Descriptive Set Theory Continuous Reductions on Polish Spaces and Quasi-Polish Spaces

Master Thesis

Louis Vuilleumier

Under the supervision of

Pr. Jacques Duparc

January 15, 2016

Abstract

A subset A ⊆ X of a topological space X is continuously reducible to a subset 1 B ⊆ X if there exists a continuous function f ∶ X → X such that f − (B) = A. In this case, we write A ≤W B. The binary relation ≤W is a quasi-order. This Master Thesis is devoted to the study of the continuous reducibility for subsets of some topological spaces. First, we study continuous reducibility on the Baire ω space ω using games. Then, we show that ≤W is very complicated on the subsets of the real line. Finally, we give a game characterization of ≤W in P(ω) equipped with the Scott topology.

Contents

Acknowledgments 1

Introduction 2

1 Prerequisites 4 1.1 Set Theory ...... 4 1.2 Topology ...... 5 1.3 Metric Spaces ...... 7 1.4 Trees ...... 9 1.5 Infinite Sequences ...... 10 1.6 Baire Property ...... 11 2 Polish Spaces 15 2.1 Definition and Examples ...... 15 2.2 Polish Subspaces ...... 19 2.3 Some Universal Properties ...... 23 2.4 Perfect Polish Spaces ...... 23 2.5 Zero-Dimensional Polish Spaces ...... 31 3 Hierarchies on Polish Spaces 34 3.1 Borel Hierarchy ...... 34 3.2 Difference Hierarchy ...... 43 3.3 Lipschitz and Wadge Hierarchies ...... 44 4 Games and Hierarchies 48 4.1 Gale-Stewart Games on ωω ...... 48 4.2 Lipschitz and Wadge Games on ωω ...... 53 4.3 Lipschitz and Wadge Hierarchy on ωω ...... 58 4.4 Construction of the Lower Levels of the Wadge Hierarchy of ωω . 70 4.5 Cantor Space ...... 76

5 Wadge Hierarchy on R 78 6 Quasi-Polish Spaces 86 6.1 Deﬁnition ...... 86 6.2 Examples ...... 88 6.3 Borel Hierarchy and Universal Property of P(ω) ...... 91 6.4 Two Views on the Wadge Hierarchy on P(ω) ...... 95 6.5 Games on P(ω) ...... 98 6.6 First Results ...... 104 Conclusion 107 Acknowledgments

First, I would like to thank Pr. Jacques Duparc who allowed me to do this Master Thesis with him. I also thank him for the exciting hours he took to allow me to enter in this new world of descriptive set theory. Then, I am especially grateful to Gianluca Basso who agreed to take some of his time to correct lots of mistakes in my work with an exceptional enthusiasm. Without him, this Master Thesis would not be as readable as it is. I would also like to thank Kevin Fournier and Yann Pequignot who took some time to clarify some mathematical points which were confusing to me and to share some of their experience with me. As this Master Thesis is also the ﬁnal point of my student life, I have a grateful thought to all my friends with whom I spent wonderful years of study. I start by William Borgeaud which, in addition to being my friend, helped me with some proofs of this work. Then, I think of Aurélien Chappuis, Léo Diserens, Gaëtan Gogniat, Clélia Liebermann, Nicolas Rossi, Matthieu Wimmer and all the others that I cannot list here. Thank you to all of you ! Finalement, je tiens à remercier toute ma famille, et tout particulièrement mes parents. Maman, Papa, plus que de m’en donner la possibilité, vous m’avez donné l’envie de comprendre le monde qui nous entoure, ainsi que tout le soutien que j’aurais pu espérer. Merci.

January 15, 2016, Louis Vuilleumier

1 Introduction

Historically, descriptive set theory is the study of the subsets of the real line. It was born with measure theory, when the question of which subsets of R are measurable, and which are not, was addressed. During the development of this new area of study, mathematicians observed that they could generalize this study to the study of subsets of more general spaces, the Polish spaces, without losing too many tools. In order to understand the measurability of subsets, it is an usual practice to hierarchize them from the simple ones to the more complex ones. A first way of doing this is to consider the σ-algebra of the Borel sets, with the hierarchy obtained by considering how far sets are from being open, i.e. which operations we need to construct them from the open sets. This hierarchy is fundamental in the development of descriptive set theory. One interesting property of each class of this hierarchy is the closure under continuous preimages. Therefore, W. Wadge [Wad84] had the idea of considering the hierarchy built from classes closed under continuous preimages. It is the Wadge hierarchy. This is natural in a topological sense since all classical topology is built from the open sets and the continuous functions. We can view two subsets A and B in some topological space X as being equivalent in the 1 Wadge hierarchy if there exist two functions f, g ∶ X → X such that f − (A) = B 1 and g− (B) = A. In this case, we say that A is continuously reducible to B and vice versa. Wadge’s famous idea was to consider games that are equivalent to continuous reducibility in the Baire space ωω of infinite sequences of integers. Using game theoretical tools, Wadge was able to give a whole, nice picture of the Wadge hierarchy on the Borel sets of the Baire space ωω. The first part of this Master Thesis is to try to understand the Wadge hierarchy on some Polish spaces and to try to justify the constructions and the choices which are made all along the study. A second part consists of trying to study a similar game characterization of continuous reducibility designed by Pr. ω Jacques Duparc on a space which is different from ω , namely P(ω) equipped with the Scott topology. In Chapter 1, we give some of the set theoretical and topological prerequisites and notations that we will need all along the work. It is in this Chapter that we define the most important space of Wadge’s work, the Baire space ωω. In Chapter 2, we define and understand the objects of study of descriptive set theory, the Polish spaces. We start by giving examples and then we give some properties of these spaces. One important property is the universality of ωω among all the zero-dimensional Polish spaces (Theorem 2.54), which are a kind of Polish spaces. Indeed, this result allows us to design games in all zero-dimensional Polish spaces. Chapter 3 is dedicated to the definitions of hierarchies on Polish spaces,

2 such as the Borel hierarchy and the Wadge hierarchy. In this Chapter, we try to motivate our deep study of the Wadge hierarchy on ωω. In Chapter 4, we come finally to the definition of the game designed by Wadge. With this game and some game theoretical tools such as Borel-determinacy (Theorem 4.10), we obtain the whole picture of the Wadge hierarchy on the Borel subsets of the Baire space. In particular, we show that this hierarchy is well-founded and has maximal antichains of size 2. Chapter 5 is dedicated to the study of the Wadge hierarchy on the real line. In this Chapter, we show that, without a game characterization, the Wadge hierarchy becomes very difficult to understand. In particular, it is not well- founded and it has infinite antichains. Finally, in Chapter 6, we try to generalize Polish spaces by considering quasi- Polish spaces. M. De Brecht [DB13] showed that it is an interesting generalization for several reasons. Among these quasi-Polish spaces, P(ω) equipped with the Scott topology is universal (Theorem 6.30). The study of the Wadge hierarchy on this space is encouraged by the game characterization of continuous function designed by Pr. Jacques Duparc. Hence, we hope this game will give us a nice picture of the Wadge hierarchy of P(ω). This study that we constructed in this work is not exhaustive and deserves more time.

3 Chapter 1

Prerequisites

In this Chapter, we introduce all the basic notions we need in the rest of this work. This is mostly notations and well known topological facts. The major part of this Chapter is covered in a ﬁrst semester course in Topology, hence, we will most of the time omit the proofs.

1.1 Set Theory

We precise that, all along this work, our framework is classical logic and we work within the theory (ZF ) of Zermelo-Fraenkel. Sometimes, we will need more assumptions on the theory, such as the Axiom of Choice (AC), or the Axiom of countable Choice (ACω). We just recall the deﬁnitions of the Axiom of Choice and of countable Choice. The Axiom of Choice guarantees the existence of a choice function on every family of nonempty sets. Hence, we can state it as:

∀S(X ∈ S → ∅ ≠ X) → ∃f ∶ S → X, (f(X) ∈ X) . X∈S The Axiom of countable Choice is the same statement applied only for countable families S. Under (ACω), we can prove the following result which we will use a lot. Proposition 1.1. A countable union of countably many sets is countable. In the rest of this work, we will always assume the Axiom of countable Choice to be true. In some cases, for example to construct pathological behaviors, we will use the Axiom of Choice. Hence, in these cases, we will say explicitly that we assume the Axiom of Choice. We have two others characterizations of the Axiom of Choice. First, it is equivalent to Zermelo’s Theorem. Theorem 1.2. Every set can be well-ordered. A well-ordering of a set is a binary relation which is total, anti-symmetric, transitive and such that every nonempty subset has a least element. We have a second statement which is equivalent to the Axiom of Choice, it is Zorn’s Lemma. For this, we need a Deﬁnition.

4 Deﬁnition 1.3. A partially ordered set (poset) is a space together with a binary relation which is reﬂexive, antisymmetric and transitive. A chain in a poset is an increasing (with respect to the binary relation) sequence of elements of the poset. Hence, we can state Zorn’s Lemma. Theorem 1.4. Let X be a poset. If every chain has an upper bound in X, then X has a maximal element. We will also need a classical result of set theory which does not require the Axiom of Choice, the Cantor-Schröder-Bernstein Theorem.

Theorem 1.5. Let A and B be two sets. Suppose f ∶ A → B and g ∶ B → A are two injective functions. Then, there exists a bijective function h ∶ A ↔ B. Let us make some notations clear.

Definition 1.6. We write ω = N = {0, 1, 2,... } for the first infinite ordinal. It is also the smallest set containing 0 and closed under successor. It will also be useful to write ω∗ = ω ∖ {0}. Moreover, we need the following definitions about set theory.

Deﬁnition 1.7. An ordinal α is of coﬁnality κ, which we write cof(α) = κ, if κ is the least ordinal such that there exists a strictly increasing sequence of ordinal (βγ )γ<κ with βγ < α for all γ < κ and sup{βγ ∶ γ < κ} = α.

1.2 Topology

Deﬁnition 1.8. Let X be a space. A topology τ on X is a subset of the power sets of X, i.e. τ ⊆ P(X), such that: - ∅,X ∈ τ, - if U, V ∈ τ, then U ∩ V ∈ τ,

- if Ui ∈ τ for all i ∈ I, then ⋃i∈I Ui ∈ τ. Deﬁnition 1.9. Let X be a space, and let τ be a topology on X. The couple (X, τ) is a topological space. Moreover, the elements of τ are the open sets of X. The complement of an open set is a closed set. We have a characterization of open sets which we will use a lot.

Proposition 1.10. A subset A in a topological space X is open if, for all x ∈ A, there exists U open such that x ∈ U ⊆ A. There is a ﬁrst important type of spaces.

Deﬁnition 1.11. A topological space X is T0 if, for all x, y ∈ X, x ≠ y, there exists an open subset U ⊆ X such that x ∈ U, y ∉ U or x ∉ U, y ∈ U. Hence, x, y are separable by the topology. For the rest of this work, if we do not need to be precise, we will say "Let X be a topological space" without specifying a particular topology.

5 Deﬁnition 1.12. Let (X, τ) be a topological space. A basis B for the topology τ is a subset B ⊆ τ such that:

- For all x ∈ X, there exists B ∈ B such that x ∈ B, - For all B1,B2 ∈ B, and for all x ∈ B1 ∩ B2, there exists B3 ∈ B such that x ∈ B3 ⊆ B1 ∩ B2. Every basis generates a topology by taking all the possible unions of elements of the basis. Let us give some deﬁnitions about some spaces.

Definition 1.13. A topological space is second countable if it has a countable basis. We also say that the space is countably based. Definition 1.14. A subset D of a topological space X is dense if it has nonempty intersection with all the open sets of X. Definition 1.15. A topological space is separable if it contains a countable dense subset. We also have the next fundamental definition in topology, which is the definition of continuous functions. The idea is to describe the functions that preserve the topology by preimage. We will see, later in this work, that the notions that are preserved under continuous preimage are of fundamental importance in descriptive set theory. Definition 1.16. Let X,Y be topological spaces. A function f ∶ X → Y is continuous if the preimage of every open set is open, 1 i.e. for all open UY ⊆ Y , f − (UY ) is open in X. This definition allows us to define a topology on a Cartesian product.

Deﬁnition 1.17. Let (Xα)α∈J be a sequence of topological spaces. We deﬁne the product topology on the Cartesian product Πα∈J Xα as the one having the following basis:

Πα J Uα ∶ Uα open in Xα, ∈ ¡ . and Uα = Xα for all except a ﬁnite number of α

The reason of this particular deﬁnition is the following:

Proposition 1.18. Let (Xα)α∈J be a sequence of topological spaces. The product topology on Πα∈J Xα makes all projection continuous, i.e. for all β ∈ J, the functions

projXβ ∶ Πα∈J → Xβ

(xα)α∈J ↦ xβ, are continuous. We have another important notion. Deﬁnition 1.19. A topological space X is compact if, for each union of open sets such that X = ⋃i∈I Ui, there exists a ﬁnite number of elements of the union X n U such that = ⋃j=1 j.

6 In topology, most of the objects and properties are defined from open subsets, since they are considered as the simple subsets in some topological sense. Moreover, open sets are closed under arbitrary union, but only under finite intersection. For this reason, we define the first kind of sets that are more complicated than the open sets.

Deﬁnition 1.20. Let (X, τ) be a topological space. We deﬁne:

Gδ = Un ∶ Un is open ¡ , n∈ω

Fσ = Fn ∶ Fn is closed ¡ . n∈ω In a few pages, we will generalize this deﬁnition, and we will see that descriptive set theory is also the study of the complexity of all such subsets of some topological space X.

1.3 Metric Spaces

Definition 1.21. Let X be a space. A metric is a function d ∶ X × X → [0, +∞) such that: - (d(x, y) = 0 ↔ x = y) for all x, y ∈ X, - d(x, y) = d(y, x) for all x, y ∈ X, - d(x, z) ≤ d(x, y) + d(y, x) for all x, y, z ∈ X. Definition 1.22. Let X be a space and d ∶ X × X → [0, +∞] be a metric. For all x ∈ X and for all r ∈ R, we define B(x, r) = {y ∈ X ∶ d(x, y) < r}. It follows that B = {B(x, r) ∶ x ∈ X, r ∈ R} is a basis of a topology, known as the metric topology, and usually written τd. Definition 1.23. Let X be a space and τ, σ be two topologies on X. τ and σ are equivalent if we have:

- for all Uτ ∈ τ, there exists Uσ ∈ σ such that Uσ ⊆ Uτ ; - for all Uσ ∈ σ, there exists Uτ ∈ τ such that Uτ ⊆ Uσ.

Deﬁnition 1.24. Let (X, τ) be a topological space. If there exists a metric d ∶ X × X → [0, +∞] on X such that τd is equivalent to τ, we say that (X, τ) is metrizable or that (X, τ) is a metric space. If we have a metric space with metric d, then d′ d is also a metric, = 1+d compatible with the topology of d, such that d′ ≤ 1. Hence, working in metric space, we can always suppose that the metric is less or equal to one. We have the following fact, which is important and which we will use a lot in this work. Proposition 1.25. Let X be a metric space. Then X is second countable if and only if X is separable.

7 Proof. If X is second countable, let {Un}n∈ω be a countable basis of nonempty sets. Using (ACω), we ﬁnd xn ∈ Un, then it is clear by deﬁnition of a basis that {xn}n∈ω is dense in X. If X is separable with dense subset (xn)n∈ω. We have the following basis: + B = {B(xn, r) ∶ n ∈ ω, r ∈ Q }.

By (ACω), B is countable since it is a countable union of countably many sets.

Now, we will define two types of functions between metric spaces. The first one, that we have already seen for arbitrary topological space, is the notion of continuous function. Definition 1.26. Let X,Y be metric spaces. A function f ∶ X → Y is continuous if, for all x ∈ X and for all > 0, there exists δ > 0 such that: f(B(x, δ)) ⊆ B(f(x), ). We have a more restrictive property of functions. Definition 1.27. Let X,Y be metric spaces. A function f ∶ X → Y is Lipschitz if, for all x, y ∈ X, we have d(f(x), f(y)) ≤ d(x, y). Of course, one can proves easily that a Lipschitz function is continuous by letting δ = . In the previous Section, we defined the notion of the product topology. Is it compatible with metric spaces? The answer is positive and is given in the next Proposition.

∗ Proposition 1.28. Let (Xn)n∈ω be a sequence of metric spaces, with metrics dn ≤ 1, respectively. The product topology on the Cartesian product is equivalent to the metric topology induced by the metric: 1 d x, y d x, y . ( ) = Q n n( ) ∗ 2 n∈ω Some metrizable spaces have a nice behavior with respect to convergent sequences. We deﬁne them. Deﬁnition 1.29. Let X be a metric space, with d as a metric. A Cauchy sequence (xn)n∈ω in X is a sequence such that:

lim d(xn, xm) = 0. n,m→∞ Definition 1.30. Let X be a metric space. X is complete if every Cauchy sequence in X has a limit in X, i.e. if there exists x ∈ X such that d(xn, x) → 0 when n → ∞ and (xn)n∈ω is Cauchy. Definition 1.31. Let X be a topological space. If there exists a metric on X such that X is complete and X is a metric space, then X is a completely metrizable space. We will see that our first object of study, namely the Polish spaces, are the separable completely metrizable spaces.

8 1.4 Trees

In this section we deﬁne objects that we will use a lot, trees.

Definition 1.32. Let A be a nonempty set, and let n ∈ ω. We define the set of finite sequence of length n of elements of A:

n A = {s = (s0, s1, . . . , sn−1) ∶ si ∈ A for all i < n}.

We also deﬁne the set of all ﬁnite sequences as:

ω n A< = A . n∈ω

We will write l(s) for the length of a sequence s. Deﬁnition 1.33. Let A be a nonempty set, and let n, m ∈ ω, with n ≤ m. n m s ∈ A is an initial segment of t ∈ A if si = ti for all i < n. In this case, we write s ⊆ t. We say that two sequences are compatible if one is an initial segment of the other.

n m Deﬁnition 1.34. Let A be a nonempty set, n, m ∈ ω and s ∈ A , t ∈ A . We deﬁne the concatenation of s and t as:

n+m sˆt = (s0, . . . , sn−1, t0, . . . , tm−1) ∈ A .

Moreover, if m = n, we deﬁne:

2n s ∗ t = (s0, t0, s1, t1,... ) ∈ A .

With these concepts, we can deﬁne trees. In a picture, a tree is simply something of the form :

0 1 2

4 7 7 12 21

1 4 0 2 5 9

The green point corresponds to the empty sequence, the red point corresponds to the sequence 0ˆ4, and the blue point corresponds to the sequence 0ˆ7ˆ2. Hence, a formal definition of a tree is the following. Definition 1.35. Let A be a nonempty set. A tree T on A is a subset of A<ω which is closed under initial segments, i.e. if s ∈ T, then, for all initial segments t of s, we have t ∈ T. The elements of T are its nodes.A branch of T of length n is a subset of T closed under initial segments and containing exactly one element of length i ω for all i < n. An infinite branch of T is an element s of A such that every initial segment of s is in T. The body of T , denoted by [T ], is the set of all its infinite branches. We also say that a tree T is pruned if all its elements have an extension in T , i.e. for all s ∈ T, there exists a ∈ A such that sâ ∈ T .

9 1.5 Inﬁnite Sequences

Let us focus on the infinite sequences, which will play a predominant role in this work. Definition 1.36. Let A be a nonempty set. We define the set of all infinite sequences as:

ω A = {s = (sn)n∈ω ∶ si ∈ A for all i ∈ ω}. We also define the set of all finite and infinite sequences as:

ω ω ω A≤ = A ∪ A< .

ω Deﬁnition 1.37. Let A be a nonempty set, and s ∈ A . For all n ∈ ω, we deﬁne sSn as the initial segment of s of length n, i.e. sSn = (s0, . . . , sn−1). The next goal is to give a topological structure to Aω and A≤ω. We need the ω following sets for all s ∈ ω< : ω Ns = {t ∈ A ∶ s ⊆ t}.

Let us start by deﬁning a topological structure on Aω. A basis of a possible topology is given by the standard open sets

<ω B = {Ns ∶ s ∈ A }.

ω This topology is compatible with the following metric on A ∶ 1 d(s, t) = , n ∈ ω the least integer s.t. sn ≠ tn. 2n+1 Hence, Aω is metrizable. Moreover, if A is countable, Aω is separable since it is metric and has a countable basis. Indeed, under (ACω), we have that a countable union of countably many sets is countable, hence:

ω ω SBS = S{s ∈ A< }S = W {s ∈ A< ∶ l(s) = n}W ≤ SωS, n∈ω where we use that there exists a bijection between ω×ω and ω. It is a well known fact and we will describe an homeomorphism between these sets later on, which n ω will witnesses the existence of a bijection. Finally, suppose (s )n∈ω ⊆ A is n m 1 Cauchy. Since lim d(s , s ) = 0, for all > 0, if k > l > log2( ), then the l ﬁrst n m n terms are the same in s and s . Taking the limit → 0, we obtain that (s ) converges in Aω. Hence Aω is a separable completely metrizable space if A is countable. We can prove, in the exact same way, with the exact same metric topology, that A≤ω is also separable and completely metrizable as long as A is countable. Now, we have a big link between trees and this topological space. Theorem 1.38. Let A be a countable set. There is a bijection between pruned trees on A and closed subsets in Aω.

10 Proof. We give immediately both functions that witness this bijection.

ω f ∶ {pruned trees on A} → {closed subset in A } T ↦ [T ],

ω g ∶ {closed subset in A } → {pruned trees on A} F ↦ TF = {sSn ∶ s ∈ F, n ∈ ω}. it is really easy to see that the functions are the inverse of one other. Moreover ω [T ] is in A because we take only inﬁnite branches, and it is closed because the complement is open, being a big union of elements from the basis. Indeed, let

ω E = {t ∈ A< ∶ t ∉ T, tS(l(t) − 1) ∈ T }, we have c ([T ]) = Nt. t∈E

Finally, TF is a tree by its definition, and it is pruned because we started with elements in Aω. Hence, we can have some intuition about the topology on Aω. Theorem 1.38 tells us that a closed set can be viewed as a pruned tree. Therefore, descending along the tree, we can always stay in the tree. An other point of view is that, if we go out of the tree, we can never go in the tree again. Hence, the choice of non-belonging to a closed set is definitive. On the other hand, if s is in some open set U, there exists n ∈ ω such that NsSn ⊆ U. This means that, after a finite amount of time, we can say if a point is in the open set. The intuition is that the choice of belonging to an open set is definitive. We will see, later on, that this intuition is really helpful to a well understanding of the subsets of Aω. Now, we state and prove a useful property of closed subsets of Aω.

ω Proposition 1.39. Let F ⊆ G be two closed subsets of A . Then there exists a continuous surjection f ∶ G → F such that f is the identity on F.

Proof. By the previous Theorem 1.38, we define F = [T1] and G = [T2]. First, we define f ∶ T2 → T1 by recursion on the length of s ∈ T1. Let f(∅) = ∅. Suppose we have s ∈ T1, and f(s) ∈ T2. If sâ ∈ T1, define f(sâ) = f(s)â. If sâ ∉ T1, just set f(sâ) = f(s)ˆb, where b is such that we do not go out of T1, which is possible since these trees are pruned. This function induces a function from [T2] −1 to [T1]. Moreover, it is continuous since f (Ns) = Ns ∪ ⋃t Nt, where t ranges over the elements which are sent to an initial segment of s.

1.6 Baire Property

In this Section, we deﬁne some topological properties of certain spaces. This will conduct us to the notion of the Baire property. This property will be helpful in our program of classifying the subsets of some spaces.

Deﬁnition 1.40. A set A ⊆ X in a topological space is nowhere dense if its ¯ closure has empty interior. This is equivalent to say that X ∖ A is dense in X.

11 Deﬁnition 1.41. A set A ⊆ X in a topological space is meager if it can be written as a countable union of nowhere dense sets. A set is comeager if its complement is meager.

We also have the properties. Proposition 1.42. The set of all meager sets contains the empty set, is closed under subsets and is also closed under countable unions. Proof. The two first statements are trivial, and the last one comes directly from the definition of a meager set as a countable union of nowhere dense sets. A family of sets having these three properties is a σ-ideal. Definition 1.43. Let X be a topological space. A subset A ⊆ X has the Baire Property (BP) if there exists an open subset U ⊆ X such that: (A ∖ U) ∪ (U ∖ A) is meager. Hence, this definition tells us that a set having the BP is an open set modulo some meager set. Hence, it can be viewed as an almost open set. Usually, we will write (A ∖ U) ∪ (U ∖ A) = A △ U, and call it the symmetric difference of A and U. We define the notion of a σ-algebra.

Deﬁnition 1.44. Let X be a space. A σ-algebra A on X is a family A ⊆ P(X) such that:

- A is nonempty; - A is closed by complements; - A is closed by countable unions. We have the important following characterization of the set of all subsets having the BP. Theorem 1.45. Let X be a topological space. The set of all subsets having the BP is the smallest σ-algebra containing the open subsets and the meager subsets.

Proof. First, it is trivial that open sets and meager sets have the BP. Moreover, any closed set F ⊆ X has the BP since F ∖ Int(F ) = ∂F is closed and nowhere dense, hence meager. Suppose now A has the BP. Hence there exists an open U such that A △ U = c c c c (A ∩ U ) ∪ (U ∩ A ) is meager. Therefore A △ U is also meager since it is the same set. Now, we have:

c c c c c c A △ Int(U ) = (A ∩ Int(U ) ) ∪ (Int(U ) ∩ A) c c c c = [(A ∩ Int(U ) ) ∪ (Int(U ) ∩ A)] ∩ U c c c c c ∪ [(A ∩ Int(U ) ) ∪ (Int(U ) ∩ A)] ∩ U c c c c c c ⊆ (A ∩ U) ∪ (Int(U ) ∩ U) ∪ (Int(U ) ∩ U ) ∪ (A ∩ U ) c c = (A △ U) ∪ (U △ Int(U )).

12 c c c Hence, using Proposition 1.42, A △ Int(U ) is meager, hence A has the BP. Suppose now that, for each n ∈ ω, An has the BP, hence there exists an open Un such that An △ Un is meager. We have (the unions and the intersections are over ω):

c c An △ Un = An ∩ ( Un) ∪ Un ∩ ( An) c c = ( An ∩ Un) ∪ ( Un ∩ An) c c = (An ∩ Un) ∪ (Un ∩ An) c c = (An ∩ Un) ∪ (Un ∩ An) = (An ∖ Un) ∪ (Un ∖ An) = (An △ Un) . Again, using Proposition 1.42, we have that a countable union of sets having the BP has the BP. It follows that the family of sets having the BP is a σ-algebra containing the meager sets and the open sets. Moreover, it is the smallest one because if A has the BP, then M = A △ U is meager. Therefore A = M △ U. Indeed, if x ∈ A ∩ U, c then x ∉ M, so x ∈ U ∖ M. If x ∈ A ∩ U , then x ∈ M, so x ∈ M ∖ U. Therefore A ⊆ U △ M. If x ∈ U ∖ M, then x ∈ A ∩ U, so x ∈ A. Finally, if x ∈ M ∖ U, then x ∈ A∖U, so x ∈ A. We have shown that A = U △M. Hence, A is in the smallest σ-algebra generated by all the open and the meager sets. Finally, we have a last characterization of sets having the BP that will be useful for us.

