Thomas Streicher recently noted:
omega + 1 is an infinite countable Ha which nevertheless is complete
True. And the dual algebra {0} + omega* with the descending copy integers is also a complete Ha (cHa). But this second example, which is non-atomic, is not fully non-atomic; that is, there are minimal gaps in the ordering. Take Q = rationals in [0,1]. This is a countable, non-complete Ha, where between any two distinct elements there is a third. Note that all these Ha's are subalgebras of the unit interval [0,1] of all reals -- which is uncountable, of course. In linearly ordered Ha's, the implication b --> c == c if b > c, and is 1 otherwise. It follows that the not-not stable elements are just 0 and 1. As I remarked once before, it is an old theorem that all countable, non-atomic Boolean algebras are isomorphic. Call this Boolean algebra F for free, for short. This isomorphism theorem is far from true for Ha's, however. Take the algebra Q x Q. This is a fully non-atomic, but it has FOUR not-not stable elements = {0,1} x {0,1}. So this is not the algebra Q. Take any other finite power, Q x Q x ... x Q. They are all different by reference to the not-not stable elements. The Ba I called F has a recursive structure (as one sees from what we know about propositional calculus) in that all the operations are recursive in terms of a simple enumeration of the elements of F. The question I want to consider is what countable Ha extend F in such a way F equals the not-not stable elements. 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.) Here is one construction. Let A be the arithmetically definable elements of C. That is, take the elements of x of C where the set {a in F | a =< x } is (first-order) arithmetically definable (in terms of the enumeration of F). I claim that A is a Ha. Of course it contains F; and, if it is a subalgebra of C, then the not-not stable elements give us an extension of F. But it is countable and the extension of F, call it G, is non-atomic; hence, the stable elements of A form a Ba G isomorphic to F. To check the subalgebra part we note for x,y in C and a in F: a =< x /\ y <==> a =< x & a =< y; a =< x \/ y <==> (for some b,c in F)[ a = b \/ c & b =< x & c =< y ]. Hence, if x,y are in A, then so are x /\ y and x \/ y. At first I had thought the recursively enumerable opens would be enough, but there is trouble with negation. We have for x in C and a in F, and with -x notating negation in C: a =< -x <==> (for all b in F)[ b =< x ==> a /\ b = 0 ]. The universal quantifier seems needed, and so {a in F | a =< -x } does not seem to be r.e. in general. But, if x is in A, then so is -x. We need to generalize this remark to implication in C, so for a in F and x,y in C: a =< x --> y <==> (for all b in F)[ b =< x ==> a /\ b =< y ]. Hence, if x,y are in A, then so is x --> y. And A is a sub-Ha of C. Note that in A, if an element x is not 0, than the interval [0,x] in A is in itself a non-Boolean cHa (a homomorphic image of A). Consider next the algebra A x F. This algebra is a countable, fully non-atomic Ha. The not-not stable elements of A come out to be G x F, which is isomorphic to F (and G). But, the element x = (0,1) has the interval [0,x] in A x F isomorphic to F. And that is Boolean. So, A and A x F are NOT isomorphic. Note, however, that A x F and (A x F) x F ARE isomorphic. Also, as I forgot to mention, A and A x A are isomorphic. Anyway, we have at least two non-isomorphic, non-Boolean Ha's with an isomorph of F as the not-not stables. And they are fully non-atomic. Are there other, non-isomorphic extensions of F? At this moment I do not see a knock-down argument. One idea, is to call the above algebra A_1 (for first-order definable opens) and consider the algebras A_n of n-th order definable opens. I think we get the same conclusions about A_n and A_n x F not being isomorphic, BUT is every A_n somehow isomorphic to A_1 (with a highly non-recursive automorphism of F)? Is that possible, or have I missed an obvious point? If the A_n are all not isomorphic, then we can probably get an uncountable collection of mutually non-isomorphic extensions of F. And, how would these relate to Sub(1) Ha's of countable topoi? Or are these considerations a red herring?