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Université Pierre et Marie Curie

École Doctorale de Sciences Mathématiques deParis Centre

Thèse de doctoratDiscipline : Mathématiques

présentée par

Rafael Zamora

Problèmes de séparation de relationsanalytiques

dirigée par Dominique Lecomte

Soutenue le 7 juillet 2015 devant le jury composé de :

M. Gabriel Debs Université Pierre et Marie Curie examinateurM. Dominique Lecomte Université Pierre et Marie Curie directeurM. Alain Louveau Université Pierre et Marie Curie examinateurM. Etienne Matheron Université d’Artois examinateurM. Julien Melleray Université Claude Bernard rapporteurM. Benjamin Miller Kurt Gödel research center examinateur

Institut de Mathématiquesde Jussieu Paris Rive Gauche4, place JussieuCase 24775252 Paris Cedex

UPMCEcole Doctorale de SciencesMathématiques de Paris Centre4 place Jussieu75252 Paris Cedex 05Boite courrier 290

Table des matières

Remerciements 7

1 Introduction 9La hiérarchie borélienne . . . . . . . . . . . . . . . . . . . . . . . . . . 9La hiérarchie de Wadge . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Introduction 15The Borel hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15The Wadge hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Preliminaries 21Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Injective reductions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Effective theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Description of Wadge classes . . . . . . . . . . . . . . . . . . . . . . . 32

4 Separability by rectangles 35First results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Topological characterizations . . . . . . . . . . . . . . . . . . . . . . . 37Separation by a Σ0

1 ×Σ01 set . . . . . . . . . . . . . . . . . . . . . . . 40

Separation by a Π01 ×Σ0

1 set . . . . . . . . . . . . . . . . . . . . . . . 42Separation by a Π0

ξ ×Π0ξ set . . . . . . . . . . . . . . . . . . . . . . . 43

Separation by a (Σ01 ×Σ0

2)σ set and by a Π01 ×Π0

2 set . . . . . . . . . 46Separation by a Π0

2 ×Σ01 set . . . . . . . . . . . . . . . . . . . . . . . 49


2 set . . . . . . . . . . . . . . . . . . . . . . . 52

5 Separability by pot(Γ) sets 55Topological characterizations . . . . . . . . . . . . . . . . . . . . . . . 55Separation by a pot(∆0

2) set . . . . . . . . . . . . . . . . . . . . . . . . 57Separation by a Dξ(Σ0

1) set . . . . . . . . . . . . . . . . . . . . . . . . 62Separation by a pot(∆(Dξ(Σ0

1))) set . . . . . . . . . . . . . . . . . . . 68Separability of oriented graphs by Γ sets . . . . . . . . . . . . . . . . . 77

References 79

5

Remerciements

J’adresse mes remerciements aux personnes qui m’ont aidé dans la réalisationde cette thèse.

Je comprends le grand travail qu’entraîne la direction d’une thèse, surtoutquand il s’agite d’un directeur dévoué : je remercie infiniment Dominique Le-comte. Sans les heures qu’il y a dédiées, cette thèse n’aurait pas été possible.

Je voudrais remercier les rapporteurs, Alexander Kechris et Julien Melleray,pour avoir lu soigneusement cette thèse. Alexander, je voudrais aussi le remercierpour ses conseils et pour la chaleur avec laquelle il m’a accueilli pendant monséjour à Caltech. Envers Julien je suis très reconnaissant pour sa longue liste decorrections qui m’a permis d’améliorer cette thèse.

Mes plus sincères gratitudes à Gabriel Debs, Dominique Lecomte, Alain Lou-veau, Étienne Matheron, Julien Melleray et Benjamin Miller pour avoir acceptéde faire partie du jury.

Une des aides les plus fondamentales a été donnée par l’Universidad de CostaRica à travers de l’OAICE. J’espère pouvoir repayer dans la même mesure.

Je remercie l’Escuela de Matemática de l’UCR pour la formation qu’ellem’a offerte. Très particulièrement, je tiens à remercier Eugenio Chinchilla etWilliam Alvarado, qui m’ont introduit dans mon domaine. Je suis très recon-naissant envers l’ensemble des professeurs de LMFI pour leurs enseignements,ainsi qu’envers les équipes d’analyse et de logique de l’IMJ pour leur accueilchaleureux pendant ces derniers quatre ans. Notablement, je voudrais aussi re-mercier le groupe de travail en théorie descriptive des ensembles pour toutes lesséances enrichissantes.

Dans un thèse il n’y as pas que les maths. Il y a aussi le stress et la fatigue.Et avec ceci, il y a la bonne chance de avoir connu un grand groupe de per-sonnes qui ont aide a soulager ces afflictions. Un groupe qu’on a déjà démontrequ’il est le meilleur groupe du monde, particulièrement dans [NG14], [BM14],[SM15], [ST14] et [JZ15]. Donc, Kuba, Ioana, Nico, Seb, Benoît, Romain, Nadja,Luis, et Shahin (le dernier c’est le meilleur, hein ?) : je vous tiendrai au coeurpour toujours. Je peux pas oublier tous les soirées passées entre bières et discus-sions avec Silvain, Shadi, Jean Phillipe, Nikita, Lulu, Benjamin, Armen, Jehane,Catherine, Clari, Alonso, Tristan, Cristobal, Barbara et Guillaume.

En Costa Rica hay mucha gente que no puedo olvidar agradecer. Nuestratransción a la vida francesa no hubiera sido posible sin Walsh y Ermelinde.Alberto, fue un placer haber estado de este lado del atlántico con vos. A todosmis compañeros del bachillerato, muchas gracias por la mate, los viajes y lacomida. A Don Victor, Doña Samaria y Sandy, muchas gracias por habermeaceptado dentro de su familia, y por haber estado ahí para Samy y para mí.

Mi familia siempre ha estado ahí, celebrando mis logros. Pero deben saber

7

8

que estos logros no hubieran sido posibles sin ellos. Muchas gracias a todos mistíos, Agus, Roberto, Diego y Juan, a sus esposas e hijos, y a mis abuelas Estelae Ileana. Adri, Telis y Tomás, les deseo las mejores de las suertes en su vida,como ustedes siempre me las han deseado a mí. A mis padres, Ileana y Rafael,les agradezco infinitamente por haberme dado todas las herramientas que henecesitado para llegar hasta aqui.

Cualquier cosa que pueda decir en este último párrafo quedará corto. Nohay forma en la que puedo expresar mi agradecimiento por todos los momentosen que ella ha estado a mi lado, en los buenos y en los malos, en los de triunfoo de fracaso, en los estresantes o en los de ocio. Cada día, el solo despertarmea su lado es un viaje en el tiempo y en el espacio. Esta tesis es ciertamente unaprueba de lo bien que ella hace en mi vida. Por esto y mucho más, me faltan laspalabras para agradecer a mi “Doctor & companion”: Samy. Gracias por todo.

Bibliographie des remerciements

NG14 N. Gastineau; Partitionnement, recouvrement et colorabilité dans les gra-phes, Doctoral thesis. (2014)

BM14 B. Monin; Higher computability and randomness, Doctoral thesis. (2014)SM15 S. Montenegro; Théorie des modèles des corps pseudo-réelsclos et pseudo-

p-adiquement clos, Doctoral thesis. (2015).ST14 S. Tavenas; Bornes inférieures et supérieures dans les circuits arithmé-

tiques, Doctoral thesis. (2014)JZ15 J. Zwolakowski; A formal approach to distributed application synthesis

and deployment automation, Doctoral thesis. (2015)

Chapitre 1

Introduction

La hiérarchie borélienneLa théorie descriptive des ensembles est l’étude des ensembles définissables

des nombres réels. La plupart des propriétés dans ce contexte dépent uniquementde certains aspects topologiques de l’espace des nombres réels. Ils peuvent doncêtre étudies dans un cadre plus général. On dit qu’un espace topologique estpolonais s’il est séparable et métrisable avec une métrique complète.

Les premiers ensembles à être étudiés ont été les ensembles boréliens, qui ontété introduits dans le cadre de la théorie de la mesure. La manière habituelle dedéfinir la classe des ensembles boréliens est par un argument descendant :c’est la plus petite tribu contenant les ensembles ouverts. Cependant, nous pou-vons aussi les définir par un argument ascendant, qui donne une meilleure idéede la définissabilité de ces ensembles. Ainsi, nous obtenons la hiérarchie bo-rélienne d’un espace polonais X en définissant, par récurrence, pour chaque0 < ξ < ω1,

Σ01(X) := A ⊆ X|A est ouvert,

Π0ξ(X) := A ⊆ X|X\A ∈ Σ0

ξ,si ξ ≥ 2, Σ0

ξ(X) := ∪i∈ωAi|∀i ∈ ω∃ν < ξ(Ai ∈ Π0ν).

Un point important dans l’étude de ces ensembles fut la découverte parSouslin d’une erreur dans un article de Lebesgue. Son erreur était de supposerque la projection d’un ensemble borélien est un ensemble borélien. Cependant, cen’est pas le cas en général. On obtient la classe des ensembles analytiques,définie par :

Σ11(X) := A ⊆ X|∃B ⊆ X × ωω B est borélien

∧ ∀x ∈ X(x ∈ A⇔ ∃y ∈ ωω(x, y) ∈ B),

et la classe des ensembles coanalytiques, définie par :

Π11(X) := A ⊆ X|X\A ∈ Σ1

1.

Pour compléter les classes boréliennes on definit les classes ambigües :

∆0ξ(X) := Σ0

ξ(X) ∩Π0ξ(X),

9

10 Chapitre 1. Introduction

et la classe analogue pour la classe des ensembles analytiques,

∆11(X) := Σ1

1(X) ∩Π11(X).

On remarque les inclusions suivantes.

Σ01 Σ0

2 · · · Σ11

∆01 ∆0

2 · · · ∆11

Π01 Π0

2 · · · Π11

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

Un résultat important sur les ensembles analytiques est le théorème suivant,dû à Luzin.

Theorem 1.1. (Luzin) Soient X un espace polonais, et A et B des sous-ensembles analytiques disjoints de X. Il existe un sous-ensemble borélien C deX tel que

A ⊆ C ∧ C ∩B = ∅. (S)

Une conséquence importante de ce théorème est le fait que A est ∆11 si et

seulement s’il est borélien. Si (S) est satisfait, nous disons que A est séparablede B par C. Une question naturelle qui suit est de savoir si nous pouvonsborner la complexité de l’ensemble C. En d’autres termes, y a-t-il un moyenpour caractériser si A est séparable de B par un ensemble dans une des classesΣ0ξ ? Ce problème a d’abord été résolu par Hurewicz dans le cas ξ = 2, puis par

Louveau et Saint Raymond dans [LSR87].

Theorem 1.2. (Louveau, Saint Raymond) Soient ξ un ordinal denombrable nonnul, S ∈ Σ0

ξ(2ω)\Π0ξ(2ω), X un espace polonais, et A, B des sous-ensembles

analytiques disjoints de X. Exactement une des assertions suivantes est satis-faite :

1. A est séparable de B par un ensemble Π0ξ(X),

2. il existe une fonction continue f : 2ω → X telle que S ⊆ f−1(A) et2ω\S ⊆ f−1(B).

Soit ∆ une classe, et ≤ une relation binaire sur ∆. Rappelons que ≤ est unquasi-ordre si ≤ est réflexive et transitive. Il existe une relation d’équivalencenaturelle associée à ≤. Nous disons que A est équivalent à B si A ≤ B etB ≤ A.

Soit C une sous-classe de ∆ dont le complémentaire ∆\C est héréditaire(c’est-à-dire, (A ∈ ∆\C ∧ B ∈ ∆ ∧ B ≤ A) → B ∈ ∆\C), et B ⊆ C. Nousdisons que B est une base pour C si, pour chaque C ∈ C, il existe B ∈ B telque B ≤ C. Une base sera d’autant plus intéressante qu’elle sera petite, donc engénéral nous allons demander qu’elle soit une ≤-antichaîne (c’est-à-dire quedeux éléments différents ne sont pas ≤-comparables).

On dit que A ∈ C est minimal dans C, si ∀B ∈ C((B ≤ A) → (A ≤

B)). Notons que tous les éléments d’une antichaîne base B pour C doivent être

minimaux. En effet, si A ∈ B et B ∈ C satisfont B ≤ A, alors il existe C ∈ B telque C ≤ B. Alors C ≤ A. Comme B est une antichaîne, A = C. Inversement, un

La hiérarchie de Wadge 11

élément minimal doit être équivalent à un et un seul élément de toute antichaînebase.

Dans ce cadre, soit Γ = (X,A,B)|X est polonais, A,B ∈ Σ11(X), A ∩ B =

∅, et ≤ le quasi-ordre défini par (X,A0, A1) ≤ (X ′, A′0, A′1) s’il existe unefonction continue f : X → X ′ telle que ∀ε ∈ 2(Aε ⊆ f−1(A′ε)). Ensuite, lethéorème 1.2 indique que (2ω, Sξ, 2ω\Sξ) est une antichaîne base pour la classedes triplets (X,A0, A1) tels que A0 n’est pas séparable de A1 par un ensembleΠ0ξ .Ce type de résultats est connu sous le nom de dichotomie de type Hurewicz :

soit un ensemble est simple, ou il réduit un exemple compliqué particulier. Nouscherchons à trouver des résultats de ce type pour différentes notions de com-plexité et différents quasi-ordres sur le produit de deux espaces polonais. Unexemple important est le théorème suivant, dû à Lecomte. Nous notons, pourtout ensemble X, Diag(X) := (x, x)|x ∈ X la diagonale de X.

Theorem 1.3. (Lecomte) Il existe une relation H0 sur 2ω telle que, pour tousespaces polonais X, Y et pour tous sous-ensembles analytiques disjoints A,Bde X × Y , exactement une des assertions suivantes est satisfaite :

1. A est séparable de B par une réunion ∪i∈ω(Ci ×Di), où Ci et Di sontdes ensembles boréliens,

2. il existe des fonctions continues f : 2ω → X et g : 2ω → Y telles queDiag(2ω) ⊆ (f × g)−1(A) et H0 ⊆ (f × g)−1(B).

On dit que C × D est un rectangle. Pouvons-nous trouver un résultatsimilaire pour les réunions finies de rectangles ? C’est une question importantequi, nous le verrons, est liée à l’étude des relations analytiques sur un espacepolonais. Le cas d’un rectangle borélien peut être facilement déduit du résultatde Luzin. Au vu du théorème 1.2, il est naturel de se demander s’il existe unedichotomie pour la séparabilité par un rectangle Σ0

ξ . Dans le chapitre 4 nousétudierons cette question. Etant donnés deux classes Γ et Γ′ des ensemblesboreliens, soit Γ× Γ′ := A× B|A ∈ ΓB ∈ Γ′. Notre théorème principal pourle chapitre 4 est le suivant.

Theorem 1.4. Soient Γ,Γ′ ∈ Σ01,Π0

1,Π02, ou Γ × Γ′ = Π0

1 × Σ02. Il existe

une antichaîne finie C contenue dans la classe des quadruplets de la forme(Z,W,C,D), où Z,W sont des espaces polonais, et C,D ⊆ Z × W sont dessous-ensembles analytiques disjoints tels que, pour tous espaces polonais X,Yet tous sous-ensembles analytiques disjoints A,B de X×Y , exactament une desassertions suivantes est satisfaite :

1. A est séparable de B par un ensemble Γ× Γ′,2. il existe (X,Y,A,B) ∈ C et des fonctions continues f : X→ X et g : Y→

Y tels que A ⊆ (f × g)−1(A) et B ⊆ (f × g)−1(B).

La hiérarchie de WadgeUn des quasi-ordres les plus étudiés sur P(ωω) est la réduction de Wadge,

définie par A ≤W B s’il existe une fonction continue f : ωω → ωω telle que A =f−1(B). On peut étendre cette approche aux espaces de dimension zéro. Martina montré dans [Mar75] que les jeux boréliens sont déterminés. Une conséquence


de ceci est que ce quasi-ordre est un bon quasi-ordre sur la classe des sous-ensembles boréliens de ωω, c’est-à-dire un quasi-ordre qui n’a pas d’antichaîneinfinie, ni de chaîne infinie strictement décroissante.

Ce quasi-ordre fournit la hiérarchie de classes de sous-ensembles boréliens deωω close par images inverses continues la plus fine. En d’autres termes, c’est lahiérarchie de complexité topologique la plus fine considérée en théorie descriptivedes ensembles.

Definition. Nous disons que Γ est une classe de Wadge s’il existe A ∈ ωωtel que Γ(X) = f−1(A)|f : X → ωω est continue, pour tout espace polonaiszéro-dimensionnel X.

Notation. Si A est un sous-ensemble d’un espace polonais X, nous écrivons¬A = X\A. Si Γ est une classe, nous écrivons Γ = ¬A|A ∈ Γ pour la classeduale de Γ. Aussi, ∆(Γ) := Γ ∩ Γ est la classe ambigüe associée à Γ. Nousdisons que Γ est auto-duale si Γ = Γ.

Nous voudrions introduire une notion analogue pour les espaces produits.Rappelons le théorème classique suivant.

Theorem 1.5. Soit (X, τ) un espace polonais, et (An)n∈ω une suite de sous-ensemble boréliens de X. Il existe une topologie polonaise τ ′ qui raffine τ telleque, pour tout entier n, l’ensemble An est ouvert-fermé dans τ ′.

On obtient comme corollaire que si B est une réunion dénombrable de rec-tangles boréliens, alors il existe des topologies polonaises τ ′X , τ ′Y , raffinent lestopologies originales sur X et Y , telles que B est ouvert pour τ ′X × τ ′Y . Unensemble ayant cette propriété est appelé potentiellement ouvert. Donc lethéorème 1.3 peut être vu comme une dichotomie pour la séparabilité des en-sembles analytiques par des ensembles potentiellement ouverts.