Proposition 1.46. Let X be a topological space, with A ⊆ X having the BP. Then, either A is meager or A is comeager on a nonempty open set U ⊆ X, i.e. U ∖ A is meager. Proof. Let M = A△U be the meager set that exists by definition of the BP for A. Suppose A is not meager, then U ≠ ∅. Hence, by Proposition 1.42, U ∖A ⊆ M is meager. Therefore, A is comeager on U. There exist some spaces not having the BP, we give a first example here. It comes from a class of sets that we will make use of to find some contradictions later on. Let us define them. ω ω Definition 1.47. A flip set is a subset F ⊆ 2 such that, for any x ∈ 2 , the membership or not to F changes if we change exactly one coordinate. Of course, since it has to be true in all 2ω, the question of whether there exists a flip set is not trivial. But, with the Axiom of Choice, we can construct such a set. That is what we do now. Proposition 1.48. Assuming the Axiom of Choice, there exists a flip set in 2ω. Proof. We recall that the Axiom of Choice is equivalent to Zorn’s Lemma. Hence, we can make use of it. ω ω First, we define an equivalence relation on 2 . For x, y ∈ 2 , we say that x ∼ y if they differ by a finite even number of digits. It is easy to see that this defines ω an equivalence relation. Hence, we can take the quotient Ω = 2 ~ ∼ . We write

13 ω ω x˙ for an element of 2 which diﬀers from x ∈ 2 by exactly one digit. We deﬁne a poset ⟨X = {E ⊆ Ω ∶ [x] ∈ E → [x˙] ∉ E}, ⊆⟩.

Let (Xi)i∈I be a chain in X. We show that B = ⋃i∈I Xi is an upper bound in X. Let [x] ∈ B, and towards a contradiction, suppose there exists [x˙] ∈ B. Therefore, there exists i0 ∈ I such that [x], [x˙] ∈ Xi0 , a contradiction. Hence, we can use Zorn’s Lemma and we get a maximal element in X, say M. Now, we consider F = m. [m]∈M We want to show that F is a flip set. First, suppose x ∈ F, then [x] ∈ M, therefore [x˙] ∉ M, which implies [x˙] ∉ F. Now, consider the second case where x ∉ F. Towards a contradiction, we suppose x˙ ∉ F. We will contradict the maximality of M by showing that M ⊊ M ∪{[x]} ∈ X. We have two cases to consider. - If [y] ∈ M, then [y˙] ∉ M by definition of M. Moreover, we make the absurd hypothesis that [y˙] = [x]. We then have that y differs by one digit from y,˙ which differs by a finite even number of digits from x, which differs by one digit from x.˙ Therefore, y and x˙ differ by an even number of digits. Hence [y] = [x˙] ∈ M, which contradicts the fact that x˙ ∉ F. So our absurd hypothesis is wrong, hence y˙ ≁ x. - It remains to show that [x˙] ∉ M, but it is trivial since x˙ ∉ F by hypothesis. Therefore, we proved that M ⊊ M ∪ {[x]} ∈ X. This contradicts the maximality of M. Therefore, x˙ ∈ F. So, F is a flip set. Finally, we show that flip sets have a first bad property. Proposition 1.49. A flip set does not have the BP. Proof. Suppose, by contradiction, that F is a flip set having the BP. Using ω Proposition 1.46, there exists a nonempty open set Ns ⊆ 2 such that F is either meager or comeager in Ns. Set n = l(s), and consider the function:

f ∶ Ns → Ns xi ↦ xi if i ≠ n, xn ↦ 1 − xn, otherwise.

This is a homeomorphism because we just ﬂip 1 coordinate. Indeed, if t ∈ Ns with l(t) < ω, we have f N N ( t) = t0ˆ...ˆtn−1ˆ(1−tn)ˆtn+1ˆ... and f −1 N N ( t) = t0ˆ...ˆtn−1ˆ(1−tn)ˆtn+1ˆ... c Moreover, since F is a ﬂip set, we have f(F ∩ Ns) = (F ∩ Ns). If F is meager c c in Ns, then F is also meager in Ns via f. So, F and F are both meager sets. c Otherwise F is comeager in Ns, then F is also comeager in Ns via f. So, as in the other case, F and F c are both meager sets. But meager sets are closed under c union, hence Ns = (F ∩ Ns) ∪ (F ∩ Ns) is meager, which is a contradiction.

14 Chapter 2

Polish Spaces

In this Chapter, we deﬁne the objects of study of descriptive set theory. His- torically, this discipline is the study of the subsets of the real line. However, we can collect certain topological particularities of the real line and work in this more general framework. That is what we propose to do here. We will mostly follow Chapter 1 of [Kec95].

2.1 Deﬁnition and Examples

These more general spaces are the Polish ones, which we can already deﬁne.

Deﬁnition 2.1. A topological space X is a Polish space if it is separable and completely metrizable. We precise that a space is Polish if there exists a metric which makes it Polish, but this could not be the case for all metrics. We will see an example. Observing the deﬁnition of a Polish space, we see that the separability condition allows us to consider Polish spaces as spaces that are not too big.

Proposition 2.2. Let X be a Polish space. Then SXS ≤ 2ℵ0 . Proof. Since X is separable, there exists a dense subset D of cardinality less or ∗ equal to SωS. Hence, by density, for all x ∈ X, there exists a sequence (xn)n∈ω ⊆ D 1 such that xn ∈ D ∩ B(x, n ). This means that every point can be characterized by a sequence of elements in D. Moreover, being metrizable, X is Hausdorﬀ, therefore a sequence has only one limit point. We ﬁnally have:

∗ ℵ0 SXS ≤ S{(xn) ⊆ D ∶ n ∈ ω }S ≤ SP(ω)S = 2 .

After this observation, we see classical examples of Polish spaces.

Example 2.3. 1. R is separable by taking balls with rational radius cen- tered at rational points. Here, we use the fact that a countable union of countably many sets is countable. This is true if we work in ZF + (ACω). Moreover, the usual metric is complete, hence R is Polish.

15 2. I = [0, 1] is also Polish for the exact same reasons as R, and the fact that it is closed, hence every sequence in I has a limit in I.

3. C is also Polish, again with the same arguments.

4. S = {x ∈ C ∶ SxS = 1}, being closed, is also Polish for the induced metric. 5. Any countable space X with the discrete topology is Polish, using the metric d(x, y) = 1 if x ≠ y, x, y ∈ X. For the next example, we need more work.

Example 2.4. (0, 1) is not complete with respect to the usual metric, since the 1 ∗ sequence ( n )n∈ω is not convergent in (0, 1). Hence, we need another metric. That will be our ﬁrst example of a space which is complete using a diﬀerent metric than the usual one. Since we know R being Polish, and R homemorphic to (0, 1) via, for example:

f ∶ (0, 1) → R 1 t ↦ 1 + 2−t It suffices to transfer the metric of R in (0, 1) via this homemorphism. Hence ′ ′ 1 1 we define a metric d . For x, y 0, 1 , let d x, y d −x , −y , where d ∈ ( ) ( ) = 1+2 1+2 1 ∗ is the usual distance on R. With this metric, ( n )n∈ω is not Cauchy anymore. Indeed, it is correspond to a sequence diverging in R. Now, we come to the crucial examples which will very helpful for us. Example 2.5. Let A be a countable space. The space Aω constructed in Section 1.5 (as an infinite product of spaces with the discrete topology) is Polish. Indeed, we showed that it is separable and completely metrizable, hence Polish.

ω Therefore, we constructed a family of examples, namely A = {A ∶ A countable}. In all these examples, two are of a capital importance. It is 2ω and ωω. Indeed, ω we will show that, if A ∈ A, then either A is homeomorphic to 2 , or it is ω homeomorphic to ω . So, these two sets are universal for A. Hence, we deﬁne. ω ω Deﬁnition 2.6. The Cantor space is the space 2 = {0, 1} . The universality comes from.

ω ω Proposition 2.7. Let A be a ﬁnite set. Then A ≅ 2 .

Proof. Let A be ﬁnite, hence A is in bijection with some {0, 1, 2, . . . , k}. We will code the elements of A into elements of 2<ω. We deﬁne:

ω f ∶ {0, 1, . . . , k} → 2< i ↦ 1⋯1 0 ±i k ↦ 1⋯1 ±k

The coding is given by the following tree.

16 0 1 0 0 1 1 1 2 0 1 k k − 1

The inverse of this function is the decoder, which is easily understandable from the tree, we denote it by δ. Let us deﬁne

ω ω g ∶ {0, 1, . . . , k} → 2

(sn)n∈ω ↦ f(x0)ˆf(x1)ˆf(x2) ....

g is injective because the coding f is one-to-one and allows us to code term to term. It is also surjective because we have an easy decoder using the tree we drew and proceeding step by step. Hence g is bijective. g g N N . is open since, by the coding, ( s) = f(s1)ˆ...ˆf(s(l(s)−1)) It remains to 1 show that g is continuous. Let us compute g− (Ns), we have two cases.

• If s(l(s)−1) = 1, the last term is well-deﬁned, and we can decode via the −1 tree term by term. We get g (Ns) = Nδ(s).

• If s(l(s)−1) = 0, we have to be more careful. We remove all the 0’s at the end of s to obtain s′. Then, if there was more than k 0’s, we add a k to s′. Doing this until we reach a point where there are not enough 0’s, we get s¯. We observe that s¯ is decodable by this construction. Let us say, without loss of generality, that it remains i 0’s. We obtain

−1 g (Ns) = Nδ(s¯ˆj). j≥i

Hence, g is a homeomorphism.

ω ω Deﬁnition 2.8. The Baire space is the space ω = {0, 1, 2, 3,... } . We have also a ﬁrst universality property.

ω ω Proposition 2.9. Let A be a countable inﬁnite set. Then A ≅ ω . Proof. Since A is countable inﬁnite, there exists a bijection b ∶ A ↔ ω. There- fore,

ω ω f ∶ A → ω s ↦ b(s0)ˆb(s1)ˆb(s2)ˆ...

This is a homeomorphism since, in both sides, we have {Ns} as a basis. Moreover, these two sets are not homeomorphic. Proposition 2.10. The Cantor space is compact and the Baire space is not compact. Hence, they cannot be homeomorphic.

17 Proof. By the ternary expression, it is a well known fact that 2ω is homeomor- ω phic to the Cantor set in [0, 1]. Hence, it is closed and bounded. Therefore 2 is compact by the Heine-Borel Theorem. The following open cover of ωω has no ﬁnite subcover.

{Nk ∶ k ∈ ω}. Hence, since being compact is a topological invariant, there is no homeomorphism between these two spaces. Moreover, they are universal also in the following sense. We will see that a large number of Polish spaces are recovered from these two spaces. Indeed, throughout this work, we will see a lot of results witnessing this idea of universality of 2ω and ωω. Moreover, we said that descriptive set theory is, historically, the study of the subsets of the real line. For this reason and the following proposition, we reserve a special importance to the study of the Baire space. Indeed, this result tells us that the Baire space is almost the real line. Proposition 2.11. The Baire space is homeomorphic to the real line minus the ω rationals, hence ω ≅ R ∖ Q. Proof. We define the homeomorphism which proves this Proposition. It uses continued fraction expansion. ω f ∶ ω → (0, 1) ∩ Irr 1 (s0, s1, s2,... ) ↦ 1 . 1 s0 1 + + 1 s1 + + 1+s + 1 2 ... The proof of the fact that f is a homeomorphism is very technical, however, the intuition that f is open, continuous and injective is pretty good. For the surjectivity, it is more complicated. However, from an irrational number, it is possible to construct an element of ωω by successive division. More details can be found in [Hen06, p.12] and in [GL13, p.101]. Because of this, in almost all books of descriptive set theory, we call the elements of the Baire space reals, and we identify R to the Baire space. However, in this work, we will never do this because we find this analogy confusing. Indeed, we will see that the hierarchical picture of their subsets is not the same at all. Moreover, we have. ω ω ω Proposition 2.12. ω × ω ≅ ω , but R × R ≇ R. Proof. We define a homeomorphism. ω ω ω f ∶ ω × ω → ω ((sn), (tn)) ↦ (sn) ∗ (tn) = (s0, t0, t1, s2,... ). The fact that f is surjective and injective comes directly from the definition. ω ω Suppose we have U ⊆ ω × ω open. We want to show that f(U) is open. Let (un) ∈ f(U). Hence, there exists (sn), (tn) such that (sn) ∗ (tn) = (un). Moreover, U being open implies the existence of k, l ∈ ω such that NsSk ×NtSl ⊆ U. Let m = max{k, l}, we get NsSm × NtSm ⊆ U. Hence:

(un) = (sn) ∗ (tn) ∈ f(NsSm × NtSm) = Ns∗tS2m ⊆ f(U).

18 So f is open. ω ω 1 Now, suppose Ns ⊆ ω , s ∈ ω< . Let (t, u) ∈ f − (Ns). Suppose moreover l(s) = n = 2k. If n is odd, just consider n + 1. We have, directly from the deﬁnition of f: −1 (t, u) ∈ NtSk × NuSk ⊆ f (Ns). Hence, f is a homeomorphism. However, suppose there exists an homeomorphism g ∶ R × R → R. Removing a point and its image gives a connected space and a non-connected one, and preserves the homeomorphism. This is impossible since connectness is preserved under homeomorphism. Finally, a last thing that makes these spaces useful is their constructions. Indeed, they are construct dynamically in the sense that their elements are sequences. For this reason, we will see that we can deﬁne games in this framework, which will be our main tool in the next chapters. Before coming to more examples, we can generalize the preceding ones in the following sense. Example 2.13. 2≤ω is Polish by the similar arguments as for 2ω. Indeed, the same basis makes it a separable completely metrizable space. Having seen the usual examples, we can construct other Polish spaces. Proposition 2.14. The completion of a separable metric space is Polish. A closed subset of a Polish space is Polish, A countable product of Polish spaces is Polish. Proof. These topological proofs can be found in [Kur66]. Hence, this implies that the following examples are Polish. n ω n ω n ω n ω n ω Example 2.15. R , R , C , C , I , I , S , S , (0, 1) , (0, 1) are Polish.

2.2 Polish Subspaces

In this Section, we will characterize the Polish subspaces of Polish spaces using topological tools. We have seen that open intervals and closed intervals in R are Polish. The question is whether or not this can be generalized to other Polish spaces. At the same time, we will show an important fact about metrizable spaces, which will, later on, allow us to have a nice hierarchy on them. Let us start by this fact.

Proposition 2.16. Let X be a metric space. If F ⊆ X is closed, then F is Gδ. Proof. We need to deﬁne the distance between a point and a set. Let x ∈ X and ∅ ≠ A ⊆ X. d(x, A) = inf{d(x, y) ∶ y ∈ A}, B(A, ) = {x ∈ X ∶ d(x, A) < }. First, we show that B(A, ) is open. −δ Let x ∈ B(A, ), hence d(x, A) = δ < . We will show that B(x, 2 ) ⊆ B(A, ). −δ +δ Let y ∈ B(x, 2 ), and take z ∈ A such that d(x, z) < 2 . We get: d(y, A) ≤ d(x, z) ≤ d(x, y) + d(y, z) < .

19 Hence B(A, ) is open. Therefore, to finish the proof, it suffices to show that: 1 F = B F, . ∗ n∈ω n c The first inclusion is obvious. For the second one, suppose x ∉ F. Since F ∗ 1 is open, there exists n ∈ ω such that B(x, n ) ∩ F = ∅. This implies directly 1 ∗ x ∉ ⋂n∈ω B(F, n ), which is Gδ. Now, we are interested in the behavior of metric space under continuity. We need some technical definitions.

Deﬁnition 2.17. Let X,Y be metric space, A ⊆ X, and f ∶ A → Y. We deﬁne:

diam(A) = sup{d(x, y) ∶ x, y ∈ A}, oscf (x) = inf{diam(f(A ∩ U)) ∶ x ∈ X, x ∈ U open }.

The second element of the Deﬁnition is the oscillation of f at x since it measures how bad the behavior of f on the open neighborhoods of x is. We will, for clarity, write f(U) instead of f(A ∩ U).

Proposition 2.18. Let X,Y be metric space, A ⊆ X, f ∶ A → Y and x ∈ A. f is continuous at x if and only if oscf (x) = 0.

Proof. If oscf (x) = 0, for all > 0, there exists some δ > 0 such that f(B(x, δ)) ⊆ B(f(x), ), which is exactly the deﬁnition of continuity at x. If oscf (x) ≠ 0, take 0 < < oscf (x). For all open neighborhood U of x, we have diam(f(U)) > . Hence, for such an , we cannot satisfy the deﬁnition of continuity at x. From this result, we also have.

Corollary 2.19. Let X,Y be metric space, f ∶ X → Y . The set of points of continuity of f is Gδ in X.

Proof. We deﬁne A = {x ∈ X ∶ oscf (x) < }. By the previous proposition, if we show that A are open and that

A 1 = {x ∈ X ∶ oscf (x) = 0}, ∗ n n∈ω we are done. We set A = {x ∈ X ∶ oscf (x) = 0}. One inclusion is obvious. For the other, suppose, by contradiction x A 1 A. Then oscf x γ 0, which is ∈ ⋂ n ∖ ( ) > > absurd. To show that A is open, just take x ∈ A. Then oscf (x) = δ < . Hence, for all γ > 0, there exists an open neighborhood U of x such that diam(f(U)) < δ + γ. Taking γ = − δ, we get diam(f(U)) < . Hence x ∈ U ⊆ A. We showed that A is open, hence the set of continuity points of f is Gδ.

Before going further, we must observe that, for all sets A ⊆ X, X metric, we ¯ have diam(A) = diam(A). This follows directly from the fact that the diameter is deﬁned by a supremum. We now have all we need for an important topological result, known as Kuratowski Theorem.

20 Theorem 2.20. Let X be a metric space, Y be completely metrizable, A ⊆ X and f ∶ A → Y continuous. Then, there exists a Gδ set G in X and a function g ∶ G → Y extending f such that: ¯ • A ⊆ G ⊆ A, • g is continuous. ¯ Proof. We define G = {x ∈ X ∶ oscf (x) = 0} ∩ A. By Proposition 2.16 and Corollary 2.19, this is a Gδ set. Moreover, since f is continuous, we get, by ¯ Proposition 2.18, that A ⊆ G ⊆ A. Suppose x ∈ G, then there exists a sequence (xn)n∈ω ⊆ A such that xn → x. We define g(x) = lim f(xn). Since f is continuous and Y is metric, we get that g(x) is well-defined. Finally, using the observation we made before the statement of this Theorem, we get, for x ∈ G:

oscg(x) ≤ diam(g(U)) ≤ diam(f(U)) ≤ diam(f(U)) → 0.

This conduct us to the conclusion of this Section, but, before stating the ﬁnal theorem, we need a lemma. This lemma is in fact a generalization of our construction in Example 2.4 which shows that open intervals are Polish in R.

Lemma 2.21. Let X be Polish, and U ⊆ X open. Then U is Polish. Proof. We need to define a new metric. Let d be the metric induced by X, which we can suppose to be d ≤ 1. We define: 1 1 d′(x, y) = d(x, y) + W − W . d(x, U c) d(y, U c) The two first conditions to be a metric are obvious. For the triangular inequality, we have 1 1 d′(x, y) = d(x, y) + W − W d(x, U c) d(y, U c) 1 1 1 1 = d(x, y) + W − + − W d(x, U c) d(z, U c) d(z, U c) d(y, U c) 1 1 1 1 ≤ d(x, z) + d(y, z) + W − W + W − W d(x, U c) d(z, U c) d(z, U c) d(y, U c) = d′(x, z) + d′(y, z).

The idea is the following. With the metric d, U ⊆ X is not complete since we can ﬁnd a sequence in U converging to the frontier. Hence, we want such a sequence to be not converging. The metric d′ allows us to move each point of the sequence away from the others to make the sequence not convergent. We have to show that this metric is compatible with the subspace topology on U, which is obtained from the metric d. Since d′ > d, any d′-open set is also d-open.

21 c For the converse, let x ∈ U open with respect to d. Then d(x, U ) = r > 0. Let 0. We choose δ 0 such that γ γ for all 0 γ δ. If y B x, δ , > > + r(r−γ) < < ≤ ∈ ( ) then: c c d(y, U ) > d(x, U ) − d(x, y) = r − δ. Therefore r d′(x, y) < δ + W W < . r(r − δ)

Hence, Bd′ (x, δ) ⊆ B(x, ). ′ It just remains to show that d is complete. Let (xn)n∈ω ⊆ U be Cauchy with respect to d′, since d < d′, it is also Cauchy with respect to d. Hence xn → x ∈ X. Since (xn) is Cauchy with respect to d′, we have, at some point, 1 1 lim 0. W c − c W = i,j→∞ d(xi,U ) d(xj,U )

Therefore, since xi ∈ U with U open, there exists > 0 such that Bd(xi, ) ⊆ U. c c Hence, d(xi,U ) ≥ , which means that d(xi,U ) is bounded away from 0, which c allows us to write d(x, U ) > 0. Hence, x ∈ U. We are now ready for the big theorem of this section.

Theorem 2.22. Let X be Polish, Y ⊆ X. Y is Polish if and only if Y is Gδ. Proof. We have two directions. Suppose ﬁrst that Y is Polish. Consider the function id ∶ Y → X. It is continuous, hence by Theorem 2.20, there exists a ¯ continuous extension of f from a Gδ set such that Y ⊆ G ⊆ Y. We show that G = Y. Let x ∈ G, there exists a sequence (xn)n∈ω ⊆ Y such that xn → x. Hence, this sequence is Cauchy, therefore it converges to an element of Y, since Y is Polish. Thus x ∈ Y. For the second direction, we want to generalize the previous Lemma 2.21. Suppose Y = ⋂n∈ω Un, with each Un open. For each n ∈ ω, by Lemma 2.21, there exists a complete metric dn ≤ 1. We deﬁne a new metric on Y : 1 ′ d (x, y) = dn(x, y). Q 2n+1 n∈ω

′ If (xn)n∈ω is Cauchy with respect to d , it is Cauchy with respect to dn. Indeed, n 1 dn < 2 + d′. Hence, in each Un, since dn is complete, there exists x ∈ Un such that xn → x. This x is unique since all the metrics are compatible with the usual one. Therefore, x ∈ ⋂n∈ω Un = Y. Now, we make a classical example which is a special case of the previous Theorem.

ω Example 2.23. The set {∃∞1} = {s ∈ 2 ∶ S{i ∈ ω ∶ s(i) = 1}S = ω} is a Gδ set in 2ω. Indeed, we have:

n ω {∃∞1} = {∃ 1} = {s ∈ 2 ∶ S{i ∈ ω ∶ s(i) = 1}S ≥ n}. n∈ω n∈ω

22 The sets of the intersection are open by the intuitive deﬁnition of open sets in ω n 2 . Indeed, once we have n 0’s, we cannot go out of {∃ 1}. More precisely, we have: ∞ {∃ 1} = Nsˆ1 n∈ω s∶l(s)=n

We can ﬁnd this result also using the preceding theorem with the function: ω ω f ∶ ω → 2 s0 s1 (sn)n∈ω ↦ 0 10 1 ... Considering the usual basis as always, this is clearly a homeomorphism on its ω image, hence f(ω ) is Polish. Indeed, injectivity is clear from the deﬁnition. Moreover, it is open since

f(Ns) = N0s0 10s1 1...10sl(s)1 t t t Finally, it is continuous since, if s = 0 0 10 1 1 ... 10 k : f −1 N N ( s) = (t0,t1,...,tk−1,l) l

ω It remains to see that f(ω ) = {∃∞1}, which is trivial.

2.3 Some Universal Properties

In this section, we just present some classical and general results which will ﬁx some ideas about which are the Polish spaces. Since these results are not really important for the rest of this work, we do not prove them. The reader can ﬁnd the proofs in [Kec95]. Theorem 2.24. Every separable metrizable space is homeomorphic to a subset ω of I .

Corollary 2.25. Polish spaces are, up to homeomorphism, all the Gδ subsets ω of I . ω ω Since R looks like I but is bigger, we can ask the question whether we ω need less than the Gδ to have all Polish spaces in R . ω Theorem 2.26. Every Polish space is homeomorphic to a closed subset of R .

2.4 Perfect Polish Spaces

Now, we come to the definitions of two types of Polish spaces. The first ones are the perfect ones and the second ones are the zero-dimensional ones. We begin with the following topological definition. Definition 2.27. Let X be a topological space. A point x ∈ X is a limit point if ∀U open s.t. x ∈ U, we have U ∖ {x} ≠ ∅. A point that is not a limit point is an isolated point, since it can be isolated from the rest of the space by some open set.

23 With the previous Definition, we can define a class of Polish spaces. Definition 2.28. A Polish space is perfect if it has only limit points. Most of the examples we saw are perfect. In particular, it is the case of n n n ω ω R , C , I , 2 , ω . However, it is not the case of all Polish spaces, since finite sets are Polish but not perfect. Hence, the study of these perfect sets is a quite general study in descriptive set theory. We use, in the next proof, a tool which is used a lot in [Kec95], which is the notion of Scheme. This corresponds to defining recursively some elements of a set. For the first time, we will see how the dynamical aspect of 2ω can be useful in descriptive set theory. Theorem 2.29. Let X be a nonempty perfect Polish space. Then X contains a copy of 2ω. The idea of the proof is to use the dynamical characterization of elements of 2ω to identify them to elements of X.

s ∅ 0 Proof. We will deﬁne the elements (x )s∈2ω . Take x = x to be some element 1 0 0 0 0 1 of X. Since X is perfect, there exists x ∈ B(x , 1)∖{x }. Set also δ = d(x , x ). Now, we deﬁne recursively on the length of s the elements xsˆ0, xsˆ1 and δsˆ0.

sˆ0 s x = x , δs xsˆ1 B xsˆ0, B xsˆ0, δsS(l(s)−1) xs , ∈ [ ( 3 ) ∩ ( )] ∖ { } sˆ0 sˆ0 sˆ1 δ = d(x , x ) where the existence of xsˆ1 is guaranteed by perfectness. The picture is the following.

x11 x00 δ0 x01 X x10

0 B(x00, δ ) B(x00, 1) 3

10 δ0 00 0 B(x , 3 ) ∩ B(x , δ )

s By construction, all the (x )s∈ω are different. Hence, to have an embedding ω f ∶ 2 → X s s ↦ x , it suffices to show that f is continuous. Indeed, the fact that f is open on its image is trivial by construction, by the definition of the induced topology and

24 by the standard basis of 2ω. 1 s Let U ⊆ X be open, and s ∈ f − (U). Hence, x ∈ U, therefore there exists > 0 s n 1 such that B(x , ) ⊆ U. It suﬃces to take n such that δ − < to get

−1 s s ∈ NsSn ⊆ f (B(x , )).

Hence, we have immediately the following consequence.

Corollary 2.30. Let X be a perfect nonempty Polish space. Then SXS = 2ℵ0 . Proof. This is a consequence of the previous Theorem 2.29 together with Propo- sition 2.2. Now, we want to classify the Polish spaces as a function of how far they are being perfect. Indeed, some spaces are almost perfect, in the sense that it suﬃces to remove a point to get a perfect Polish space.

ω ω ω Example 2.31. We take the following subset of 2 , {0 } ∪ {1ˆ2 }.

ω ω By removing just the leftmost branch (corresponding to 0 ∈ 2 ), we get the ω perfect subset {1ˆ2 }. We see that we can generalize this idea of removing points to other less perfect Polish spaces.

Example 2.32. We have the following iterated version. Suppose we have the n m ω following subset {0 1 0 ∶ n ∈ ω, m ∈ ω∗} of 2≤ , which is Polish by Example 2.13. In the following picture, this corresponds to the set of leaves of the tree.

⋅

If we take only limit points, which is the same as removing isolated points, we n ω obtain the subsets {0 1 ∶ n ∈ ω}.

25 ⋅

ω If we do the same a second time, we get the subset {0 }.

⋅

This is just an element of 2≤ω, hence isolated. Therefore, if we iterate one more time, we get the empty set which is trivially perfect. So, we can say that this set was three iteration far from a perfect subset. This process gives us the following deﬁnition.

Definition 2.33. Let X be a Polish space. We define the Cantor-Bendixson derivatives by transfinite recursion.

0 • X = X, α 1 α α • X + = {x ∈ X ∶ x limit point of X }, λ α • X = ⋂α∈λ X , for an ordinal α and a limit ordinal λ.

We observe that this deﬁnition corresponds exactly to the successive steps in Example 2.32. But we also have examples with inﬁnitely many nonempty Cantor-Bendixson derivatives.

Example 2.34. Let Tn be the full binary tree of length n. We construct the following subsets of 2≤ω.

