Modern descriptive set theory. Jindřich Zapletal Czech Academy of Sciences University of Florida

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1 Modern descriptive set theory Jindřich Zapletal Czech Academy of Sciences University of Florida

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3 Contents 1 Introduction 1 2 Polish spaces Basic definitions Production theorems Polish groups Universal objects Natural classes of mathematical objects form Polish spaces Countable groups Separable Banach spaces Borel sets, analytic sets Borel hierarchy Projective hierarchy Wadge hierarchy Borel and analytic sets Separation theorems Uniformization Coding of Borel sets Examples Borel sets Analytic sets Effective theory Borel equivalence relations Examples Ideal equivalences Isomorphisms of structures Group actions and orbit equivalences Constructing the map The map description id E E iii

4 iv CONTENTS K σ C E S E Σ G max Some proofs Determinacy Games: basic definitions Basic determinacy results Applications to abstract analysis Perfect set property Baire category Lebesgue measure and capacities Superperfect set theorem Continuous reducibility Hausdorff measures Full determinacy Models of determinacy Well-ordered cardinals Non-well-ordered cardinals Periodicity theorems Inner models

5 Chapter 1 Introduction I wrote these notes as the text for a topics course in set theory at University of Florida in Spring The intended audience is a mix of graduate students specializing in mathematical logic, topology and abstract analysis. The course seeks to expose them to the basic ideas behind infinitary games, determinacy and their uses in these parts of mathematics. I expected the students would have previous exposure to very basic topology and set theory; they should understand notions such as topological space and cardinality. Other than that, there are no prerequisites. A necessary part of a rigorous development of the theory of infinitary games is the study of hierarchies of complexity for subsets of Polish spaces. The first chapter is a very rudimentary introduction to Polish spaces, expected to take no longer than three weeks. The second chapter introduces the Borel, projective and Wadge hierarchies, again expected to take no longer than three weeks. The third chapter defines infinitary games and states the key determinacy theorems. I find that there is no time to prove Borel determinacy within the confines of a semester-long course, and I suspect the proof would go well beyond the attention span of my intended audience; therefore I treat it as a black box. A student interested in set theory should certainly go through the proof. The fourth chapter then introduces a number of infinitary games relevant to set theory, topology and analysis. Since this is the main topic of the course, I hope to reserve at least six weeks for this chapter. The notation follows the set theoretic standard. Ordinal numbers, including natural numbers, are treated as von Neumann ordinals, and so α β indicates that α is an ordinal smaller than β. The letter c denotes the size of the continuum. 1

6 2 CHAPTER 1. INTRODUCTION

7 Chapter 2 Polish spaces 2.1 Basic definitions In this section, I will introduce the most basic concepts used in this textbook. The reader is assumed to be familiar with most of them. Definition A topological space is a pair X, O, where X is a set and O is the topology, a subset of P(X) including 0, X, and closed under finite intersections and arbitrary unions. The topology will be often clear from the context and not mentioned at all. The sets in the topology are called open, their complements are closed, sets which are both open and closed are called clopen. Topologies are often generated by a collection O gen of sets that we want to declare to be open. Just let O to be the closure of the set O gen on finite intersections and arbitrary unions. The category of topological spaces comes equipped with continuous functions and homeomorphisms. A map f : X Y is continuous if preimages of open sets are open. It is a homeomorphism if it is one to one, onto, and both it and its inverse are continuous. The continuity concept brings another common way of generating a topology O on a set X uses a collection F of maps into a topological space Y : O will be the smallest topology that makes all the functions in F continuous. In other word, O is generated by the preimages of open sets under functions in the collection F. Topological spaces come in many different flavors. In this book, we will be interested in Polish spaces, a very familiar kind of topological spaces that find uses in most parts of mathematics. Definition A topological space X is separable if it has a countable dense set, i.e. a set intersecting every nonempty open set. Definition A metric on a set X is a map d from X 2 to nonnegative reals such that d(x, y) = d(y, x), d(x, y) = 0 x = y and d(x, y) d(x, z) + d(z, y). 3

8 4 CHAPTER 2. POLISH SPACES The metric is complete if every Cauchy sequence has a limit. The associated metric topology is generated by open balls, sets of the form B(x, ε) = {y X : d(x, y) < ε}, for x X and ε > 0. Definition A topological space X is Polish if it is separable and completely metrizable, i.e. there is a complete metric on X that generates the topology on X. The nature of the metric generating a given Polish topology is mostly irrelevant in our considerations. There can be many very different metrics generating the same topology. One context in which the existence of a suitable metric becomes relevant is the Polish groups. Example A countable set with the discrete topology. Example The Cantor space 2 ω with the minimum difference metric. Just let d(x, y) = 2 n where n = (x, y) is the least number such that x(n) y(n). The topology is generated by sets of the form O t = {x 2 ω : t x}, where t 2 ω ranges over all finite binary sequences. Note that if O 2 ω is open and x O, then there must be a number n ω such that O x n O. The Cantor space possesses two important characteristics: it is zero-dimensional (this means that the topology is generated by clopen sets) and compact (every cover by open sets has a finite subcover). To prove compactness, assume for contradiction that C is an open cover of 2 ω with no finite subcover. By induction on n find bits b n 2 such that the sets O tn cannot be covered by finitely many elements of C, where t n = b 0, b 1,... b n 1. In the end, let x 2 ω be the sequence given by x(n) = b n and choose a set O C such that x O. Since the set O is open, there must be a number n ω such that O x n O. This contradicts the induction hypothesis at n though. Example The Baire space ω ω with the minimum difference metric. The Baire space is again zero-dimensional, but it is not compact. For example, the cover C consisting of sets O n : n ω does not have a finite subcover. In fact, The Baire space cannot be covered by countably many compact subsets. Example The real line with the Euclidean metric. The real line is not zero-dimensional; the only clopen subsets are 0 and R. It can be decomposed into two zero-dimensional subsets, such as the rationals and irrationals. An example of a Polish space that cannot be decomposed into countably many zero-dimensional subsets is the Hilbert cube, see below. The real line is not compact. However, it is locally compact: every point has a neigborhood whose closure is compact. Example Every separable Banach space with the norm metric.

