Theorems on Ultrafilters

Home , Filter (mathematics), Ultrafilter, Ultrafilter (set theory)

Leiden University

Bachelor project

Theorems on ultraﬁlters

Author: Supervisor: Tycho Neve Dr. K.P. Hart

{1, 2, 3}

{1, 2} {1, 3} {2, 3}

{1} {2} {3}

∅

August 25, 2013 Contents

1 Fundamental Matters 3

2 Independent sets and the number of ultraﬁlters 7

3 Nonlinearity of the Rudin-Keisler ordering 10

4 The existence of good ultraﬁlters 14

5 Appendix (3-set Lemma) 17

Preface

In this thesis we explore some interesting theorems about ultrafilters. We start by defining filters and ultrafilters on a set X as a collection of subsets of X, satisfying properties in such a way that we can intuitively think of the sets in the filter as being ‘large’. The notion of (ultra)filters and many of the results we will encounter are abstract and set- theoretic (e.g., some proofs use infinite combinatorics). However, filters are known to be applicable in a wide variety of mathematical areas, most notably in topology and model theory. Ideals of Boolean rings are the dual notion of filters on these sets, as an example of an occurence in algebra. Also, in functional analysis they may apply as a method of convergence, similar to the notion of nets (more specifically, for every filter base an associated net can be constructed). Despite these interesting applications, we will focus more on properties of ultrafilters themselves, but sometimes it will be hinted why it is useful to get a specific result. For example, knowing the number of ultrafilters on a set (see section 2) provides enough information to easily compute the number of topologies on these sets. In a different view, we can define a natural topology on the set of all ultrafilters on a given set X, this is called the Stone-Cechˇ compactification of X. Using ultrafilters on X is not the only method to obtain this topolo- gical space. This compactification (denoted by βX) has a very rich mathematical structure, for which we refer the reader to [3] for more details. In the first section we will give some fundamental definitions and examples, followed by a few basic, yet important corollaries and theorems. In the next part, we study independent sets and derive a method to compute the number of ultrafilters on any given set. The third section is concerned with the Rudin-Keisler ordering on ultrafilters; the main objective here is to prove that the ordering is non-linear. More specifically, we will indicate two non-comparable uniform ultrafilters. Finally, we devote the last section to a proof of the existence of a good, countably incomplete ultrafilter on any infinite set. The latter has some useful consequences in model theory, particularly because they make ultrapowers saturated. Preliminary knowledge required to read this thesis includes naive set theory, e.g., ordinals, cardinals, transfinite induction/recursion and the corresponding notation. We assume the

1 Axiom of Choice (AC), but will not excessively use cardinal arithmetic. The following is an example of a statement that I did not find necessary to prove, indicating the overall level: “The family [κ]<ℵ0 of finite subsets of an infinite cardinal number κ has cardinality κ.”1 Before we start, I would like to thank my supervisor K.P. Hart for many enlightening con- versations about the subject and for his general support throughout this year. While not concretely referred to anywhere in this thesis, the source [1] has proven to be most useful in the last two sections, therefore many credits are directed to this article.

1The proof is just |κ<ℵ0 | = S [κ]n = P |[κ]n| = κ · ℵ = κ, where [κ]n denotes the set of (ﬁnite) n∈ℵ0 n∈ℵ0 0 subsets of κ which have exactly n elements.

2 1 Fundamental Matters

Informally, given a set X, we would like to provide a way of deciding whether a subset Y ⊂ X is ‘large’. Deﬁnition 1.1. Let X be a set. A (proper) ﬁlter on X is a nonempty collection F ⊂ P(X) such that

1. ∅ ∈/ F ; 2. If A, B ∈ F then A ∩ B ∈ F ; 3. If A ∈ F and A ⊂ B ⊂ X then B ∈ F . First let us examine a few elementary consequences of this definition. By an induction argument it easily follows from (2) that a filter is closed under finite intersections. Also, from (3) it follows that X is a member of any filter on X. Note that P(X) is not a filter on X. However, it satisfies (2) and (3). Therefore it is sometimes called the improper filter on X, explaining the optional word ‘proper’ in the definition. Conversely, if F is a collection of subsets of X containing the empty set and satisfying (2) and (3), it follows from (3) that F = P(X), i.e., F is improper. Let us continue by defining the most central objects in this thesis. Definition 1.2. Let F be a proper filter on a set X. Then F is an ultrafilter on X if there is no proper filter G on X such that F ( G . In other words, if we denote by Ω(X) the set of all filters on X, then (Ω(X), ⊂) is a partially ordered set and ultrafilters are its maximal elements. If X is understood, we will simply speak of ‘(ultra)filters’ instead of ‘(ultra)filters on X’. Frequently used symbols for ultrafilters are U and V . There are many equivalent ways to define ultrafilters and throughout the thesis, the following is used frequently. Proposition 1.3. Let F be a proper filter on a set X. Then F is an ultrafilter if and only if for every A ⊂ X either A ∈ F or X\A ∈ F . Proof. Suppose F is not an ultrafilter. Let G be a filter such that F ( G and take A ∈ G \F . Then X\A/∈ F , for otherwise we would have ∅ = A ∩ (X\A) ∈ G by (2). Conversely, if there is A ⊂ X such that neither A nor its complement X\A is a member of F , it suffices to point out a proper filter G which extends F . Take G = {G ⊂ X :(∃F ∈ F ) F ∩ A ⊂ G}. First we need to show that G is actually a filter: 1. If ∅ would be a member of G then there is F ∈ F such that F ∩ A = ∅. But then F ⊂ X\A, so X\A ∈ F because F is closed under supersets, contradiction. We conclude ∅ ∈/ G .

2. If G1,G2 ∈ G then there are F1,F2 ∈ F such that F1 ∩ A ⊂ G1 and F2 ∩ A ⊂ G2. Now F1 ∩F2 ∈ F because F is closed under (ﬁnite) intersections and (F1 ∩F2)∩A ⊂ G1 ∩G2. Consequently, G1 ∩ G2 ∈ G .

