Existence of a minimal continuous map on $S^1 \times [0, 1]$ First please see the question:
Let $X = S^1 \times [0, 1]$. Does there exist a continuous map $f: X \rightarrow X$ satisfying, for any point $x \in X$, $\{f^{(n)}(x)\}_{n=1}^\infty$ is dense in $X$? Here we denote $f^{(n)}=f \circ f \circ ... \circ f$ by the composition of $f$ for  $n$ times. (EDIT: Here "for any point" means "for all points", "$\forall$")
Note that if $X = S^1 \times S^1$ is the torus, then the answer is "yes": just let $f(a, b)=(ae^{2\pi ip}, be^{2 \pi iq})$ where $p, q$ are rationally independent irrational numbers and we see that for any $x \in X$, $\{f^{(n)}(x)\}_{n=1}^\infty$ is dense in $X$, by the property of irrational numbers.
Note also that if $X=[0, 1] \times [0, 1]$, then the answer is "no". From the Brouwer's fixed point theorem we see that given any continuous $f : X \rightarrow X$ we can find a fixed point $t \in X$, and $f^{(n)}(t)=t$ for all $n$.
I am trying to show that given a continuous $f: S^1\times [0, 1] \rightarrow S^1 \times [0, 1]$ ,  there always exists some $a \in X$ such that the closure of $\{f^{(n)}(a)\}_{n=1}^\infty$ is a closed curve but not the whole of $X$. I don't know if this is true for all $f$, but it works for many examples. The simplest example is as follows: let $f(r, b)=(re^{2\pi ip}, b)$, where $p$ is an irrational number.
I have not found any other clues. Maybe this question requires some knowledge in dynamical systems or algebraic topology, which is not what I am familiar with.
Any hint would be greatly appreciated!
 A: Consider the map:
$$g: \Bbb{T}^2 \to \Bbb{T}^2, \quad (r, s) = \big(r e^{2\pi i p}, r e^{2\pi i q}\big)$$
With $p, q$ irrationally independent. The sequence $\{g^{(n)}\}$ is nothing but:
$$\{ g^{(n)} \} = \{ \big(re^{2\pi in p}, r e^{2\pi i (np + q)} \big)\}$$
You can see that much like the other sequence this sequence is dense in $\Bbb{T}^2$, because $np + q$ and $p$ are never rationally dependent and the second entry never repeats.
Now a continuous map preserves limit points, so that if $D$ is dense in $X$, then $\phi(D)$ is dense in $\phi(X)$. Consider the map:
$$f: X \to X, \quad (r, b)=(re^{ip}, \Re[re^{iq}])$$
This map induces the same sequence except the second entry is the cosine of the angle. But the map:
$$\phi: \Bbb{T}^2 \to X, \quad \big(e^{i\theta_1}, e^{i\theta_2} \big) = \big(e^{i\theta_1}, \Re[e^{i\theta_2}] \big)$$
is clearly continuous and onto. Since every point of $\Bbb{T}^2$ is a limit point of the $g$ sequence, every point of $X$ is a limit point of the $f$ sequence. So $f$ is the required map, and it is valid for any initial point $x\in X$ as the argument does not depend on $r$.