26 T2

Tn+1

After n iteration of the Cantor-Bendixson process, the full binary tree Tn van- ished. Hence, after ω iteration of this process, there only remains the following n ω ω subset {0 ∶ n ∈ ω} ∪ {0 } of 2≤ ∶

⋅

n Finally, we see that, in this space, {0 } is isolated by Ns ∖Nsˆ0. Hence, after one ω more iteration, there only remains {0 }, which vanishes in the next iteration. We have also the following fact about these iterations. Lemma 2.35. In each Cantor-Bendixson iteration, we remove at most a countable amount of points. Proof. If a point is removed, it means that it is isolated. Hence, there exists a basis element which witnesses this property. Since the space is separable, there is at most a countable number of points that we remove at each step. For the next Theorem, we need the following general fact about second countable spaces. Theorem 2.36. Let X be a second countable space. If (Fα)α<β is a strictly decreasing transﬁnite sequence of closed sets, then β is countable.

Proof. Suppose {Un ∶ n ∈ ω} is a basis of X. We deﬁne:

N ∶ {F ⊆ X ∶ F closed } → P(ω) F ↦ N(F ) = {n ∶ Un ∩ F ≠ ∅}.

27 c c We observe that, since F is open, we have F = ⋃n∈ω{Un ∶ n ∉ N(F )}. Hence, c c F ≠ G → F ≠ G → {Un ∶ n ∉ N(F )} ≠ {Un ∶ n ∉ N(G)} → N(F ) ≠ N(G).

Therefore, N is injective. Moreover, by deﬁnition, it is clear that F ⊆ G implies N(F ) ⊆ N(G). By injectivity, we get that F ⊊ G implies N(F ) ⊊ N(G). Finally, suppose (Fα)α<β is a strictly decreasing transﬁnite sequence of closed sets, then N(Fα)α<β is a strictly decreasing sequence in P (ω) with respect to inclusion. Hence, β is countable.

This gives easily the following theorem about Polish spaces. Theorem 2.37. Let X be a Polish space. There exists a countable ordinal α0 such that for all ordinals α ≥ α0, we have α α X = X 0 . Proof. This follows directly from the fact that, taking only limit points, we create closed subsets, and that an intersection of closed sets is closed too. Hence, Cantor-Bendixson derivative creates only closed subsets. Therefore, by the previous Theorem 2.36, we have our result. Thus, at some point, we only have limit points, hence a perfect subset of the original set, which can possibly be the empty set. This suggests the following deﬁnition.

Deﬁnition 2.38. The smallest possible α0 from the previous theorem is called the Cantor-Bendixson rank of X. It gives us the perfect kernel of X, which is Xα0 .

In Example 2.32, the Cantor-Bendixson rank is 3, with no nonempty perfect subset. In Example 2.34, the Cantor-Bendixson rank is ω + 2. Our next goal is to characterize the elements of this perfect kernel. We observe, in the Example 2.32, that being a limit point is not suﬃcient to be in the perfect kernel. We need a much stronger topological characterization of these points. We begin by the following observation for limit points.

Lemma 2.39. Let X be a metrizable space. If x ∈ X is a limit point, every open neighborhood has cardinality at least SωS. Proof. Let x ∈ X be a limit point and U an open neighborhood of x. There exists > 0 such that B(x, ) ⊆ U. Since x is limit, there exists x′ ≠ x such that x′ ∈ B(x, ). If we repeat this process with = d(x, x′) we get inﬁnitely many diﬀerent points in U. Hence SUS ≥ SωS. The stronger topological characterization is the following.

Deﬁnition 2.40. Let X be a topological space. A point x ∈ X is a condensation point if every open neighborhood of X has uncountable cardinality.

To get our topological characterization of elements in the perfect kernel, we need two lemmas.

28 Lemma 2.41. Let X be a Polish space, and let P ⊆ X be a perfect subspace, α then P ⊆ X 0 . Hence, the perfect kernel is the biggest perfect subset of a Polish space.

Proof. We proceed by induction on α ≤ α0. The initial case is simply P ⊆ X. α For the successor case, suppose P ⊆ X . Since P is perfect, it contains only α α 1 limit points, hence P ⊆ {x ∈ X ∶ x is limit} = X + . For the limit case: it is obvious since we just take the intersection of all the previous steps.

Lemma 2.42. Let X be a Polish space, and P ⊆ X be perfect. Then P = {x ∈ P ∶ x is a condensation point of P }. Proof. An inclusion is obvious. For the second one, let x ∈ P, and U be an open neighborhood of x in P . Since P is perfect, it is closed in X, hence by Theorem 2.22, P is perfect Polish. By the same Theorem 2.22, we get that U is Polish and perfect using the induced topology. So SUS = 2ℵ0 by Corollary 2.30, which means x is a condensation point. We are now ready for the characterization. Theorem 2.43. Let X be a Polish space. The perfect kernel of X is the set of all its condensation points. Proof. It is obvious, from Lemma 2.39 and the deﬁnition of a condensation point, that K = {x ∈ X ∶ x is a condensation point} is perfect. Moreover, using successively Lemma 2.41 and Lemma 2.42, we get:

α α K ⊆ X 0 ⊆ {x ∈ X 0 ∶ x is a condensation point} ⊆ K.

Moreover, as a consequence, we have the following Cantor-Bendixson Theorem.

Theorem 2.44. Let X be a Polish space. Then X can be uniquely written as X = P ∪ C, where P is a perfect subset of X, C is countable and open. Proof. The existence of such a decomposition follows from the Cantor-Bendixson derivatives, the fact that a perfect subset is closed and Lemma 2.35, hence we α α take X = X 0 ∪ (X ∖ X 0 ). Indeed, Lemma 2.35 together with the fact that a α countable union of countably many sets is countable implies X ∖ X 0 is countable. For uniqueness, suppose X = P ∪ C.P is perfect, hence Lemma 2.41 implies α α P ⊆ X 0 . Let x ∈ C and suppose by contradiction that x ∈ X 0 . Then every neighborhood has uncountable cardinality, in particular C, which is a contra- α α diction. Hence C ⊆ (X ∖ X 0 ). Since X = P ∪ C, this implies that P = X 0 and α C = X ∖ X 0 . Looking at the proof, we have the following corollaries.

Corollary 2.45. We have P ∩ C = ∅, hence C = ⋃{U ∶ SUS ≤ ω}.

29 Proof. The first statement follows from the definition of P and C has complement of each other. For the second statement, the first inclusion follows from the fact that C is countable and open. Suppose now that x ∉ C, then x is in the perfect kernel, hence all its open neighborhoods have uncountable cardinality. This implies x ∉ ⋃{U ∶ SUS ≤ ω}. Corollary 2.46. Any uncountable Polish space X contains a homeomorphic copy of 2ω, hence it has cardinality 2ℵ0 .

Proof. Since X is uncountable, Xα0 is nonempty by Theorem 2.44. Hence X contains a perfect subspace, which contains a Cantor space 2ω by Theorem 2.29. Hence the cardinality is 2ℵ0 . To end this Section about perfect Polish spaces, we give an example which use them and the Axiom of Choice. we will see that this set will have pathological properties later on. First, we deﬁne these sets.

ω c Deﬁnition 2.47. A Bernstein set is a set B ⊆ ω such that B and B do not contain any nonempty perfect subset and intersect all the perfect subsets.

To construct a Bernstein set in R, we need the following Lemma.

Lemma 2.48. There exist exactly 2ℵ0 many nonempty perfect subsets in R. Proof. Since we have a countable basis for R, an open set corresponds to a sequence in 2ω. Hence, there are at most 2ℵ0 open sets. Closed sets being complements of open ones, there are at most 2ℵ0 closed sets. Therefore, there are at most 2ℵ0 perfect sets in R. Moreover, we can easily construct 2ℵ0 perfect sets taking [0, x], x ∈ R+. We observe that this proof is also correct in the Baire space ωω. Indeed, it suffices to view the Baire space as a subset of the interval [0, 1] via the continuous ω fraction expansion f, and to take the perfect subsets as [0, x]∩f(ω ), x ∈ [0, 1]∩ R ∖ Q. Now, we are ready to construct our Bernstein set using the Axiom of Choice. The Axiom of Choice is equivalent to Zermelo’s Theorem which says that every set can be well-ordered. ℵ Hence, let (Pα)α<2 0 be a well-ordering of the nonempty perfect subsets ω ℵ ℵ of ω . We define two increasing sequences of subsets (Aα)α<2 0 and (Bα)α<2 0 0 recursively on α < 2ℵ . First, let A0 = {a} and B0 = {b} where a ≠ b and a, b ∈ P0, which is possible since P0 is perfect. For the recursion, suppose we have already constructed the sets Aα,Bα, for α < β, and that Aα and Bα are countable for 0 all α < β. Since Pβ is perfect and nonempty, it has cardinality 2ℵ . Moreover, since we have the Axiom of Choice, a countable union of countably many sets is countable, hence ⋃α<β Aα is countable, and the same is true for the Bα’s. Hence, we can find

′ ′ ⎛ ⎛ ⎞⎞ ′ ′ a , b ∈ Pβ ∖ Aα ∪ Bα , a ≠ b . ⎝ ⎝α<β α<β ⎠⎠

30 Then, set:

′ Aβ = Aα ∪ {a }, α<β ′ Bβ = Bα ∪ {b }. α<β

Hence, Aβ and Bβ are still countable, and our recursion follows. ℵ ℵ Finally, our construction yields A = ⋃α<2 0 Aα and B = ⋃α<2 0 Bα disjoint and both intersect all the perfect subsets of ωω. Hence, both are Bernstein sets.

2.5 Zero-Dimensional Polish Spaces

In this Section, we study a second kind of Polish spaces. This is the place where ω we will see the big difference between R and ω . Definition 2.49. A topological space X is connected if the only clopen subsets of X are ∅ and X itself. Remark 2.50. This definition is obviously equivalent to the usual one telling that a space is connected if it cannot be written as a disjoint union of two nonempty open subsets.

It is a well known fact that R is connected. At the other end of the picture, we have the spaces that are not connected at all. Deﬁnition 2.51. A topological space X is zero-dimensional if it is Hausdorﬀ and has a basis of clopen subsets.

With the standard open sets in 2ω and ωω, we see that these spaces are zero-dimensional. Indeed, we have, for ωω:

c Ns = NsS(i−1)ˆa. i

31 X n B′. B B′ B , say = ⋃i=1 i Deﬁne recursively i = i ∖⋃j

n X = + Bi. i=1

0i−1ˆ1 0n−1 Using the following tree, we set Bi = C for 1 ≤ i ≤ n − 1 and Bn = C .

1 0

B1 1 0

B2 1

B3 0 1

Bn Bn−1

Now, since Bi are clopen, we have that Bi is also perfect, nonempty, compact, Polish and zero-dimensional. Hence we can iterate our process infinitely many times, making sure the diameter of the Cs become arbitrarily small, and that the subcover has always cardinality at least 2. s s ω In this construction, we have defined (C )s∈ω such that liml(s)→∞ diam(C ) = 0. Finally, since every Cs is nonempty and our space is metrizable, we can define xs as: s n s C S = {x }. n∈ω To finish the proof, it suffices to show that the following function is a homeomorphism:

ω f ∶ 2 → X s s ↦ x f is surjective since we always take a covering of the preceding set, it is in- s jective because liml(s)→∞ diam(C ) = 0 and X is Hausdorﬀ. It is open since s ω 1 f(Ns) = C is clopen, s ∈ 2< . Finally, if U is open in X, let s ∈ f − (U). s sSn Since liml(s)→∞ diam(C ) = 0, there exists n ∈ ω such that C ⊆ U. Hence −1 sSn s ∈ NsSn ⊆ f (C ). So f is continuous. We have an equivalent result about the Baire space, which is the Alexandrov- Urysohn’s Theorem.

Theorem 2.53. The Baire space ωω is the unique, up to homeomorphism, nonempty, Polish, zero-dimensional space, such that all its compact subsets have an empty interior.

Proof. We have already seen that ωω is Polish, nonempty and zero-dimensional. ω Moreover, let A be a set with nonempty interior. Hence Ns ⊆ A for some s ∈ ω< . Hence, A cannot be compact since we have the following open cover of A that has no inﬁnite subcover:

ω A = (A ∩ (ω ∖ Ns)) ∪ Nsˆn. n∈ω

32 Suppose now that X has all these properties. Then we will construct a s ∅ Scheme (C )s∈ωω in X. Set C = X. Since X is non-compact, there exists an inﬁnite cover of X consisting of clopen sets which has no ﬁnite subcover. As in the previous proof, we can make sure that this cover is a disjoint one. Since X is Polish, we have a partition into clopen subsets:

X = + Un. n∈ω n Just deﬁne C = Un. Moreover, since Un is clopen, it is Polish, nonempty, zero-dimensional and such that any compact set has empty interior. Hence, we can iterate this process making sure that the diameter of the Cs’s becomes arbitrarily small. Hence, as in the proof of the previous Theorem, we deﬁne:

ω f ∶ ω → X s s ↦ x ,

s sSn where {x } = ⋂n∈ω C . For the exact same reasons as in the proof of the previous Theorem, this is a homeomorphism. Finally, we end up this part with the following important theorem, which justiﬁes, once again, the deep study of the Baire space ωω. Theorem 2.54. Let X be a zero-dimensional Polish space. Then X is homeomorphic to a closed subspace of ωω.

Proof. Going along the preceding proof, we observe that we can also write X = *n∈ω Un, except that taking the disjoint union may create empty sets. Indeed, we use the hypothesis about the open sets having nonempty interior in order to find a cover of every element of our Scheme which is infinite. In this Theorem, we do not have this assumption, therefore it is possible that we need to consider the empty set in order to create an infinite cover into clopen sets. We will see that it is not a serious problem if we consider not the whole ωω, but s a closed subset of it. So go along the process of defining a Scheme (C )s∈ωω , and set ω s n D = {s ∈ ω ∶ C S ≠ ∅}. n∈ω Define

f ∶ D → X s s ↦ x ,

s sSn where {x } = ⋂n∈ω C . As always, injectivity and surjectivity follow from the construction. The choice of D together with the fact that X is Polish allow us to say that f is well-deﬁned. Moreover, f is open and continuous for the same exact reason as previously. It remains to show that D is closed. ω Let (xn)n∈ω ⊆ D → x ∈ ω . Then, using the fact that f is continuous, we have that (f(xn))n∈ω is Cauchy. Thus, since X is Polish, there exists y ∈ X such s ω x n that f(xn) → y ∈ X. Since C is closed for all s ∈ ω< , and f(xn) ∈ C S , we get x n y ∈ C S for all n ∈ ω. Hence, x ∈ D, and since f is continuous, f(x) = y.

33 Chapter 3

Hierarchies on Polish Spaces

We said, in the introduction, that this work is about giving hierarchies on a class of topological spaces. In the previous Chapters, we described and studied deeply this class of topological spaces, the Polish ones. Now, we come to the deﬁnitions of the hierarchies on topological spaces. We start by deﬁning them in a general framework. As we can see with the previous results on Polish spaces, we can focus on the Baire space ωω.

3.1 Borel Hierarchy

We start by the very ﬁrst hierarchy on a topological space. This study comes from measure theory, when the question of which sets are good, in the sense of measurable, and which sets are not, is addressed. To have it in mind, and understand where our theory comes from, we recall the deﬁnition of a measure.

Deﬁnition 3.1. Let X be a space. A σ-algebra A on X is a subset A ⊆ P(X) such that:

• A is nonempty, • A is closed by complements, • A is closed by countable unions. Deﬁnition 3.2. A measurable space (X, A) is a space together with a σ- algebra on it.

Deﬁnition 3.3. Let (X, A) be a measurable space, where X is a space and A ⊆ P(X) a σ-algebra. A function µ ∶ A → [0, +∞) from the measurable sets to [0, +∞) is a measure when the following conditions are satisﬁed:

• µ(∅) = 0,

• If (En)n∈ω are some pairwise disjoint elements in A, then

µ En = Q µ(En). n∈ω n∈ω

34 A well known measure on R is given by µ′([a, b]) = b − a. It is the Lebesgue measure which coincides with the intuitive notion of a volume. Of course, the ﬁrst question which comes in mind is the existence of a nonzero measure on the σ-algebra P(X) which extends the Lebesgue measure. Indeed, such a measure would make the study of which are the measurable sets unin- teresting. Luckily, we have a good example of a non Lebesgue-measurable set. It is from such examples that descriptive set theory was born. Our example is the Bernstein sets. As always with pathological sets, it is the Axiom of Choice which gives such a bad behavior. Before going into this example, we need a well known general fact about the Lebesgue measure.

Lemma 3.4. Let µ′ be the Lebesgue measure on R. If A ⊆ R is countable, then µ′(A) = 0.

Proof. Let > 0 and let A ⊆ R be countable. We can write A = (an)n∈ω. For all n ∈ ω, consider Un = ai − 2i+2 , ai + 2i+2 . By the second condition to be a measure, we obtain the result,

 ′ ′ ′ µ (A) ≤ µ Un ≤ µ (Un) = ≤ → 0. Q Q 2i+1 n∈ω n∈ω n∈ω

We will also make use of the Regularity Theorem of the Lebesgue measure which can be found in [SS05, p.21].

Theorem 3.5. Let A be a Lebesgue-measurable set. For all > 0, there exists F ⊆ R closed and U ⊆ R open with F ⊆ A ⊆ U such that µ′(U ∖ F ) < . Now, we can show that a Bernstein set is not measurable. Proposition 3.6. A Bernstein set is not Lebesgue-measurable.

c Proof. Suppose B is a Lebesgue-measurable Bernstein set. Since B ∪ B = R, one of these two sets has a strictly positive measure. Without loss of generality, suppose it is B. Therefore, by the Regularity Theorem of the Lebesgue measure, B contains a closed set F with strictly positive measure. This means that F is closed, and uncountable. Therefore, F is uncountable and Polish and, by the Cantor-Bendixson Theorem (Theorem 2.44), it contains a perfect subset. Hence, we have a contradiction to the fact that Bc intersects all perfect subsets. Hence, measure theory needs to understand more about the sets that are contained in a σ-algebra. Since the most natural and useful sets in topology are the open ones, it is quite natural to consider the σ-algebra that contains them all. It is the σ-algebra of the Borel sets, which is an important and fundamental object in descriptive set theory. Deﬁnition 3.7. Let X be a topological space. The Borel sets of X are the ones that are contained in the smallest σ-algebra B(X) containing the open sets. We have a ﬁrst important consequence. Proposition 3.8. Any Borel set has the BP.

35 Proof. This comes directly from the σ-algebra deﬁnition of Borel sets and Theorem 1.45.

The smallest σ-algebra containing some D ⊆ P(X) is obviously the one generated by the elements of D. Hence, it is E(D) ⊆ P(X) obtained by closing under complements and countable unions the elements in D. We want to put a hierarchy on the Borel sets. Indeed, some are, in a topological sense, easy to obtain, for example the open sets (because they are given by the topology). Other sets are more difficult to find, as is the case of the sets that are obtained after some, maybe lots of, countable unions. We give the precise definition now: Definition 3.9. Let X be a topological space. We define by transfinite recursion the Borel hierarchy on X:

0 • Σ1(X) = {U ⊆ X ∶ U is open }, 0 c 0 • Πα(X) = {A ⊆ X ∶ A ∈ Σα}, Σ0 X A X A Π0 , α α . • α( ) = {⋃n∈ω n ⊆ ∶ n ∈ αn where n < } 0 0 0 We also deﬁne ∆α(X) = Σα(X) ∩ Πα(X). The next goal is to understand the shape of this hierarchy. We ﬁrst restrict to the case of metrizable spaces. Hence, we can use Proposition 2.16. We have the following Lemma.

Lemma 3.10. Let X be a metric space. For all α < β < ω1, we have:

0 0 0 Σα(X) ∪ Πα(X) ⊆ ∆β(X).

Proof. This proof is by transﬁnite recursion on α < ω1. We prove the inclusion 0 for Πα. In this proof, for more clarity, we write Γ instead of Γ(X). By taking 0 complement, the result will trivially hold for Σα. In fact, it suﬃces to show that Π0 ∆0, Π0 ∆0 Π0 ∆0 , λ α λ. 1 ⊆ 2 α ⊆ α+1 and α ⊆ λ with limit ordinal and < 0 0 Π1 ⊆ ∆2: 0 Suppose A ∈ Π1. By Proposition 2.16, we know that every closed subset of 0 0 a metric space is Π2, hence A ∈ Π2. Since it is trivial, by taking a countable 0 0 0 union of A’s, that A ∈ Σ2, we get Π1 ⊆ ∆2. 0 0 Πα ⊆ ∆α+1: 0 Suppose A ∈ Πα. It is also trivial, with the same countable intersection A Σ0 . A A as the previous one, that ∈ α+1 Moreover, we can write = ⋂n∈ω n A Σ0 , β α. A ∆0 . with n ∈ β < By the induction hypothesis, we have n ∈ β+1 So

36 A Σ0 . n ∈ β+1 Finally:

0 A = An,An ∈ Σβ, β < α n∈ω A ,A ∆0 , β α = n n ∈ β+1 < n∈ω c c A ,A ∆0 , β α = n n ∈ β+1 < n∈ω c Ac ,A ∆0 , β α = n n ∈ β+1 < n∈ω c B ,B ∆0 , β α = n n ∈ β+1 < n∈ω c B ,B Π0 , β α. = n n ∈ β+1 < n∈ω Ac Σ0 , A Π0 Hence can be written as a α+1 so ∈ α+1, which is what we want. 0 0 Πα ⊆ ∆λ, with λ limit and α < λ: A Π0 , A A Σ0 , α α 1 If ∈ α taking a countable union of ’s, we have ∈ α+1 < + < λ. Hence, by the Deﬁnition of limit levels, taking suitable unions and 0 0 0 intersections, we get A ∈ Σλ ∩ Πλ = ∆λ.

Finally, by construction and by the preceding Lemma, we get that:

0 0 0 B(X) = Σα = Πα = ∆α. α<ω1 α<ω1 α<ω1 Hence, the hierarchy is the one in the following picture.

0 0 0 0 Σ Σ Ση Σ ⎫ 1 ⊆ 2 ν ⎪ 0 ⊆ 0 ⊆ 0 ⊆ 0 ⊆ ⎪ ∆1 ∆2 ⋯ ∆η ⋯ ∆ν ⋯ ⎬ B(X) ⊆ ⊆ ⊆ ⊆ ⎪ Π0 ⊆ Π0 Π0 Π0 ⎪ 1 2 η ν ⎭⎪

Figure 3.1: Borel hierarchy on metrizable spaces

where each level contains every level that is on the left of it, and where η ≤ ν. By construction, it follows that this hierarchy has at most ω1 levels. But, there is a lot of things that we do not know yet. For example, does this hierarchy collapse at some point? Is it possible to reﬁne this hierarchy with another hierarchy, which has also a mathematical interest, and which has more levels? Is it possible to create a hierarchy on B(X) which is not the same at all, but which makes sense too from a mathematical point of view? The rest of this work will answer some of these questions. We ﬁrst take note on some facts about the classes in the Borel hierarchy.

37 Proposition 3.11. Let X be a metric space and Γ be a class in the Borel hierarchy.

• Γ is closed under finite unions and finite intersections; • Γ is closed under continuous preimage; 0 • if Γ = Σα(X) for some α < ω1, then Γ is closed under countable unions; 0 • if Γ = Πα(X) for some α < ω1, then Γ is closed under countable intersections; 0 • if Γ = ∆α(X) for some α < ω1, then Γ is closed under complements. Proof. The only thing to check is the statement about the closure under continuous preimage. Indeed, all the other things follow directly from recursive definition of the Borel hierarchy, and the fact that open subsets are closed under countable unions. Let f ∶ X → X be a continuous function. The only things we need are −1 c −1 c −1 −1 f (A) = f (A ) and f (⋃n∈ω An) = ⋃n∈ω f (An). 1 c c f − (A ) = {x ∈ X ∶ f(x) ∈ A } = {x ∈ X ∶ f(x) ∉ A} c = {x ∈ X ∶ f(x) ∈ A} 1 c = f − (A) .

−1 f An = {x ∈ X ∶ f(x) ∈ An} n∈ω n∈ω = {x ∈ X ∶ ∃n ∈ ω, f(x) ∈ An} = {x ∈ X ∶ f(x) ∈ An} n∈ω −1 = f (An). n∈ω

We will see, in a few pages, that the stability by continuous preimage will be a fundamental tool for the refinement of this hierarchy. The first thing we want to show is that this hierarchy has indeed ω1 levels. That is, that each class strictly increases the classes on the left of it. Of course, this is not true for every metrizable space. Indeed, if we take a finite set with the discrete, metric topology, the hierarchy cannot have ω1 levels. Indeed, there is only a finite number of subsets, hence the hierarchy cannot grow ω1 times. In this case, we need more assumption on the ambient space, which is indeed required to be an uncountable Polish space. To prove this will take some time, and we need the definition of a universal set. The idea is to code all the elements of a given class inside a set having a good behavior. Definition 3.12. Let X and Y be sets, and let Γ be a class of the Borel hierarchy. A set U ⊆ Y ×X is Y -universal for Γ(X) if the following conditions are satisfied:

• U ∈ Γ(Y × X),

38 • {Uy ∶ y ∈ Y } = Γ(X), where Uy = {x ∈ X ∶ (y, x) ∈ U}. Hence, the set U is not too complicated since it is in Γ(Y × X). However, it contains the code of every element in Γ(X) using projection functions. Let us focus on the existence of such particular sets. Theorem 3.13. Let X be a separable metrizable space. ω 0 0 For all α ≥ 1, there exists a 2 -universal set for Σα(X) and for Πα(X). ω 0 Proof. First, we construct a 2 -universal set U for Σ1(X). Since X is separable and metrizable, we know that it is second countable. Hence, we have a countable ω basis {Un}n∈ω. We deﬁne U ⊆ 2 × X by letting:

(s, x) ∈ U ↔ x ∈ {Un ∶ s(n) = 1}. n∈ω This means that s codes x by saying in which open set x lives. This code is unique since our space is metrizable, hence T0. ω 0 Let us show that U is 2 -universal for Σ1(X). There are two things to show. First, we show that U is open. Indeed, let (s, x) ∈ U. Fix k ∈ ω. We have that

N U s(0) U s(1) U s(k−1) sSk × 0 ∪ 1 ∪ ⋅ ⋅ ⋅ ∪ k−1 is an open neighborhood of (s, x) which is contained in U. Therefore U is open. ω 0 It remains to show that {Us ∶ s ∈ 2 } = Σ1(X). This follows directly from the fact that each open set is a union of elements of the basis, and each union of elements of the basis is an open set. 0 Now, we observe that, if U is universal for Σα(X), then ∼ U = {X ∖U ∶ U ∈ U} is 0 universal for Πα(X). This follows directly from the property of universal sets. Hence, we constructed universal sets for the ﬁrst level of the Borel hierarchy, now, we will climb the hierarchy step by step. β ω 0 Suppose U is 2 -universal for Πβ, for β < α. We ﬁx a bijection

b ∶ ω × ω → ω.

Take an increasing sequence (γn)n∈ω such that

sup{γn + 1 ∶ n ∈ ω} = α.

This is possible since cof(α) ≤ ω, for all α < ω1. ω ω Moreover, for all s ∈ 2 , we deﬁne (s)n ∈ 2 , by

(s)n(m) = s(b(n, m)).

ω Finally, we deﬁne U ⊆ 2 × X by:

γn (s, x) ∈ U ↔ ∃n ∈ ω ((s)n, x) ∈ U .

