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From: Arturo Magidin on 25 Apr 2010 15:08
On Apr 25, 12:47 am, 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"?

"Countable" means "either finite, or denumerably infinite".

--
Arturo Magidin
From: tmoon on 25 Apr 2010 17:02
On Apr 25, 2:11 am, Ray Vickson <RGVick...(a)shaw.ca> wrote:

> 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.

Thanks! It took me a while to chew on this, but now \[ sigma ring
\subset ring \] makes sense.
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