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A. Blass (2010)
Combinatorial Cardinal Characteristics of the Continuum
James Monk, Robert Bonnet, S. Koppelberg (1989)
Handbook of Boolean Algebras
J. Baumgartner, P. Dordal (1985)
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In the book Cardinal Invariants on Boolean Algebras by J. Donald Monk many such cardinal functions are defined and studied. Among them several are generalizations of well known cardinal characteristics of the continuum. Alongside a long list of open problems is given. Focusing on half a dozen of those cardinal invariants some of those problems are given an answer here, which in most of the cases is a definitive one. Most of them can be divided in two groups. The problems of the first group ask about the change on those cardinal functions when going from a given infinite Boolean algebra to its simple extensions, while in the second group the comparison is between a couple of given infinite Boolean algebras and their free product. Keywords Cardinal invariants · Boolean algebras · Simple extensions · Free products Mathematics Subject Classification 03E05 · 03G05 1 Introduction A way to study the structure of a given Boolean algebra is through the so called cardinal invariants, which can be described, somehow loosely, as bounds, either upper or lower, to the size of certain types of its substructures, be they algebraic, combinatorial or topological. They are mainly generalizations of cardinal invariants of topological spaces, as Boolean algebras always are, or of the characteristics of the continuum, which are usually defined on the Boolean algebra P (ω) /Fin, i.e. the quotient of the power set of ω modulo the ideal of the finite subsets of ω. Of all these possibilities, in The author is supported by CONACYT scholarship 926448. This work is mainly based on the author’s masters degree thesis under the supervision of Michael Hrušák. Mario Jardón Santos dantiam@ciencias.unam.mx Centro de Ciencias Matemáticas, UNAM, Campus Morelia, Xangari, Apartado Postal 61-3 , 58089 Morelia, Michoacán, México 123 M. J. Santos this text we will focus on a half dozen of them. The reader is assumed to have basic knowledge of Boolean algebras, as can be found in the first parts of [4]. Let (A, +, ·, −, 0, 1), usually abbreviated as A, be an infinite Boolean algebra with < as its order relation. The set of all non-zero elements of A will be denoted A .A subset I ⊆ A is an ideal if 1 ∈ / I , it is downward closed and a +b ∈ I , for all a, b ∈ I . A partition of A is a non-empty subset X ⊆ A such that for all a, b ∈ X, a ·b = 0, and such that for all c ∈ A there exists x ∈ X such that c · x = 0. A (decreasing) + −1 tower is a non-empty subset X ⊆ A well-ordered by < whose product is 0, i.e. whose only lower bound is 0. A centered family is a non-empty subset X ⊆ A such <ω + that a = 0 for all F ∈ [X] . If there exists a ∈ A such that a ≤ x,for a∈F all x ∈ X, it is said that a is a pseudointersection of X. The algebra A is said to be + + atomless if for all a ∈ A there exists b ∈ A such that b < a.If A is atomless, a splitting family of A is a non-empty subset X ⊆ A such that for all a ∈ A there exists x ∈ X such that a · x = 0 = a ·−x. From these kinds of subfamilies of a given infinite Boolean algebra A one can define the following cardinal characteristics: a (A) := min {|X|| X ⊆ A is an in f inite partition} t (A) := min {|X|| X ⊆ Ais a tower } p (A) := min {|X|| X ⊆ A is a centered f amily with no pseudointersection} s (A) := min {|X|| X ⊆ Ais a splitting f amily} . For the notions of centered families and pseudointersections there exist the dual notions of families with the finite union property, i.e. X ⊆ A such that a = 1, a∈F <ω for all F ∈[X ] , and pseudounion, i.e. an upper bound not equal to 1. A p-family is a family with the finite union property with no pseudounion. Similarly one can define increasing towers. Thus the definitions of both p(A) and t(A) can be restated: t (A) := min {|X|| X ⊆ Ais an increasing tower } p (A) := min {|X|| X ⊆ Ais a p − family} . Since every infinite partition is a p-family and every decreasing tower is a centered family with no pseudointersection, it immediately follows that p (A) ≤ a (A) , t (A). It is also easy to verify that every maximal centered subfamily of a splitting family is a centered family with no pseudointersection. So, p (A) ≤ s (A). Notice also that, when not ∞, t(A) is a regular cardinal. These small cardinal characteristics are generalizations of well-known cardinal characteristics of the continuum, when instead of A we have P ω /Fin,simply ( ) denoted a, t, p and s. For more information on these cardinals see [2]. Other larger cardinal invariants derive from the following concepts. A subset X ⊆ + + A is dense if for all a ∈ A , there exists x ∈ X such that x ≤ a. A subset X ⊆ A is If A is an infinite Boolean algebra, an infinite partition of A can be found through Axiom of Choice. If P is said partition, the set {−x | x ∈ P} is a centered family with no pseudointersection. So a(A) and p(A) are well defined. If A is also atomless, then A is a splitting family, and also s(A) is well defined on atomless Boolean algebras. However, not every infinite Boolean algebra has a tower. In this case t(A) =∞. 