Uniformly sweeping out does not imply mixing
In addition to his well-known ``middle thirds'' rank one weakly mixing stacking transformation, Chacon studied another such transformation \(T\) defined inductively by cutting an interval into two equal intervals, putting a spacer above the right hand interval and stacking to obtain a tower of height 3, and then continuing this same procedure repeatedly. Explicitly, it can be defined in terms of \(n\)-blocks as follows: Let \(B_ 1= (0)\) and let \(s\) denote a spacer, and define \(B_{n+1}= B_ n B_ n s\). The resulting rank one map was shown by Chacon to be weakly mixing, but not mixing. Denote the resulting dynamical system by \((X,{\mathcal F},\mu,T)\). \textit{N. A. Friedman} and \textit{J. L. King} [Isr. J. Math. 73, No. 3, 281- 288 (1991; Zbl 0754.28010)] showed that \(T\) is lightly mixing (i.e., \(\forall A,B\in {\mathcal F}\); \(\mu(A)> 0\), \(\mu(B)> 0\), \(\liminf \mu(T^ n A\cap B)> 0)\), not partially mixing (\(T\) is partially mixing if \(\exists\beta> 0\) such that \(\forall A,B\in {\mathcal F}\); \(\liminf \mu(T^ n A\cap B)\geq \beta\mu(A)\mu(B))\), and not lightly 2-mixing. In this paper the author shows that \(T\) is uniformly sweeping out (i.e., \(\forall A\in {\mathcal F}\), \(\mu(A)> 0\) and \(\varepsilon> 0\), \(\exists N> 0\) such that \(\mu\Bigl(\bigcap^ n_{i=1} T^{k_ i} A\Bigr)> 1-\varepsilon, \forall k_ 1< k_ 2< \cdots < k_ N)\). It is known that not lightly 2-mixing implies not sweeping out of order 2, so that uniformly sweeping out does not imply sweeping out of order 2, in contrast to Kalikow's theorem which says that rank one mixing implies mixing of all orders.
rank one weakly mixing stacking transformation
scientific article; zbMATH DE number 561197