## Is there such a thing as the sigma-completion of a Boolean algebra?

Hi all,

Suppose that $\mathcal{B}$ is a Boolean algebra. It there a way to extend $\mathcal{B}$ to a smallest Boolean algebra $\mathcal{B}'$ that contains an isomorphic copy of $\mathcal{B}$ and is countably complete, i.e. every countable subset of $\mathcal{B}'$ has a least upper bound in $\mathcal{B}'$? By "smallest" I mean that the inclusion $i: \mathcal{B} \hookrightarrow \mathcal{B}'$ has the obvious universal property, i.e. for every homomorphism $f$ from $\mathcal{B}$ to a countably complete Boolean algebra $\mathcal{C}$ there exists a unique homomorphism $g: \mathcal{B}' \to \mathcal{C}$ such that $g \circ i = f$ (it would be nice if $g$ turned out to commute with countable sups too). If no such $\mathcal{B}'$ exists, is there some other useful definition of "smallest" countably complete Boolean algebra containing $\mathcal{B}$?

If it makes any difference, I'm mostly interested in the special case where $\mathcal{B}$ is a direct limit of a sequence of finite Boolean algebras.

Edit: Thanks very much for the replies, it's a shame I can only mark one as the answer. It will take me a while to absorb the various references I've been given, so if I run into difficulty I'll bump the thread with an edit.

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The short answer is "yes", and it's a special case of a much, much more general theorem on relatively free algebraic constructions.

In other language, you are asking whether the underlying functor from countably complete Boolean algebras to Boolean algebras has a left adjoint. The more general question is whether, given a homomorphism $\phi: S \to T$ between two monads on $Set$, the evident underlying functor

$$Set^\phi: Set^T \to Set^S$$

from the category of $T$-algebras to the category of $S$-algebras has a left adjoint. For this I'll direct your attention to this nLab article.

Of course, we have to know that countably complete Boolean algebras can be described as algebras of a monad on $Set$, but this too follows from general theory. I'll refer you to another nLab article for this; the article is not complete but it should give the idea. The upshot is that for any algebraic theory with only a small set of operations of each arity, there is a corresponding monad on $Set$ whose algebras are the models of the theory. The general constructions go back to work in the sixties, due to Lawvere, Linton, and others.

Edit: I'll remark that had you said "complete" instead of "countably complete", then the answer would have been no. In fact, the underlying functor from complete Boolean algebras to sets has no left adjoint; this is mentioned for instance in Categories for the Working Mathematician. But in your case, the theory is generated by a set of operations and equations, and all is well.

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 ... but the link I mention below gives a completion functor with a useful adjointness property, albeit not the most obvious one. – Neil Strickland Nov 24 2010 at 23:31 Apropos of what, Neil? Are you referring to the corollary at the bottom of the page? It is not a morphism of complete Boolean algebras, although it is I reckon an answer to one of OP's questions (where the extension need not preserve countable joins). – Todd Trimble Nov 24 2010 at 23:51 @Todd: yes, I am referring to the Corollary at the bottom of Johnstone's page 108 and the associated remarks on page 109, which cover essentially the same ground as your discussion with Joel below. – Neil Strickland Nov 25 2010 at 10:39

Every Boolean algebra $\mathbb{B}$ embeds densely in its completion as a Boolean algebra, which is a complete Boolean algebra (more than just countably complete). The completion $\bar\mathbb{B}$ can be constructed as the regular open algebra, the set of all regular open subsets of $\mathbb{B}-\{0\}$, where the topology is generated by the basic open sets consisting of lower cones. A set is regular open if it is the interior of its closure.

Indeed, the completion operation is extremely general, and is used pervasively in set theory in the context of the forcing technique. Every separative partial order $\mathbb{P}$, and this includes any Boolean algebra (minus $0$), embeds densely in its regular open algebra, and this is always a complete Boolean algebra. This fact is the main connection between the poset-based account of forcing and the Boolean-algebra based account of forcing.

The wikipedia link above lists several universal properties of this completion.