123 Questions on cardinal invariants... irredundant if x is not algebraically generated by X \ {x }, for all x ∈ X.If X is not irredundant, it will be called redundant. We can define π (A) := min {|X|| X ⊆ Ais dense} Irr (A) := sup {|X|| X ⊆ Ais irredundant } . Of these concepts it is known that: Theorem 1 (McKenzie, [4], Proposition 4.23.) Every maximal irredundant family of a Boolean algebra generates a dense subalgebra. An immediate consequence of this proposition is that ω ≤ π (A) ≤ Irr (A) for all infinite Boolean algebra A. In [5] there are several results on these cardinal functions, both concerning their relation to other functions and their behaviour on different kinds of Boolean algebras (although there it is written tow instead of t and spl instead of s). In that book several questions are asked about these cardinal invariants. In this text an answer, be it partial or complete, is given to some of them. Here a list of these problems is provided, with the same enumeration as in [5]. – Problem 7. Does A ≤ B imply that a(B) ≤ a(A)? – Problem 8. Is it true that for all infinite BAs A, B one has a(A ⊕ B) = min(a(A), a(B))? – Problem 37. Does A ≤ B imply that tow (A) ⊆ tow (B) or tow(B) ≤ s spect spect tow(A)? – Problem 38. ArethereBAs A, B such that A ≤ B and tow(B)< tow(A)? – Problem 45. ArethereBAs A, B such that A ≤ B and p(B)< p(A)? – Problem 46. Is it true that for all infinite BAs A, B we have p(A ⊕ B) = min(p(A), p(B))? – Problem 48. Is p(A) = min{|X|: X is a maximal rami f ication set in A}? – Problem 52. Is spl(A ⊕ B) = min(spl(A), spl(B)) for atomless A, B? – Problem 70. ArethereBAs A, B such that A ≤ B and π(A)>π(B)? – Problem 71. ArethereBAs A, B such that A ≤ B and π(A)>π(B)? – Problem 86. Can one have Irr (A)< Irr (B) for A ≤ B or A ≤ B? s m 2 Simple and minimal extensions If A ⊆ B are two Boolean algebras with the same minimum and maximum elements, and if the operations of B restricted to A coincide with the operations of A (or, usually, if A is isomorphic to such a substructure of B), it is said that A is a subalgebra of B, or that B is an extension of A. This fact is generally denoted A ≤ B. If A ≤ B, it is said that B is a simple extension of A, if there exists x ∈ B such that B = A (x ), which means that B is the algebra generated by x and all the elements of 123 M. J. Santos A. This fact will be denoted A ≤ B. It is said that B is a minimal extension of A if any algebra C such that A ≤ C ≤ B is either equal to A or equal to B. This fact is denoted A ≤ B.If A ≤ B, it follows easily A ≤ B. m m s This section is about the changes, or the lack thereof, that can happen to the cardinal functions defined in the Introduction when they pass from a Boolean algebra to their simple and minimal extensions. 2.1 p-families, infinite partitions and towers Considering the cardinal characteristics defined in the previous section and minimal extensions we have the following result. Theorem 2 Let A and B be infinite Boolean algebras. If A ≤ B, then p (B) ≤ p (A), t (B) ≤ t (A) and a (B) ≤ a (A). This fact can be found in [5] (Propositions 3.34, 4.36, 4.54). In this book the author asks if this result can be extended to the case when B is a simple extension of A and also if the inequalities can be strict in the case of B being a minimal extension of A (Problems 7, 37, 38, 44, 45). For all these questions in this section affirmative answers are given. Firstly Theorem 2 will be generalized to simple extensions. Some ideals and their quotients will be important for this. Definition 1 Let A and B be Boolean algebras such that A ≤ B.If x ∈ B define the ideal on A below x as follows: A x := {a ∈ A | a ≤ x }. Recall that for an ideal I ⊆ A the quotient A/I is the Boolean algebra on the equivalence classes defined by the relation a ∼ b iff a b = (a·(−b))+(b·(−a)) ∈ I , with the operations induced by those of A. This means that [a] +[b] =[a + b] , I I I [a] ·[b] =[a·b] , and −[a] =[−a] , where [a] is the equivalence class of a, for all I I I I I I a, b ∈ A. Also recall that the simple product A × B, whenever A and B are Boolean algebras, refers to the Boolean algebra on the set A × B, with (1, 1) as maximum element, (0, 0) as minimum element, and the operations defined coordinatewise. Lemma 1 Let A be a Boolean algebra and suppose that A (x ) is a simple extension of A. If I := A x and I := A −x, then A (x ) (A/I ) × (A/I ). 0 1 0 1 On the face of Lemma 1, if we are working on simple extensions of Boolean algebras and some of their cardinal characteristics, it would be useful to know the behavior of said cardinal characteristics on simple products. Fortunately we have the following lemma. Lemma 2 Let A and B be two infinite Boolean algebras. Then { } – p (A × B) = min p (A) , p (B) , – t A × B = min {t A , t B } and ( ) ( ) ( ) – a (A × B) = min {a (A) , a (B)}. 