On peut étendre la définition d’ensemble potentiellement ouvert aux autresclasses de Wadge de boréliens. Lecomte a étendu le théorème 1.3 aux autresclasses de Wadge potentielles dans [Lec13]. Aussi, Debs a montré que dans cetteextension, on peut avoir des fonctions injectives pour les classes de boréliensnon-ambigûes contenant ∆0

3. En utilisant ce resultat, Lecomte à montré dans[Lec14] une dichotomie impliquant une notion de comparaison plus forte quecelle donnée par le théorème 1.3, sous certaines conditions naturelles.

Rappelons qu’un sous-ensemble d’un espace produit est localement dé-nombrable si toutes ses sections sont dénombrables. Cette définition est toutà fait naturelle et robuste. En particulier, on peut trouver des exemples de rela-tions localement dénombrables non séparables par un ensemble potentiellementdans Γ, pour Γ contenue dans Σ0

2.Cependant, les elements des antichaînes bases trouvée par Lecomte dans

[Lec13] ne sont pas localement dénombrables. Si nous exigeons injectivité, cesantichaînes bases ne fonctionnent pas. Cela fait partie des raisons pour lesquellesle résultat de Debs ne s’applique pas à ces classes. Nous nous concentrons surl’étude de ces classes manquantes. En particulier, nous avons trouvé de nouvellesantichaînes bases constituées d’exemples localement dénombrables.

Louveau a donné dans [Lou83] une description de toutes les classes de Wadgede boréliens en utilisant quelques opérations ensemblistes. Pour les cas que nousétudions, nous avons seulement besoin d’introduire une nouvelle opération. Rap-pelons que, pour chaque ordinal ν, il existe un ordinal limite λ, un entier n et

La hiérarchie de Wadge 13

ε ∈ 2, tous uniques, tels que ν = λ+ 2n+ ε, de sorte que nous pouvons étendrela congruence modulo 2 à tous les ordinaux.

Definition. (Hausdorff, Kuratowski) Soient X un espace polonais, 0 < ξ < ω1,et (Aν)ν<ξ une suite croissante de sous-ensembles de X. On définit

Dξ((Aν)ν<ξ) :=⋃

ν 6≡ξ mod 2Aν\

(∪θ<ν Aθ

).

Soit Γ une classe de sous-ensembles de X. La classe des ξ-differences desensembles dans Γ est définie par

Dξ(Γ) := Dξ((Aν)ν<ξ)|(Aν)ν<ξ suite croissante de sous-ensembles dans Γ.

Une conséquence de la description de Louveau est que les classes de Wadgecontenues dans Σ0

2 sont tous de la forme Dξ(Σ01), Dξ(Σ0

1), ou ∆(Dξ(Σ01)). Nous

montrons les dichotomies suivantes.

Theorem 1.6. •Soit Γ ∈ Dξ(Σ01)|0 < ξ < ω1∪Dξ(Σ0

1)|0 < ξ < ω1∪∆02.

Il existe une antichaîne finie C contenue dans la classe des pairs de la forme(C,D), où C,D sont des relations analytiques disjointes sur 2ω, telle que pourtout espace polonais X et toutes relations analytiques disjointes A0, A1 sur Xtelles que A0 ∪ A1 est localement dénombrable ou a symétrisation acyclique,exactement une des assertions suivantes est satisfaite,

1. A0 est séparable de A1 par un ensemble potentiellement dans Γ,2. il existe (C0, C1) ∈ C et des injections continues f, g : 2ω → X tels que,

pour tout ε ∈ 2, Cε ⊆ (f × g)−1(Aε).•Soit Γ ∈ ∆(Dξ(Σ0

1))|0 < ξ < ω1. Il existe une antichaîne finie C contenuedans la classe des pairs de la forme (C,D), où C,D sont des relations analy-tiques disjointes sur 2ω, telle que pour tout espace polonais X et toutes relationsanalytiques disjointes A0, A1 sur X telles que A0 ∪ A1 est contenue dans unerelation potentiellement fermé et localement dénombrable (ou à symétrisationacyclique), exactement une des assertions suivantes est satisfaite,

1. A0 est séparable de A1 par un ensemble potentiellement dans Γ,2. il existe (C0, C1) ∈ C et des injections continues f, g : 2ω → X tels que,

pour tout ε ∈ 2, Cε ⊆ (f × g)−1(Aε).

Pour démontrer ce théorème, nous allons introduire une nouvelle notion quigénéralise les relations localement dénombrables et les relations de symétriséacyclique.

Un résultat important que Louveau a obtenu dans [Lou83] est que la descrip-tion des classes ∆(Dθ(Σ0

ξ)) lorsque θ est successeur est différent de la descrip-tion lorsque θ est limite. Nous retrouvons une telle différence dans la taille desantichaînes du théorème 1.6. Nous montrons que pour les classes auto-duales,∆(Dθ(Σ0

1)), l’antichaîne a taille 3 si θ est successeur, et elle a taille 1 si θ estlimite. Cela contraste avec d’autres dichotomies de ce type. Par exemple, dansle cas de Dξ(Σ0

1), toutes les antichaînes ont taille 1.Les graphes orientés sont des relations qui n’intersectent pas leur inverse.

Dans [Lou], Louveau a étudié la complexité des graphes orientés. En particulier,il a prouvé la dichotomie suivante.


Theorem 1.7. (Louveau) Soit ξ = 1, 2. Il existe un graphe orienté borélien Gξsur 2ω tel que, pour tout espace polonais X et tout graphe orienté analytique Rsur X, exactement une des assertions suivantes est satisfaite :

1. G est séparable de G−1 par une relation potentiellement dans ∆0ξ,

2. il existe une fonction continue f : 2ω → X telle que Gξ ⊆ (f × f)−1(G).

Dans le chapitre 5, nous allons généraliser ce résultat à toutes les classes dela hiérachie borélienne, et toutes les classes de Wadge de boréliens.

Chapter 2

Introduction

The Borel hierarchyDescriptive set theory is the study of definable sets of reals. Most of the

properties one studies in this context depend at most on certain topological as-pects of the real line. As such, they can be generalized to a more general setting.We say that a topological space is Polish if it is separable and metrizable witha complete metric.

The first sets to be studied where the Borel sets, which were introduced inthe context of measure theory. The usual way to define the class of Borelsets is by a top-bottom argument: it is the smallest σ-algebra containing theopen sets. However, we can also define them by a bottom-top argument, whichprovides more light on the definability of these sets. Thus, we obtain the Borelhierarchy of a Polish space X by defining recursively, for every 0 < ξ < ω1,

Σ01(X) := A ⊆ X|A is open,

Π0ξ(X) := A ⊆ X|X\A ∈ Σ0

ξ,if ξ ≥ 2, Σ0

ξ(X) := ∪i∈ωAi|∀i ∈ ω∃ν < ξ(Ai ∈ Π0ν).

One important point in the study was the discovery by Suslin of a mistakein a paper by Lebesgue. The mistake was that Lebesgue claimed that theprojection of a Borel set is a Borel set. However, this is not generally the case.We obtain the class of analytic sets, defined by:

Σ11(X) := A ⊆ X|∃B ⊆ X × ωω Borel ∀x ∈ X(x ∈ A⇔ ∃y ∈ ωω(x, y) ∈ B),

and the class of co-analytic sets, defined by:

Π11(X) := A ⊆ X|X\A ∈ Σ1

1,

Completing the Borel classes, are the so called ambiguous classes:

∆0ξ(X) := Σ0

ξ(X) ∩Π0ξ(X),

∆11(X) := Σ1

1(X) ∩Π11(X).

We point out the following inclusions, that hold for any Polish space.

15

16 Chapter 2. Introduction

Σ01 Σ0

2 · · · Σ11

∆01 ∆0

2 · · · ∆11

Π01 Π0

2 · · · Π11

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

⊆

One important result regarding analytic sets is the following theorem, dueto Luzin.

Theorem 2.1. (Luzin) Let X be a Polish space, and A and B be disjointanalytic subsets of X. There is a Borel subset C of X such that

A ⊆ C ∧ C ∩B = ∅ (S)

An important consequence of this theorem states that A is a ∆11 set if and

only if it is Borel. Whenever (S) holds, we say that A is separable from B byC. A natural question that follows, is whether we can bound the complexity ofthis set C. In other words, is there a way to characterize whether A is separablefrom B by a set in a Borel class Σ0

ξ . This was first answered by Hurewicz in thecase ξ = 2, and it was expanded by Louveau and Saint Raymond in [LSR87].

Theorem 2.2. Let 0 < ξ < ω1, S ∈ Σ0ξ(2ω)\Π0

ξ(2ω), X be a Polish space, andA and B be disjoint analytic subsets of X. Exactly one of the following holds:

1. A is separable from B by a Π0ξ(X) set.

2. There is a continuous function f : 2ω → X such that S ⊆ f−1(A) and2ω\S ⊆ f−1(B).

Let ∆ be a class, and ≤ be a binary relation on ∆. Recall that ≤ is a quasi-order if it is reflexive and transitive. Notice there is a natural equivalencerelation associated with ≤. We say that A is equivalent to B if A ≤ B andB ≤ A.

Let C be a subclass of ∆ whose complement ∆\C is hereditary (i.e., (A ∈∆\C ∧B ∈ ∆ ∧B ≤ A)→ B ∈ ∆\C), and let B ⊆ C. We say that B is a basisfor C if, for every C ∈ C, there is B ∈ B such that B ≤ C. A basis will be asinteresting as it is small, so in general, we will ask for it to be a ≤-antichain(i.e., any two elements are not ≤-comparable).

We say that A ∈ C isminimal in C, if ∀B ∈ C((B ≤ A)→ (A ≤ B)

). Notice

that every element of an antichain basis B for C must be minimal. Indeed, ifA ∈ B, and B ∈ C satisfies B ≤ A, then there is C ∈ B such that C ≤ B. ThenC ≤ A. As B is an antichain, A = C. Conversely, a minimal element must beequivalent to one and only one element of any antichain basis.

In this frame, let Γ = (X,A,B)|X is Polish, A,B ∈ Σ11(X), A ∩ B = ∅,

and ≤ be the quasi-order defined by (X,A0, A1) ≤ (X ′, A′0, A′1) if there is acontinuous function f : X → X ′ such that ∀ε ∈ 2(Aε ⊆ f−1(A′ε)). ThenTheorem 2.2 states that (2ω, Sξ, 2ω\Sξ) is an antichain basis for the class oftriples (X,A0, A1) such that A0 is not separable from A1 by a Π0

ξ set.This type of result is known as Hurewicz-like dichotomies: either a set is

simple, or it reduces a specific complicated example. We will be interested in

The Wadge hierarchy 17

finding results of this type for different notions of complexity and different quasi-orders on the product of two Polish spaces. An important example of this is thefollowing theorem by Lecomte. We write, for any set X, Diag(X) := (x, x)|x ∈X for the diagonal of X.

Theorem 2.3. (Lecomte) There is a relation H0 on 2ω such that, for any Polishspaces X,Y and any disjoint analytic subsets A,B of X ×Y , exactly one of thefollowing holds:

1. A is separable from B by a set of the form ∪i∈ω(Ci ×Di), where Ci andDi are Borel sets.

2. There are continuous functions f : 2ω → X and g : 2ω → Y such thatDiag(2ω) ⊆ (f × g)−1(A) and H0 ⊆ (f × g)−1(B).

We say that C × D is a rectangle. Can we find a similar result for finiteunions of rectangles? This is an important question that, we will see, is related tothe study of analytic relations on a Polish space. The case of one Borel rectanglecan be easily deduced from Lusin result. After Theorem 2.2, it is natural toask whether there is a dichotomy for the separability by a Σ0

ξ rectangle. InChapter 4 we will study this question. Given two classes Γ and Γ′ of Borel sets,let Γ × Γ′ := A × B|A ∈ ΓB ∈ Γ′. Our main theorem in Chapter 4 will bethe following.

Theorem 2.4. Let Γ,Γ′ ∈ Σ01,Π0

1,Π02, or Γ × Γ′ = Π0

1 × Σ02. There is a

finite antichain C contained in the class of quadruples of the form (Z,W,C,D),where Z,W are Polish spaces, and C,D ⊆ Z ×W are disjoint analytic subsets,such that, for any Polish spaces X,Y and any disjoint analytic subsets A,B ofX × Y , exactly one of the following holds:

1. A is separable from B by a Γ× Γ′ set.2. There are (X,Y,A,B) ∈ C and continuous functions f : X → X and

g : Y→ Y such that A ⊆ (f × g)−1(A) and B ⊆ (f × g)−1(B).

The Wadge hierarchyOne of the most studied quasi-orders on P(ωω) is Wadge reducibility, defined

by A ≤w B if there is a continuous function f : ωω → ωω such that A = f−1(B).One can extend this to zero-dimensional spaces. Martin showed in [Mar75] thatBorel games are determined. A consequence of this is that this quasi-orderis a well quasi-order on the Borel sets, i.e. a quasi-order that has no infiniteantichains, nor infinite strictly descending chains.

Also, this quasi-order produces the finest hierarchy of classes of Borel subsetsof ωω closed by inverse images by continuous functions. In other words, it is thefinest hierarchy of topological complexity considered in descriptive set theory.

Definition. We say that Γ is a Wadge class if there exists A ∈ ωω such thatΓ(X) = f−1(A)|f : X → ωω is continuous for any zero-dimensional Polishspace X.

Notation. If A is a subset of a Polish space X, we write ¬A = X\A. If Γ is aclass, we write Γ = ¬A|A ∈ Γ for the dual class of Γ. ∆(Γ) := Γ ∩ Γ is theambigous class associated with Γ. We say that Γ is self-dual if Γ = Γ.


We would like to introduce an analogous notion for product spaces. Recallthe following classical theorem.

Theorem 2.5. Let (X, τ) be a Polish space, and (An)n∈ω a sequence of Borelsubsets of X. There is a Polish topology τ ′ refining τ such that, for all n ∈ ω,the set An is clopen in τ ′.

As a corollary, a countable union B of Borel rectangles has the property thatthere are Polish topologies τ ′X , τ ′Y refining the original topologies on X and Ysuch that B is open in τ ′X × τ ′Y . A set with this property is called potentiallyopen. So Theorem 2.3 can be rewritten as a dichotomy for separability ofanalytic sets by potentially open sets.

One can extend the definition of potentially open to the other Wadge classesof Borel sets. Lecomte extended Theorem 2.3 to the other potential Wadgeclasses in [Lec13]. Also, Debs found that in this extension, one can have in-jective functions, for the Borel non-ambigous classes containing ∆0

3. This al-lowed Lecomte to show in [Lec14] a dichotomy involving a notion of comparisonstronger than that provided by Theorem 2.3, under some natural conditions.

Recall that a set of a product space is locally countable if all its sectionsare countable. This definition is quite natural and robust. In particular, onecan find examples of locally countable relations not separable by a potentiallyΓ set, for Γ contained in Σ0

2.However, the antichain bases found by Lecomte in [Lec13] are not locally

countable. If we demand injectivity, these antichain bases do not work. This ispart of the reason Debs result doest not work for these classes . We focus onthe study of these missing classes. In particular, we found new antichain basesconsisting of locally countable examples.

Louveau provided in [Lou83] a description of all Wadge classes of Borel setsusing a handful of operations on sets. For the cases we study, we only need tointroduce one new operation. Recall that for each ordinal ν there are uniqueλ limit, n finite and ε ∈ 2 such that ν = λ + 2n + ε, so that we can extendcongruence modulo 2 to all ordinals.

Definition. (Hausdorff, Kuratowski) Let X be a Polish space, 0 < ξ < ω1, and(Aν)ν<ξ be an increasing sequence of subsets of X. We define

Dξ((Aν)ν<ξ) :=⋃

ν 6≡ξ mod 2Aν\

(∪θ<ν Aθ

).

Let Γ be a class of subsets of X. We define the class of ξ-differences ofsets of Γ by

Dξ(Γ) := Dξ((Aν)ν<ξ)|(Aν)ν<ξ an increasing sequence of sets in Γ.

A consequence of Louveau’s description, is that Wadge classes contained inΣ0

2 are all of the form Dξ(Σ01), Dξ(Σ0

1), or ∆(Dξ(Σ01)). We were then able to

find the following dichotomy.

Theorem 2.6. •Let Γ ∈ Dξ(Σ01)|0 < ξ < ω1∪Dξ(Σ0

1)|0 < ξ < ω1∪∆02.

There is a finite antichain C contained in the class of pairs of the form (C,D),where C,D are disjoint analytic relations on 2ω, such that for any X Polishspace and A0, A1 disjoint analytic relations on X such that A0 ∪ A1 is locallycountable or has an acyclic symmetrization, exactly one of the following holds:

The Wadge hierarchy 19

1. A0 is separable from A1 by a potentially Γ set.2. There are (C0, C1) ∈ C and f, g : 2ω → X injective continuous such that

Cε ⊆ (f × g)−1(Aε) for ε ∈ 2.•Let Γ ∈ ∆(Dξ(Σ0

1))|0 < ξ < ω1. There is a finite antichain C contained inthe class of pais of the form (C,D), where C,D are disjoint analytic relationson 2ω, such that for any X Polish space and A0, A1 disjoint analytic relationson X such that A0 ∪ A1 is contained in a locally countable (or has an acyclicsymmetrization) potentially closed relation, exactly one of the following holds:

1. A0 is separable from A1 by a potentially Γ set.2. There are (C0, C1) ∈ C and f, g : 2ω → X injective continuous such that

Cε ⊆ (f × g)−1(Aε) for ε ∈ 2.

