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FINITE COMBINATIONS OF BAIRE NUMBERS

arXiv:math/9209207v1 [math.LO] 16 Sep 1992FINITE COMBINATIONS OF BAIRE NUMBERSAvner LandverAbstract. Let κ be a regular cardinal.

Consider the Baire numbers of the spaces(2θ)κ for various θ ≥κ. Let l be the number of such different Baire numbers.

Modelsof set theory with l = 1 or l = 2 are known and it is also known that l is finite. Weshow here that if κ > ω, then l could be any given finite number.The Baire number of a topological space with no isolated points is the minimalcardinality of a family of dense open sets whose intersection is empty.

The Bairenumber (also called the Nov´ak number [V]) of a partial order is the minimal car-dinality of a family of dense sets that has no filter [BS] (i.e. no filter on the givenpartial order intersecting all these dense sets non-trivially).

Fnκ(θ, 2) is the collec-tion of all partial functions p : θ →2 such that |p| < κ, and is partially orderedby reverse inclusion. For κ regular and θ ≥κ we consider the spaces (2θ)κ whosepoints are functions from θ to 2 and a typical basic open set is {f : θ →2 | p ⊂f}where p ∈Fnκ(θ, 2).

We denote the Baire number of (2θ)κ by nθκ. It is not hardto see that nθκ is also the Baire number of Fnκ(θ, 2).

Let us now list some knownfacts (see [L] §1).Facts. Let κ be a regular cardinal and let θ ≥κ.

Then1. κ+ ≤nθκ ≤2κ.2.

If 2<κ > κ, then nθκ = κ+.3. If θ1 ≤θ2, then nθ2κ ≤nθ1κ and therefore {nθκ : θ ≥κ is a cardinal} is finite.4.

If θ1 ≤θ2 and nθ2κ = θ1, then nθ1κ = θ1.5. If θ = n2κκ , then θ is the unique cardinal with nθκ = θ and for every θ1 ≥θ,nθ1κ = θ.1991 Mathematics Subject Classification.

54A35, 03E35, 54A25, 54E52.Key words and phrases. Baire numbers, product spaces (2θ)κ, product forcing, diamond.Tt bAMS T X

A. Miller [M] proved that cof(nωω) > ω but also produced a model for cof(nω1ω ) =ω. In this model |{nθω : θ ≥ω is a cardinal}| = 2.

Similar models for κ > ω can befound in [L]. In his above mentioned paper, Miller uses a countable support productof Fnω(ω, 2) to increase nωω without changing the value of nω1ω(and hence gettingnω1ω < nωω).

This idea will be used next to prove the following theorem.Theorem. Let κ > ω be a regular cardinal.

If ZFC is consistent, then for every1 ≤l ∈ω, ZFC is consistent with |{nθκ : θ ≥κ is a cardinal}| = l.This answers ([L] 1.6) for κ > ω. We do not know whether the Theorem is truefor κ = ω.

Before we turn to the proof of the theorem, we will need the followinglemma which is due to Miller. The proof of the lemma is a forcing argument thatuses ♦κ.The use of ♦’s in forcing arguments originated in [B]; for other sucharguments see [Ka], [L] and [L1].Definition.

For the cardinals κ, θ, λ, let Qκ(θ, λ) be the product of λ many copiesof Fnκ(θ, 2) with support of cardinality ≤κ. A condition q ∈Qκ(θ, λ) is a functionwith dom(q) ∈[λ]≤κ and such that for every α ∈dom(q), q(α) ∈Fnκ(θ, 2).

Thepartial ordering is defined by putting q ≤p if and only if dom(q) ⊃dom(p) and forevery α ∈dom(p), q(α) ⊃p(α). If {qα : α < γ} ⊂Qκ(θ, λ) have a lower bound inQκ(θ, λ), then let us denote the largest lower bound by Vα<γ qα.Lemma.

Let κ > ω be a regular cardinal such that ♦κ holds. Let λ, θ ≥κ becardinals.

Let Q = Qκ(θ, λ). Then forcing with Q over V has the following property:for every function f : κ →V in the extension there is a set A ∈V such that(|A| = κ)V and range(f) ⊂A (in particular, forcing with Q preserves κ+).Proof of the lemma.

Assume thatq0 ⊩Q “τ : κ →V ”.Let M be an elementary substructure of the universe such that |M| = κ, M isclosed under sequences of length < κ (i.e. for every α ∈κ, αM ⊂M), and suchthat q0, Q, λ, θ, κ, τ are all in M. Notice that every set in M that has cardinality≤κ is also a subset of M. Therefore, if q ∈Q ∩M, then q ⊂M.2

Let L = {λξ : ξ < κ} = M ∩λ, and T = {θξ : ξ < κ} = M ∩θ. For every ξ < κ,let Lξ = {λδ : δ < ξ}, and Tξ = {θδ : δ < ξ}.

