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The question is in the title. The motivation behind the question is as follows: there are plenty of references about profinite groups and profinite completions of groups. It seems that their not exactly a wealth of references about profinite sets and profinite completions of sets.

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What do you mean by the profinite completion of a set? –  Qiaochu Yuan Mar 2 '10 at 21:33
I guess one could define the profinite completion of a set as the projective limit of all finite sets endowed with a (surjective) map from the given set. –  Leonid Positselski Mar 2 '10 at 21:55
Let us first say what a profinite set is. This is a compact Haussdorf totally disconnected topological space. We may form the category of profinite spaces where the morphisms are continuous maps between them. Their is a forgetfull functor from profinite sets to sets that forgets the topology. Profinite completion is the left adjoint to this functor. –  jackie boy Mar 2 '10 at 22:01
The projective limit definition would amount to the same thing. –  jackie boy Mar 2 '10 at 22:02
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4 Answers

up vote 2 down vote accepted

Profinite sets are just another name for compact totally disconnected topological spaces. I think this is (essentially) explained somewhere in Bourbaki's books on general topology.

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What's an \ell-space? –  Leonid Positselski Mar 2 '10 at 22:09
An $\ell$-space is the terminology used by Bernstein for a locally compact, totally disconnected, Hausdorff (maybe this last adjective is redundant) topological space. It seems that fpqc has been reading up on $p$-adic groups lately - I'll have to talk to DeBacker about this! ;) –  Marty Mar 2 '10 at 22:39
@fpqc: this is not Twitter. If you have something to say, please use whole sentences, and explain how your comment is relevant (are you pointing out an error in Leonid's answer? giving an alternate description? etc.) –  Reid Barton Mar 2 '10 at 22:44
Leonid, you either forgot "Hausdorff" or you were being a bit merciless with your terminology :-) Also: another name for profinite spaces is Stone spaces. –  Tom Leinster Mar 3 '10 at 0:58
@Leonid: Only algebraic geometers make that distinction. Do you say quasilocal for "has a unique maximal ideal" and local for "quasilocal + noetherian" like a commutative algebraist? I don't understand why AG people like to make it an issue. We realize that all of the standard texts use "quasicompact", but I doubt that using "quasicompact" instead of "compact" is ever going to catch on outside of AG. Also, it's silly because Hausdorff spaces pretty much never come up in AG. –  Harry Gindi Mar 3 '10 at 14:13
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I agree with some of the comments that "profinite set" is not a standard term. But you can certainly look at the category of pro-(finite sets). In other words, begin with the category $Set_f$ of finite sets and functions. Then one can form a category $Pro(Set_f)$ as the projective completion of the category $Set_f$; it is the full subcategory of the category of functors from $Set_f$ to $Set$, consisting of objects isomorphic to projective limits of systems of finite sets. In other words, the objects of $Pro(Set_f)$ are (not necessarily representable) functors from $Set_f$ to $Set$, which are inductive (viewing finite sets via Yoneda as functors from $Set_f$ to $Set$ switches arrow directions) limits of representable functors from $Set_f$ to $Set$. I'm sure one should be careful about some smallness/universe issues to make this precise.

The category $Pro(Set_f)$ is equivalent to the category of compact totally disconnected topological spaces. This elaborates on Leonid's answer.

The reference for this somewhat highbrow answer is a paper of Gaitsgory and Kazhdan, in GAFA, titled "Representations of algebraic groups over a 2-dimensional local field".

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More precisely, objects of $Pro(Set_f)$ are covariant functors from $Set_f$ to $Set$ which are filtered inductive (rather than projective) limits of representable functors. –  Leonid Positselski Mar 2 '10 at 22:27
Whoops - I forgot about the Yoneda reversal. I think it's correct now. Thanks Leonid! –  Marty Mar 2 '10 at 22:33
This is a little highbrow for my taste. [Marty is an old friend of mine, so I know he can take being called "highbrow" with a smile.] Suppose that $S$ is a countably infinite set. Its profinite completion is...? –  Pete L. Clark Mar 3 '10 at 2:34
Excuse my high brows :) I didn't say anything about profinite completions -- I don't know a definition of "the profinite completion of a set". Given a set $X$, endowed with the discrete topology, perhaps the Stone-Cech compactification $\beta X$ satisfies an appropriate universal property to be called a "profinite completion of $X$". I wouldn't use this terminology though! "Profinite completion" should only be used for groups, I think. –  Marty Mar 3 '10 at 4:11
What I am calling the profinite completion of a set is almost the same as for a group, except everytime the word group is used, you replace it the with the word set. Now the Stone Cech compactifaication of a (discrete) space misses the words, "totally disconnected". Wikipedia states (so a grain of salt) that the stone cech compactification happens to be totally disconnected. So ths notion of profinite completion seems to be the stone cech compactification. I would like to make a remark on the terminology. Any concrete category with filtered limits has a "profinite completion" functor. –  jackie boy Mar 3 '10 at 6:03
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Johnstone's book Stone Spaces is a suitable reference for this and much more. The book by Ribes and Zalesskii Profinite Groups also has good information. Algebre et Théories Galoisienne by Doady and Doady also gives a nice accounting.

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See http://math.harvard.edu/~waffle/boolean.pdf for notes on Boolean algebras and the fact that they are essentially (anti-)equivalent to totally disconnected compact Hausdorff spaces.

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By the way, these notes were written by an active MO user. –  David Corwin Jul 4 '10 at 12:15
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