123 Questions on cardinal invariants... Proofs of both lemmas can be found in [5], Propositions 2.28, 3.36, 4.37 and 4.55. If b ∈ A,the set A b := {a ∈ A | a ≤ b} with 0 as minimum element, b as maximum element, the product and sum of A, and the complement restricted to b is a Boolean algebra. We say that X ⊆ A b is a p-family (resp. infinite partition, resp. tower) below b if it is so in the Boolean algebra A b. We now proceed to answer the questions. Lemma 3 Let A be an infinite Boolean algebra and I ⊆ A be an ideal. Suppose that A/ I is infinite and that p (A) < p (A/I ) (resp. a (A) < a (A/I ), resp. t (A) < t (A/I )). If {a | α< κ} is a p-family in A (resp. infinite partition, resp. inc. tower) of minimum + κ size, then there exist b ∈ A and E ∈ [κ] such that {a · b | α ∈ E } is a p-family (resp. infinite partition, resp. tower) below b consisting of elements of I . Proof Case 1. Suppose that [a ] =[1] . α I I α<κ <ω Since A/I has no p-families of size κ, it follows that there exists F ∈ κ such [ ] that [a ] =[1] . Hence b := − a is an element of I.Itfollows α I I α α∈F α∈F immediately that a · b is an element of I , for all α< κ.Let E be the set of all α< κ such that a · b = 0. If E were a finite subset of E such that b = a · b,it α α α∈E would follow that 1 = a , which is a contradiction. α∈E ∪F We have that b = a · b, since otherwise we would have a non-zero element α∈E c ≤ b such that a · c = 0 for all α ∈ E, and hence for all α< κ, contradicting the fact that {a | α< κ} is a p-family of A. It follows that {a · b | α ∈ E } is a p-family α α below b. Finally, if we had |E | <κ, then {a · b | α ∈ E}∪{b} wouldbea p-family of A of size less that p(A). Therefore |E|= κ. (When dealing with infinite partitions <ω there also exists F ∈ [κ] such that [a ] =[1] . Define b := − a α I I α α∈F α∈F and E := κ\F.If {a | α< κ} is a tower, there exists α< κ such that if α ≤ β< κ, then a =[1] . In this case b := −a and E := κ\α.) β I α Case 2. Suppose that [a ] =[1] . α I α<κ There exists b ∈ A\I such that [b] · [a ] =[0] , i.e. b · a ∈ I , for all α< α I α I I κ. Consider E := {α< κ | a · b = 0}.If E were a finite subset of E such that b = b · a , it would follow that b ∈ I , which is a contradiction. Therefore α∈E {a · b | α ∈ E } is a p-family below b and, as was observed in case 1, the size of E is κ (resp. infinite partition, resp. contains a tower). Theorem 3 Let A and B be infinite Boolean algebras. If A ≤ B, then p (B) ≤ p (A), a (B) ≤ a (A) and t (B) ≤ t (A). Proof From Lemma 1, it is known that there exist I and I ideals of A, such that 0 1 I ∩ I = {0} and B (A/I ) × (A/I ). We have two cases. 0 1 0 1 Case 1. The size of, say, A/I is finite. In this case p(B) = p(A/I ). For proving this take {c | α< p(A/I )},a p-family 1 α 1 in A/I .Itfollows that {([1] , c ) | α< p(A/I )} is a p-family in B. Therefore 1 I α 1 p(B) ≤ p(A/I ). 123 M. J. Santos Now take λ< p(A/I ) and P ={(d , c ) | α< λ}⊆ B such that 1 α α (d , c ) = ([1] , [1] ) α α I I 0 1 α∈F <ω <ω for all F ∈[λ] . Since A/I is finite, there exists F ∈[λ] such that for all α< λ there exists β ∈ F such that d = d . Therefore α β d = d . α β α<λ β∈F If d =[1] , then ( d , [1] ) is a pseudounion of P witnessing that that β I β I β∈F 0 β∈F 1 P is not a p-family in B.If d =[1] , it follows that c =[1] , for all β I α I β∈F 0 α∈F 1 <ω F ∈[λ] .If c ∈ A/I is a pseudounion of {c | α< λ}, then ([1] , c) witnesses 1 α I that P is not a p-family in B. Therefore λ< p(B). We conclude that p (B) = p (A/I ) (analogously for the other two functions). Suppose that κ = p (A) < p (A/I ).Let {a | α< κ} be a p-family in A, and let 1 α b and E be as given by Lemma 3 with I = I . Since A/I is finite, there exist α< β 1 0 in E such that b · a ∼ b · a , though b · a = b · a . Hence b · a b · a = 0 and α I β α β α β lies both in I and I , which is a contradiction. Therefore, p (A) ≥ p (A/I ) = p(B). 0 1 1 The proof is analogous for the other two cardinal functions. Case 2. Both the size of A/I and the size of A/I are infinite. 0 1 In this case p (B) = min {p (A/I ) , p (A/I )}, (resp. with the other two functions). 0 1 Suppose that κ = p (A) < p (A/I ) , p (A/I ).Let {a | α< κ} be a p-family in 0 1 α A (resp. infinite partition). If E and b are as given by Lemma 3 with I = I,it follows that {a · b | α ∈ E } ∪{−b} is a p-family in A (resp. infinite partition) whose elements, but possibly one of them, lie in I . Applying Lemma 3 to this p-family { } (partition) and I , it follows that I ∩ I = 0 , which is a contradiction. Therefore, 1 0 1 { } p (A) ≥ min p (A/I ) , p (A/I ) = p(B) (resp. for the function a). 