9 2.2. PRODUCTION THEOREMS Production theorems Later on, I will have to verify that various sets form Polish spaces with a topology that is naturally derived from the context. This may not be quite easy. In this section, I will list a number of theorems that produce more complicated Polish spaces from simpler ones. Theorem If X is a Polish space and Y X is a G δ set, then Y with the inherited topology forms a Polish space. Here, a G δ set is one that is equal to the intersection of countably many open sets. Note that in Polish spaces, this includes all closed sets. If F X is closed then F = q O q where q ranges over positive rationals and O q is the open set of all points with distance < q from F. Proof. Suppose X is a Polish space with a complete metric d. Let me start with the case of an open set O X. To show that O with the inherited topology is Polish, first note that it is separable as any dense set of X is also dense in O. To find the metric, let F = X \ O and let e(x, y) = d(x, F ) 1 d(y, F ) 1. The function e satisfies the triangle inequality. Let d (x, y) = d(x, y) + e(x, y). This is a metric on O compatible with the topology. I will show that d is complete. If x n : n ω is a Cauchy sequence in d, then it is Cauchy in d as well and it has a limit x. The sequence d(x n, F ) 1 : n ω is Cauchy in the reals and so it must be convergent. In particular, it is bounded and so the numbers d(x n, F ) : n ω must be bounded away from zero. Since d(x, F ) = lim n d(x n, F ), it follows that x O and the set O is completely metrized by d. Now on to the case of a general G δ set. Suppose A = n O n is a countable intersection of open sets. Find a complete metric d n on each set O n using the previous paragraph; without loss of generality d n 1 for each n ω. Now let d (x, y) = Σ n 2 n d n (x, y). This is a complete metric on the set A. In fact, a subset A X of a Polish space is Polish in the inherited topology if and only if it is G δ, [?] Theorem Example The Cantor middle set with topology inherited from [0, 1] is a Polish space, in fact homeomorphic to the Cantor space 2 ω. Example The irrationals with topology inherited from R form a Polish space. This space is homeomorphic to the Baire space ω ω. Theorem If X n : n ω are Polish spaces then Π n X n with the product topology forms a Polish space. Example The higher-dimensional Euclidean spaces, as well as the Hilbert cube [0, 1] ω with product topology. Definition (The hyperspace) Let X be a Polish space. The hyperspace K(X) consists of all compact subsets of X with Vietoris topology. This topology

10 6 CHAPTER 2. POLISH SPACES is generated by open sets of the form {K : K U} and {K : K U = 0} as U ranges over all open subsets of X. The topology is generated by the Hausdorff metric: d(k, L) = max{e(x, L), e(y, K) : x K, y L}, where e is a complete metric on X and e(x, L) = min{e(x, y) : y L}. Theorem K(X) is a Polish space. If X is compact, so is K(X). Proof. To begin, the space K(X) is separable: if D X is a countable dense set, then collection of all finite subsets of D is a countable dense subset of K(X). To prove that the Hausdorff metric d is complete, let K n : n ω be a Cauchy sequence in it. Let K = n (closure of m>n K m). This is certainly a closed set; I will argue that it is compact and that it is a limit of the sequence. In order to prove the compactness of K in the original metric e on X, it is just enough to show that K is totally bounded, that means for every ε > 0 there are finitely many ε-balls covering K. Just choose n large enough so that d(k m, K n ) < ε/2 for every m > n, and cover the compact set m n K m with finitely many ε/2-balls. The ε-balls with the same centers will cover the set K. To see that the set K is indeed the limit of the sequence, let ε > 0 be a real number. Find a number n ω such that d(k m, K n ) < ε/2 for every m > n and argue that d(k, K m ) < ε for every m > n. Definition If X, Y are Polish spaces and X is compact, then C(X, Y ) is the set of all continuous functions from X to Y, with the topology induced by uniform convergence. Write C(X) for C(X, R). It may seem that the uniform convergence topology depends on the metric one chooses for the space Y, but this is in fact not true. A complete metric generating this topology is given by d(f, g) = max{e(f(x), g(x)) : x X}, where e is a complete metric on Y. To see the independence of the resulting topology on the choice of the metric e, note that a sequence of functions f n C(X, Y ) : n ω converges to f C(X, Y ) if and only if for every sequence x n X : n ω of points in the space X with limit x X, and every increasing sequence m n : n ω of natural numbers it is the case that the points f n (x mn ) : n ω converge to f(x) in the space Y. Theorem If X is a compact Polish space and Y is Polish, then C(X, Y ) is a Polish space. Proof. The completenes of the metric d is immediate. Suppose that f n C(X, Y ) : n ω is a Cauchy sequence of functions in the metric d. Then for every point x X, the sequence f n (x) Y : n ω is Cauchy in the metric e and therefore has a limit, call it f(x). It is easy to see that f is continuous and f is the limit of the functions f n : n ω. To see the separability, fix a metric c on the space X. Use the compactness of X to find a finite 2 n -net X n X for every number n ω. Also fix a countable basis O of the space Y. Note that every continuous function on a compact space is in fact uniformly continuous, and for numbers n, m ω consider the set

