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Home , Almost, Cardinality, Disjoint sets

OF MAXIMAL DISJOINT FAMILIES

FABIAN KAAK

1. Introduction A family A of of ω is said to be almost disjoint if the inter- section of any two distinct sets in A is finite. If instead we required the intersection to be empty, we would get a rather uninteresting no- tion, as any such family is countable. This is easy to see, as we could inject it into ω by picking one in each of the . Of course, if we tried doing this for an almost disjoint family, we would not necessarily get different elements for different sets, as we could be picking an element in their finite, but not necessarily empty, intersec- tion. We might still think that such a family should be small, but as the following theorem shows, this could not be further from the truth. Theorem 1. There is an almost disjoint family of cardinality 2ω.

Proof. Let (xn)n<ω be an injective enumeration of all finite subsets of ω. For X ⊆ ω, let AX = {n ∈ ω | xn = X ∩(max(xn)+1)}, i.e. n ∈ AX if and only if X end extends xn. Then A = {AX | X ⊆ ω} is an almost disjoint family of size 2ω. Indeed, let X 6= Y be subsets of ω. Then, without loss of generality, there is some m ∈ X\Y . As the

(xn)n<ω is injective, there is some N < ω such that for every n > N we have m < max(xn)+1. Thus there is no n > N such that n ∈ AX ∩AY , as otherwise xn = X ∩ (max(xn) + 1) 6= Y ∩ (max(xn) + 1) = xn.  As every of an almost disjoint family is again almost disjoint, this also tells us that there is an almost disjoint family of every cardi- nality less than or equal to 2ω. This completely describes the cardinals, that can be the size of an almost disjoint family. Now we consider max- imal almost disjoint families, that is, almost disjoint families such that there is no subset of ω that can be added while preserving almost dis- jointness. We will also call these mad families for short. Zorn’s Lemma tells us that every almost disjoint family is contained in a maximal al- most disjoint family. In particular, this means that there is a maximal

Date: February 9, 2021. 1 2 FABIAN KAAK almost disjoint family of cardinality 2ω. The main goal now is to show that the of of mad families is more interesting.

2. Main Theorem Towards this goal, we will show that if we have a set of cardinals in the ground model, then, under some restrictions on this set, we can find a generic extension in which it is the set of cardinalities of mad families. This result is due to Blass, see [1, Theorem 9]. Theorem 2. Assume GCH. Let C be a set of uncountable cardinals such that: (1) C is closed, (2) ℵ1 ∈ C, (3) if κ ∈ C has countable cofinality, then κ+ ∈ C and (4) all uncountable cardinals less than or equal to |C| are in C. Then there is a extension, with the same cardinals, in which 2ω = max(C) and there is a mad family of cardinality κ if and only if κ ∈ C, for all κ ≤ 2ω. Proof. We will construct a forcing similar to Cohen forcing that, for each cardinal κ ∈ C, adds a mad family of cardinality κ. For a cardinal S κ, define Iκ = {(κ, α) | α < κ} and I = κ∈C Iκ. Let P be the forcing whose conditions are functions p: F × n → 2, where n < ω and F ⊆ I is finite. For two conditions p: F × n → 2 and p0 : F 0 × n0 → 2, define p ≤ p0 if and only if p ⊇ p0 and (1) for all n0 ≤ k < n and distinct (κ, η), (κ, ξ) ∈ F 0, at least one of p(κ, η, k) and p(κ, ξ, k) is 0. The first thing we want to show is that this forcing has the countable chain condition, as this gives us that no cardinals are collapsed in the forcing extension. Lemma 1. P has the ccc. Proof. Let A ⊆ P be an . We will find two compatible elements in A. First we use the ∆-system lemma to find an uncountable B ⊆ A such that {dom(p) | p ∈ B} is a ∆-system with root R. As there are only finitely many functions from R to 2, there has to be an uncountable C ⊆ B of conditions that are equal on R. Find two different conditions p: F × n → 2 and p0 : F 0 × n → 2 in C. These conditions are compatible as p ∪ p0 extends them as a , and (1) is trivial if both conditions are defined on the same n. Thus A is not an antichain and therefore P has the ccc.  CARDINALITY OF MAXIMAL ALMOST DISJOINT FAMILIES 3

Let G be a P-generic filter over V . Now we want to prove that, for every κ ∈ C, there is a mad family of cardinality κ in V [G]. Define Aκ,α = {n ∈ ω | ∃p ∈ G p(κ, α, n) = 1} for κ ∈ C and α < κ, and let ˙ Aκ,α be a name for it.