0 1 (When {a | α< κ} is an increasing tower, we take b and E corresponding to I α 0 and we are in one of the following two cases: 1. [b · a ] = [b] or α∈E I I 1 1 2. [b · a ] = [b] . α∈E I I 1 1 Repeating the proof of Lemma 3 we get a similar contradiction.) Now an example of these inequalities being strict will be given. The following lemma tells us when a simple extension is a minimal one. Lemma 4 Let B := A (x ) be a simple extension of a Boolean algebra A. Then B is a minimal extension of A iff Smp , the ideal generated by A x and A −x, is either equal to A or a maximal ideal of A. For a proof of this lemma see [5], Proposition 2.32. Recall that an ideal I ⊆ A is maximal if for any ideal J such that I ⊆ J ⊆ A, it follows that I = J . Theorem 4 There exist A and B infinite Boolean algebras such that A ≤ B and t (B) < t (A). 123 Questions on cardinal invariants... Proof Let {X | n <ω} be a set of disjoint copies of βω\ω and p be an element of X . Define X := X , 0 n n<ω U := {a ∈ P (X ) | (∀n <ω)(a ∩ X ∈ clop (X )) ∧ n n (∃n <ω)(∀m ≥ n)(a ∩ X = X ) ∧ p ∈ a ∩ X } n n 0 and I := {a ∈ P X | ∀n <ω a ∩ X ∈ clop X ∧ ( ) ( )( ( )) n n ∃n <ω ∀m ≥ n a ∩ X =∅ ∧ p ∈ / a ∩ X }, ( )( )( ) n 0 where P(X ) is the power set of X and clop(X ) is the family of clopen subsets of X . n n Take A := I ∪ U as a set algebra, i.e. as a subalgebra of P(X ). To verify that A is indeed a Boolean algebra take a, b ∈ A.If b ∈ U , then a ∪ b ∈ U . If both a, b ∈ I , then a ∪b ∈ I . Similarly it can be verified that A is closed under intersections. Besides a ∈ I iff X \ a ∈ U . By its definition I is a maximal ideal of A. Claim ω< t(A). Proof Let C := {a | n <ω} be a strictly increasing family of elements from A.If C ⊆ I , then p ∈ / X ∩ a . Since clop(βω \ ω) has no countable towers, there 0 n n<ω exist a, b ∈ clop (X ) such that 1. p ∈ b 2. X ∩ a ⊂ a, 0 n n<ω 3. b ∩ a =∅ and 4. a ∪ b = X . Then a ∪ b ∪ X is an element of A not equal to X which contains each 0<n<ω element of C, and so this last set does not form a tower. If C is not subset of I , without loss of generality, it can be supposed that C ⊆ U . Then there exists m <ω such that for all n, k <ω,if k ≥ m, then a ∩ X = n k X . Therefore, there exists k < m such that for all n <ω, a ∩ X X , i.e. k n k k {a ∩ X | n <ω} is a strictly increasing family in clop (X ). Since this is an algebra n k k with no countable towers, there exists a ∈ clop (X ) such that a ∪ X k n n∈ω\{k} witnesses that C is not a tower and the claim has been proved. Let x := X and B := A (x ). Since x is not element of A, B is a simple n∈ω\{0} extension of A, not equal to A. Furthermore, it is a minimal extension of A. Indeed, both A x and A −x are respectively {y ∈ I | y ∩ X =∅} and {y ∈ I | y ⊂ X } . 123 M. J. Santos It is easy to verify that Smp = I . Since I is a maximal ideal of A, it follows that A ≤ B. Trivially ⎧ ⎫ ⎨ ⎬ X | n <ω ⎩ ⎭ k≤n is a countable tower of B. Then ω = t (B) < t (A). Lemma 5 For any infinite Boolean algebra A, t (A) = ω iff a (A) = ω iff p (A) = ω. Proof Observe that if {a | n <ω} is a p-family in some Boolean algebra A, then {b | n <ω}, where b := a for each n <ω,isatowerin A;if {b | n <ω} n n i n i ≤n is an increasing tower in A, then {c | n <ω}, where c := b · (− b ) for each n n n i i <n n <ω, is an infinite partition of A; and if {c | n <ω} is an infinite partition of A,it is a p-family in A. From this lemma and the previous theorem this corollary follows. Corollary 1 There exist A and B Boolean algebras, A ≤ B such that p (B) < p (A) and a (B) < a (A). 2.2 Dense and irredundant families Opposite to what was concluded at the end of the previous subsection, the cardinal function π is fixed on minimal extensions. Theorem 5 Let A and B be infinite Boolean algebras. If A ≤ B, then π A = π B . ( ) ( ) See [5], Proposition 6.2. On the other hand there are cases where A ≤ B, and π (A) <π (B). Take for example A = P (ω) and B = P (ω) /Fin × P (ω) / {∅}. Trivially we have π (A) = ω, and taking as witness any almost disjoint family of P (ω) /Fin of size c it is verified that π (B) = c. Now, answering negatively to Problem 71, it will be proved impossible to have the opposite inequality. Definition 2 If I ⊆ A is an ideal, define π (I ) as the minimum size of a dense subset of I , i.e. a set D ⊂ I \{0} such that for all x ∈ I \{0} there exists y ∈ D such that y ≤ x. Lemma 6 If I , J ⊆ A are ideals such that I ∩ J ={0}, then π(A/I ) ≥ π(J ). Proof Suppose that D := {x | α< π(A/I )} ⊆ A\I , is a set of representatives of a dense family of A/I . Define D := {x ∈ D |∃y ∈ Jsuch that [x ] ≤ [y ] }. α α α α I I For each x ∈ D ,fix y and define z := x · y . Since x ∈ / I , while x · (−y ) ∈ I , α α α α α α α α for any x ∈ D , it follows that z ∈ / I . In particular z = 0, for all x ∈ D .The setof α α α α 123 Questions on cardinal invariants... such z is dense on J . Indeed, if y ∈ J \{0}, there exists x ∈ D, such that [x ] ≤ [y] . α α α I It follows that x ∈ D . Suppose that z y. Then 0 = z · (−y) ≤ x · (−y) ∈ I . α α α α But z ∈ J , and thus 0 = z · (−y) ∈ I ∩ J which is a contradiction. Therefore, α α z ≤ y. Since we got a dense subset of J of at most π(A/I ) many elements, it follows that π(A/I ) ≥ π(J ). Theorem 6 Let A and B be infinite Boolean algebras. If A ≤ B, then π (A) ≤ π (B). Proof Remember that if A ≤ B, then there exist I and I ideals of A such s 0 1 that I ∩ I = {0} and B (A/I ) × (A/I ). It follows easily that π (B) = 0 1 0 1 max {π (A/I ) ,π (A/I )}. 