Borel fieldsDate: 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? 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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