3 3. If G ∈ F , take F ∈ F such that F ∩ A ⊂ G. Now G ⊂ H ⊂ X implies F ∩ A ⊂ H, hence H ∈ G for any H containing G. Because F ∩ A ⊂ F whenever F ∈ F , we have F ⊂ G . Also, A ∈ G \F (take, for example, F = X in the definition of G ). Hence, G is a proper filter on X strictly containing F . This proposition shows that ultrafilters actually ‘choose’ between subsets and their complements on a set. Certain intuitive properties are achieved by this selection, e.g. ultrafilters induce a 2-valued measure on the set. Generally, filters can be generated by certain collections of subsets. In the previous proof, we implicitly used the notion of a generated filter already. In fact, any collection of subsets enjoying the finite intersection property generates a filter as we will see in a moment. Definition 1.4. Let X be a set and A ⊂ P(X) a collection of subsets. Then A has the finite intersection property (FIP) if any finite intersection of sets in A is nonempty. In other words, A has (FIP) if and only if for every natural number n ∈ N and any sequence {A } ⊂ we have T A 6= . k k6n A k6n k ∅ Clearly, filters have (FIP) because they do not contain the empty set. Also, if A has (FIP) then it can not contain the empty set itself. Note that T A might be empty if A is infinite. T For example, A = {N\{n} : n ∈ N} ⊂ P(N) has (FIP) but A = ∅. Definition 1.5. Let X be a set and A ⊂ P(X) a collection of subsets. The (im)proper filter generated by A is the set \ hA i = {F ⊂ P(X): F ⊃ A and F is a(n) (im)proper filter on X} .

So hA i is the intersection of all (im)proper ﬁlters on X that contain A . Here follows an equivalent characterization that is often easier to work with. Proposition 1.6. Let X be a set and A ⊂ P(X) a collection of subsets. Then

hA i = {F ⊂ X :(∃n ∈ N)(∃A1,...,An ∈ A ) A1 ∩ · · · ∩ An ⊂ F }. Proof. Let us denote the set on the right hand side in the proposition by G . To prove G ⊂ hA i, take G ∈ G and let F be an (im)proper filter on X containing A . Because G ∈ G there are A1,...,An ∈ A such that A1 ∩ · · · ∩ An ⊂ G. Since A ⊂ F we have A1 ∩ · · · ∩ An ∈ F . Hence G ∈ F because F is closed under supersets. Since F was arbitrary, we have G ∈ hA i and therefore G ⊂ hA i. Conversely, by definition of hA i it suffices to show that G is an (im)proper filter on X containing A . Let G1,...,Gn ∈ G . Then there exist, for every 1 6 i 6 n, an ni ∈ N and a i T i sequence {A }1 j n ⊂ such that A ⊂ Gi. We have j 6 6 i A 16j6ni j

\ \ i \ Aj ⊂ Gi, 16i6n 16j6ni 16i6n and because the intersection on the left is ﬁnite, this yields G1 ∩ · · · ∩ Gn ∈ F . Closure of F under supersets is trivial as in proposition 1.3.

4 Clearly, if A has (FIP) in the context above then ∅ ∈/ hA i. Thus, hA i is a proper filter on X if and only if A has (FIP). Let us look at some natural examples of (ultra)filters and definine free, principal and non- principal filters. Let X be a set. •{ X} is called the trivial filter on X. For nonempty X this is a proper filter, but it is an ultrafilter (if and) only if X is a singleton set.

• For any subset ∅ 6= A ⊂ X, the set {Y ⊂ X : A ⊂ Y } is the filter generated by A; it equals h{A}i. If A is a singleton subset, i.e., A = {a} for some a ∈ X, then the principal filter generated by A consists of the subsets containing a. In this case, h{A}i is easily verified to be an ultrafilter by making use of proposition 1.6.

• The Fréchet filter on X, denoted by FRX , is the collection of all subsets with finite complement, i.e. cofinite subsets. If X is finite then FRX = P(X) is not a proper filter. For this reason it is only defined for infinite X in most texts. When X is infinite, clearly ∅ ∈/ FRX and if A, B are finitely complemented in X, then X\(A ∩ B) = X\A ∪ X\B is finite as well. Supersets of elements in FRX have an even smaller complement, so FRX is indeed a filter on infinite X. From now on, we will always assume X to be infinite if we work with the Fréchet filter on X. T • A filter F on X is free if F = ∅. Since filters are closed under finite intersections, no filter on a finite set is free. Indeed, any filter on a finite set is generated by the intersection of its members. T The Fréchet filter on X is free. Otherwise, take a ∈ FRX and take any Y ∈ FRX . Then a ∈ Y , but a∈ / Y \{a} ∈ FRX , contradiction. Hence, FRX is also non-principal. Another interesting property is that every free filter contains the Fréchet filter: let F be a free filter on X and let Y ∈ FRX . For x ∈ X\Y , take Fx ∈ F such that x∈ / Fx. Since X\Y is T finite, the intersection F := x∈X\Y Fx is a member of F and F ⊂ Y by construction. Thus Y ∈ F . If we assume (AC), however, the Fréchet filter is never an ultrafilter on X, because in that case we can write X = X1 t X2 where |X1| = |X2| = |X|. As a consequence, neither X1 nor its complement, X2, is a member of FRX . Note that there are non-principal filters that are not free either. The filter F = h{N\{2k} : T k ∈ N}i provides an example: we have F = N\E where E denotes the set of even natural numbers. Any F ∈ F has a finite complement, so N\E/∈ F . Thus, F is neither free nor principal. An interesting question is whether there exist non-principal ultrafilters on an infinite set. The (positive) answer to this requires the Axiom of Choice in the form of Zorn’s lemma. Under this assumption, we will prove that any collection with (FIP) can be extended to an ultrafilter. On the other hand, this so-called ultrafilter lemma does not imply (AC). Indeed, it is equivalent to the Boolean prime ideal theorem (BPIT), a well-known intermediate point between the axioms of Zermelo-Fraenkel set theory (ZF) and the theory that augments (ZF) with the Axiom of Choice, (ZFC).