Notice that Lξ, Tξ ∈M.For every α ∈κ we define a function Bα : Q →℘(α × α) as follows:(ξ, η) ∈Bα(q) ⇐⇒[λξ ∈dom(q) ∧θη ∈dom(q(λξ)) ∧q(λξ)(θη) = 1].Notice that for every α ∈κ, Bα ∈M (because Lα, Tα ∈M).Now, let us fix a ♦κ-sequence I = {Iξ : ξ < κ} on κ × κ.Notice that forevery ξ < κ, Iξ ∈M.We are now ready to construct a decreasing sequence{qα : α < κ} ⊂Q ∩M, below q0, that satisfies the following conditions:(1) α < β =⇒qβ ≤qα. (2) (∀α < κ) Lα ⊂dom(qα).

(3) α < β =⇒qβ ↾Lα = qα ↾Lα. (4) If α ∈κ is a limit ordinal, then qα = Vβ<α qβ.

(Notice that qα ∈Q ∩Mbecause {qβ : β < α} ∈M. )(5) Given qα let us define qα+1.Case (i): There exist r ≤qα such that for every ξ < α, dom(r(λξ)) = Tα,and Bα(r) = Iα, and r decides τ ↾α.

In this case, the same is true inM. Hence there are rα, tα ∈M such that rα ≤qα, and for every ξ < α,dom(rα(λξ)) = Tα, and Bα(rα) = Iα, andrα ⊩Q “τ ↾α = tα”.Let qα+1 be defined as follows: qα+1 = (qα ↾Lα) ∪(rα ↾(dom(rα) \ Lα)).Case (ii): ¬ (case (i)).

Let qα+1 ≤qα be any extension in M that satisfies(2) and (3), and let tα = ∅.Finally, define q = Vα<κ qα. By (1) and (3) of the construction, q ∈Q.

By (2),dom(q) = L. Let A = Sα<κ range(tα). We claim thatq ⊩Q “range(τ) ⊂A”.Assume not.

Let s ≤q, and δ ∈κ be such that s ⊩Q “τ(δ) /∈A”. Let us define adecreasing sequence {sα : α < κ} in Q that satisfies the following conditions:(1) s0 = s.3

(2) (∀α < κ) sα decides τ ↾α. (3) If α < κ is a limit ordinal, then sα = Vβ<α sβ.

(4) (∀α < κ)(∀ξ < κ) Tα ⊂dom(sα(λξ)).Now let B = {(ξ, η) ∈κ × κ : sη+1(λξ)(θη) = 1}. Notice that for every α < κ,B ∩α × α = Bα(sα).

Let C = {α < κ : (∀ξ < α) dom(sα(λξ)) = Tα}; C is aclub. In addition, S = {α < κ : B ∩α × α = Iα} is stationary.

Pick α ∈C ∩Ssuch that α > δ. Then sα witnesses that case (i) of part (5) in the construction of{qα : α < κ} holds (i.e.

r = sα). So, we are given rα, tα ∈M such that rα ≤qα,and rα ⊩Q “τ ↾α = tα”.

Hencerα ⊩Q “τ(δ) ∈A”.But sα ≤rα, and sα ≤s, and this implies the desired contradiction.□Proof of the theorem. Since the theorem is trivial for l = 1, let us assume thatl ≥2.

Start with a model V of ZFC + GCH + ♦κ. Letκ ≤θ1 < θ2 < · · · < θlbe cardinals with θi ̸= κ+, and θl = θ+l−1, and such that if θi ̸= κ, then cof(θi) > κ.Letλ1 > λ2 > · · · > λl = θlbe cardinals with λ1 = λ+2 and such that cof(λi) > κ+.Let Qi = Qκ(θi, λi) .

Let us force withP = Q1 × · · · × Ql−1.By the GCH, the partial orders Fnκ(θi, 2) all have the κ+.c.c. ([K] VII 6.10).Therefore, P is (isomorphic to) a product of κ+.c.c.

partial orders with support ofsize ≤κ. Now use a delta system lemma and the Erd¨os-Rado theorem ((2κ)+ →(κ+)2κ) to show that P is κ++.c.c.

([K] VIII(B7)), and hence P preserves cardinals≥κ++. Clearly, P is κ-closed and therefore cardinals ≤κ are preserved.

Finally,by the Lemma, κ+ is preserved as well.Let G be a P-generic filter over V . Let θ ̸= κ+ be a cardinal with κ ≤θ ≤θl.Let i be the minimal such that θ ≤θi.

