Topological dynamical system question - For a point y in a space, find a point whose orbit converges to y

Question

Let $\alpha$ be irrational, and define a map on the torus $T : \mathbb{T}^2 \to \mathbb{T}^2$ to be such that $T(x_1, x_2) = (x_1 + \alpha \mod1,\ 2 x_2\mod 1)$.

Define metric $d$ on the torus to be the usual metric (i.e. the length of the shortest path on the torus).

Given a point $(y_1, y_2) \in \mathbb{T}^2$, construct a point $(x_1, x_2 ) \in \mathbb{T}^2$ such that there exists a subsequence $n_k \to + \infty$ such that $$\lim_{k\to \infty}d(T^{n_k}(x_1, x_2), (y_1, y_2)) = 0.$$

My thoughts so far: There exists a point $x_1 \in \mathbb{T}$ whose orbit under the irrational rotation is dense in $\mathbb{T}$ (in fact every point satisfies this). Also, we can find a point $x_2 \in \mathbb{T}$ whose orbit is dense under the doubling map. Thus we can find a subsequence of the orbit of $x_1$ which converges to $y_1$, and similarly for $x_2$ and $y_2$. But I'm not sure how to combine these subsequences to get a sequence $n_k$ so that $T^{n_k}(x_1, x_2) \to (y_1, y_2).$

Answer 1

You need to do it a bit more carefully, not just letting $x_2$ be any point where the doubling map has dense orbit.

From fixed $x_1$, we get a sequence $n_j$ such that $x_1+n_j\alpha\to y_1$. We want to extract a subsequence $n_{j_k}$ and $x_2$ such that $2^{n_{j_k}}x_2\to y_2$. Then $n_{j_k}\to+\infty$ and $T^{n_{j_k}}(x_1,x_2)\to(y_1,y_2)$.

To do this, we choose $j_k$ inductively that poses compatible constraints on $x_2$.

• Start with $j_1=1$, and choose an interval $I_1\subset\mathbb{T}^1$ such that $2^{n_{j_1}}I_1=\overline{B}(2^{-2},y_2)$ ($\overline{B}(r,y_2)$ is the closed ball of radius $r$, centre $y_2$). This choice ensures $d(2^{n_{j_1}}x, y_2)\leq 2^{-2}<2^{-1}$ for all $x\in I_1$
• Choose $j_2>j_1$ so that $2^{n_{j_2}}I_1=\mathbb{T}$. Choose interval $I_2\subset I_1$ such that $2^{n_{j_2}}I_2=\overline{B}(2^{-3},y_2)$. So $d(2^{n_{j_2}}x,y_2)\leq 2^{-3}<2^{-2}$ for all $x\in I_2$
• Choose $j_3>j_2$ so that $2^{n_{j_3}}I_2=\mathbb{T}$. Choose interval $I_3\subset I_2$ such that $2^{n_{j_3}}I_3=\overline{B}(2^{-4},y_2)$. So $d(2^{n_{j_3}}x,y_2)\leq 2^{-4}<2^{-3}$ for all $x\in I_3$
• At the $k$-th stage, choose $j_k>j_{k-1}$ so that $2^{n_{j_k}}I_{k-1}=\mathbb{T}$. Choose interval $I_k\subset I_{k-1}$ such that $2^{n_{j_k}}I_k=\overline{B}(2^{-k-1},y_2)$. So $d(2^{n_{j_k}}x,y_2)\leq 2^{-k-1}<2^{-k}$ for all $x\in I_k$.

We have $I_k$ as nested sequence of nonempty closed subsets of the compact $\mathbb{T}^1$, so $\bigcap_k I_k\neq\varnothing$. Thus for (the unique) $x_2\in\bigcap_{k=1}^\infty I_k$, we have $2^{n_{j_k}}x_2\to y_2$ as desired.

Answer 2

I think that your insights about dense orbits is a good way to convince yourself of what is going on, but I think the shortest path between you and a solution will take us via a different path.

Firstly, all orbits under an irrational rotation are dense, so lets fix $x_1 = y_1$, $n_0 = 0$, and we'll construct the remainder of the series of powers and $x_2$ later.

Let $$\lim_{i \to \infty} \frac{p_i}{q_i} = \alpha$$ for natural number $\{p_i\}_{i\in \mathbb{N}}$ and $\{q_i\}_{i\in \mathbb{N}}$ and let $\epsilon_i = |\alpha - p_i/q_i|$. Note that $$T^{q_i} ( x_1, b ) = ( y_1 + q_i \alpha , b' ) = (y_1 + q_i (p_i/q_i + \epsilon_i), b') = (y_1 + q_i \epsilon_i, b').$$ So, we need something stronger than simply $\epsilon -> 0$.

Dirichlet's approximation to the rescue!

https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem

This allows us to assume that $\epsilon_i < 1/q_i^2$, which gives us $T^{q_i} ( x_1, b )$ within $1/q_i$ of $(y_1, y_2)$ in the first coordinate (a natural selection for the convergents are those provided by the continued fraction convergents, and they have this property).

Now, we need to understand what the action on the second coordinate is doing. I claim that if $b$ has a binary representation as $0. b_1 b_2 b_3 \ldots$ then $T(a , b) = (a', 0.b_2 b_3 b_4 \ldots)$ which is to say that it is acting like the shift on the binary coordinates. If $y_2 = 0.b_1 b_2 b_3 b_4 \ldots$ then we define a new sequence. For convenience let $q_0 = 0$ and for $q_i < n \leq q_{i+1}$, define $c_n = b_{n - q_i}$ and let $x_2$

Now, let $T^{q_i}( x_1, x_2 ) - (y_1, y_2) = (e_{i, 1}, e_{i,2})$. $e_{i,1}$ is bounded by $1/q_i$ and the first $q_{i+1}$ binary places of the expansions of the second torus coordinates agree, so $|e_{i,2}| < 2^{-q_{i+1}}$.