# Hausdorff ultrafilters

I know that Ramsey ultrafilters are Hausdorff ($\mathcal{U}$ is Hausdorff iff for every $f,g:\mathbb{N}\rightarrow\mathbb{N}$ $f(\mathcal{U})=g(\mathcal{U})$ then $f\cong_\mathcal{U} g$ $\;$). So if we assume the continuum hypothesis then there exist Hausdorff ultrafilters. I'm wondering this: if we assume continuum hypothesis then there exists a Hausdorff ultrafilter not Ramsey?

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So an ultrafilter is Hausdorff is whenever two functions, when extended to $P(\mathbb N)$ act the same on $\mathcal U$ it implies they only differ by a null set? –  Asaf Karagila May 9 '11 at 17:35
$f(\mathcal{U}):=\{X\subset\mathbb{N}:f^{-1}(X)\in\mathcal{U}\}\;\;\;\;$ and $f\cong_\mathcal{U} g\;\;$ means that the differ by a null set. –  Jacob Fox May 9 '11 at 18:27

Under the continuum hypothesis, or even Martin's Axiom, there are Hausdorff ultrafilters that are not selective. The following is a good survey by Fremlin of many related matters:

http://www.essex.ac.uk/maths/people/fremlin/n09102.pdf

The construction, under CH, of such ultrafilters goes back to Andreas Blass, in the 1970s.

The harder question of whether the existence of Hausdorff ulrafilters can be proved in ZFC was settled not many years ago. The answer is no. Shelah is one of the authors.

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Is there an easier article than essex.ac.uk/maths/people/fremlin/n09102.pdf maybe the Blass one? The preprint in which Shelah proved the existence of a model withouth Hausdorff ultrafilters contained an error and it has been retired, so it is still an open problem. –  Jacob Fox May 9 '11 at 18:16
@Jacob Fox: Will try to think of one. The Shelah paper retraction is new to me! Am a lot out of touch, my last paper on the subject was about $30$ years ago. –  André Nicolas May 9 '11 at 20:03