11 2.3. POLISH GROUPS 7 C m,n of all functions f C(X, Y ) such that x 0, x 1 X, c(x 0, x 1 ) < 2 n implies d(f(x 0 ), f(x 1 )) < 2 m. Thus for every function f C(X, Y ) and every number m ω there is a number n ω such that f C m,n. Now for every n ω and every function F : X n O, if there is a function f C(X, Y ) such that f(x) F (x) for every x X n, then choose one such a function f = f F. The collection of all such functions f F is countable and dense in C(X, Y ). Definition A Borel probability measure on a Polish space X is a function φ : B(X) [0, 1] such that φ(0) = 0, φ(x) = 1, A B φ(a) φ(b), and (countable additivity) if A n : n ω are pairwise disjoint Borel sets then φ( n A n) = Σ n φ(a n ). Definition If X is a Polish space then P (X) is the set of all Borel probability measures on X. The topology on P (X) is the one making all maps µ f dµ continuous, where f ranges over all continuous bounded real valued functions on X. Theorem If X is Polish then so is P (X). If X is compact Polish then so is P (X). Theorem (Banach) Suppose that X is a separable Banach space. The unit ball of the dual space in the weak* topology is a compact Polish space. The dual space is the space of all linear functionals f : X R. The weak* topology is the smallest one making all functions of the following form continuous: if x X then it generates a function F x : X R by F (f) = f(x). The norm on the dual space is given by f = sup{f(x) : x X, x = 1} where x is normed using a norm on X. The dual space may not be separable, but its unit ball is. 2.3 Polish groups Definition A group G, with topology τ is Polish if τ is a Polish topology on G and it is compatible with the group operation: the map x, y xẏ from G 2 to G is continuous and so is the map x x 1. Example Every countable group with discrete topology is Polish. Example R n is a Polish group with addition. So is T with multiplication. Example If X is a compact Polish space then H(X), the group of its homeomorphisms with composition operation, is Polish. Verify that it is a G δ subset of the Polish space C(X, X) and that the composition operation is continuous in the inherited topology. Example If µ is a Borel probability measure on X, then the group of measure-preserving transformations with composition is Polish.

12 8 CHAPTER 2. POLISH SPACES Example If d is a compatible metric on a Polish space X, then the group Iso(X) of isometries with composition is Polish. The topology is generated by functions f f(x) : x X. Example The unitary group, the group of unitary operators (linear isometries) on an infinite dimensional separable Hilbert space with composition, is Polish. The topology is generated by functions f f(x) : x Hilbert space. Example S, the group of permutations of ω with topology inherited from ω ω with the operation of composition, is Polish. The Polish groups generate a whole field of mathematical inquiry. I will quote only the most pressing problems of this field. Given a group G,, find a criterion for the existence of topology that makes it a Polish group. Given a Polish group G, find a criterion for existence of a compatible complete left-invariant metric. If G, H are Polish groups and π : G H is a group homomorphism, must π be continuous? Most issues though have to do with the notion of Polish actions. Definition If G is a Polish group and X is a Polish space, a Polish action of the group G on X is a continuous map a : G X X such that x X 1 x = x and g(hx) = (gh)x. In this case, we call X a G-space. Example Every Polish group acts on itself by conjugation. Example Let X be a Polish space and H(X) be its group of homeomorphisms. Then H(X) acts on X by g x = g(x). If X is compact, then H(X) is a G δ subset of the Polish space C(X, X) and therefore Polish. Example Let G be a countable group, considered with discrete topology. Then G acts on 2 G by shift: g x(h) = x(gh). Again, the Polish group actions generate many important problems in mathematics. I will state two concepts with examples. Definition A Polish group is amenable if every continuous action on a compact space admits an invariant Borel probability measure. The group is extremely amenable if every continuous action on a compact space has a fixed point. The compact spaces in this definition do not have to be Polish; restricting attention to Polish actions leads to a related, interesting, but not identical notion. Both amenability and non-amenability have a number of equivalent restatements; I chose the formulations that are most closely related to the work in this section. Which groups are amenable or extremely amenable and which are not? Every countable abelian group is amenable, while the free group on two generators is not; the latter fact leads to the paradoxical decomposition of the unit ball. The unitary group is extremely amenable, while S is not. Learn more in [?].

13 2.4. UNIVERSAL OBJECTS Universal objects Theorem Every uncountable zero-dimensional compact Polish space is homeomorphic to 2 ω. Theorem Every uncountable Polish space contains a homeomorphic copy of 2 ω. Proof. First, perform the Cantor-Bendixon analysis to remove a countable set of points from the Polish space X so that the closed remainder has no isolated points. Replace the space X with this closed remainder. By tree induction on t 2 ω build nonempty basic open sets O t X in such a way that O t has diameter at most 2 t t s implies the closure of O s is a subset of O t the closures of O t : t 2 n are pairwise disjoint. For every x 2 ω let f(x) =the unique element of n O x n. Then f : 2 ω X is a homeomorphism of the Cantor space and a closed subset of X. Theorem Every compact Polish space is a continuous image of the Cantor space. Proof. This is just a composition of several observations. First, f : 2 ω I defined by f(x) = Σ n 2 n 1 x(n) is a continuous surjection of 2 ω onto I. Thus f ω : 2 ωω I ω defined by f( x(n) : N ω ) = f(x(n)) : n ω is a continuous surjection as well. The spaces 2 ωω and 2 ω are homeomorphic. Every compact Polish space X is homeomorphic to a compact subspace of I ω, so there is a surjection of a closed subset of 2 ω onto X. Finally, there is a continuous surjection of Cantor space onto any of its closed subsets. Theorem Every Polish space X which is not a countable union of compact sets contains a closed copy of the Baire space. Proof. Let Y = X \ {O : O X : O basic open and coverable by countably many compact sets}. Y X is closed, therefore Polish, and no nonempty open subset of it can be covered by countably many compact sets. By tree induction on t ω <ω build nonempty basic open sets O t Y so that O t has diameter at most 2 t t s implies that the closure of O s is a subset of O t for every t ω <ω there is a real number ε > 0 such that the sets O t n : n ω are pairwise at least ε away from each other.