5 Theorem 1.8. (Ultrafilter lemma) Let X be a set and suppose A ⊂ P(X) has (FIP). Then there is an ultrafilter U on X which contains all of A . Proof. We will apply Zorn’s lemma to the set B consisting of all proper filters on X containing A , partially ordered by set inclusion. Note that B is nonempty because hA i ∈ B. S S Let C be a chain in B. We will prove C ∈ B. First of all, ∅ ∈/ C since it is not included S in any element of C . If A, B ∈ C , then there are C1,C2 ∈ C such that A ∈ C1 and B ∈ C2. Since C is a chain, we have C1 ⊂ C2 without loss of generality. Consequently, A and B are S elements of C2 and since C2 is a filter, we have A ∩ B ∈ C2 ⊂ C . It is a trivial matter to verify that S C is closed under supersets, so we have S C ∈ B indeed. This union is obviously an upper bound of C in B. According to Zorn’s lemma, B has maximal elements. Let U be a maximal element of B. If F ⊃ U is a filter, then A ⊂ F . Hence F ⊂ U by maximality of U and we have U = F . So U is an ultrafilter and it contains all of A . This theorem directly implies the existence of free ultrafilters on infinite X by extending A = {X\{x} : x ∈ X} to an ultrafilter, for example.

6 2 Independent sets and the number of ultraﬁlters

This section is devoted to introduce the concept of independent sets and families of large oscillation. After defining them and proving their existence, we will use them to compute the number of ultrafilters on a given set. From now on, we will look at ultrafilters on cardinal numbers. Under the assumption of the Axiom of Choice, every set X can be identified with a cardinal number κ by means of a bijective map. Thus, the ultrafilters on X and κ are essentially the same in this situation. Definition 2.1. Let α, β and κ be cardinals and let S ⊂ βα. The family S is a family of κ-large oscillation if the following condition is satisfied: if λ < κ, {fζ : ζ < λ} are distinct elements of S , and {ηζ : ζ < λ} are elements of β, then there is ξ ∈ α such that fζ (ξ) = ηζ for all ζ < λ.

In this thesis we only need families of ℵ0-large oscillation. κ Theorem 2.2. Let κ be an inﬁnite cardinal. Then there is a family S ⊂ κ of ℵ0-large oscillation such that |S | = 2κ. Proof. We deﬁne

F := {(F, G, s): F ∈ [κ]<ℵ0 ,G ∈ P(P(F )) and s ∈ κG}.

(2|F |) If |F | < ℵ0 then |P(P(F ))| = 2 < ℵ0 6 κ. Also, since every G ∈ P(P(F )) is finite, we have |κG| = κ for G as such. Thus |F | = κ. Therefore, it is sufficient (by identifying F κ F with κ) to find S ⊂ κ such that |S | = 2 and S is of ℵ0-large oscillation. We define F f : P(κ) → κ by A 7→ fA where fA : F → κ is given by ( s(A ∩ F ) if A ∩ F ∈ G fA(F, G, s) = 0 if A ∩ F/∈ G.

To see that f is injective, let A 6= B ∈ P(κ). Then there is x ∈ A 4 B, say x ∈ A\B. Take F = {x}, G = {F } and s(F ) = 1. Then

fA(F, G, s) = s(A ∩ F ) = s(F ) = 1 6= 0 = fB(F, G, s), hence fA 6= fB. κ Therefore, the set S = {fA : A ∈ P(κ)} has cardinality 2 . It remains to prove that S is a family of ℵ0-large oscillation. Let n < ℵ0, {Am : m < n} be a family of distinct subsets of κ and {ηm : m < n} a subset of κ. For ` < m < n we choose a`,m ∈ A` 4 Am and we set F = {a`,m : ` < m < n}. We deﬁne G = {Am ∩ F : m < n}, and a function s : G → κ by s(Am ∩ F ) = ηm. This function is well-deﬁned, for if ` 6= m, then a`,m is a member of only one of A` ∩ F and Am ∩ F , which shows that A` ∩ F 6= Am ∩ F . Clearly, (F, G, s) ∈ F and for m < n we have

fAm (F, G, s) = s(Am ∩ F ) = ηm. More generally, for cardinals λ and κ, we can show the existence of a 2κ-power subset of κκ of λ-large oscillation if a certain condition is satisﬁed. This condition is that κ can be written

7 as the sum of all κξ, where ξ < λ. This assumption is vital for F to be of cardinality κ in the proof above (see [3, p. 76] for details). Note that the condition is automatically satisfied when λ = ℵ0. Another, closely related concept will be needed in the next section. Definition 2.3. Let S ⊂ P(X). Then S is said to be independent (over X) if for any n, m ≥ 0 and any finite sequence A1,...,An,B1,...,Bm ∈ S of distinct sets,

A1 ∩ · · · ∩ An ∩ (X\B1) ∩ · · · ∩ (X\Bm) 6= ∅. The existence of independent sets is a fairly straight consequence of theorem 2.2; Corollary 2.4. Let κ be an inﬁnite cardinal. Then there is an independent S ⊂ P(κ) of cardinality 2κ.

κ Proof. Let T be a family of ℵ0-large oscillation of functions κ → {0, 1} such that |T | = 2 . −1 Put S = {f ({0}): f ∈ S }. Let A1,...,An,B1,...,Bm ∈ S be distinct. Then there −1 −1 are distinct f1, . . . , fn and g1, . . . , gm in T such that Ai = fi ({0}) and Bj = gj ({0}) for 1 6 i 6 n and 1 6 j 6 m. Since T is of ℵ0-large oscillation, there is η ∈ κ such that −1 fi(η) = 0 and gi(η) = 1 for 1 6 i 6 n and 1 6 j 6 m. Using that κ\Bj = gj ({1}) for all 1 6 j 6 m, we see A1 ∩ · · · ∩ An ∩ (κ\B1) ∩ · · · ∩ (κ\Bm) 6= ∅, because η is an element of the intersection. Hence, S is independent over κ. Finally, assume f 6= g ∈ S . Then there is ξ ∈ κ such that f(ξ) 6= g(ξ). Since ξ ∈ f −1({0}) 4 g−1({0}), these κ preimages are distinct. Thus, |S | = 2 . We denote by U(X) the set of all ultraﬁlters on X.

(2κ) Theorem 2.5. Let κ be a cardinal. Then |U(κ)| = κ if κ < ℵ0, and |U(κ)| = 2 if κ ≥ ℵ0. Proof. If κ is a finite cardinal then every ultrafilter U on κ is generated by a singleton subset {a} (a ∈ κ). To see this, assume the contrary. Then for all a ∈ κ, κ\{a} ∈ U . Since κ is finite, the empty intersection of all κ\{a} is an element of U , which is impossible. So |U(κ)| 6 κ. If a 6= b then h{a}i= 6 h{b}i, so we have |U(κ)| ≥ κ as well. (2κ) For infinite κ, |U(κ)| 6 2 is clear from U(κ) ⊂ P(P(κ)). To prove the other inequality, we will exhibit a family of 2(2κ) distinct ultrafilters on κ.

κ Let S be a family of ℵ0-large oscillation of functions κ → {0, 1} such that |S | = 2 . For C ∈ P(S ), we set

B(C ) = {f −1({0}): f ∈ C } ∪ {f −1({1}): f ∈ S \C }.

If B1,...,Bn ∈ B(C ) then there are f1, . . . , fn ∈ S and δ1, . . . , δn ∈ {0, 1} such that −1 Bi = fi ({δi}) for all 1 6 i 6 n. Because S is a family of ω-large oscillation, there is η ∈ κ such that f (η) = δ for all 1 i n. This shows T B 6= , so ( ) has the ﬁnite i i 6 6 16i6n i ∅ B C intersection property. Thus, for all C ∈ P(S ) we can extend B(C ) to an ultraﬁlter U (C ) on κ. If C1 6= C2 ∈ P(S ) then there is f ∈ C1\C2, without loss of generality. For such f we −1 −1 have f ({0}) ∈ B(C1) ⊂ U (C1) and f ({1}) ∈ B(C2) ⊂ U (C2), hence U (C1) 6= U (C2) because f −1({0}) = κ\f −1({0}).

8 κ Since |P(S )| = 2(2 ), the subfamily

{U (C ): C ∈ P(S )}

(2κ) (2κ) (2κ) of U(κ) has cardinality 2 and this shows that |U(κ)| ≥ 2 , i.e., |U(κ)| = 2 . (2κ) Corollary 2.6. If |X| = κ ≥ ℵ0, then the number of topologies on X is also 2 . This cardinal being an upper bound follows in the same way as for ultrafilters. Since U ∪ {∅} is κ a topology on X whenever U is an ultrafilter on X, 2(2 ) is a lower bound as well. For finite X, however, computing the number of possible topologies on X is more involved (it is equal to the number of pre-orders2 on the set, see [6] for a relatively new article that proves this claim and uses graph theory in these computations).

2 A preorder on X is a relation 6 ⊂ X × X that is reﬂexive and transitive.

9 3 Nonlinearity of the Rudin-Keisler ordering

In this section we will study an ordering on ultrafilters, defined independently by M.E. Rudin and H.J. Keisler around 1970. The fundament of this definition is a rather categorical concept between collections of ultrafilters. Definition 3.1. Let X and Y be infinite sets and let f : X → Y be a function. Then we −1 define f∗ : U(X) → U(Y ) by U 7→ {A ⊂ Y : f (A) ∈ U }.

Thus, if we identify U with a 2-valued measure on X, f∗(U ) is the induced measure on Y in the usual measure-theoretic sense. However, we will not examine these measure properties in this thesis. Proposition 3.2. Let X,Y,Z be inﬁnite sets and let f : X → Y and g : Y → Z be functions. Then (g ◦ f)∗ = g∗ ◦ f∗. Proof. Let U ∈ U(X). Then we have

−1 (g ◦ f)∗(U ) = {A ⊂ Z :(g ◦ f) (A) ∈ U } = {A ⊂ Z : f −1(g−1(A)) ∈ U } −1 = {A ⊂ Z : g (A) ∈ f∗(U )} = (g∗ ◦ f∗)(U ).

Definition 3.3. (Rudin-Keisler order) Let X and Y be infinite sets and let U ∈ U(X), V ∈ U(Y ). Then U 6RK V if there exists a function f : X → Y such that U = f∗(V ). More often than not, the subscript ‘RK’ is omitted. This relation is transitive by proposition 3.2, which implies that the relation ≈ defined by

U ≈ V ⇔ U 6 V and V 6 U is an equivalence relation. The following theorem (see the appendix) shows that ≈ is a reasonable notion of equivalence: Theorem 3.4. Let X and Y be inﬁnite sets and let U ∈ U(X) and V ∈ U(Y ). Then

1. U ≈ V if and only if there are A ∈ U , B ∈ V and f : X → Y such that V = f∗(U ) and f A is injective and onto B;

2. If U ≈ V and |X| = |Y | then there is a bijection f : X → Y such that V = f∗(U ) −1 and U = (f )∗(V ).

The generalized Fréchet filter on an infinite set X is FRX = {A ⊂ X : |X\A| < |X|}. Note that we use the same notation as in section 1. An ultrafilter U on a set X is said to be uniform if |A| = |X| for all A ∈ U . Proposition 3.5. Let X be an infinite set. Then an ultrafilter U on X is uniform if and only if it contains FRX .

10 Proof. Suppose U does not contain FRX . Then there is A ∈ FRX \U , so X\A ∈ U but |X\A| < |X|. Therefore, U is not uniform. Conversely, if U is not uniform, take A ∈ U such that |A| < |X|. Then |X\(X\A)| = |A| < |X|, so X\A ∈ FRX \U . Hence, U does not contain FRX . The following proposition justifies the choice of specifying our investigation in these uniform ultrafilters. Proposition 3.6. Let X be an infinite set. If U ∈ U(X) is not uniform and A ∈ U is of least cardinality, then U ≈ U ∩ P(A) ∈ U(A). Proof. It is a trivial matter to verify that U ∩ P(A) is an ultrafilter on A. Next, we prove U 6 U ∩ P(A). Let f : A → X be the inclusion of A into X. Then

−1 f∗(U ∩ P(A)) = {Y ⊂ X : f (Y ) ∈ U ∩ P(A)} = {Y ⊂ X : Y ∩ A ∈ U ∩ P(A)} = U .

