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.


**Edit 2:** Bumping with followup question, please see my answer below.



Is there such a thing as the sigma-completion of a Boolean algebra?
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](http://ncatlab.org/nlab/show/Lawvere+theory#AdjCatAlgs). 


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](http://ncatlab.org/nlab/show/algebraic+theory) 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. 


