Uniformly sweeping out does not imply mixing
Terrence Adams · Illinois Journal of Mathematics · 1993
IntroductionLet T be an invertible measure preserving transformation on a measure space that is isomorphic to the unit interval with Lebesgue measure.It was shown in [F1] that if T is mixing, then T is uniformly sweeping out (see 2 for definitions).A sequential counterexample to the converse was given in [F2] where a transformation was constructed that is not mixing on a sequence but is uniformly sweeping out on the sequence.In [C], Chacon constructed another example of a rank one transformation that is weakly mixing but not mixing that is different from Chacon's transfor- mation [F3, 86-89].In [FK] the example in [C] was shown to be lightly mixing, not partially mixing, and not lightly 2-mixing which implies not sweeping out of order 2.Our purpose is to show the transformation T in [C] is uniformly sweeping out.Thus T is rank one, not partially mixing, uniformly sweeping out, but not sweeping out of order 2.This is in contrast to Kalikow's theorem which states that rank one mixing implies 2-mixing [KA].We also note that it is not difficult to construct a partially mixing transfor- mation that is not uniformly sweeping out.It was shown in [FT] that (2k 1)-mixing implies uniformly sweeping out of order k, k > 1.Thus mixing of all orders implies uniformly sweeping out of all orders.Concerning the converse, we do not know if uniform sweeping out of all orders implies mixing.