Refinement properties and extensions of filters in Boolean algebras

Abstract

We consider the question, under what conditions a given family A A in a Boolean algebra B \mathcal {B} has a disjoint refinement. Of course, A A cannot have a disjoint refinement if A A is a dense subset of an atomless B \mathcal {B} , or if B \mathcal {B} is complete and A A generates an ultrafilter on B \mathcal {B} . We show in the first two sections that these two counterexamples can be the only possible ones. The third section is concerned with the question, how many sets must necessarily be added to a given filter in order to obtain an ultrafilter base.