To show the other inequality, we take an arbitrary a ∈ A and deﬁne g : X → A by g(x) = x if x ∈ A and g(x) = a if x∈ / A. We need to prove the equality {Y ⊂ A : g−1(Y ) ∈ U } = U ∩ P(A). Suppose there is Y ⊂ A such that g−1(Y ) ∈ U but Y/∈ U . Then X\Y ∈ U , and so is g−1(Y ) ∩ A ∩ (X\Y ) = {x ∈ A\Y : g(x) ∈ Y } = ∅, contradiction. For the other inclusion, let Y ∈ U ∩ P(A) and suppose g−1(Y ) ∈/ U . Then its complement is a member −1 of U and so is Y ∩ (X\g (Y )) = {x ∈ Y : ¬((∃y ∈ Y )g(y) = x)} = ∅, contradiction.

We denote by Uu(X) the set of uniform ultraﬁlters over X. The main proof in this section reveals that 6RK restricted to Uu(X) is not a linear relation, i.e.

Theorem 3.7. If X is inﬁnite, there are U , V ∈ Uu(X) such that U 6 V and V 6 U .

Proof. We must construct U , V ∈ Uu(X) such that V 6= f∗(U ) and U 6= f∗(V ) for every function f : X → X. So we must have, for every such f,

(∃A ∈ U )(X\f −1(A) ∈ V ) and (∃B ∈ V )(X\f −1(B) ∈ U ). (∗)

κ Say |X| = κ. Fix an enumeration {fη : η < 2 } of all functions X → X. Using transfinite recursion, the construction will be carried out over the ordinals η < 2κ. We shall construct two increasing sequences of filters {Fη}η<2κ , {Gη}η<2κ and take U , V to be ultrafilters extending their respective unions. To be precise, we will insure that the following five properties hold for all η < 2κ:

1. Fη and Gη are ﬁlters on X;

2. If ξ < η, then Fξ ⊂ Fη and Gξ ⊂ Gη;

3. F0 = G0 = FRX ; S S 4. If η is a limit ordinal, we set Fη = ξ<η Fξ and Gη = ξ<η Gξ; −1 −1 5.( ∃A ∈ Fη+1)(X\fη (A) ∈ Gη+1) and (∃B ∈ Gη+1)(X\fη (B) ∈ Fη+1). Conditions (1)–(4) would not necessarily present a problem, but (5) may become impossible at some stage η. For example, if η is a limit ordinal, the construction before stage η determines what Fη and Gη must be, and it might happen that they are already ultraﬁlters (which

11 cannot be extended anymore) and that Fη = (fη)∗(Gη). To prevent such failure, we enlist the aid of a generalized concept of independent sets from section 2. Definition 3.8. If S ⊂ P(X) and F is a filter on X, then S is independent modulo F if and only if whenever n, m ≥ 0 and A1,...,An,B1,...Bm are distinct elements of S , we have (X\A1) ∪ · · · ∪ (X\An) ∪ B1 · · · ∪ Bm ∈/ F . Note that S is independent over X (see section 2) if and only if S is independent modulo {X}. If S 6= ∅ is independent modulo F , then F is not an ultrafilter because for any S ∈ S , we would have S/∈ F and X\S/∈ F (see proposition 1.3). We formulate the next property in a lemma. Lemma 3.9. If S is independent modulo F and A ⊂ S , then S \A is independent modulo hF ∪ A i. Proof. We first need to show that F ∪ A enjoys (FIP), so that G := hF ∪ A i is a filter. Let A1,...,An ∈ A and F ∈ F . If A1 ∩ · · · ∩ An ∩ F = ∅, then F ⊂ (X\A1) ∪ · · · ∪ (X\An), so (X\A1) ∪ · · · ∪ (X\An) ∈ F as a superset of F ∈ F . But this is impossible since S is independent modulo F . To continue, we will show that S \A is independent modulo G . If we assume the contrary, then there are B1,...,B`,C1,...,Cm ∈ S \A (distinct), A1,...An ∈ A and F ∈ F such that (X\B1) ∪ · · · ∪ (X\B`) ∪ C1 ∪ · · · ∪ Cm ⊃ F ∩ A1 ∩ · · · ∩ An.

By ﬁrst taking complements and then intersecting with A1 ∩ · · · ∩ An on both sides, we get

B1 ∩ · · · ∩ B` ∩ (X\C1) ∩ · · · ∩ (X\Cm) ∩ A1 ∩ · · · ∩ An ⊂ X\F. Taking complements again and using that F is closed under supersets shows that the complement of the left-hand side is a member of F , which is a contradiction since all Ai,Bj and Ck are distinct and S is independent modulo F . Furthermore, an independent set S ⊂ P(X) of cardinality 2κ exists by corollary 2.4 and it can be taken to be independent modulo FRX . To see this, we take g : X → X such that |g−1({x})| = κ for all x ∈ X. The set S 0 = {g−1(S): S ∈ S } is easily seen to be of κ cardinality 2 . Take S1,...,Sn,T1,...,Tm ∈ S to be distinct. Then

h −1 −1 −1 −1 i X\ (X\g (S1)) ∪ · · · ∪ (X\g (Sn)) ∪ g (T1) ∪ · · · ∪ g (Tm) is just −1 g (S1 ∩ · · · ∩ Sn ∩ (X\T1) ∩ · · · ∩ (X\Tm)), where the intersection is nonempty, since S is independent. Therefore, the cardinality of the inverse image is κ and −1 −1 −1 −1 (X\g (S1)) ∪ · · · ∪ (X\g (Sn)) ∪ g (T1) ∪ · · · ∪ g (Tm) ∈/ FRX . 0 So S is independent modulo FRX indeed.