Let us show that(∗)nθκ = λi.4

Notice that (∗) suffices for the proof of the theorem since it in particular shows thatnθlκ = θl and therefore by fact 5, (∗) implies that(∀θ ≥θl) nθκ = λl.In the remaining case where θ = κ+, (∗) implies that nκ+κ= λ1 or nκ+κ= λ2.Therefore, (∗) implies that {nθκ : θ ≥κ is a cardinal} = {λi : 1 ≤i ≤l}.Let us first show that nθκ ≥λi. By fact 4, we may assume that 1 ≤i < l. Noticethat since P is κ-closed, Fnκ(θ, 2) is absolute and has cardinality θ<κ ≤θi < λi.By the product lemma, we may view forcing with P as forcing with the productQ{Qj : 1 ≤j < l and j ̸= i} ×Qi.

Now, by the definition of Qi and since θ ≤θi, itis easy to see that any collection of < λi many dense subsets of Fnκ(θ, 2) in V [G],has a filter.Finally we show that nθκ ≤λi. Notice that if i = 1, then this is clear because(2κ = λ1)V [G] (to see this use a counting nice names argument ([K] VII)).

So letus assume that i > 1 and hence θ ≥κ++. In addition we may assume that θ isregular (otherwise, if θ is singular, then it suffices to prove that nθ++i−1κ≤λi sincenθκ ≤nθ++i−1κ).Let us now view forcing with P as forcing with S × R, whereS = Qi × · · · × Ql−1andR = Q1 × · · · × Qi−1.Notice that if i = l, then R = P and S is the trivial partial order.

Let H be an S-generic filter over V , and K be an R-generic filter over V [H] such that V [H ×K] =V [G]. For every a : θ →2 with |a| = κ let us defineDa = {t ∈Fnκ(θ, 2) : (∃ξ ∈dom(a)) t(ξ) ̸= a(ξ)}.In V [H], define D = {Da | a: θ →2 and |a| = κ}.

D is a collection of dense subsetsof Fnκ(θ, 2) and |D| = λi (because (2κ = θκ = λi)V [H]). Let us show that D hasno filter in V [G].5

Assume, by way of contradiction, that F ∈V [G] is a filter for D. Assume withoutloss of generality that⊩S×R “F is a filter for D”.Let τ be a P-name for S F. It suffices to find (s, r) ∈S × R and an S-name π suchthats ⊩S “[π : θ →2 and |π| = κ and r ⊩R “π ⊂τ”]”.We now work in V . For every ξ ∈θ, let (sξ, rξ) ∈S × R and uξ ∈2 be such that(sξ, rξ) ⊩“τ(ξ) = uξ”.Consider {rξ : ξ ∈θ}.

Since θ ≥κ++ and θ is regular, we may use the delta systemlemma to get X ∈[θ]θ such that {dom(rξ) : ξ ∈X} form a delta system with aroot ∆. Now, since |Fnκ(θi−1, 2)| = θi−1 < θ and |∆| ≤κ, there exists Y ∈[X]θsuch that {rξ : ξ ∈Y } all agree on ∆(i.e.

(∀ξ, η ∈Y ) rξ ↾∆= rη ↾∆).Consider {sξ : ξ ∈Y }. Since S is κ++.c.c.

there exists s′ ∈S and a name σ withs′ ⊩S “σ = {ξ ∈Y : sξ ∈Γ} and |σ| = θ”,where Γ is the canonical name for the S-generic filter. By the Lemma, there existsA ∈[Y ]κ and s ≤s′ such thats ⊩S “|σ ∩A| = κ”.Let π be an S-name for the function whose domain is σ ∩A and such that for everyξ ∈σ ∩A, π(ξ) = uξ.

Let r = S{rξ : ξ ∈A}. Then r ∈R (because A ⊂Y andA ∈V ), ands ⊩S “[π : θ →2, and |π| = κ, and r ⊩R “π ⊂τ”]”.□Remark 1.

If κ = ω, then it is known that P (defined as in the proof of the Theorembut for κ = ω) collapses ω1 ([K] VIII(E4) and [M] p. 280), and (assuming CH) isℵ2.c.c. What one needs in order to get the argument of the Theorem to go throughfor the case κ = ω, is the following: if σ is a set in the extension that is unboundedin (ω2)V , then there exists a countable set A in V such that A ∩σ is infinite.

Thisis false by the following Proposition.6

Proposition. Let λ ≥ω, and θ > ω be cardinals.

Let Q = Qω(θ, λ). Then forcingwith Q adds a set σ ⊂θ, that is unbounded in θ, and such that if A is a countable(in V ) ground model subset of θ, then A ∩σ is finite.Proof.

For every n ∈ω, let gn be the n’th generic function (i.e. gn : θ →2, andgn(α) = 1 if and only if there exists p in the Q-generic filter such that p(n)(α) = 1).Let σ be the set defined in the extension by σ = {α ∈θ : (∀n ∈ω) gn(α) = 1}.Since θ ≥(ω1)V , and the supports of members of Q are countable, it is not hardto see that σ is unbounded in θ.

