# Fields of sets in which, if the l.u.b. of a subset exists at all, it is the union of the subset

I am learning about boolean algebras and how they can be represented as fields of sets. Stone's representation theorem tells us that every boolean algebra is isomorphic to a field of sets. Consider an isomorphism from an arbitrary boolean algebra to a field of sets. A subset of the original boolean algebra will have a least upper bound iff its image does, but the image of the least upper bound of the subset may be strictly bigger than the union of the images of the elements of the subset. (Indeed, this always happens with Stone's representation unless the least upper bound of the subset is actually the least upper bound of a finite subset of the subset.) It would seem natural to consider the following property of boolean algebras: being isomorphic to a field of sets in such a way that this phenomenon does not occur. However, I have not seen the matter discussed this way in any books. (Instead, the focus is on $\kappa$-completeness and related notions.) Is there a reason for this neglect? Or am I mistaken that the notion is neglected?

Note that, if we call the property I am asking about P, having P is equivalent to being isomorphic to a field of sets that has the property that all of its least upper bounds are unions. So we might (as in the title of this question) just consider that property of fields of sets instead of P.

• One issue with property P is that every complete boolean algebras with this property is isomorphic to power set algebra which aren't very interesting. – hot_queen Nov 13 '13 at 19:18
• Granted, for forcing applications it is natural to complete boolean algebras, but I am interested in the general case. – user39080 Nov 14 '13 at 0:44