To continue the proof, we keep a large family of sets Sη, independent modulo Fη and Gη. This prevents the ﬁlters on every level of the construction from being ultraﬁlters. Thus, in addition to (1)–(5), we arrange for the following to hold for all η < 2κ:

12 6. Sη is independent modulo Fη and Gη; κ 7. If ξ < η < 2 , then Sξ ⊃ Sη; κ 8. |Sη| = 2 ; T 9. If η is a limit ordinal, we set Sη = ξ<η Sξ;

10. Sη\Sη+1 is ﬁnite. κ Note that (8) is assured by (9) and (10), because we have |S0| = 2 by the previous argument. κ For limit ordinals η < 2 , Sη will be independent modulo Fη and Gη: assuming the contrary, we ﬁnd, without loss of generality, ζ < η and distinct sets A1,...,An,B1,...,Bm ⊂ X such that for all ξ < η we have A1,...,An,B1,...,Bm ∈ Sξ and

(X\A1) ∪ · · · ∪ (X\An) ∪ B1 ∪ · · · ∪ Bm ∈ Fζ .

This is not possible because Sζ is independent modulo Fζ . The following lemma glues everything together; by applying it twice (once to both filters) in every stage, we can actually carry out the construction. Lemma 3.10. Let H and K be filters over X. Let T ⊂ P(X) be infinite and independent modulo H and K . Let f : X → X. Then there are filters H 0 ⊃ H , K 0 ⊃ K , and a family T 0 ⊂ T such that T 0 is independent modulo H 0 and K 0, T \T 0 is finite, and, for some B ∈ H 0, X\f −1(B) ∈ K 0. Proof. Fix A ∈ T . We distinguish two cases: 1. T \{A} is independent modulo hK ∪ {X\f −1(A)}i. We can take T 0 = T \{A}, K 0 = hK ∪{X\f −1(A)}i, H 0 = hH ∪{A}i, and B = A. Note that T 0 is independent modulo H 0 by a straightforward application of lemma 3.9. −1 2. T \{A} is not independent modulo hK ∪ {X\f (A)}i. Then then are distinct A1, ...,An,B1,...,Bm ∈ S \{A} and K ∈ K such that

−1 (X\A1) ∪ · · · ∪ (X\An) ∪ B1 ∪ · · · ∪ Bm ⊃ K ∩ (X\f (A)).

Hence, −1 A1 ∩ · · · ∩ An ∩ (X\B1) ∩ (X\Bm) ∩ K ⊂ f (A). 0 0 0 Take K = hK ∪ {A1,...,An,X\B1,...,X\Bm}i, H = hH ∪ {X\A}i, T = −1 −1 T \{A, A1,...,An,X\B1,...,X\Bm} and B = X\A. Then X\f (X\A) = f (A) ∈ K 0. By using an argument similar to the proof of lemma 3.9, we can see that T 0 is 0 0 independent modulo K and H . κ κ What we have now are sequences {Fη : η < 2 } and {Gη : η < 2 } satisfying all conditions S S previously stated. Extend F = η<2κ Fη and G = η<2κ Gη to ultraﬁlters U and V , respectively. Let f : X → X. Then f appears in the enumeration of all functions X → X we κ started out with, so there is η < 2 such that f = fη. If we take A and B as in condition (5), they obviously suﬃce for (∗) as well. Finally, proposition 3.5 implies that U , V ∈ Uu(X) because they both contain FRX . This concludes the proof of theorem 3.7.

13 4 The existence of good ultraﬁlters

The interest of good ultrafilters in model theory is that they make ultrapowers saturated. If A and B are two infinite structures of cardinality less than or equal to κ and U is a good, countably incomplete ultrafilter on κ, then, as Keisler showed in [4], the ultrapowers Aκ/U and Bκ/U have power 2κ and they are κ+-saturated. Thus, if 2κ = κ+ (this is (GCH) at level κ), and A and B are elementarily equivalent, then Aκ/U and Bκ/U are isomorphic. Here we will define and prove the existence of a good, countably incomplete ultrafilter. For details about ultrapowers and saturation in model theory, the reader is referred to [2] and [5]. To understand the main proof in this section, we need to be familiar with the concept of cofinality. Definition 4.1. Let α be a limit ordinal. The cofinality of α, written as cf (α), is the least ordinal β such that there is an increasing β-sequence {αξ}ξ<β with limξ→β αξ = α. We list a few easy-to-prove consequences and examples here;

• cf (ω + ω) = cf (ℵω) = ω; • cf (cf (α)) = cf (α) for any limit ordinal α; • cf (2κ) > κ for any infinite cardinal κ. Proving the last is a little more involved, see corollary 5.12 in [5, p. 54]. We need the following to define whether an ultrafilter is good or not. Definition 4.2. Let X be a set and let A , B ⊂ P(X). A function p:[X]<ℵ0 → A is • multiplicative, if p(s ∪ t) = p(s) ∩ p(t) for all s, t ∈ [X]<ℵ0 and

• monotone, if p(s) ⊃ p(t) whenever s ⊂ t ∈ [X]<ℵ0 .

<ℵ <ℵ <ℵ If p:[X] 0 → A and q :[X] 0 → B, we write p 6 q iff for all s ∈ [X] 0 we have p(s) ⊂ q(s). Definition 4.3. Let X be a set. An ultrafilter U on X is called good iff whenever <ℵ <ℵ p:[X] 0 → U is monotone, there is a multiplicative q :[X] 0 → U such that q 6 p.