Lemma 2. For each κ ∈ C, Aκ = {Aκ,α | α < κ} is a mad family of cardinality κ.

Proof. Let κ ∈ C. We start by showing that Aκ is almost disjoint. Let η, ξ < κ. By genericity, there is some p: F × n → 2 ∈ G such that (κ, η), (κ, ξ) ∈ F . Towards a contradiction, we assume that there is 0 m > n in Aκ,η ∩ Aκ,ξ. Thus there are q, q ∈ G such that q(κ, η, m) = 1 and q0(κ, ξ, m) = 1. As G is a filter, there is an r ∈ G which extends p, q and q0. In particular, r(κ, η, m) = r(κ, ξ, m) = 1, but the definition of r ≤ p implies that only one of these can be one, a contradiction. Hence Aκ,η and Aκ,ξ are almost disjoint. Now we need that Aκ is maximal. We assume that it is not. Thus there is an infinite x ∈ V [G] such that x ∩ Aκ,α is finite for every α < κ. We pick a namex ˙ for x such that every condition involved in it has domain contained in J × ω, where J is a . To see that this is possible, pick for every n ∈ ω a maximal antichain Bn of conditions forcing “n ∈ x˙”. As P has the countable chain condition, Bn is countable, therefore the name {(ˇn, p) | n < ω, p ∈ Bn} is as desired. There is some condition p: F × n → 2 ∈ G forcing “x ˙ is ˙ infinite and ∀α < κ Aκ,α ∩ x˙ is finite”. As F is finite, we can assume that F ⊆ J. Because J is countable and κ is uncountable, we can ˙ pick η ∈ κ \ J. Then p forces “Aκ,η ∩ x˙ is finite”. Thus we can find m < ω and a condition p0 : F 0 × n0 → 2 extending p that forces ˙ 0 0 “Aκ,η ∩ x˙ ⊆ m”. We can extend p by zeros to increase n , so we can assume that n0 ≥ m, and we can trivially increase m to get m = n0. Let X = {α < κ | (κ, α) ∈ J ∩ F 0}. Then p0 forces that there is some k > n0 such that ˙ (2) k ∈ x˙ ∧ ∀ξ ∈ X k∈ / Aκ,ξ.

Indeed, for any ξ ∈ X, x ∩ Aκ,ξ is finite and X is finite, therefore S 0 x ∩ ξ∈X Aκ,ξ is finite and, as x is infinite, there is such a k > n . Now we can strengthen p0 to get a condition q : H × n00 → 2 that forces (2) for a specific k > n0. We are almost at our desired contradiction. What we would like to do now is add q(κ, η, k) = 1 to q, as then 0 0 n < k ∈ x ∩ Aκ,η ⊆ n . The problem is that (κ, η, k) could already be in the domain of q and map to zero. The solution is that we can restrict q to a smaller domain while keeping the important properties for the contradiction. As this problem cannot happen if n00 ≤ k, we can 4 FABIAN KAAK assume that k < n00, to avoid some minor annoyance later on. Define 00 00 00 00 F = H ∩ J and p = q  F × n . Claim. p00 forces (2) for the same k. Proof. Assume it does not. Then there is r ≤ p00 forcing the negation. All the conditions in (2) can be arranged to have domains contained ˙ in J × ω. Forx ˙ this is clear and for “k ∈ Aκ,ξ” we can use the name {k,ˇ f | f : {κ} × {ξ} × (k + 1) → 2 and f(κ, ξ, k) = 1}. Thus we can assume that r has domain contained in J × ω. Let r0 be a condition that as a function extends r ∪ q with zeros; the length does not matter, as adding zeros can not destroy (1). This condition extends both r and q as functions. To see (1), we notice that q and p00 ≥ r agree on J and r is not defined outside of J. So r0 forces (2) and its negation, a contradiction.  Now we can go on to define the condition that will yield the desired contradiction. Let us define s:(F 00 ∪F 0 ∪{(κ, η)})×n00 → 2 by making s agree with p0 on F 0 × n0 and p00 on F 00 × n00, setting s(κ, η, k) = 1 and setting s at all remaining elements of its domain to zero. This is a function, and therefore a condition, since p0 ∪ p00 ⊆ q and neither p0 nor p00 assign a value to (κ, η, k). It clearly extends p00 as a function, and as they have the same length, it also extends p00 as a condition. But we will also see that s extends p0. As functions this is clear. To get it as conditions, we need that for any j with n0 < j ≤ n00 and (λ, α) 6= (λ, β) in F 0, not both s(λ, α, j) and s(λ, β, j) are one. If they are, then neither can be given by the first clause in the definition of s and not both can be given by the second clause, as p00 agrees with q, an extension of p0. Thus one has to be given by the second clause and the other one must be (κ, η, k). So j = k, λ = κ, without loss of generality β = η and (κ, α) ∈ F 0. Thus p00(κ, α, k) = 1 for (κ, α) ∈ F 0 ∩ F 00 = F 0 ∩ (H ∩ J) ⊆ F 0 ∩ J. But then α ∈ X and the claim says that p00(κ, α, k) = 0, a contradiction. Thus s extends both 0 00 0 ˙ 0 00 p and p . So, like p , s forces “Aκ,η ∩ x˙ ⊆ n ” and, like p , s forces ˙ “k ∈ x˙”. Also s(κ, η, k) = 1, which means s forces “k ∈ Aκ,η”, but ˙ 0 0 then “k ∈ Aκ,η ∩ x˙ ⊆ n ” and k > n , a contradiction.  It remains to show that, for every cardinal κ > ω not in C, there is no mad family of size κ. This will also yield that 2ω = max(C), as the largest mad family is of size 2ω. Assume there is a mad family {Xα | α < λ} of cardinality λ > ω not in C in V [G]. Because of the countable chain condition, we can assume that all names of reals we are working with contain only countably many conditions. In particular, for every namex ˙, we can find a set J such that CARDINALITY OF MAXIMAL ALMOST DISJOINT FAMILIES 5