It only remains to show that U is universal. First, we show that, if we ﬁx n ∈ ω,

ω ω C ∶ 2 → 2 s ↦ (s)n

39 ω 1 is a continuous function. Let an open set U ⊆ 2 , and let x ∈ C− (U). Then, there exists m ∈ ω such that x x N U. C( ) = ( )n ∈ (x)nSm ⊆

1 ′ ′ − ′ Let m = max1≤k≤m b(n, k). We show that x ∈ NxSm ⊆ C (U). Let y ∈ NxSm , we have

C(y) = (y)n = (y(b(n, 1)), y(b(n, 2)), . . . , y(b(n, m)),... ) = (x(b(n, 1)), x(b(n, 2)), . . . , x(b(n, m)),... ) = ((x)n(1), (x)n(2),..., (x)n(m),... ) N U. ∈ (x)nSm ⊆

Therefore C is continuous. This implies that ω ω C × id ∶ 2 × X → 2 × X is continuous. Moreover,

γn U = {(s, x) ∶ ∃n ∈ ω ((s)n, x) ∈ U } γn = {(s, x) ∶ ((s)n, x) ∈ U } n∈ω 1 γn = (C × id)− (U ). n∈ω

γn 0 ω Finally, by induction hypothesis, we have U ∈ Πβ(2 × X) for some β < α. 0 ω 0 ω Hence U is a countable union of Πβ(2 ×X), with β < α, hence U is Σα(2 ×X). To complete the proof, it remains to prove the second property of universal set.

γn Us = {x ∈ X ∶ ∃n ∈ ω ((s)n, x) ∈ U } γn = {x ∈ X ∶ ((s)n, x) ∈ U }. n∈ω

By the induction hypothesis, ﬁxing n = n0 ∈ ω, we have

γ ′ γn0 n 0 {x ∈ X ∶ (s , x) ∈ U } = Us′ ∈ Πβ(X) for some β < α. To conclude, it suﬃces to observe that Us is a countable union 0 0 of Πβ(X) with β < α, therefore Us is Σα(X). Now, we use a usual method of obtaining a contradiction, known as diago- nalization.

Theorem 3.14. Let X be an uncountable Polish space. 0 0 For all α, we have Σα(X) ≠ Πα(X). ω Proof. By Corollary 2.46, X uncountable implies that 2 ⊆ X. Moreover, we have 0 ω 0 ω 0 ω 0 ω Γα(2 ) = Γα(X)S2 and Πα(2 ) = Πα(X)S2 , 0 ω ω 0 where Σα(X)S2 = {A∩2 ∶ A ∈ Σα(X)}. Indeed, this follows from the deﬁnition by the open sets of these classes, and by the deﬁnition of the induced topology on subsets.

40 0 0 Suppose, by contradiction, that Σα(X) = Πα(X) for some α < ω1. We have: 0 ω 0 ω 0 ω 0 ω Σα(2 ) = Σα(X)S2 = Πα(2 ) = Πα(X)S2 .

ω 0 ω By the previous Theorem, let U be 2 -universal for Σα(2 ). ω We deﬁne A ⊆ 2 : s ∈ A ↔ (s, s) ∉ U. 0 ω 0 ω ω Since U is universal for Σα(2 ), we have U ∈ Σα(2 × 2 ). We show that the following function is continuous:

ω ω ω f ∶ 2 → 2 × 2 s ↦ (s, s)

ω ω ω 1 Let U ⊆ 2 × 2 be open. If U ∩ f(2 ) = ∅, then f − (U) = ∅ is open. ω 1 Otherwise, U ∩ f(2 ) ≠ ∅. Let s ∈ f − (U). It means that (s, s) ∈ U, so there −1 exists k ∈ ω such that NsSk × NsSk ⊆ U. Therefore, s ∈ NsSk ⊆ f (U). So f is c −1 0 ω continuous. We get that A = f (U) ∈ Σα(2 ), since this class is closed under continuous preimages by Proposition 3.11. Hence, using the absurd hypothesis, we get 0 ω 0 ω A ∈ Πα(2 ) = Σα(2 ). Finally, we ﬁnd a contradiction because U is universal, hence there exists ω s0 ∈ 2 such that Us0 = A. But:

s0 ∈ A → (s0, s0) ∉ U → s0 ∉ Us0 = A ☇ and

s0 ∉ A → (s0, s0) ∈ U → s0 ∈ Us0 = A ☇.

From this Theorem and its proof, we have some corollaries which allow us to better understand what are the levels in the Borel hierarchy. Corollary 3.15. Let X be an uncountable Polish space. 0 0 0 0 For all α, we have Σα(X) ∖ Πα(X) ≠ ∅, and Πα(X) ∖ Σα(X) ≠ ∅. Proof. Observing the proof of the Theorem 3.14, we see that A deﬁned as 0 ω 0 ω above is in Πα(2 ) ∖ Σα(2 ) via f and the contradiction. Exchanging the role 0 ω 0 ω ′ 0 ω 0 ω of Σα(2 ) and Πα(2 ), we can also ﬁnd A ∈ Σα(2 ) ∖ Πα(2 ). Corollary 3.16. Let X be an uncountable Polish space. α, ∆0 X Σ0 X ∆0 X , For all we have α( ) ⊊ α( ) ⊊ α+1( ) and the same is true with 0 0 Πα(X) in lieu of Σα(X). 0 0 Proof. The previous Corollary gives an immediate proof of ∆α(X) ⊊ Σα(X). For the second statement, we have the following:

A Π0 X Σ0 X A ∆0 X Σ0 X . ∈ α( ) ∖ α( ) → ∈ α+1( ) ∖ α( )

Corollary 3.17. Let X be an uncountable Polish space. 0 0 For each limit ordinal λ < ω1, we have ⋃α<λ Σα(X) ⊊ ∆λ(X).

41 Proof. Since λ < ω1 and λ limit, there exists an increasing sequence (αn)n∈ω lim α λ, α λ n ω, A Σ0 X Π0 X such that n→∞ n = n < . For all ∈ take n ∈ αn ( ) ∖ αn ( ) as in the proof of Theorem 3.14. Using the inclusion in the Borel hierarchy and the deﬁnition of limit levels, it is easy to see that 0 An ∈ ∆λ. n∈ω 0 ′ Suppose, by contradiction, that ⋃n∈ω An ∈ ⋃α<λ Σα. Hence, there exists α < λ 0 such that ⋃n∈ω An ∈ Σα′ (X), which contradicts limn→∞ αn = λ and the deﬁnition of the An’s. Hence, the Borel hierarchy on uncountable Polish spaces is strictly increasing, which means that it does not collapse. So, we know exactly the shape of this hierarchy as well as its length. Corollary 3.18. The length of the Borel hierarchy on any uncountable Polish space is ω1.

0 0 0 0 Σ Σ Ση Σ ⎫ 1 ⊊ 2 ν ⎪ 0 ⊊ 0 ⊊ 0 ⊊ 0 ⊊ ⎪ ∆1 ∆2 ⋯ ∆η ⋯ ∆ν ⋯ ⎬ B(X) ⊊ ⊊ ⊊ ⊊ ⎪ Π0 ⊊ Π0 Π0 Π0 ⎪ 1 2 η ν ⎭⎪

Figure 3.2: Borel hierarchy on uncountable Polish spaces

We have, in the Borel hierarchy of uncountable Polish spaces, a representa- 0 0 ω tive for each Σα(X) and Πα(X) given by these 2 -universal sets. We prove, 0 ω now, that these universal sets cannot exists for ∆α(2 ) under some assumptions. Lemma 3.19. Let Γ be a class closed under complements, closed under continuous preimage and X be a metrizable space. Then, there exists no X-universal set for Γ(X). Proof. Towards a contradiction, suppose there exists a X-universal set U for Γ(X). Then, U ∈ Γ(X × X) and Ux ∈ Γ(X) for all x ∈ X. Deﬁne: f ∶ X → X × X x ↦ (x, x). As previously, since we are in a metrizable space, this function is continuous. 1 Indeed, suppose U is open in X × X. We can suppose f − (U) nonempty, oth- 1 erwise, we are done. Let x ∈ f − (U). Hence, (x, x) ∈ U, therefore, there exists 1 > 0 such that B(x, ) × B(x, ) ⊆ U. Then, we obtain x ∈ B(x, ) ⊆ f − (U). Let us deﬁne a set A ⊆ X ∶ x ∈ A ↔ (x, x) ∉ U. c 1 Since Γ is closed under continuous preimage, we get A = f − (U) ∈ Γ(X), and since X is universal and Γ closed under complements, we get the existence of s ∈ X such that A = Us. But:

s ∈ A → (s, s) ∉ U → s ∉ Us = A ☇

42 and s ∉ A → (s, s) ∈ U → s ∈ Us = A ☇.

ω 0 ω Hence, there exists no 2 -universal sets for all classes of the form ∆α(2 ).

3.2 Diﬀerence Hierarchy

In the previous section, we have seen a natural way to construct elements in both 0 0 Σα(X) and Πα(X) for a topological space X by taking countable unions and complements of open sets. However, we have not seen an equivalent construc- 0 tion for ∆α(X). That is what we propose to do here, using a new topological construction. Actually, we will just give the properties and the construction without proving them. Indeed, the proofs come from technical arguments of topology and can be found in [Kec95].

Deﬁnition 3.20. Let X be a topological space, β an ordinal, and (Aγ )γ<β an increasing sequence of subsets of X. We deﬁne:

Dβ((Aγ )γ<β) = {x ∈ X ∶ the least α such that x ∈ Aα has parity opposite to that of β}.

Example 3.21. We give some examples of this construction.

• D2((A0,A1)) = A1 ∖ A0,

• D3((A0,A1,A2)) = (A2 ∖ A1) ∪ A0,

• Dω((Aγ )γ<ω) = ⋃α<ω(A2α+1 ∖ A2α),

...... A2 A1 A0

• Dω+1((Aγ )γ<ω+1) = A0 ∪ ⋃α<ω(A2α ∖ A2α−1) ∪ (Aω ∖ ⋃α<ω Aα).

Aω ...... A2 A1 A0

43 Deﬁnition 3.22. Let X be a topological space, Γ a class in the Borel hierarchy and β an ordinal. We deﬁne:

Dβ(Γ) = {D ⊆ X ∶ ∃(Aα)α<β ⊆ Γ(X) s.t. D = Dβ((Aα)α<β)}, and ˇ c Dβ(Γ) = {D ∶ D ∈ Dβ(Γ)}. We have the following ﬁrst observations about these new construction.

Proposition 3.23. Let X be a metrizable space. For all 1 < α, β < ω1, we have that : 0 • the class Dβ(Σα) is closed under continuous preimage; ˇ 0 0 0 • Dβ(Σα) ∪ Dβ(Σα) ⊆ Dβ+1(Σα); ˇ 0 0 • if X is an uncountable Polish space, then Dβ(Σα) ≠ Dβ(Σα). Hence, by the previous Proposition, there is a lot of similarities between the Borel and the Difference hierarchy. Moreover, it is clear that each level in the Borel hierarchy is also a level in the Difference hierarchy. Therefore, the Difference hierarchy is just a refinement of the Borel hierarchy where we add, between each level of the Borel hierarchy, ω1 levels. Finally, in the case of uncountable Polish spaces, these levels are distinct, hence the Difference hierarchy, as a refinement of the Borel hierarchy, does not collapse. Finally, we get the main result of this Section which shows that there exists a ∆0 X . topological construction from the open sets for the classes α+1( ) The proof of the next statement can be found in [Kec95, p.176].

Theorem 3.24. Let X be a Polish space. Then, for all 1 ≤ α < ω1, we have ∆0 Σ0 . α+1 = Dβ( α) β<ω1 3.3 Lipschitz and Wadge Hierarchies

In the previous sections, we have seen that the classes inside the Borel and the Diﬀerence hierarchy are stable by continuous preimage. Is it just a coincidence or is it a real interesting property that we can use? The natural topological answer to this question is the second one. Indeed, all classical topology theory is based on the notion of continuity. Hence, we are going to use this idea, but, in parallel with continuous functions, we will use Lipschitz ones, in order to make things easier, as we will see.

Deﬁnition 3.25. Let X,Y be two topological spaces. We say that A ⊆ X is Lipschitz-reducible to B ⊆ Y if there exists a Lipschitz function f ∶ X → Y 1 reducing A to B, i.e. such that f − (B) = A. If A ⊆ X is Lipschitz-reducible to B ⊆ Y, we write A ≤L B. If this is not the case, we simply write A ≰L B. Let us analyze this notion of Lipschitz-reducibility. From now on, we will consider the space X = Y. Indeed, this allows us to compare the subsets of X via ≤L. This symbol suggests two things. First, it suggests that it induces a quasi-order on the subsets of X. Indeed, it is reﬂexive via the identity function

44 which is Lipschitz, and it is transitive since the composition of two Lipschitz function is also Lipschitz, hence:

−1 −1 A ≤L B ≤L C → f (C) = B, g (B) = A 1 1 1 → A = g− (f − (C)) = (f ○ g)− (C), → A ≤L C, for f, g Lipschitz functions. The second thing that is suggested by this symbol is that it acts as a com- parison. In this framework, ≤L compares the complexity of two sets. Indeed, A ≤L B means that the set B is at least as complicated as A. This comes from the fact that the problem of x being a member of A can be reduced to the problem of f(x) being a member of B:

1 f(x) ∈ B → x ∈ f − (B) → x ∈ A, 1 f(x) ∉ B → x ∉ f − (B) → x ∉ A.

Thus, the quasi-order induced by ≤L measures the complexity of a set modulo Lipschitz functions. Now, we can go further with the study of ≤L. Having this quasi-order implies the following deﬁnitions.

Deﬁnition 3.26. Let A, B ⊆ X. We say that A and B are Lipschitz comparable if A ≤L B or B ≤L A. If it is not the case, A and B are Lipschitz incomparable and we write A ⊥L B. We say that A is Lipschitz equivalent to B, written A ≡L B, if A ≤L B and B ≤L A.

Of course, we also write A

Deﬁnition 3.27. Let A ⊆ X a topological space. We deﬁne the Lipschitz degree of A as: [A]L = {B ⊆ X ∶ A ≡L B}.

Since we have a quasi-order, this notion of degree is well-defined and ≤L also induces a quasi-order on the Lipschitz degrees. We can now define one of our object of study. Definition 3.28. The Lipschitz hierarchy is the classification of the Lipschitz degrees into different levels which respect the quasi order ≤L. Of course, all we have just done can be done with continuous functions instead of Lipschitz ones.

Deﬁnition 3.29. Let X,Y be two topological spaces. We say that A ⊆ X is Wadge-reducible, or continuously reducible, to B ⊆ Y if there exists a 1 continuous function f ∶ X → Y reducing A to B, i.e. such that f − (B) = A. If A ⊆ X is Wadge-reducible to B ⊆ Y, we write A ≤W B. If this is not the case, we simply write A ≰W B.

45 From this and the fact that the identity function is continuous, and that the composition of two continuous functions is also continuous, we can, as for Lipschitz-reducibility, deﬁne the notions of Wadge comparable, Wadge equivalent, Wadge incomparable, Wadge hierarchy and all the symbols Pointclass Wadge hierarchy of a topological space X is an increasing hierarchy of subset of P(X), where A ⊆ X occupies the level

{B ⊆ X ∶ B ≤W A}.

Hence, this Pointclass Wadge hierarchy consists of all the initial segments of the Wadge hierarchy. For uncountable Polish spaces, this hierarchy is the following picture, which is a reﬁnement of Figure 3.2.

0 0 0 Σ1 Σ2 Σα 0 ... 0 ... 0 ... ∆1 ∆2 ∆α 0 0 0 Π1 Π2 Πα

Figure 3.3: Pointclass Wadge hierarchy on the Borel sets of uncountable Polish spaces

The Pointclass Wadge hierarchy is a refinement of the Borel hierarchy and of the Difference hierarchy because these classes are closed under continuous preimage. Moreover, since Lipschitz functions are continuous but the converse does not hold, we obtain that the Lipschitz hierarchy is itself a refinement of the Wadge hierarchy. These definitions of reducibilities are very clear, but they are not easy to use. Indeed, it is, in general, difficult to find continuous functions between two topological spaces that reduces some set to another one. Hence, the Lipschitz and the Wadge hierarchies are not studied with these notions, but with analo- gous ones, namely the Lipschitz and Wadge games. That is what we are going to do later on, in the next chapter. We describe some more interesting notions about the Pointclass Wadge hierarchy.

Deﬁnition 3.31. Let X be a Polish space, and Γ(X) a class of sets in this space. A ⊆ X is Γ-hard if B ≤W A for all B ∈ Γ(X). If, moreover, A ∈ Γ(X), we say that A is Γ-complete.

46 Hence, A is complete for a class Γ if A is exactly in the corresponding level in the Pointclass Wadge hierarchy. Our goal is to understand the Wadge hierarchy. Hence, we can describe this hierarchy in two different ways. The first one is to consider the Wadge hierarchy as we first defined it, hence as a quasi-order. Hence, in order to describe it, the better way is to be able to find a representative at each level of this quasi-order. This work has been done in [Dup01] for all the Wadge hierarchy on Borel sets. On the ordinals, the addition, the multiplication and the exponentiation are well known operations. Also, we will see, in the next chapter, that we can associate to each level of the hierarchy an ordinal, by the well-ordering of this hierarchy (Theorem 4.31). In the paper [Dup01], the idea is to associate to the ordinal operations described above some corresponding operations on sets. Hence, if we decompose an ordinal α into these basic operations, using for example the Cantor normal form, we can, from ∅ and by the corresponding operations on sets, find a set at level α in the Wadge hierarchy. Of course, this needs some work, and we will not get into more details here. The second point of view is the one of the Pointclass Wadge hierarchy. It suggests, in order to describe it, to find an operation on the open sets which 0 0 describes every level of the hierarchy. Indeed, the classes Σα(X) and Πα(X) were described by some operation on open sets. Moreover, we saw in the Section 0 3.2 that there exist also some operations describing the classes ∆α(X). Working in second countable T0 spaces, we can associate to each point a sequence in 2ω corresponding to the open sets of the basis in which the point is. Hence, the complements, the unions, and the differences correspond to some boolean combination on the sequence 2ω. The question is whether or not there exists a boolean combination corresponding exactly to each level of the Wadge hierarchy on the Borel subsets of ωω. The answer is yes, and these operations are described in [Lou12].

47 Chapter 4

Games and Hierarchies

In this Chapter, we try to ﬁnd a diﬀerent simpler characterization of Lipschitz and Wadge reducibility. To do so, we consider games. Since games consist of constructing dynamically elements in some space, we consider games on the spaces ωω. Of course, there will be an equivalent game in 2ω. Moreover, since, by Theorem 2.54, all zero-dimensional spaces are homeomorphic to closed subset of ωω, our game can be generalized in all zero-dimensional spaces.

4.1 Gale-Stewart Games on ωω

Finally, we define the following famous tools, which are games on the Baire space ωω. We precise that a game with complete perfect information is a game where all the players are aware of what has been played before, of what each player can play at every time of the game, and of what the winning condition of the game is. ω Definition 4.1. Let A ∈ P(ω ). We define the Gale-Stewart game G(A) as the following game. G(A) is a two players game with complete perfect information. The players are I and II. Each player plays successively an element of ω, starting from I. I wins the game if the resulting sequence is in A. Otherwise, II wins. Schematically, we draw: I II

x1 I wins if and only if y1 (x0, y0, x1, y1,... ) ∈ A. x2

y2 ⋮ ⋮

We observe that a run of the game can be viewed as a descent along a branch of the tree ωω. This point of view will be useful since it captures the dynamical

48 aspect of the game. As an example, we have a run of a game in the following picture.

In a game, a strategy is a rule that tells a player what he has to do. Math- ematically, we get the next deﬁnition. Deﬁnition 4.2. A strategy for I is a subtree σ of ωω such that:

• for all x ∈ σ such that l(x) = 2k for some k ∈ ω, there exists a unique y ∈ ω such that xˆy ∈ σ; • for all x ∈ σ such that l(x) = 2k + 1 for some k ∈ ω, we have xˆy ∈ σ for all y ∈ ω. This deﬁnition gives the following picture.

We observe that this definition gives rise to a nonempty pruned tree which simulates exactly the choices for I knowing what has been played before. We can, of course, define a strategy for II in the same way, adapting the indices. ω Moreover, if I has a strategy σ, and II plays some y ∈ ω , we write σ ∗ y for the run of the game which is played if I follows his strategy and II plays y. Definition 4.3. We say that a strategy σ is a winning strategy for a Player A if, by following σ, and regardless of what the opponent is playing, Player A wins the game. This Definition is equivalent to the fact that

ω {σ ∗ y ∶ y ∈ ω } ⊆ A, if σ is a winning strategy for I. If σ is winning for II, this condition is:

ω c {x ∗ σ ∶ x ∈ ω } ⊆ A ,

Deﬁnition 4.4. A game is determined if one of the players has a winning strategy.

49 It is clear that both players cannot have a winning strategy at the same time, otherwise both players will win the game, which is impossible by the winning condition. ω Definition 4.5. A class Γ of subsets of ω is determined if, for all A ∈ Γ, the game G(A) is determined. We come now to the difficult question of determinacy. The question is, which are the classes that are determined? Indeed, it is not difficult, with the Axiom of Choice, to find subsets of the Baire space which are not determined, we give some examples. Example 4.6. Consider the game G(A), where A is a Bernstein set. Suppose ω I has a winning strategy σ. Consider X = {σ ∗ y ∶ y ∈ ω }. Clearly, X is closed since it is a pruned tree and we have Theorem 1.38. Moreover, σ being a strategy implies that every point has a nonempty neighborhood which contains another point. Indeed, it suffices to see the strategy as a tree, which is exactly the set X. Hence, X is perfect. But, since σ is a winning strategy, we get X ⊆ A, which is impossible since A is a Bernstein set. Hence, I has no winning strategy. The same reasoning applied to a winning strategy for II and Ac also leads to a contradiction. Hence, G(A) is not determined. Therefore, the existence of a Bernstein set in a class contradicts the determinacy of this class. There also exists another beautiful example which also uses the Axiom of Choice. This example is about strategies. Example 4.7. First, there are exactly 2ℵ0 strategies by player. Indeed, a strategy can be viewed as a subtree of ωω, hence there are at most 2ℵ0 strategies. Moreover, every element of ωω gives rise to a strategy which plays the element, without regarding the other player’s moves. Using the Axiom of Choice, we can well-order the strategies for I, say by ℵ ℵ (σα)α<2 0 , and the strategies for II, say by (τα)α<2 0 . We construct two increas- 0 ing sequences Xα and Yα, for α < 2ℵ . We start with X0 = Y0 = ∅. Suppose we have already constructed Xα,Yα for all α < β, and that all these sets are countable. Moreover, suppose that Xα is disjoint from Yα for all α < β. Using the Axiom of Choice, we get that ⋃α<β Xα is countable, and the same is true for the ω ω Yα’s. Since ω is uncountable, there exists b ∈ ω such that σβ ∗ b ∉ ⋃α<β Xα. Just set yβ = σβ ∗ b, and Yβ = ⋃α<β Yα ∪ {yβ}. We can do the same reversing ℵ Xβ and Yβ where we choose a ∗ τβ ∉ ⋃α<β Yα. Finally, set X = ⋃α<2 0 Xα and ℵ Y = ⋃α<2 0 Yα. ω c For all α, by construction, there exists b ∈ ω such that σα ∗b ∈ Y ⊆ X . Since we enumerated all the strategies, I has no winning strategy in G(X). Moreover, ω for all α, there exists a ∈ ω such that a ∗ τα ∈ X, hence II has no winning strategies in G(X). Thus, G(X) is not determined. Finally, we give a last example of a not determined game. As for the Bern- stein set, we saw that a flip set can be construct from the Axiom of Choice, and that it has the bad property that it does not have the BP. We show that, if F is a flip set, then G(F ) is not determined. This time, we will use Banach- Mazur games (GBM (F )) instead of Gale-Stewart games. The game is the same except that the players can play elements in 2<ω instead of integers. By a bijection between these two sets, which is possible since they are both countable, Banach-Mazur games and Gale-Stewart games are equivalent.

50 Example 4.8. Towards a contradiction, suppose GBM (F ) is determined. First, suppose that I has a winning strategy σ. We construct a winning strategy for II in GBM (F ). This will be our contradiction. We will play simultaneously two times the game GBM (F ). In the first one, I plays according to his strategy σ. The sequence played are σi, respectively ti for I, respectively II. In the second game, we construct a winning strategy for II. The sequence played are xi, respectively yi for I, respectively II. Now, we describe the strategy for II in the second game. II makes sure to play t0 longer than x0. Therefore, II can play y0 such that σ0ˆt0ˆσ1 and x0ˆy0 have the same length but differ exactly in an odd number of digits. Then, by a back and forth argument, we define all the moves of II in both games by copying what I plays in the other game. That is, we have t1 = x1, y1 = σ2, t2 = x2, y2 = σ3,.... Schematically, we get, I II I II

σ0ˆt0ˆσ1 and x0ˆy0 σ0 have the same length x0 t0 but diﬀer exactly in an y0 σ1 odd number of digits. t1 = x1 x1 σ2 y1 = σ2 t2 = x2 x2 σ3 y2 = σ3 t3 = x3 x3

I wins II wins

Therefore, since the two results diﬀer in exactly an odd number of digits, and since σ is a winning strategy and F is a ﬂip set, we constructed a winning strategy in the second game for II, a contradiction. In the second case, if we suppose that II has a winning strategy, the argument is very similar, hence we just give the schematic proof.

I II I II

t0ˆσ0 and x0 t0 have the same length x0 σ0 but diﬀer exactly in an t1 = y0 y0 odd number of digits. σ1 x1 = σ1 t2 = y1 y1 σ2 x2 = σ2 t3 = y2 y2

II wins I wins

51 Therefore, the game GBM (F ) is not determined. Hence, the game G(F ) is not determined. As for the Bernstein sets, the existence of a ﬂip set in a class contradicts the determinacy of this class.

Hence, determinacy of a class is not a trivial thing. We have the following usual examples of class determinacy. Closed-determinacy is the hypothesis that all games with closed subsets of the Baire space as winning condition are determined. Borel-determinacy is the hypothesis that all games with Borel subsets of the Baire as winning condition space are determined. More generally, Γ-determinacy is the hypothesis that all A ∈ Γ are determined. In 1953, Gale and Stewart proved Closed-determinacy and Open-determinacy in [GS53].

ω Theorem 4.9. Let U ⊆ ω be an open subset. Then the game G(U) is determined. Proof. A position in a run of a game is not loosing for a player if the other player has no winning strategy from this point. Suppose U is open and I has no winning strategy. We will construct a winning strategy for II. Since I has no winning strategy, the position ∅ is not losing for II. Hence, whatever I plays, say a ∈ ω, there exists at least one element b ∈ ω such that aˆb is not losing for II. Otherwise, there will be a winning strategy for I starting by a. We can iterate this process. We observe that, to make the choice of b at each step, we need the Axiom of countable Choice. Suppose that the game generates ω x ∈ ω . If x ∈ U, since U is open, there exists n ∈ ω such that NxSn ⊆ U. Hence, from the position xSn, I has a trivial winning strategy by playing whatever, a contradiction. Hence, x ∉ U and we constructed a winning strategy for II. The proof of closed determinacy is the same except we suppose that II has no winning strategy. In 1975, Martin proved Borel-determinacy in [Mar75].

ω Theorem 4.10. Let B ∈ B(ω ) be a Borel subset. Then the game G(B) is determined. The proof of this result is very long and consists of reducing the problem to the Closed-determinacy of Gale and Stewart. The reader can ﬁnd it in [Kec95, p.140]. The Axiom of Determinacy (AD) asserts that all games are determined. Although it cannot be true if we assume the Axiom of Choice (by Example 4.6 Example 4.7, and Example 4.8), it is quite natural to consider the possibility that, for all games, one of both players has a winning strategy. Indeed, all the games known to be non-determined make a strong use of the Axiom of Choice in their construction. Hence, no one has be able, yet, to really describe constructively a set which is not determined. Hence, the Axiom of Determinacy and the Axiom of Choice are incompatible. However, the Axiom of Determinacy implies the Axiom of countable Choice on ωω.

Proposition 4.11. Assuming (AD), for all countable families of nonempty subsets of ωω, there exists a choice function.

52 ω Proof. Suppose (Xn)n∈ω is a countable family of nonempty subsets of ω . Consider the game G(⋃k∈ω Ak), where:

Ak = {(sn)n∈ω ∶ s0 = k, (s2n+1)n∈ω ⊆ Xk}.

