---

Bounding and dominating number

arXiv:math/9304204v1 [math.LO] 26 Apr 1993Bounding and dominating numberof families of functions on NClaude Laflamme∗Department of Mathematics and StatisticsUniversity of CalgaryCalgary, AlbertaCanada T2N 1N4AbstractWe pursue the study of families of functions on the natural numbers,with emphasis here on the bounded families. The situation being morecomplicated than the unbounded case, we attack the problem by classi-fying the families according to their bounding and dominating numbers,the traditional scheme for gaps.

Many open questions remain.1IntroductionOver the years, the notion of gaps of functions (or sets) of natural numbers hasplayed an important role in the application of Set Theory to different branchesof mathematics, see for example [13] for a survey. It should thus not come asa surprise that families of functions in general might have an impact.

It was infact shown recently in [2] that the structure of directed unbounded families offunctions had an influence on several problems and in [6] an application of non-directed unbounded families was made. In the papers [2, 4, 7, 8], a sufficientlyprecise description of these families was given to address these questions.The next step is to consider bounded families and provide a similar descrip-tion.

Such families are no more than generalizations of the classical notion of(linearly ordered) gaps and the situation appears quite complex; indeed, notonly all partial orders of size ≤ℵ1 embedd as bounded families of functions butunbounded families themselves reflect as bounded ones. We have tried in thispaper to classify the bounded families according to their bounding and dominat-ing numbers, a criterion much weaker than cofinal equivalence for example, butwhich seemed a good starting point but in fact many open questions remain.∗This research was partially supported by NSERC of Canada.1980 AMS Subj.

Class. (1985 Revision) Primary 03E35; Secondary 04A20.Key words and phrases.

Gaps, ultrafilter, forcing1

The ultimate results we hope to achieve are to describe the families thatcan be built from ZFC alone with enough details to be useful for applications;forcing is used to verify that ZFC has been exhausted, i.e. that no other familiescan be built from ZFC.We express our warm thanks to Stevo Todorcevic for discussions on thetopic.2Notation and PreliminariesWe write ω for the set of natural numbers, ωω for the set of all functions on ωand ω↑ω for the set of non-decreasing (monotone) functions.

We often use f ≤gto abbreviate (∀n)f(n) ≤g(n), and f ≤∗g for (∀∞n)f(n) ≤g(n) and similarlyfor < and <∗; here “∀∞n” means “for all but finitely many n” and similarly“∃∞n” means “there exists infinitely many n”. Also important is the orderingf ≺g defined by limn g(n) −f(n) = +∞.

As ⟨ωω, ≺⟩embedds in ⟨ω↑ω, ≺⟩, weshall be interested essentially in the latter structure and we shall assume for theremaining of this paper that we deal with monotone functions only.We use F, G, H to denote families of functions. Further, we shall assume thatthe families considered are closed downward under ≤∗; this simplifies greatlythe discussion without interfering with the results.

In particular the boundingand dominating numbers that we define are the same for a given family or itsdownward closure, and further it suffices to present the generators to describea family.Given two families of functions H ⊆F, we say that H is unbounded in F if:(∀f ∈F)(∃h ∈H) h ≮∗f,and H is said to dominate F if(∀f ∈F)(∃h ∈H) f ≤∗h.We shall in practice prove slightly more, for example that f ≺g instead of onlyf ≤∗g or that lim supn f(n) −g(n) = +∞instead of just f ≮∗g; the reasonis that often for applications two functions are identified if they differ only by afixed natural number. Indeed we shall work mostly with the ≺ordering.A family dominating ωω is usually called a dominating family, and one un-bounded in ωω is simply called an unbounded family.We define the bounding number b(F) of a family F asb(F) = min{|H| : H ⊆F is unbounded in F}and the dominating number d(F) asd(F) = min{|H| : H ⊆F dominates F}.2

b=b(ω↑ω) is the usual bounding number and d=d(ω↑ω) the dominating number.The infinite subsets of ω are denoted by [ω]ω, the standard ordering is A ⊆∗B if A \ B is finite. We shall be interested in almost disjoint families, that isfamilies of infinite sets with pairwise finite intersections.

By fixing a bijectionfrom the rationals and ω and considering for each irrational number a sequenceof rationals converging to it, we see that there is an almost disjoint family ofsubsets of ω of size c, the continuum. Typical functions that will interest us areof the form next(−, X) for some X ∈[ω]ω, wherenext(n, X) = the smallest element of X greater than or equal to n,and similarly for the function last(−, X).An ultrafilter is a proper family of infinite sets closed under finite intersec-tions, supersets, and maximal with respect to those properties; in particular itmust contain X or ω\X for any X ⊆ω, and must be nonprincipal.

We use U, Vto denote ultrafilters. We write χ(U) for the minimal cardinality of a collectiongenerating the ultrafilter U, and u for the least cardinality of a family of setsgenerating any ultrafilter.

A Pκ-point is an ultrafilter U with the property thatany κ decreasing sequence from U has a lower bound in U.3Unbounded FamiliesWe shall consider in this section three sorts of unbounded (closed downward)families (of monotone functions).Definition 3.11. The D-class (the dominating class): F ∈D iffF is dominating.2.

