# Examples for “nice” Boolean algebras that are not complete or not atomic

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).

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 $Set$.

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:

What are nice examples for Boolean Algebras that are not complete or not atomic?

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".

I admit that nice 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).

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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.

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.

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You are (of course) right with your correction, and I have removed the mistake from the question. Thanks for that and thanks for the example you pointed out. –  Niemi Feb 25 '12 at 16:45

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.

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.

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I believe what you mention is just the free Boolean algebras on countably many generators. –  Niemi Feb 25 '12 at 16:18
Yes, that is right, and I see that you mentioned it that way. Meanwhile, forcing is full of atomless Boolean algebras, and if you don't complete them, then generally they won't be complete. Take any separative partial order; it maps densely into its regular open algebra, which is a complete Boolean algebra, but it is very rare for this to be the same as the Boolean algebra generated by the image of the partial order itself. Thus, we get numerous non-complete atomless Boolean algebras, which are as natural and nice as the partial orders are. –  Joel David Hamkins Feb 26 '12 at 5:48
An attractive concrete description of this Boolean algebra is: the set of all periodic subsets of $\mathbb{Z}$, under the ordinary set-theoretic operations. –  Nik Weaver Sep 19 at 2:05

An example of an atomic but not complete boolean algebra is the set of all finite unions of cartesian products of two sets (with the restriction that these sets are subsets of some fixed "universal" set, to eliminate set-theory paradoxes) with join and meet being set-theoretic union and intersection.

Atoms are cartesian products of one-element sets.

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The 3-volume "Handbook of Boolean Algebras" abounds with examples of Boolean algebras, but of course only few of them are "nice".

• You may have a look at Koppelberg's chapter 6 in volume 1, "Special classes of BAs", and also at several chapters in volume 3 describing some of these classes in more detail. This may challenge or sharpen your idea of "nice".
• Generalizing your question one could also ask for nice constructions of Boolean algebras (products, free products, quotients, iterated quotients, interval algebras, completions, etc); some of them will yield nice algebras.
For example, dividing any BA $B$ by the ideal $At(B)$ generated by the atoms will yield a new BA $B':=B/At(B)$; this new algebra $B'$ will often be atomless (e.g. if you start with a power set BA, as mentioned by Andreas Blass). If not, you can continue this process of dividing, even transfinitely often.
• Monk's chapter 12 in volume 2 describes several methods for getting "many" BAs with particular properties. For example, starting with any linear order $L$, the interval algebra $I(L)$ is defined as the subalgebra of $P(L)$ (the power set algebra) generated by the half open intervals $[a,b)$ in $L$. Sufficiently different/nice linear orders will give nonisomorphic/nice BAs.
For $L=\mathbb N$ you will get the (highly atomic) finite-cofinite subalgebra of $P(\mathbb N)$, and for $L=\mathbb Q$ you will get the countable atomless BA, mentioned by Joel Hamkins.
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