To prove this theorem, we would introduce a new notion that generalizesrelations that are locally countable and relations that have acyclic symetrization.

An important result Louveau obtains in [Lou83] is that the description of theclasses ∆(Dθ(Σ0

ξ)) when θ is successor is different from the description when θis limit. We recover such a difference in the size of the antichains from Theorem2.6. We show that for the self-dual classes ∆(Dξ(Σ0

1)), the antichain has size 3if ξ is sucessor, and has size 1 if it is limit. This contrasts with other dichotomiesof this type. For example, in the case of Dξ(Σ0

1) sets all antichains have size 1.Oriented graphs are relations that do not intersect their inverse. In [Lou],

Louveau studies the complexity of oriented graphs. In particular, he proved thefollowing dichotomy.

Theorem 2.7. (Louveau) Let ξ = 1, 2. There is a Borel oriented graph Gξ on2ω such that for any Polish space X and any analytic oriented graph R on X,exactly one of the following holds:

1. G is separable from G−1 by a potentially ∆0ξ set.

2. There is a continuous function f : 2ω → X such that Gξ ⊆ (f×f)−1(G).

We will provide a generalization of this result to all Borel classes, and to allWadge classes of Borel sets (see Theorem 5.26).


Chapter 3

Preliminaries

The notation on this manuscript follows [Kec95]. In particular, if s ∈ A<ω,then Ns := α ∈ Aω|s v α. In some cases we will work with several differentsets for A, in which case we will write Ns(Aω).

RelationsBy a relation on X we understand a subset R of X2. We say that R is in a

class Γ we mean that R ∈ Γ(X2). The study of relations has been a central partof descriptive set theory in at least the last 30 years. We first note the followingnotation.

We will work with several directed graphs and their symmetrizations. Wewill use the following definitions for these concepts, which can be stated for anyrelation.

Definition. Let R be a relation on X.1. R−1 := (y, x) ∈ X2|(x, y) ∈ R,2. s(R) := R ∪R−1,3. An R-path is a finite sequence (x0, ..., xp) of elements of X such that∀i < p(xi, xi+1) ∈ R. We say R is connected if for any x, y ∈ X thereis a path (x0, ..., xp) with x0 = x and xp = y.

4. If (x0, ..., xp) is an injective path with (xp, x0) ∈ R and p ≥ 2 we saythat (x0, ..., xp, x0) is an R-cycle (we omit R when there is no chance ofconfusion). We say R is acyclic if it has no cycles.

5. R is s-acyclic if s(R) is acyclic. R is s-connected if s(R) is.6. If R is a relation on 2ω, then B(R) := (0α, 1β)|(α, β) ∈ R is its bi-

partite version.7. R is an equivalence relation if R is transitive, symmetric and reflexive.8. R is a directed graph if it is irreflexive. It is a graph if it is irreflexive

and symmetric. It is an oriented graph if it is irreflexive and antisym-metric.

9. The horizontal section of R at x is the set Rx := y ∈ X|(x, y) ∈ R.The vertical section of R at y is the set Ry := x ∈ X|(x, y) ∈ R.

21

22 Chapter 3. Preliminaries

We say R is locally countable if both Rx and Ry are countable for anyx, y ∈ X.

One of the main areas has been the study of equivalence relations. Forexample, the following theorem by Silver generalizes the Perfect Set theoremfor Borel sets.

Theorem 3.1. (Silver) Let X be a Polish space and E a co-analytic equivalencerelation on X. Exactly one of the following holds:

1. E has at most ω equivalence classes.2. There is a continuous injective function f : 2ω → X such that ∆(2ω) =

(f × f)−1(E).

Notice that from 2 one can define a quasi-order on the class of pairs (X,E)such that X is a Polish space and E is a definable equivalence relation on X.We say that (X,E) reduces to (X ′, E′) and write (X,E) ≤B (X ′, E′) (E ≤B E′

if there is no chance of confusion), if there is a Borel function f : X → X ′

such that E = (f × f)−1(E′). If the function is continuous, we write (X,E) ≤c(X ′, E′). If the function is injective we either write (X,E) vB (X ′, E′) or(X,E) vc (X ′, E′) accordingly. This is known as Borel reducibility and it hasbeen central on the study of relations on Polish spaces.

Some other work was done by Glimm and Effros in the 1960’s and laterexpanded by Harrington, Kechris and Louveau in the following dichotomy in[HKL90]. Let E0 := (α, β) ∈ 2ω × 2ω|∃n ∈ ω∀m > nα(m) = β(m).

Theorem 3.2. (Harrington, Kechris, Louveau) Let X be a Polish space, andE a Borel equivalence relation on X. Exactly one of the following holds:

1. E ≤B ∆(2ω).2. E0 ≤B E.

Moreover, 1 is equivalent to1’. There is a Polish topology σ extending that of X such that E is closed in

(X2, σ × σ).

Notice that 1′ defines a new class of Borel sets of X2. First, we recall thefollowing notation.

Notation. 1. Suppose that for each Polish space X, we have defined asubset Γ(X) ⊆ P(X). We define Γ := ∪Γ(X)|X is Polish.

2. If Γ is a class, we let Γ(X) := A ⊆ X|A ∈ Γ.

Definition. (Louveau) Let X and Y be Polish spaces and Γ be a class of Borelsets. We define the class pot(Γ)(X×Y ) as the class of Borel subsets A of X×Ysuch that there are Polish topologies τX and τY refining the topologies of X andY respectively, such that A ∈ Γ(X × Y, τX × τY ).

One important point here is that A must be in Γ for the new product topol-ogy, since we can refine any Polish topology to another Polish one to con-vert a Borel set into a clopen set. Thus, for example, the diagonal of an un-countable Polish space X is not pot(Σ0

1). Else, since τX is second countable,∆(X) = ∪i∈ω(U0

i × U1i ), for U0

i , U1i open in τX . However, the only rectangles

contained in the diagonal are the singletons (x, x), contradicting uncountabil-ity. Another result we will use several times is the following.

Relations 23

Lemma 3.3. Let Γ be a Wadge class of Borel sets or ∆0ξ for 0 < ξ < ω1, (X, τ)

be a Polish space and A ∈ pot(Γ)(X ×X). There is a Π02 dense subset H of X

such that A ∩H2 ∈ Γ(H2).

Proof. Let τ ′ be a Polish topology that refines de topology of X and such thatA ∈ Γ(X ×X, τ ′ × τ ′). Notice that any O ∈ τ ′ is Borel in τ by [Kec95, 15.4],so O has the Baire property by [Kec95, 8.22]. Consider the identity functionid : (X, τ)→ (X, τ ′). By [Kec95, 8.38], there is H a dense Π0

2 set such that id|His continuous. In particular (id|H× id|H)−1(A) = A ∩H2 is in Γ(H ×H).

Unfortunately, the combination of symmetry and transitivity proved to betoo powerful. For example, Lecomte showed in [Lec07] the following negativeresult for other classes of relations.

Theorem 3.4. (Lecomte) There is a perfect ≤B-antichain (Rα)α∈2ω made oflocally countable relations which are ≤B-minimal among non-pot(Π0

1) Borel re-lations. Moreover, this is still true if we restrict this to the following subclasses:

- directed graphs (i.e., irreflexive relations),- graphs (i.e., irreflexive and symmetric relations),- oriented graphs (i.e., irreflexive and antisymmetric relations),- quasi-orders,- strict quasi-orders,- partial orders,- strict partial orders.

So any basis must contain an element that reduces and is reduced by anelement of this antichain. Later, Lecomte and Miller proved in [LM08] that infact, there is no antichain basis for the usual notions of comparison.

Lecomte showed that we can fix this problem if we change the notion ofcomparison. In [Lec93] he showed that one can find a basis for a rectangularversion of Borel reducibility, if we also restrict ourselves to a Borel subset. Letus consider the following notion of comparison introduced by Kechris, Soleckiand Todorcevic in [KST99].

Definition. - (Kechris, Solecki, Todorcevic) Let X,X ′ be Polish spaces,and A,A′ analytic relations on X,X ′ respectively. We write (X,A) c(X ′, A′) if there is a continuous function f : X → X ′ such that A ⊆(f × f)−1(A′). We call this function a continuous homomorphismfrom A to A′.

- For ε ∈ 2, let Xε, Yε be Polish spaces, and Aε, Bε ⊆ Xε×Yε. We say that(X0, Y0, A0, B0) is reducible to (X1, Y1, A1, B1) if there are continuousfunctions f : X0 → X1 and g : Y0 → Y1 such that

A0 ⊆ (f × g)−1(A1)

andB0 ⊆ (f × g)−1(B1).

In this case, we write (X0, Y0, A0, B0) . (X1, Y1, A1, B1).


In Chapter 4 we will work specifically with the second quasi-order. We canreformulate Theorem 2.3 as follows.

Theorem 3.5. (Lecomte) There is a relation H0 on 2ω such that, for any Polishspaces X,Y and any Borel subsets A,B of X × Y , exactly one of the followingholds:

1. A is separable from B by a pot(Π01) set.

2. (2ω, 2ω,H0,∆(2ω)) . (X,Y,A,B).

In fact, H0 is a very specific directed graph, which was first introduced byKechris, Solecki and Todorcevic. In 2<ω, we fix for each n a sequence sn suchthat |sn| = n and for all t ∈ 2<ω there is n ∈ ω with t ⊆ sn. A set of sequencesthat satisfy this last property is said to be dense.

Let us see that we can find a set of sequences with these properties. Letb : ω → 2<ω be the bijection defined by the strict total well-order s < t if|s| < |t| or (|s| = |t| ∧ ∃q s|q = t|q ∧ s(q) < t(q)), i.e., ∅, 0, 1, 00, 01, 10, 11, ...Notice that this bijection satisfies |b(n)| ≤ n. We define then sn = b(n)0n−|b(n)|.So, if t ∈ 2<ω, then there is n such that t = b(n) and thus t ⊆ sn.

The graph introduced by Kechris, Solecki, Todorcevic is

G0 := (α, β) ∈ 2ω × 2ω|∃n sn v α, sn v β∧α(n) 6= β(n) ∧ ∀m > n(α(m) = β(m)).

And its directed form is

H0 := (sn0γ, sn1γ) ∈ 2ω × 2ω|n ∈ ω ∧ γ ∈ 2ω.

Obviously, the definition of these relations depends on the choice of sn butfor most applications, any choice of sn that satisfies our hypothesis will generateinterchangable relations. The graph G0 plays an important role in descriptivegraph combinatorics, which is the study of definable graphs. Notice first thatthis graph is Borel.

We say that a graph (or directed graph) G on X has a coloring with κcolors (for κ a cardinal), if there is a set Y of cardinality κ and a functionc : X → Y such that G ⊆ (c× c)−1(¬∆(Y )). If X is a Polish space we say thatG has a Borel (or continuous, or Γ-mesurable) coloring with κ colors ifthere is a Polish space Y and a Borel (or continuous, or Γ-mesurable) functionc : X → Y such that G ⊆ (c× c)−1(¬∆(Y )).

The (Borel, continuous, Γ-mesurable respectively) chromatic number is thesmallest κ such that G has a (Borel, continuous, Γ-mesurable) coloring with κcolors. We write χ(G) = κ (χB(G), χc(G), χΓ(G), respectively).

It is easy to see that G0 and H0 have a coloring with 2 colors. However,they do not have a Borel countable coloring.

Problaby the most important property of G0 is the following theorem byKechris, Solecki, Todorcevic.

Theorem 3.6. (Kechris, Solecki, Todorcevic) Let X be a Polish space, and Gbe an analytic graph on X. Exactly one of the following holds:

1. G has a countable Borel coloring.2. G0 c G.

Relations 25

Let G be a relation on X, and A be a subset of X. We say it is G-discreteif G ∩ A2 = ∅. Suppose that G has a countable coloring. Consider Ay =c−1(y) for y ∈ Y . These sets form a partition of X and each Ay is G-discrete.Moreover, if c were a Borel (continuous) coloring, then Ay will be a Borel (open,respectively) set.

Conversely, suppose that (Ak)k∈ω is a partition of X in Borel (open) setssuch that each Ak is G-discrete. Let c : X → ω be defined by c(x) = k if andonly if x ∈ Ak. Then c is a countable Borel (continuous) coloring.

In other words, we have shown that G has a Borel (continuous) countablecoloring if and only if ∆(X) is separable from G by a countable union of Borel(open) squares that are G-discrete.

Thus, we see how there is an interaction between Borel colorings and unionsof Borel rectangles. In fact, notice how Theorem 3.5 is related to Theorem 3.6for the separability by countable unions of Borel rectangles. From these twotheorems, we can ask the following questions.

Question 3.7. Let k ∈ ω be non-zero.1. Is there a finite .-antichain basis for the non-separability of analytic sets

by a union of k Borel rectangles?2. If k > 1 is there a finite B-antichain basis for the class of analytic

directed graphs without a Borel coloring with k colors?3. Is there a finite .-antichain basis for the non-separability of analytic sets

by any finite union of Borel rectangles?4. Is there a finite B-antichain basis for the class of analytic directed

graphs without any finite Borel coloring?

The case k = 1 in question 1 is probably folklore, but we will provide aproof. On the other hand, Miller proved in [Mil] that Question 2 has no positivesolution when k = 2. More specifically, he proved the following theorem.

Theorem 3.8. (Miller) Let X be a Polish space, and G be an analytic directedgraph on X with no Borel coloring with 2 colors. If B is a basis for the classof analytic graphs H with no Borel colorings with 2 colors such that H B G,then the partial order (R<ω,w) embeds into (B,B).

One can extend this result to separability by two Borel rectangles. Thisshows that Question 1 has no positive solution for k = 2.

Another approach is to consider bounded complexities for the rectangles orthe colorings. Lecomte and Zeleny in [LZ14] started the study of ∆0

ξ-measurablecolorings and of separation of analytic sets by countable union of rectangles inΣ0ξ . In particular, they proved the conjecture below (also proposed in [LZ14])

when ξ = 1, 2.

Notation. Let be Γ,Γ′ be classes of sets. Then Γ×Γ′ := A×B|A ∈ Γ, B ∈ Γ′,and Γσ := ∪n∈ωAn|An ∈ Γ. We also use π0 and π1 for the projection on thefirst and second coordinate respectively.

Conjecture 3.9. (Lecomte, Zeleny) For each 0 < ξ < ω1 there are Polishspaces Xξ,Yξ and analytic subsets Aξ,Bξ of Xξ × Yξ such that for all Polishspaces X,Y and analytic disjoint A,B ⊆ X × Y , exactly one of the followingholds:


1. B is separable from A by a (Σ0ξ ×Σ0

ξ)σ set.2. (Xξ,Yξ,Bξ,Aξ) ≤ (X,Y,B,A).

From this, we obtain several natural questions.

Question 3.10. Let k ∈ ω be non-zero, and 0 < ξ < ω1.1. Is there a finite antichain basis for the non-separability of analytic sets

by a union of k rectangles in Σ0ξ?

2. If k > 1 is there a finite antichain basis for the class of analytic directedgraphs without a ∆0

ξ coloring with k colors?3. Is there a finite antichain basis for the non-separability of analytic sets

by a finite union of rectangles in Σ0ξ?

4. Is there a finite antichain basis for the class of analytic directed graphswithout any finite ∆0

ξ coloring?

We will obtain a positive solution for Question 1 above in the case k = 1and ξ ≤ 1. In fact we will extend this to several different types of rectangles.This leads to Theorem 2.4, our main result in Chapter 4.

The proof of Theorem 2.4 depends on a certain topological condition on theprojections of the sets A and B, so we lose some information about these sets.For this reason, the examples in our antichain basis will need to have a certaindegree of discreteness, which we introduce by adding a copy of ω in a precise way.So to construct these examples, recall the definition of topological direct sum.The disjoint union of Xii∈I is the set tiXi := (i, x) ∈ I × ∪Xi|x ∈ Xi.

Definition. Let Xii∈I be an indexed family of open sets. The direct sum⊕Xi is the topological space consisting of the disjoint union tXi with the topol-ogy generated by the canonical injections πi : Xi → tXi (i.e. πi(x) = (i, x)).In other words, the open sets are sets of the form ∪i∈Ii × Oi for Oi open inXi.

When there is no chance of confusion, we will denote by x = (i, x). Alsoin this case and abusing notation, we will denote by · the respective canonicalinjection.

Injective reductionsWe saw that there is no antichain basis for the class of non-potentially closed

relations and the quasi-order of Borel reducibility. Among similar lines, thefollowing holds.

Theorem 3.11. Let 0 < ξ < ω1. There is a Borel subset S of 2ω × 2ω suchthat the following holds:

1. (Debs-Lecomte) Let X,Y be Polish, and A,B be disjoint analytic subsetsof X × Y . Exactly one of the following holds:(a) A is separable from B by a pot(Π0

ξ) set.

(b) (2ω, 2ω, S, S\S) . (X,Y,A,B).2. (Debs) Moreover, one cannot replace S\S with ¬S in (b).

Injective reductions 27

In this theorem, the authors solved the problem by introducing acyclicity.Debs also showed that in (b) one can have the injectivity of the functions in-volved, if ξ > 2. Using acyclicity, Lecomte solved in [Lec14] the problem of thereduction on the whole plane.

Theorem 3.12. (Lecomte) Let Γ 6= Γ be a Borel class.

1. There is a vc-concrete finite antichain basis for non-pot(Γ) Borel rela-tions with s-acyclic closure.

2. Let Γ = Π02, then there is a vc-concrete finite antichain basis for non-

pot(Γ) Borel locally countable s-acyclic relations.

The proof of this result relies heavily on finding injective versions of part 1of Theorem 3.11. In other words, we restrict the quasi-order to only injectivecontinuous functions.