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But this is not a free construction that the OP was asking about (see his universal property). There is a theorem due to Gaifman and Hales that free complete Boolean algebras do not exist. – Todd Trimble Nov 24 2010 at 23:46
Every homomorphism of $\mathbb{B}$ to a complete boolean algebra $\mathbb{C}$ extends uniquely to a homomorphism of $\bar\mathbb{B}$ into $\mathbb{C}$. Isn't this what we should want? – Joel David Hamkins Nov 24 2010 at 23:55
We're both right. :-) You are answering a question where one doesn't demand that the extension preserve arbitrary joins, but be just a homomorphism of Boolean algebras, yes? I was referring to morphisms of complete Boolean algebras, where the condition is that the extension preserve arbitrary joins. This is where Gaifman-Hales comes in. – Todd Trimble Nov 25 2010 at 0:03
@Todd : I never knew this was due to Gaifman (and Hales). Any chance you know a reference? – Andres Caicedo Nov 25 2010 at 0:17
What I am saying is that if $f$ respects those infinitary joins that exist in $\mathbb{B}$, an obvious necessary requirement, then the extension of $f$ to the completion $\bar\mathbb{B}$ will respect all infinitary joins. And this seems to be exactly what the OP wants, right? – Joel David Hamkins Nov 25 2010 at 2:49

You can also look at III.3.11 in Johnstone's book on Stone Spaces:

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This question was answered in topological terms by J. Vermeer in The smallest basically disconnected preimage of a space. Topology Appl. 17 (1984), no. 3, 217–232. See here for a review and here for the paper.

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Hi all,

I've just come up with something that's relevant to the question I asked here. I'm bumping the thread partly in case anyone else cares, and partly in case (as is more likely) I've made an error and someone can point it out. Anyway: I believe I can prove that the $\sigma$-algebra generated by a Boolean algebra $A$ (in the sense of Todd's answer, i.e. the image of a left adjoint to the forgetful functor from $\sigma$-algebras to Boolean algebras) has a rather natural representation, namely as the $\sigma$-field generated by the double dual of $A$, i.e. the smallest $\sigma$-field containing all the clopen subsets of the dual space of $A$. Here is the proof:

Let $A$ be a Boolean algebra, let $A^\star$ be its dual Boolean space and $A^{\star \star}$ the dual algebra of its dual space, i.e. the set of clopen subsets of $A^\star$. Let $\bar{A}$ be the $\sigma$-algebra of Baire sets in $A^\star$, i.e. the $\sigma$-field of subsets of $A^\star$ generated by $A^{\star \star}$. Let $\alpha: A \cong A^{\star \star}$ be the canonical isomorphism, and let $\eta: A \to \bar{A}$ be the composition of $\alpha$ with the inclusion.

Suppose given a $\sigma$-algebra $B$ and a homomorphism (of Boolean algebras) $h: A \to B$. Define $B^\star$, $B^{\star \star}$, $\bar{B}$ and $\beta: B \cong B^{\star \star}$ as before. By Theorem 41, p. 376 of [1], $B^\star$ is a $\sigma$-space, i.e. the closure of every open Baire set is open. By Theorem 42, p. 381, there is a $\sigma$-homomorphism $\phi: \bar{B} \to B^{\star \star}$ such that $\phi$ maps ever clopen set to itself.

$\beta h \alpha^{-1}$ is a homomorphism $A^{\star \star} \to B^{\star \star}$, so by duality there is a unique continuous function $f: B^\star \to A^\star$ such that $f^{-1} P = \beta h \alpha^{-1} (P)$ for every $P \in A^{\star \star}$. It is easy to see that $f^{-1} S$ is a Baire set whenever $S$ is, so define

$f^\star : \bar{A} \to \bar{B}$; $S \mapsto f^{-1} S$.

$f^\star$ is clearly a $\sigma$-homomorphism. Let

$\bar{h} \equiv \beta^{-1} \phi f^\star: \bar{A} \to B$.

Then one may check, using the defining property of $f$ and the fact that $\phi$ maps clopen sets to themselves, that $\bar{h} \eta = h$. The uniqueness of $\bar{h}$ with this property follows from the fact that the range of $\eta$ generates $\bar{A}$.

So there's the alleged proof; I can't see anything wrong with it but the result strikes me as being "too good to be true", and if it is true then I'm surprised I didn't see any reference to it online before I started this thread. So I'll be grateful if anyone can spot a mistake.

[1] Steven Givant and Paul Halmos, Introduction to Boolean Algebras, Springer 2009

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 The unit of the Stone adjunction $\beta$ is not in general $\sigma$-continuous, so why is the factorization $\overline{h}$ $\sigma$-continuous? – G. Rodrigues Mar 20 2012 at 20:45