14 10 CHAPTER 2. POLISH SPACES For every x ω ω let f(x) =the unique element of n O x n. Then f : ω ω Y is a homeomorphism of the Baire space and a closed subset of Y. Theorem The Baire space contains a closed copy of any zero-dimensional Polish space. Proof. Let X be a zero dimensional Polish and d 1 a complete metric on it. By tree induction on t ω <ω build clopen sets O t X so that O 0 = X, s t O t O s, and O s i O s j = 0 whenever i j the diameter of O s is 2 s. To do this, given the set O t, cover it with countably many clopen sets P i : i ω of diameter 2 t 1 and let O t i = O t P i \ j i P j. Note that it can happen that some of the clopen sets will be empty. In the end, let C = {x ω ω : n ω O x n 0}. This is a closed set. Let f : C X be defined by f(x) =the unique point in n O x n. Check that this is a homeomorphism. Theorem Every Polish space is a continuous image of the Baire space ω ω. A word of warning: continuous images of the Baire space can be much more complex than just Polish spaces. In fact, every analytic set is a continuous image of the Baire space, see below. Proof. The proof is a composition of two observations: every Polish space is a continuous bijective image of a closed subset of the Baire space, and every nonempty closed subset of the Baire space is a continuous image of the whole Baire space. The latter is easy; I will concentrate on the former. Let X be a Polish space with a complete metric d 1. By tree induction on t ω <ω build F σ -sets A t ω so that A 0 = X, s t A t A s, and A s i A s j = 0 whenever i j the diameter of A s is 2 s A s = i A s i = i Ās i. Let me first argue that this is indeed possible. Suppose that A s has been constructed. First, write A s = n C n as an increasing union of closed sets. Note that for every number n ω, D n = C n+1 \ C n is an F σ set. Write D n = m Em n as a union of countably many closed sets of diameter 2 s 1. Let Fn m = En m \ k m Ek n and observe that Fn m is an F σ set. Now let A s i : i ω list the set Fn m : n, m ω in some way. This concludes the induction step.

15 2.4. UNIVERSAL OBJECTS 11 Once the induction is complete, consider the set C = {x ω ω : n A x n 0}, argue that C ω ω is closed and the function f : C X, given by f(x) =the unique element of the set n A x n, is a continuous bijection. The only nontrivial point is that the set C ω ω is closed. Suppose that x n : n ω are points in D converging to some point x ω ω. Note that f(x n ) : n ω is a Cauchy sequence in the space X, write y for its limit. Now for any number m ω, y Āx m since all but finitely many points f(x n ) belong to A x m. By the last item of the induction hypothesis, this means that for all m ω, y A x m, and so x C and f(x) = y. Theorem Every Polish space is homeomorphic to a G δ Hilbert cube [0, 1] ω. subset of the Proof. Let X be a Polish space with a complete metric d 1. Let x n : n ω be a dense set, and define a map f : X [0, 1] ω by f(x) = d(x i, x) : i ω. Verify that f is a continuous injection, its inverse is continuous, and its range is G δ. Theorem Every Polish space is homeomorphic to a closed subset of R ω. If one wants to embed Polish spaces with a metric into a universal metric space, a new concept, that of Urysohn space, is useful. This space has no known realization among pre-existing mathematical objects. It is characterized by several of its properties, and it has several rather abstract constructions. Definition A metric space is ultrahomogeneous if every isometry between two of its finite subsets can be extended to the isometry of the whole space to itself. It is universal if it contains an isometric copy of every complete separable metric space. There are many ultrahomogeneous spaces. The discrete countable metric space with distance given by d(x, y) = 1 if x y is clearly ultrahomogeneous. A much more sophisticated example is the unit ball of a separable infinite-dimensional Hilbert space [?, Chapter IV, paragraph 38]. There are also many universal metric spaces, such as C([0, 1], R) by the Banach-Mazur theorem [?]. However, there is exactly one space that satifies both of these properties at once: Theorem There is exactly on up to isometry, Polish ultrahomogeneous universal metric space, called the Urysohn space U. Proof. Both the existence and uniqueness present a challenge. I will outline the idea of Vershik [?]. Urysohn and Katětov [?] present other important ways to construct the space. Consider the space X of all metrics on ω with rational distances. This is a closed subset of Q ω ω and as such it is a Polish space. There is a dense G δ set B X such that every two metrics from B are isometric. In other words, there is such a thing as a generic countable metric space with rational distances. The Urysohn space U is the completion of this generic countable metric space.

