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Reading this documment: www.math.ucla.edu/~tao/preprints/compactness.pdf, I got interested in the following thing: "One can also use compactifications to view the continuous as the limit of the discrete; for instance, it is possible to compactify the sequence Z/2Z, Z/3Z, Z/4Z, etc. of cyclic groups, so that their limit is the circle group T = R/Z.". Could you give me a point of start to understand what idea of compactification is being used there? Where could I find an sketch of proof for that fact?

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Is it possible that the circle can be viewed as the inductive limit of the ciclic groups in some appropriate category of topological groups? –  Gian Maria Dall'Ara Feb 7 '10 at 14:02
    
Q/Z is a colimit, not a limit, of the finite cyclic groups. –  Qiaochu Yuan Feb 7 '10 at 16:33
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@QY: In many areas of mathematics, including algebra, one says "direct limit" and "inverse limit" -- not "colimit" and "limit" -- for this type of construction. –  Pete L. Clark Feb 7 '10 at 18:55
    
Or even better, projective and inductive. The way I keep them all straight is by remembering that pushouts are colimits, since they push out to the bottom right, so they're left=>right is direct. right=>left =inverse limit = limit. –  Harry Gindi Feb 7 '10 at 20:26
    
I guess it's also worth pointing out that "limit" as Tao is using it is not quite the same as "limit" in the category-theoretic sense, which is the sense several of us have been using. –  Qiaochu Yuan Feb 7 '10 at 21:12
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Let's use the abbreviation $\mathbb{Z}_n=\mathbb{Z}/n\mathbb{Z}$, and give it a metric $d(u,v)=n^{-1}\min\{|u-v-kn|\colon k\in\mathbb{Z}\}$. For each $n$ and $k$, multiplication by $k$ embeds $\mathbb{Z}_n$ isometrically into $\mathbb{Z}_{kn}$. The resulting inductive limit is a metric space whose completion is the circle $\mathbb{R}/\mathbb{Z}$. I suspect this is what Tao had in mind, but I could of course be wrong.

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A little less abstractly, you can think of everything in sight as subgroups of the unit circle in $\mathbb{R}^2$. Then the direct limit (colimit, the categorists would say) of the groups $\mu_n$ of $n$th roots of unity is the group $\mu$ of all roots of unity, which is dense in the unit circle and therefore its completion is the unit circle. –  Pete L. Clark Feb 7 '10 at 18:52
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I'd like to clear up something that came up in the comments. There are two natural ways to fit the finite cyclic groups together in a diagram. One is to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, m | n$ given by sending $1$ to $1$. This gives a diagram (inverse system) whose limit (inverse limit) is the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$. This diagram also makes sense in the category of unital rings, since they also respect the ring structure, giving the profinite integers the structure of a commutative ring.

This is not the diagram relevant to understanding the circle group. Instead, one needs to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, n | m$ given by sending $1$ to $\frac{m}{n}$. This is the diagram relevant to understanding the cyclic groups as subgroups of their colimit (direct limit), which is, as I have said, $\mathbb{Q}/\mathbb{Z}$. And this group, in turn, compactifies to the circle group in whichever way you prefer.

(These two diagrams are "dual," though, something which I learned recently when I was asked to prove on an exam that $\text{Hom}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}}$. Just observe that $\text{Hom}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \mathbb{Z}/n\mathbb{Z}$ and that contravariant Hom functors send colimits to limits!)

Edit: Let me also say something about the precise meaning of "compactification" here. A compactification of a space $T$ is an embedding $T \to X$ into a compact Hausdorff space $X$ with dense image. The embedding being considered here is the obvious one from $\mathbb{Q}/\mathbb{Z}$ to $\mathbb{R}/\mathbb{Z}$, and the fact that it has dense image is essentially what the word "completion" also means. Compactifications are not unique, but it's possible that there is a sense in which as a topological group $\mathbb{R}/\mathbb{Z}$ is the "most natural" compactification of $\mathbb{Q}/\mathbb{Z}$. But I don't know too much about topological groups.

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Harry, you are the only person in this discussion who's brought up the profinite integers. What makes you think the OP is asking a question about the profinite integers? –  Qiaochu Yuan Feb 7 '10 at 20:40
    
My mistake. I've given you a +1 to compensate for my incorrectness. –  Harry Gindi Feb 7 '10 at 20:47
    
Okay, then I can also remove my comment :) –  user717 Feb 7 '10 at 21:47
    
When thinking in terms of topologies on groups, you have to be careful. If one were not "gunning for the circle", it would seem natural -- and is certainly the correct thing to do in many instances -- to put the direct limit topology on $\mathbb{Q}/\mathbb{Z}$, i.e., the discrete topology. This makes it into a complete topological group, so it is certainly not dense in some other topological group like the circle. I think that Tao's remark is rather vague (there is no such thing as a "continuous space"), but M. Emerton is probably on the right track in his Gromov-Hausdorff response. –  Pete L. Clark Feb 7 '10 at 22:45
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I haven't looked at the link, but it seems likely that the author of said link is discussing convergence in the Hausdorff topology. In this context the idea of taking limits is due to Gromov, I believe. (There is a wikipedia entery on Gromov--Hausdorff convergence which seems to be the relevant one.) It is a common technique in parts of geometric topology and geometric group theory.

If you ask another question on Gromov--Hausdorff convergence, I'm sure it would draw the attention of the (at least) several experts on the topic who I know read MO.

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Right, but you still need to metrize the finite cyclic groups according to their embeddings into the Euclidean unit circle, i.e., so that the minimum distance between distinct points approaches $0$ as $n$ approaches infinity. So the "Gromov" part of Gromov-Hausdorff doesn't seem to be buying us anything here: you could just say that $\mu_n$ converges to $S^1$ in the Hausdorff metric on compact subsets of $\mathbb{R}^2$. –  Pete L. Clark Feb 7 '10 at 23:47
    
Dear Pete, Thanks; that sounds right. –  Emerton Feb 8 '10 at 2:34
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