(a) J is countable and (b) all conditions inx ˙ have domain contained in J × ω. ˙ Fix names Xα and such sets Jα, for every Xα. Let µ be the largest element in C below λ. As C is closed, this definition makes sense, and property (3) ensures that µ has uncountable cofinality. The strategy now is to find, in the ground model, a set M ⊆ κ of cardinality µ such that {Xα | α ∈ M} is a mad family, in the exten- sion. This yields a contradiction, as {Xα | α < κ} is a strictly bigger almost disjoint family. We construct M recursively as S M , where σ<ω1 σ (M ) is an increasing continuous sequence of sets of cardinality less σ σ<ω1 than or equal to µ. As µ ≥ ℵ1, it follows that |M| ≤ µ. As µ ≥ ℵ0, a straightforward induction shows that, in order to ensure that |Mσ| ≤ µ for all σ < ω1, it is sufficient to ensure that |Mσ| ≤ µ =⇒ |Mσ+1| ≤ µ for all σ < ω1. Let M0 = ∅. Assume we already constructed Mσ, for some given σ < ω1, and let S S us construct Mσ+1. Let S = α<λ Jα ∪ κ≤µ,κ∈C Iκ. We have |S| = λ, since the Jα are countable and Iκ has cardinality κ, for every κ ∈ C. Let K = S J ∪ S I . Clearly |K | ≤ |M | × ω + µ = µ. σ α∈Mσ α κ≤µ,κ∈C κ σ σ We say a set J ⊆ I is a σ-support if it satisfies (a) and

(c) for all κ ∈ C, if (J ∩ Iκ) \ Kσ is nonempty, then it is infinite. We say a σ-support supports a namex ˙ if, additionally, property (b) holds. Let Gσ be the group of permutations of I that fix Kσ pointwise and map Iκ into itself, for every κ ∈ C. Let Gσ act as a group of automorphisms on P, by g(p)(g(κ, ξ), k) = p(κ, ξ, k).

It is well known that Gσ also acts on the of P-names and preserves the forcing relation. We say two σ-supports J, J 0 are σ-equivalent if 0 there is g ∈ Gσ such that g[J] = J . Notice that the ¯ of a σ-support J only depends on J = {κ ∈ C | (J ∩ Iκ) \ Kσ 6= ∅} 0 ¯ ¯0 and J ∩ Kσ. Indeed, let J, J be σ-supports with J = J and J ∩ Kσ = 0 ¯ 0 J ∩Kσ. Then, for every κ ∈ J, we have (J∩Iκ)\Kσ and (J ∩Iκ)\Kσ are infinite, thus of cardinality ℵ0, so we can find a between them. We also have that, for κ ∈ J¯, κ > µ, so there is a bijection between 0 Iκ \ (J ∪ Kσ) and Iκ \ (J ∪ Kσ). Combine all these and the 0 identity on Kσ to get a g ∈ Gσ such that g[J] = J . By assumption (4), ¯ there are only |C| ≤ µ many options for J and only |Kσ| ≤ µ many options for J ∩ Kσ. Thus there are only µ many equivalence classes of σ-supports. Let J 0 be a σ-support. We want to construct a σ-support J, with 0 J ∩ S = J ∩ Kσ, in the same equivalence class as J . For κ ∈ C, if 6 FABIAN KAAK

