?

Yes, "the number of operations implied by finite unions" is
smaller than the "number of operations implied by
countable unions". That's exactly why every sigma-algebra
is an algebra!

You're looking at something backwards.

Say X is strange if X likes green tomatoes.
Say X is curious if X like all tomatoes.

Every curious X is strange, precisely _because_
the number of green tomatoes is less than the
number of tomatoes.



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