0 1 Applying Lemma 6 to I and I , we get that π(A/I ) ≥ π(I ) and that π(A/ 0 1 0 1 I ) ≥ π(I ). We can also apply it to 1 0 I := {a ∈ A |∀x ∈ I ∀y ∈ I a · x = 0 = a · y}, 0 1 which is an ideal such that I ∩ I ={0}= I ∩ I , to conclude that π(I ) ≤ π(A/ 0 1 I ), π(A/I ). Since at least one of these last two cardinals is infinite, we conclude that 0 1 π(I ) + π(I ) + π(I ) ≤ π(A/I ) + π(A/I ) = π(B). 0 1 0 1 Now we prove that the left side of the last inequality is ≥ π(A).Take D ⊆ I , 0 0 D ⊆ I and D ⊆ I , dense subsets of minimum size of the respective ideals. Take 1 1 x ∈ A . If there exists, for some i < 2, some non-zero y ∈ I such that y · x = 0, i i i then there exists z ∈ D such that z ≤ x. If this is not the case, then x ∈ I and there exists z ∈ D such that z ≤ x.So D ∪ D ∪ D is a dense subset of A. 0 1 Then π (I ) + π (I ) + π I ≥ π (A). Finally we conclude that π(A) ≤ π(B). 0 1 Finally, we state that Irr is not moved by simple extensions, giving an answer to problem 86. In order to prove this we need the following theorem (see [5], Theorem 8.4.). Theorem 7 Irr (A × B) = max {Irr (A) , Irr (B)} for all infinite Boolean algebras A and B. Theorem 8 Let A and B be infinite Boolean algebras. If A ≤ B, then Irr A = ( ) Irr (B). Proof Since any irredundant subset of A is an irredundant set of B it follows that Irr (A) ≤ Irr (B). We also know that there exist two ideals I and I of A such that 0 1 I ∩ I = {0} and B (A/I × A/I ). 0 1 0 1 Theorem 7 tells us that Irr (B) = max {Irr (A/I ) , Irr (A/I )}.Wlog Irr (B) = 0 1 Irr (A/I ). We will consider two cases. Case 1. Irr (A/I ) is a successor cardinal. Suppose that Irr (A/I ) = κ and take 0 0 + + [a ] | α< κ an irredundant family of A/I .If a | α< κ were redundant, α 0 α + + we would have α< κ such that a is generated by a | α = β ∈ κ , while α β [a ] is not generated by a | α = β ∈ κ , which is clearly a contradiction. α β So κ ≤ Irr (A). 123 M. J. Santos Case 2. Irr (A/I ) is a limit cardinal. Suppose that Irr (A/I ) = λ and take 0 0 κ< λ.Let [a ] | α< κ be an irredundant family of A/I . As in the previous α I 0 case {a | α< κ} is irredundant, and it follows that κ ≤ Irr (A). Since λ is limit, we conclude that λ ≤ Irr (A). In either case Irr (B) ≤ Irr (A) and the equality is verified. 3 Free products Definition 3 (Free product) If A and B are two Boolean algebras, their free product, denoted A ⊕ B, is an algebra C such that there exist A , B ≤ C, such that A = A , B = B , C =A ∪ B := a · b | n <ω, a ∈ A , b ∈ B i i i i i <n and for all a ∈ A \ {0} and all b ∈ B \ {0}, a · b = 0. Given two Boolean algebras A and B, this algebra exists and is unique up to isomorphisms. ∼ ∼ In topological duality we have that if A = clop (X ) and B = clop (Y ),for some zero-dimensional compact Hausdorff spaces A and B, then A ⊕ B = clop (X × Y ). This algebra consists of sets of the form a × b ,for some n <ω, where i i i <n a ∈ clop(X ) and b ∈ clop(Y ), for all i < n. It is worth observing that i i a × b = ( a \ a ) × ( b ). i i i j i i <n ∅=J ⊆n i ∈J j ∈ / J i ∈J Therefore we can always assume without loss of generality either that the a ’s are disjoint, or that the b ’s are disjoint. More basic information on free products can be found in [4], volume 1, chapter 4. From now on A and B will be two infinite Boolean algebras isomorphic to (and often interchanged with) the algebra of clopen sets of two zero-dimensional compact Hausdorff spaces X and Y . Theorem 9 Let A and B be two infinite Boolean algebras. Then – p (A ⊕ B) ≤ min {p (A) , p (B)}, – t (A ⊕ B) ≤ min {t (A) , t (B)}, – a (A ⊕ B) ≤ min {a (A) , a (B)} and, if A and B are atomless, – s (A ⊕ B) ≤ min {s (A) , s (B)} . Proof Take {a | α< p(A)}⊆ A a centered family with no pseudointersection. It follows that {a × Y | α< p(A)} is a centered family of A ⊕ B. If it had a + + pseudointersection, wlog of the form a ×b with a ∈ A and b ∈ B , then a wouldbea pseudointersection of the a ’s, which is a contradiction. Therefore {a ×Y | α< p(A)} α α has no pseudointersection and p(A ⊕ B) ≤ p(A). Analogously p(A ⊕ B) ≤ p(B). For the other cardinal functions the proofs are analogous. 123 Questions on cardinal invariants... Considering these simple inequalities a natural question to arise is whether equality holds in any of the four cases or any of them can be strict. Monk asks these questions (Problems 8, 46, 52) for p, a and s, and claims to have an affirmative answer for t. Here we give an affirmative answer for p and s, and provide other, perhaps better, proof for t.For a we give a lower bound. Theorem 10 s (A ⊕ B) = min {s (A) , s (B)}, for atomless infinite Boolean algebras A and B. Proof Suppose that κ< min {s (A) , s (B)} and that ⎧ ⎫ ⎨ ⎬ α α C := c := a × b | α< κ i i ⎩ ⎭ i <n is a subset of A ⊕ B. It follows that neither a | α< κ, i < n is a splitting family of A, nor b | α< κ, i < n is a splitting family of B.Let a and b non-empty witnesses of this fact, which means that: α α 1. ∀α< κ ∀i < n , either a ∩ a =∅ or a ⊂ a , and i i α α 2. ∀α< κ ∀i < n , either b ∩ b =∅ or b ⊂ b . i i The set a × b witnesses that C is not a splitting family of A ⊕ B. Indeed, take α< κ and suppose that (a × b) ∩ c =∅. Then there exists i < n such that α α α α α α (a × b) ∩ a × b =∅. But thus a ∩ a =∅ and b ∩ b =∅, from whence follows i i i i α α that a ⊂ a and b ⊂ b and hence that (a × b) ⊂ c . Therefore for all α< κ i i either a × b and c are disjoint or a × b is subset of c . We finally conclude that α α κ< s (a ⊕ b). Recall that a maximal centered family U ⊆ A is called an ultrafilter and that if U is an ultrafilter of A and a ∈ U , then there exists i < n such that a ∈ U . i i i <n Theorem 11 p (A ⊕ B) = min {p (A) , p (B)}. Proof Suppose that κ< min {p (A) , p (B)} and that ⎧ ⎫ ⎨ ⎬ α α C := c := a × b | α< κ i i ⎩ ⎭ i <n is a centered family in A ⊕ B. Extend C to an ultrafilter U of A ⊕ B. So for all α< κ α α α α there is i < n such that a × b is an element of U . Hence a × b | α< κ α α i i i i α α α α α α is a centered family and both a | α< κ and b | α< κ are centered families i i α α in A and B respectively. Let a and b be some pseudointersections of each family. So α α a × b is a pseudointersection of a × b | α< κ and therefore of C. It follows i i α α that κ< p (A ⊕ B). The proof provided here is shorter and uses set-theoretic notation which seems to the author more intuitive when dealing with free products of Boolean algebras. 123 M. J. Santos Implicit in the last proofs is the fact that if we have in A ⊕ B a centered family with no pseudointersection (resp. splitting) is because its projection on one coordinate is also a centered family with no pesudointersection (resp. is also splitting). So these structures on A ⊕ B strongly “inherit” the behavior they have on A and B. α α On the other hand, if on A ⊕ B we have c := a × b ,for α< κ, which i <n i i form a strictly decreasing family, and choose, through and ultrafilter, some i < n α α for each α, as in the last proof, we can be sure only of taking on each coordinate a centered family, not precisely a decreasing family, and hence any hypothesis regarding t(A) or t(B) cannot be immediately used. For this reason, even when next result is analogous to both the previous ones, its proof is considerably less trivial. Theorem 12 t (A ⊕ B) = min {t (A) , t (B)}. Proof Let κ be a regular cardinal less than min {t (A) , t (B)} and ⎧ ⎫ ⎨ ⎬ α α C := e := a × b | α< κ i i ⎩ ⎭ i <n be a strictly decreasing family of A ⊕ B. It will be proved that C has a pseudointer- section. First suppose that κ = ω. Since ω< t (A) and ω< t (B), from Lemma 5 it follows that ω< p (A) and ω< p (B). Therefore, from Theorem 11, we conclude that ω = κ< p (A ⊕ B) ≤ t (A ⊕ B) and that C has a pseudointersection. Now suppose that ω< κ. Suppose also that for all α< κ and i < j < n we have α α that b ∩ b =∅. Since κ is a regular uncountable cardinal, we may also suppose that i j there exists n <ω such that n = n , for all α< κ. It will be proved inductively that for all n <ω such a strictly decreasing family has a pseudointersection. The basic step, when n = 1, is trivial. Suppose that n is greater than 1 and that the claim is already proved for all m ∈ n\ {0}.If α< κ and I is a nonempty subset of n, define α α α c := a \ a I i i i ∈I i ∈n\I and α α d := b . I i i ∈I Clearly for all α< κ α α α α c × d = a × b = e . I I i i I ∈P(n)\{∅} i <n Theorem4.40in[5], states this fact, although it mainly refers to the spectrum of towers of the Boolean algebras in question, i.e. the set of regular cardinals κ such that there is a tower of size κ in the respective Boolean algebra. Notice that the following proof also serves for proving said theorem. Instead of beginning with κ< min{t(A), t(B)} begin with κ/ ∈ (t (A) ∪ t (B)) and everything else follows. spec spec 123 Questions on cardinal invariants... Observealsothat forallα< β < κ and J , I ∈ P(n)\{∅},if c ∩c =∅,itfollows that J I β β β α α α d ⊆ d . Indeed, if there exist x ∈ c ∩ c and y ∈ d \d , we have that (x , y) ∈ e . J I J I J I α α α α α On the other hand (x , y)/ ∈ c × d , because y ∈ / d , and (x , y)/ ∈ c × d for I I I I I α α α I ∈ P(n)\{∅, I }, because x ∈ c and c ∩ c =∅. Therefore (x , y)/ ∈ e . But this I I I means that e e , which is a contradiction. β α Since the sets of the form c , I ∈P(n)\{∅} where α< κ, form a decreasing family of non-empty sets, and therefore a centered family extendable to an ultrafilter, for all α< κ there exists I ∈ P (n) \ {∅} such that α α c | α< κ is a centered family. So d | α< κ is a decreasing family. It follows I I α α that ⎧ ⎫ ⎨ ⎬ α α C := a × b | α< κ i i ⎩ ⎭ i ∈I is a decreasing family of non-empty sets. Wlog we can suppose that there exists a nonempty I ∈ P(n) such that I = I for all α< κ. Notice that { a | α< κ} is i ∈I i a centered family. If I = n, there exists m < n such that |I|= m. So, by induction hypothesis, C has a pseudointersection, and hence C also has a pseudointersection. If I = n,wehave two cases: Case 1. There exists α< κ such