There may be no Hausdorff ultrafilters

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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.

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Logic

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http://salesiana.dossiersoluciones.com/handle/123456789/69084

Consulte el texto completo en el siguiente enlace:

https://arxiv.org/abs/math/0311064
 http://arxiv.org/abs/math/0311064
 in: {Logic Colloquium 2004} (2008) 18--32

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arXiv