Theorem 3.13. (Lecomte)

1. There are relations S0 and S1 on 2ω, such that for any Polish space X,and any disjoint analytic relations A and B on X such that A is s-acyclic,exactly one of the following holds:

(a) A is separable from B by a pot(Π02) set.

(b) There is a continuous injective function f : 2ω → X such that S0 ⊆(f × f)−1(A) and S1 ⊆ (f × f)−1(B).

2. Let X be a Polish space, and A and B be disjoint analytic relation on X.Then, the following are equivalent:

(a) There is an s-acyclic relation R such that A∩R is not separable fromB ∩R by a pot(Π0

1) set.(b) There are continuous injective functions f, g : 2ω → X such that

H0 ⊆ (f × g)−1(A) and Diag(2ω) ⊆ (f × g)−1(B).

He also provides some similar theorems by substituting s-acyclic with locallycountable.

In [Lec13], Lecomte proves the following generalization of Theorem 3.11 toall Wadge classes of Borel sets.

Theorem 3.14. (Lecomte) Let Γ be a Wadge class of Borel sets, or the class∆0ξ for some countable non zero ordinal ξ. Then, there are Borel relations S0

and S1 on 2ω such that for any Polish spaces X,Y and for any disjoint analyticsubsets A0, A1 of X × Y , exactly one of the following holds:

1. A0 is separated from A1 by a pot(Γ) set.2. (2ω, 2ω,S0,S1) . (X,Y,A,B).

In fact, all the examples found by Lecomte in the previous theorem arecontained in the set of branches of a particular tree on 2 × 2. The followingdefinition can be found on [Lec09]. Let <>: ω2 → ω be the Cantor bijection,i.e., < n,m >= Σk≤n+mk+m. If n ∈ ω, we denote by (n)0, (n)1 its inverses by<>, i.e. the unique numbers such that < (n)0, (n)1 >= n.


Definition. (Lecomte)1. Let F ⊆ (2× 2)<ω. We say F is a frame if(a) ∀l ∈ ω∃!(sl, tl) ∈ F ∩ (2× 2)l,(b) ∀p, q ∈ ω∀w ∈ 2<ω∃N ∈ ω(sq0w0N , tq1w0N ) ∈ F and (|sq0w0N | −

1)0 = p,(c) ∀l > 0∃q < l∃w ∈ 2<ω(sl, tl) = (sq0w, tq1w).

2. The set T generated by a sequence (sl, tl)l∈ω of elements of (2× 2)<ω isdefined by

T := (s, t) ∈ (2× 2)<ω|s = ∅ ∨(∃q ∈ ω∃w ∈ 2<ω(s, t) = (sq0w, tq1w)

).

Notice that T is a tree if the sequence is a frame.3. Let T be the set generated by a sequence (sl, tl)l∈2ω . The oriented graph

generated by T is the oriented graph on 2<ω × 2<ω defined by GT :=(sl0γ, tl1γ)|l ∈ 2ω ∧ γ ∈ 2ω.

We will encounter several sequences that satisfy a. This is particularly im-portant because of the following proposition.

Proposition 3.15. Let F = (sl, tl)|l ∈ ω be a subset of (2× 2)<ω such that∀l ∈ ω(sl, tl) ∈ F ∩ (2× 2)l. If T is the set generated by this sequence, then

a. T ∩ (2× 2)n is s-acyclic and s-connected for any n ∈ ω,b. GT is s-acyclic.

Proof. Observe that b is a consequence of a. Indeed, notice that (sl0γ, tl1γ) ∈GT if and only if (sl0w, tl1w) ∈ T ∩ (2 × 2)l+1+|w| for any w v γ. Therefore,any path (in particular, cycle) in s(GT ) has a corresponding path (respectively,cycle) in s(T ∩ (2× 2)n) for some n big enough, since paths are finite.

So let us prove that T ∩ (2 × 2)n is s-acyclic and s-connected by inductionon n. For n = 0, T ∩ (2× 2)n = (∅, ∅) which is s-acyclic and connected. Forn = 1, T ∩ (2× 2) = (0, 1), which is s-acyclic and connected.

Now, let n = m + 1. To see that T ∩ (2 × 2)n is s-connected, let s, t ∈ 2n.If s, t end with the same coordinate, by the induction hypothesis, we only copythe path that goes from s|m to t|m. Else, suppose that s(m) = 0, then thereis a path by the induction hypothesis from s to sm0. Similarly, there is a pathfrom t to tl1. This creates the required path.

To see that s(T ∩ (2× 2)n) is s-acyclic, suppose by contradiction that thereis an s(T ∩ (2× 2)n) cycle, (u0, u1, . . . , uk, u0). For every i ≤ k, there are l andw such that (ui, ui+1) is equal to (sl0w, tl1w) or (tl1w, sl0w). If w 6= ∅ for allk, there is a corresponding s(T ∩ (2× 2)m) cycle obtained by removing the lastcoordinate of every sequence. If not, then there is one and only one i such that(ui, ui+1) is equal to (sm0, tm1) or (tm1, sm0), as cycles are injective. We cansuppose, by reindexing, that i 6= k and that (ui, ui+1) = (sm0, tm1). Noticethis will be the only place where the last coordinate can change. If j ≤ i, thenuj(m) = 0 and if i < j, then uj(m) = 1. Therefore, as (uk, u0) ∈ s(T ∩(2×2)n),it must be equal to (tm1, sm0). This cannot be the case, since k ≥ 2.

We note that in the case of non self-dual clases, S1 = dT e\S0. In particularthey are not locally countable. However, for all Wadge classes of Borel sets

Injective reductions 29

included in Σ02 there are locally countable examples. Since we are demanding

injectivity, we will need to find new examples.As mention in the introduction, we were able to find a proof that unifies the

cases of s-acyclicity and locally countable. The idea is in the style of the proofof the following theorem by Kechris, Solecki and Todorcevic, proved in [KST99].

Theorem 3.16. (Kechris, Solecki, Todorcevic) Let X be a Polish space, and Gan analytic graph on X that is either acyclic or locally countable. Exactly oneof the following holds:

1. G has a countable Borel coloring.2. There is a continuous injective function f : 2ω → X such that G0 ⊆

(f × f)−1(G).

To prove this, they generalized both notions of acyclicity and local countabil-ity to a technical notion that they call almost acyclicity. We want to go a stepfurther and generalize this notion, so that it can be applied to our particularcases. In particular, it will generalize both s-acyclicity and local countability.

Definition. i. Let R be an analytic relation on X. We say R is quasi-acyclic if there is a sequence (Rn)n∈ω of analytic relations with R = ∪Rnsuch that, for any x 6= y ∈ X, z, x1, . . . , xk, y1, . . . , yk ∈ X, n1, . . . , nk ∈ω, and ε1, . . . , εk ∈ 1,−1, if(a) x1 . . . xk, y1, . . . yk ∩ x, y, z = ∅,(b) (x, z) ∈ s(R) and (z, y) ∈ s(R),(c) (x, x1) ∈ Rε1

n1, (y, y1) ∈ Rε1

n1, and for all i < k, (xi, xi+1) ∈ R

εi+1ni+1 ,

(yi, yi+1) ∈ Rεi+1ni+1 ,

then xk 6= yk.ii. Let R be a quasi-acyclic relation with witnesses Rn. Two s(R)-paths

(x, x1, . . . , xk) and (y, y1, . . . , yl) are said to be q.a.-equivalent if theysatisfy c for some n1, . . . , nk, ε1, . . . , εk.Graphically, what we obtain is the following situation.

x x1 xk−1 xk

z

y y1 yk−1 yk

Rε11 Rεkmk

6=

Rε11 Rεkk

s(R)

s(R)

Proposition 3.17. 1. If R is an s-acyclic analytic relation in X, then Ris quasi-acyclic.

2. If R is a locally countable analytic relation, then R is quasi-acyclic.

Proof. 1. Let Rn = R. Notice then that if xk = yk, we will have twodifferent paths of length at least two from x to y.

2. The proof is similar as the one after [KST99, Definition 6.5].


Theorem 2.6 can then be reformulated in the following way.

Theorem 3.18. •Let Γ ∈ Dξ(Σ01), Dξ(Σ0

1),∆02. There are disjoint analytic

relations A0,A1 on 2ω such that, for any Polish space X and disjoint analyticrelations A0, A1 on X such that A0 ∪ A1 is quasi-acyclic, exactly one of thefollowing holds:

1. A0 is separable from A1 by a pot(Γ) set,2. there are f, g : 2ω → X injective continuous such that Aε ⊆ (f ×

g)−1(Aε), for ε ∈ 2.•Let 0 < ξ < ω1. There is a finite C formed of pairs (C,D), where C,D are

disjoint analytic relations on 2ω, such that for any X Polish space and disjointanalytic relations A0, A1 on X such that A0∪A1 is contained in a quasi-acyclicpotentially closed relation, exactly one of the following holds:

1. A0 is separable from A1 by a potentially ∆(Dξ(Σ01)) set.

2. There are (C0, C1) ∈ C and f, g : 2ω → X injective continuous such thatCε ⊆ (f × g)−1(Aε), for ε ∈ 2.

Effective theoryWe will use effective descriptive set theory to solve our questions, even

though most of our results are strictly of a classical nature. The basic ideais that the effective theory provides a refinement of the classical one. It is avery powerful tool to prove classical results. Let us recall some basic definitionsand notation. We will not define the recursive sets of Nk, but the basic idea isthat they are sets whose membership can be decided by a computable method.

1. A recursively presented Polish space, is a Polish space X with adistinguished dense sequence (rn)n∈ω and a compatible complete distanced : X → R+ such that the relations

(i, j, k, l) ∈ P ⇔ d(ri, rj) ≤l

k + 1 ,

(i, j, k, l) ∈ Q⇔ d(ri, rj) <l

k + 1 ,

are recursive. Using this, there is a canonical way to enumerate a topo-logical basis of X, which we will denote by N(X,n)n∈ω.

2. We define the class of semirecursive sets as the class of sets of theform ∪n∈SN(X,n), where S is a semirecursive set of N. We denote thisclass by Σ0

1(X). Notice the choice of a lightface symbol. We can define,for α ∈ ωω, a relativized version of this, Σ0

1(α)(X), by considering thesection at α of a semirecursive subset of X × ωω.

3. We obtain also a lightface version of the Borel hierarchy:

Π0n(X) := X\A|A ∈ Σ0

n

and

Σ0n+1(X) := A|∃B ∈ Π0

n(ω ×X)∀x ∈ X(x ∈ A⇔ ∃n ∈ ω(n, x) ∈ B).

Effective theory 31

4. There is a lightface version of the analytic, co-analytic and Borel sets:

Σ11(X) := A|∃B ∈ Π0

1(ωω ×X)(x ∈ A⇔ ∃α ∈ ωω(α, x) ∈ B),

Π11(X) := X\A|A ∈ Σ1

1,

∆11(X) = Σ1

1(X) ∩Π11(X).

5. There are relativized versions for Γ = Σ0n,Π0

n,Σ11, which we denote by

Γ(α) for α ∈ ωω.6. All these classes are countable. Also, they form a non trivial hierarchy,

even if X is countable.7. If Γ is Σ0

n, Π0n, Σ1

1, or Π11 and Γ(α) is its lightfaced relativized version,

then Γ = ∪α∈ωωΓ(α).The following, while not a formal statement, is a principle that, in general,

is not dificult to verify. It rests on the fact that the properties and theorems weuse are not dependent of the choice of α when we relativized.

The relativization principle. Most results remain true if we replace a light-face class with its relativized version.

This is stronger than it might look. We did not provide concrete recursivepresentations of the common Polish spaces, but they have some. However, notall Polish spaces might have one. But, Fact 7 above and the relativization prin-ciple imply that we might as well suppose that they do. This will be extensivelyused through out this manuscript.

Recall that ωCK1 is the supremum of all recursive ordinals, i.e., the ordinalsthat are the order type of a recursive well order of N.

Now, we will use the lightface classes to introduce some topologies that refinethe original topology of a recursively presented Polish space X. This topologieshave been used extensively to solve different separation problems on recursivelypresented Polish spaces. Let τ1(X) denote the topology of X.

1. The Gandy-Harrington topology ΣX is the topology generated bythe Σ1

1 sets. We fix a distance dGH inducing this topology.2. We denote the topology generated by the ∆1

1 sets by ∆X . We also denotethe topology ∆X ×∆X on X2 by ∆2

X .3. For 1 < ξ < ωCK1 we write τξ for the topology generated by the Σ1

1∩Π0<ξ

subsets of X, where Π0<ξ := ∪ν<ξΠ0

ν .We obtain the following facts about these topologies.

Proposition 3.19. Let X be a recursively presented Polish space.1. All these topologies refine τ1(X).2. (Louveau) (X,∆X) is a zero-dimensional Polish space.

3. For every Σ11 subsets A,B of X and every Σ1

1 subset C of X2, if C∆2X ∩

(A×B) 6= ∅, then C ∩ (A×B) 6= ∅

Proposition 3.20. Let X be a recursively presented Polish space. Then thereis a concrete Σ1

1 set ΩX ⊆ X that satisfies the following properties.1. ΩX is dense and open in ΣX .


2. (ΩX ,ΣX) is a zero-dimensional Polish space such that.3. For any A ∈ Σ1

1(X), A ∩ ΩX is a clopen subset of (ΩX ,ΣX)4. If Y is another recursively presented Polish space, ΩX×Y ⊆ ΩX × ΩY .

The topologies τξ were introduced by Louveau in [Lou80a], to serve as a wayto check separability of Σ1

1 sets by a Σ0ξ set on recursively presented spaces.

Theorem 3.21. (Louveau) Let 0 < ξ < ωCK1 , X be a recursively presentedspace, and A,B be disjoint Σ1

1 subsets of X. The following are equivalent:1. A is separable from B by a Σ0

ξ subset.

2. A ∩Bτξ = ∅.

We also note that, as proved in [Lou80a], if B is Σ11, then B

τξ is also Σ11.

Description of Wadge classesAs stated in the Introduction, Louveau found a description for all Wadge

classes of Borel sets. However, we will require a slightly different description,which was developed by Louveau and Saint Raymond in [LSR87].

Definition. (Louveau, Saint Raymond) Let 0 < ξ < ω1, and Γ,Γ′ be twoclasses of sets. Then

A ∈ Sξ(Γ,Γ′)⇔ A =⋃n∈ω

(An ∩ Cn) ∪(B\(∪nCn)

),

for some sequence An ∈ Γ, B ∈ Γ′, and a sequence (Cn) of pairwise disjoint Σ0ξ

sets.

We will now define the notion of second type description for the non self-dualWadge classes of Borel sets. A description u will be an element of ωω1 (which isidentified with (ωω1 )ω through the usual bijection 〈·〉).

Definition. (Louveau, Saint Raymond) The relations “u is a second typedescription” (u ∈ D) and “u describes Γ” (Γu = Γ), are the least relationssatisfying the following properties:

1. If u = 0∞, then u ∈ D and Γu = ∅.2. If u = ξ1v with v ∈ D and v(0) = ξ, then u ∈ D and Γu = Γv.3. If u = ξ2〈up〉 such that ξ ≥ 1, up ∈ D, and up(0) ≥ ξ, then u ∈ D and

Γu = Sξ(⋃p≥1 Γup ,Γu0).

Theorem 3.22. (Louveau, Saint Raymond) Let Γ be a non self-dual Wadgeclass of Borel sets. Then there is a u ∈ D such that Γ(ωω) = Γu(ωω).

Most importantly, for the classes of differences, we can obtain an u precise,by induction on its construction.

1. As D1(Σ0ξ) = Σ0

ξ = Sξ(ωω, ∅), u = ξ2〈vn〉 works if v0 = 0 andvn = 010∞ for n > 0.

2. As Dθ+1(Σ0ξ) = Sξ(Dθ(Σ0

ξ),Σ0ξ)), u = ξ2〈vn〉 works if v0 describes Σ0

ξ

and vn describes Dθ(Σ0ξ) for n > 0.

Description of Wadge classes 33

3. If θ is limit, then Dθ(Σ0ξ) = Sξ(

⋃nDθn(Σ0

ξ), ∅), where (θn) is a se-quence cofinal in θ. Thus u = ξ2〈vn〉 works if v0 = 0 and vn describesDθn(Σ0

ξ) for n > 0.Note how the description of Dθ(Σ0

ξ) is different for the limit and successorcasses. We will use these descriptions to obtain a topological characterizationof separability of Σ1

1 sets by a pot(Γ) set. This difference produces another dif-ference between the topological characterizations for the case pot(∆(Dθ(Σ0

1))).This causes the difference in the antichain basis for these cases that we statedin the introduction.


Chapter 4

Separability by rectangles

The contents of this chapter have been published in [Zam15].

First resultsThe results here are all consequences of classical results.

Proposition 4.1. Let X,Y be Polish spaces, and let A,B be Σ11 subsets of

X × Y . The following are equivalent:1. A is not separable from B by a ∆1

1 ×∆11 set.

2. (π0[A] × π1[A]) ∩ B 6= ∅, so in particular A is not separable from B byan arbitrary rectangle.

3.(2, 2, (0, 0), (1, 1), (0, 1)

). (X,Y,A,B).

Proof. (3⇒ 1). Suppose that A is separable from B by C×D. Then, f−1(C)×g−1(D) must also separate (0, 0), (1, 1) from (0,1). This is clearly absurd.

