From: ara@zurich.ai.mit.edu (Allan Adler)
Subject: Re: THANKS! (for Hints on Countable Sigma Algebras)
Organization: M.I.T. Artificial Intelligence Lab.
	<ARA.92Sep24221451@camelot.ai.mit.edu> <a_rubin.717435306@dn66>
Date: Sat, 26 Sep 1992 07:13:41 GMT


The notion of a Boolean sigma algebra is defined in a number of places,
e.g. Halmos's book Lectures on Boolean Algebras, which is where I learned
what I know about them and which is a wonderful book. Unfortunately, in
analysis courses, one is often only taught about sigma algebras of
subsets of a given set under union, intersection and complement.

One defines a Boolean ring to be a commutative ring with unity in which
every element is idempotent. (Actually, the other hypotheses imply
the ring is commutative). The notion of a Boolean ring and the notion
of a Boolean algebra are equivalent. From a Boolean ring, one defines
the meet operation /\ to be the product in the ring and the join operation
\/ by x \/ y = x+y+xy. /\ and \/ are supposed to be like intersection and
union. One also defines complement ' by x'=1+x. Conversely, from a Boolean
algebra, with operations /\, \/ and ', one defines a Boolean ring in which
the multiplication is /\, the addition is defined by x+y = (x /\ y')\/(y/\x'),
ie. symmetric difference.

One has a natural partial ordering R on a Boolean algebra defined by
x R y if xy=x. (I find it easier to use the multiplication in the ring
instead of writing /\).

If every countable subset S of a Boolean algebra B has a least upper bound
in the sense of the partial ordering R, then we say that B is a sigma algebra.

That is what I meant by an abstract sigma algebra. Similarly, one defines
a complete Boolean algebra to be one in which every subset has a least upper
bound.

Since the Boolean algebra is also a ring, one can speak of ideals. One defines
a sigma ideal to be an ideal which is closed under taking countable least upper
bounds. For example, the Borel subsets  of the unit interval form a Boolean
sigma algebra Bor. The Borel sets of measure zero form a sigma ideal MZ.
The Borel sets of first category form another sigma ideal Ba. When you
mod out an ideal from a Boolean algebra, you get another Boolean algebra.
When you mod out a sigma ideal from a sigma algebra, you get another sigma
algebra.

Miraculously, in the case of Bor and MZ or Ba, you get more than a sigma
algebra for a quotient. Bor/MZ and Bor/Ba are in fact complete Boolean
algebras. Halmos gives a proof that these complete Boolean algebras
are not isomorphic. According to Sierpinski, there is (assuming the continuum
hypothesis) a permutation of the reals (or of the unit interval) mapping
MZ bijectively onto Ba, but this permutation therefore doesn't preserve Bor.

Complete Boolean algebras such as Bor/MZ are actually useful. For example,
such a complete Boolean algebra (I forget whether it is on the unit interval or
on the product space 2^(aleph1) or on 2^(aleph2)) is the Boolean algebra used 
to prove the independence of the continuum hypothesis, by taking that Boolean 
algebra as an algebra of truth values and using it to build up a Boolean valued
model of set theory. In that model, the continuum hypothesis has truth
value less than 1 (in fact, 0, if I remember right) and therefore
is not a theorem. I'm remembering this from the paper of Scott and Solovay,
which I looked at 30 years ago and no longer own. If you have a copy you
wouldn't mind sending me, please let me know. At the end, they even talk
about using the space 2^ON, where ON is the class of all ordinal numbers,
which they remark causes the set of reals to become larger than all bounds.
In view of that, I would guess that the one mentioned above for scuttling the 
continuum hypothesis is 2^(aleph2) and that it causes the continuum to equal 
aleph2. But that is a guess.

Anyway, to get back to the original question, it was pointed out to me
that it is obvious that there is a countable subset S of any infinite
sigma algebra B such that the elements of S are pairwise disjoint. 
In fact, if there are infinitely many atoms, them you can just use
them. If there are finitely many atoms, list them first and let b1 denote
the complement of their union. Since b1 is not an atom, let c1 be a nonzero
element less than b1. Let b2 denote the complement of c1 in b1. Then b2
is also not an atom so find nonzero c2 less than b2 and let b3 denote the
complement of c2 in b2. And so on.. one obtains the infinite disjoint
sequence c1,c2,c3,... Now, for every subset K of the set of positive
integers, one gets an element of B, namely the least upper bound of
{ ci | i in K} and these elements are all distinct. Therefore, an infinite
sigma algebra is always uncountable.

Allan Adler
ara@altdorf.ai.mit.edu