that β β 1 0 a ⊆ a i i i <n i <n for all α< β <β <κ. It is clear that there exists a ∈ A such that a ⊆ a 0 1 i <n i for all α< β < κ.If b ∈ B is a pseudointersection of b | α< κ ,itfollows i <n i that a × b is a pseudointersection for C. Case 2. For all α< κ there exist α< β <β <κ such that 0 1 β β 1 0 a a . i i i <n i <n β β 0 1 Observe that this means that there exists j < n such that b ∩ b =∅ for all i < n. j i β β 0 1 Indeed, suppose that for all j < n there exists i < n such that b ∩ b =∅ and j i β β 1 0 hence such that a ⊆ a (as it was observed when dealing with the d’s and c’s). Take i j β β β β 1 1 0 0 x ∈ a and j < n. Since a ⊆ a for some i < n, it follows that x ∈ a .We i <n i i j j As witnessed by this observation, observations (5), (6) and (7) in the proof of Theorem 4.40 in [5] heavily influenced the present proof. 123 M. J. Santos β β 1 0 conclude that a ⊆ a , which is contradiction. From this observation it i <n i <n i i β β follows that b ⊆ b for all β ≤ β< κ. i <n i i ∈n\{ j } i With this idea it is easy to inductively construct a cofinal set {β | α< κ} ⊆ κ and a sequence { j | α< κ} such that ⎧ ⎫ ⎨ ⎬ β β α α a × b | α< κ i i ⎩ ⎭ i ∈n\{ j } is a decreasing family. Applying the inductive hypothesis to this family we get a pseudointersection of C. Now take two disjoint elements of (A ⊕ B) ,say a × b and c × d. There is nothing preventing, say, a and c from not being disjoint, being enough that b and d are disjoint. Hence it can be observed that given some infinite disjoint family of A ⊕ B its projection to either coordinate must not be, not even something close to, a disjoint family. Nevertheless, it can be noticed also that if we restrict the same disjoint family to some subfamily such that, say, projecting to the second coordinate give us a centered family, then necessarily in the first coordinate we get a disjoint family. With this thoughts the following result was obtained. Theorem 13 min {min {a (A) , a (B)} , max {p (A) , p (B)}} ≤ a (A ⊕ B). Proof Take ω ≤ κ< min {min {a (A) , a (B)} , max {p (A) , p (B)}},sowlogwecan assume that κ< a (A) , p (B).Let P := {c | α< κ} be a disjoint family of A ⊕ B. Since each c can be replaced by the disjoint union of finitely many sets of the form i i a × b , as was observed at the beginning of this section, and κ is an infinite cardinal, α α we may suppose that each c := a × b . We will prove that P is not a partition of α α α A ⊕ B. We consider two cases. ≥ω <ω Case 1. For all E ∈ [κ] there exists F ∈ [E] such that b ⊆ b for all α β β∈F α ∈ E. Let F be such a finite subset for κ = E. Suppose that for some n <ω we have already defined F , for all i < n.Let F be such a finite subset for κ\ F .So i n i i <n recursively define a sequence {F | n <ω} of finite, pairwise disjoint subsets of κ such that if m < n <ω, then b ⊆ b . α α α∈F α∈F n m Extending the centered family ⎧ ⎫ ⎨ ⎬ b | n <ω , ⎩ ⎭ α∈F to an ultrafilter, it follows easily that we can choose for each n <ω some α ∈ F n n such that b | n <ω is a centered family. 123 Questions on cardinal invariants... Extend the set {α | n <ω} to D, a maximal subset of κ such that {b | α ∈ D} is n α a centered family. Since for all α< β elements of D, c and c are disjoint, it follows α β that {a | α ∈ D} is an infinite set of disjoint elements of A. Sinceκ< a (A), there exists a ∈ A witnessing that {a | α ∈ D} is not a partition. Also, κ< p (B), which means that there exists b ∈ B , such that b ⊆ b , for all α ∈ D. Hence a × b witnesses that P is not a partition of A ⊕ B. ≥ω <ω Case 2. There exists E ∈ [κ] such that for all F ∈ [E] there exists α ∈ E such that b b . α β β∈F Suppose that E is a maximal subset of κ with this property. It follows that {b | α ∈ κ\E } is a centered family in B. To prove it take α , ..., α ∈ κ \ E. Because α 0 n of the maximality of E,if i ≤ n, there exists F , a finite subset of E, such that b ⊆ b ∪ b . α β α α∈E β∈F If F := F , it follows that for all i ≤ n i ≤n b ⊆ b ∪ b . α β α α∈E β∈F Also, by hypothesis, we know that there exists α ∈ E such that b b . α β β∈F Hence ∅ = b \ b ⊆ b for all i ≤ n.Itfollows that {b | α ∈ κ\E } is a α β α α β∈F i centered family. If κ \ E is infinite, as in the previous case, it can be proved that P is not a partition of A ⊕ B. <ω Suppose that |κ \ E | <ω. By hypothesis we know that for all F ∈ [E] , Y = b . Since κ< p (B), it follows that there is b ∈ B such that b ∩ b =∅ for all β α β∈F α ∈ E. Clearly if a = X, then the set (X \ a ) × b would witness α α α∈κ\E α∈κ\E that P is not a partition of A ⊕ B. Suppose that this is not the case, so a = X. α∈κ\E Since κ \ E is finite, we may suppose that it is equal to {α | i ≤ n} for some n <ω. As it was observed when proving that {b | α ∈ κ\E } is a centered family, there exist <ω α ∈ E and F ∈[E ] such that ∅ = b \ b ⊆ b for all i ≤ n. In particular β α β∈F i b ∩ b =∅ for all i ≤ n.Also a ∩ a =∅ for some j ≤ n, which would imply α α α α i j that c ∩ c =∅. This is a contradiction, since P is supposed to be a disjoint family. α α So a = X and we conclude