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These are really two questions, but the second presupposes the first.

First, let $( B_i )_{i\in I}$ be an arbitrary family of Boolean algebras. I want to directly form a product of them that is like the product topology on the product of their Stone spaces, ideally without using AC (or BPI, pun not intended). I haven't worked with Boolean algebras as such before--no doubt this will show in what I say--and my little literature search finds at least one author doing products of Boolean algebras by passing to Stone spaces, which I don't want to do.

I am thinking something along the lines of the space of finite formal joins of finite formal meets. If I can't find anything, I guess what I'll do will be something like define the "cylinder elements" as finite sets $\{(i_1,a_1),...,(i_n,a_n)\}$ of index-element pairs, with $a_j \in B_{i_j}$ and the $i_j$ all distinct, and then take the space of finite sets of cylinder elements, modulo an appropriate equivalence relation. But I don't want to reinvent the wheel or multiply nonstandard notation. I'd much rather be able to cite something and use whatever notation is standard. A good feature from my point of view of the above partially-described construction is that if $I\subset J$, then the product of $(B_i)_{i\in I}$ is a subalgebra of the product of $(B_i)_{i\in J}$--I'd ideally like a construction that does that.

The second thing is that I need a result showing that if I have a (finitely-additive) probability measure $P_i$ on each $B_i$ (satisfying the obvious Boolean algebra analogues of the finitely-additive probability axioms), then I can get the natural measure on the product measure. (In the above notation, we will want $P(\{\{(i_1,a_1),...,(i_n,a_n)\}\}) = \Pi_{j=1}^n P_{i_j}(a_{i_j})$.) Again, this can't be hard to prove, but I don't want to spend space writing up a proof of this if I can just find a citation.

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3 Answers

up vote 2 down vote accepted

This is called the "direct sum" or "internal sum" of Boolean algebras. See Introduction to Boolean Algebras by Givant and Halmos, p. 427 for the abstract definition and p. 432 for the concrete description you want (the elements of the internal sum are finite joins of finite meets of elements and complements of elements from the disjoint union of the summands).

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Since the opposite of the category of Stone spaces is equivalent to the category of Boolean algebras, the product of Stone spaces would have to correspond to the coproduct of Boolean algebras. Thus what you want is to construct the coproduct of a family of Boolean algebras directly, but without invoking Stone duality.

This is just general (categorical) universal algebra. The theory of Boolean algebras is a finitary algebraic theory, so on general grounds you can construct the coproduct of a family $\{B_i\}_{i \in I}$ by taking a filtered colimit over the system of finite coproducts. Usually (and more generally for commutative rings) this is called an (infinite) tensor product:

$$\bigotimes_{i \in I} B_i = \mathrm{colim}_{\mathrm{finite}\; F \subset I} \otimes_{i \in F} B_i$$

More concretely, this colimit is simply a set-theoretic union of finite tensor products of Boolean algebras, construed as commutative algebras over the ring $\mathbb{Z}/(2)$. This description matches that in Nik's answer.

For the moment I'll stop here, although I might come back to address the second query on probability measures.

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Hm, for some reason someone downvoted this. If I said something wrong, I would like to have it explained to me. –  Todd Trimble Feb 25 '13 at 1:55
I gave the downvotes that everyone is crying about. You did not explain how to extend measures on Boolean algebras to measures on the free products. Also, even though the category of Boolean algebras is isomorphic to the category of Boolean rings, it makes little sense to consider Boolean algebras in terms of rings. The Handbook of Boolean Algebras does not mention Boolean rings after the first section of the first chapter. –  Joseph Van Name Feb 27 '13 at 2:07
Crying about, Joseph? Anyway, I like to consider Boolean algebras in terms of Boolean rings. Notice for example that the Stone space can be understood in terms of the spectrum (in the algebraic geometry sense) of the Boolean ring; it is pertinent to remark here that Stone began his work by establishing the equivalence between Boolean algebras and Boolean rings and working with the space of maximal ideals thereof. It makes little sense to me that you think this POV makes little sense! –  Todd Trimble Feb 27 '13 at 3:12
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First of all, I would like to say that I do not see any good reason not to use Stone spaces in order to describe the coproduct of Boolean algebras. The product of Stone spaces is the most natural way to represent free products of Boolean algebras. However, besides Stone spaces, there are a couple other ways to represent the free product of Boolean algebras.