16 12 CHAPTER 2. POLISH SPACES Theorem (Uspenskij) Every Polish group is homeomorphic to a closed subgroup of H([0, 1] ω ) and a closed subgroup of Iso(U). Theorem Every separable Banach space is isomorphic via a linear isometry with a closed subset of C([0, 1]). 2.5 Natural classes of mathematical objects form Polish spaces Countable groups The class of countable groups can be made into a Polish space in various ways. The two principal approaches are via a universal object and via a generic construction. For the first approach use the following well-known theorem: Fact Every countable group is isomorphic to a subgroup of F 2, the free group on two generators. Now equip F 2 with the discrete topology, 2 F2 with the product topology, and show that C = {G F 2 : G is a group} is a compact set. The set C inherits Polish topology from 2 F2 and in natural sense consists of countable groups, and contains an isomorphic copy of every countable group. The generic construction approach proceeds differently. Let ω 3 be equipped with Polish topology and 2 ω3 be equipped with the product topology. Now argue that the set D = {x 2 ω3 : x is a characteristic function of a group operation with 0 playing the role of the unit element} is a G δ set. Thus it inherits a Polish topology from the space 2 ω3, and it in natural sense consists of infinite countable groups, and it contains an isomorphic copy of every countable infinite group Separable Banach spaces

17 Chapter 3 Borel sets, analytic sets 3.1 Borel hierarchy Definition Given a topological space X, T, there is the smallest σ- algebra of subsets of X containing all open sets. Namely, by transfinite induction on α ω 1 define pointclasses Σ 0 α, Π 0 α so that: The Σ 0 0, Π 0 0 are left undefined, Σ 0 1 = T, the open sets, Π 0 1 =the closed sets. Σ 0 α collects exactly all countable unions of sets in β α Π0 β, Π0 α collects exactly the complements of all sets in Σ 0 α; or, restated, Π 0 α collects all countable intersections of sets in β α Σ0 β. In the end, let B = α ω 1 Σ 0 α. The sets in the pointclass B are called Borel subsets of X; the hierarchy of pointclasses Σ 0 α, Π 0 α is called the Borel hierarchy. The ambiguous classes Π 0 α Σ 0 α are denoted by 0 α. The classes Π 0 α, Σ 0 α are said to be mutually dual. Note that the enumeration of the hierarchy pointclasses begins with 1. For historical reasons, some mathematicians refer to Σ 0 2 sets as F σ, to the Π 0 2 sets as G δ, and then continue towards the more complex classes with F σδ, G δσ... The fact that the letters Π and Σ are boldface has a meaning, there are also the lightface pointclasses, but we will not use them in this book. The basic closure properties of the Borel pointclasses are recorded in the following basic proposition. Proposition The following is true for any class Γ {Π 0 α, Σ 0 α, 0 α : α ω 1 }: 1. Γ is closed under continuous preimages 2. Γ is closed under finite intersections and finite unions. 13

18 14 CHAPTER 3. BOREL SETS, ANALYTIC SETS Moreover, the Σ classes are closed under countable unions and the Π classes are closed under countable intersections. A set is Π 0 α if and only if its complement is Σ 0 α. Proof. Induce on α. The Borel sets are not closed under continuous images. They are closed under one-to-one continuous images though, see section??. The Borel hierarchy on Polish spaces closes off at ω 1, and not earlier. This is in fact a nontrivial statement, and to prove it I will need the notion of a universal set. Definition Let X, Y be topological spaces. A set A X Y is universal Π 0 α if it is Π 0 α and for every Π 0 α set B X there is a point y Y such that B = {x X : x, y A}. Similarly for Σ 0 α and other pointclasses defined in this book. Note that since the Π 0 α sets are closed under continuous preimages, the horizontal sections of a Π 0 α set A X Y are again Π α 0 : consider the continuous functions f : X Y, f(x) = x, y for various points y Y. Proposition Let X be a Polish space and α ω 1 be a countable ordinal. There is a universal Π 0 α subset of X ω ω. There is also a complete Σ 0 α subset of X ω ω. Proof. Induce on α. Corollary For every ordinal α ω 1, Π 0 α 0 α Σ 0 α. Proof. Suppose that A ω ω ω ω is a universal Π 0 α set. Let B = {x ω ω : x, x / A} ω ω. Then the set B is Σ 0 α it is the preimage of the complement of the set A under the continuous function x x, x. On the other hand, it is not Π 0 α, since if it were, there would be an x ω ω such that B = {y ω ω : y, x A} by the completeness of the set A. However, this is impossible: try to decide whether x B or not! Theorem If X and Y are Polish spaces then there is a Borel bijection f : X Y. Since Borel preimages and one-to-one images of Borel sets are again Borel, this means that the algebras B(X) of Borel subsets of uncountable Polish spaces X are all isomorphic. Thus this algebra can be called a really fundamental mathematical object. 3.2 Projective hierarchy Definition Given a topological space X, T, the hierarchy of projective sets is defined simultaneously for all spaces X ω ωn where n ω.