(1 ⇒ 2). We argue by contradiction. In this case, π0[A] is disjoint fromE := x ∈ X|∃y ∈ π1[A]

((x, y) ∈ B

). Since these sets are analytic, Lusin’s

Theorem gives a Borel set C which separates π0[A] from E.Note that

(C×π1[A]

)∩B = ∅. Therefore, π1[A] is disjoint from y ∈ Y |∃x ∈

C((x, y) ∈ B

). Again, Lusin’s Theorem gives a Borel set D which separates

them. Thus, C ×D separates A from B.(2 ⇒ 3). Let (x, y) ∈ (π0[A] × π1[A]) ∩ B. There is y′ ∈ Y (respectively

x′ ∈ X) which witnesses the fact that x ∈ π0[A] (resp. y ∈ π1[A]). Define

f(e) =x if e = 0,x′ if e = 1,

g(e) =y if e = 1,y′ if e = 0.

Finally, f × g is clearly the required reduction.

Note that, conversely, one could also deduce Lusin’s Theorem from Propo-sition 4.1. Indeed, if two analytic sets C,D are disjoint, then C ×D is disjointfrom the diagonal. Proposition 4.1 gives a Borel set separating C from D. Thisimplies that these results are equivalent.

35

36 Chapter 4. Separability by rectangles

Now, we would like to answer the following question: can we find a Hurewicz-like test to decide if a particular Borel set in the plane is a rectangle of aparticular complexity? As we will see, this is in fact just an easy applicationof the previously mentioned result by A. Louveau and J. Saint Raymond in[LSR87]. For each 0 < ξ < ω1, let Sξ ∈ Π0

ξ(2ω)\Σ0ξ(2ω). Define

Aξ = (0α, 1∞)|α ∈ Sξ ∪ (1∞, 0α)|α ∈ Sξ,

Bξ = (0α, 0α)|α /∈ Sξ.

Proposition 4.2. Let 0 < ξ < ω1, X,Y be Polish spaces, and A be a Borelsubset of X × Y . Exactly one of the following must hold:

1. A is a Σ0ξ rectangle.

2. (2ω, 2ω,Aξ,Bξ) . (X,Y,A,¬A).

Proof. The exactly part comes from the fact that Aξ cannot be separated fromBξ by a Σ0

ξ rectangle; which in turn is because Sξ is not separable from ¬Sξ bya Σ0

ξ set.So suppose that A is not a Σ0

ξ rectangle. There are two cases.First case: A is not a rectangle. In this particular case, (π0[A] × π1[A]) ∩

¬A 6= ∅. So take (x, y) in this intersection. Pick x′ ∈ X, y′ ∈ Y ′ such that(x, y′), (x′, y) ∈ A. The following functions define the required reduction f × g:

f(α) =x if α(0) = 0,x′ if α(0) = 1,

g(α) =y if α(0) = 0,y′ if α(1) = 1.

Second case: A = Π0[A] × Π1[A]. Then, at least one of the sides must notbe a Σ0

ξ set. Without loss of generality, suppose that it is Π0[A]. This gives acontinuous function f : 2ω → X such that

Sξ ⊆ f−1(Π0[A])

¬Sξ ⊆ f−1(¬Π0[A])

Choose (x, y) ∈ A. Note that(f(α), y

)∈ A for all α ∈ Sξ. Define f : 2ω →

X and g : 2ω → Y by:

f(eα) =f(α) if e = 0,x if e = 1,

g(α) = y.

So in both cases we obtain a reduction.

In Section 4, we will obtain C for open rectangles. We note that usually, theantichain for separation of two disjoint analytic sets does not differ a lot fromthe special case when these two analytic are complements of each other. This isnot the case here, the antichain not only being quite different, but it will havetwo elements instead of one.

Topological characterizations 37

Topological characterizationsRecall from the Preliminaries that the topologies τξ were introduced by

Louveau to characterize separability of analytic sets by Σ0ξ sets on recursively

presented spaces. Lecomte and Zeleny found in [LZ14] a similar statement forcountable unions of Σ0

ξ rectangles. We give a further generalized statement,which follows with the exact same proof as the original one.

Lemma 4.3. (Lecomte, Zeleny) Let 0 < ξ, ξ′ < ωCK1 , X,Y be recursivelypresented Polish spaces, and A,B be disjoint Σ1

1 subsets of X×Y . The followingare equivalent:

1. A is separable from B by a (Σ0ξ ×Σ0

ξ′)σ set.

2. A ∩Bτξ×τξ′ = ∅.

We can deduce from this a dual condition for the separability by one Π0ξ

rectangle.

Lemma 4.4. Let 0 < ξ, ξ′ < ωCK1 , X,Y be recursively presented Polish spaces,and A, B be disjoint Σ1

1 subsets of X × Y . The following are equivalent:1. A is separable from B by a (Π0

ξ ∩ Σ11)× (Π0

ξ′ ∩ Σ11) set.

2. A is separable from B by a (Π0ξ ∩∆1

1)× (Π0ξ′ ∩∆1

1) set.3. A is separable from B by a Π0

ξ ×Π0ξ′ set.

4. B ∩ π0[A]× π1[A]τξ×τξ′ = ∅.

Proof. (1 ⇒ 2). Let C ×D ∈ (Π0ξ ∩ Σ1

1)× (Π0ξ′ ∩ Σ1

1) such that it separates Afrom B. Then, C is a Π0

ξ which separates π0[A] from π0[(X ×D) ∩ B]. Sincethese two sets are Σ1

1, by [Lou80a, Theorem B] there is C ′ ∈ Π0ξ ∩ ∆1

1 whichseparates Π0[A] from π0[(X × D) ∩ B]. From this, we obtain that Π1[A] isseparated from π1[(C ′×Y )∩B] by D. Since D is Π0

ξ′ , by [Lou80a, Theorem B]we obtain D′ ∈ Π0

ξ ∩∆11, which separates these two sets. In particular, C ′×D′

separates A from B.(2⇒ 3). It is obvious.(3 ⇒ 4). Note that if A is separable from B by a Π0

ξ × Π0ξ′ set, then so

is π0[A] × π1[A]. It follows that B will be separable from π0[A] × π1[A] by a(Σ0

ξ × Σ0ξ′)σ set, namely the complement of the Π0

ξ ×Π0ξ′ set. Finally, apply

Lemma 4.3.(4⇒ 1). It is the fact that Cτξ is a Π0

ξ ∩Σ11, if C is Σ1

1, as shown in [Lou80a,Lemma 7].

We would like to find a similar topological characterization for the separa-bility by a set in some of the other classes Γ × Γ′. We start with a sufficientcondition for the separability by a Σ0

ξ ×Σ0ξ′ set.

Lemma 4.5. Let 0 < ξ, ξ′ < ωCK1 , X,Y be recursively presented Polish spaces,and A,B be disjoint subsets of X×Y . If A is not separable from B by a Σ0

ξ×Σ0ξ′

set, then at least one of the following holds.1. There is x ∈ π0[A] such that, for every τξ-open neighborhood U of x,

there is y ∈ π1[A] such that, for every τξ′-open neighborhood V of y,(U × V ) ∩B 6= ∅.


2. There is y ∈ π1[A] such that, for every τξ′-open neighborhood V of y,there is x ∈ π0[A] such that, for every τξ-open neighborhood U of x,(U × V ) ∩B 6= ∅.

Proof. Suppose that neither 1. nor 2. holds. Since τξ(X) and τξ′(Y ) havecountable basis Unn∈ω and Vmm∈ω respectively, for each x ∈ π0[A] we canfind a nx that witnesses the negation of 1. We can find respectively a my thatwitnesses the negation of 2. for each y ∈ π1[A]. Let N := nx|x ∈ π0[A] andM := my|y ∈ π1[A].

Now, for each y ∈ π1[A] there exists an m′y ∈ ω such that, for all n′ ≤ my inN , (Un′×Vm′y )∩B = ∅ and y ∈ Vm′y ⊆ Vmy . Indeed, each n

′ ∈ N is a nx for somex, and so, we only need to take Vm′y ⊆ Vmy as a basic open neighborhood of y inthe intersection of all witnesses for each n′ ≤ my in the respective negation of1. Likewise, for each x ∈ π0[A] there is a n′x which satisfies the dual statement.Set N ′ := n′x|x ∈ π0[A] and M ′ := m′y|y ∈ π1[A].

Finally, we claim that (∪n∈N ′Un) × (∪m∈M ′Vm) separates A from B. Itobviously contains A. To see that it does not intersect B suppose it does, sothat there are n′ ∈ N ′ and m′ ∈ M ′, such that (Un′ × Vm′) ∩ B 6= ∅. We mustthen have n′ = n′x for some x and m′ = m′y for some y. In particular, we cansuppose that nx ≤ my (the other case is similar). Then, Un′x×Vm′y ⊆ Unx×Vm′y ,by our choice of n′x. However, by the definition of m′y the right side should notintersect B.

We would like to point out that in Lemma 4.5, there are no hypothesis onthe complexity of A nor B. Also, when ξ = ξ′ = 1, the effective hypothesis onthe spaces is useless, Lemma 4.5 still holds with general Polish spaces, not onlyrecursively presented Polish spaces. In fact, we can obtain a different version inthe open case, if we add some hypothesis on the complexity of A and B.

Lemma 4.6. Let X,Y be recursively presented Polish spaces, and A,B be dis-joint Σ1

1 subsets of X × Y . The following are equivalent:1. A is not separable from B by a Σ0

1 ×Σ01 set.

2. A is not separable from B by a (Σ01 ∩∆1

1)× (Σ01 ∩∆1

1) set.3. At least one of the following holds:(a) There is x ∈ π0[A] such that, for every τ1-open neighborhood U of x,

there is y ∈ π1[A] such that, for every τ1-open neighborhood V of y,(U × V ) ∩B 6= ∅.

(b) There is y ∈ π1[A] such that, for every τ1-open neighborhood V of y,there is x ∈ π0[A] such that, for every τ1-open neighborhood U of x,(U × V ) ∩B 6= ∅.

Proof. (1.⇒ 2.). It is obvious.(2. ⇒ 3.). As in the previous Lemma, we will show the contrapositive

statement. The proof is basically the same, but one needs to be sure that thechoices can be made effectively. For this, we fix basis (Un) and (Vm) for X andY respectively, made of Σ0

1 sets.Note then that, for all x ∈ Π0[A], there is n ∈ ω which satisfies the following:

x ∈ Un ∧ ∀y ∈ Π1[A]∃m ∈ ω(y ∈ Vm ∧ (Un × Vm) ∩B = ∅

).

Topological characterizations 39

Note this statement is Π11 in (x, n), so by the ∆-selection principle (see [Mos80,

4B.4]), we can choose nx in a ∆11 way. By a similar argument, we can choose my

in a ∆11 way. This implies that N := nx|x ∈ Π0[A] and M := my|y ∈ Π1[A]

are Σ11.

Now, for each y ∈ π1[A] there exists an m′ ∈ ω such that,

y ∈ Vm′ ∧ Vm′ ⊆ Vmy ∧ ∀n′ ≤ my

(n′ ∈ N → (Un′ × Vm′) ∩B = ∅

).

Again, note that this statement is Π11 in (y,m′), so we can choose m′y in a ∆1

1way. A similar argument allows us to choose n′x. This shows that N ′ := n′x|x ∈Π0[A] and M ′ := m′y|y ∈ Π1[A] are Σ1

1.So with the same argument as in the proof of Lemma 4.5, (∪n∈N ′Un) ×

(∪m∈M ′Vm) is a (Σ01 ∩Σ1

1)× (Σ01 ∩Σ1

1) set that separates A from B. We claimthat this is enough. In fact, following the proof of Lemma 4.4, by applying[Lou80a, Theorem B] twice we can obtain a (Σ0

1∩∆11)×(Σ0

1∩∆11) set separating

A from B.(3. ⇒ 1.). Again, we prove the contrapositive statement. If A is separated

from B by a Σ01 ×Σ0

1 set, then each of the sides of this rectangle will witnessthe negation of 1. and of 2. respectively.

We note that (1.⇒ 2.) and (2.⇒ 3) can be generalized with a similar proofto the Σ0

ξ ×Σ0ξ′ cases.

We would like to find a general condition for the separabilty by a Π0ξ ×Σ0

ξ′

set. For now, we consider the cases where either ξ or ξ′ is equal to 1.

Lemma 4.7. Let 0 < ξ < ωCK1 , X,Y be recursively presented Polish spaces,and A,B be disjoint Σ1

1 subsets of X × Y .1. The following are equivalent:(a) A is separable from B by a Π0

1 ×Σ0ξ set.

(b) A is separable from B by a (Π01 ∩∆1

1)× (Σ0ξ ∩∆1

1) set.(c) For all y ∈ π1[A] there is a τξ-open neighborhood V of y such that

(π0[A]× V ) ∩B = ∅.

2. The following are equivalent:(a) A is separable from B by a Π0

ξ ×Σ01 set.

(b) A is separable from B by a (Π0ξ ∩∆1

1)× (Σ01 ∩∆1

1) set.(c) for all y ∈ π1[A] there is an open neighborhood V of y such that

(π0[A]τξ × V ) ∩B = ∅.

Proof. (1.)((a)⇒ (b)). Suppose that we can separate A from B by a Π01 ×Σ0

ξ

set, say C × D. In particular, D separates π1[A] from E := y ∈ Y |∃x ∈π0[A](x, y) ∈ B. Note that E is Σ1

1. By [Lou80a, Theorem B], we obtain aset D′ ∈ Σ0

ξ ∩ ∆11 which separates Π1[A] from E. This implies that π0[A] is

separable from x ∈ X|∃y ∈ D′(x, y) ∈ B by π0[A]. Thus, again by [Lou80a,Theorem B] we obtain a C ′ ∈ Π0

1 ∩∆11 such that C ′ ×D′ separates A from B.


((b)⇒ (c)) Suppose that we can separate A fromB by a (Π01∩∆1

1)×(Σ0ξ∩∆1

1)set, say C × D. In particular, D separates π1[A] from E := y ∈ Y |∃x ∈π0[A](x, y) ∈ B. Note that E is Σ1

1. By Theorem 3.21, π1[A] is contained in

V = ¬y ∈ Y |∃x ∈ π0[A](x, y) ∈ Bτξ,

which is τξ-open. Therefore, for each y ∈ π1[A], y is in V and π0[A] × V doesnot intersect B, as desired.

((c)⇒ (a)). Let (Vn)n∈ω be a basis for τξ. Then, for each y ∈ π1[A], we canfind a ny ∈ ω such that Vny witnesses (c). Then,

π0[A]×⋃

y∈π1[A]

Vny

is a Π01 ×Σ0

ξ that separates A from B.(2.) Note ((b)⇒ (a)) is obvious. We will show ((a)⇒ (c)) and ((c)⇒ (b)).((a)⇒ (c)). Suppose that we can separate A from B by a Π0

ξ ×Σ01 set, say

C × U . Each y ∈ π1[A] is in U so there is a a Σ01 neigborhood U ′ of y with

U ′ ⊆ U . In particular, (C × U ′) ∩ B = ∅. This means that π0[A] is separablefrom x ∈ X|∃y ∈ U ′(x, y) ∈ B by C, a Π0

ξ set. Again, by Theorem 3.21, weobtain (π0[A]

τξ × U ′) ∩B = ∅.((c) ⇒ (b)). Let (Vn)n∈ω be a basis for τ1 consisting of Σ0

1 sets. Then, foreach y ∈ π1[A], we can find a ny ∈ ω such that Vny witnesses (c). We claimthat we can choose this in a ∆1

1 way. In fact, “Vn ∩ π1[(π0[A]τξ × Y ) ∩B] = ∅”

is a Π11 property in n, so that we can apply the ∆-selection principle. Then

π0[A]τξ ×

⋃y∈π1[A]

Vny

is a (Π0ξ ∩ Σ1

1) × (Σ01 ∩ Σ1

1) that separates A from B. By applying [Lou80a,Theorem B] twice as before, we can find C ′×D′ ∈ (Π0

ξ ∩∆11)× (Σ0

1∩∆11) which

separates A from B.

Note that in the case ξ = 1, we can get a stronger version in the non-effectivecase. We only require the spaces to be Polish, and A,B to be disjoint. In fact,if A is separable from B by a Π0

1 ×Σ01 set, then, for each y ∈ π1[A], take V as

the open side that witnesses the separation.

Separation by a Σ01 ×Σ0

1 setWe we first provide our C for the problem of separability by a Σ0

1 ×Σ01 set,

since this contrasts a lot with the other results. In particular, this is the onlycase where our antichain basis has two elements.

Definition. A left-branching scheme of a zero-dimensional space X is afamily of non-empty clopen subsets F := Fm,ε|m ∈ ω, ε ∈ 2 such that

1. F0,ε|ε ∈ 2 is a partition of X, and2. Fm+1,ε|ε ∈ 2 is a partition of Fm,0.

We say that it converges if diam(Fm,0)→ 0 as m→∞.

Separation by a Σ01 ×Σ0

1 set 41

Note that if a left-branching scheme on a complete space converges then∩m∈ωFm,0 is a singleton.

In [LZ14], the authors define, for ξ < ω1, a ξ-disjoint family as a family ofsets which are Π0

ξ and pairwise disjoint, where by convention Π00 = ∆0

1.Consider a 0-disjoint partition Fn for ωω. In each Fn, consider a left-

branching scheme Fnm,ε|m ∈ ω, ε ∈ 2. Finally, for each F 0m,1, consider a

0-disjoint partition Cml |l ∈ ω. Define the following sets:

A := Diag(⋃n∈ω

⋂m∈ω

Fnm,0),

B0 :=⋃

n,m∈ω

(Cnm × Fn+1

m,1),

B1 :=⋃

n,m∈ω

(Fn+1m,1 × Cnm

).

