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 <a href="http://en.wikipedia.org/wiki/Diagram_%28category_theory%29">diagram</a>.  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 <a href="http://en.wikipedia.org/wiki/Limit_%28category_theory%29">limit</a> (inverse limit) is the <a href="http://en.wikipedia.org/wiki/Profinite_group#Profinite_completion">profinite completion</a> $\hat{\mathbb{Z}}$ 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 <a href="http://en.wikipedia.org/wiki/Embedding#General_topology">embedding</a> $T \to X$ 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.