Prove that symbolic dynamics is topological mixing

Question

Let $ \Sigma _{2} $ be the product space $ \prod _{-\infty} ^{\infty} {X}$, where $X$ is the discrete space $\{0,1\}$. and $f$ is the shift operator: $$ (f(x))_{i}=x_{i+1}$$ I want to prove that $f$ is topological mixing, I can prove that $f$ is topological transitive. so for every open set $U,V$ there is $n$, such that $f^{n}(U)\cap V \ne \emptyset$ but how can I find a $N$, such that $f^{m}(U)\cap V \ne \emptyset$ for every $m>N$?

Answer 1

This has to do with how open sets in $\Sigma_2$ are defined: they are generated by sets of the form

$$ U:=\{\mathbf{x}: x_i\in U_i: \; \forall |i|<k, U_i\subset\{0,1\} \} $$ i.e., $U$ is the set of all sequences where some of the indices, a finite set of them around 0, have been specified.

Let $U$, $V$ be sets of this form and say $U$ specifies indices $-k,-k+1,\ldots,0,\ldots,k$ and $V$ specifies all indices $-l,\ldots,0,\ldots,l$.

Then for $N> k+l+1$ we have that $f^N(U)$ specifies indices less than $-l$. And $V$ still specifies indices greater than $-l$. So $f^N(U)\cap V$ is nonempty since each set specifies a disjoint set of indices.