ULTRAFILTERS AND HOW TO USE THEM BURAK KAYA Abstract. These are the lecture notes of a one-week course I taught at the Nesin Mathematics Village in S¸irince, Izmir,_ Turkey during Summer 2019. The aim of the course was to introduce ultrafilters, prove some fundamental facts and cover various constructions and results in algebra, model theory, topology and social choice theory that use ultrafilters. Contents 1. Big sets and small sets 1 2. Fantastic beasts and where to find them 2 3. Ultraproducts andLos'stheorem 5 4. Topologizing ultrafilters and Stone-Cech˘ compactification 9 5. Ultralimits and amenability of Z 11 6. Principal ultrafilters and dictators 11 References 11 The main purpose of this course is to introduce ultrafilters and their basic ap- plications in various fields of mathematics. This course is intended for third and fourth year undergraduate students and beginning graduate students. Although the necessary background from algebra, topology, set theory and model theory was briefly reviewed in class, it is not included in these lectures notes. We refer the reader to [Hun80], [Mun00], [Jec03], [SS17] and [Hod93] for the basic definitions and facts necessary for these lecture notes. 1. Big sets and small sets In this section, we shall introduce ultrafilters on sets, prove some basic facts about them and show their existence. Let X be a non-empty set. A filter on the set X is a collection of sets F ⊆ P(X) such that •; 2= F and X 2 F, • if A 2 F and A ⊆ B ⊆ X, then B 2 F, and • if A 2 F and B 2 F, then A \ B 2 F. The most basic example of a filter is the trivial filter F = fXg. For any non-empty subset X^ ⊆ X, the collection F = fA ⊆ X : X^ ⊆ Ag is also a filter which is called the principal filter generated by X. We shall next make some observations about filters which immediately follow from the definition. The intersection of a collection of filters on X is also a filter on X and the union of a collection of filters which are pairwise ⊆-comparable is 1 2 BURAK KAYA a filter on X. Filters, being closed under intersection of two sets, have the finite intersection property1. It turns out that collections with finite intersection property can be extended to filters as well. Proposition 1. Let X be a non-empty set and C ⊆ P(X) be a non-empty set with the finite intersection property. Then there exists a filter F ⊇ C on X. Proof. It is an exercise to the reader to check that the collection + F = fA : 9n 2 N 9C1;C2;:::;Cn 2 C C1 \ C2 \ :::Cn ⊆ Ag is a filter on X containing C as a subset. A filter U on the set X is said to be an ultrafilter 2 if it satisfies that • A 2 U or X − A 2 U for every A 2 P(X). An ultrafilter on a set can be considered as a tool to split the power set of that set into two: \Big sets" and \small sets". Big sets are those that are in the ultrafilter3. Indeed, in order for the reader to understand this motivation, we expect the reader to check that an ultrafilter U on a set X canonically induces a finitely additive measure µU : P(X) ! f0; 1g and vice versa. Let X be a non-empty set and x 2 X. Then, since x 2 A or x2 = A for every A ⊆ X, the principal filter generated by fxg is an ultrafilter and is called the principal ultrafilter generated by x. It is also clear that any ultrafilter which is also principal has a generating set that is a singleton and has to be of this form. Before we proceed, let us prove a basic but important fact. Proposition 2. Any ultrafilter on a non-empty finite set is principal. Proof. Let X be a non-empty finite set and U be an ultrafilter. Consider the set T A = U2U U. Since X is finite, so is the ultrafilter U. It then follows from the properties of a filter that A 2 U and hence A 6= ;. (Up to now, we have only used that U is a filter and indeed shown that any filter on a non-empty set is principal whose generating set is the intersection of all sets in the filter.) Let a 2 A. Being an ultrafilter, we know that either fag 2 U or X − fag 2 U. If it were the case that X − fag 2 U, then we would have a 2 A ⊆ X − fag which is a contradiction. Thus, fag 2 U. We claim that U is generated by a. If it were not, then there would be B 2 U such that fag * B in which case fag \ B = ; 2 U which is a contradiction. Therefore, U is generated by a and is principal. 