On Nov 26, 2007, at 10:09 AM, Dana Scott wrote, recall in an earlier posting:
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.)
The phrase "fully non-atomic" was not well chosen. Let us use "atomless" to mean "no minimal non-zero elements", and let us say a Ha is "gapless" if we cannot have elements a < b with [a, b] = {a, b} (i.e. nothing strictly between). An atomless Ba is always gapless. Now the cHa C above, as I pointed out in the previous message, DOES have gaps, even though it is atomless. (The Ba F is atomless, and it generates C by taking unions of clopen sets to make opens. Thus, no open set could be an atom in C.) Take any point t in the Cantor set 2^N. Let b = 2^N and let a = 2^N \ {t}. Clearly, a is a dense open set and [a, b] is a gap. By removing one pont at a time, we can have a whole sequence of dense open sets a_0 < a_1 < ... < a_n with each [a_i, a_(i+1)] being a gap. Note that negation in C gives --a_i = b, since in topological lattices double negation is interior-of-closure. In general, in any Ha which F generates, if a < b and [a, b] is a gap, then b =< --a. Because if not, then b /\ -a is non-zero. By the generation, there must be a non-zero e in F with e =< b /\ -a. Thus, e /\ a = 0. Because F is atomless, we can write e = f \/ g, with two disjoint, non-zero elements of F. But then c = a \/ f is an element strictly between a and b. This comment shows that gaps, if they exist are somewhat limited. But, C has many gaps, and in general an interval [a, --a] might be quite large. Remember, assuming a = --a for all a makes the Ha Boolean. So, I have not really made much progress in answering how F might generate a countable, non-boolean Ha. I am guessing there are many non-isomorphic ways this can happen.