Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
    
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

The paper“On the Existence of Nonprincipal Arithmetical Ultrafilters on ω” was appeared in ACTA MATHEMATICA SINICA CHINESE SERIES Mar.2006,49(2): 283-288. In this paper it is proved that the statement (Q) implies the existence of nonprincipal arithmetical (i.e. Hausdorff ) ultrafilters on ω. Statement (Q): If B ⊂ P(ω) has the sfip (strong finite intersection property) and |B| < 2^ω, then there is a Q-point q ⊃ B. Statement (Q) is strongly weaker than Statement (R). Statement (R): If B ⊂ P(ω) has the sfip (strong finite intersection property) and |B| < 2^ω, then there is a Ramsey ultrafilter r ⊃ B. Fangting Wang

share|improve this answer
    
I see you have decided not to edit your above answer, even though this appears to be a (somewhat) cleaned up form of it. I also note that you do not provide a link of any sort to this paper, but only mention a couple of results. As far as I can tell this is the MathSciNet review for this paper, and this is the Zentralblatt review. I note that this paper appears to be only available in Chinese, which may be less than helpful to most math.SE users. –  Arthur Fischer Jan 15 '13 at 14:37

On the Existence of Nonprincipal Arithmetical Ultrafilters on ω Fangting Wang Department of Mathematics University of Science and Technology of China Hefei, Anhui, 230026 P.R.China Email: ftwang@ustc.edu.cn January 8, 2013 Abstract Each nonprincipal arithmetical ultrafilter p ∈ βω−ω is associated with a simple arithmetical model Np= {f(p) : f ∈ωω} ⊃ N, and a positive real number is just an equivalence class of finite fractions made by arithmetical ultrafilters (instead of natural numbers), just like the ancient Greek’s idea. It is well known that MAcountableimplies the statement (Q): 1 (Q) If B ⊂ P(ω) has the sfip (strong finite intersection property) and |B| < 2ω, then there is a Q-point q ⊃ B. In this paper we will prove that the statement (Q) implies our Hypoth- esis: There exist nonprincipal arithmetical ultrafilters on ω. Key words: the Stone-ˇCech compactification on ω, Q-point, arithmetical ultrafilter, arithmetical model MR(2000): 03E35, 03E65, 54D80, 54D35, 03C62 1 Introduction βω, the Stone-ˇCech compactification of the discrete space ω, consists of all ultrafilters on ω. For p ∈ βω and f ∈ ωω, the image of p in βω under the Stone extention f : βω → βω is given by f(p) = {a ⊂ ω : f−1[a] ∈ p}. For f,g ∈ ωω and p ∈ βω, f =pg means that {n ∈ ω : f(n) = g(n)} ∈ p. An ultrafilter p ∈ βω is called arithmetical [1], if for all f,g ∈ ωω, (1) f(p) = g(p) ↔ f =pg. 2 (The nonprincipal arithmetical ultrafilter was called ∗-point in [2].) For all f,g ∈ ωω and all ultrafilters p ∈ βω we always have (2) f =pg → f(p) = g(p), but the converse of (2) may not be true [2]. Because of (2), an ultrafilter p is arithmetical iff for all f,g ∈ ωω, (3) f(p) = g(p) → f =pg. We are interested in this class of ultrafilters since each arithmetical ulrtrafilter p is associated with a simple arithmetical model Np: Np= {f(p) : f ∈ ωω}. In Npthe operations +,· and h(∈ ωω) are defined naturally: f(p) + g(p) = (f + g)(p), f(p) · g(p) = (f · g)(p), h(f(p)) = (h ◦ f)(p). Npis a structure of the language L = {0,+,·} ∪ {f : f ∈ ωω}. We have the Transfer between Npand N (the standard arithmetical model): Transfer For any L∪ {p} -sentence ϕ(p), Np|= ϕ(p) ↔ {n ∈ ω : N |= ϕ(n)} ∈ p. 3 Specially, for any L-sentence ϕ, Np|= ϕ ↔ N |= ϕ. The proof of the Transfer is completed by a simple induction on the length of ϕ(p). The first step is the property (1) which the arithmetical ultrafilter p has. Is there a point in βω − ω which is arithmetical? Our Hypothesis (AR) is: (AR) There exist nonprincipal arithmetical ultrafilters on ω. A point q ∈ βω − ω is called a Q-point, if for every partition (ai)i∈ωof ω into pairwise disjoint finite subsets there exists b ∈ q with |b ∩ ai| ≤ 1 for all i ∈ ω. We write sfip for “strong finite intersection property”, and let (Q) be the statement: (Q) If B ⊂ P(ω) has the sfip and |B| < 2ω, then there is a Q-point q ⊃ B. Then we have: Main Theorem The statement (Q) implies the Hypothesis (AR). Before we prove the main theorem, we list some well-known results about the existence of Q-points: 1◦MAcountableimplies the statement (Q), and the word “Q-points” in the statement (Q) can be replaced by “selective (minimal, or Ramsay) ultrafilters” 4 (cf.[3], Th.2). 2◦If there exists a dominant family (inωω) of cardinality ω1, then there exists a Q-point [4]. 3◦the assumption of the existence of Q-point is weaker than MAcountable, since MAcountableimplies that the cardinality of each dominant family (inωω) must be 2ω[5]. 2 Lemmas The following lemma 1 improves the proposition 2 in [2]. Lemma 1 For any f,g ∈ ωω and p ∈ βω , if f(p) = g(p) and (f → g)p, then f =pg, where (f → g)pmeans that ∃a ∈ p ∀x,y ∈ a (f(x) = f(y) → g(x) = g(y)). Proof Assume f(p) = g(p) and (f → g)p. It suffices to show that both f h(ni). Hence ∀ni∈ f[ω]∃k ∈ ω(hk(ni) / ∈ f[ω]). Now let A = {ni∈ f[ω] : h(ni) ∈ g[ω] − f[ω]}, B = {ni∈ f[ω] − A : the least k such that hk(ni) / ∈ f[ω] is even}, C = f[ω] − A − B. The remaining part is the same as in the case f

share|improve this answer
3  
This appears to be a poorly executed copy-and-paste job from an article (of yours, if I am not mistaken). Please consider editing this to make its connection to the question posed clear (and also to make it readable). –  Arthur Fischer Jan 8 '13 at 13:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.