The S-class (the superperfect class): F ∈S iffi) (∃h)(∀f ∈F)(∃∞n )[f(n) ≤h(n)]ii) (∃g)(∀f)[(∃∞n )f(n) ≤g(n) →f ∈F].3. The U-class (for an ultrafilter U): F ∈U iffi) (∃h)(∀f ∈F)[{n : f(n) ≤h(n)} ∈U].ii) (∃g)(∀f)[{n : f(n) ≤g(n)} ∈U →f ∈F].These three classes are easily seen to be distinct and we have shown in [7, 8]the relative consistency of any unbounded family of functions falling into oneone these classes; that is in ZFC alone, no other unbounded family of functionscan be obtained and even a single ultrafilter of your choice may be used for allmembers of the 3rd class.

This fulfills our original motivation for unbounded3

families, in other words these descriptions are sufficiently detailed to provideanswers to many general mathematical problems (see [2, 4, 6, 8]).The reason to pursue their studies here is their influence on bounded fam-ilies as we will see in the next section. So we now discuss the bounding anddominating number of families in these three classes.

If F is dominating, thenb(F) =b, the usual bounding number and d(F) =d, the usual dominating num-ber. Although it is possible to make structural distinctions between dominatingfamilies, applications have not made them yet necessary to analyze.

If we de-mand that our families be closed downward under ≤∗, then there is only onedomimnating family, namely ω↑ω.We now turn to the S-class. In [10], Kechris showed that any unboundedBorel family must contain a superperfect tree, and we showed in [7] that anynon-dominating family containing a superperfect tree belongs to the S-class.Proposition 3.2 If F is in the S-class, then b(F)= 1 or 2 and d=c.

Furtherthese two values of b are attainable.Proof:3.2.1: We first show that b(F)≤2 if F belongs to the S-class.The point is that a bounding number of at least 3 means that the family isdirected; it thus suffices to show that a directed family sastisfying 2ii) in theS-class is dominating, a contradiction.So let F belong to the S-class and witnessed by g and h as in definition 3.1.Fixing any p ∈ω↑ω, define a sequence of integers by π0 = 0 and more generallysuch that g(πn+1) > p(πn). If we now let Xi = {π2n+i : n ∈ω} for i = 0, 1and define fi(n) = g(next(n, Xi)) ∈F, then max{f0(n), f1(n)} ≥p(n) for eachn.

Since p was arbitrary, we see that F must be dominating if directed, i.e. ifb(F)≥3.3.2.2: We now show that d(F) =c if F belongs to the S-class.Again, let F belong to the S-class and witnessed by g and h. Choose alsoan increasing sequence of integers X = {πn : n ∈ω} such that g(πn+1) >h(πn) for each n. Now for any infinite Y ⊆X, consider the function fY (n) =g(next(n, Y )); this function must belong to F as it is equal to g infinitely often.Observe however that for p ∈F such that fY ≤∗p,Sp = {n : p(n) ≤h(n)} ⊆∗S(Y ) =[{(πn−1, πn] : πn ∈Y }.Since moreover Y ∩Y ′ =∗∅→S(Y ) ∩S(Y ′) =∗∅, choosing an almost disjointfamily of infinite subsets of X of size c shows that c functions are necessary todominate F.3.2.3: An example of F in the S-class with b(F)=1 (and d(F)=c).Fix any unbounded function g ∈ω↑ω and let F = {f ∈ω↑ω : (∃∞n)f(n) ≤g(n)} = {g(next(−, X)) : X ∈[ω]ω}.

Since g itself is unbounded in F (andbelongs to F), we have b(F)=1.4

3.2.4: An example of F in the S-class with b(F)=2 (and d(F)=c).Consider the identity function id(n) = n and for X ∈[ω]ω lethX(n) = next(n, X) + |Xc ∩n|and finally put F = {hX : X ∈[ω]ω}.The fact that F belongs to the S-class is witnessed by the functions g = idand h(n) = 2n.Then b(F)≤2 and it thus suffices to show that no singlemember of F is unbounded. But given hX ∈F, choose an infinite Y ⊆X suchthat X \ Y is also infinite.

Then for each N and n large enoughhY (n)= next(n, Y ) + |Y c ∩n|≥next(n, X) + |Y c ∩n|≥next(n, X) + |Xc ∩n| + N= hX(n) + N.Thus hX ≺hY and hence b(F)=2. The proof of 3.2.2 will actually give you twospecific functions unbounded in F. This completes the proof of Proposition 3.2.✷We now turn our attention to members of the U-class.Proposition 3.3 For any ultrafilter U, λ ∈{1, 2, ω} and χ(U) ≤κ ≤c, thereis a family F in the U-class such that b(F)= λ and d(F)= κ.Proof: First choose two functions g, h ∈ω↑ω and an increasing sequence ofintegers ⟨πn : n ∈ω⟩such that:1. g ≺h2.

limn h(n + 1) −[h(n) + n] = +∞3. (∀n) h(πn) ≥g(πn+2)4.