By (AD), this game is determined, and since I has no winning strategy, otherwise some Xk is empty, II has a winning strategy σ. Hence, the choice function f X nˆ0ω σ X . is given by ( n) = (( ∗ )2n+1)n∈ω ∈ n

4.2 Lipschitz and Wadge Games on ωω

Our next goal is to apply these game theoretical tools to study hierarchies on the Baire Space. For this, we need to construct games that are, in some way, equivalent to the notion of reducibility. That is the next step. We start by an investigation of Lipschitz-reducibility. ω Deﬁnition 4.12. Let A, B ∈ P(ω ). We deﬁne the Lipschitz game GL(A, B) as the following game. GL(A, B) is a two players game with complete perfect information. The players are I and II. Each player plays successively an element of ω, starting from I. In this way, each player produces a sequence in ωω. Let ω ω us say that I produces x = (xn) ∈ ω and II produces y = (yn) ∈ ω . II wins the game if and only if (x ∈ A ↔ y ∈ B). Schematically, we get:

IA IIB

x1 II wins if and only if y1 ((xn) ∈ A ↔ (yn) ∈ B). x2

y2 ⋮ ⋮

(xn) (yn)

ω We can deﬁne a strategy as subtree of ω as in the game G(A). Now, we deﬁne a construction that allows us to turn a strategy for I into a strategy for II.

Deﬁnition 4.13. A strategy σ for I in the game GL(A, B) is turned into a c strategy τ for II in the game GL(B,A ) if we deﬁne τ in the following way: • for all x ∈ τ such that l(x) = 2k for some k ∈ ω, we have xˆy ∈ σ for all y ∈ ω; • for all x ∈ τ such that l(x) = 2k + 1 for some k ∈ ω, there exists a unique y ∈ ω such that xˆy ∈ τ, and this unique y is the one played by I when I follows his strategy σ and xS(2k) is played in the game GL(A, B).

53 In the previous Definition, x is an initial segment of (xn)∗(yn) in a run of the game. Moreover, with this definition, we see that the strategy τ does not take into account the last element played by I, we will say that this strategy is one move delay. In the rest of this work, when we say that I in the game GL(A, B) c becomes II in the game GL(B,A ), we mean that he turned his strategy as I into a strategy for II. Moreover, we observe that, in the previous Definition, if σ c is winning in GL(A, B), then τ is winning in GL(B,A ). Indeed, the sequences produced are the same and the winning condition and the roles of the players has been reversed. We now reach a crucial point of this construction.

Proposition 4.14. Let GL(A, B) be a Lipschitz game. II has a winning strategy if and only if A is Lipschitz-reducible to B. Proof. In this proof, we use directly the Deﬁnition 1.27 of Lipschitz functions. Suppose II has a winning strategy τ. We deﬁne a function:

ω ω f ∶ ω → ω x ↦ f(x) = y, where y is played by II in the run x ∗ τ. This function is well-defined by the ω definition of strategy. We show that it is Lipschitz. Suppose x, x′ ∈ ω be such ′ 1 ′ that d(x, x ) = 2n+1 . This implies that xi = xi for all i < n. Hence, using the strategy τ, the answers of II are the same for the n first moves, i.e. f(x)i = ′ ′ 1 f(x )i for all i < n. This means d(f(x), f(x )) ≤ 2n+1 , hence f is Lipschitz. Moreover, x ∈ A if and only if f(x) ∈ B since τ is winning and f is constructed from τ. Suppose now that we have a Lipschitz function as in the statement of the Proposition. We construct a winning strategy for II in GL(A, B) by induction on the length of the tree of the strategy. For the initial case, we have that both players play the empty sequence. Suppose, for the successor step, that I already ω ω played s ∈ ω< and II answered with t ∈ ω< . If the next move of I is x0, at ω 1 the end of the game, we will have d(x, sˆx0ˆ0 ) ≤ 2n+2 , where n = l(s). Hence, ω 1 since the function f is Lipschitz, we have d(f(x), f(sˆx0ˆ0 )) ≤ 2n+2 . Hence, ω the answer of II is simply f(sˆx0ˆ0 )n+1, which is known since we know the function f. Hence, we have constructed a strategy, say τ, for II in the game 1 GL(A, B). Finally, if I plays x, II plays f(x). Therefore, since f − (B) = A, the winning condition of the game is satisfied. We have seen that Wadge’s study is about continuous reducibility. Hence, our next goal is to construct a continuous equivalent to the Lipschitz game. The game we are going to construct is very similar to the Lipschitz one, except we need to capture the fact that, in a continuous function, the dependence of the famous and δ is not constant, as in the Lipschitz functions.

ω Definition 4.15. Let A, B ∈ P(ω ). We define the Wadge game GW (A, B) as the following game. GW (A, B) is a two players game with complete perfect information. The players are I and II. Each player plays successively starting from I. Although I continues to play elements of ω, II is allowed to skip his turn or to play elements of ω. We just require II to play an infinite sequence, and not a finite one. In this way, each player produces sequences in ωω. Let us

54 ω ω say that I produces x = (xn) ∈ ω and II produces y = (yn) ∈ ω . II wins the game if and only if (x ∈ A ↔ y ∈ B). Schematically, we get:

IA IIB

x1 II wins if and only if ((xn) ∈ A ↔ (yn) ∈ B). x2

y1 ⋮ ⋮

(xn) (yn)

As usual, we can deﬁne strategies, but with this new extra move for II, we need to be careful. We use the letter s to simulate the skip move of II.

ω Deﬁnition 4.16. A strategy for I is a subtree σ of {ω ∪ {s}} such that: • for all x ∈ σ such that l(x) = 2k for some k ∈ ω, there exists a unique y ∈ ω such that xˆy ∈ σ, • for all x ∈ σ such that l(x) = 2k + 1 for some k ∈ ω, we have xˆy ∈ σ for all y ∈ {ω ∪ {s}}. Of course, adapting the indices and the alphabets, we can deﬁne a strategy for II. We now prove the link between Wadge reducibility and games.

Proposition 4.17. Let GW (A, B) be a Wadge game. II has a winning strategy if and only if A is Wadge reducible to B. Proof. In this proof, we use the Deﬁnition 1.26 of a continuous function for metric spaces. Suppose II has a winning strategy τ. We deﬁne a function:

ω ω f ∶ ω → ω x ↦ f(x) = y, where y is played by II in the run x ∗ τ. This function is well-deﬁned by the ω deﬁnition of strategy. We show that it is continuous. Let Ns ⊆ ω , we have 1 ω 1 to show that f − (Ns) is open for any s ∈ ω< . We can suppose t ∈ f − (Ns), otherwise we are done. There exists n ∈ ω such that s is answered by II when 1 tSn is played by I. Indeed, t ∈ f − (Ns), implies f(t) ∈ Ns, hence the answer of −1 II to t passes through s. Hence, we get f continuous since t ∈ NtSn ⊆ f (Ns). Moreover, x ∈ A if and only if f(x) ∈ B since τ is winning and f is constructed from τ.

55 Now, suppose we have a continuous function f as in the statement of the Proposition. We construct a winning strategy for II in the game GW (A, B) by recursion on the length of the tree of the strategy. We just compute the first step, the other ones being equivalent, by just changing the starting conditions. ω ω 1 Suppose I plays x0. Consider x0ˆ0 ∈ ω . Let = 2 . If δ > 0 in the definition of 1 ω ω continuity is δ ≤ 2 , set y0 = f(x0ˆ0 )0. Otherwise, consider x0ˆx1ˆ0 and look for δ, and so on and so forth until we find y0. If this process never stops, it 1 1 means that, for = 2 , there exists no δ > 0 such that f(B(x, δ)) ⊆ B(f(x), 2 ), where x is played by I. Hence, f is not continuous at x, a contradiction. Once we constructed y0, we restart with what both players already played, and with ′ = 2 . Therefore, II produces an infinite sequence too. Finally, if I plays x, II plays f(x), hence the winning condition is guaranteed. We can generalize all the games described above by playing not in the Baire space, but in some arbitrary zero-dimensional Polish space. Using the fact that every zero-dimensional Polish space is homeomorphic to a closed subet of the Baire space, and using the fact that the closed subsets of the Baire space are exactly the pruned trees on ω. We have the next Definition, which is a generalization of Gale-Stewart’s games.

ω Deﬁnition 4.18. Let A ⊆ [T ], where T ⊆ ω is a pruned tree. We deﬁne the game G(A, T ) as the following game. G(A, T ) is a two players game with complete perfect information. The players are I and II. Each player plays successively an element of ω, starting from I, and with the restriction that they have to stay in T . I wins the game if the resulting sequence is in A. Schematically, we draw: I II

x1 I wins if and only if y1 (x0, y0, x1, y1,... ) ∈ A. x2

y2 ⋮ ⋮

Since we start with a pruned tree, it is always possible for a player to play some element, hence the restriction condition does not affect the development of the game. We say that the tree T is the tree of legal positions of the game. The notion of strategy, and that of winning strategy, are defined as above, taking care of respecting the legal positions tree. In a similar way, we can define the Lipschitz game and the Wadge game with a legal position tree. Here, we have to make some important remarks. We can, now, justify our deep study of the Baire space as a general theory for all zero-dimensional Polish spaces. Indeed, the game we just defined (Definition 4.18) allows us to generalize all the future results on the different hierarchies. Indeed, pruned trees on ω and ωω share all the properties we will use about these games. Hence, constructing the hierarchies via games on ωω implies constructing the hierarchies on every

56 zero-dimensional Polish space. A second remark which is as important as the preceding one and that also justiﬁes our study is the following Proposition.

Proposition 4.19. Let X be a zero-dimensional Polish space, and A ⊆ X. ω Then, there exists B ⊆ ω such that A ≡W B. Proof. Since X is zero-dimensional, it is homeomorphic to a pruned tree [T ] by Theorem 2.54. Hence, by Proposition 1.39, there exists a continuous surjection ω 1 f ∶ ω → [T ], where f is the identity on [T ]. Just set B = f − (A). Hence B ≤W A. ω Moreover, by the identity function id ∶ [T ] → ω which is continuous, we have 1 id− (B) = B ∩ A = A, hence A ≤W B. By this result, if we want to understand the Wadge hierarchy on a zero- dimensional Polish space, it suﬃces to study the Wadge hierarchy on ωω. This is what we are going to do for the rest of this Chapter. The following fact is rarely mentioned in books because it is considered trivial. We prefer, to be complete, to mention it and to prove it. To do so, we need the following notation. Notation 4.20. Let A and B be sets. We deﬁne:

∗ ω A ∩ B = {x = (xn) ∈ ω ∶ (x2n) ∈ A and (x2n+1) ∈ B}.

ω ω Proposition 4.21. If A, B ∈ B(ω ), then A ∩∗ B ∈ B(ω ). Proof. We deﬁne

ω ω ω f ∶ ω → ω × ω

(xn) ↦ ((x2n), (x2n+1))

1 We have f − (A × B) = (A ∩∗ B), f continuous, since it is the inverse of the ∗ function, which is a homeomorphism, as seen in the proof of the Proposition 2.12. Hence, since the product of Borel sets are exactly the Borel sets of the product, and since Borel sets are closed under preimage, we get our result. The preceding proposition allows us to make use of Borel determinacy in our Lipschitz game framework. Indeed, we have also: ω Proposition 4.22. Let A, B ∈ B(ω ). Then GL(A, B) is determined.

Proof. It suﬃces to observe that the game GL(A, B) is equivalent to the game c c G((A ∩∗ B) ∪ (A ∩∗ B )), which is determined by Borel determinacy and the previous Proposition. We have the same results for Wadge games, but we need one more Lemma. ω Lemma 4.23. If A ∈ B(ω ), then s ω s A = {x ∈ {ω ∪ {s}} ∶ x∖ ∈ A},

s ω where x∖ is obtained from x by deleting every occurrence of s, is also in B(ω ).

57 ω Proof. Let D be the subspace of (ω ∪ {s}) whose sequences are still inﬁnite when we omit the letter s. We use the function: ω f ∶ D → ω , s x ↦ x∖ . Since we have:

−1 n0 n1 nl(t)−1 f (Nt) = ⋅ ⋅ ⋅ s t0s t1 . . . s tl(t)−1, n0<ω nl(t)−1<ω 1 s s f is continuous. Moreover, it is clear that f − (A) = A , so A is Borel. Hence, we have the result of determinacy for Wadge games. ω Proposition 4.24. Let A, B ∈ B(ω ). Then GW (A, B) is determined. Proof. It suﬃces to observe that the game GW (A, B) is equivalent to the game s c s c G((A ∩∗ B ) ∪ (A ∩∗ (B ) )), which is determined by Borel determinacy and the previous Proposition. With this result, we can use this very strong game theoretical tool (Borel determinacy) in our framework of Wadge games to understand the Wadge hierarchy. The big argument that Borel-determinacy gives is that, if a player has no winning strategy, the other has a winning strategy. Hence, if A

4.3 Lipschitz and Wadge Hierarchy on ωω

We precise that, from now on, we will work exclusively in the Borel subsets of ωω. However, if we supposed Γ-determinacy, all the results we have about the Borel subsets would certainly be true for the sets in Γ. Hence, one thing we can always keep in mind is that our restriction on Borel sets is just for clarity, but working in a stronger theory will certainly implies our results at higher levels. Now, we can prove one of the most important result in Wadge’s work, namely Wadge’s Lemma. We just precise that, by the definition of the Lipschitz game and of the Wadge game, a strategy in the Lipschitz game gives rise to a strategy in the Wadge game. Moreover, this strategy is winning if the original one (in the Lipschitz game) is winning. ω Lemma 4.25. Let A, B ∈ B(ω ). c If A ≰W B, then B ≤W A . Proof. Since A ≰W B, we have that II has no winning strategy in GW (A, B), hence no winning strategy in GL(A, B) either. By Borel-determinacy, this tells us that I has a winning strategy, say σ, in GL(A, B). We now turn this strategy for I into a strategy τ for II using Definition 4.13. It suffices to prove that this c strategy is a winning strategy in GL(B,A ) to conclude. c ω Suppose that, in GL(B,A ), I plays a sequence x ∈ ω . Because of the one move ω delay of τ, II plays exactly the same y ∈ ω that he had played, as I, in the ω game GL(A, B) (answering x ∈ ω too). Hence, since I has a winning strategy in GL(A, B), we get (x ∈ A ↔ y ∉ B), which means II has a winning strategy c c in GL(B,A ). Therefore, II has a winning strategy in GW (B,A ).

58 Remark 4.26. As we can see in the proof of Wadge’s Lemma, we only make use of Lipschitz games. Hence, Wadge’s Lemma is also true for ≤L . With Wadge’s Lemma, we can already get some conclusions about the shape of the quasi-orders induced by ≤W and ≤L. Proposition 4.27. There do not exist three incomparable Borel subsets of ωω, in both ≤L and ≤W .

Proof. Suppose A ⊥W B and B ⊥W C. Hence, A ≰W B and B ≰W C. By c c c Wadge’s Lemma, B ≤W A and C ≤W B. Hence, by transitivity, we get C ≤W c A , which is of course equivalent to C ≤W A. The proof is the same for ≤L making use of Wadge’s Lemma for this quasi-order. This result implies that the maximal antichains in both these two quasi- ω ω orders are of size 2. Moreover, since in both cases, ∅ ⊥L ω and ∅ ⊥W ω , the maximal antichains are of size exactly 2. Moreover, the next result shows us what is the relation between the elements of an antichain of size 2.

ω Proposition 4.28. Let A, B ∈ B(ω ). If A is incomparable with B, then A ≡W Bc.

Proof. Since A ⊥W B, we have both A ≰W B and B ≰W A. Wadge’s Lemma c c c gives B ≤W A and A ≤W B . Hence, A ≡W B . Again, since this result is just a corollary of Wadge’s Lemma, it is also true in the Lipschitz hierarchy. Finally, we have a last Proposition which follows from the Wadge’s Lemma.

ω c Proposition 4.29. Let A, B ∈ B(ω ). Suppose A

Of course, this is also true in the Lipschitz hierarchy using Wadge’s Lemma in this framework. With these results, we can deﬁne two kind of sets.

ω Deﬁnition 4.30. A subset A ∈ P(ω ) is a self-dual set if it is Wadge- equivalent to its complement. If this is not the case, the set is a non-self-dual set.

A degree in the Wadge hierarchy is self-dual if all its elements are self-dual. Of course, if it is not the case, the degree is non-self-dual. These notions are well-deﬁned since the degrees are just equivalence classes. Hence, in the Wadge hierarchy, we have two kinds of levels. By Proposition 4.28, the ﬁrst kind, occupied by the self-dual degrees, are comparable with any other set. The second kind, occupied by non-self-dual degrees, are comparable

59 to any other set, except the ones that are equivalent to their complement, again by Proposition 4.28. If we identify the non-self-dual degrees with their complement, we obtain a linear ordering on the Wadge degrees. Hence, we say that the Wadge degrees are semi-linear ordered. The next natural question when studying a linear ordering is the following: is it well-founded? Martin gave the answer to this question in the following beautiful result.

ω Theorem 4.31. On the Borel subsets of ω , the quasi-orders given by ≤W and ≤L are well-founded semi-linear orderings. Proof. Since the Lipschitz hierarchy is a refinement of the Wadge hierarchy, it suffices to write the proof for the Lipschitz hierarchy. Indeed, an infinite descending chain in the Wadge hierarchy is also infinite descending in the Lipschitz hierarchy. Taking the converse, we have what we want. Suppose, towards a contradiction, that there exists a strictly infinite descending chain: ⋯

By all the previous results, we have that I has a winning strategy in GL(Ai,Ai+1), σ0, G A ,Ac , σ1 say i and also in the game L( i i+1) say i . We construct a Martin-Monk ω diagram, which is the following picture, for ∈ 2 .

0 1 2 3 σ0 σ1 σ2 σ3 I II I II I II I II ⋯

0 1 2 3 x0 x0 x0 x0 1 2 3 4 ⋯ x0 x0 x0 x0 0 1 2 3 x1 x1 x1 x1 1 2 3 4 ⋯ x1 x1 x1 x1 0 1 2 3 x2 x2 x2 x2 1 2 3 4 ⋯ x2 x2 x2 x2 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮

Figure 4.1: A Martin-Monk diagram

0 c 1 In this diagram, A = A and A = A. Moreover, the games are played with I (i) playing according to the corresponding winning strategy σi . Deﬁne

ω 0 X = { ∈ 2 ∶ (xn)n∈ω() ∈ A0},

0 where (xn)n∈ω() is the element played by I in the first game of the Martin- ω Monk diagram corresponding to ∈ 2 . ω First, we prove that X is a flip set. Indeed, suppose , ′ ∈ 2 differ in exactly one digit, say the number k ∈ ω. Then, the strategy used in the game number k changes, but not the others ones. Moreover, using the fact that I wins all l l games, we have x () = x (′) for all l > k. Indeed, all these sequences do not change if the Martin-Monk diagram is made according to or ′. Then, we have

60 k k x () ∈ Ak ↔ x (′) ∉ Ak, since the strategy is chosen in this fashion. For the previous ones, the same strategies are used. Hence, for l < k, we have: ′ ′ xl Al xl−1 Al−1 xl ′ A l xl−1 ′ A l−1 . ( ) ∈ l → ( ) ∈ l−1 ↔ ( ) ∈ l → ( ) ∈ l−1 0 0 Therefore, by recursion, x () ∈ A0 ↔ x (′) ∉ A0. This proves that X is a flip set. Now, we prove that the function: ω ω f ∶ 2 → ω 0 ↦ (xn)(). ω is continuous. Let Ns be open in ω . We can suppose that there exists ∈ −1 −1 f (Ns), otherwise, we are done. Suppose l(s) = n. We prove NSn ⊆ f (Ns). ′ Let ∈ NSn. The n first digits are the same, hence the upper triangle of width n in the Martin-Monk diagram corresponding to ′ is the same as the one cor- 0 0 ′ responding to . Hence, the n first digits of (xn)() and (xn)( ) are the same. 1 Hence, ′ ∈ f − (Ns). Therefore, f is continuous. Finally, we get our contradiction. Indeed, being that X is a flip set, X does not have the BP by Proposition 1.49. Moreover, Borel sets are closed under 1 continuous preimage, A0 is Borel and f − (A0) = X. Therefore, X is Borel and has the BP by Proposition 3.8, a contradiction. Hence, we can associate to each Wadge degree an ordinal, which gives its height in the Wadge hierarchy, hence its complexity. ω Definition 4.32. We define, for X ∈ B(ω ) ∶ ω 1. if X ∈ {∅, ω }, then dW (X) = 1; ω 2. if X ∉ {∅, ω }, then dW (X) = sup{dW (Y ) + 1 ∶ Y

• nÂi = {nˆs ∶ s ∈ Ai}, for each n ∈ ω; • Ai ⊕ Aj = ⋃n∈ω ((2n)Âi ∪ (2n + 1)Âj) ; • >i∈ω Ai = ⋃i∈ω iÂi. c c c We have easily that (A ⊕ B) = A ⊕ B . We saw that the first level of the hierarchy is occupied by two incomparable ω sets, namely ∅ and ω , by considering the constant functions, which are continuous. The question is, what is the next level? We first show a more general result, which we can prove easily with our game characterization of Lipschitz reducibility. But, before going into these proofs, we make a trivial observation. c c The The game GL(A, B) is exactly the same as the game GL(A ,B ) as one can see by looking at the winning condition. The same observation is also true with the Wadge game. We will also need the following Definition.

61 ω Deﬁnition 4.34. Let A ⊆ ω . For all n ∈ ω, we deﬁne:

ω A[n] = {s ∈ ω ∶ nˆs ∈ A}.

We can view A[n] in the following way. If a player is in charge of the set A at the beginning of a game, he is in charge of A[n] after he played n ∈ ω.

Proposition 4.35. If [A]L is a non-self-dual degree, then the level just above c is occupied by a self-dual degree, which is [A ⊕ A ]L. Proof. We have three things to show:

c B = A ⊕ A is self-dual: c We construct a winning strategy for II in GL(B,B ). For the ﬁrst move, if I plays k ∈ ω, then II plays k + 1. Hence, the game is just GL(A, A). So II just copies what I plays to win.

c c A, A

c There is no B such that A

We know what the level just above a non-self-dual one is. The next proposition describes the level just above a self-dual one. Again, the game characterization allows us to ﬁnd a nice proof.

Proposition 4.36. If [A]L is a self-dual degree, then the level just above is occupied by a self-dual degree, which is [0ˆA]L.

62 Before the proof of this proposition, we observe that a self-dual degree never contains the empty sets, nor the whole space. Proof. Again, we have three things to show: 0Â is self-dual: c We construct a winning strategy for II in the game GL(0Â, (0Â) ). We c ω c ∗ observe that (0Â) = ⋃n∈ω nˆω ∪ 0Â . c If I plays 0, then II plays 0 and the game becomes GL(A, A ), for which II has a winning strategy since A is self-dual. If I plays something else than 0, II just plays 0 and then something that is in A, which is possible by the observation made just before the proof.

There is no B such that A

Hence, we have the picture of what happens at all successor levels, but what about the limit ones? Suppose we have a limit ordinal λ. Since, after each degree, self-dual or not, there is a self-dual degree, we always have an inﬁnite number of self-dual degrees before a limit ordinal. Proposition 4.37. Let λ be a limit ordinal such that cof(λ) = ω, then, the level λ of the Lipschitz hierarchy is occupied by a self-dual degree, which is of the form [>i∈ω Ai]L.

Proof. Since cof(λ) = ω, there exists a strictly increasing sequence (αi)i∈ω whose limit is λ. For each αi, take a set Ai in the level αi of the Lipschitz hierarchy. Set A = >i∈ω Ai. We observe that, with our construction, we get

A0 i∈ω Ai . If I plays k ∈ ω, then II plays k + 1 ∈ ω. Then, we are in the game GL(Ak,Ak+1) for which II has a winning strategy.

63 Ai

We can also prove the converse of the last Proposition, for this, we need the following Lemma. Lemma 4.38. Let A be a self-dual subset of the Baire space. Then, for all n ∈ ω, A[n]

Proof. We start by the easy part. Of course, in the game GL(A[n],A), II has a winning strategy by playing n ∈ ω and then copying what I plays with one move delay. Moreover, nÂ[n] ≤L A. Indeed, in the game GL(nÂ[n],A), a winning strategy for II is to copy what I plays if I’s first move is 0, and to play something out of A otherwise, which is possible since A is self-dual. Therefore, we have A[n]

We are ready to prove the converse of the previous Proposition. Proposition 4.39. Let λ be a limit ordinal such that cof(λ) > ω, then, the level λ of the Lipschitz hierarchy is occupied by a non-self-dual degree. Proof. We show the contraposite of this proposition. Hence, suppose that a self-dual set A occupies the level λ. We will construct a countable sequence of sets from A[n] which is unbounded below A. We define first B0 = A[0]. Then we define: ⎧ ⎪A[n+1] if Bn n∈ω A[n] = A, by definition. Moreover >n∈ω A[n] ≤L >n∈ω Bn. Indeed, a winning strategy for II comes from copying the first move of I and then use the winning strategy in the game GL(A[n],Bn). Therefore, we get

A ≤L ? A[n] ≤L ? Bn ≤L A. n∈ω n∈ω

Hence, >n∈ω Bn witnesses the countable coﬁnality of λ.

64 Hence, by the two previous proposition, what was missing to describe the Lipschitz hierarchy on the Borel subsets of the Baire space is proved.

Theorem 4.40. Let λ be a limit ordinal. The level λ of the Lipschitz hierarchy is occupied by a self-dual degree if and only if cof(λ) = ω. Hence, we can draw the Lipschitz hierarchy on the Borel sets of the Baire space.

[∅]L

⋯ ⋯ ⋯

ω [ω ]L cof(λ) = ω cof(λ) > ω

Figure 4.2: Lipschitz hierarchy on the Borel sets of the Baire space

We described the Lipschitz hierarchy on the Baire space. The next goal is to describe the Wadge hierarchy. First, we observe that, since a Lipschitz function is always a continuous function, the Lipschitz hierarchy is a reﬁnement of the Wadge hierarchy. Indeed, suppose two subsets have the same Lipschitz degree, they also have the same Wadge degree, with the same functions as witnesses. Hence, there are no « new » levels in the Wadge hierarchy. On the other hand, since there exist continuous functions that are not Lipschitz, it is possible that some group of degrees of the Lipschitz hierarchy becomes a single degree. That is the study we now carry out. By all we already know about the Lipschitz hierarchy, some results will be easy to compute. Indeed, we have the following Proposition. The proofs are not exactly the same since II has the extra move of skipping his turn in the game.

Proposition 4.41. If [A]W is a non-self-dual degree, then the level just above c is occupied by a self-dual degree, which is [A ⊕ A ]W . Proof. We proceed as previously, by checking three things.

c B = A ⊕ A is self-dual: c A winning strategy for II in GW (B,B ) consists of playing k + 1 for the ﬁrst move when I played k. Then, it suﬃces to copy what I does.

c c A ,A

c There is no B such that A

65 Suppose this strategy requires to play an even number ﬁrst, then, it means that II has no winning strategy in GW (A, B), otherwise, skipping against σ leads to a winning strategy, which is impossible. Hence, I has a winning strategy in GW (A, B), which implies B

However, the next result show that the Lipschitz and the Wadge hierarchies are two diﬀerent things.

Proposition 4.42. If [A]W is a self-dual degree, then [A]W ≡W [0ˆA]W .

Proof. We construct winning strategies for II in the games GW (A, 0Â) and GW (0Â, A). For the first one, II just has to play 0 and then copies I with one move delay. For the second game, if I starts by playing something else than 0, II just constructs something which is not in A, which is possible since A is self-dual, hence not ωω. If I starts by playing 0, then II just skips on the first move, and then copies I. Hence, the ω levels of the Lipschitz hierarchy after a self-dual degree are contained in the same Wadge degree. To have a stronger result, we need the following Lemma.

Lemma 4.43. Suppose there exists a sequence (An)n∈ω such that An ≤W A for all n. Then >n∈ω An ≤W A.