Theorem 4.8. Let X,Y be Polish spaces, and A,B be disjoint subsets of X×Y .At least one of the following holds:

i) A is separable from B by an open rectangle.ii) There is i ∈ 2 such that (ωω, ωω,A,Bi) . (X,Y,A,B).

If moreover, the left-branching scheme Fnm,e|m ∈ ω, e ∈ 2 converges for eachn ∈ ω, then exactly one of the previous holds.

Proof. Suppose that A is not separable from B by an open rectangle. Then,1. or 2. of Lemma 4.5 must hold. Suppose that 1. holds, we will show that(ωω, ωω,A,B0) reduces to (X,Y,A,B). We note that the same argument willgive us the reduction for i = 1 when 2. holds.

Find x ∈ π0[A] which satisfies 1. and let y ∈ Y such that (x, y) ∈ A. It iseasy to see that, for each n,m ∈ ω, there are (xn, yn) ∈ A and (xnm, ynm) ∈ Bsuch that,

(a) for all sequences (mn)n∈ω, xn(mn) → x as n→∞,(b) ynm → yn as m→∞.We define f : ωω → X by:

f(α) :=

xn if α ∈ Fn+1,xnm if α ∈ Cnmx if α ∈ ∩l∈ωF 0

l,0.

Similarly, define g : ωω → Y by:

g(α) :=

yn if α ∈ ∩l∈ωFn+1

l,0ynm if α ∈ Fn+1

m,1y if α ∈ F0.

It is easy to see that these maps are well defined.We first show the continuity of f . Notice that we only need to check the

continuity in ∩l∈ωF 0l,0, since each other part is clopen. So suppose that α ∈

∩l∈ωF 0l,0 and αk → α. For each l there is K such that αk ∈ F 0

l,0 for k > K. Wecan suppose that αk /∈ ∩l∈ωF 0

l,0, since otherwise f(αk) = x. Then f(αk) = xlkmk ,for some lk greater than l.


We claim that lk diverges to ∞. Indeed, lk ≥ l if αk ∈ F 0l,m, and for each l

this holds for k big enough. Then, xlkmk → x, as required. To check continuityfor g, we only need to verify it in ∩l∈ωFn+1

l,0 , and this is done similarly. Checkingthat f × g is a reduction is routine.

Suppose that each scheme converges. We will show that A is not separablefrom B0 by an open rectangle. So let U, V be open subsets such that A ⊆ U×V .Note that ∩F 0

l,0 = x, and F 0l,0 is a basis at x made of clopen neighborhoods.

Then, there is an l such that F 0l+1,0 ⊆ U . Note that ∩m∈ωF l+2

m,0 = yl ⊆ V , sothere is an m such that F l+2

m,0 ⊆ V . Then Cl+1m+1×F

l+2m+1,1 ⊆ B0∩

(F 0l,0×F

l+2m,0)⊆

B0 ∩(U × V

). Thus, A is not separable from B by an open rectangle.

This shows that at most 1. or 2. must hold, as in the proof of Proposition4.1.

One can obtain an example satisfying our conditions by setting Fn = N(n),

Fnm,ε =Nnm+2 if ε = 0,⋃k 6=nNnm+1k if ε = 1.

and Cml = N0m+1(l+1). It is routine to show they satisfy the hypothesis in theconstruction of A and B, and that each scheme is converging.

We can in fact shrink a bit our minimal examples, as some easy verificationcan show they stay non separable.

A := (n∞, n∞)|n ∈ ω,

B0 := (0n+1(m+ 1)∞, (n+ 1)m+10∞)|n,m ∈ ω,

B1 := ((n+ 1)m+10∞, 0n+1(m+ 1)∞)|n,m ∈ ω,

We would also note that this is the best we can do.

Proposition 4.9. (ωω, ωω,A,B0) and (ωω, ωω,A,B1) are .-incomparable.

Proof. Suppose that (ωω, ωω,A,B0) . (ωω, ωω,A,B1), the other side being sim-ilar. Then, for any n,m there are kn,m and ln,m such that f(0n+1(m+ 1)∞)= (kn,m + 1)ln,m+10∞, and g((n+ 1)m+10∞) = 0kn,m+1(ln,m + 1)∞.

As f(0n+1(m+ 1)∞)→ f(0∞), when n →∞, f(0∞) = (K + 1)∞ for someK ∈ ω. Then, there is N ∈ ω, such that for n ≥ N and any m ∈ ω, f(0n(m +1)∞) ∈ NK+1, by continuity. However 0K+1(ln,m + 1)∞ = g((n+ 1)m+10∞),but as m → ∞, the left side cannot converge to something of the form (K ′)∞and the right side converges to something of this form.


1 setThis case has some things in common with the open case. In particular, the

construction uses the same families of sets. Let Fn be a 0-disjoint partitionof ωω, and, for each n, Fnm,ε be a left-branching scheme. Instead of the 0-disjoint partitions of F 0

m,1, we consider for each n ∈ ω, a 0-disjoint partitionDn

m|m ∈ ω of Fn+1. We now define the following sets:

A :=((∩l∈ωF 0

l,0)× (∩l∈ωF 0l,0))⋃(

∪m,n∈ω (Fn+1m,1 ×Dn

m)),

Separation by a Π0ξ ×Π0

ξ set 43

B :=⋃n∈ω

((∩l∈ωFn+1

l,0 )× F 0n,1).

Proposition 4.10. Let X,Y be Polish spaces, and A,B be disjoint subsets ofX × Y . At least one of the following holds:

1. A is separable from B by a Π01 ×Σ0

1 set.2. (ωω, ωω,A,B) . (X,Y,A,B).

If moreover each scheme converges, then exactly one of the previous holds.

Proof. Suppose that A is not separable from B by a Π01 ×Σ0

1 set.By Lemma 4.7, there is y ∈ π1[A] such that for each open neighborhood V

of y, (π0[A]× V ) ∩B 6= ∅. Let x ∈ X with (x, y) ∈ A.Let (Vn) be a decreasing neighborhood basis at y. For each n ∈ ω, choose

(xn, yn) ∈ (π0[A]× Vn) ∩B.We note that yn → y as n→∞. Also, since xn ∈ π0[A], there is a sequence

(xnm, ynm) ∈ A, such that xnm → xn as m→∞.Define the following functions:

f(α) =

x if α ∈ F0,xn if α ∈ ∩l∈ωFn+1

l,0 ,

xnm if α ∈ Fn+1m,1 .

g(α) =

y if α ∈ ∩l∈ωF 0

l,0,

yn if α ∈ F 0n,1,

ynm if α ∈ Dnm.

The proof that this is in fact a well defined reduction is the same as that ofTheorem 4.8. By Lemma 4.7, if the scheme converges, A is not separable fromB by a Π0

1 ×Σ01 set.

Using the same families as in the open case, one can obtain a more concreteexample. Choose also, Dn

m = N(n+1)m. We shrink the examples as in theprevious case, to get the following sets, which define a .-minimal example fornon separation by Π0

1 ×Σ01 set.

A := (0∞, 0∞) ∪ ((n+ 1)m+10∞, (n+ 1)m∞

)|n,m ∈ ω,

andB :=

((n+ 1)∞, 0n+11∞

)|n ∈ ω.


ξ setLecomte and Zeleny used Lemma 4.3 to prove Conjecture 3.9 in the cases

ξ = 1, 2. As one can expect from the proof of Lemma 4.4, this conjectureprovides a weaker dichotomy for the case of one Π0

ξ rectangle.

Lemma 4.11. If Conjecture 3.9 is true, then for every 0 < ξ < ω1, for everyPolish spaces X,Y , and for every disjoint analytic A,B ⊆ X × Y , exactly oneof the following holds:


1. A is separable from B by a Π0ξ ×Π0

ξ set.2. There are continuous functions f : Xξ → X and g : Yξ → Y such that

π0[Aξ] ⊆ f−1(π0[A]),π1[Aξ] ⊆ g−1(π1[A]),

Bξ ⊆ (f × g)−1(B).

Proof. For the exactly part, we note that if C × D separates A from B, withC,D in Π0

ξ , then f−1[C] × g−1[D] separates π0[Aξ] × π1[Aξ] from Bξ, whichcannot be the case, because of Conjecture 3.9.

Now, by relativization, we can suppose that X,Y are recursively presentedPolish spaces, and that A,B are Σ1

1 subsets of X × Y . Suppose that A is notseparable from B by a Π0

ξ ×Π0ξ set. Lemmas 4.4, 4.3, and Conjecture 3.9 give

our functions.

We can improve this lemma by finding an actual .-minimum example forthe cases ξ = 1, 2, which coincidentally are the cases where Conjecture 3.9 isproved. In both cases, we need an additional hypothesis on Aξ, which can beeasily fulfilled, as we will see below. First, we introduce some notation

Let ξ ∈ 1, 2. Suppose that (Xξ,Yξ,Aξ,Bξ) satisfy Conjecture 3.9. Supposethat Aξ has countable projections, so let αnn∈ω and βnn∈ω be enumerationsof the first and second projections respectively. We define

X′ξ := Xξ ⊕ ω,Y′ξ := Yξ ⊕ ω,A′ξ :=

(αn, n) ∈ X′ξ × Y′ξ|n ∈ ω ∪ (n, βn) ∈ X′ξ × Y′ξ|n ∈ ω

,

B′ξ :=

(α, β) ∈ X′ξ × Y′ξ|(α, β) ∈ Bξ.

We obtain the following theorem.

Theorem 4.12. Let ξ ∈ 1, 2. Let X,Y be Polish spaces, and A,B ⊆ X × Ybe disjoint analytic subsets. Exactly one of the following holds:

1. A is separable from B by a Π0ξ ×Π0

ξ set.2. (X′ξ,Y′ξ,A′ξ,B′ξ) . (X,Y,A,B).

Proof. Applying the previous lemma to A′ξ and B′ξ we can see that A′ξ is notseparable from B′ξ (using the canonical embeddings from Xξ into X′ξ, and Yξinto Y′ξ), so the exactly part follows.

Suppose that A is not separable from B by a Π0ξ × Π0

ξ set. Lemma 4.11gives auxiliary continuous functions f ′ : Xξ → X and g′ : Yξ → Y . Now,f ′(αn) ∈ π0[A], so there is yn ∈ Y such that (f ′(αn), yn) ∈ A. Similarly, we canobtain xn ∈ X, such that (xn, g′(βn)) ∈ A. We then define f : X′ξ → X, andg : Y′ξ → Y by:

f(α) = f ′(α)f(n) = xn


ξ set 45

and

g(β) = g′(β)g(n) = yn

These are continuous maps, and f × g is a reduction, by Lemma 4.11 andthe choice of xn and yn.

In order to be more concrete, we would like to give particular instances forthese .-minimum examples. These can be obtained directly from [LZ14]. Werecall the general form of these examples, and then give a particular examplethat satisfies our condition on the projections of Aξ.

Proposition 4.13. 1. (Lecomte-Zeleny) Let X and Y be zero-dimensionalPolish spaces, and let C0

i |i ∈ ω and C1i |i ∈ ω be 0-disjoint families

of X and Y respectively. Then if B ⊆(X\(⋃

i∈ω C0i

))×(Y\(⋃

i∈ω C1i

))is not separable from A ⊆

⋃i∈ω C

0i × C1

i by a (Σ01 ×Σ0

1)σ set, then theysatisfy Conjecture 3.9 for ξ = 1.

2. There are X, Y, A and B which satisfy the hypothesis in 1. such that Ahas countable projections.

Proof of 2. Let X := Y := 2ω. It is easy to see that Cεn := N0n1 defines a 0-disjoint family. Then B := (0∞, 0∞) and A := (0n1∞, 0n1∞);n ∈ ω satisfy2.

We obtain from this proposition and Theorem 4.12 the following minimumexample for non separability by closed rectangles:

A′ = (0n1∞, n)|n ∈ ω ∪ (n, 0n1∞)|n ∈ ω,

B = (0∞, 0∞).

We now consider Π02 rectangles. The following definition was introduced in

[LZ14].

Definition. (Lecomte-Zeleny) Let 1 ≤ ξ < ω1. A ξ-disjoint family (Cεi )(ε,i)∈2×ωof subsets of a 0-dimensional Polish space W is said to be very comparing iffor each natural number q, there is a partition (Opq )p∈ω of W into ∆0

ξ sets suchthat, for each i ∈ ω,

1. if q < i, then there is a piq ∈ ω such that C0i ∪ C1

i ⊆ Opiqq ,

2. if q ≥ i and ε ∈ 2, then Cεi ⊆ C2i+εq ,

3. if (ε, i) ∈ 2× ω, then⋃r≥i⋂q≥r O

2i+εq = Cεi .

Proposition 4.14. 1. (Lecomte-Zeleny) Let (Cεi )(ε,i)∈2×ω be a very com-paring 1-disjoint family of subsets of a 0-dimensional Polish space W. LetX ⊆ W\

⋃i∈ω(C1i

),Y ⊆ W\

(⋃i∈ω C

0i

),B ⊆ Diag(W\

(⋃(ε,i)∈2×ω C

εi

)),

and A ⊆⋃i∈ω C

0i ×C1

i such that B is not separable from A by a (Σ02×Σ0

2)σset. Then they satisfy Conjecture 3.9 for ξ = 2.

2. There are X, Y, A and B which satisfy the hypothesis in 1. such that Ahas countable projections.


Proof of 2. Take W := 3ω, and Cεi = θ(i)εα|α ∈ 2ω, where θ is an enu-meration of s ∈ 3<ω|s = ∅ ∨ s(|s| − 1) = 2. This was shown in [LZ14]to be a very comparing family. By taking B := ∆

(W\(⋃

(ε,i)∈2×ω Cεi

)), and

A := (θ(i)0∞, θ(i)1∞)|i ∈ ω, we obtain 2.

Then we obtain our examples for non separability by Π02 rectangles:

A′ = (θ(n)0∞, n)|n ∈ ω ∪ (n, θ(n)1∞)|n ∈ ω,

B′ = (α, α)|α ∈ B.

Separation by a (Σ01 × Σ0


2set

One would like to know if the argument in the previous section extends toother classes of the type Γ×Γ′. Even if Conjecture 3.9 holds for ξ > 2, the firstthat is clear is that if Γ = Γ′ = Π0

ξ , then we cannot extend the definition rightbefore Theorem 4.12. Indeed, if A has countable projections, then the productof its projections will be a Π0

ξ ×Π0ξ set which separates A from B.

So another option is to see what happens if Γ 6= Γ′. In particular, we cansee what happens in the case Π0

1 ×Π02. Following the same argument as in the

previous section, it is enough to prove the natural extension of Conjecture 3.9for (Σ0

1 ×Σ02)σ.

Fix a Σ02 subset F of 2ω, let (Fn) be an increasing sequence of closed sets

such that F = ∪n∈ωFn. For each s ∈ 2<ω let ns := minn|Fn ∩Ns 6= ∅, whenthis exists. One can in fact suppose F is dense, so that ns will always exist. LetD := s ∈ 2<ω|s = ∅ ∨ ns 6= nsm, where sm := s||s|−1 for s 6= ∅ and sm := s ifs = ∅. With this in mind, set:

XF := 3ω\F, YF := 2ω,

BF := Diag(2ω\F ),

AF := (α, β)|∃s ∈ D(s2 v α ∧ s v β ∧ β ∈ Fns).

Theorem 4.15. Let F be as above, X,Y be Polish spaces, and A,B be disjointanalytic subsets of X × Y . At least one of the following holds:

1. B is separable from A by a (Σ01 ×Σ0

2)σ set.2. (XF ,YF ,BF ,AF ) . (X,Y,B,A).

If moreover F is meager in 2ω, then exactly one of the previous must hold.

Proof. Suppose that B is not separable from A by a (Σ01 ×Σ0

2)σ set.Note that, without loss of generality, we can suppose that X,Y are recur-

sively presented Polish spaces, and that A,B are Σ11. Then, by Lemma 4.3,

N := B ∩Aτ1×τ2 is not empty.For each s ∈ 2<ω\∅, let s− := s|max|t| |t@s∧t∈D, so that s− is either

empty or the last proper initial segment t of s where nt changed value, and set∅− := ∅.

We construct, for each t ∈ 3<ω and s ∈ 2<ω,i. points xt ∈ X, ys ∈ Y ,

Separation by a (Σ01 ×Σ0


2 set 47

ii. Σ01 sets Xt ⊆ X and Ys ⊆ Y ,

iii. a Σ11 set Vs ⊆ X × Y if s ∈ D.

We will require these sets to satisfy the following conditions, for each t ∈ 3<ωand s ∈ 2<ω:

1. xt ∈ Xt, ys ∈ Ys, and (xs, ys) ∈ Vs for s ∈ D2. diam(Xt) < 2−|t|, diam(Ys) < 2−|s|, and diamGH(Vs) < 2−|s| if s ∈ D

3.

Xs ⊆ Xs− if s ∈ D\∅Xt ⊆ Xtm if t ∈ 3ω\DYsε ⊆ Ys if ε ∈ 2Vs ⊆ Vs− if s ∈ D

4. Vs ⊆ N ∩ Ω ∩ (Xs × Ys)

5. if s ∈ D and sε /∈ D, then

(xs2, ysε) ∈ A ∩ Ω ∩ (Xs × Ys)ysε ∈ π1[Vs]

6. xs2t = xs2, and, if s /∈ D, and sm /∈ D, then ys = ysm .So suppose that these are already constructed. We will construct our func-

tions f : XF → X and g : YF → Y . If α ∈ 2ω then we define

g(α) =⋂n∈ω

Yα|n

and as usual the map g defined this way is continuous, and yα|n → g(α).For f , there are two cases. If there is i ∈ ω such that α(i) = 2, let i0 be

the smallest. By conditions 2 and 3, one can define f(α) := ∩n>i0Xα|n , andxα|n → f(α).

