Consider the invertible extension of the circledoubling map $T(x)=2x \pmod 1$, the new system can be represented as $X=\{(x_k)\  \ x_{k+1}=T(x_k)\}$ (see GTM 259 M. Eindiedler & T. Ward Exercise 2.1.9) dynamically. Another representation can be described algebraically (see also GTM259 2.1.9) as the quotient of $\mathbb{R}×\mathbb{Q}_2$. A third representation may be the dual of 2adic rational field. My question is how to image it intuitively?

1$\begingroup$ Probably the best thing is to think about it in $2$adic expansion, as twosided sequences with prescribed "right tails" (which comes from the original onesided $\times 2$ system), in the proper inverse limit construction. As the topology or shape, the fibers of this extension over the onesided system are compact, so you don't lose much dealing with that system instead of the onesided one. I think you might want to look also in Schmidt's book  Dynamical Systems of Algebraic Origin. $\endgroup$ – Asaf Sep 2 '13 at 12:45

1$\begingroup$ I see the natural way of extension as just some way of adding data to a point so that you can see where you came from. If you're doing things like the Birkhoff ergodic theorem, averaging a function that "doesn't depend on the past data", you get exactly the same conclusions you did before. $\endgroup$ – Anthony Quas Sep 2 '13 at 14:51

1$\begingroup$ In the paper MR1307296 Hubbard, John H.; ObersteVorth, Ralph W.: Hénon mappings in the complex domain. I. The global topology of dynamical space. Inst. Hautes Études Sci. Publ. Math. No. 79 (1994), 5–46, there is an exhaustive discussion of solenoidal mappings. The intuition behind them is that derivatives of such mappings preserve certain families of cones and are expanding/contracting in specific directions. $\endgroup$ – Margaret Friedland Sep 2 '13 at 18:29
You can visualize a solenoid as part of a simple 3D dynamical system. The Wikipedia page on solenoids has excellent illustrations. If you want something even more concrete, and have some "Silly Putty" handy, you can make your own solenoid by repeating the process shown in this picture. After a few repetitions, if you're careful, you'll be holding a decent approximation to a solenoid.
In my opinion, the best way to understand the solenoid intuitively is using nested intersections.
Let $\mathbb{T}_3 \overset{def}{=} D^2 \times S^1$ where $S^1 = [0,1] \mod 1$ and $D^2 = \{z \in \mathbb{C} \mid \leftz\right < 1\}$. Consider $f:\mathbb{T}_3 \to \mathbb{T}_3$ given by $$f(z,\theta) = (\dfrac{1}{10}z + \dfrac{1}{2}e^{i \theta},2\theta \mod 1).$$
Note that $f$ is well defined because $$\left \dfrac{z}{10} + \dfrac{1}{2}e^{i \theta}\right \leq \left\dfrac{z}{10}\right + \dfrac{1}{2}\lefte^{i\theta}\right \leq \dfrac{1}{2} + \dfrac{1}{10} \leq 1.$$ Actually this implies that $f(\overline{\mathbb{T}_3}) \subset \mathrm{int} \mbox{ } \mathbb{T}_3$.
This map gives you an extension of $2x \mod 1$. Then the solenoid is the nested intersection $\Lambda = \mathop{\bigcap}\limits_{n=0}^\infty f^n(\mathbb{T}_3)$, which is the attractor of the dynamical system $(\mathbb{T}_3, f)$. Moreover $\Lambda$ is the set where $f$ is bijective. Observe that you have a contraction in one direction and expansion in the other. Actually the map acts in the following way: Note $S^1$ spins twice inside $\mathbb{T}_3$, and the discs $D^2 \times \{\theta\}$ go to $D^2 \times \{2\theta\}$, i.e.: $$f(D^2 \times \{\theta\}) \subset D^2 \times \{2\theta\}.$$
Here you can see some pictures: http://www.matcuer.unam.mx/~aubin/vista/index4.html
This book can be illustrative as well: C. Christenson and W. Voxman, Aspects of topology. 2nd Edition, BSC Associates, Moscow, Idaho, USA, 1988, pp. 167170. Also, it contains the inverse limit construction that @Asaf mentioned.

$\begingroup$ Oh this is a nice construction, I haven't seen it until now. I'm guessing that the $\times n$ generalization should be obvious right? I still find the algebraic construction more flexible and natural, mainly because it turns out naturally from harmonic analysis (think about the dual group of $\mathbb{Q}$ with the adeles for example, or just try to find a fundamental domain for the $\times 2$ diagonal action on $\mathbb{R}\times \mathbb{Q}_{2} \backslash \mathbb{Z}[1/2]$). This adelic language is indeed confusing at start, but it pays to learn it in the long run. $\endgroup$ – Asaf Sep 3 '13 at 9:12

1$\begingroup$ Yes, the generalisation $\times n$ is easy. You just need to consider $n \theta \mod 1$ and there you go. Also, you can consider any $\lambda \in (0,1)$ instead of $\frac{1}{10}$. Possibly, the algebraic construction is more flexible and natural, but I think that it is harder to visualise. $\endgroup$ – Rafael Alcaraz Barrera Sep 3 '13 at 10:01