Proof. We construct a simple winning strategy for II in the game GW (>n∈ω An,A). If I plays k ∈ ω, the ﬁrst move of II is just to skip, and then use the winning strategy of II in the game GW (Ak,A). Hence, we have the following stronger result.

Proposition 4.44. For α < ω1, every sequence of α consecutive self-dual degrees in the Lipschitz hierarchy is contained in a single Wadge degree.

Proof. Suppose we have such a sequence of Lipschitz degrees ([Aα]L)α<ω1 . We proceed by induction on α. If α = 1, the sequence is just one element, hence, since the Lipschitz hierarchy is a refinement of the Wadge hierarchy, the sequence is contained in exactly one Wadge degree. Now, suppose we already have [Aβ]W = [A]W for all β < α. Since α < ω1, we have either cof(α) = 1, or cof(α) = ω. If cof(α) = 1, α is a successor ordinal, hence, since the sequence we consider is built from consecutive degrees, the Lipschitz degree is of the form [Aα]L = [0Âα−1]L. By Proposition 4.42, we get [Aα]W = [A]W . If cof(α) = ω, Proposition 4.37 implies that Aα ≡L >n∈ω An, where An n∈ω An, and A ≡W An ≤W Aα for all n ∈ ω. By the previous Lemma, we get Aα ≡W >n∈ω An ≤W A. Therefore, Aα ≡W A, and we are done. So we know what happens to a consecutive sequence of self-dual degrees. Let us start the analysis of the limit levels of the Wadge hierarchy. For countable cofinality, the result is the same as the Lipschitz one.

66 Proposition 4.45. Let λ be a limit ordinal such that cof(λ) = ω, then the level λ of the Wadge hierarchy is occupied by a self-dual degree, which is of the form [>n∈ω An]W .

Proof. Suppose we have a strictly increasing sequence of Wadge degrees ([An]W )n∈ω, such that An is in the level αn of the hierarchy, where (αn)n∈ω witnesses the countable coﬁnality of λ. We have, as always, three things to show.

A = >n∈ω nˆAn is self-dual: c The winning strategy for II in the game GW (A, A ) is just to play k + 1 to answer k in the ﬁrst move, and then use the winning strategy for II in GW (Ak,Ak+1).

For all n ∈ ω, there is no B such that An

However, for uncountable coﬁnality, our proof used the fact that, for a self- dual degree, we have A[n]

The last goal of this Section is to compute the length of the Lipschitz and the Wadge hierarchy. We use the proof of [And07]. We start by deﬁning a particular ordinal: ω Θ = sup{α ∶ ∃f ∶ ω ↠ α}.

67 [∅]W

⋯ ⋯ ⋯

ω [ω ]W cof(λ) = ω cof(λ) > ω

Figure 4.3: Wadge hierarchy on the Borel sets of the Baire space

Hence, Θ is the supremum of all the ordinals that are surjective images of the Baire space ωω. Now, we need two lemmas in the proof of the next theorem. Lemma 4.47. There exist exactly 2ℵ0 continuous functions from ωω to ωω. ω ω Proof. We deﬁne A ⊆ {φ ∶ ω< → ω< } by:

φ ∈ A ↔(s ⊆ t → φ(s) ⊆ φ(t)) lim l(φ(x)Sn) = ∞. n→∞ We show that there exists a bijection

ω ω ω ω {f ∶ ω → ω continuous} ↔ {φ ∶ ω< → ω< , φ ∈ A} f ↦ φf fφ ← φ [ We deﬁne

ω ω fφ ∶ ω → ω x ↦ φ(xSn). n∈ω This function is well-deﬁned since φ ∈ A. We show that it is continuous. Let −1 x ∈ fφ (Ns), therefore there exists m ∈ ω such that φ(xSm) ∈ Ns. Therefore, −1 x ∈ NxSm ⊆ fφ (Ns).

ω For the second direction, we start by an observation. Suppose that f ∶ ω → ω ω ω is continuous. Let x ∈ ω , for all m ∈ ω, there exists m′ ∈ ω such that 1 ′ − x ∈ NxSm ⊆ f (Nf(x)Sm), since f is continuous. We need this observation in order to construct an increasing function φf from f. We deﬁne

<ω <ω φf ∶ ω → ω ω s ↦ f(sˆ0 )Sms, where ms = sup{m ∈ ω ∶ m′ ≤ l(s)}. With this deﬁnition, we have, as we want, 1 N ′ f − N ω . sSm ⊆ ( f(sˆ0 )Sms ) Therefore,

ω s t t N ′ f t N ω φ t φ sˆ0 m φ s , ⊆ → ∈ sSm → ( ) ∈ f(sˆ0 )Sms → ( ) ⊆ ( )S s = ( )

68 where we use that m′ ≤ l(s) ≤ l(t) for the last implication. Therefore, φf ∈ A. Moreover, it is easy to see that both functions we have just deﬁned are injective. Therefore, by the Cantor-Schröder-Bernstein Theorem (Theorem 1.5), ω ω ω ω there exists a bijection {f ∶ ω → ω continuous} ↔ {φ ∶ ω< → ω< , φ ∈ A}. Now, we will compute the cardinality of these sets. First, every constant function ω ω from ω to itself is continuous, hence the cardinality is at least Sω S = 2ℵ0 . Moreover, since ω<ω is countable, we have:

ω ω ω ω ω S{φ ∶ ω< → ω< , φ ∈ A}S ≤ S{φ ∶ ω< → ω< }S ≤ Sω S = 2ℵ0 . Hence, the proof is complete. By the previous Proposition, we can label the continuous functions from ωω to ωω with some element of ωω. Hence, the set of all continuous functions can be seen as ω {fx ∶ x ∈ ω }. Lemma 4.48. Let

ω ω J ∶ P(ω ) → P(ω ) A ↦ {0ˆx ∶ fx(0ˆx) ∉ A} ∪ {1ˆx ∶ fx(1ˆx) ∈ A}. ω c For all A ⊆ ω , we have A, A

Proof. Suppose, towards a contradiction, that J(A) ≤W A. Hence, there exists x ωω f −1 A J A . some 0 ∈ such that x0 ( ) = ( ) This is a contradiction since

fx0 (0ˆx0) ∉ A ↔ 0ˆx0 ∈ J(A) ↔ fx0 (0ˆx0) ∈ A.

By the same argument on 1ˆx0, we get a contradiction if we suppose J(A) ≤W Ac. Therefore, we can compute the length of the Wadge hierarchy with a beautiful proof. Theorem 4.49. The length of the Wadge hierarchy is Θ. ω Proof. Let A ⊆ ω . Suppose A is in the level α of the Wadge hierarchy. We ω write SSASSW = α for this. Let β ≤ α. Take B ⊆ ω such that SSBSSW = β. Then, B A, x ωω f −1 A B. ≤W hence there exists 0 ∈ such that x0 ( ) = We deﬁne

ω f ∶ ω → α + 1 −1 x ↦ SSfx (A)SSW . By the preceding observations on an arbitrary B, we obtain that f is surjective. Therefore, ω sup{SSASSW ∶ A ⊆ ω } ≤ Θ. It remains to show the other inequality. Let α < Θ. We will construct a strictly increasing sequence of Wadge degrees which will witness that α ≤ ω ω sup{SSASSW ∶ A ⊆ ω }. Therefore, we will obtain Θ ≤ sup{SSASSW ∶ A ⊆ ω }. ω Fix some function f ∶ ω ↠ α. This function exists since α < Θ. For all β < α, deﬁne ω Aβ = J({x ∗ y ∈ ω ∶ f(x) < β ∧ y ∈ Af(x)}),

69 where J is the function from Lemma 4.48. ω If β < γ < α, take x0 ∈ ω such that f(x0) = β. Consider the function ω ω g ∶ ω → ω y ↦ x0 ∗ y.

It is continuous, by a similar argument to one that we already carried out in ω this work. Moreover, it witnesses the fact that Aβ ≤W {x∗y ∈ ω ∶ f(x) < γ ∧y ∈ −1 ω Af(x)}. Indeed, we need to show that g ({x ∗ y ∈ ω ∶ f(x) < γ ∧ y ∈ Af(x)}) = Aβ. We have

−1 ω x ∈ g ({x ∗ y ∈ ω ∶ f(x) < γ ∧ y ∈ Af(x)}) → g(x) = x0 ∗ x, f(x0) < γ ∧ x ∈ Af(x)

→ x ∈ Af(x) = Aβ. −1 ω x ∉ g ({x ∗ y ∈ ω ∶ f(x) < γ ∧ y ∈ Af(x)}) → g(x) = x0 ∗ x, f(x0) ≥ γ or x ∉ Af(x)}

→ x ∉ Af(x) = Aβ.

Where the last line comes from β < γ. By Lemma 4.48, we get

ω Aβ ≤W {x ∗ y ∈ ω ∶ f(x) < γ ∧ y ∈ Af(x)} ω

4.4 Construction of the Lower Levels of the Wadge Hierarchy of ωω

In the previous sections, some proofs used constructions to describe some levels of the Wadge hierarchy on the Baire space ωω. In this Section, we play the games to describe the ﬁrst levels of the hierarchy and try to give some intuition on these levels. We precise that we want to describe the Wadge hierarchy viewed as a quasi-order, and not the Pointclass Wadge hierarchy. Therefore, we want to give a representative in each of the lower levels. This Section will certainly be less formal than the previous ones, but it has the advantage of being more intuitive. We think that this intuition is a really important part of this work for two reasons. The ﬁrst one is that this motivates the construction we made, and the second one is that this intuition may help to generalize the Wadge hierarchy to other spaces in the next chapters. Hence, we dedicated a whole Section to this intuitive work.

70 First, it is trivial, using the constant functions, which are continuous, that the first level of the Wadge hierarchy is occupied by two non-self-dual degrees, ω namely [∅]W and [ω ]W . To win a Wadge game, if I am I and I am in charge of a set A and my opponent is in charge of a set B, my goal is to end the game if in A if and only if II does not end the game in B. Therefore, I need to force him playing something in B and finally I play something not in A or the contrary. Then, my goal is to make my opponent believe that I will end in A although I will in fact end in Ac. That is the idea we will make use of. Of course, all these notions will become clearer once we really play the games. In a Wadge game, if I am in charge of ∅, I can never, during the game, make my opponent believe that I will enter in the set I am in charge of. Therefore, we can represent this set as . To the contrary, if I am in charge of ωω, it is sure that I will be in the set at the end of the game, hence we represent this by the symbol . Now, we have seen that the level following a non-self-dual level occupied by c c the degrees of some sets A and A is simply A ⊕ A . Therefore, a representative ω of the second level is ⋃n∈ω(2n)ˆω . This means that the first element we play chooses one of the two preceding Wadge degrees. Therefore, we can represent this as

A set represented by this symbol is a clopen set. Indeed, it is open since, once we are in, we can not go out, and this is exactly how we deﬁned an open set. Moreover, it is closed since, if we choose not to be in, we can not reenter in it before the end of the game, as the pruned tree representation of a closed set suggests. Now, since clopen sets are just the intersection of the class of open sets and the class of closed sets, we hope that the next level is occupied by these classes. This is obvious by the preceding observation and the fact that open sets are preserved by continuous preimage. Therefore, the next level is occupied by the proper open and closed sets, i.e. the ones that are not clopen. To get a better ω idea, we give an example of a proper closed set, namely {0 }. This is clearly not open since every open neighborhood contains more than one point, and this is closed since the complement is

N0nˆa. ∗ n∈ω a∈ω Therefore, if I am in charge of this set, I can make my opponent believe that I will stay in my set by playing 0’s, and then just change my mind and go out of the set by playing something else, hence we can represent this set as

We observe that this correspond exactly to the idea of a closed set seen as a pruned tree. Namely, we can always stay in, and once we go out, we can not change our decision. Of course, we can make the same development with the open sets, which leads to

71 Now, as an example, let us simulate the games which allow us to construct the ﬁrst levels of the hierarchy we already described. For this, we will describe strategy using colors representing diﬀerent runs of the game. We think that the pictures are self-explanatory, hence, we do not give more details on this. Let us show that

 

II wins I wins

 Let us show that is self-dual. For this, we observe that the complement is I II

 , , II wins

Then, we show that

 , , II wins I wins

Finally, we show that is non-self-dual.

72 I II

 I wins

The next question is, what are the next levels? The intuition gives a natural answer which is the correct one. Indeed, after a non-self-dual level ... n times we have the self-dual level ... ... n times

And, after such a self-dual level, we have a non-self-dual level which is simply ... n + 1 times

Therefore, we can easily draw the ω ﬁrst levels of the hierarchy recursively starting from and . The next question which comes in mind is what kind of sets are ... If we think topologically, this is a set where we can go in and out a ﬁnite number of time, hence, it is like

Therefore, the intuition here is that the ω first levels of the Difference hierarchy correspond exactly to the ω first non-self-dual levels of the Wadge hierarchy. Of course, to be precise, we should exhibit a more detailed proof, but as we said earlier, this Section is devoted to the intuition about these games. Nevertheless, we give a concrete example of such a set. Consider

(N0n ) ∖ (N0nˆ1m ) ∪ N0nˆ1mˆ2l . ∗ ∗ ∗ n∈ω n,m∈ω n,m,l∈ω

0 ω This is clearly a D3(Σ1(ω )). Moreover, we can think of this set as having the following natural shape.

73 Clearly, descending along the tree, the best strategy a Player in charge of such a set has is of the form

Hence, the ﬁrst ω levels are now known, so we are interested in the next one. By the previous Section, we have seen that such a level is occupied by a set of the form >i∈ω Ai where (Ai)i∈ω is strictly increasing in the Wadge hierarchy, with limit level corresponding to the level we try to reach. Hence, with what we did earlier, we have the following kind of set in such a level

 

This kind of set is trivially self-dual by playing the game. Indeed, player II can always make his first choice knowing what I made as a choice, hence he can easily choose a more powerful set to be in charge of. ... We saw that the first non-self-dual sets are easily represented by We can generalize this by saying that the level α < ω1 of the Difference hierarchy 0 Dα(Σ1) is occupied by α

which is used as follows. A player in charge of such a set has two moves. Either he stays in the side he is, or he can move to the other side and strictly decrease the ordinal. This second move can be used as long as the ordinal is greater than 0. It is easy to see that the greater the ordinal is, the greater the corresponding set’s power is. Moreover, ... n times is equivalent to

74 n

 

Therefore, we generalize our construction of ω non-self-dual degrees to ω1 non-self-dual degrees. 0 ω Finally, we reach the level ω1, which corresponds to the level Σ2(ω ) of the Borel hierarchy. We construct a set which is at this level, therefore we construct 0 ω ∞ ω a Σ2(ω )-complete set. Consider the set ∃ 0 = {x ∈ ω ∶ Si ∶ x(i) = 0S ≥ ω} . This 0 ω is in Π2(ω ) since we can write

n ∃∞0 = ∃ 0, n∈ω n ω where ∃ 0 = {x ∈ ω ∶ Si ∶ x(i) = 0S ≥ n} is clearly open. Hence, the set ∃<∞0 = ∞ c 0 ω (∃ 0) is Σ2(ω ). Let us show that it is complete. 0 ω Consider A ∈ Σ2(ω ). Since a closed set corresponds to a pruned tree, we can write A = ⋃n∈ω Tn, where Tn is a pruned tree for all n ∈ ω. We describe a winning strategy for II in the game GW (A, ∃<∞0). We consider recursively the trees Ti. Starting from T0, at each move, I has two moves. In the ﬁrst case, I stays in the tree T0, then II plays a 1. In the second case, I goes out of T0, II plays 0. Once I goes out of T0, we consider T1. There are two possibilities. The ﬁrst one is that I is already out of T1, in this case, II plays a 0 and considers T2 for the next move. In the other case, I is still in T1 and II plays as we explained previously for T0. The game is played following the preceding moves that we described, going recursively into the trees Ti. Schematically, we get (where the color red represents the fact that we go out of a tree):

<∞ I⋃ Tn II∃ 0

T0 1

T0 0

T1 0

T2 1

T2 0 ⋮ ⋮

Let us show that the strategy we described is winning. There are two cases. If I plays something in A, there exists a tree Ti in which I stays, therefore, at some point, II plays only 1’s and so II plays something in ∃<∞0. In the second case, I plays something not in A, therefore, each time I went out of a tree Ti, II played a 0. So II plays inﬁnitely many 0’s, so he plays something which is not in ∃<∞0. Hence, the strategy is winning. <∞ 0 ω Therefore, ∃ 0 is Σ2(ω )-complete. Hence, we described the lower levels of the Wadge hierarchy. We will try to use all these intuitive tools later on when we will try to generalize the Wadge hierarchy to other more general spaces as the Polish ones.

75 4.5 Cantor Space

We described the Lipschitz and the Wadge hierarchies on the Baire space. Now, we want the Wadge hierarchy on the Cantor space. The biggest change comes from the topological diﬀerence between these two spaces, which is the compactness of the Cantor space. The compactness of 2ω actually makes just one construction impossible, this is the construction >n∈ω An. All the game tools we used are still true in this setting since 2ω is a pruned tree in ωω, therefore we can design a game in this setting. Hence, all the proofs that do not need the construction >n∈ω An remain true in the Cantor space setting. Therefore, for the Cantor space 2ω, we have, as it is said in [And07]:

1. Wadge’s Lemma holds for both ≤L and ≤W .

2. Both ≤L and ≤W are well-founded. ω 3. The lowest level is occupied by ∅ and 2 .

4. In ≤L, all the successor degrees are self-dual. 5. Theorem 4.46 holds in 2ω.

6. In ≤W , the self-dual levels and the non-self-dual levels alternate. 7. The length of the Wadge and of the Lipschitz hierarchies are Θ. What changes in the Cantor space occurs then in the limit levels. It is actually impossible to construct self-dual limit levels. Therefore, we have the following things which change, 1. Every limit level in both the Wadge and the Lipschitz hierarchy are non- self-dual. Therefore, we can draw both hierarchies.

[∅]L

⋯ ⋯ ⋯

ω ω ω ω [2 ]L

Figure 4.4: Lipschitz hierarchy on the Borel sets of the Cantor space

76 limit levels [∅]W

⋯ ⋯ ⋯

ω [2 ]W

Figure 4.5: Wadge hierarchy on the Borel sets of the Cantor space

77 Chapter 5

Wadge Hierarchy on R

Every book on descriptive set theory starts by saying that this area of mathematics is the study of the real line. Hence, a natural way to carry on the study we are doing here is to consider Wadge reducibility on the real line. That is the study we intend to do in this Chapter. With the real line, the problem is that we cannot deﬁne a game characterization of the continuous function. Hence, all the results of game theory that we made strong use of are not helpful anymore. That is the reason why this hierarchy has a completely diﬀerent shape from the ones before. We will see that Wadge’s Lemma and well-foundness will fail in this setting. To prove this, we use the construction described in [Ike10, Chapter 5].

Definition 5.1. Let X be a space and A, B ∈ P(X). We define A ⊆fin B if A ∖ B is finite. We also define A ⊥fin B ↔ (A ⊈fin B ∧ B ⊈fin A.) We first give the statement of the key result of this Chapter, which will allow us to get all the conclusions we want about the Wadge hierarchy on the real line.

Theorem 5.2. There exists an embedding i ∶ (P(N), ⊆fin) → (P(R), ≤W ). This big theorem will be a consequence of several lemmas and propositions that we now prove.

Lemma 5.3. Let a < b < c < d and e < f < g be real numbers, and let h ∶ R → R be a continuous function such that:

h(b) ∈ [e, f); h([a, b)) ∩ [e, f) = ∅; h([b, c)) ∩ [f, g) = ∅; h([c, d)) ∩ [e, f) = ∅; h([b, c)) ⊉ (e − , e). for all > 0. Then h(b) = e, h(c) = f and h([b, c)) = [e, f).

78 Proof. The following drawing gives exactly the proof we need.

a b c d

Lemma 5.4. There exist increasing sequences (aα)α<ωω , (bα)α<ωω and (cn)n<ω such that:

- aα < bα < aα+1; - sup{aα ∶ α < λ} < aλ; n - sup{aα ∶ α < ω } < cn < aωn ; for each ordinal α, each limit ordinal λ and n ∈ ω. To make thing clearer, we precise that, in this Lemma, ωω does not refer to the Baire space but to an ordinal.

Proof. First, observe that arctan is a bijection between [0, 1) and [0, ∞) which ω preserves the order. Then, observe that every α < ω can be written as

n n−1 α = αnω + αn−1ω + ⋅ ⋅ ⋅ + α1ω + α0.

Finally, observe that it is easy to give a partition of [0, ∞) into ω intervals [n, n + 1) for n ∈ ω. We will construct

ω f ∶ {α ∶ α < ω } → R α ↦ aα, with f that preserves the order. ω n n−1 Let α < ω , α = αnω + αn−1ω + ⋅ ⋅ ⋅ + α1ω + α0. Consider the interval [αn, αn + 1). By arctan, this is in bijection (with respect to the order) with [0, ∞). Hence, via this bijection, we can take a countable inﬁnite partition of [αn, αn +1). Now, consider the interval number αn−1 of this partition. Again, we ω can use a bijection, and so on and so forth. Hence, for all α < ω , we generate an interval which is disjoint from all the others. Now, just let aα be the midpoint of this interval. We generated a sequence (aα)α<ωω with the required property. aα aα+1 ω + For the sequence (bα)α<ω , just let bα = 2 . 1 c , c sup n a a n . Finally, for the sequence ( n)n<ω just let n = 2 ( α<ω α + ω ) This is well-deﬁned and has the required properties since we construct the aα’s from the midpoints of disjoint intervals.

79 Now, we deﬁne the subsets of R that will allow us to prove the bad behavior of the Wadge hierarchy on R. Lemma 5.5. The function

i ∶ P(N) → P(R) X ↦ [aα, bα) ∪ {cn ∶ n ∉ X} α<ωω is an embedding.

Proof. Suppose i(X) = i(Y ), this means that {cn ∶ cn ∈ i(X)} = {cn ∶ cn ∈ i(Y )}, therefore X = {n ∶ n ∈ X} = {n ∶ n ∈ Y } = Y . To get a better idea of the subsets we consider, we can draw the following picture.

b0 b1 bω bω+1 c c 0 1 ...... a0 a1 aω aω+1

bω⋅2 bω⋅2+1 bωn bωn+1 cn ... ...... n n aω⋅2 aω⋅2+1 aω aω +1

Lemma 5.6. In the settings of Lemma 5.5, if X,Y ∈ P(N) are such that X ⊆ Y, then i(X) ≤W i(Y ).

Proof. We deﬁne a continuous function f ∶ R → R that witnesses i(X) ≤W i(Y ). For this, we need to ﬁx some points:

aγ = sup{aα ∶ α < γ},

dn ∈ (cn, aωn ). These points are well-defined by the previous construction of our sequences (aα)α<ωω , (bα)ωω and (cn)n<ω. By definition of the embedding i, X ⊆ Y implies that i(Y ) ⊆ i(X). Therefore, we have i(X) ∖ i(Y ) = {cn ∶ n ∈ Y ∖ X}. We define:

f ∶ R → R − t ↦ t if t ∉ [aωn , aωn ] where n ∈ Y ∖ X, − − [aωn , cn] ↦ [aωn , aωn ] linearly if n ∈ Y ∖ X, [cn, dn] ↦ [aωn , dn] anti-linearly if n ∈ Y ∖ X, [dn, aωn ] ↦ [dn, aωn ] linearly if n ∈ Y ∖ X. We precise that linearly means here that the function is bijective and forms a straight line with a positive slope. And anti-linearly means that the function is bijective and forms a straight line with a negative slope. Hence, for n ∈ Y ∖ X, we have: −1 f ([aωn−1 , bωn−1 ] ∪ ⋅ ⋅ ⋅ ∪ [aωn , bωn ]) = ([aωn−1 , bωn−1 ] ∪ ⋅ ⋅ ⋅ ∪ cn ∪ [aωn , bωn ]) . The picture is the following for n ∈ Y ∖ X.

80 i(Y )

bωn

aωn

− aωn ...

− bωn 1+1

− aωn 1+1

bωn−1

a n−1 − − ω bωn 1 bωn 1+1 bωn ... a n−1 a n−1 − n ω ω +1 aωn cn dn aω i(X)

It is now obvious, by deﬁnition of this function, that it is continuous. More- over, it clearly witnesses i(X) ≤W i(Y ) by construction. We now have the two propositions which show that our embedding respects both orders.

Proposition 5.7. In the settings of Lemma 5.6, if X ⊆fin Y, then i(X) ≤W i(Y ).

Proof. Since X ⊆fin Y, we have that X ∖ Y is finite. Hence, we can define n = max{k ∶ k ∈ X ∖ Y } + 1. For all k ≥ n, we have (k ∈ X → k ∈ Y ). Hence, this reduces to the case of Lemma 5.6. Therefore, it remains to define our function on (−∞, aωn ], with aωn ↦ aωn to have a continuous function. The key idea of this part is that we have enough intervals between aωn−1 and a− a a− n ωn to send all the intervals between 0 and ω onto them. Indeed, the order n n 1 n type of {α ∶ α < ω } is the same as the order type of {α ∶ ω − ≤ α < ω }. Hence, using the bijection between ordinals which witnesses the equality of the order type, we construct a bijection between the intervals that we consider. c a n−1 a− Now, it suffices to observe that there is no n between ω and ωn , hence, our previous construction of Lemma 5.6 can be applied here. Hence we can construct a function that witnesses i(X) ≤W i(Y ) by prolonging in a continuous way with a constant function on the left and with the identity function on the right. To get a better idea, we also have the following picture.

81 i(Y )

bωn

aωn

a− ωn ...

bωn−1 Using the construction of Lemma 5.6

aωn−1 ... a0 a− cn aωn n b0 ωn dn bω i(X) dn−1

Proposition 5.8. In the settings of Lemma 5.6, if i(X) ≤W i(Y ), then X ⊆fin Y.

Proof. Suppose that h ∶ R → R is a witness of i(X) ≤W i(Y ). First, we show that aα’s are mapped on intervals. Otherwise, suppose h(aα) = cn. By continuity of h and by the fact that h reduces i(X) to i(Y ), we get h(bα) = cn, which is absurd. ω Therefore, there exists α0 < ω such that h(a0) ∈ [aα0 , bα0 ). Also let n0 be such n0 that α0 < ω . Now, we prove by induction on α, the following things:

h(aα) = aα+α0 ,

h(bα) = bα+α0 ,

h([aα, bα)) = [aα+α0 , bα+α0 ).

Hence, we prove that we map intervals on intervals, without making any gap.

Therefore, ﬁxing h(a0) = aα0 , we only have one way of sending the intervals by h. This means that everything we do is very rigid.

