TWO REAL ULTRAFILTERS ON ω

ALAN DOW AND JINYUAN ZHOU

Abstract. We prove the following results concerning the space ω ∪{p}, where p is an ultrafilter on ω:

(i) There are ultrafilters p on ω with the property that every compactification of ω ∪ {p} contains a copy of βω.

(ii) There is an ultrafilter p on ω with a strong Noetherian base, i.e. every member of the base is contained, mod finite, in only finitely many others.

1. Introduction

We prove several results which relate to the space ω∪{p}, where p is an ultrafilter on ω. In section 2, we prove, in ZFC, that there are ultrafilters p on ω with the property that every compactification of ω ∪ {p} contains a copy of βω, thus answering a question raised by van Mill in [4]

In section 3, we construct an ultrafilter p for which there exists an almost disjoint refinement I of p and a base B for p such that for each subset I′ ∈ [I]^{<𝔠}, there is a member of p which is almost disjoint with each member of I′. Additionally, we have that each I ∈ I is almost disjoint from all but finitely many members of B. The last property implies that B is a strong Noetherian base.

2. Large Ultrafilter

It is asked in van Mill [4] if there is an ultrafilter p on the countable discrete space ω such that the space ω ∪ {p} (as a subspace of βω) has the property that every compactification contains a copy of βω. Let us temporarily call such ultrafilter a large ultrafilter. Ryll-Nardzewski and Telgársky [5] showed that if p is a simple P-point then the space ω ∪ {p} has a scattered compactification. It was also shown by Rajagopolan that p is not large if it contains a tower since space ω ∪ {p} has a compactification whose remainder is an ordinal (and there exist such p in any model of Martin’s Axiom) while Malyhin has shown that, in the Cohen model, every ultrafilter is large. In this section we show (in ZFC) that there are many large ultrafilters.

First of all, we give the following combinatorial lemma. It is certainly straightforward but for the reader’s convenience, we provide a standard Δ-system proof. Note, however, that it is not totally trivial since 𝔠 may be singular.

Lemma 2.0.1 (Folklore). Suppose {φ_{α} : α ∈ 𝔠} is a sequence such that φ_{α} is a function into 2 with finite domain F_{α} ⊆ 𝔠 and α ∈ F_{α} for all α < 𝔠. Then there exists J ∈ [𝔠]^{𝔠} such that {φ_{α}, {⟨β, 1⟩} : α ∈ G, β ∈ H} is a set of mutually compatible functions, for any disjoint subsets G, H of J, Proof. If 𝔠 is a regular cardinal, let I ∈ [𝔠]^{𝔠} be such that {F_{α} : α ∈ I} is a Δ-system with root R and such that φ_{α} ↾ R all agree.

1991 Mathematics Subject Classification. 04A20,54C25.

Key words and phrases. compactification, βω, embedding, Noetherain base, almost disjoint refinement.

If 𝔠 is singular apply the Δ-system lemma to {F_{α} : α < 𝔠} as follows. Choose an increasing sequence κ_{ξ} (ξ < cf(𝔠)) of regular cardinals which are cofinal in 𝔠 (each of which is greater than cf(𝔠)). Let, for each ξ < cf(𝔠), I_{ξ} ∈ [κ_{ξ} −sup{κ_{ζ} : ζ < ξ}]^{κ_{ξ}} be such that {F_{α} : α ∈ I_{ξ}} is a Δ-system with root R_{ξ} and for α, β ∈ I_{ξ}, φ_{α} and φ_{β} agree on R_{ξ}. We can apply the Δ-system lemma to the R_{ξ} and by re-indexing a subset we may assume that {R_{ξ} : ξ ∈ cf(𝔠)} forms a Δ-system with root R and, again, that all the φ_{α} agree on R for α ∈ ⋃_{ξ<cf(𝔠)} I_{ξ} . For each ξ, choose a subset I′_{ξ} of I_{ξ} with cardinality κ_{ξ} such that

(F_{α} − R_{ξ}) ∩ [( ⋃_{ζ<cf(𝔠)} R_{ζ} ) ∪ (⋃{F_{β} : β ∈ ⋃_{ζ<ξ} I_{ζ}})]

is empty. Let I = ∪_{ξ<cf(𝔠)}I′_{ξ}.

It is easy to see that for each finite subset G of I, ⋃_{α∈G} φ_{α} is a function. Now, take a subset J of I with size 𝔠 such that, for each α ∈ J, J ∩ dom φ_{α} = {α}. J has the required properties. □

Fix any function f from ω onto a dense subset of 2^{𝔠}. Extend f to a map from βω onto 2^{𝔠}. It is well known that we can find a closed subset K of βω for which f ↾ K maps K irreducibly onto 2^{𝔠}, i.e. F [K] = 2^{𝔠} but for any proper closed subset J ⊂ K, f[J] ≠ 2^{𝔠}. See 3.1.C.(a) of [2]

Theorem 2.0.2. Let K be as defined above, then any point p in K has the property that every compactification of ω ∪ {p} contains a copy of βω.