0 (J ∩ Iκ) \ Kσ 6= ∅, then |Iκ \ S| = κ. So we can find a permutation of 0 Iκ that fixes Kσ pointwise and maps J ∩ (S \ Kσ) out of S. Combining all these permutations and extending arbitrarily we get g ∈ Gσ such that g[J 0] is as desired. Pick a set of representatives of σ-supports J, with J ∩ S = J ∩ Kσ. We will call the so picked elements the standard σ-supports. For a fixed σ-support J, there are only countably many conditions that have domain contained in J × ω. Every namex ˙ supported by J can be specified by giving, for every n < ω, a maximal antichain of conditions decidingx ˙(n). Thus, by CH, there are only ℵ1 many names that are supported by J, up to names that add the same real. So there are only µ many names that have a standard σ-support. For each namex ˙, we can find a countable set A(x ˙) ⊆ λ such that ˙ ˙ “∃α ∈ A(x ˙)x ˙ ∩ Xα is infinite” is forced. As “∃α < λ x˙ ∩ Xα is infinite” ˙ is forced, the set D = {p ∈ P | ∃α < λ p “x ˙ ∩ Xα is infinite”} is dense open. Thus we can pick a maximal antichain of P contained in D. Let A(x ˙) be the set of all ordinals α such that some element ˙ of this antichain forces “x ˙ ∩ Xα is infinite”. Because of the countable chain condition, A(x ˙) is countable, and since the antichain is maximal, ˙ “∃α ∈ A(x ˙)x ˙ ∩ Xα is infinite” is forced. Let Mσ+1 be the of all the A(x ˙), for namesx ˙ that are supported by a standard σ-support. As there are only µ many names supported by standard σ-supports, we have |Mσ+1| = µ. Notice that, for every α ∈ Mσ, Jα is a σ-support, ˙ and since Jα ⊆ Kσ, we have that Xα has standard σ-support. Thus Mσ ⊆ Mσ+1. This concludes the construction of M. It remains to show that {Xα | α ∈ M} is a mad family. Clearly it is an almost disjoint family. Letx ˙ be a name of a real and let J satisfy (a) and (b). Let K = S K , and σ<ω1 σ choose σ < ω1 such that K ∩ J = Kσ ∩ J. By enlarging J if necessary, we can assume that J is a σ-support. Then there is g ∈ Gσ such that g[J] is a standard σ-support. Find h ∈ Gσ+1 with h  J = g  J. This is possible since J does not intersect Kσ+1 \ Kσ. It is clear that h[J] is a standard σ-support and supports h(x ˙). Thus A(h(x ˙)) ⊆ Mσ+1 ⊆ M, ˙ so “∃α ∈ M h(x ˙) ∩ Xα is infinite” is forced. As h fixes Jα pointwise, ˙ ˙ ˙ for every α ∈ Mσ, we have h(Xα) = Xα. So “∃α ∈ M h(x ˙) ∩ h(Xα) is infinite” is forced. Since h preserves the forcing relation, it is forced ˙ that “∃α ∈ M x˙ ∩ Xα is infinite”. So {Xα | α ∈ M} is a mad family, which yields the contradiction and finishes the proof.  CARDINALITY OF MAXIMAL ALMOST DISJOINT FAMILIES 7

3. Optimality of the result We will conclude with a short discussion about the necessity of the assumptions. As there is a mad family of cardinality 2ω, we need that max(C) exists and equals 2ω in the forcing extension. An easy diagonal- ization argument shows that all cardinals in C have to be uncountable. It is also clear that all the other assumptions have been used in our proof. Still, it is possible to weaken all of them, while preserving the conclusion, as Shelah and Spinas showed in [2].

References [1] Andreas Blass. Simple cardinal characteristics of the . In 63–90, Israel Math. Conf. Proc., 6, Bar-Ilan Univ., Ramat, pages 63–90, 1993. [2] Saharon Shelah and Otmar Spinas. Mad spectra. The Journal of Symbolic Logic, 80(3):901–916, 2015.