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### Borel fields

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Date: 08/10/97 at 11:01:42
From: Suzanne Mooney
Subject: Borel fields

My textbook defines an ALGEBRA as a subset, say C of some set X such
that

(i) A union B is in C whenever A and B are;
(ii) the complement of A is in C whenever A is in C; and
(iii) A intersection B is in C whenever A and B are.

It then goes on to define a "sigma-algebra," or a Borel field, as an
algebra such that every union of a countable collection of sets in the
algebra is itself again in the algebra.

Now, from the definition given for an algebra, I don't understand why
EVERY algebra would not be a Borel field, since, for every Asub1,
Asub2, Asub3, Asub4...., say, in the algebra, then Asub1 union Asub2
must be in it, and then (Asub1 union Asub2) union Asub3 must be in it,
and so forth, until you get the result that the union of all of them
is in it.

So my question is, what makes the distinction for a Borel field, that
is, how could you have an algebra such that every union of a countable
collection of sets in it would NOT again be in it?
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Date: 08/11/97 at 14:57:26
From: Doctor Ceeks
Subject: Re: Borel fields

Hi,

Here is an example of an algebra which is not a Borel field:

Let C be the collection of subsets of real numbers such that
either the subset has only a finite set of real numbers in it or
its complement has only a finite set of real numbers in it.

I claim this is an algebra. It's probably easier if you check
this yourself, but here's a quick (not really meant to be read)
argument to see why:

(i) Let A and B be in C.
If A and B are finite sets of real numbers, then so is
A union B. If A is the complement of a finite set of real
numbers, then since the complement of (A union B) is smaller
than the complement of A, and so A union B is in C.

(ii) By construction of C, the complement of any set in C is
also in C.

(iii) Let A and B be in C. If A and B are both complements of finite
sets of reals, then A intersect B, being the complement of the
union of the complements of A and B, is also the complement of
a finite set, and so in C. If A is finite, then the intersection
of A and B is contained in A, and so is finite, hence in C.

However, C is not a Borel field.

To see this, let AsubK be the set which contains the single number K.
Then the countable union of the AsubK consists of the positive
integers, which is not finite, and its complement is not finite
either.

-Doctor Ceeks,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/

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Associated Topics:
High School Sets

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