(∀n) πn+1 −πn ≥2n.Also fix an almost disjoint family A = ⟨Aα : α

let P(X) = ∪{[π2n−1, π2n) : X ∩[π2n, π2n+1) ̸= ∅.With these preliminaries we are ready to build the desired families.3.3.1: We build an F in the U-class such that b(F)=1 and d(F)= κ whereχ(U) ≤κ ≤c.For X ∈U such that X ⊆E and any α < κ, let5

f Xα (n) = h(next(n, [Acα ∩P(X)] ∪X))and put F = {f Xα : X ⊆E, X ∈U, α < κ}. Observe that g(next(−, X)) ≤f Xαfor any α and that if X, α are given, then f Xα (n) = h(x) for any x ∈X andtherefore g, h witness that F belongs to the U-class.As h is unbounded in F, we conclude readily that b(F)=1.

We must nowshow that the dominating number is κ. If B is a base for the ultrafiler U, thenH = {f Xα : X ∈B, α < κ}clearly dominates the family F and therefore d(F)≤χ(U) · κ = κ.

On the otherhand, fix a family H ⊆F of size less than κ, and fix some ordinal β ∈κ notmentionned in any indexing of the functions from H. But if X ∈U, X ⊆Eand α ̸= β, then f Xα does not dominate f Eβ . Indeed the set Acα ∩P(X) \ Acβis infinite as otherwise P(X) ∩Aβ ⊆∗Aα, and since P(X) is an infinite unionof intervals of the form [πn, πn+1), P(X) ∩Aβ is infinite and thus Aβ ∩Aα isinfinite contradicting that A is an almost disjoint family.But now for any x ∈Acα ∩P(X) \ Acβ, f Xα (x) = h(x) and f Eβ (x) ≥h(x + 1)and as limn h(n+1)−h(n) = +∞we get lim supn f Eβ (n)−f Xα (n) = +∞as well.Therefore, no member of H dominates the function f Eβ ∈F and we concludethat d(F)≥κ and thus d(F)= κ.3.3.2: We build an F in the U-class such that b(F)=2 and d(F)= κ for anyχ(U) ≤κ ≤c.For X ∈U such that X ⊆E and any α < κ, letf Xα (n) = h(next(n, [Acα ∩P(X)] ∪X)) + |Xc ∩n|and put F = {f Xα : X ⊆E, X ∈U, α < κ}.

For any f Xα ∈F and x ∈X,wehave f Xα (x) ≤h(x) + x and therefore g and h′(n) = h(n) + n witness that Fbelongs to the U-class.We first show that the bounding number is 2. No f Xα itself is unbounded inF since choosing Y ∈U such that X \ Y is infinite, we get for each N and nlarge enoughf Xα (n)= h(next(n, [Acα ∩P(X)] ∪X)) + |Xc ∩n|≤h(next(n, [Acα ∩P(Y )] ∪Y )) + |Xc ∩n|≤h(next(n, [Acα ∩P(Y )] ∪Y )) + |Y c ∩n| −N= f Yα (n) −Nand therefore f Xα ≺f Yα .However, we claim that for any α ̸= β, the pair {f Eα , f Eβ } is unbounded inF.

To verify this, we consider any f Xγ∈F, without loss of generality α ̸= γ.For any x ∈Acγ ∩P(X) \ Acα which is infinite, we have f Xγ (x) ≤h(x) + x andf Eα (x) ≥h(x + 1). But as lim supn h(n + 1) −[h(n) + n] = +∞by assumption6

we get lim supn f Eα (n) −f Xγ (n) = +∞as well. The fact that d(F)= κ is provedas in the previous example.3.3.3: We build an F in the U-class such that b(F)= ω and d(F)= κ whereχ(U) ≤κ ≤c.For any a ∈[κ]<ω and X ∈U such that X ⊆E we letf Xα (n) = h(next(n, [\α∈aAcα ∩P(X)] ∪X)) + |Xc ∩n|and put F = {f XX ⊆E, X ∈U, a ∈[κ]<ω}.

Observe that for any such f Xaand x ∈X ⊆E, f Xa (x) = h(x) + |Xc ∩x| ≤h(x) + x and therefore g andh′(n) = h(n) + n again witness that F belongs to the U-class.Our first task is to show that F is directed and therefore b(F)≥ω. But givenf Xa and f Yb , put c = a∪b and choose Z ∈U such that Z ⊆X ∩Y and X ∩Y \Zis infinite; then f Zc ≻f Xa , f Yb .

To show now that b(F)≤ω, choose A ∈[κ]ω andwe prove that the collection H = {f Eα : α ∈A} ⊆F is unbounded in F. So letus fix f Xa∈F and choose β ∈A \ a. Now the set [Tα∈a Acα ∩P(X)] \ Acβ isinfinite as otherwise we would obtain P(X)∩Aβ ⊆∗Sα∈a Aα, and as P(X)∩Aβis infinite Aβ would have infinite intersection with some Aα contradicting thatA is an almost disjoint family.