19 3.2. PROJECTIVE HIERARCHY 15 A set A X ω ωn is Σ 1 1 if there is a closed set C X ω ωn+1 such that A = p(c) = { x, a : b ω ω x, a, b C}. A set is Π 1 1 if it is complement is Σ 1 1. A set A X ω ωn is Σ 1 n+1 if there is a Π 1 n set B X ω ωn+1 such that A = { x, a : b ω ω x, a, b B}. A set is Π 1 n+1 if it is complement is Σ 1 n+1. If a set is in the collection n Σ1 n then it is called projective. Σ 1 1 sets are frequently called analytic, Π 1 1 sets are called co-analytic. The ambiguous classes Π 1 n Σ 1 n are denoted by 1 n. Proposition The following is true for all the superscript 1 pointclasses. closure under countable unions closure under countable intersections closure under continuous preimages. Moreover, both closed and open sets are analytic, and Σ 1 n, Π 1 n 1 n+1. Proof. Induce on n. The superscript 1 Σ pointclasses are closed under continuous images. prove this only for analytic sets. I will Proposition Let X be a Polish space. A set A X is analytic if and only if it is a continuous image of the Baire space. Proof. Suppose that A X is an analytic set. There is a closed set C X ω ω such that A = p(c). The set C with the inherited topology is a Polish space and as such is a continuous image of the Baire space. The set A in turn is a continuous image of the set C. On the other hand, if f : ω ω X is a continuous function then the set C = { x, y X ω ω : f(y) = x} is a closed set whose projection is the set rng(f). Corollary A continuous image of an analytic set is an analytic set. Proof. Let A X be an analytic set and g : A Y be a continuous function to a Polish space Y. The set A is a continuous image of the Baire space, A = rng(f). Therefore the set rng(g) Y is a continuous image of the Baire space as well, rng(g) = rng(g f), and by the Proposition it is analytic. Proposition?? implies that every Borel set is analytic. The opposite implication does not hold. To produce a counterexample, I will need a universal analytic set. Proposition Let X be a Polish space. There is a universal analytic set for X.

20 16 CHAPTER 3. BOREL SETS, ANALYTIC SETS Proof. Let C 2 ω (2 ω X) be a universal closed set for 2 ω X. Its projection into the first and third coordinates is the universal analytic set for X. Corollary There is an analytic set which is not Borel. Proof. A diagonalization argument. Suppose A 2 ω 2 ω is the universal analytic set. Suppose for contradiction that it is Borel. Then the set B = {x 2 ω : x, x / A} is also Borel, therefore analytic, and must be indexed as a section of the set A, B = A x for some point x 2 ω. Consider the question whether x B or not. Both answers yield a contradiction. 3.3 Wadge hierarchy There is a much finer hierarchy of sets than both Borel and projective hierarchies. It takes its simplest form on subsets of the Baire space ω ω. Definition Suppose that X, Y are Polish spaces and A X, B Y are sets. Say that A is Wadge reducible to B (A W B) if there is a continuous function f : X Y such that x A f(x) B. Proposition (Wadge s lemma) For Borel sets A, B ω ω, either A W B or B W (ω ω \ A). Under additional set theoretic hypotheses this is true for projective subsets of the Baire space as well, and under certain alternatives to the Axiom of Choice it is true for all subsets of the Baire space period. We will prove this proposition in Chapter??. The relation W is clearly a preorder, and its equivalence classes (where A W B A W B B W A) are called the Wadge degrees. The previous proposition shows that the structure of Wadge degrees on Borel (or projective) sets is very simple; it is essentially a well-order. The Borel classes introduced earlier form initial segments of this ordering by Proposition??. The Wadge s preorder should be viewed as rating Borel sets in terms of complexity. Given A, B X, how hard is it to verify the validity of x A or x B, for a given point x X? If A W B, then the continuous reduction reduces the former problem to the latter, and so the latter should be viewed as more complex. 3.4 Borel and analytic sets Separation theorems Theorem (Lusin separation theorem) Let X be a Polish space and A, B X its two disjoint analytic sets. Then there are disjoint Borel sets A, B X such that A A, B B.

21 3.4. BOREL AND ANALYTIC SETS 17 Proof. Call sets C, D X separated if there are disjoint Borel sets separating them. Note that if C = n C n and D = n D n and for every pair n, m ω of natural numbers the sets C n, D m are separated, then even the sets C, D X themselves are separated. Now suppose for contradiction that A, B X are two disjoint analytic sets which are not separated. Let f : ω ω A and g : ω ω B be continuous surjections as guaranteed by the previous Proposition. For every number n ω let A n = {f(x) : x ω ω, x(0) = n} and B n = {g(x) : x ω ω, x(0) = n}. By the previous paragraph, there must be numbers n 0, m 0 such that the sets A n0, B n0 are not separated. Proceeding by induction, produce two sequences n 0, n 1, n 2... and m 0, m 1, m 2... of natural numbers such that for every i ω the sets Āi = {f(x) : x ω ω, j i x(j) = n j } A and B i = {g(x) : x ω ω, j i x(j) = m j } B are not separated. Consider the points r = f(n 0, n 1, n 2... ) A and s = g(n 0, n 1, n 2... ) B. Since r s, there are disjoint open neighborhoods O r, O s separating the two points. Since the maps f, g are continuous, there must be a number i ω such that Āi O r and B i O s. However, this contradicts the non-separation of the sets Āi, B i. Corollary (Suslin s theorem) Let X be a Polish space. A set A X is Borel if and only if it is both analytic and coanalytic. Proof. On one hand, suppose that a set A X is both analytic and coanalytic. The disjoint analytic sets A, X \ A can be separated by Borel sets by Lusin separation, but the only sets separating them must be A and X \ A again. Thus the set A X is Borel. On the other hand, Proposition?? immediately implies that every Borel set is analytic. It follows that every Borel set is coanalytic as well because its complement is Borel and therefore analytic. Corollary A Borel one-to-one image of a Borel set is Borel. Proof. For simplicity I will deal with Borel functions with domain ω ω, the general case can be easily reduced to this one. Let f : ω ω X be an injective Borel function. Its range is clearly analytic; to prove that it is Borel, I must argue that it is coanalytic and apply Suslin s theorem. For every number n ω, the sets f O t : t ω n are pairwise disjoint analytic sets, since the function f is injective. Let P t f O t : t ω n be pairwise disjoint Borel sets obtained by Lusin s separation. Repeat this procedure for every number n ω and make sure that t s implies P s P t. Now x is in the range of f iff for every n, x t ω n P t, and moreover for every y ω ω, if n x P y n then x = f(y). This is a coanalytic condition Uniformization Definition Suppose that A X Y is a set. A uniformization of the set A is a partial function f : X Y such that its graph is a subset of A and whenever x X is a point such that the section A x is nonempty, f(x) is defined.

