CARDINALITY OF MAXIMAL ALMOST DISJOINT FAMILIES FABIAN KAAK 1. Introduction A family A of subsets of ! 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 ! by picking one element in each of the disjoint sets. 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 2!. Proof. Let (xn)n<! be an injective enumeration of all finite subsets of !. For X ⊆ !, let AX = fn 2 ! j xn = X \(max(xn)+1)g, i.e. n 2 AX if and only if X end extends xn. Then A = fAX j X ⊆ !g is an almost disjoint family of size 2!. Indeed, let X 6= Y be subsets of !. Then, without loss of generality, there is some m 2 XnY . As the enumeration (xn)n<! is injective, there is some N < ! such that for every n > N we have m < max(xn)+1. Thus there is no n > N such that n 2 AX \AY , as otherwise xn = X \ (max(xn) + 1) 6= Y \ (max(xn) + 1) = xn. As every subset 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 2!. 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 ! 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 2!. The main goal now is to show that the set of cardinalities 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) @1 2 C, (3) if κ 2 C has countable cofinality, then κ+ 2 C and (4) all uncountable cardinals less than or equal to jCj are in C. Then there is a forcing extension, with the same cardinals, in which 2! = max(C) and there is a mad family of cardinality κ if and only if κ 2 C, for all κ ≤ 2!. Proof. We will construct a forcing similar to Cohen forcing that, for each cardinal κ 2 C, adds a mad family of cardinality κ. For a cardinal S κ, define Iκ = f(κ, α) j α < κg and I = κ2C Iκ. Let P be the forcing whose conditions are functions p: F × n ! 2, where n < ! 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) 2 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 uncountable set. We will find two compatible elements in A. First we use the ∆-system lemma to find an uncountable B ⊆ A such that fdom(p) j p 2 Bg 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 function, 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 2 C, there is a mad family of cardinality κ in V [G]. Define Aκ,α = fn 2 ! j 9p 2 G p(κ, α; n) = 1g for κ 2 C and α < κ, and let _ Aκ,α be a name for it. Lemma 2. For each κ 2 C, Aκ = fAκ,α j α < κg is a mad family of cardinality κ. Proof. Let κ 2 C. We start by showing that Aκ is almost disjoint. Let η; ξ < κ. By genericity, there is some p: F × n ! 2 2 G such that (κ, η); (κ, ξ) 2 F . Towards a contradiction, we assume that there is 0 m > n in Aκ,η \ Aκ,ξ. Thus there are q; q 2 G such that q(κ, η; m) = 1 and q0(κ, ξ; m) = 1. As G is a filter, there is an r 2 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 2 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 × !, where J is a countable set. To see that this is possible, pick for every n 2 ! a maximal antichain Bn of conditions forcing \n 2 x_". As P has the countable chain condition, Bn is countable, therefore the name f(ˇn; p) j n < !; p 2 Bng is as desired. There is some condition p: F × n ! 2 2 G forcing \x _ is _ infinite and 8α < 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 2 κ n J. Then p forces \Aκ,η \ x_ is finite". Thus we can find m < ! 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 = fα < κ j (κ, α) 2 J \ F 0g. Then p0 forces that there is some k > n0 such that _ (2) k 2 x_ ^ 8ξ 2 X k2 = Aκ,ξ: Indeed, for any ξ 2 X, x \ Aκ,ξ is finite and X is finite, therefore S 0 x \ ξ2X 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 2 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 × !. Forx _ this is clear and for \k 2 Aκ,ξ" we can use the name fk;ˇ f j f : fκg × fξg × (k + 1) ! 2 and f(κ, ξ; k) = 1g. Thus we can assume that r has domain contained in J × !. 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.