But now for x ∈[Tα∈a Acα ∩P(X)] \ Acβ,f Xa (x) = h(next(x, [\α∈aAcα ∩P(X)] ∪X)) + |Xc ∩x| ≤h(x) + x,f Eβ (x) = h(next(x, Acβ ∩P(E)] ∪E)) + |Ec ∩x| ≥h(x + 1).As limn h(n+ 1)−[h(n)+ n] = +∞, we get that lim supn f Eβ (n)−f Xa (n) = +∞as well and H is indeed unbounded in F.The verification the the dominating number is κ is again very similar to thefirst example.This completes the proof of Proposition 3.3.✷Corollary 3.4 For any λ ∈{1, 2, ω} and u ≤κ ≤c, there is a family F in theU-class such that b(F)= λ and d(F)= κ.The next problem is whether we can construct a family F in the U-classwith an uncountable bounding number. We show that this requires a P-pointand therefore, in view of Shelah’s consistency result [14] that there might be nosuch P-points, we cannot construct such families in ZFC alone.Proposition 3.5 If F in the U-class has an uncountable bounding number,then there is a finite-to-one map m such that m(U) is a P-point.Proof: Fix functions g and h witnessing that F belongs to the U-classand define a sequence of integers such that π0 = 0 and more generally such7

that g(πn+1) > h(πn). We may assume without loss of generality that E =Sn[π2n, π2n+1) ∈U as the other case is analogous.

Now for any X ∈U, ifX ⊆E, any f ∈F with g(next(−, X)) ≤∗f must satisfyS(f) = {n : f(n) ≤h(n)} ⊆∗T (X) =[{[π2n−1, π2n+1) : X ∩[π2n, π2n+1) ̸= ∅}Now define a map m ∈ω↑ω by m”[π2n−1, π2n+1) = n and consider V = m(U).Certainly V is a (non principal) ultrafilter as m is finite-to-one. To show itis actually a P-point, let {Yn : n ∈ω} ⊆V be given and consider the setsXn = m−1(Yn) ∩E ∈U.

Since we are assuming that the bounding number ofF is uncountable, fix a function f ∈F such that g(next(−, Xn)) ≤∗f for eachn. Therefore S(f) ⊆∗T (Xn) for each n and thus m(S(f)) ⊆∗m(T (Xn)) ⊆Yn.Since moreover S(f) ∈U and therefore m(S(f)) ∈m(V), the proof is complete.✷Under the existence of P-points or more generally Pκ points, one can easilyconstruct members of the U-class with bounding number κ by fixing some g ∈ω↑ω and defining F = {g(next(−, X)) : X ∈U}.

Thus in general we have:Proposition 3.6 There is a Pκ ultrafilter if and only if there if a family F inthe U-class with bounding and dominating number κ.We can also deduce from the proof of Proposition 3.5 that d(F)≥u for anyfamily F in the ultrafilter class, but I do not know if d(F)≥χ(U) whenever Fbelongs to the U-class with witness U.4Bounded FamiliesLet F be a bounded family and letF↓= {g ∈ω↑ω : (∀f ∈F) f ≺g}Certainly F↓is nonempty as F is bounded and the pair (F, F↓) forms a gap inthe sense that there is no h ∈ω↑ω such that(∀f ∈F)(∀g ∈F↓) f ≺h ≺g.To make a first distinction between bounded families, we make the followingdefinition.Definition 4.1b↓(F)= min{|H| : H ⊆F↓and H is unbounded in F↓in the reverse order }= min{| H |: ¬(∃g ∈F↓)(∀h ∈H) g ≺h}8

We loosely call b↓(F) and for that matter F ↓depending on the context theupper bound of F and we will classify the families according to this cardinalb↓(F) which takes either the value 1 or an infinite regular cardinal; notice thatthe value 2 cannot occur here. Observe also that if H ⊆F↓is unbounded in F↓in the reverse order as above, then the pair (F, H) is also a gap.

Much work hasbeen done on gaps (F, H) for which both F and H are linearly ordered by ≺;in particular gaps (F, G) for which b(F)=d(F). Such gaps are usually qualifiedas (b(F), b↓(F)) gaps.

Here we will work in a more general situation.4.1Bounded families with a countable upper boundUnbounded families have much influence on the bounded ones; we can use theresults of §3 to construct families with countable upper bounds and variousbounding and dominating numbers.Proposition 4.2 There are families F with countable upper bounds, that isb↓(F) = 1 or ω, such that:1. b(F)=b and d(F)=d.2. b(F)=1 or 2 and d(F)=c.3.

b(F)=1, 2 or ω and u≤d(F) ≤c.Proof: The goal of the proof is to build families F with the same boundingand dominating number as the families from §3; we fix for our constructions thefunctions g(n) = n2 and more generally for ℓ∈ωgℓ(n) = n2 −ℓlog n or gℓ(n) = n2 −ℓdepending on the context.We shall build gaps (F, {gℓ: ℓ∈ω}) giving us familes F with upper bounds1 or ω, depending on which collection {gℓ: ℓ∈ω} one chooses, and with theapropriate bounding and dominating numbers.Observe first that irrespective of the collection we choose, we havegℓ+1(n + 1) ≥gℓ(n)for each n and ℓ; this will make our verifications easier.Now if H is anyunbounded family and h ∈H, letfh(n) = gm(n) if h(m −1) < n ≤h(m)and put F(H) = {fh : h ∈H}. In this case we have :Claim 4.3 b(H)=b(F(H)) and d(H)=d(F(H)).9

