How does the irrational rotation generate the Cantor set? In tha book Davidson K R. C*-algebras by example[M]. American Mathematical Soc., 1996, p244, there is a sentence I find it is hard to understand.

Consider the Cantor set X obtained from the unit circle T=R/Z by
  introducing cuts at the point n$\theta$ for all n in Z. More
  precisely, replace each such point by a left and right limit point.

Who can tell me the details? How does the irrational rotation with $\theta$ generate the Cantor set?
 A: This construction is called "Denjoy blowup". It takes some time and effort to prove that it works, one can do so by enumerating a countably-infinite subset $A$ of $S^1$ (the orbit of the rotation) and inductively blowing-up each point $a_n$ of $A$, resulting in the sequence
$$
... \to C_n \to C_{n-1} \to ...\to C_1\to C_0=S^1
$$ 
each map $q_n: C_n\to C_{n-1}$ is the "blow-down" when you collapse an arc $A_n\subset C_n$ (each arc is a closed subset of the circle) to a point (which will eventually project to the point $a_n$ in $C_0$). It easy to convince yourself that each $C_n$ is homeomorphic to $S^1$. Let $C$ be the inverse limit of this sequence of circles $C_n$. (As every inverse limit it comes equipped with continuous maps $p_n: C\to C_n$, so that $p_{n-1}=q_n \circ p_n$). It takes further effort to show that the inverse limit $C$ is still homeomorphic to $S^1$. Now, consider the preimage $\tilde A$ of $A$ in $C$: It is a countable union of disjoint sets $\tilde A_n$, each homeomorphic to the closed unit interval. Let $K$ be the union of all end-points of the intervals $\tilde A_n$ and the set $C\setminus \tilde A$. Now, you have to check that $K$ is compact and metrizable (clear since it is a closed subset of the circle), totally-disconnected (also clear since the set $S^1\setminus A$ is totally disconected), perfect (has not isolated points); the latter takes a bit of work, but the key is that the set $S^1\setminus A$ is perfect. Now, you quote a theorem from the general topology that every metrizable, compact, totally-disconnected, perfect topological space $K$ is homeomorphic to the standard Cantor set.  
Edit. For details see  Proposition 12.2.1 in the book 
"Introduction to the Modern Theory of Dynamical Systems" by A.Katok and B.Hasselblatt. 
A: I interpret it this way: The Cantor set is usually constructed from a unit interval by cutting away successive thirds. There is no essential difference in the actual width of the cut-away, although the fact that your cut takes away a point (a closed set) instead of an open interval might cause troubles. Anyhow, there is no difference whether you do the cuts systematically, or "randomly", as long as each contiguous interval is eventually cut. And there is no problem in cutting not exactly in the middle. What you get the end is still a set of infinite sequences of $\{0, 1\}$ which is completely separable but not discrete.
Going back to your circle, the first cut will make it into essentially a line. Each cut after that will cut one of the intervals you have, and the irrationality of $\theta$ guarantees that each interval will eventually get cut.
So the fact that you cut away closed points rather than open intervals is the only thing here that would create something un-Cantor. But if the textbook says it makes a Cantor set I'm inclined to believe it.
A: This is how I interpret this statement, but without having the book in front of me, I can't be sure if I'm correct or not. 
If you take a point $x$ on the circle $\mathbb{R}/\mathbb{Z}$ and remove it, you end up with a non-compact space (homeomorphic to an open interval). If you wanted to compactify the space again, there are several ways you could do it. One is of course to add the point $x$ back in and get back the circle. On the other hand, if we think of $(\mathbb{R}/\mathbb{Z})\smallsetminus\{x\}$ as an open interval, we can instead add two points at the "ends" to get a closed interval. I think this is what the author wants you to do, except that instead of just doing it at one point $x$, you are supposed to do it at every point $x = n\theta$. So you can think of this as a funny way of compactifying the space $(\mathbb{R}/\mathbb{Z})\smallsetminus\{n\theta : n\in \mathbb{Z}\}$.
To see that the space $X$ you obtain in the end is a Cantor set, you should prove that $X$ is a compact metrizable space that is perfect, nonempty, and totally disconnected. It is a theorem that any such space is homeomorphic to the Cantor set. 
The nice thing about this construction is that rotation by $\theta$ map on $\mathbb{R}/\mathbb{Z}$ induces a map on $X$ that (I think) is a homeomorphism with interesting dynamical properties. I don't think it would be obvious how to construct such a map if one started with the usual definition of the Cantor set....
