Woke up in the middle of the night and realized I said something obviously wrong. (Not the first time, and won't be the last, I'm afraid.) On Nov 25, 2007, at 10:03 PM, Dana Scott wrote:
Now, F can also be considered as the clopens of the Cantor space 2^N. The cHa of all opens is fully non-atomic, but it is uncountable. Call it C for Cantor. The not-not stable elements of C are the so- called regular open sets. They form an uncountable cBa which is the completion of F. But we want to ask if there are there interesting countable subalgebras of C? We note that elements of C are determined by the elements of F they contain. (In fact, the cHa C is isomorphic to the lattice of ideals of F.)
Well, it is true that the stable elements of C form the completion of F. And, every element of C is a sup of elements of F, so C is atomless in the sense of having no minimal non-zero elements. And, the stables of C form an atomless cBa. BUT -- and here is my oversight -- C does have gaps, and so the cHa is NOT fully non-atomic. Think of the Cantor set as a subspace T of the unit interval. There is a blank from 1/3 to 2/3, if we make the construction via the middle-third process. This means that [0,1/3) meet T is open in T, but [0,1/3] meet T = [0,2/3) meet T is both open and closed. This gives a gap between two opens in the cHa C. So C is not fully gapless. This gap also exists in the subalgebra A of arithmetically definable opens. Aarrgghh. Is there a fix? Can we take a quotient in the category of Ha's that closes the gaps? Maybe, and maybe not. In countable Ba's, dividing by the ideal generated by the atoms can result in a quotient algebra that still has atoms. Aarrgghh! I will have to think further. Rats!