Proof. Let X be any compactification of ω∪{p}. We have to show that X contains a copy of βω. By the maximality of βω, it suffices to show that some continuous image of X contains a copy of βω. We will construct a continuous mapping h : X → ℝ^{𝔠} such that ran(h) contains 2^{𝔠}. Since 2^{𝔠} contains a copy of βω , we will be done.

Let g be the map from βω onto X which is the identity map on ω ∪{p}. Without loss of generality, we can assume f(p) is the constantly 0 function in 2^{𝔠}. For each α ∈ 𝔠, let A_{α} ∈ p be chosen so that A_{α} = ω ∩ f^{←}([⟨α, 0⟩]) where [⟨α, 0⟩] denotes the basic clopen subset of 2^{𝔠} of all functions which are 0 at α.

Since ω ∪ {p} is a subspace of X, there is a neighbourhood U_{α} of p in X such that U_{α} ∩ ω ⊂ A_{α} – i.e. p is not in the closure of ω − A_{α}. Using the continuity of g, there is a B_{α} ∈ p such that the closure of g[B_{α}] is contained in U_{α}. Then B_{α}^{∗} ∩ K is a clopen subset of K which, by irreducibility of f ↾ K, will map to a subset of 2^{𝔠} with non-empty interior. Fix a finite set F_{α} ⊂ 𝔠 and a function φ_{α} : F_{α} ↦ 2 such that [φ_{α}] ⊂ f[B_{α}^{∗} ∩ K]. It follows that α ∈ F_{α} because

f(B_{α}^{∗} ∩ K) ⊂ f(A_{α}^{∗} ∩ K) = [⟨α, 0⟩] .

Applying Lemma 2.0.1 to {φ_{α}, α ∈ 𝔠}, we get a set J of size 𝔠 as in the Lemma.

Claim 2.1. The family {⟨B_{α}, ω − A_{α}⟩ : α ∈ J} is a J × 2-independent matrix, i.e. if G, H are disjoint finite subsets of J, then

⋂_{α∈G} B_{α} ∩ ⋂_{β∈H} ω − A_{α}

is not empty.

Proof of the Claim 2.1: Let G, H be disjoint finite subsets of J. It suffices to show that ⋂_{α∈G} B_{α}^{∗} ∩ K − ⋃_{β∈H} A_{β}^{∗} ∩ K is not empty. We will instead show that f(⋂_{α∈G} B_{α}^{∗} ∩ K − ⋃_{β∈H} A_{β}^{∗} ∩ K) is not empty. Since f ↾ K is irreducible onto 2^{𝔠},

f( ⋂_{α∈G}B_{α}^{∗} ∩ K − ⋃_{β∈H}A_{β}^{∗} ∩ K) = ⋂_{α∈G}f(B_{α}^{∗} ∩ K) − ⋃_{β∈H}f(A_{β}^{∗} ∩ K) .

Recall that f[B_{α}^{∗}∩K] contains [φ_{α}] and that f[A_{α}^{∗}∩K] is equal to [⟨α, 0⟩]. Therefore we have

⋂_{α∈G} f[B_{α}^{∗} ∩ K] − ⋃_{β∈H} f[A_{β}^{∗} ∩ K] ⊇

⊇ ⋂_{α∈G}[φ_{α}] − ⋃_{β∈H}[⟨β, 0⟩] ⊇

⊇ [ ⋃_{α∈G} φ_{α} ∪ {⟨β, 1⟩ : β ∈ H}]

Since φ = ⋂_{α∈G} φ_{α} ∪ {⟨β, 1⟩ : β ∈ H} is a function, the last set in the above formula is not empty, which finishes the proof of the claim.

To finish the proof, fix, for each α ∈ J, a function h_{α} ∈ C^{∗}(X) such that h_{α}[g(B_{α})] = {0} and h_{α}[X − U_{α}] = {1}. Let h = Π_{α∈J}h_{α} . Since X − U_{α} contains ω − A_{α}, it follows immediately from Claim 2.1 and the compactness of 2^{𝔠} that h is a mapping of X into ℝ^{J} whose image contains 2^{𝔠} . □

3. A Special Ultrafilter

Definition 3.0.3. Given a family F of subsets of some set S. A family I is called a refinement of F if for each F in F , there is an I in I such that I ⊆ F. If I is an almost disjoint family we say I is an almost disjoint refinement of F.