Proof: It suffices to prove that for all h1, h2 ∈H, we haveh1 ≤∗h2 ifffh1 ≤∗fh2.To verify this, suppose first that h1(m) ≤h2(m) for all m ≥M and fix n ≥h2(M).Choose first m such thath1(m −1) < n ≤h1(m)and ℓsuch thath2(ℓ−1) < n ≤h2(ℓ).Observe that we must have ℓ≤M and thusfh1(n) = gm(n) ≤gℓ(n) = fh2(n).Suppose now for the other direction that fh1(n) ≤fh2(n) for all n ≥h2(N)and fix n ≥N; we show that h1(n) ≤h2(n). But if for the sake of a contradictionwe have h2(n) < h1(n), pick ℓ≥n + 1 such that h2(ℓ−1) < h1(n) ≤h2(ℓ).Thenfh1(h1(n)) = gn(h1(n))andfh2(h1(n)) = gℓ(h1(n)) ≤gn+1(h1(n)) < gn(h1(n))and we obtain the desired contradiction.

This proves the claim.The Proposition is now proved by replacing H by the apropriate families of§3. Actually, to obtain F(H) ⊆ω↑ω, we should first replace the families H byH′ = {h(n) + n : h ∈H} for example to ensure that we have strictly increas-ing functions; observe that this does not affect the bounding and dominatingnumber.✷There is however more than just reflecting unbounded families to boundedones, indeed let us see how close we are.

Let F be a family of functions and{gn : n ∈ω} a collection such that F ≺gn+1 ≤∗gn for each n, and assumewithout loss of generality that gn+1(k) + 1 ≤gn(k) for each k and n.Forf ≺{gn : n ∈ω}, we definehf(n) = max{k : gn(k) ≤f(k)}and put H(F) = {hf : f ∈F}. The following proposition, due to Rothberger,shows that unbounded families are always involved somehow.Proposition 4.4 (Rothberger) The pair (F, {gn : n ∈ω}) is a gap if and onlyif H(F) is an unbounded family.10

Proof: Suppose first that the family H(F) is bounded, say by h; we mightas well assume that n < h(n) < h(n + 1) for each n. Define a function p by:p(j) = gm(j) where m is the smallest integer such that h(m + 1) > j.As j increases, m increases as well and therefore p ≺gm for each m. Now forany f ∈F, and therefore for hf ∈H, choose N large enough so that(∀n ≥N) hf(n) < h(n).Hence for all m ≥h(N), if we let ℓ≥N be as large as possible such thatm ≥h(ℓ), we obtain:m ≥h(ℓ) > hf(ℓ)and thereforef(m) < gℓ(m) = p(m).We conclude that F ≺p ≺{gn : n ∈ω} and thus the pair (F, {gn : n ∈ω}) isnot a gap.For the other direction, since we have the implicationf ≤∗f ′ →hf ≤∗hf ′,we conclude readily that H(F) is bounded if the pair (F, {gn : n ∈ω}) is not agap.✷Corollary 4.5 d(F) ≥b for any F with countable upper bound.Since f ≤∗f ′ →hf ≤∗hf ′, we also obtainCorollary 4.6 d(H(F)) ≤d(F) and b(F) ≤b(H(F)) unless b(H(F)) = 1 inwhich case b(F) ≤2.This allows us to extend Proposition 4.2 as follows.Proposition 4.7 Let H be any unbounded family and λ ≤b a regular (infinite)cardinal. Then there is a family F with countable upper bound such thatb(F) = min{λ, b(H)}and d(F) = d(H).Proof: To simplify the calculations, we fix the functions gk(n) = nn −knfor k ∈ω and an increasing sequence of sets ⟨Xα : α < λ⟩such that Xβ \ Xα isinfinite whenever α < β; this is guaranteed by λ ≤b.Without loss of generality, we may assume that each h ∈H is strictly increasing,that h(n) > n for each n and that the range is included in X0.

Now for h ∈Hand α < λ, definefh,α(n) = gm(last(n, Xα)) + |Xα ∩n| where h(m −1) < n ≤h(m)and put F = {fh,α : h ∈H, α < λ}. As H(F) = H, we conclude from Corollary4.6 that (F, {gk : k ∈ω}) is a gap and that b(F) ≤b(H) + 1 and d(F) ≥d(H).11

Claim 4.8 b(F) ≤λ.Proof: Fix h ∈H and let S = {fh,α : α < λ}. We show that S (⊆F) isunbounded in F. Indeed, fix any h′ ∈H and any α < λ and consider any β,α < β < λ; we claim that fh,β(n) ≥fh′,α(n) for infinitely many n, indeed onalmost all x ∈Xβ \ Xα.

