Examples for "nice" Boolean algebras that are not complete or not atomic - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:28:45Z http://mathoverflow.net/feeds/question/89487 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89487/examples-for-nice-boolean-algebras-that-are-not-complete-or-not-atomic Examples for "nice" Boolean algebras that are not complete or not atomic Sebastian 2012-02-25T14:39:15Z 2012-02-25T16:25:54Z <p>A Boolean algebra may, or may not, be complete (i.e, any set of elements has a sup and an inf) or atomic (i.e., every element is a sup of some set of atoms). </p> <p>Boolean Algebras that are complete as well as atomic (also called CABAs) are of course precisely those that are isomorphic to some power set (equipped with the obvious choices for the operations), or equivalently stated, those that form a category dually equivalent to \$Set\$. </p> <p>The category of all Boolean algbras, however, is well-known to be equivalent to the category of Stone spaces (compact totally disconnected Hausdorff spaces) with continuous morphisms. Thus, for a Boolean algebra (of infinite cardinality), it is a very special case to be complete and atomic. My question is: </p> <blockquote> <p>What are <strong>nice</strong> examples for Boolean Algebras that are not complete or not atomic?</p> </blockquote> <p>Please understand that I do not look for any kind of example (so the emphasis lies on the word "nice"). I am, for instance, aware that looking at free BAs would lead to such an example, and I also know the classic example of the BA that is formed by all finite and confinite sets of integers. Also, as mentioned above, I know how the Stone Duality transforms Stone spaces into Boolean algebras, so please don't simply say "the clopen subsets of a that-and-that Stone-Space form a Boolean algebra".</p> <p>I admit that <strong>nice</strong> is a somewhat vague notion. What I mean are Boolean Algebras that arise naturally (except those I have already mentioned) and are of special interest for some reason (yes, I know that this formulation is not vague at all).</p> http://mathoverflow.net/questions/89487/examples-for-nice-boolean-algebras-that-are-not-complete-or-not-atomic/89489#89489 Answer by Joel David Hamkins for Examples for "nice" Boolean algebras that are not complete or not atomic Joel David Hamkins 2012-02-25T15:10:48Z 2012-02-25T15:10:48Z <p>There is up to isomorphism a unique countably infinite atomless Boolean algebra (by a back-and-forth argument), making this algebra highly canonical. But it cannot be complete, since every infinite Boolean algebra has an infinite antichain, and so by completeness must have size at least continuum.</p> <p>Meanwhile, there are numerous non-complete atomless Boolean algebras, and one could list dozens of examples. Many interesting examples arise in connection with forcing in set theory.</p> http://mathoverflow.net/questions/89487/examples-for-nice-boolean-algebras-that-are-not-complete-or-not-atomic/89491#89491 Answer by Andreas Blass for Examples for "nice" Boolean algebras that are not complete or not atomic Andreas Blass 2012-02-25T15:36:37Z 2012-02-25T15:36:37Z <p>First of all, let me point out an error in the question. It is not true that "the Boolean algebras that are not atomic or not complete are precisely those that are carried to non-discrete Stone spaces via the Stone Duality." If \$X\$ is an infinite set, then, even though the power set algebra \$P(X)\$ is atomic and complete, its Stone dual is not discrete. Specifically, its Stone dual is not \$X\$ but rather the Stone-Cech compactification of (discrete) \$X\$. The point is that the dual equivalence between the category of CABAs and the category of sets is not (a restriction of) Stone duality.</p> <p>One naturally occurring "not complete or not atomic" Boolean algebra is the quotient of the algebra of Lebesgue measurable subsets of the reals modulo the ideal of measure-zero sets. This is complete but not atomic. If you just take the Lebesgue measurable sets (and don't divide by any ideal), you get an algebra that is atomic but not complete. Finally, for algebras that are neither atomic nor complete, Joel has given one of the most natural examples, but since you already mentioned it (under the guise of "free algebra") in the question, another "nice" example is the quotient of the power set of the natural numbers modulo the ideal of finite sets.</p>