22 18 CHAPTER 3. BOREL SETS, ANALYTIC SETS Under the Axiom of Choice, every set can be uniformized. However, often we are interested in reasonably definable uniformizations as opposed to the arbitrary unpredictable ones produced by the Axiom of Choice. Suppose the set A is Borel (analytic, coanalytic... ). Does it have a Borel (analytic, coanalytic,... ) uniformization? The answer to this question is surprisingly nuanced. that cannot be uni- Proposition There is a closed set C 2 ω ω ω formized by an analytic function. Theorem (Kondo) Every coanalytic set has a coanalytic uniformization. Theorem Every Borel set with countable vertical sections has a Borel uniformization. In fact, it is a union of a countable collection of Borel functions. There are other important uniformization theorems for Borel sets. Theorem Every Borel set with vertical sections of mass > ε has a Borel subset with compact vertical sections of mass > ε. Theorem Every Borel set with meager vertical sections can be covered by a union of countably many Borel sets, each with closed nowhere dense vertical sections Coding of Borel sets Theorem Let X be a Polish space. There are a coanalytic set A 2 ω, analytic set B 2 ω X and coanalytic set C 2 ω X such that 1. for every point y A, B y = C y holds; 2. for every Borel set D X there is a point y A such that D = B y = C y. Thus, the vertical sections of the set B or C indexed by points in A are exactly all Borel subsets of X. 3.5 Examples Borel sets the collection of finite subsets of a compact space X is Σ 0 2 complete. the collection of nowhere dense compact sets is Π 0 2-complete. If µ is a probability measure then the collection of null compact sets is Π 0 2-complete. the set {x [0, 1] ω : x 0} is Π 0 3 complete. for every natural number n, the set {f C(T) : f C n (T)} is Π 0 3- complete. This is the set of all functions whose n-th derivative is continuous.

23 3.5. EXAMPLES Analytic sets There are many examples of analytic non-borel sets in practice. Before we delve into details, it is important to note that a powerful additional set theoretic axiom (the assumption that there is a measurable cardinal) implies that there is only one Wadge degree of such sets in ω ω. Thus, on a practical level, these sets must be continuously reducible to each other, if they are subsets of ω ω, and in other Polish spaces they will be reducible to each other via Borel functions. In fact, this is exactly what we will do: start with the universal analytic set and then extend our list of examples by bi-reducing other sets to sets already on the list. Note that as a practical consequence, the longer the list of examples gets, the easier it is to verify that a given set is non-borel analytic. More precisely, all analytic, non-borel sets identified below will be complete, in the sense of the following definition: Definition Let X be a Polish space. An analytic set A X is complete if for every zero-dimensional Polish space Y and every analytic set B Y, B W A. Clearly, an analytic set is non-borel if and only if its complement is a coanalytic non-borel set. Whether people prefer to look at a set or its complement is mostly a matter of historical development of a given field. I will now give examples of analytic and coanalytic non-borel sets. A tree on ω is a subset of ω <ω closed on initial segment. The tree is illfounded if it has an infinite branch, it is wellfounded otherwise. The set of all illfounded trees on ω is a complete analytic subset of P(ω <ω ). The set {K K[0, 1] : K contains an irrational} is analytic complete. A set x ω is a difference set if there is a set y Z such that x = { n m : n, m y}. The collection of all difference sets is complete analytic subset of P(ω) by a result of Schmerl [?]. isomorphism between countable structures with one binary relation is complete analytic. The coanalytic sets A X typically come with a coanalytic rank. This is a function φ from A into the ordinals such that there are an analytic relation a on X and a coanalytic relation c on X such that for x, y A x a y and x c y coincide with φ(x) φ(y) and A is downwards closed in both a and c. The following examples illustrate this slippery notion in a definite context. The collection uf countable compact subsets of 2 ω (or any other uncountable Polish space) is a complete coanalytic subset of K(2 ω ). The Cantor- Bendixson rank forms a coanalytic norm. The collection of scattered countable linear orders is complete coanalytic. A linear ordering is scattered if it contains no copy of the rationals. The Hausdorff rank forms a coanalytic norm.

24 20 CHAPTER 3. BOREL SETS, ANALYTIC SETS The collection of all separable Banach spaces with separable dual is complete coanalytic. The collection of all functions in C([0, 1]) which are differentiable everywhere is complete coanalytic. The set of all closed sets of uniqueness in K(T). Here, a set C T is a set of uniqueness if every trigonometric series converging to zero pointwise outside C is in fact zero. [?] There are naturally occurring universal analytic sets. They can be recovered from the following theorems. Theorem (Clemens) If X, d is a Polish space with a compatible metric, then its set of distances e X = {d(x, y) : x, y X} R + 0 is analytic. Moreover, every analytic subset of R + 0 can be obtained in this way. 3.6 Effective theory