Now let p ∈Q, and A ∈[θ]ℵ0. Let us find q ≤psuch that q ⊩“|A ∩σ| < ℵ0”.

We may assume that dom(p) ⊃ω.Let A∗= {α ∈A : (∃n ∈ω) α /∈dom(p(n))}. Notice that A \ A∗is finite.

Forevery K ∈[ω]<ℵ0 define a(K) = {α ∈A∗: (∀n /∈K) α ∈dom(p(n))}; a(K) isfinite. Fix {αi : i ∈ω} an enumeration of A∗.We now construct {qi : i ∈ω} ⊂Q, {ni : i ∈ω} ⊂ω, and {Fi : i ∈ω} finitesubsets of A∗that satisfy the following conditions:(1) q0 ≤p and for every i ∈ω, qi+1 ≤qi.

(2) For every i ∈ω, qi ↾(λ \ {nk : k ≤i}) = p ↾(λ \ {nk : k ≤i}). (3) For every i ∈ω, Fi ⊂Fi+1, and Fi ⊃a({nk : k ≤i}).

(4) Si∈ω Fi = A∗. (5) i < j =⇒qj ↾{nk : k ≤i} = qi ↾{nk : k ≤i}.

(6) For every i ∈ω and every α ∈Fi, qi ⊩“α /∈σ”.Stage 0: Pick n0 ∈ω with α0 /∈dom(p(n0)). Let F0 = a({n0}) ∪{α0}.

Defineq0(n0) by:q0(n0)(α) = 0α ∈F0p(n0)(α)α /∈F0 and α ∈dom(p(n0)).Stage i+1: If αi+1 ∈Fi, then ni+1 = ni, Fi+1 = Fi, and qi+1 = qi. Otherwise,by (3), αi+1 /∈a({nk : k ≤i}).

Therefore, we can pick ni+1 /∈{nk : k ≤i} suchthat αi+1 /∈dom(p(ni+1)). By (2), αi+1 /∈dom(qi(ni+1)) as well.

Let Fi+1 =Fi ∪a({nk : k ≤i + 1}) ∪{αi+1}. Define qi+1(ni+1) by:qi+1(ni+1)(α) = 0α ∈Fi+1 \ Fiqi(ni+1)(α)α /∈Fi+1 \ Fi and α ∈dom(qi(ni+1)).Notice that α ∈Fi+1 \ Fi implies that either α = αi+1, or α ∈a({nk : k ≤i + 1}) \ a({nk : k ≤i}), and in either of these cases α /∈dom(qi(ni+1)).7

Finally, let q = Vi∈ω qi. By (2) and (5), q ∈Q and clearly, q ≤p.

By (4) and(6), q ⊩“A∗∩σ = ∅”.□Remark 2. In the extension of the above Proposition we also have: σ is an un-bounded subset of θ, and if x ∈[σ]ℵ0, then (ω1)V is countable in V [x].

This istrue because Q is ℵ2.c.c., and thus there is X ∈V with |X| = ℵ1 and X ⊃x.Now one can enumerate X, in V , in type (ω1)V , and x must be unbounded in thisenumeration since otherwise it would be contained in a countable ground modelset.Finally, we would like to mention that the Lemma implies that, the Proposition,stated for κ > ω (rather than ω), is false.References[B] J. Baumgartner, Almost-disjoint sets, the dense set problem, and partition calculus, Ann.Math. Logic 10 (1976), 401–439.[BS]B.

Balcar and P. Simon, Handbook of Boolean Algebras, vol. 2 (J.D.

Monk and R. Bonnet,eds.

), North-Holland, 1989, pp. 333–386.[Ka]A.

Kanamori, Perfect-Set Forcing for Uncountable Cardinals, Ann. Math.

Logic 19 (1980),97–114. [K] K. Kunen, Set Theory, North-Holland, 1980.

[L] A. Landver, Baire Numbers Uncountable Cohen Sets and Perfect-Set Forcing, J. SymbolicLogic 57 (1992), no. 3 (to appear).[L1]A.

Landver, Singular Baire Numbers and Related Topics, Ph.D. Thesis, The University ofWisconsin, Madison, 1990. [M] A. Miller, The Baire Category Theorem and Cardinals of Countable Cofinality, J. SymbolicLogic 47 (1982), no.

2, 275–288. [V] B. Veliˇckovi´c, Jensen’s □Principles and the Nov´ak Number of Partially Ordered Sets, J.Symbolic Logic 51 (1986), no.

1, 47–58.Department of Mathematics, The University of Kansas, Lawrence, KS 66045E-mail address: landver@kuhub.cc.ukans.edu8

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