For fix such an x, if h′(m −1) < x ≤h′(m), thenfh′,α(x)= gm(last(x, Xα)) + |Xα ∩x|≤gm(x −1) + x= (x −1)x−1 −m(x −1) + xand if h(ℓ−1) < x ≤h(ℓ) thenfh,β(x)= gℓ(last(x, Xβ)) + |Xβ ∩x|≥gℓ(x) = xx −ℓx.As ℓ, m ≤x, we get fh,β(x) ≥fh′,α(x) for almost all such x’s. This proves theclaim.Claim 4.9 b(F) ≥min{λ, b(H)}.Proof: Let S ⊆F, |S| < min{λ, b(H)}, and fix ζ < λ, T ⊆H such thatS ⊆{fh,α : h ∈T , α < ζ} and |T | < b(H).Therefore choose an h′ ∈H such that h <∗h′ for any h ∈T and we show thatfh,α <∗fh′,ζ for all h ∈T and α < ζ, and thus S is bounded in CF.Choose first N ∈Xα such that Xα \ N ⊆Xζ and fix n ≥N; if m is suchthath′(m −1) < n ≤h′(m),and ℓsuch thath(ℓ−1) < n ≤k(ℓ)we obtain, with x = last(n, Xζ),fh,α(n)= gℓ(last(n, Xα)) + |Xα ∩n|≤gℓ(last(n, Xζ)) + |Xα ∩n|= gℓ(x) + |Xα ∩n|= xx −ℓx + |Xα ∩n|andfh′,ζ(n)= gm(last(n, Xζ)) + |Xζ ∩n|= gm(x) + |Xζ ∩n|= xx −mx + |Xζ ∩n|.But m ≤ℓ(for n large enough) and as Xζ \ Xα is infinite, we get fh,α(n)

This proves the claim.Finally, as we already know that d(F) ≥d(H), we must show the reverseinequality. But F is generated by λ×d(H) =d(H) functions, and the proof iscomplete.✷12

This gives an idea of what can be done in terms of bounded families withcountable upper bound, they all involve unbounded families by proposition 4.4,but this is only very partial information and a lot of freedom remains.4.2Families with upper bound ω1One of the surprising construction in ZFC is a gap build by Hausdorffwhich hasbounding and dominating number ω1. Lusin build one with bounding number1 and dominating number ω1; it is this construction that we will adapt toproduce gaps with various bounding and dominating numbers.Although inboth Hausdorff’s and Lusin’s construction the upper bound is at most ω1, I donot know if could be ω; if b> ω1, they certainly cannot.Proposition 4.10 For each λ ∈{1, 2, ω} and ω1 ≤κ ≤c, there is a family Fwith upper bound at most ω1 such that b(F)=λ and d(F) = κ.Proof:We first build {fα : α < ω1} and {gα : α < ω1} such that:1: (∀α < β) fα + id ≺gβ ≺gα, where id is the identity function id(n) = n.2: (∀α)(∀a ∈[ω1 \ {α}]<ω) lim supn fα(n) −maxγ∈a{fα(n) + n} = +∞.3: (∀α)(∀n) fα(n) ≤gα(n).4: (∀α < β)(∃n) fα(n) > gβ(n).Let us first observe that this construction, essentially due to Lusin, will giveus a gap.Claim 4.11 The collection ⟨{fα : α < ω1}, {gα : α < ω1}⟩is a gap.Proof: Suppose on the contrary that {fα : α < ω1} ≺h ≺{gα : α < ω1}for some function h. Choose X ∈[ω1]ω1 and n so thata: (∀α, β ∈X) fα ↾n = fβ ↾n and gα ↾n = gβ ↾n.b: (∀m ≥n) fα(m) ≤gβ(m).Thus (∀α < β ∈X)(∀k)fα(k)= fβ(k) ≤gβ(k)if k < nfα(k)≤gβ(k)if k ≥nBut this contradicts requirement 4.✷If we can accomplish this construction, we put F1 = {fα : α < ω1}, G = {gα :α < ω1} and we get a gap (F, G) with b(F1) = 1 and d(F1) = ω1.

Choosing13

functions 0 ≺hn ≺hn+1 ≺id and using F2 = {fα + hn : α < ω1, n ∈ω},we obtain a family with b(F2) = 2 and d(F2) = ω1.Finally, we let Fω ={maxα∈a{fα} + hn : a ∈[ω1]<ω, n ∈ω} we obtain a family with b(F) = ω andd(F) = ω1. To obtain familes with various dominating number, fix for examplef0 and choose a set X = {xn : n ∈ω} such that f0(xn+1) > f0(xn) + xn and letA = {Aα : α < κ} ⊆P(X) an almost disjoint family.

Assume further that weactually have 0 ≺2hn ≺2hn+1 ≺id. Now for β < κ, definef β0 (n) = max{f0(n), f0(last(n, Aβ)) + 12last(n, Aβ)}.Notice that for β ̸= β′, if xn+1 ∈Aβ \ aβ′,f β0 (xn+1)= f0(xn+1) + 12xn+1f β′0 (xn+1)≤max{f0(xn+1), f0(xn) + xn}= f0(xn+1)and therefore f β0 (xn+1) −f β′0 (xn+1) ≥xn+1 and hence for each mlim supkf β0 (k) −[f β′0 (k) + hm(k)] = +∞.We can then replace f0 in the above families by {f β0 : β < κ} to obtain familieswith dominating number κ.The construction proceeds by induction on α, that is we start withf0(n) = n and g0(n) = n2Now assume that we have already defined the functions {fξ : ξ ∈α} and{gξ : ξ ∈α} such that:2.1: (∀β < γ < α) fβ + id ≺gγ ≺gβ.2.2: (∀β < α)(∀a ∈[α \ {β}]<ω) lim supn fβ(n) −maxγ∈a{fγ(n) + n} = +∞.2.3: (∀β < α)(∀n) fβ(n) ≤gβ(n).2.4: (∀β < γ < α)(∃n) fβ(n) > gγ(n).and we proceed to build fα and gα in countably many steps.