that P is not a partition. α∈κ\E All cases considered, it follows that κ< a (A ⊕ B). The complicated statement of the previous theorem can be translated as follows: If there exists a disjoint family of A ⊕ B which disproves the equality for a in Theorem 9, its size must be at least as big as either p (A) or p (B). We conclude this section pointing at a possibility for answering Problem 8. Sim- plifying to the case where A = B, it follows from Theorem 13 that if we wanted to get an example of a (A ⊕ A) < a (A), it would be necessary that p (A) < a (A).Now take the better known case where A = P(ω)/Fin. It is known to be consistent that p < a (see [1], [3]). Hinting at a possible negative answer to Problem 8, the following question arises: Question 1 Is it consistent that p = a(P(ω)/Fin ⊕ P(ω)/Fin)< a? 123 M. J. Santos 4 Other questions In this last section, as a kind of a appendix, we will briefly deal with a couple of questions that, though related to the cardinal invariants defined in the first section, remain somehow detached from those dealt with in sections 2 and 3. The first asks about the possibility of an alternative definition of p. That possible definition is related to the so called ramification sets. Definition 4 Let A be a Boolean algebra. X ⊂ A is said to be a ramification set if for all a, b ∈ X either a ≤ b, b ≤ a,or a · b = 0. Problem 48 asks: Is p (A) = min {|X|| Xis maximal ramif ication setof A}? Here this question is answered negatively. The algebra which proves this is an interval algebra. Definition 5 Let L be a linearly ordered set. The interval algebra of L, denoted Intalg (L), is the set algebra on L generated by the intervals of the form [a, b), where a ∈ L ∪ {−∞} and b ∈ L ∪ {∞}. It is easy to verify that for all linearly ordered set L, its interval algebra is the set of all [a , b ), where n <ω, a ∈ L ∪ {−∞}, b ∈ L ∪ {∞}, a < b and b < a i i i i i i i j i <n for all i < j < n. Take A := Intalg (ω ).The set {[n, n + 1) | n <ω} ∪ {[ω, ∞)} witnesses that a (A), and hence that p (A), is equal to ω. Now suppose that R := {c | n <ω} is ramification set of A. Each c is of the n n n n n form [α ,β ). Suppose that β < ∞ for all n <ω.Take i ≤m i i m n n β := sup{β | n <ω}. Clearly the interval [β, ∞), being disjoint to every element of R, witnesses that R is not a maximal ramification set. On the other hand, suppose that there exists n <ω such that β =∞. Then 0 m take n n n α := sup({α | n <ω}∪{β | n <ω ∧ β < ∞}). m m m n n n n n n n Take n <ω.If β =∞, then [α, ∞) is subset of [α ,β ) ⊆ c .If β < ∞, then m m m m n n n n n n [α ,β )< [α, ∞), and hence [α, ∞) is disjoint to c . Therefore [α, ∞) witnesses m m n n that R is not a maximal ramification set. We conclude that the minimum size of a maximal ramification set defines a cardinal function other than p. The following question introduces a kind of extension other than those of the Sect. 2. 123 Questions on cardinal invariants... Definition 6 Let A and B two Boolean algebras such that A ≤ B. It is said that A is σ -embedded in B, A ≤ B, if for all b ∈ B the ideal A b is σ -generated. Recall that an ideal I of some Boolean algebra A is σ -generated if there exists X ∈[I ] such that for all a ∈ I there exists x ∈ X such that a ≤ x. The following theorem answers negatively to Problem 70: Are there BAs A, B such that A ≤ B and π(A)>π(B)? Theorem 14 Let A and B two infinite Boolean algebras. If A ≤ B, then π (A) ≤ π (B). Proof A family X ⊆ B \{1} will be said to be cofinal in B if for all b ∈ B \{1} there exists x ∈ X such that b ≤ x. It is clear that X ⊆ B \{1} is cofinal in B iff {−x | x ∈ B} is dense in B. Therefore there exists X a cofinal family in B of size π (B).For x ∈ X take C ∈[A] , a set witnessing that A x is σ -generated. Take a ∈ A \{1}. Since X is cofinal in B, there exists x ∈ X such that a ≤ x.It follows that a is an element of A x. Hence there exists y ∈ C such that a ≤ y. Therefore Y := C x ∈X is cofinal in A. Since {−y | y ∈ Y } is a dense set of A of size π B · ω = π B ,it ( ) ( ) follows that π (A) ≤ π (B). Declarations Conflict of interest The author declares that they have no conflict of interest. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. Baumgartner, J.E., Dordal, P.: Adjoining dominating functions. J. Symbol. Logic 50(1), 94–101 (1985) 2. Blass, A.: Combinatorial cardinal characteristics of the continuum. In: Foreman, Matthew, Kanamori, Akihiro (eds.) Handbook of Set Theory. 2, 3, vol. 1, pp. 395–489. Springer, Dordrecht (2010) 3. Brendle, J.: Forcing and the structure of the real line: the Bogotá lectures. Lecture notes (2009) 4. Koppelberg, S.: Handbook of Boolean Algebras. North Holland (1989) 5. Monk, J.D.: Cardinal Invariants on Boolean Algebras, 2nd edn. Birkhauser (2014) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Archive for Mathematical Logic – Springer Journals
Published: Nov 1, 2023
Keywords: Cardinal invariants; Boolean algebras; Simple extensions; Free products; 03E05; 03G05
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