To explain this, let us prove that every ultrafilter on a set X of size ℵ0 is good, as an example. <ℵ0 It suffices to consider X = ℵ0. Let U be an ultrafilter on ℵ0 and let p:[ℵ0] → U be <ℵ0 monotone. For a finite subset F ⊂ ℵ0 we set n(F ) = min{n ∈ ℵ0 : F ⊂ n} ∈ [ℵ0] , and <ℵ0 we define q :[ℵ0] → U by q(F ) = p(n(F )). Since F ⊂ n(F ) and p is monotone we have <ℵ0 q(F ) = p(n(F )) ⊂ p(F ) for F ∈ [ℵ0] , i.e., q 6 p. To prove that q is multiplicative, let <ℵ0 F,G ∈ [ℵ0] . Then n(F ∪ G) = n(F ) ∪ n(G) = max{n(F ), n(G)}, and hence

q(F ∪ G) = p(n(F ∪ G)) = p(max{n(F ), n(G)}) = p(n(F )) ∩ p(n(G)) = q(F ) ∩ q(G), since p is monotone. Another property we will need is completeness.

14 Definition 4.4. Let α and κ be cardinals. A filter F on α is κ-complete if T A ∈ F + whenever A ⊂ F and |A | < κ. A filter is countably incomplete if it is not ℵ0 -complete.

Note that every filter is ℵ0-complete. Now we can state the main theorem of this section: Theorem 4.5. On every infinite X there is a good, countably incomplete ultrafilter. In this section, we use a notion of independence slightly different from the independent sets in section 3. The following defines independent sets of functions, with respect to a filter. Definition 4.6. Let X be a set, F a filter over X and S ⊂ XX = {f : X → X}. Then S is independent from F iff, whenever f1, . . . , fn are distinct members of S and x1, . . . , xn ∈ X,

X\{x ∈ X : fi(x) = xi for all 1 6 i 6 n} ∈/ F .

We say S is independent iﬀ S is independent from {X}.3 If S is independent and ∅ ⊂ Y ⊂ X, then {f −1(Y ): f ∈ S } is easily seen to be an independent family of sets. Also, if S is independent and inﬁnite, S is independent from X κ FRX . The existence of an independent set S ⊂ X of power 2 follows directly from theorem 2.2 again, which will prove to be important at the base of the recursion we will carry out.

Proof of 4.5. Let {Aη}η<2κ enumerate P(X) and let {pη}η<2κ enumerate all monotone functions [X]<ℵ0 → P(X) in such a way that each monotone p:[X]<ℵ0 → P(X) is listed 2κ times. To prove the theorem, we construct sequences {Fη}η<2κ and {Sη}η<2κ which satisfy the following for all η < 2κ:

X 1. Fη is a ﬁlter over X, Sη ⊂ X and Sη is independent from Fη;

2. For ξ < η, Fξ ⊂ Fη and Sξ ⊃ Sη; κ 3. |Sη| = 2 ; S T 4. If η is a limit ordinal, Fη = ξ<η Fξ and Sη = ξ<η Sξ;

5. Sη\Sη+1 is ﬁnite; 6. is generated by sets B (n < ℵ ) such that T B = ; F0 n 0 n<ℵ0 n ∅

7. Either Aη ∈ Fη+1 or X\Aη ∈ Fη+1;

<ℵ0 <ℵ0 8. If pη :[X] → Fη, then there is a multiplicative q :[X] → Fη+1 such that q 6 pη.

S <ℵ0 By (7), U := η<2κ Fη will be an ultraﬁlter. If p:[X] → U is monotone, then, since κ κ cf 2 > κ, ran(p) ⊂ Fξ for some ξ < 2 . Applying (8) to some η > ξ such that pη = p ℵ0 shows that there is a multiplicative q :[X] → Fη+1 ⊂ U such that q 6 p. Thus, U will be good. Condition (6) insures that U will be countably incomplete. To make (6) hold, take κ S0 ∪ {f} to be independent and of power 2 . Take Bn = {x ∈ X : n < f(x) < ℵ0} and F0 = h{Bn : n < ℵ0}i.

3The reason I used the words ‘independent from’, diﬀerent from the source, is to distinguish between the independent sets from the previous section and independent families of functions w.r.t. a ﬁlter.

15 Conditions (1)–(4) will take care of themselves. To get (5), (7) and (8), we apply 4.6 and 4.7 at each stage η < 2κ; Lemma 4.7. If S is independent from F and A ⊂ X, there are S 0 ⊂ S and F 0 ⊃ F such that S 0 is independent from F 0, S \S 0 is ﬁnite and either A or X\A is a member of F 0. Proof. We distinguish two cases. 1. S is independent from hF ∪ {A}i. Take S 0 = S and F 0 = hF ∪ {A}i.

2. S is not independent from hF ∪ {A}i. Let f1, . . . , fn be distinct members of S and let x1, . . . , xn ∈ X such that

X\{x ∈ X : fi(x) = xi for all 1 6 i 6 n} ∈ hF ∪ {A}i. 0 Let S = S \{f1, . . . , fn} and

0 F = hF ∪ {x ∈ X : fi(x) = xi for all 1 6 i 6 n}i. 0 Note that X\A ∈ F . Lemma 4.8. If S is independent from F and p:[X]<ℵ0 → F is monotone, then there are S 0 ⊂ S , F 0 ⊃ F , and a multiplicative q :[X]<ℵ0 → F 0 such that S 0 is independent from 0 0 F , S \S is ﬁnite, and q 6 p. Proof. Fix g ∈ S . Let S 0 = S \{g}. For each t ∈ [X]<ℵ0 we let ( 0, if s 6⊂ t qt(s) = p(t), if s ⊂ t.

<ℵ0 Let tx (x ∈ X) enumerate [X] . Let

[ −1 0 q(s) = {qtx (s) ∩ g ({x})} and F = hF ∪ ran(q)i. x∈X

ℵ It is easily veriﬁed that q 6 p. To show that q is multiplicative, take r, s ∈ [X] 0 . We have

[ −1 q(r ∪ s) = {qtx (s ∪ r) ∩ g ({x})} x∈X [ −1 = {p(tx) ∩ g ({x})}

x∈X:r,s⊂tx " # " # [ −1 [ −1 = {p(tx) ∩ g ({x})} ∩ {p(tx) ∩ g ({x})}

x∈X:r⊂tx x∈X:s⊂tx " # " # [ −1 [ −1 = {qtx (r) ∩ g ({x})} ∩ {qtx (s) ∩ g ({x})} x∈X x∈X = q(r) ∩ q(s).