25 Chapter 4 Borel equivalence relations Let X be a Polish space. An equivalence E on X is Borel (or analytic, coanalytic, etc.) if it is a Borel subset of the Polish space X 2 with the product topology. Most fields of mathematics come with an underlying equivalence of objects. Topology has homeomorphisms of spaces, group theory has isomorphisms of groups, ergodic theory has conjugacy of measure preserving transformations etc. Frequently, the most basic underlying problem is to be able to decide whether given two objects are equivalent. In our context, we will deal with the situation where the objects of the field form a Polish space in some natural sense, and the equivalence relation is Borel or analytic. Thus general topology with large spaces and homeomorphism is out; the unitary operators on a separable Hilbert space with conjugacy equivalence are in. The main reason I included the previous chapter in this textbook is that in it I showed that in various contexts, the collection of studied objects forms a Polish space, and the equivalence in question is analytic or Borel. This chapter will be concerned with comparison of various equivalence relations. Borel (analytic etc.) equivalence relations are ordered by the relation of reducibility. Definition Let E, F be equivalence relations on Polish spaces X, Y. E is reducible to F, E F, if there is a Borel map f : X Y such that x 0, x 1 X x 0 Ex 1 f(x 0 )F f(x 1 ). It is not difficult to see that is a partial quasi-order. One can consider competing notions of reducibility where the reducing map is continuous, or one-to-one, or analytic etc. However, the Borel reducibility is most frequently studied. Note that E F and F E does not imply E = F or even that E, F are Borel isomorphic. The reducibility should be viewed as a rating of mathematical problems according to their difficulty. For example, consider the equivalence relation E of homeomorphism of two-dimensional manifolds in R 4, and the equivalence relation F of isomorphism of countable groups. The reducibility E F in 21

26 22 CHAPTER 4. BOREL EQUIVALENCE RELATIONS essence means that it is possible, given a two-dimensional manifold in R 4, to compute in a reasonably effective fashion a countable group (such as a homotopy group of some sort) which characterizes the manifold up to homeomorphism. The computation reduces the problem of checking whether two manifolds are homeomorphic to the problem whether their associated groups are isomorphic. 4.1 Examples There are many examples of Borel and analytic equivalence relations. Some of them arise in pre-existing mathematical context, others are invented as points of reference for the theory we are developing right now Ideal equivalences Definition An ideal on ω is a collection of sets closed under subset and union and containing the empty set. Example The Fréchet ideal is the collection of all finite subsets of ω. Example The summable ideal. Let a I iff Σ n a 1/n + 1 <. Example The density zero ideal. For a set a ω let ud(a), the upper density of a be defined as limsup of the numbers a n /n : n ω. Let I = {a : ud(a) = 0}. I will view ideals as subsets of P(ω) with its Polish topology. Thus I can speak of F σ ideals, or Borel or analytic ideals. The summable ideal is F σ, while the density zero ideal cannot be even enclosed in a nontrivial F σ -ideal. If I is an ideal then the relation E I on P(ω) given by xe I y (x, y) I is an equivalence relation. If I is Borel then so is E I Isomorphisms of structures Suppose L is a language of first-order logic, and φ is a sentence of that logic. The collection of all countable models of the language L that satisfy the sentence φ can be presented as a Polish space. For example, let L = { } be a language containing one binary function, and φ be the conjunction of all axioms of group theory. Let M = {A ω 3 : x y!z x, y, z A}. This can be identified with the collection of all models of L on ω. It is a G δ subset of P(ω 3 ), and therefore a Polish space. The set M φ = {A M : A = φ} is a Borel subset of P(ω 3 ) and therefore can be equipped with a Polish topology. The natural equivalence relation on M is that of isomorphism. It is in general analytic, but in some instances it is Borel.

27 4.1. EXAMPLES Group actions and orbit equivalences Definition Suppose X is a Polish G-space. Whenever x X, the set {g x : g G} is an orbit of the action. The orbit equivalence on the space X is defined by xey g G g x = y, which is to say that x, y belong to the same orbit. The orbit equivalence relations are a priori analytic, but many of them are Borel. Example The Vitali equivalence relation on the reals is the orbit equivalence of the action of Q on R by addition. Example Every model isomorphism equivalence relation is the orbit equivalence of an action of S. Example (Feldman-Moore theorem) Every Borel equivalence relation with countable classes is generated by a Polish action of a countable group. Example Let I be an ideal on ω and E I be its associated equivalence relation on P(ω): xe I y x y I. This is an orbit equivalence relation of the action of I on P(ω), where I is equipped with the operation and it acts by. But when exactly is it the case that there is a Polish topology on I that makes this into a Polish action? Example Define an equivalence E L2 on the space of cntinuous bounded functions from R to R by setting fe L2 g x y L 2. This is an orbit equivalence relation of the action of L 2. Example Let E Kσ be the equivalence relation on Z ω given by xe Kσ y x y is polynomially bounded. One can consider the group G of those functions in Z ω whose sequence of absolute values is polynomially bounded, and act with G on ω ω by coordinatewise addition. However, there is no topology on G that will make this into a Polish action. Example Let E be the equivalence relation of isometry between Polish spaces. To put it into our context, recall that every separable complete metric space is up to isometry a closed subset of the Urysohn metric space U. Consider the group G = Iso(U) of all isometries of U with composition. This is a Polish group with a suitable topology, and it acts on the space X of all closed subsets of U by h x = h x. Consider the orbit equivalence E. Clearly, if two points x, y X are in the same orbit then they are isometric, but it is not true that two isometric points x, y must be in the same orbit, since there may be a difficulty extending the isometry between x, y into an isometry of the whole space U. A result of Katětov [?] gives a Borel map π : X X such that π(x) is canonically isometric to x, and any isometry g of two points x, y X there is an isometry h of the space U extending g. Thus the isometry relation on X is reducible to the orbit equivalence relation E.

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