As α is countable,we list α × [α]<ω as {⟨αk, ak⟩: k ∈ω}, {fβ : β < α} as {f k : k ∈ω} and{gβ : β < α} as {gk : k ∈ω}.At stage N, suppose that we have fα ↾n and gα ↾n for some n, such that:3.1: (∀k < N)(∃m < n) fαk(m) −maxγ∈(ak∪{α})\{αk}{fγ(m) + m} ≥N,14

3.2: (∀k < N)(∃m < n) fα(m) −maxγ∈ak{fγ(m) + m} ≥N,3.3: (∀m < n) fα(m) ≤gα(m),3.4: (∀k < N)(∃m < n) f k(m) > gα(m).We will also ensure that for m ≥n3.5: maxk n such that:1. f N(m0) −maxk N.2.

f N(m0) > gα(n −1).Then we define, for n ≤m ≤m0,gα(m)= maxk mi such that:1. fαi(pi) −maxγ∈ai\{αi}{fγ(pi) + pi} ≥N + 1.2. maxγ∈ai\{αi}{fγ(pi)} ≥fα(mi).Then we define, for mi ≤m ≤pi,fα(m)= fα(mi)gα(m)= maxk pN large enough so that for all m ≥n′, we havegα(pN)≤max{maxγ∈∪k

I do not knowif there is always such a family F with large dominating number, say d(F) =cfor example.15

4.3Families with upper bound bIn view of Rothberger’s result and the fact that the smallest size of an unboundedfamily in ω↑ωis b, it is not at all surprising that this cardinal has a role to playin bounded families. We have the following result.Proposition 4.12 For any λ ∈{1, 2, ω}, and λ ≤κ ≤c, there is a family Fwith upper bound b such that b(F)= λ and d(F)= κ.Proof: We provide a general construction which will work for all values ofλ and κ.Fix an increasing unbounded family ⟨hα : α < b⟩and let f(n) = n2.

Nowfor ℓ∈ω definefℓ(n) = n2 + ℓlog(n), and thus fℓ≺fℓ+1and for α < b putgα(n) = fm(n) = n2 + m log(n) if hα(m −1) < n ≤hα(m).These functions are technically not in ω↑ω because of the log function, but onecould easily take instead the smallest integer greater than or equal to thesevalues. Notice thatfℓ≺fℓ+1 ≺gβ ≺gα for all α < β and ℓ.Claim 4.13 For all X ∈[ω]ω, the pair ⟨{fℓ↾X : ℓ∈ω}, {gα ↾X : α < b}⟩isa gap.Proof of the claim: Suppose otherwise that there is a function h such that(∀ℓ)(∀α) fℓ↾X ≺h ≺gα ↾X.Then we definep(n) = min{x ∈X : (∀y ∈X \ x)h(y) > fn(y)}It now suffices to show that hα ≤∗p for each α to obtain a contradiction.But fix N large enough so that(∀x ∈X) x ≥N →h(x) < gα(x)So for each n with p(n) ≥N we must have p(n) ≥hα(n) as well; indeed, ifx = p(n) < hα(n), we geth(x) = h(p(n)) > fn(x).16

Now choose m ≤n such that hα(m −1) < x ≤hα(m), thengα(x) = fm(x) ≤fn(x)and therefore gα(x) < h(x), a contradiction. This proves the claim.✷Now let A = {Aα : α < κ} be an almost disjoint family and for each α < κand ℓ∈ω, letf ′α,ℓ(k) = fℓ(last(n, Aα))Then certainly f ′α,ℓ≺f ′α,ℓ+1.

Further, if α ̸= β and ℓ, k are given, pick n ∈Aβ \ Aα, and thusf ′β,ℓ(n) = fℓ(n) = n2 + ℓlog(n)but as last(n, Aα) ≤n −1 we obtainf ′α,k(n) ≤(n −1)2 + k log(n −1) = n2 −2n + 1 + k log(n −1)and hence lim supn f ′β,ℓ(n) −f ′α,k(n) = +∞.If we now let F = {f ′α,ℓ: α < κ, ℓ∈ω}, we obtain a family with upperbound b, bounding number 2 and dominating number κ × ω = κ.On the other hand if we let F = {max{f ′α,ℓ: α ∈a} : a ∈[κ]<ω, ℓ∈ω},we obtain a family with again upper bound b but bounding number ω anddominating number κ.Moreover, if we had used the functions fℓ(n) = n2 + ℓinstead, then thefamily F = {f ′α,0 : α < κ} would constitute a family with upper bound b,bounding number 1 and dominating number κ.This completes the proof of Proposition 4.12.✷The obvious question now is whether we can have a family with upper boundb and uncountable bounding number; we shall see that there is no such family inthe Mathias model and hence such familes cannot be constructed in ZFC alone.5Models with few families of functionsWe shall be interested in two forcing notions.Definition 5.1Mathias forcing M1 = {⟨a, A⟩: a ∈[ω]<ω, A is an infinite subset of ω disjointfrom a} equipped with the ordering⟨a, A⟩≤⟨b, B⟩iffA ⊆B, b ⊆a, and a \ b ⊆B.17