This concludes the proofs of 4.7 and 4.5.

16 5 Appendix (3-set Lemma)

In order to prove Theorem 3.4, we need a combinatorial lemma concerning a specific parti- tioning of sets. In fact, this lemma is by far more interesting than the proof of the theorem itself, in my opinion. We will apply the Compactness Theorem from Model Theory, which states that a theory has a model if every finite set of axioms in the theory has a model. The latter is often easier to prove by an induction argument, as will be demonstrated below. More generalized versions of the lemma are provable, for which we refer to [7]. Theorem (3-set lemma). Let X be an infinite set and let h: X → X be a function. Then there are pairwise disjoint X ,X ,X ,X ⊂ X with X = S X such that h(x) = x if 0 1 2 3 i63 i x ∈ X0 and h(Xi) ∩ Xi = ∅ if i = 1, 2, 3. 0 Proof. Let X0 = {x ∈ X : h(x) = x} and X := X\X0. We consider the language L with 0 h as the only function letter, for every x ∈ X an individual constant symbol cx (where cx and cy are distinct iff x and y are distinct) and three unary predicate letters A1, A2 and A3. We define a theory K of L by adding, for every x ∈ X0, the following axioms:

• A1(cx) ∨ A2(cx) ∨ A3(cx); h i •¬ (A1(cx) ∧ A2(cx)) ∨ (A1(cx) ∧ A3(cx)) ∨ (A2(cx) ∧ A3(cx) ;

• Ai(cx) ⇒ Ai+1 mod 3(h(cx)) ∨ Ai+2 mod 3(f(cx)) (i = 1, 2, 3). 0 Now if K has a model χ, then we can identify χ with X and interpret A1,A2 and A3 as the subsets X1,X2 and X3 we were looking for. To prove the existence of a model, we show that every finite subset of axioms has a model, which proves the claim because of the compactness theorem [8]. For a finite subset S of axioms, let F S ⊂ X0 be the set consisting of all x ∈ X0 such that there S is an axiom in S which demands a property of cx. Obviously, F is finite and it does not contain fixed points of f. We will use induction on the cardinality of F S to show that there S S S S S S S S S S are F1 ,F2 ,F3 ⊂ F , pairwise disjoint, such that F = F1 ∪ F2 ∪ F3 , and Fi ∩ f(Fi ) = ∅ (i = 1, 2, 3). Note that we implicitly use induction on the size of S. The case |F S| = 0 is trivial. Let m ≥ 1 and assume that for every subset S such that F S has cardinality m − 1, F S can be partitioned as desired. Let F T be of cardinality m. Construct T a graph Γ whose vertices are the elements of F . The vertices g1 and g2 are connected in Γ if and only if g1 = h(g2) or g2 = h(g1). Since h is a function, m is an upper bound for the amount of edges in F T . So there exists a vertex g ∈ F T which is connected with at most 2 T T T other vertices. By the induction hypothesis, F \{g} is the union of three sets G1 ,G2 and T T G3 such that Gi is totally disconnected (i = 1, 2, 3). It follows that there is 1 6 j 6 3 such T T that g is not connected with any vertex in Gj , so g can be added to Gj without making new T T T T connections. So we take Fj = Gj ∪ {g} and Fi = Gi if i 6= j. This completes the induction step.

17 Now we can prove the following theorem, which we recall from section 3: Theorem 3.4. Let X and Y be inﬁnite sets and let U ∈ U(X) and V ∈ U(Y ). Then

1. U ≈ V if and only if there are A ∈ U , B ∈ V and f : X → Y such that V = f∗(U ) and f A is injective and onto B;

2. If U ≈ V and |X| = |Y | then there is a bijection f : X → Y such that V = f∗(U ) −1 and U = (f )∗(V ). Proof.

1. Suppose U ≈ V . Take f : X → Y and g : Y → X such that f∗(U ) = V and g∗(V ) = U . Let h = g ◦f and k = f ◦g. Let X0,X1,X2,X3 ⊂ X and Y0,Y1,Y2,Y3 ⊂ Y be pairwise disjoint such that X = S X , Y = S Y , h X = id , k Y = id i63 i i63 i 0 X0 0 Y0 and Xi ∩ h(Xi) = Yi ∩ k(Yi) = ∅ for all 1 6 i 6 3.

Next we will show X0 ∈ U . If Xi ∈ U for i ∈ {1, 2, 3} then h(Xi) ∈ h∗(U ). But h(Xi) ∩ Xi = ∅, so U 6= h∗(U ). We conclude Xi ∈/ U . So X1 ∪ X2 ∪ X3 ∈/ U and X0 ∈ U . In a similar fashion, we infer Y0 ∈ V .

For x ∈ X0 we have (f ◦ g)(f(x)) = f((g ◦ f)(x)) = f(k(x)) = f(x), so f(x) ∈ Y0. The same calculation shows that f(y) ∈ X0 whenever y ∈ Y0. So f X0 and g Y0 are each others inverse. Now A = X0 and B = Y0 satisfy the properties in the theorem. For the other implication, it suffices to show the existence of a function g : Y → X with g∗(V ) = U . If A and B are finite then U and V are principal ultrafilters, so they are generated by singleton subsets. More specifically, if there is x ∈ X such that U consists of all subsets that contain x, then V is just {C ∈ Y : f(x) ∈ C}, so clearly U ≈ V . If A and B are infinite, let x ∈ X and define g : Y → X by g(y) = f −1(y) if y ∈ B and g(y) = x if y∈ / B. 2. Take A, B and f as in part 1. If A and B are finite, the claim is obvious. If they are infinite, then there are subsets A0,A1 ⊂ A such that A = A0 ∪ A1, A0 ∩ A1 = ∅, |A| = |A0| = |A1| and A0 ∈ U . Now |X\A0| = |Y \f(A0)|. Take a function π : X → Y such that π(X\A0) = Y \f(A0) and π A0 = f A0. Then π∗(U ) = f∗(U ) = V and −1 π∗ (V ) = U .

18 References

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