Otherwise, (nα|i) diverges, since otherwise α ∈ F . In this case, by conditions2 and 3, we can define f(α) := ∩i∈ωX(α|i)− .

We note that on the open set α|∃i(α(i) = 2) the function will be continuousas usual. So suppose (nα|i) diverges. Let (αk) be a sequence in XF whichconverges to α. Given an open neighborhood of f(α), we can find s v α suchthat s ∈ D and Xs is contained in such a neighborhood. Now, for big enoughk, s v αk. If nαk|i changes infinitely often, then by definition f(αk) ∈ Xs. Ifit does not change infinitely often, there is a last k0 such that s v αk|k0 andαk|k0 ∈ D. In particular, Xαk|l ⊆ Xs if l ≥ k0, and then, applying condition 3and the definition of f(αk), f(αk) ∈ Xs.

Now, we need to show that f × g is a reduction. If (α, α) ∈ BF , thennα|i changes infinitely often. Define F (α) = ∩V(α|i)− ⊆ B. By conditions 1and 2, (x(α|k)− , y(α|k)−) → F (α) in the Gandy-Harrington topology. Since thistopology refines the product topology,

(f(α), g(α)

)= F (α) ∈ B.

If (α, β) ∈ AF , then condition 6 and the definition of our maps imply thatthere is an s ∈ D such that

(f(α), g(β)

)= (xs2, ysε), which belongs to A by

condition 5.Now, we will show that the construction is possible. We will construct

xt, ys, Xt, Ys, Vs by induction on the lengths of t and s. For the length 0, sinceN is a non-empty Σ1

1 set, N ∩ Ω 6= ∅, there is (x∅, y∅) ∈ N ∩ Ω. We find Σ01

neighborhoods of good diameter, X∅ and Y∅. As

(x∅, y∅) ∈ N ∩ Ω ∩ (X∅ × Y∅),


and this set is Σ11, we can find another Σ1

1 neighborhood V∅ with good diameterin the Gandy-Harrington topology, such that V∅ ⊆ N ∩ Ω ∩ (X∅ × Y∅). Theysatisfy the required conditions.

So suppose that we have constructed xt, ys, Xt, Ys, Vs for all s ∈ 2p, t ∈ 3p.For each k ∈ 3 and ε ∈ 2, we will construct the respective points and sets foreach finite sequences tk, sε of length p+ 1 by cases.

The first case is when t /∈ 2<ω. In this case, by condition 6, we must simplycopy the point xtk := xt and shrink the Σ0

1 neighborhood Xt to a new one ofgood diameter and which satisfies Xtk ⊆ Xt.

The second case is when s ∈ D. Define ysε = ys if sε ∈ D and xsε = xs ifs ∈ 2<ω. Note also that (xs, ys) ∈ (Xs × Ys) ∩ (X × π1[Vs]) ∩N . In particular(Xs×Ys)∩ (X ×π1[Vs])∩A

τ1×τ2 6= ∅. So there exists (x, y) ∈ (Xs×Ys)∩ (X ×π1[Vs]) ∩A ∩ Ω. Define xs2 = x and ysε = y if sε /∈ D.

In all of these cases xsk ∈ Xs and ysε ∈ Ys, so shrink the neighborhoods Xs

and Ys to Σ01 neighborhoods Xsk and Ysε of good diameter. Finally, if sε ∈ D,

since (xsε, ysε) ∈ Vs ∩ (Xsε × Ysε) ∩ Ω, we can also shrink Vs to a new Σ11 set

Vsε ⊆ Vs ∩ (Xsε×Ysε) of good diameter. It is easy to check that these elementsand sets satisfy all our conditions.

The third and final case is when s ∈ 2<ω\D. If sε /∈ D, we can definexsε = xs. If ε 6= 2, let ysε = ys. Shrink Xs and Ys accordingly. Clearly, ifsε ∈ 2<ω, then condition 6 is satisfied.

Since s ∈ 2<ω, note that ys = ys−ε′ ∈ π1[Vs− ] ∩ Ys for ε′ ∈ 2 such thats−ε′ v s. Then, π1[Vs− ] ∩ Ys 6= ∅. So take (x, y) ∈ Vs− ∩ (Xs− × Ys) ∩ Ω.If sε ∈ D, define xsε = x and ysε = y. Then shrink the neighborhoods Xs−

and Ys to new neighborhoods Xsε and Ysε. Finally find a Σ11 neighborhood

Vsε ⊆ Vs− ∩ (Xsε×Ysε)∩Ω. It is clear that these objects satisfy our conditions.So the construction is possible.Now suppose that F is meager. We will show that BF is not separable from

AF , so that we can only have at most one of our options, like in previous cases.So suppose BF ⊆ ∪i∈ω(Ui × Vi), with Ui ∈ Σ0

1(3ω\F ) and Vi ∈ Σ02(2ω). In

fact, we can suppose Vi ∈ Π01(2ω), for each i ∈ ω. First, note that 2ω\F ⊆

∪i∈ω(Ui ∩ Vi). Since F is meager, there is i such that Ui ∩ Vi is not meager in2ω. So there exists s ∈ 2<ω such that Ns(2ω) ⊆ Ui ∩ Vi.

Note that there is s v s ∈ 2<ω such that Ns(3ω\F ) ⊆ Ui. Thus, if there isno s v s′ ∈ D, then Ns(2ω) ⊆ Fns since Fns is closed. This contradicts the factthat F is meager. Finally, take α ∈ Ns′2(3ω\F ) and β ∈ Fns′ ∩Ns′(2

ω), so that(α, β) ∈ (Ui × Vi) ∩ AF .

Proposition 4.16. 1. Let Fn be closed subsets of 2ω such that their unionF is dense. If X is a Polish subspace of XF ,Y is a Polish subspace ofYF , B ⊆ BF ∩ (X× Y), A ⊆ AF ∩ (X× Y), and B is not separable fromA by a (Σ0

1 ×Σ02)σ set in X× Y, then we can replace (XF ,YF ,BF ,AF )

with (X,Y,B,A) in Theorem 4.15.2. There are X, Y, A and B which satisfy the hypothesis in 1. such that A

has countable projections.

Proof. Let Ψ : ω → s ∈ 2<ω|s = ∅ ∨ s(|s| − 1) = 1 a bijection such thatΨ−1(s) ≤ Ψ−1(t) if s v t. Take Fn = Ψ(m)0∞|m ≤ n. In particular,D = Ψ(m)|m ∈ ω. We obtain the following sets:


1 set 49

X := 3ω\α ∈ 2ω;∀∞i, α(i) = 0,Y := 2ω

B := (α, α) ∈ X× Y|∃∞i α(n) = 1

A := (Ψ(m)2α,Ψ(m)0∞)| m ∈ ω ∧ α ∈ 3ω

One can then define A = (Ψ(m)20∞,Ψ(m)0∞)| m ∈ ω. It is routine tocheck they remain non separable.

Corollary 4.17. Let X,Y be Polish spaces, and A,B be disjoint analytic subsetsof X × Y , exactly one of the following holds:

1. A is separable from B by a Π01 ×Π0

2 set.2. There are continuous functions f : X→ X and g : Y→ Y such that:

π0[A] ⊆ f−1(π0[A]),π1[A] ⊆ g−1(π1[A]),

B ⊆ (f × g)−1(B).

If X′,Y′,A′,B′ are defined exactly as in the previous section, then we copythe same proof to get the following Corollary.

Corollary 4.18. Let X,Y be Polish spaces, and A,B be disjoint analytic subsetsof X × Y . Exactly one of the following holds:

1. A is separable from B by a Π01 ×Π0

2 set.2. (X′,Y′,A′,B′) . (X,Y,A,B).

So our antichain basis C is defined by the following objects.

X′ :=X⊕ ω,Y′ :=Y⊕ ω,B′ :=

(α, α) ∈ X′ × Y′|∃∞j(α(j) 6= 0)

,

A′ :=

(Ψ(m)20∞,m) ∈ X′ × Y′|m ∈ ω∪

(m,Ψ(m)0∞) ∈ X′ × Y′|m ∈ ω.


1 setThis is an interesting case, as it combines most of the methods used in the

previous sections. First, we need a lemma, reminiscent of Theorem 4.15.

Lemma 4.19. Let W be a zero-dimensional Polish space, S ∈ Σ02(W), X,Y

be recursively presented spaces, and C ⊆ X, U ⊆ Y and B ⊆ X × Y , Σ11 sets.

If (Cτ2 × U) ∩ B 6= ∅, then there are continuous functions f : W → X andg : W\S → U such that:

S ⊆ f−1(C)

Diag(W\S) ⊆ (f × g)−1(B).


Proof. Let N := (Cτ2 × U) ∩ B, and Cnn∈ω a 1-disjoint family, such thatS = ∪n∈ωCn.

Fix a basis of clopen subsets of W, Ns|s ∈ 2<ω such that Ns ⊆ Nt if t v s,and Ns ∩ Nt = ∅ if s is not compatible with t. In particular, for each x ∈ W,there is a unique α ∈ 2ω such that x ∈ ∩Nα|k.

As in the proof of Theorem 4.15 define ns := minn|Ns ∩ Cn 6= ∅, D :=s ∈ 2<ω|s = ∅∨ns 6= nsm, and s− := s|maxn<|s||s|n∈D if s 6= ∅, and s− := sif s = ∅.

For each s ∈ 2<ω and each t ∈ D, construct the following:i. xs ∈ X and yt ∈ Y ,ii. Σ0

1 sets Xs ⊆ X and Yt ⊆ Y ,iii. a Σ1

1 set Vt ⊆ X × Y .We ask these sets to satisfy the following conditions, for each s ∈ S and

t ∈ D:1. xs ∈ Xs, yt ∈ Yt, and (xt, yt) ∈ Vt2. diam(Xs) < 2−|s|, diam(Yt) < 2−|t|, and diamGH(Vt) < 2−|t|

3.

Xsε ⊆ Xs if ε ∈ 2,Yt ⊆ Yt− if t ∈ D\∅Vt ⊆ Vt− if t ∈ D

4. Vt ⊆ N ∩ Ω ∩ (Xt × Yt)5. if t ∈ D and tε /∈ D, then xtε ∈ C ∩ π0[Vt] ∩Xt

6. xs = xsm if s /∈ D ∧ sm /∈ D.Suppose that these objects have been constructed. If α ∈W, define

f(α) :=⋂α∈Ns

Xs.

And if α /∈ ∪Cn, then there is an increasing sequence (sk) in D such thatα ∈ ∩k∈ωNsk , so define, by condition 3,

g(α) :=⋂k∈ω

Ysk ,

F (α) :=⋂k∈ω

Vsk .

Note that these functions are continuous, xsk → f(α) and ysk → g(α) forany strictly increasing sequence (sk) (with sk ∈ D in the second case) such thatα ∈ ∩k∈ωNsk .

Note that, by condition 6, if α ∈ Cn, and sk is an increasing sequence suchthat α ∈ Nsk , then f(α) = xsk ∈ C, for k big enough.

If α /∈ ∪Cn, then (xsk , ysk) → F (α) in the Gandy-Harrington topology, soin the usual topology as well. In particular, (f(α), g(α)) = F (α) ∈ Vsk ⊆ B,and g(α) ∈ U .

To construct the points, we proceed by induction of the length of the se-quences, as usual. Note there is (x∅, y∅) ∈ N ∩ Ω, since N is a non-empty Σ1

1set. Choose Σ0

1 neighborhoods X∅, Y∅ of small diameter. We can also choose V∅a Σ1

1 neighborhood of (x∅, y∅) of small diameter contained in N ∩ (X∅×Y∅)∩Ω.


1 set 51

Suppose that everything is constructed for s such that |s| ≤ l. We proceedby cases.

Suppose first that s ∈ D and ε ∈ 2. If sε ∈ D, define (xsε, ysε) := (xs, ys).We only need to shrink the neighborhoods to respective Xsε, Ysε and Vsε, andall conditions will be met. Note that xs ∈ C

τ2 ∩Xs ∩ π0[Vs]. Since Xs ∩ π0[Vs]is τ2-open, there is x ∈ C ∩ Xs ∩ π0[Vs] ∩ Ω. So, if sε /∈ D, define xsε = x.Shrink the neighborhood Xs to one of good diameter, and they will satisfy therelevant conditions.

Suppose now that s /∈ D. If sε /∈ D, by condition 6, we must define xsε := xs.Again, shrink the neighborhood, and they will satisfy the relevant conditions.Note there is ε′ ∈ 2, such that s−ε′ v s. It is clear by applying condition 6enough times that xs−ε′ = xs. In particular, xs ∈ Xs ∩ π0[Vs− ]. Thus, thereis x ∈ Xs ∩ π0[Vs− ], and y ∈ Ys− such that (x, y) ∈ Vs− . If sε ∈ D, note(sε)− = s−, so define (xsε, ysε) := (x, y). Find neighborhoods Xsε, Ysε, Vsεsuch that Vsε ⊆ (Xsε×Ysε)∩Vs− ⊆ (Xs×Ys−)∩Vs− . These will clearly satisfyour conditions.

Thus we can construct all our objects.

Now, consider a zero-dimensional Polish space W, and let Fn,ε be a leftbranching scheme. In each Fn,1, consider a 1-disjoint family Cnmm∈ω. Fixa dense countable αn,mk |k ∈ ω subset of Cnm. Define the following zero-dimensional Polish spaces:

X :=(W\(∩n∈ωFn,0)

)⊕ ω,

Y :=(W\(∪n,m∈ωCnm)

)⊕ ω.

If <, ,> is a bijection from ω3 onto ω, then define, in X× Y,

A :=

(0, α)|α ∈ ∩n∈ωFn,0⋃

(αn,mk , < n,m, k >)|n,m, k ∈ ω,

B :=

(α, α) ∈ X× Y|α /∈ (∩n∈ωFn,0) ∪ (⋃

n,m∈ωCnm).

Proposition 4.20. Let X,Y be Polish spaces, and A,B be disjoint analyticsubsets of X × Y . At least one of the following must hold:


1 set.2. (X,Y,A,B) . (X,Y,A,B).

If, in addition, Fn,ε converges and ∪m∈ωCnm is dense and meager inFn,1, then at most one of the previous can hold.

Proof. Suppose that A is not separable from B by a Π02×Σ0

1 set. As usual, wecan assume X,Y to be recursively presented, and A,B to be Σ1

1.By Lemma 4.7, there is y ∈ π1[A] such that for every open neighborhood V

of y,(π0[A]

τ2 × V)∩B 6= ∅.

We fix a decreasing basis of neighborhoods Vn at y and x ∈ X such that(x, y) ∈ A. Note that, for every k ∈ ω, Nn :=

(π0[A]

τ2 × Vn)∩B is not empty.

By Lemma 4.19, we can find, for each Vn, continuous functions fn : Fn,1 → Xand gn : Fn,1\(∪m∈ωCnm)→ Vn such that

1. fn(α) ∈ π0[A] if α ∈ ∪m∈ωCnm,


2. (fn(α), gn(α)) ∈ B, if α ∈ Fn,1\(∪m∈ωCnm).So, for each αn,mk , there is a yn,m,k such that (fn(αn,mk ), yn,m,k) ∈ A. Definef : X→ X and g : Y→ Y by:

f(ε, α) =

x if ε = 1,fn(α) if ε = 0 & α ∈ Fn,1.

g(ε, α) =

yn,m,k if ε = 1 & α =< n,m, k >,gn(α) if ε = 0 & α ∈ Fn,1,y if ε = 0 & α ∈ ∩Fn,0.

The continuity of f is clear, since Fn,1 is a partition of W\(∩n∈ωFn,0) intoclopen sets. For g, the continuity must only be checked in ∩Fn,0. If αk → α,then we can assume that αk ∈ Fnk,1 for some nk which must increase as k tendsto infinity. In that case, note that g(αk) ∈ Vnk , so that it converges to y = g(α).

It is routine to check that f × g is as a reduction.To show the second part, we show A is not separable from B. So suppose

that A ⊆ U × V with U ∈ Π02(X) and V ∈ Σ0

1(Y). We note that ∩Fn,0 = y,and that Fn,0 is a family of clopen neighborhoods of y. Therefore, there isn ∈ ω such that Fn,0 ⊆ π0[V ], where π0 is the canonical projection from W⊕ ωto W.

Also, note that ∪Cn+1m is a dense set in Fn+1,1 ∩π0[U ]. Then Fn+1,1 ∩π0[U ]

is a Π02 dense set in Fn+1,1. Since ∪Cn+1

m is meager in Fn+1,1, there is α ∈(Fn+1,1\ ∪m∈ω Cn+1

m ) ∩ π0[U ]. In particular (α, α) ∈ (U × V ) ∩B.

We are ready to give an example of a basis for this case. Take W = 2ω,Fn,ε = N0nε. Let (sm)m∈ω be an enumeration of all finite sequences in 2<ωwhich do not end with 0. Let Cnm = 0n1sm0∞, so that αn,mk = 0n1sm0∞.It is easy to see these sets satisfy everything that is needed. In particular, ourexamples are

A :=

(0, 0∞)⋃

(0n1sm0∞, < n,m, k >)|n,m, k ∈ ω,

B :=

(α, α)|∃∞i(α(i) = 1).


2 setLet Fn|n ∈ ω be a 0-disjoint partition of ωω, P = αn|n ∈ ω be a

countable dense and co-dense subset in F0, and Fn+1m,ε a left-branching scheme

in Fn+1. Let X := ωω\P and Y := ωω\(∪n∈ω ∩m∈ω Fn+1m,0 ). Define the following

sets in X× Y:A := Diag(F0\P) ∪∆

(∪n,m∈ω Fn+1

m,1),

B := (α, αn)|(α ∈ ∩m∈ωFn+1m,0 ) ∧ (n ∈ ω).