There has been many interesting results about almost disjoint refinements of filters on ω, see section 4 of [1] for a survey. The following Proposition gives an ultrafilter which has a quite unexpected almost disjoint refinement that does not follow from the constructions in [1]. A Noetherian space is a space with a base B such that for each U ∈ B, the set {V ∈ B: U ⊆ V } is finite (see [3]). A base B for an ultrafilter p on ω is a Noetherian base (strong Noetherian base) if for each A ∈ B, the set {B ∈ B: A ⊆ B} ({B ∈ B: A ⊆^{∗} B}) is finite. Malykin has asked if there exists an ultrafilter with a Noetherian base. The base B in the ultrafilter p in the following proposition is actually strong Noetherian. Thus we give a positive answer to Malykin’s question. In addition the property (i) of the filter p is quite unexpected.

Proposition 3.0.4. There is an ultrafilter p on ω which has an almost disjoint refinement I and a base B such that

(i) for each subset I′ ∈ [I]^{<𝔠} there is a member of p which is almost disjoint with each member of I′, and

(ii) for each B ∈ B and I ∈ I either B ∩ I is finite or B almost contains I and I is almost disjoint with all but finitely many members of B.

Remark 3.0.5. It is straightforward to see that the property (ii) in Proposition 3.0.4 implies the base B is a strong Noetherian base. A careful analysis of the construction in [1] will reveal that the almost disjoint refinements constructed by that method will fail to satisfy property (ii) for any base, i.e. for any base, there will be some member of the refinement almost contained in infinitely many members of the base.

The following Lemmas give the proof of the Proposition 3.0.4.

Lemma 3.0.6. If F is a filter base on ω such that, for each F ∈ F , there is an infinite I ⊆ F such that {F′ ∈ F : I ∩ F′is infinite} is finite, then there is an ultrafilter p on ω which extends F and satisfies the following property

F or each F ∈ p , there exists I ∈ [F]^{ω}

such that {F′ ∈ F : I ∩ F′ is infinite } is finite (†)

Proof. Let [ω]^{ω} = {Y_{α} : α ≤ 𝔠}. We will build inductively an increasing sequence {F_{α} : α ∈ 𝔠} of filters on ω such that

(i) F ⊆ F_{0}

(ii) For each α ≤ 𝔠, F_{α} satisfies (†)( with p taken to be F_{α}).
(iii) For each α ∈ 𝔠, either Y_{α} ∈ F_{α+1} or ω \ Y_{α} ∈ F_{α+1},

Let F_{0} be the filter generated by F. Suppose we have constructed {F_{β} : β < α} which satisfies (i)-(iii). If α is a limit ordinal, let F_{α} = ⋃_{β<α} F_{β}. If α is the successor of β, we have to build F_{β+1} which extends F_{β} and satisfies (ii) and (iii).

We say a set Y ⊆ ω is F_{β}-acceptable if for each F ∈ F_{β} there exists an I ∈ [F ∩ Y]^{ω} such that {F′ ∈ F : F′ ∩ I is infinite} is finite.

Claim 3.1. There exists a Y ∈ {Y_{β}, ω \ Y_{β}} which is F_{β}-acceptable.

Proof of Claim 3.1: Suppose not. Then there exist F_{1} and F_{2} in F_{β} such that for each I ∈ [ω]^{ω}, if I ⊆ F_{1} ∩ Y_{β} or I ⊆ F_{2} \ Y_{β}, then {F′ ∈ F : F′ ∩ I is infinite} is infinite. Since F_{β} satisfies (†), there exists an I_{0} ⊆ F_{1} ∩F_{2} such that {F′ ∈ F : F′ ∩ I_{0} is infinite} is finite. Without loss of generality, we can assume I_{0} ∩ Y_{β} is infinite. Then I_{0} ∩ Y_{β} ∈ [ω]^{ω} and I_{0} ∩ Y_{β} ⊆ F_{1} ∩ Y_{β} and {F′ ∈ F : F′ ∩ I ∩ Y_{β} is infinite} is finite. This contradicts the choice of F_{1}.

Now take a Y ∈ {Y_{β}, ω \ Y_{β}} which is F_{β}-acceptable. Let F_{β+1} be the filter generated by F_{β} ∪ {Y }. Obviously, F_{β+1} satisfies (ii) and (iii). Finally, we let p = F_{𝔠}. It is easy see that p works. □

Lemma 3.0.7. Under the condition of the Lemma 3.0.6, if |F| = 𝔠, and p is obtained from Lemma 3.0.6, then Proposition 3.0.4 holds for p.