The initial case α = 0 is Lemma 5.3 with a0 − 1 < a0 < b0 < a1 and aα0 < bα0 < aα0+1. For the successor case α = β + 1, we use a ﬁrst time Lemma 5.3 with aβ < bβ < aβ+1 < bβ+1, aβ+α0 < bβ+α0 < aβ+α0+1 and the induction hypothesis. We get h(aβ+1) ∈ [bβ+α0 , aβ+α0+1). Hence, we can use a second time Lemma 5.3 with bβ < aβ+1 < bβ+1 < aα+2, bβ+α0 < aβ+α0+1 < bβ+α0+1, which gives the successor case. We observe that, since we are in the successor case, no cn enters the picture here, hence our argument is correct. For a limit ordinal α, it is more complicated. First, we observe that, by conti- h a− a− . nuity, we get ( α) = α+α0

82 α ωn, c h a− a− - If ≠ since there is no n in the picture, we use ( α) = α+α0 to conclude. Indeed, since h is a reduction of i(X) to i(Y ), we obtain − h(aα) = aα+α0 using Lemma 5.3 with aα < aα < bα < aα+1 and aα+α0 < bα+α0 < aα+α0+1. n − - If α = ω with n ∈ X, it means that i(X)∩(aωn , aωn ) = ∅. We have, by con- h h i Y a− n , a n tinuity of and the fact that is a reduction, that ( )∩( ω +α0 ω +α0 ) = − ∅. Hence, it suﬃces to use Lemma 5.3 with aωn < aωn < bωn < aωn+1 and n n n aω +α0 < bω +α0 < aω +α0+1 to conclude. n − - Finally, if α = ω with n ∉ X, it means that i(X) ∩ (aωn , aωn ) = {cn}. We n have two cases to consider, which are whether or not cα0+ω is in i(Y ). Both cases are given in the following picture, using the fact that h is a continuous reduction.

i(Y )

n bω +α0 cn+α0 ∉ i(Y ) cn+α0 ∈ i(Y )

n aω +α0

cn+α0

a− ωn+α0

a− cn aωn n ωn bω i(X)

ω Therefore, for all α < ω we have proved the following.

h ∶ aα ↦ aα+α0 ;

bα ↦ bα+α0 ;

[aα, bα) ↦ [aα+α0 , bα+α0 ). Moreover, observing the argument of the limit case, we have

n ∈ X → cn ∉ i(X) − → i(X) ∩ (aωn , aωn ) = ∅ − i Y a n , a n → ( ) ∩ ( ω +α0 ω +α0 ) = ∅ n → logω(ω + α0) ∈ Y

Hence we get (X ∖ Y ) ⊆ {0, 1, . . . , n0} which ﬁnally implies X ⊆fin Y. Proof. The two previous propositions conclude the proof of Theorem 5.2. Corollary 5.9. In the settings of Theorem 5.2, we have

X ⊆fin Y ↔ i(X) ≤W i(Y ),

X ⊥fin Y ↔ i(X) ⊥W i(Y ), and X ⊊fin Y ↔ i(X)

83 Proof. The ﬁrst statement is exactly the Theorem. For the second statement, we use the ﬁrst.

X ⊥fin Y ↔ X ⊈fin Y ∧ Y ⊈fin X ↔ i(X) ≰W i(Y ) ∧ i(Y ) ≰ i(X) ↔ i(X) ⊥W i(Y ).

And for the last statement.

X ⊊fin Y ↔ X ⊆fin Y ∧ Y ⊈fin X ↔ i(X) ≤W i(Y ) ∧ i(Y ) ≰W i(X) ↔ i(X)

Now, we will make use of what we proved to show the bad behavior of the Wadge hierarchy on the real line. Corollary 5.10. The Wadge hierarchy on the real line is ill-founded. Proof. By Corollary 5.9, we only need to find a strictly decreasing infinite chain in (P(N), ⊆fin). We define, for all n ∈ ω ∖ {0},

n An = {(2 ) ⋅ k ∶ k ∈ ω}.

n That is, An is the set of all multiples of 2 . It is easy to see that:

A1 ⊋fin A2 ⊋fin A3 ⊋fin ⋯ which concludes the proof. Corollary 5.11. The Wadge hierarchy on the real line has an infinite antichain. Proof. Again, by Corollary 5.9, we just need to find an infinite antichain in (P(N), ⊆fin). For all prime number p, we define,

k Bp = {p ∶ k ∈ ω}.

Since there are infinitely many prime numbers, and clearly all these subsets are infinite and disjoint, so we have an infinite antichain:

B2 ⊥fin B3 ⊥fin B5 ⊥fin ⋯

We also observe that the previous construction is carried out in the lower levels of the Borel hierarchy of the real line. Indeed, we have the following. Corollary 5.12. The bad news about the Wadge hierarchy on the real line occurs already at levels corresponding to diﬀerences of two open sets.

84 Proof. We just need to show that, for all A ∈ P(ω), we have

0 DA = [aα, bα) ∪ {cn} ∈ D2(Σ1(R)). ω α<ω n∈A

Let A ∈ P(ω), we deﬁne

ω n ΩA = {λ < ω ∶ λ limit, λ = ω for some cn ∈ A}.

′ ω And let ΩA = {λ < ω ∶ λ limit, λ ∉ ΩA}. We obtain

bα + aα+1 DA = , bα+1 α<ωω 2 sup b a α<λ α + λ ∪ , bλ ′ 2 λ∈ΩA sup b c α<λ α + n , b ∪ 2 λ λ∈ΩA

bα + aα+1 ∖ , aα α<ωω 2 sup b a α<λ α + λ ∪ , aλ ′ 2 λ∈ΩA sup b c α<λ α + n , c c , a . ∪ 2 n ∪ ( n λ)¡ λ∈ΩA

0 Therefore, we have DA ∈ D2(Σ1(R)) as required. A generalization of these results can be found in [Sch12]. The author proves the next very strong result.

Theorem 5.13. Let X be a Polish space. The following are equivalent: • X is zero-dimensional, • (B(X), ≤W ) is semi-linearly ordered, • (B(X), ≤W ) contains no infinite antichains, • (B(X), ≤W ) contains no uncountable antichains, • (B(X), ≤W ) is well quasi-ordered, i.e. it contains no infinite antichains and no infinite strictly decreasing sequence.

85 Chapter 6

Quasi-Polish Spaces

We can summarize what we did in the previous chapters in the following way. First, we defined the Polish spaces, then we made a deep study of some particular Polish spaces which led us to the study of the zero-dimensional Polish spaces. An important observation here was the universality of the Baire space ωω among these zero-dimensional Polish spaces. After that, we described some hierarchies that can be studied on Polish spaces, namely the Borel hierarchy, the Difference hierarchy, the Lipschitz hierarchy and finally the Wadge hierarchy. For the last two hierarchies, we used some game theoretical tools in order to get a precise description of these hierarchies on the zero-dimensional Polish spaces. In this Chapter, we propose a generalization of the whole construction we made for some more general spaces than the Polish ones. In fact, we will not enter into the same detailed study as we did for the Polish spaces for several reasons. The first is that a lot of the results we will state are just technical generalization of what we did before. The second is a lack of time which does not allow us to complete this study properly. Finally, we prefer to focus on the last part of this work which is the design of a game-theoretical characterization of the Wadge reducibility in this more general framework. What is interesting with games is that it is easier to compare subsets, especially in the lower levels. Since we saw that bad behaviors occurs already at very low levels in R, we hope this game-theoretical characterization to help us understanding these levels. Since we will not enter into to many details, we will mostly refer to other papers for the technical proofs. A very good paper which is at the origin of this new study of more general spaces is [DB13]. In this Chapter, we will use it a lot as a reference to construct and to describe these spaces. We will also try to motivate the game-theoretical characterization that we will establish in the last part of this Chapter.

6.1 Deﬁnition

First, we deﬁne the objects of study of this Chapter. We saw that Polish spaces are metric ones. The idea here is to generalize the study to more than metric spaces, hence we need something similar and less restrictive than a metric.

Deﬁnition 6.1. Let X be a space. A quasi-metric is a function d ∶ X × X → [0, ∞) such that:

86 - (d(x, y) = d(y, x) = 0 ↔ x = y), for all x, y ∈ X, - d(x, y) ≤ d(x, z) + d(z, x), for all x, y, z ∈ X. We observe that a quasi-metric is just a metric which is not required to be symmetric. From now on, we suppose that we have a quasi-metric d on a space X. From a quasi-metric as for a metric, we have an induced topology given by the open balls, i.e. for all > 0 and for all x ∈ X:

Bd(x, ) = {y ∈ X ∶ d(x, y) < }.

It is really easy, as in the case of a metric, to see that this family of sets defines a topology. We denote this topological space as (X, d). This topology, called the quasi-metric topology, is not always Hausdorff, but it is at least T0. We recall this definition. Definition 6.2. Let X be a topological space. X is T0 if, for all x, y ∈ X, there exists an open neighborhood U containing x and not y, or y and not x. Hence, the big difference in the topology of metric spaces and quasi-metric spaces is the separation axiom, which is T2 (Hausdorff) in the case of a metric, and only T0 in the case of a quasi-metric. However, from a quasi-metric, we can construct a corresponding metric, which will be denoted by d,ˆ and which ˆ is defined by d(x, y) = max{d(x, y), d(y, x)}. We have of course that Bdˆ(x, ) ⊆ Bd(x, ). Now, to generalize the Polish spaces, we need a definition of a complete quasi-metric. There are several different possible definitions, but we make the same choice as [DB13] which gives good properties. Let us start by defining a Cauchy sequence in a quasi-metric space.

Deﬁnition 6.3. Let (X, d) be a topological space where d is a quasi-metric. A sequence (xn)n∈ω ⊆ X is Cauchy if, for all > 0, there exists N ∈ ω such that for all N ≤ n ≤ m, we have d(xn, xm) < . Obviously, we have the deﬁnition of a complete quasi-metric space.

Definition 6.4. A topological space (X, d) where d is a quasi-metric is complete if every Cauchy sequence in X converges to an element in X with respect ˆ to d. This means that, if (xn)n∈ω is a Cauchy sequence, there exists x ∈ X such ˆ that d(xn, x) → 0 when n → ∞. Definition 6.5. A topological space X is completely quasi-metrizable if it is compatible with some complete quasi-metric. We are now able to define the objects of study of this Chapter.

Deﬁnition 6.6. A topological space X is quasi-Polish if it is completely quasi- metrizable and countably based.

87 6.2 Examples

In this Section, we give examples of quasi-Polish spaces. Of course, since a metric is in particular a quasi-metric, all Polish spaces are quasi-Polish. Hence, we are interested in the other quasi-Polish spaces. First, we show two examples of topological spaces with a quasi-metric.

Example 6.7. The ﬁrst simple example is the Sierpinski space S = {0, 1} with open subsets {∅, {1}, {0, 1}}. The quasi-metric is given by d(0, 1) = 0, and d(1, 0) = 1. It is trivial that this is a quasi-metric which is complete and that the space is countably based. Therefore, the Sierpinski space is quasi-Polish. This is an useful example for ﬁnding counter-example. The second example is, as we will see later on, an universal one.

Example 6.8. Consider P(ω) with the following metric. Let X,Y ∈ P(ω), we define n d(X,Y ) = sup{2− ∶ n ∈ X ∖ Y }. It is easy to see that this defines a quasi-metric. Let us show that it is complete and countably based. Suppose we have a Cauchy sequence (Xn)n∈ω. Hence, for all > 0, there exists N such that for all N ≤ n ≤ m, we have d(Xn,Xm) < . We will define 1 recursively sets Ln ∈ P(ω) for n ∈ ω. We start by setting L0 = ∅. For = 2− , consider, XN and set L1 = XN ∩ {0, 1,...,N}. Now, suppose we have already −i defined Li−1. Consider = 2 and XN . Just let Li = XN ∩ {0, 1,...,N}. By construction, Li in an increasing sequences. Moreover, let L = ⋃i∈ω Li ∈ P(ω). dˆ L, X 2−i 0. We have that ( N2−i ) ≤ → Hence, we have a complete metric. Now, to show that it is countably based, it suffices to show that we have the following countable basis

+ {Bd(F, ) ∶ ∈ Q ,F ﬁnite}.

n Let A ∈ P(ω), > 0, and take n ∈ ω such that 2− < . Let

ASn = A ∩ {0, 1, . . . , n}.

n If C ∈ Bd(ASn, 2− ), then for all k < n + 1, we have k ∈ C ↔ k ∈ A. This implies n C ∈ Bd(A, ). Therefore Bd(ASn, 2− ) ⊆ Bd(A, ). Therefore, we showed that P(ω) with the preceding quasi-metric is countably based and completely quasi-metrizable, hence it is quasi-Polish. We ﬁnish with the example which motivated the article [DB13]. Indeed, in this paper, we have the following paragraph. We precise that the properties and objects cited in this paragraph will be presented later in this Chapter.

« For example, we will see that the class of quasi-Polish spaces is in general enough to contain both the Polish spaces and the countably based locally compact sober spaces, hence all ω-continuous domains, but is not too general as demonstrated by the fact that every quasi-Polish space is sober and every metrizable quasi-Polish space is Polish. »

88 Therefore, our next goal is to deﬁne ω-continuous domains. First, we recall some topological deﬁnitions.

Definition 6.9. (P, ≤) is a partially ordered set (poset) if ≤ is a reflexive, antisymmetric and transitive binary relation. A subset D ⊆ P is directed if it is nonempty and all pair of elements in D have an upper bound in D. We say that P is a directed complete poset if it is a poset such that every directed subset has a supremum sup D in P . We want to give a topology to a directed complete poset. This is done in the next Definition.

Deﬁnition 6.10. Let P be a directed complete poset and suppose x, y ∈ P. We say that x is way below y, written x ≪ y, if every directed D ⊆ P such that sup D exists and y ≤ sup D, there exists d ∈ D such that x ≤ d. We deﬁne ↡ x = {y ∈ P ∶ y ≪ x}, and ↟ x = {y ∈ P ∶ x ≪ y}. A subset U ⊆ P is Scott open if:

x ∈ U ∧ x ≤ y → y ∈ U, and D ⊆ P directed such that sup D exists and sup D ∈ U, then D ∩ U ≠ ∅. The Scott open sets are a basis of a topology known as the Scott topology. A subset B ⊆ P is a domain theoretic basis for P if for every x ∈ P, the set B ∩ ↡ x contains a directed subset D such that sup D = x. Finally, we can describe the ω-continuous domains. Definition 6.11. A directed complete poset P is a ω-continuous domain if it has a countable domain theoretic basis. The name of a « domain theoretic basis » is justified by the following Propo- sition. Proposition 6.12. Suppose P is an ω-continuous domain with B as a domain theoretic basis, then ↟ x is Scott open for all x ∈ P and {↟ x ∶ x ∈ B} is a countable topological basis for the Scott topology. In general, the fact that ω-continuous domain are quasi-Polish spaces is not easy to show, it is done in [DB13]. Theorem 6.13. Every ω-continuous domain is quasi-Polish. However, the principal case we deal with is more easy to see. That is what we do now. Let us consider P(ω) ordered by the inclusion. It is clearly a poset. Suppose we have a directed subset D. We first observe that, in this setting, the supremum is the union. So sup D = X ⊆ ω. X∈D

Indeed, ⋃X∈D X is clearly an upper bound. Moreover, it is the smallest one because every subset which is strictly included in ⋃X∈D X is not an upper bound anymore. Hence, it is a directed complete poset. Finally, if we write P<∞(ω) for all the ﬁnite subsets of ω, it suﬃces to show that it is countable and that it is a

89 domain theoretic basis. The fact that it is countable comes from (ACω). Let us show that it is a domain theoretic basis. Suppose X ∈ P(ω). Consider D = {X′ ∈ P<∞(ω) ∶ X′ ⊆ X}. It is obvious that sup D = X. The first inclusion comes from the definition of D, the second comes from the fact that the supremum is in this case the union, as we said earlier. Moreover, by definition, D ⊆ P<∞(ω). Finally, we want to show that D ⊆ ↡ X. If Y ∈ D, we have that Y is some finite subset of X. Suppose that D′ is a directed subset such that X ⊆ sup D′. We have

y ∈ Y → y ∈ Y ⊆ X ⊆ sup D′ = d′ ′ ′ d ∈D ′ → ∃dy ∈ D s.t. y ∈ dy ′ → Y ⊆ dy ⊆ B ∈ D y∈Y ′ ′ (since D is directed, there exists B ∈ D such that dy ⊆ B). y∈Y

Therefore, Y ∈ ↡ X. Hence, we showed that, for all X ∈ P(ω), there exists a directed subset D ⊆ P(ω) ∩ ↡ X such that sup D = X. This implies that P<∞(ω) is a domain theoretic basis. Therefore, using the previous Proposition, the open subsets of the Scott topology of P(ω) are just ↟ F for F ∈ P<∞(ω). We want to understand these sets.

Proposition 6.14. For all F ∈ P<∞(ω), we have ↟ F = {X ∈ P(ω) ∶ F ⊆ X}. Proof. Let X ∈ ↟ F. Consider the directed subset D = {X′ ∈ P<∞(ω) ∶ X′ ⊆ X}. Clearly, X = sup D. Since F ≪ X, there exists d ∈ D such that F ⊆ d ⊆ X. Now, let X ∈ P(ω) be such that F ⊆ X. Let D be a directed subset such that F ⊆ X ⊆ sup D = ⋃Y ∈D Y. Therefore, for all f ∈ F, there exists Yf ∈ D such that f ∈ Yf . Since F is ﬁnite and D is directed, there exists B ∈ D such that

F ⊆ Yf ⊆ B ∈ D. f∈F

Hence, X ∈ ↟ F. Now, we have a last Proposition that shows that this space is actually a quasi-Polish space. We recall that we deﬁned a quasi-metric on this space at the beginning of this Section.

Proposition 6.15. In the space P(ω), the Scott topology and the topology induced by the quasi-metric are the same. Proof. Let F be a ﬁnite subset of ω. We have

− max F − max F X ∈ Bd F, 2 → d(F,X) < 2 → (n ∈ F ∖ X → n > max F ) → F ∖ X = ∅ → F ⊆ X → X ∈ ↟ F.

90 max F Therefore Bd F, 2− ⊆ ↟ F. n Let Bd(A, ). We saw that there exists n ∈ ω such that Bd(ASn, 2− ) ⊆ Bd(A, ). We have

X ∈ ↟ ASn → ASn ⊆ X → ASn ∖ X = ∅ → d(ASn,X) = 0 → X ∈ Bd(ASn, 0). n Therefore, ↟ ASn ⊆ Bd(ASn, 0) ⊆ Bd(ASn, 2− ) ⊆ Bd(A, ).

6.3 Borel Hierarchy and Universal Property of P(ω)

We have seen that the Borel hierarchy for Polish spaces has a simple shape. This was done by transfinite recursion using the fact that closed subsets are Gδ. The problem here is that, in quasi-Polish spaces, not all closed subsets are Gδ. Indeed, in the Sierpinski space, {0} is closed but not Gδ. Therefore, we need an other definition for the Borel hierarchy, namely, the following. 0 Definition 6.16. Let (X, τ) be a topological space, we define Σα(X) by transfinite recursion on 1 ≤ α < ω1. 0 Σ1(X) = τ,

Σ0 X B B′ B ,B′ Σ0 , β α , α 1. α( ) = ( n ∖ n) ∶ n n ∈ βn n < ¡ for > n∈ω 0 0 0 Of course, we can define Πα(X) as the complement of Σα(X) and ∆α(X) 0 0 as the intersection of Σα(X) and Πα(X) as we already did earlier. Once again, we define the Borel subsets of X as: 0 B(X) = Σα(X). α<ω1 The first important fact which justifies this new Definition is that every open 0 set is ∆2(X). Indeed, let U ⊂ X be open. Then, 0 U = U ∖ ∅ ∈ Σ2(X), and c 0 0 U = X ∖ U ∈ Σ2(X) → U ∈ Π2(X). 0 In particular, with this definition of Σα(X), we get that, in the Sierpinski space S, 0 {1} = S ∖ {0} ∈ Σ2(S). We can also prove, as in the case of Polish spaces, that the Borel hierarchy has the same simple shape. Indeed, we have the following similar results as in the case of Polish spaces.

Proposition 6.17. Let X be a topological space. For all α < β < ω1, we have: 0 0 0 Σα(X) ∪ Πα(X) ⊆ ∆β(X).

91 0 0 0 Proof. We just prove that Σα(X) ⊆ ∆β(X). Let A ∈ Σα(X), then, since ∅,X ∈ 0 Σ1(X), we have 0 A = A ∖ ∅ ∈ Σβ(X), and c 0 A = X ∖ A ∈ Σβ(X). 0 Therefore, it follows that A ∈ ∆β(X). We also have, as in the case of Polish spaces and using a similar proof, the following result. Proposition 6.18. Let X be a topological space and Γ be a class in the Borel hierarchy. • Γ is closed under finite unions and finite intersections; • Γ is closed under continuous preimage; 0 • if Γ = Σα(X) for some α < ω1, then Γ is closed under countable unions; 0 • if Γ = Πα(X) for some α < ω1, then Γ is closed under countable intersections; 0 • if Γ = ∆α(X) for some α < ω1, then Γ is closed under complements. The second important fact is that this Definition coincides with the Defi- nition 3.9 on metrizable space. Hence, on Polish spaces, both definitions give the same hierarchy. This is due to the two following propositions. The first Proposition is about the second level of the Borel hierarchy in the case of metrizable spaces. We precise that, from now on, we will use the new Definition of 0 0 0 Σα(X), Πα(X) and ∆α(X). In case we want to consider Definition 3.9, we will explicitly say that we consider a Polish topology and not a quasi-Polish one. 0 Proposition 6.19. If X is a metrizable, then every A ∈ Σ2(X) is Fσ. 0 Proof. Let X be metrizable and let A ∈ Σ2(X). Then, there exist open subsets ′ Un,Un such that: ′ A = (Un ∖ Un). n∈ω Moreover, since X is metrizable, we use Proposition 2.16 to write

Un = Fn,j, j∈ω with Fn,j being closed for all n, j ∈ ω. Therefore, we have

′ A = (Un ∖ Un) n∈ω

′ = Fn,j ∖ Un n∈ω j∈ω ′ = (Fn,j ∖ U n). n,j∈ω

′ ′ c To complete the proof, we just observe that every Fn,j ∖ Un = Fn,j ∩ (Un) is closed. The second Proposition is about the next levels on every topological space.

92 0 Proposition 6.20. Let X be is a topological space and α > 2 If A ∈ Σα(X), B Π0 X , β α, then there exists i ∈ βi ( ) i < such that:

A = Bi. i∈ω

0 Proof. Let X be a topological space and let A ∈ Σα(X) for some α > 2. Then, we can write ′ A = (Dn ∖ Dn), n∈ω D ,D′ Σ0 β α. for some n n ∈ βn with n ≤ Moreover, by Proposition 6.17, we can suppose βn ≥ 2. Therefore, we can write

′ Dn = (Gn,j ∖ Gn,j), j∈ω

G ,G′ Σ0 γ β for some n,j n,j ∈ γn,j with n,j < n. Hence, we have

′ ′ A = ((Gn,j ∖ Gn,j) ∖ Dn) n,j∈ω ′ ′ = (Gn,j ∖ (Gn,j ∪ Dn)). n,j∈ω

To complete the proof, we just use Proposition 6.17 to observe that Gn,j ∈ Σ0 X Π0 X G′ D′ c Π0 X . γn,j ( ) ⊆ βn ( ) and that ( n,j ∪ n) ∈ βn ( ) Now, we are interested in the link between the Borel hierarchy corresponding to a quasi-metric d and the Borel hierarchy corresponding to the metric dˆ. Actually, the link between these concepts is simple and allows us to have some good behavior in the quasi-Polish context. 0 Proposition 6.21. Let (X, d) be a quasi-metric space, then Bdˆ(x, ) ∈ Σ2(X, τd). An other interesting and useful result which gives a link between d and dˆ.

Proposition 6.22. Let (X, d) be a quasi-metric. (X, d) is countably based if ˆ and only if X, d is separable. The last two propositions are proved in [DB13]. They give us the conﬁr- mation that our considerations generalize the previous ones on all the Borel subsets. 0 Theorem 6.23. If (X, d) is a quasi-Polish space, then τdˆ ⊆ Σ2(X, τd). There- 0 0 fore, Σα(X, τd) = Σα(X, τdˆ) for all α ≥ ω. In particular, B(X, τd) = B(X, τdˆ).

Proof. This is a consequence of the fact that Bdˆ(x, ) ⊆ Bd(x, ), hence τd ⊆ τdˆ, of the Proposition 6.20 and of the Proposition 6.21. As an example that this result is optimal, we use a result of [Sel06] which shows the following. Consider P(ω) with the quasi-metric we already defined. Then the corresponding metric dîs the metric we defined on 2ω for Polish spaces. The result is the following. 0 0 0 Proposition 6.24. For all n < ω, Σn(P(ω), τd) ⊈ Πn(P(ω), τdˆ) and Πn(P(ω), τdˆ) ⊈ Σ0 ω , τ . n+1(P( ) d)

93 Therefore, we can show that Theorem 6.23 is optimal. 0 0 Proposition 6.25. In P(ω), we get Σn(P(ω), τdˆ) ≠ Σn(P(ω), τd) for all n < ω. 0 0 Proof. Towards a contradiction, suppose Πn(P(ω), τdˆ) = Πn(P(ω), τd). By the previous Proposition, we get a contradiction, namely

Π0 ω , τ Π0 ω , τ Σ0 ω , τ . n(P( ) d) = n(P( ) dˆ) ⊈ n+1(P( ) d) This is a contradiction to the shape of the Borel hierarchy on Polish spaces. Now, with all these results, we have some easy but important consequences about quasi-Polish spaces. Proposition 6.26. Every uncountable quasi-Polish space has cardinality 2ℵ0 . Proof. Let d be a quasi-metric for a quasi-Polish space X. By Proposition 6.22, ˆ ˆ X, d is separable. Moreover, since a Cauchy sequence in X, d is obviously a ˆ Cauchy sequence in (X, d), we get that X, d is completely metrizable, hence Polish. Therefore, X has cardinality 2ℵ0 . We recall that the Borel hierarchy does not collapse if each level strictly increases all the levels below it. Theorem 6.27. If X is an uncountable quasi-Polish space, then the Borel hierarchy on X does not collapse. Proof. Let d be a quasi-metric for a quasi-Polish space X. Suppose, towards a 0 0 contradiction that Σβ(X) = Πβ(X) for some β < ω1. We proved in the preceding ˆ proof, that X, d is Polish, hence the Borel hierarchy on this metric does not α max β, ω A Σ0 X, τ Π0 X, τ . collapse. Let us consider = { } and take ∈ α+1( dˆ)∖ α+1( dˆ) We get, using Theorem 6.23, that

A Σ0 X Σ0 X Σ0 X, τ Π0 X, τ . ∈ α+2( ) = α( ) ⊆ α( dˆ) ⊆ α+1( dˆ) Hence, we have a contradiction. Hence, the study of the Borel hierarchy on the quasi-Polish spaces makes sense and is interesting as in the case of Polish spaces. This is a really a good starting point for us. As in the Polish case, we have a simple similar characterization of subspaces of quasi-Polish spaces that are quasi-Polish. The reader can ﬁnd the proof of this result in [DB13]. Theorem 6.28. A subspace A of a quasi-Polish space X is quasi-Polish if and 0 only if A ∈ Π2(X).

We also have a nice Proposition about countably based T0-spaces.

Proposition 6.29. Every countably based T0-space embeds in P(ω).

Proof. Let X be a countably based T0 space with basis (Un)n∈ω. We deﬁne

f ∶ X → P(ω) x ↦ {n ∈ ω ∶ x ∈ Un}.

94 Since X is T0, we have that f is injective. It remains to show that it is a homeomorphism on its image. Let F ∈ P∞(ω), we have

−1 f (↟ F ) = {x ∈ X ∶ x ∈ Un for all n ∈ F } = Un. n∈F We also have that

f(Um) = {{n ∈ ω ∶ x ∈ Un} ∶ x ∈ Um} = {A ∈ P(ω) ∶ m ∈ A ∧ A = {n ∈ ω ∶ x ∈ Un}} = {A ∈ P(ω) ∶ m ∈ A ∧ A ∈ f(X)} = f(X) ∩ ↟ {m}. Therefore, our proof is complete.

From the two previous results, we get easily the universality of P(ω). Theorem 6.30. A space is quasi-Polish if and only if it is homeomorphic to 0 some A ∈ Π2(P(ω)). This result justiﬁes our future deep study of this particular space, as we did with the Baire space ωω in the case of Polish spaces. We ﬁnish this Section with the Theorem which proves that the quasi-Polish spaces are not too general. Hence, we hope not losing too many properties by studying them. Again, the proof is in [DB13]. Theorem 6.31. A metrizable space is Polish if and only if it is quasi-Polish.

