Regular Subalgebras of Complete Boolean Algebras

It is proved that the following conditions are equivalent: (a) there exists a complete, atomless, σ–centered Boolean algebra, which does not contain any regular, atomless, countable subalgebra, (b) there exists a nowhere dense ultrafilter on ω. Therefore the existence of such algebras is undecidable in ZFC. In ”forcing language” condition (a) says that there exists a non– trivial σ–centered forcing not adding Cohen reals A subalgebra B of a Boolean algebra A is called regular whenever for every X ⊆ B, supB X = 1 implies supA X = 1; see e.g. Heindorf and Shapiro [6]. Clearly, every dense subalgebra is regular. Although every complete Boolean algebra contains a free Boolean algebra of the same size (see the Balcar-Franek Theorem; [2]), not always such an embedding is regular. For instance, if B is a measure algebra, then it contains a free subalgebra of the same cardinality as B, but B cannot contain any infinite free Boolean algebra as a regular subalgebra. Indeed, measure algebras are weakly σ-distributive but free Boolean algebras are not, and a regular subalgebra of a weakly σ-distributive one is again σ-distributive. Thus B does not contain any free Boolean algebra. On the other hand, measure algebras are not σ-centered. So, a natural question arises whether there exists a σ-centered, complete, atomless Boolean algebra B without regular free subalgebras. Since countable atomless Boolean algebras are free and every free Boolean algebra contains a countable regular free subalgebra, it is enough to ask whether B contains a countable regular subalgebra. In the paper we prove that such an algebra exists iff there exists a nowhere dense ultrafilter. The research of the second author was partially supported by the Basic Research Foundation of the Israel Academy of Sciences and Humanities. This publication has Number 640 in S. Shelah’s list. February 28, 1998 1