Proof. Let p = {a_{α} : α < 𝔠} and F = {d_{α} : α < 𝔠}, such that if α ≠ β, then d_{α} ≠ d_{β}. For each F ∈ [ω]^{ω}, denote by I_{F} the set {I ∈ [F]^{ω} : {F′ ∈ F : F′ ∩I is infinite} is finite}. Thus Lemma 3.0.6 says that for each member F of p, I_{F} is not empty. We will construct, by induction, {b_{α} : α < 𝔠} ⊆ p and {I_{α} : α < 𝔠} such that

(i) I_{α} ⊆ b_{α} ⊆ a_{α} ∩ d_{α}, (ii) ∀ξ < α, I_{ξ} ∩ b_{α} is finite, (iii) ∀α < 𝔠{ξ < α : I_{α} ⊆ b_{ξ}} = {ξ < α : I_{α} ∩ b_{ξ} is infinite } is finite, (iv) {F ∈ F : I_{α} ∩ F is infinite } is finite. Suppose first that we have constructed the above sequences, then it is straightforward to see that I and B satisfies the Proposition 3.0.4, where I = {I_{α} : α < 𝔠} and B = {b_{α} : α < 𝔠}. Let b_{0} be a_{0} ∩ d_{0} and I_{0} be any element of I_{b_{0}}. At stage α, we have to find b_{α} and I_{α} which satisfies (i) to (iv). By condition (iv) and the fact that |F| = 𝔠, we can find F ∈ F such that F ∩ I_{ξ} is finite for each ξ < α. Let b_{α} = a_{α} ∩ d_{α} ∩ F. Take any I ∈ I_{b_{α}}. Since for each ξ ∈ α, b_{ξ} ∩ I ⊆ d_{ξ} ∩ I, there are only finitely many ξ such that b_{ξ} ∩ I is infinite. It is easy to find a subset I_{α} of I such that, for each of those finitely many b_{ξ}’s which has infinite intersection with I, either it contains I_{α} or it is disjoint from I_{α}. □

Lemma 3.0.8. There exists a filter base, F, on ω such that |F| = 𝔠 and for each F ∈ F there exists an infinite set I ⊆ F such that {F′ ∈ F : F′ ∩ I is infinite} is finite.

Proof. We will identify ω with the set P = ⋃_{n∈ω} P(^{n}2). For each E ∈ [^{ω}2]^{<ω}, let X_{E} = ⋃_{n∈ω}{A ⊆ ^{n}2: E ↾n ⊆ A} and I_{E} = {E ↾n: n ∈ ω}.

For finite subsets E and E′, it is easy to see that I_{E} ⊆ X_{E} and X_{E∪E′} = X_{E} ∩ X_{E′}. Moreover, if E′ ⊈ E, then, there is an m such that for all n > m, E′ ↾n ⊈ E ↾n. Therefore, I_{E} ∩ X_{E′} ⊆ {E ↾ n: n ≤ m} is finite. Let F = {X_{E} ⊆ P : E ∈ [^{ω}2]^{<ω}}. Clearly F is a filter base on P with cardinality 𝔠 and for each X_{E} in F, I_{E} is an infinite subset of X_{E} which is almost disjoint from all but finitely many members of F. □

Remark 3.0.9. If we take p as in Proposition 3.0.4, the space ω ∪ I ∪ {p} with the usual topology is quite interesting: the first property ensures that it has tightness 𝔠 while the second property says that it is a 0-dimensional scattered space.

References

[1] Bohuslav Balcar and Petr Simon. Disjoint refinement. In J.K. Monk and R. Bonnet, editors, Handbook of Boolean Algebra, chapter 9. Elsevier Science Publishers, B.V., 1989.

[2] Ryszard Engelking. General Topology. Państwowe Wydawnictwo Naukowe (Polish Scientific Publishers), 1977.

[3] V. I. Malykhin. On Noetherian spaces. In P. S. Alexandroff, editor, Seminar on General Topology, pages 51–59. Izd-vo Moskov. Univ., 1981.

[4] Jan van Mill. An introduction to βω. In Kenneth Kunen and Jerry E. Vaughan, editors, Handbook of Set-Theoretic Topology, chapter 11. Elsevier Science Publishers B.V., 1984.

[5] Czesław Ryll-Nardzewski and R. Telgársky. On the scattered compactification Bull Acad Polon Sci, Sér Sci Math Astronom et Phys 18, pages 233-234, 1970.

Department of Mathematics, York University, 4700 Keele Street, North York, Ontario, Canada M3J 1P3

   E-mail address: Alan Dow:  Alan.Dow@@mathstat.yorku.ca
   E-mail address: Jinyuan Zhou: Jinyuan.Zhou@@mathstat.yorku.ca