2. Fantastic beasts and where to find them What about ultrafilters on infinite sets? Are there any non-principal ultrafilters on infinite sets? As we shall see soon, the answer to this question is affirmative. However, we will not be able to \explicitly see" any of these non-principal ultra- filters. In order to prove the existence of non-principal ultrafilters, we need to introduce a very special filter. 1 + A collection C of sets is said to have the finite intersection property if for every n 2 N and C1;C2;:::;Cn we have that C1 \ C2 \···\ Cn 6= ;. 2Our definition corresponds to the special case of that definition where one considers the Boolean algebra (P(X); \; [; ;; X) 3The notion of a filter has a dual notion, namely, the notion of an ideal. Ideals contain the empty set, are closed ⊆-downwards and with respect to finite unions. Small sets are those that are in the corresponding ideal. ULTRAFILTERS AND HOW TO USE THEM 3 Let X be an infinite set. The collection F = fA ⊆ X : X−A is finiteg consisting of cofinite subsets of X is a filter called the Fr´echetfilter on X. The Fr´echet filter is clearly non-principal. As will be shown in the next proposition, whether an ultrafilter is non-principal is determined by whether it contains the Fr´echet filter as a subset. Proposition 3. Let U be an ultrafilter on an infinite set X. Then U is non- principal if and only if it contains the Fr´echetfilter. Proof. Assume that U is principal, say, U = fA ⊆ X : a 2 Ag. Then, since fag 2 U, we have that X − fag 2= U. On the other hand, X − fag is cofinite. Thus U does not contain the Fr´echet filter. Assume that U does not contain the Fr´echet filter. Then there exists a cofinite A ⊆ X such that A2 = U and hence X − A 2 U. Enumerate the set X − A, say, X − A = fx1; x2; : : : ; xng. If it were that X − fxig 2 U for every 1 ≤ i ≤ n, then we would have n \ X − fxig = X − fx1; x2; : : : ; xng 2 U i=1 which is a contradiction, as the intersection of this set with X − A is empty. There- fore, there exists 1 ≤ i ≤ n such that X − fxig 2= U and hence fxig 2 U. Then, as in the proof of Proposition 2, we necessarily have fA ⊆ X : xi 2 Ag = U. Thus, in order to show the existence of non-principal ultrafilters, we just need to find one that contains the Fr´echet filter. How are we supposed to do that? Let us take the Fr´echet filter on an infinite set X. The Fr´echet filter is clearly not an ultrafilter. We can fix this by extending the Fr´echet filter as transfinitely going through all elements of P(X) and appropriately adding each element or its complement. Although such processes are usually done by transfinite recursion, we are not expecting every person taking this course to be familiar with transfinite recursion and hence we will handle this procedure using Zorn's lemma. Theorem 1 (Tarski). Let X be a non-empty set and F^ ⊆ P(X) be a filter. Then there exists an ultrafilter U on X such that F^ ⊆ U. Proof. Let P = fF : F is a filter and F^ ⊆ Fg and consider the partially ordered S set (P; ⊆). Given a chain C ⊆ P, as pointed out before, the set C is a filter on X containing F^ and is an upper bound for C. Therefore, (P; ⊆) satisfies the hypotheses of Zorn's lemma and has a maximal element, say, U is a maximal element of (P; ⊆). We claim that U is indeed an ultrafilter. Assume towards a contradiction that it is not. Then there exists A ⊆ X such that A2 = U and X − A2 = U. Consider the collection C = U [ fAg. We claim that C has the finite intersection property. Let X1;X2;:::;Xn 2 C. We split into cases. • Case 1: Assume that Xi 2 U for every 1 ≤ i ≤ n. Then X1\X2\· · ·\Xn 2 U ⊆ C since U has the finite intersection property. • Case 2: Assume that Xi 2= U for some 1 ≤ i ≤ n. By replacing these sets without changing their intersection, we can assume without loss of gener- ality that X1 = A and X2;:::;Xn 2 U. As U has the finite intersection property, we have that X2 \···\ Xn 2 U. It follows that any super- set of X2 \···\ Xn is in U. On the other hand, X − A2 = U and hence X2 \···\ Xn * X − A, that is, A \ X2 \···\ Xn 6= ;.