Matet forcing M2 = {⟨a, A⟩: a ∈[ω]<ω, A is an infinite set of pairwise dis-joint finite subsets of ω} equipped with the ordering⟨a, A⟩≤⟨b, B⟩iffb ⊆a, a \ band members of A are finite unions of elements of B.We use M1 and M2 to denote the models obtained from a model of CH by anℵ2 iteration with countable support of the (proper) partial orders M1 and M2respectively.It is known from [3] that M2 satisfies u

By Proposition 4.7, we get families with countable upper bound, boundingnumber ω1 or ω2 and dominating number ℵ2 by using H = ω↑ω. There areno such families with d(F) = ω1 by Corollary 4.5.For families with upperbound ω1 or ω2 =b, Propositions 4.10, 4.12 and Hausdorff’s result providea general context, although I do not know if M1 has a family with upperbound and bounding number ω1, and dominating number ℵ2; there is howeveran unbounded family F in M1 with b(F) = ω1 and d(F) = ω2.The twoPropositions above 5.2 and 5.3 justify our remark of §4.3 that no gaps withupper bound b has uncountable bounding number in M1.

Indeed, a standardargument would force such a family F to reflect to some Fα = F ∩M1[Gα] forsome α < ω2 where b(Fα) =b↓(Fα) = ω1 and be equivalent in this model toa linearly ordered (ω1, ω1) gap. Since this gap would be preserved to M1, weobtain b↓(F) = ω1, a contradiction.In M2 we have a little more:Proposition 5.4 In M2 there are no <∗-increasing or <∗-decreasing chainsof length ω2.18

Proof: It suffices to prove that there are no increasing chains of size ω2. LetF be such a chain.

If F is unbounded, it would have to belong to one of the 3classes described in §3. Clearly F cannot be dominating as b would then have tobe ω2 in this model; F cannot belong either to the S-class as b(F) = ω2 > 2.

Iffinally F would belong to the U-class, then Proposition 3.6 would provide us witha Pℵ2-point which do not exist in M2 by [4]. Therefore F must be bounded andthe above preservation results show that b↓(F) is countable; then Rothberger’sresult, Proposition 3.2, produces an unbounded <∗-increasing chain H(F) ofsize ω2 which we have just showed does not exist.✷This provides an alternative model to Theorem 3.1 of Shelah and Steprans[16] showing the failure of Nyikos’ axiom 6.5.

Indeed the above shows that anyfamily has an unbounded susbset of size at most ω1 and since NCF holds in M2as it follows from u

For λ = ω1, Proposition 4.2 again givesus F countable upper bound, bounding number ω1 and dominating number ℵ2;as u=ℵ1

Now for families with upper bound ω1, therethose with bounding number λ = 1,2 or ω and dominating number between λand ω2 by Propositions 4.10 and 4.12 as b= ω1. Hausdorff’s result provides onewith bounding and dominating number ω1 and again I do not know if there isone with (upper bound ω1) bounding number ω1 and dominating number ω2.There are no families with upper bound ω2.References[1] A. Blass, Near coherence of filters I. Notre Dame J.

Formal Logic 27 (1986),579-591. [2] A. Blass, Near coherence of filters II.

Trans. Amer.

Math Soc. 300 (1987),557-581.

[3] A. Blass, Applications of superperfect forcing and its relatives, in Set Theoryand its Applications, Lecture Notes in Math 1401 (1989), Springer Verlag. [4] A. Blass and C. Laflamme, Consistency results about filters and the numberof inequivalent growth types, J. Symb.

Logic 54 (1989), 50-57. [5] A. Blass and S. Shelah, There maybe simple Pℵ1 and Pℵ2-points and theRudin-Keisler ordering may be downward directed.

Ann.Pure Appl. Logic33 (1987), 213-243.19

[6] W. Just and C. Laflamme, Classifying sets of measure zero with respect totheir open covers. Trans.

Amer. Math Soc.

321 (1990), 621-645. [7] C. Laflamme, Equivalence of families of functions on the natural numbers.Trans.

Amer. Math Soc.

330 (1992), 307-319. [8] C. Laflamme, Some possible covers of measure zero sets.

Coll. Math.

63(1992), 211-218. [9] K. Kunen, Set theory.

North Holland, Amsterdam 1980. [10] A. Kechris, On a notion of smallness for subsets of the Baire space.

Trans.Amer. Math Soc.

229 (1977), 191-207. [11] R. Laver, Linear orders in (ω)ω under eventual dominance.

Logic Collo-quium 78, North Holland, Amsterdam, 1979. [12] M. Rabus, Tight gaps in P(ω).

To appear, 1992. [13] M. Scheepers, Gaps in (ωω, ≺).

To appear, 1992. [14] S. Shelah.

Proper Forcing. Lecture Notes Math 940 (1982), Springer Verlag.

[15] S. Shelah, On cardinal invariants of the continuum, in “Axiomatic SetTheory”, Amer. Math.

Soc., 31 (1984). [16] S. Shelah and J. Steprans, Maximal chains in ωω and ultrapowers of theintegers.

To appear, 1992.20

---

출처: arXiv:9304.204 • 원문 보기