6.4 Two Views on the Wadge Hierarchy on P(ω)

Now, we reach a sensitive point of the study of hierarchies on quasi-Polish spaces. Indeed, our goal is to study continuous reductions on these spaces. We decide to focus on the space P(ω) which is universal among all quasi-Polish spaces. In the literature, the study of hierarchies on P(ω) is mostly done via the hierarchies on the Baire space ωω. That is the case in [Tan81] and [Wad84]. However, we will not achieve a good description of the Wadge hierarchy obtained via continuous reduction with this technique. Indeed, via the Baire space, it is impossible to have four incomparable sets. However, we will prove later in this work that, if we consider continuous reduction on P(ω), we can ﬁnd four such spaces. Hence, this technique, although it gives good results for other hierarchies, does not capture all the subtleties of the Wadge hierarchy on P(ω). Therefore, we have two points of views that are completely diﬀerent about the Wadge hierarchy on P(ω). ω First, suppose that we have a function ρ ∶ ω → P(ω). We can consider continuous reducibilty on the preimages of ρ. For example, let A, B ⊆ P(ω), we could say that A is continuously reducible to B if there exists a continuous ω ω 1 1 1 function f ∶ ω → ω such that f − (ρ− (B)) = ρ− (A). But, as explained earlier, with this point of view, the maximal size of an antichain is 2, which does not correspond to the second point of view. In the second point of view, we consider direct continuous reducibility. Let A, B ⊆ P(ω), we will say that A is continuously reducible to B if there exists a

95 1 continuous function such that f ∶ P(ω) → P(ω) such that f − (B) = A. We will see, later on, that this point of view allows us to find an antichain of size 4. We start by investigating the first point of view. For this, we need a definition.

ω Deﬁnition 6.32. A partial continuous function ρ ∶ S ⊆ ω → X is an admissi- ω ble representation of X if for all partial continuous functions f ∶ R ⊆ ω → X, ω ω there exists a partial continuous function g ∶ T ⊆ ω → ω such that f = ρ ○ g. In a picture, we have

ωω ∀f continuous

X ∃g continuous ωω

ρ continuous

The idea is that, using an admissible representation, we can simulate every continuous function f just via some coding g of the Baire space. An important result is that admissible representations can be used to characterize quasi-Polish spaces. Theorem 6.33. A countably based space X is quasi-Polish if and only if there ω is an admissible representation ρ ∶ S ⊆ ω → X such that dom(ρ) is Polish. We also have another similar result. Theorem 6.34. A countably based space X is quasi-Polish if and only if there ω is a total admissible representation ρ ∶ ω → X. The proofs of the previous results are quite diﬃcult and can be found in [DB13]. However, as an example, we can deﬁne the enumeration map

ω δ ∶ ω → P(ω) s ↦ {x ∈ ω ∶ ∃n ∈ ω, s(n) = x + 1}

We show that it is a total admissible representation. Proposition 6.35. The enumeration map is a total admissible representation.

Proof. First, we prove that it is continuous. Suppose F ∈ P<∞(ω),F = {a1, . . . , an}. Let Σ be the set of all permutations of {1, . . . , n}, we have: δ−1 F N (↟ ) = b1âσ(1)ˆb2âσ(2)ˆ...ˆbnâσ(n) <ω bi∈ω σ∈Σ

96 Hence, δ is continuous. ω Now, we show that δ is an admissible representation. Let f ∶ S ⊆ ω → P(ω) be a continuous function. We deﬁne ω g ∶ S → ω ⎧ ⎪i + 1 if f(NsSi) ⊆ ↟ {i} s ↦ t where ti = ⎨ ⎪0 otherwise. ⎩⎪ We also need (n ∈ f(s) ↔ ∃m ∈ ω, f(NsSm) ⊆ ↟ {n}). 1 1 Suppose n ∈ f(s), then f(s) ∈ ↟ {n}. So, s ∈ f − (f(s)) ⊆ f − (↟ {n}). This last set is open since f is continuous, hence there exists an open neighborhood of s 1 included in f − (↟ {n}). This means exactly that there exists m ∈ ω such that f(NsSm) ⊆ ↟ {n}. Now, suppose ∃m ∈ ω, f(NsSm) ⊆ ↟ {n}. We have s ∈ NsSm, hence f(s) ∈ f(NsSm) ⊆ ↟ {n}. Therefore, we have n ∈ f(s). Finally, we can show that g has the required property. Let v ∈ S. We have δ ○ g(v) = {n ∈ ω ∶ ∃m ∈ ω, g(v)(m) = n + 1}

= {n ∈ ω ∶ ∃m ∈ ω, f(NvSm) ⊆ ↟ {n}} = {n ∈ ω ∶ n ∈ f(v)} = f(v). Therefore, as required, we have f = δ ○ g. Hence, as we already know, using Theorem 6.34, we have that P(ω) is quasi- Polish. The next result shows that the Difference hierarchy has a good behavior with respect to admissible representations. The proof of this result is in [DB13]. ω Theorem 6.36. Let X be a countably based T0 space, ρ ∶ S ⊆ ω → X an admissible representation. For all countable ordinals 0 < α, θ < ω1 and A ⊆ X, we have 0 −1 0 A ∈ Dθ(Σα(X)) ↔ ρ (A) ∈ Dθ(Σα(dom(ρ))). Using this result, we can generalize the Hausdorff-Kuratowski Theorem. Theorem 6.37. Let X be a quasi-Polish space, if 0 < θ < ω1, then ∆0 X Σ0 X . θ+1( ) = Dα( θ( )) 1≤α<ω1 0 Proof. Let S ∈ ∆θ 1(X). Since X is quasi-Polish, there exists a total admissible + ω representation ρ ∶ ω → X. Since ρ is continuous and the classes in the Borel ρ−1 S ∆0 ωω . hierarchy are closed under continuous preimage, we have ( ) ∈ θ+1( ) By the Hausdorff-Kuratowski Theorem for Polish spaces (Theorem 3.24), there exists α < ω1 such that −1 0 ω ρ (S) ∈ Dα(Σθ(ω )). 0 Using the previous Theorem, we obtain S ∈ Dα(Σθ(X)). The other inclusion is true for any topological space by definition of the 0 0 0 classes Σα(X), Πα(X) and ∆α(X). Therefore, although it is not possible to capture all the details of the Wadge hierarchy using admissible representations (as explained earlier), they are a powerful tool in the description of some hierarchies of quasi-Polish spaces.

97 6.5 Games on P(ω)

In this Section, we will try to understand more directly the Wadge hierarchy of P(ω), hence the second point of view presented earlier. For this, we give a game characterization of continuous functions in P(ω). Indeed, we saw, all along this work, how powerful games are. The games and strategies that we will define are an idea of Pr. Jacques Duparc. I also thank Yann Pequignot which gave me his personal notes [DP15] on this topic. We will define two different games which will characterize continuous functions on P(ω). We recall that we consider the Scott topology on P(ω).

Deﬁnition 6.38. Let f ∶ P(ω) → P(ω). We deﬁne a game G∞(f). This is a two player game, I and II, with complete perfect information. Both players play successively, starting from I. I plays an increasing sequence of elements Xn in P<∞(ω), but II plays an increasing sequence of elements Yn in P(ω). Therefore, I constructs X = ⋃n∈ω Xn and II constructs Y = ⋃n∈ω Yn. II wins the game if f(X) = Y . Schematically, we draw: I II

X0 ⊆ Y0

⊆ <∞ X1 Xn ∈ P (ω),Yn ∈ P(ω) ⊆ Y1 ⊆ X2 ⊆ Y2 II wins if and only if ⋮ ⋮ f(X) = Y.

X = ⋃n∈ω Xn, Y = ⋃n∈ω Yn

Now, we deﬁne strategies in this game.

Definition 6.39. An ultrapositional strategy for II in a game G∞(f) is an increasing function σ ∶ P<∞ → P(ω), i.e. such that for all X0,X1 ∈ P<∞(ω) with X0 ⊆ X1, we have σ(X0) ⊆ σ(X1). Definition 6.40. An ultrapositional strategy is winning for II if, following this ultrapositional strategy and whatever I plays, II wins the game G∞(f). We need a first Lemma. But, before that, just observe that, for all X ∈ P(ω), X = ⋃F ⊆X finite F. Indeed, x ∈ X implies {x} ⊆ X, and x ∈ ⋃F ⊆X finite F implies the existence of a finite F0 ⊆ X with x ∈ F0 ⊆ X. Therefore, we have the following result.

Lemma 6.41. Every increasing function f ∶ P<∞(ω) → P(ω) induces a continuous extension, which is given by ˆ f ∶ P(ω) → P(ω) X ↦ f(F ). F ⊆X ﬁnite

98 ˆ Proof. First, we verify that f is an extension of f. Let X ∈ P<∞(ω), then:

ˆ f(X) = f(F ) = f F = f(X). F ⊆X ﬁnite F ⊆X ﬁnite

It remains to show that fˆ is continuous. ˆ 1 Let F ∈ P(ω) be ﬁnite, and X ∈ f − (↟ F ).

ˆ−1 ˆ X ∈ f (↟ F ) → f(X) ∈ ↟ F ˆ ′ → F ⊆ f(X) = f(F ) ′ F ⊆X finite → since F finite, there exists a finite F0 ⊆ X such that F ⊆ f(F0).

ˆ 1 We want to show that ↟ F0 is an open neighborhood of X included in f − (↟ F ). Let Y ∈ ↟ F0, then: ˆ ˆ F ⊆ f(F0) ⊆ f(F0) ⊆ f(Y ), where the last inclusion comes directly from the deﬁnition of f.ˆ Therefore, ˆ ˆ 1 ˆ f(Y ) ∈ ↟ F, which means exactly that ↟ F0 ⊆ f − (↟ F ). So, f is continuous. Of course, considering {F ∈ P<∞(ω) ∶ F ⊆ X} or considering just an increas- <∞ ing sequence (Xn)n∈ω ⊆ P (ω) whose limit is X is the same. Therefore, if we have a function f ∶ P<∞(ω) → P(ω) and X ∈ P(ω). For any increasing sequence <∞ (Xn)n∈ω ⊆ P (ω) whose limit is X, we have ˆ f(X) = f(Xn). n∈ω We need a second Lemma.

<∞ <∞ Lemma 6.42. The restriction fSP (ω) ∶ P (ω) → P(ω) of a continuous function f ∶ P(ω) → P(ω) is increasing. Proof. First, we observe that

X ⊆ Y ↔ ∀F ﬁnite (X ∈ ↟ F → Y ∈ ↟ F ).

Indeed, the ﬁrst inclusion is obvious by transitivity of ⊆ and from the fact that X ∈ ↟ F means F ⊆ X. For the second one, just observe that x ∈ X ↔ {x} ⊆ X. Therefore: x ∈ X → X ∈ ↟ {x} → Y ∈ ↟ {x} → x ∈ Y. A last observation is that the smallest open neighborhood of some X ∈ P<∞(ω) is simply ↟ X. Therefore, suppose X ⊆ Y, with X,Y ∈ P<∞(ω) and let F be ﬁnite and such that f(X) ∈ ↟ F. Then: 1 f(X) ∈ ↟ F → X ∈ f − (↟ F ) 1 → ↟ X ⊆ f − (↟ F ) 1 → Y ∈ f − (↟ F ) → f(Y ) ∈ ↟ F.

<∞ Therefore, f(X) ⊆ f(Y ), hence fSP (ω) is increasing.

99 Therefore, we can state and prove a Proposition explaining why we designed such a game.

Proposition 6.43. Let f ∶ P(ω) → P(ω) be a function. Then f is continuous if and only if II has an ultrapositional winning strategy in G∞(f). Proof. Suppose f is continuous. By the previous Lemma, the restriction <∞ fSP (ω) is increasing. Hence, it is an ultrapositional strategy. Suppose II uses this strategy and I plays a sequence (Xn)n∈ω which leads to X. The result of the game for II is:

f(Xn) = f Xn = f(X). n∈ω n∈ω Therefore, we constructed a winning ultrapositional strategy for II in the game G∞(f). Now, suppose II has a winning ultrapositional strategy σ. Suppose I plays a sequence (Xn)n∈ω which leads to X. II answers by:

f(X) = σ(Xn) = σˆ(X). n∈ω

Therefore, f = σˆ is continuous. We have a second game which will allow us to characterize continuous functions. It is really similar except that II is only allowed to play elements in P<∞(ω). Hence, this game seems to be more symmetric.

Deﬁnition 6.44. Let f ∶ P(ω) → P(ω). We deﬁne a game G<∞(f). This is a two player game, I and II, with complete perfect information. Both players play successively, starting from I. I plays an increasing sequence of elements Xn in P<∞(ω), and II plays an increasing sequence of elements Yn in P<∞(ω). Therefore, I constructs X = ⋃n∈ω Xn and II constructs Y = ⋃n∈ω Yn. II wins the game if f(X) = Y . Schematically, we draw: I II

X0 ⊆ Y0

⊆ <∞ X1 Xn,Yn ∈ P (ω) ⊆ Y1 ⊆ X2 ⊆ Y2 II wins if and only if ⋮ ⋮ f(X) = Y.

X = ⋃n∈ω Xn, Y = ⋃n∈ω Yn

Now, we deﬁne strategies. We want this strategy to be fair. To explain this clearly, we need an observation about strategies of II in the game G∞(f). Suppose that σ is a strategy for II in G∞(f). Then, if, from some point, I plays always the same set Xn, by the increasing condition on σ, from this point, the

100 answer of II will always be σ(Xn) = Yn. Therefore, to win, at every time of the development of the game, II must play the image of the set played by I, otherwise I has just to stop increasing his sequence. In our second game G<∞(f), the situation is completely different. Indeed, if we consider the same definition for a strategy, and if there exists Xn ∈ P<∞(ω) such that f(Xn) ∈ P∞(ω), II can never win since he has to play finite subsets of ω. That is the reason why we need to change our definition of a strategy in this setting. The good way to do this is to allow II to increase his sequence as long as the game progresses.

Deﬁnition 6.45. A positional strategy for II in G<∞(f) is a function σ ∶ P<∞ × ω → P<∞(ω) such that:

∀X,Y ∈ P<∞(ω), ∀m, n ∈ ω ((X ⊆ Y ∧ m ≤ n) → σ(X, m) ⊆ σ(Y, n)). Again, a positional strategy is winning if it leads II to the victory whatever I plays. This game is equivalent to the previous one. To show this, we need an observation. Suppose (Xn)n∈ω ⊆ P(ω), we need that

(Xn ∩ {0, 1, . . . , n}) = Xn. n∈ω n∈ω

The ﬁrst inclusion is trivial. For the second one, suppose x ∈ ⋃n∈ω Xn. There exists n0 ∈ ω such that x ∈ Xn0 . Therefore, x ∈ Xn0 ⊆ ω. So x ∈ ω, therefore, there exists m ∈ ω such that x < m. Let M = max{n0, m} + 1. Since the sequence is increasing, we obtain

x ∈ (XM ∩ {0, 1,...,M}) ⊆ (Xn ∩ {0, 1, . . . , n}). n∈ω Using this observation, we have the next result.

Proposition 6.46. II has a winning positional strategy in G<∞(f) if and only if II has a winning ultrapositional strategy in G∞(f). Proof. Suppose II has a winning ultrapositional strategy σ. We deﬁne

τ ∶ P<∞(ω) × ω → P<∞ (X, n) ↦ σ(X) ∩ {0, 1, . . . , n}.

This is a positional strategy since, if X ⊂ Y, with X,Y ∈ P<∞(ω) and m ≤ n, with m, n ∈ ω, we get:

τ(X, m) = σ(X) ∩ {0, 1, . . . , m} ⊆ σ(Y ) ∩ {0, 1, . . . , n} = τ(Y, n).

Moreover, τ is winning since, if I plays a sequence (Xn)n∈ω leading to X, the answer of II is

τ(Xn, n) = σ(Xn) ∩ {0, 1, . . . , n} = σ(Xn) = f(X). n∈ω n∈ω n∈ω Conversely, suppose τ is a winning positional strategy for II. We deﬁne

σ ∶ P<∞ → P(ω) X ↦ τ(X, n). n∈ω

101 This is an ultrapositional strategy since, if X ⊆ Y, with X,Y ∈ P<∞(ω), we have

σ(X) = τ(X, n) ⊆ τ(Y, n) = σ(Y ). n∈ω n∈ω

Moreover, it is winning since, if I plays a sequence (Xn)n∈ω leading to X, the answer of II is

σ(Xn) = τ(Xn, k) = τ(Xn, n) = f(X). n∈ω n,k∈ω n∈ω

Hence, we have the following equivalence.

Theorem 6.47. Let f ∶ P(ω) → P(ω). The following are equivalent. 1. f is continuous,

2. II has a winning ultrapositional strategy in G∞(f),

3. II has a winning positional strategy in G<∞(f). Proof. This is an easy consequence of Proposition 6.43 and Proposition 6.46.

From the beginning of this Section, we defined games that characterize continuous functions. But we want to consider continuous reductions on subsets of P(ω). Therefore, as in the case of the Wadge games, we want to define games about subsets, and not about functions. That is what we propose to do now. We just need a slight modification. Indeed, we see that the function f is just taken into account for the winning condition. Therefore, even if we do not have any function, we can play the game. That is what we use now.

Deﬁnition 6.48. Let A, B ⊆ P(ω). We deﬁne the game G∞(A, B) as follows. The game is played exactly in the same way as G∞(f) except that we change the winning condition. This time, II wins the game if and only if ((X ∈ A) ↔ (Y ∈ B)).

Since ultrapositional strategy in G∞(f) is completely independent of f, the deﬁnition of such a strategy does not change. The same is also true for the notion of a winning ultrapositional strategy. Now, we want to link the winning ultrapositional strategy in G∞(A, B) with the notion of continuous reducibilty. That is the next result.

Proposition 6.49. Let A, B ⊆ P(ω). Then, A ≤W B if and only if II has a winning ultrapositional strategy in the game G∞(A, B).

Proof. Suppose A ≤W B, then there exists a continuous function f ∶ P(ω) → 1 P(ω) such that f − (B) = A. We deﬁne

σ ∶ P<∞(ω) → P(ω) X ↦ f(X).

102 <∞ Therefore, σ = fSP (ω). By Lemma 6.42, this is an increasing function, hence an ultrapositional strategy. Now, suppose I plays a sequence (Xn)n∈ω which leads to X. II answers with

σ(Xn) = f(Xn) = f Xn = f(X). n∈ω n∈ω n∈ω Therefore, since (X ∈ A ↔ f(X) ∈ B), we have that σ is a winning ultrapositional strategy. Conversely, suppose that σ is a winning ultrapositional strategy for II in G∞(A, B). By Lemma 6.41, σˆ is continuous. Moreover, suppose I plays a sequence (Xn)n∈ω which leads to X. II answers with

σ(Xn) = σˆ(X). n∈ω

Therefore, (X ∈ A ↔ σˆ(X) ∈ B), and σˆ witnesses A ≤W B. With this game characterization of continuous reducibility, we hope to have a powerful tool in the investigation of the Wadge hierarchy in P(ω). Of course, a similar adaptation to the one of the game G∞(f) can also be made in the case of G<∞(f). For completeness, that is what we do now.

Deﬁnition 6.50. Let A, B ⊆ P(ω). We deﬁne the game G<∞(A, B) as follows. The game is played exactly in the same way as G<∞(f), except that we change the winning condition. This time, II wins the game if ((X ∈ A) ↔ (Y ∈ B)).

Again, (winning) positional strategy in G<∞(f) is completely independent of f, hence the deﬁnition of such a strategy does not change. We can also link the notion of continuous reducibility with the positional winning strategy in G<∞(A, B). Proposition 6.51. Let A, B ⊆ P(ω). Then, II has a winning ultrapositional strategy in the game G∞(A, B) if and only if II has a winning positional strategy in the game G<∞(A, B).

Proof. Suppose II has a winning ultrapositional strategy σ in G∞(A, B). Fol- lowing this strategy, II has a winning strategy in G∞(σˆ). By Theorem 6.47, II has a winning positional strategy in G<∞(σˆ). This strategy is also winning in the game G<∞(A, B). The same argument also gives the converse implication. Therefore, we have a really nice equivalence.

Theorem 6.52. Let A, B ⊆ P(ω). The following are equivalent.

1. A ≤W B,

2. II has a winning ultrapositional strategy in G∞(A, B),

3. II has a winning positional strategy in G<∞(A, B). Proof. This is an easy consequence of Proposition 6.49 and Proposition 6.51.

Hence, the next idea is to investigate the Wadge hierarchy of P(ω) with these games, as we did for the Wadge hierarchy of ωω with the Wadge game.

103 6.6 First Results

We can start our investigation of the Wadge hierarchy of P(ω). We have a ﬁrst important result which shows that there is no self-dual set in P(ω). Proposition 6.53. Let A ⊆ P(ω), then A is non-self-dual. c Proof. Let A ⊆ P(ω). We show that II has no winning strategy in G<∞(A, A ). Towards a contradiction, suppose II has a winning positional strategy σ. Con- sider the run of the game where I plays the sequence

(∅, σ(∅), σ(σ(∅)),... ).

This corresponds to the game where I starts by playing ∅ and then copies what II plays. Of course, we get X = Y, which contradicts the fact that σ is winning. Schematically, this is I II

∅ ⊆ Y0 ⊆ Y0 ⊆ Y1 ⊆ Y1 ⊆ Y2 ⋮ ⋮

X = ⋃n∈ω Yn, Y = ⋃n∈ω Yn

Hence, we have a general statement for the Wadge hierarchy. Now, we will go in a more detailed study. First, we try to understand the open sets.

Proposition 6.54. All open sets in P(ω) are Wadge equivalent. Proof. Let U be an open set. Then, there exists P ⊆ P<∞(ω) such that

U = ↟ F. F ∈P

We show that U ≡W ↟ {0}. In the game G∞(U, ↟ {0}), the winning strategy for II is to play ∅ until I plays Xn such that there exists F ∈ P,F ⊆ Xn. At this moment, II plays {0} and wins the game, whatever is played after that. If I never plays such a set Xn, II ends up the game with Y = ∅ and wins the game. In the game G∞(↟ {0},U), the strategy for II is the same. Indeed, he just waits for I to play Xn such that 0 ∈ Xn, and then II plays some F ∈ P. If I never plays such a set Xn, II ends up the game with Y = ∅ and wins the game. We said earlier in this Chapter that there exist four incomparable subsets of P(ω) with respect to ≤W . To show this, we ﬁrst need a Lemma.

104 Lemma 6.55. In P(ω), {ω} ≡W P∞(ω).

∞ Proof. We just construct a winning strategy for II in both games G∞({ω}, P (ω)) ∞ and G∞(P (ω), {ω}). In the ﬁrst game, II plays the initial segment played by I. Formally,

σ(Xn) = {0, 1, . . . , m} ∶ {0, 1, . . . , m} ∩ Xn = {0, 1, . . . , m}. m∈ω

In the second game, II plays the initial segments of ω whose length is SXnS, formally, σ(Xn) = {0, 1,..., SXnS}. It is really easy to observe that these strategies are winning.

We need a major observation about the games G∞(A, B). We will use this observation a lot in the next proof. In the game G∞(A, B), suppose that II has a winning ultrapositional strategy σ. As I plays some Xn ∈ A, II plays directly some Yn ∈ B. Otherwise, σ(Xn) ∉ B, and I just plays Xm = Xn for all m ≥ n. Therefore, since the strategy is ultrapositional, σ(Xm) = Yn for all m ≥ n and II does not win the game. The same argument implies that, if Xn ∉ A, then σ(Xn) ∈ A. Now, we can prove the result. c Proposition 6.56. In P(ω), the subsets {{0}}, {{0}} , P∞(ω) and P<∞(ω) form an antichain for ≤W . Proof. Since we already know that there is no self-dual subsets, we just need to show that {{0}} ⊥W P∞(ω) and {{0}} ⊥W P<∞(ω). Actually, we show that II cannot have winning ultrapositional strategies in the four corresponding games. We also use that {ω} ≡W P∞(ω) as the previous Lemma allows us. In the game G∞({{0}}, {ω}), suppose σ is a winning ultrapositional strategy for II. By the observation before the statement of the Proposition, we have σ({0}) = ω. Therefore, if I starts with X0 = {0}, we have necessarily Y0 = ω = Y. To win the game, I just has to play X1 = {0, 1}. Hence, σ is not winning. By Theorem 6.52, it follows that {{0}} ≰W {ω}. In the game G∞({ω}, {{0}}), suppose σ is a winning ultrapositional strategy for II. We have necessarily σ({0, 1, . . . , n}) ≠ {0}. If, for all n ∈ ω, we have σ({0, 1, . . . , n}) = ∅, we obtain that

σ({0, 1, . . . , n}) = ∅, n∈ω which contradicts the fact that σ is winning. If there exists n0 ∈ ω such that σ({0, 1, . . . , n0}) ∉ {∅, {0}}, we obtain that

σ({0, 1, . . . , n}) ≠ {0}, n∈ω which contradicts the fact that σ is winning. Therefore, it suﬃces for I to play Xn = {0, 1, . . . , n}. By Theorem 6.52, we have {ω} ≰W {{0}}. <∞ In the game G∞({{0}}, P (ω)), suppose σ is a winning ultrapositional strategy for II. Then σ(∅) = A ∈ P∞(ω), and σ({0}) = B ∈ P<∞. This contradicts the fact that σ is increasing. By Theorem 6.52, it follows that {{0}} ≰W P<∞(ω).

105 <∞ Finally, in the game G∞(P (ω), {{0}}), suppose σ is a winning ultrapositional strategy for II. Then σ(F ) = {0} for all F ∈ P<∞(ω). Therefore,

σ({0, 1, . . . , n}) = {0}, n∈ω which contradicts the fact that σ is winning. Indeed, if I plays Xn = {0, 1, . . . , n}, II does not win the game. By Theorem 6.52, we have P<∞(ω) ≰W {{0}}. To be sure that the previous Proposition contradicts the capacity of admissible representations to give a good Wadge hierarchy, we need to show that these four sets are Borel. Since we can deﬁne them without the Axiom of Choice, we already now that they are not too complicated. However, as we saw in R, it is interesting to know at which level the bad behaviors occur. We have

0 {∅} = P(ω) ∖ ↟ {n} ∈ Σ2(P(ω)), n∈ω 0 {∅, {0}} = P(ω) ∖ ↟ {0, n} ∈ Σ2(P(ω)), ∗ n∈ω 0 {{0}} = {∅, {0}} ∖ {∅} ∈ Σ3(P(ω)), and <∞ 0 P (ω) = (↟ (F ∪ {n})∖ ↟ F ) ∈ Σ2(P(ω)). <∞ F ∈P (ω) n∈(ω∖F ) Therefore, it would be interesting to play the games at lower levels since, already at these levels, the hierarchy diﬀers from the one on the Baire space ωω.

106 Conclusion

The goal of this Master Thesis was to achieve a greater understanding of descriptive set theory, and especially of continuous reducibility. Using powerful game theoretical tools, we were able to understand the whole picture of the Wadge hierarchy on the Borel subsets of the Baire space ωω, which is given in the following nice picture.

[∅]W

⋯ ⋯ ⋯

ω [ω ]W cof(λ) = ω cof(λ) > ω

We also explained why such a picture is also the right one for all zero- dimensional Polish spaces. However, it is not the case for all Polish spaces as we saw with the real line. Indeed, in this space, there exist infinite antichains and infinite descending chains. Finally, we studied quasi-Polish spaces, in which P(ω) is universal. Using a Wadge-like game characterization of continuous function, we were able to show that the picture of the Wadge hierarchy on the Borel subsets of P(ω) is different from the one on the Borel subsets of the Baire space ωω. There are several ways of continuing this Master Thesis. A first well known way is to consider different types of reducibility other than the continuous one. For example, for each ordinal α < ω1, we can consider Baire class α functions −1 0 which are all the functions f ∶ X → X such that f (U) ∈ Σα(X) for all open sets U ⊆ X. A second way is to understand more deeply how the Wadge hierarchy on R is different. Indeed, we showed that bad behaviors occur already at level 0 D2(Σ1(R)). One question is whether or not these bad behaviors occur at each higher level? Another question is whether or not a slight modification of the binary relation ≤W can annihilate these bad behaviors? Or does such a slight modification annihilate the bad behaviors at each level? In this case, it would maybe be possible to have an alternative Wadge-like hierarchy which is nice. Finally, the study of continuous reducibility on quasi-Polish spaces remains mostly open. The question here is whether or not the games designed by Pr. Jacques Duparc allow us to reach nice conclusions, such as a nice picture of the Borels subsets of P(ω) equipped with the Scott topology. But that is another story we hope to tell in the future...

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