First of all, if you represent Boolean algebras in terms of algebras of sets, then the free product is easy to describe. Assume that $(X_{i},\mathcal{A}_{i})$ is an algebra of sets for $i\in I$. Let $X=\prod_{i\in I}X_{i}$ and let $\pi_{i}:X\rightarrow X_{i}$ be the projection mappings. Let $\mathcal{A}$ be the algebra of sets over $X$ generated by sets of the form $\pi_{i}^{-1}[R]$ for $i\in I$. Then $\mathcal{A}$ is the free product of the Boolean algebras $\mathcal{A}_{i}$. Furthermore, every element in $\mathcal{A}$ can be written as a finite disjoint union of sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ for $R_{1}\in\mathcal{A}_{1},...,R_{n}\in\mathcal{A}_{n}$ (This is because the sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ form a semialgebra of sets).

Also, the free product of two Boolean algebras is the Boolean power of those two Boolean algebras (i.e. $A\ast B\simeq A^{B}$ where $A^{B}$ denotes the Boolean power). The Boolean power $A^{B}$ is the set of all continuous functions $f:S(B)\rightarrow A$ where $A$ is given the discrete topology and $S(B)$ is the Stone space of $B$. The Boolean power $A^{B}$ can also be described as the direct limit $^{\lim}_{\longrightarrow}A^{|p|}$ where $p$ ranges over the set $\mathbb{P}_{\omega}(B)$ of finite partitions of $B$ ordered by reverse refinement. In other words, $p\leq q$ if for each $a\in q$ there is a $b\in p$ with $a\leq b$. The transitional mappings $A^{|p|}\rightarrow A^{|q|}$ are the natural surjections whenever $p\leq q$. See the book A Course in Universal Algebra by Stanley Burris and H.P. Sankappanaver for more details on the Boolean power.

Now, in order to effortlessly put a finitely additive measure on the product of Stone spaces, we will need to use Stone duality and a couple results from measure theory. In essense we use the product of countably additive measures on $\sigma$-algebras to give us the product of finitely additive measures on Boolean algebras.

Assume that for $i\in I$, $B_{i}$ is a Boolean algebra and $\mu_{i}$ is a finitely additive measure on $B_{i}$. Then by the Caratheodory Extension Theorem, there is a Baire measure $\nu_{i}$ on the Stone space $S(B_{i})$ such that $\nu_{i}(\{\mathcal{U}\in S(B_{i})|a\in\mathcal{U}\})=\mu_{i}(a)$ for $a\in B_{i}$. Let $\nu$ be the product measure of the measures $\nu_{i}$ on the product space $\prod_{i\in I}S(B_{i})$. Then we get a measure on the free product $\ast_{i\in I}B_{i}$ simply by restricting $\nu$ to the clopen sets of $\prod_{i\in I}S(B_{i})$.

Also, the free product of Boolean algebras is completely described by the following property. Assume that $B$ is a Boolean algebra and $B_{i}\subseteq B$ is a subalgebra for $i\in I$. The system of subalgebras $(B_{i})_{i\in I}$ is said to be independent if whenever $i_{1},....,i_{n}\in I$ are distinct elements and $b_{i}\in B_{i},b_{i}>0$ for $i\in\{i_{1},...,i_{n}\}$, then $b_{i_{1}}\wedge...\wedge b_{i_{n}}>0$. The Boolean algebra $B$ is the free product of the Boolean algebras $(B_{i})_{i\in I}$ if and only if the Boolean algebras $B_{i}$ are independent and $B$ is generated by $\bigcup_{i\in I}B_{i}$.

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Right, but aren't you using AC when you represent Boolean algebras in terms of algebras of sets? –  Alexander Pruss Feb 24 '13 at 21:28
That is affirmative. I will need some form of choice to represent all Boolean algebras as algebras of sets. The axiom of choice usually makes our lives much easier. –  Joseph Van Name Feb 25 '13 at 0:34
You're also using the axiom of choice to take the product $\prod X_i$ (and have it be nontrivial). –  Nik Weaver Feb 25 '13 at 17:45
I always try to use the axiom of choice as much as possible. –  Joseph Van Name Feb 25 '13 at 18:00
Actually, in the application I wanted, the $B_i$ would all be the same, so the nontriviality of $\Pi X_i$ wouldn't need Choice. –  Alexander Pruss Feb 25 '13 at 18:28
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