There may be no Hausdorff ultrafilters
Logic
2007-05-23 v1
Abstract
An ultrafilter U is Hausdorff if for any two functions f,g mapping N to N, f(U)=g(U) iff f(n)=g(n) for n in some X in U. We will show that it is consistent that there are no Hausdorff ultrafilters.
Keywords
Cite
@article{arxiv.math/0311064,
title = {There may be no Hausdorff ultrafilters},
author = {Tomek Bartoszynski and Saharon Shelah},
journal= {arXiv preprint arXiv:math/0311064},
year = {2007}
}