Timeline for Does the "measure-preserving property" commute with ultralimits ?

8 events

when toggle format	what		by	license	comment
May 10, 2012 at 19:40	vote	accept	js21
May 10, 2012 at 19:40	vote	accept	js21
May 10, 2012 at 19:40
May 10, 2012 at 15:51	comment	added	Nik Weaver		Let $f_n \in l^\infty({\bf Z})$ be the bi-infinite sequence which is zero between $-n$ and $n$, and one outside that interval. This sequence converges weak* to the zero sequence, but each $f_n$ evaluates to 1 along any nonprincipal ultrafilter. So evaluating on any nonprincipal ultrafilter cannot be a normal linear functional on $l^\infty({\bf Z})$.
May 10, 2012 at 15:44	comment	added	Jon Bannon		@Nik: I suspected so. How does one get that every nonprincipal ultrafilter isn't normal (a reference will do)?
May 10, 2012 at 15:35	answer	added	Nik Weaver		timeline score: 5
May 10, 2012 at 15:25	comment	added	Nik Weaver		A non-normal pure state on $l^\infty({\bf Z})$ is the same as a non-principal ultrafilter on ${\bf Z}$.
May 10, 2012 at 13:35	comment	added	Jon Bannon		There should exist a non-normal pure state on l^{\infty}(\mathbb{Z}), and that should provide you with a counterexample. I hope this helps.
May 10, 2012 at 10:35	history	asked	js21	CC BY-SA 3.0