# Strong (topological) mixing and cofinite set

For a topological dynamical system $(X,T)$ (X is a compact Hausdorff space, and T is a continuous map from X to X), it is called strong mixing if For any nonempty open set U and V, $$N(U,V):=\{n\in \mathbb{Z}_+:U\cap T^{-n}V\neq \emptyset\}$$ (Here $\mathbb{Z}_+$ is the set of non-negative integers) is cofinite in $\mathbb{Z}_+$.

My Question is, if we have a cofinite set A in hand, can we find a strong mixing system (X,T) and two nonempty open set U, V in X such that N(U,V)=A?

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This is incomplete; did you mean to say that it’s strong mixing if $N(U,V)$ is cofinal in $\Bbb Z_+$ for all nonempty open $U$ and $V$? – Brian M. Scott Mar 20 '12 at 2:32
Yes, sorry to miss some words. – Siming Tu Mar 20 '12 at 5:19

## 1 Answer

The answer is yes.

I prefer to write $\Bbb N$ for your $\Bbb Z_+$. Let $X=\{0,1\}^{\Bbb N}$ with the product topology, where $\{0,1\}$ is given the discrete topology; $X$ is a Cantor set and is certainly compact and Hausdorff. Let $V=\{x\in X:x_0=0\}$; $V$ is open in $X$. Let

$$T:X\to X:\langle x_k:k\in\Bbb N\rangle\mapsto\langle x_{k+1}:k\in\Bbb N\rangle\;,$$

the left shift operator; $T$ is essentially a projection map, so it’s continuous. Note that for any $n\in\Bbb N$, $T^{-n}[V]=\{x\in\Bbb N:x_n=0\}$.

Now let $F\subseteq\Bbb N$ be finite, let $A=\Bbb N\setminus F$, and let $U_A=\{x\in X:x_k=1\text{ for all }k\in F\}$. Then

$$U_A\cap T^{-n}[V]=\{x\in X:x_n=0\text{ and }x_k=1\text{ for all }k\in F\}\;,$$

which is empty iff $k\in F$. Clearly $N(U_A,V)=A$.

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