Theorem 4.21. Let X,Y be Polish spaces and A,B be disjoint analytic subsetsof X × Y . At least one of the following holds:


2 set.2. (X,Y,A,B) . (X,Y,A,B).

If, in addition, all left-branching schemes converge, at most one of theprevious holds.


2 set 53

Proof. As usual, we can suppose X,Y to be recursively presented spaces, andA,B to be Σ1

1 sets. Suppose they are not separable. By a Lemma 4.7, there isa y ∈ Π1[A] such that for any τ2-open neighborhood V , (Π0[A] × V ) ∩ B 6= ∅.In other words, Π1[A] ∩ π1[(Π0[A]× Y ) ∩B]

τ26= ∅.

Note also that P is a Σ02 subset of F0, so it is the union of a 1-disjoint family.

Applying lemma 4.19, let f0 : F0\P → X and g0 : F0 → Y be the respectivecontinuous functions.

Note that for each n, g0(αn) ∈ π1[(Π0[A]×Y )∩B]. Thus, there are xn ∈ Xand for each m, (xnm, ynm) ∈ A, such that (xn, g0(αn)) ∈ B and xnm → xn asm→∞.

With all these fixed, define the functions f : X→ X and g : Y→ Y by:

f(α) =

f0(α) if α ∈ F0,xn if α ∈ ∩m∈ωFn+1

m,0 ,

xnm if α ∈ Fn+1m,1 ,

g(α) =g0(α) if α ∈ F0,ynm if α ∈ Fn+1

m,1 .

These are obviously continuous and it is easy to see f × g is a reduction.Suppose that the schemes converge, and that A ⊆ C × V with C closed and

V a Σ02 set. Note that for any n, if α = ∩m∈ωFn+1

m,0 , then α ∈ π0[A], so α ∈ C.On the other hand, π1[A] ∩ F0 = F0\P which is not Σ0

2. In particular, there isαn ∈ P ∩ V . This shows that (C × V ) ∩ B is not empty.

Let Fn = Nn, P = α ∈ N0|∀∞i α(i) = 0, Fn+1m,0 = N(n+1)0m+1 and

Fn+1m,1 = ∪k 6=0N(n+1)0mk. In X× Y defineA := ∆

(α|(α(0) = 0 ∧ ∃∞i(α(i) 6= 0)

)∨ ∃n,m, k((n+ 1)0m(k + 1) v α)

)and,

B :=(

(n+ 1)0∞, αn)|n ∈ ω

.

So this provides us a concrete example for a basis in this case.


Chapter 5

Separability by pot(Γ) sets

Topological characterizationsLet A0, A1 be disjoint Σ1

1 sets of (ωω)2, δ ∈ 2, ξ < ω1. Let us write τL for thetopology ∆2

ωω . We recall that Louveau showed in [Lou80b] that this topologyis Polish. Set ∩θ<0F

δθ = (ωω)2. We define inductively F δξ by:

F δξ =

Aδ ∩⋂θ<ξ F

δθ

τL if ξ ≡ 0 mod 2,A1−δ ∩

⋂θ<ξ F

δθ

τL if ξ 6≡ 0 mod 2,

Let α ∈ ωω and n ∈ ω. We denote by (α)n the element of ωω defined by(α)n(p) := α(< n, p >). Recall from the last part of Chapter 3 the set D ofsecond type descriptions.

Lemma 5.1. Let ξ be a non-zero countable ordinal, ε ∈ 2 such that ξ ≡ εmod 2, and A0, A1 be disjoint Σ1

1 subsets of (ωω)2. If A0 is not separable fromA1 by a pot(Dξ(Σ0

1)) set, then F εξ 6= ∅.

Proof. Suppose that there is a universal set U for the Π11 sets for which there is

a recursive map A : ωω → ωω such that

UA(r) = U(r)0 ∪⋃p≥1¬¬U(r)p

τL.

This was shown to exist by Lecomte in [Lec13, Lemma 6.6]. He also showed in[Lec13, Lemmas 6.7 & 6.13] that there exists a function r : D×(∆1

1∩ωω)2 → ωω

such that if ¬Ur(u,a1,a0) = ∅ then ¬Ua1 is separable from ¬Ua0 by a pot(Γu).We want to apply this for Γu = Dξ(Σ0

1). The main properties from r thatwe need are the following.

1. If u = 0, then r(u, a0, a1) = a1,2. if u = 11w, then r(u, a0, a1) = a0,3. if u = 12〈vn〉 and rp = r(vp, a0, a1), then r(u, a0, a1) = r(v0, b0, b1),

where bi = A(〈ai, r1, r2, ...〉).If A0 = ¬Ua0 , A1 = ¬Ua1 , and u a description of Dξ(Σ0

1), we will show byinduction on ξ that F εξ = ¬Ur(u,a1,a0)

τL , for ε ≡ ξ mod 2.

55

56 Chapter 5. Separability by pot(Γ) sets

•Case 1: ξ = 1. Then u = 12〈vn〉, where v0 = 0, and vn = 010∞ for n > 0.Therefore, rn = a0 for any n > 0, r(v0, b1, b0) = b0 with b0 = A(a0, a1, a1, . . .).Thus Ub0 = Ua0 ∪ ¬¬Ua1

τL . Then ¬Ub0 = A0 ∩A1τL .

•Case 2: ξ is limit. Fix a cofinal sequence (θn) in ξ such that θn ≡ ξmod 2. Recall that by the observation after Theorem 3.22, u = 12〈vn〉, wherev0 = 0, and vn is a description for Dθn(Σ0

1) for n > 0, Therefore, if rn =r(vn, a1, a0), then F εθn = ¬Urn

τL by the induction hypothesis. On the otherhand r(v0, b1, b0) = b0. But, b0 = A(〈a0, r1, r2, ...〉), so that

Ub0 = Ua0 ∪⋃p≥1¬¬Urp

τL,

as required.•Case 3: ξ = ξ′ + 1. Then u = 12〈vn〉, where v0 is a description for Σ0

1, andvn is a description for Dξ′(Σ0

1) for n > 0. So by the induction hypothesis, ifrn = r(vn, a1, a0), then F εξ′ = ¬Urn

τL for all n > 0 (there is a double inversionof the superscript, one because the parity of ξ′ is different than that of ξ andthe other one because there is a complement, so that the roles of A0, A1 areexchanged).

By Case 1 applied to b0 and b1, Ur(v0,b1,b0) = Ub0 ∪ ¬¬Ub1

τL .But bi = A(ai, r1, r2, . . .). Therefore,

Ub0 = Ua0 ∪⋃n≥1¬¬Urn

τL = Ua0 ∪ ¬F εξ′ ,

Ub1 = Ua1 ∪⋃n≥1¬¬Urn

τL = Ua1 ∪ ¬F εξ′ .

If r = r(u, a1, a0),

¬Ur = ¬Ua0 ∩ F εξ′ ∩ ¬Ua1 ∩ F εξ′τL = A0 ∩ F εξ′ ,

because F εξ′ = A1 ∩⋂θ<ξ Fθ

τL ⊆ A1 ∩A1 ∩⋂θ<ξ Fθ

τLτL⊆ A1 ∩ F εξ′

τL (sincethe parity of ξ′ is different from that of ε).

Finally, ¬UrτL = A0 ∩ F εξ′

τL = F εξ , as required.

We also obtain a similar one for the pot(∆(Dξ(Σ01))). Set ∩θ<0Gθ = ωω.

We define inductively Gξ by:

Gξ = ⋂

θ<ξ Gθ if ξ is limit,A0 ∩Gθ

∆2X ∩A1 ∩Gθ

∆2X if ξ = θ + 1,

Lemma 5.2. Let ξ be a non-zero countable ordinal, and A0, A1 be disjoint Σ11

subsets of (ωω)2. If A0 is not separable from A1 by a pot(∆(Dξ(Σ01))) set, then

Gξ 6= ∅.

Proof. Suppose that u is a description for Dξ(Σ01).

•Case 1: ξ = θ + 1. Let u be a description for Dθ(Σ01)

With the notation of the previous proof, set R(u, a1, a0) := ¬Ur(u,a1,a0).Lecomte showed in [Lec13, Lemma 6.36] that

R′(u, a1, a0) = R(u, a1, a0)τL ∩R(u, a0, a1)

τL

Separation by a pot(∆02) set 57

is not empty. By the previous Lemma, R(u, a1, a0)τL = F εθ and R(u, a0, a1)

τL =F 1−εθ .•Case 2: ξ is limit. Let θn be cofinal sequence in ξ,and vn be a description

for Dθn(Σ01).

Lecomte showed in [Lec13, Lemma 6.36] that

R′(u, a1, a0) =⋂n∈ω

R(un, a1, a0)τL

is not empty. By the previous Lemma, R(un, a1, a0)τL = F εnθn , where θn = εn.

It remains to show, by induction on ξ, that

Gξ =F 0θ ∩ F 1

θ if ξ = θ + 1,∩F εnθn if ξ is limit.

Suppose that ξ = 1. Then G1 = A0τL ∩A1

τL = F 00 ∩ F 1

0 .Suppose that ξ = θ + 2. Then Gθ+2 = A0 ∩Gθ+1

τL ∩ A1 ∩Gθ+1τL , by the

induction hypothesis, Gθ+2 = A0 ∩ F 0θ ∩ F 1

θ

τL ∩A1 ∩ F 0θ ∩ F 1

θ

τL . Notice that ifθ is even, F 0

θ ⊆ A0τL and F 1

θ ⊆ A1τL , so that Gθ+2 = A0 ∩ F 1

θ

τL ∩A1 ∩ F 0θ

τL =F 0ξ ∩ F 1

ξ . A similar argument shows the result for θ odd.Suppose that ξ = θ + 1, with θ limit. Fix a sequence (θn) of even ordinals,

which is cofinal in θ. Then, Gθ+1 = A0 ∩GθτL ∩ A1 ∩Gθ

τL . By the inductionhypothesis, Gθ =

⋂F 0θn. Therefore Gθ+1 = A0 ∩

⋂F 0θn

τL ∩ A1 ∩⋂F 0θn

τL =F 0θ ∩ F 1

θ .Suppose that ξ is limit. Fix a sequence (θn) of infinitely many even ordinals

and infinitely many odd ordinals, which is cofinal in θ. Then Gθ = ∩Gθn+1 =∩(F 0

θn∩ F 1

θn), by the induction hypothesis. Note that F εν ⊆ F εν′ if ν ≤ ν′, so

that⋂F εθn =

⋂θn≡εmod 2

F εθn , as required.

Separation by a pot(∆02) set

Let Sε := (α, β) ∈ E0|∃n∀m > n(α(m) = β(m))∧(α(n) 6= β(n)∧(n−1)0 ≡ε mod 2. We fix a frame F := (sl, tl)|l ∈ ω, and let T be the tree generatedby this frame.

Lemma 5.3. S0 ∩ dT e is not separable from S1 ∩ dT e by a pot(∆02) set.

Proof. Suppose by contradiction that they are separable by a pot(∆02) set. By

Lemma 3.3 there is H a dense Π02 subset of 2ω such that S0 ∩ dT e ∩ H2 is

separable from S1 ∩ dT e ∩H2 by C a ∆02(H2) set.

But notice that S0∩dT e∩H2 is dense in dT e∩H2. Let (α, β) ∈ dT e∩H2, andO0, O1 be open such that (α, β) ∈ O0 × O1. Then, there is q ∈ ω and w ∈ 2<ωsuch that (α, β) ∈ (Nsp0w ×Ntp1w) ⊆ O0 ×O1. By properties of the frame, wecan take q, such that (q)0 is even, and Nsq0 ⊆ Nsp0w and Ntq1 ⊆ Ntp1w. Finally,we claim that there is γ such that (sq0γ, tq1γ) ∈ H. Else, let f : Nsq0 → Ntq1be defined by f(sq0γ) = tq1γ. Notice that f is an homeomorphism. However,(Nsq0 ∩H)∩ f−1(Ntq1 ∩H) = ∅ which contradicts the fact that H is comeager.

Similarly, one can prove that S1 ∩ dT e ∩H2 is dense in dT e ∩H2. But then,both C and (dT e∩H2)\C would be ∆0

2(dT e∩H2) dense sets, a contradiction.


Theorem 5.4. Let X be a Polish space and A0, A1 be disjoint analytic relationson X such that A0 ∪A1 is quasi-acyclic. Exactly one of the following holds:

1. A0 is separable from A1 by a pot(∆02) set.

2. There is f : 2ω → X injective continuous such that for ε ∈ 2, Sε ∩ dT e ⊆(f × f)−1(Aε) for all ε ∈ 2.

Proof. Notice they cannot both hold simultaneously by the previous proposition.Notice that there are is a countable partition of X2\Diag(X) in rectangles

of the form K0 ×K1, with K0,K1 disjoint Borel sets. Indeed, there is a Borelisomorphism h : X → 2ω. The sets Ns0 × Ns1, for s ∈ 2<ω, form a partitionof (2ω)2\Diag(2ω). By taking the inverse images of this partition by h× h, weobtain our Borel rectangles.

Notice that A0 ∩ Diag(X) and A1 ∩ Diag(X) are disjoint analytic sets inDiag(X). So A0 ∩Diag(X) is separable from A1 ∩Diag(X) by a Borel subset ofDiag(X). By [Lou80a, Theorem 6], this Borel subset must be pot(Π0

1). Thus,if A0 is not separable from A1 by a pot(∆0

2) set, then there is K0 ×K1 in thepartition such that A0 ∩ (K0 ×K1) is not separable from A1 ∩ (K0 ×K1) by apot(∆0

2) set. Since these sets are contained in A0 and A1, a reduction for themwould also be a reduction for A0 and A1. Thus, we can assume A0 and A1 arecontained in K0 ×K1.

By refining the topology, we can suppose that X is a zero-dimensional Polishspace. Furthermore, by Theorem 3.14, there are relations S0 and S1 of (2ω) notseparable by a pot(∆0

2) set and a continuous function h : 2ω → X such thatSε ⊆ (h×h)−1(Aε) for all ε ∈ 2. Again, we can suppose that Aε = (h×h)(Sε).Also, in [Lec13] Lecomte proved that these sets must be Σ0

2, so in particular Kσ

(i.e., an union of compact sets). Thus, their direct images are also Kσ.Therefore, Aε = ∪C2n+ε, for Cn pairwise disjoint closed sets. Let (Rn)n∈ω

be a witness for the fact that A0∪A1 is a quasi-acyclic relation.We can supposethat X is a recursively presented Polish space, and that the Aε’s and Rn’sare Σ1

1. Suppose also that the relation (x, y) ∈ Cn is ∆11 in (x, y, n). For

F ∈ Π01(X2,∆2

X), let F ′ := F ∩A0∆2X ∩ F ∩A1

∆2X . We define, recursively,

F0 := X2

Fξ+1 := F ′ξFλ :=

⋂ξ<λ Fξ if λ is limit

Recall that (X,∆2X) is Polish by Proposition 3.19. As each Fξ is closed in

this space, there is θ < ω1 such that Fθ = Fθ+1, by the Lindelöf property.Notice that X = Fθ t

⊔ξ(Fξ\Fξ+1). Let Hξ = Fξ ∩A1

∆2X . We claim that

Fθ 6= ∅. Else, ∪(Fξ\Hξ

)would be a ∆0

2(X2,∆2X) which separates A0 from A1.

Indeed, to see that A0 is contained in it suppose that x ∈ A0. Then thereis ξ such that x ∈ Fξ\Fξ+1. By contradiction, if x were in Hξ, then x ∈(Fξ ∩A0) ∩ Fξ ∩A1

∆2X ⊆ Fξ+1. Thus, A0 ⊆ ∪(Fξ\Hξ).

It remains to check that tFξ\Hξ does not intersect A1. Else, let x ∈ A1 ∩(Fξ\Hξ) for some ξ. Then x ∈ Fξ ∩ A1 ⊆ Hξ. So A1 does not intersects∪(Fξ\Hξ). Finally, notice that Fξ and Hξ are ∆2

X -closed sets, so that ∪(Fξ\Hξ)is ∆0

2 in ∆2X .


We claim that Fθ is Σ11. The proof uses the theory developed in [Mos80,

Section 7C]. Let ϕ be the set relation defined by

ϕ(x, y, P )⇔ (x, y) /∈ (¬P )′

Notice that ϕ is monotone (i.e., if P ⊆ Q and ϕ(x, y, P ), then ϕ(x, y,Q))and thus operative. Also, it is Π1

1 on Π11. We recall some definitons from [Mos80,

Chapter 7]. Given a operative set relation φ(x, y, P ), we define its iterates φξrecursively by φξ(x, y) if and only if φ(x, y, x′ : ∃η < ξ φη(x′, x)). We defineφ∞ by φ∞(x, y) if and only if ∃η φη(x, y).

By [Mos80, Theorem 7C.8], ϕ∞ ∈ Π11. An induction shows that ϕξ(x, y) is

equivalent to (x, y) /∈ Fξ+1. In particular, ϕ∞ is equivalent to ¬Fθ.We also show the following property for each ε ∈ 2:

∀q ∈ ω∀U, V ∈ Σ11(Fθ ∩ (U × V ) 6= ∅ ⇒

∃n(2n+ ε ≥ q ∧ Fθ ∩ C2n+ε ∩ (U × V ) 6= ∅)) (*)

In other words, for each q, the set I := Fθ ∩ (∪2n+ε≥qC2n+ε) is Σ2X -dense in

Fθ. So fix q and suppose that ε = 0, the other case being similar, and let usprove first that I is ∆2

X -dense. So let U, V ∈ ∆11, such that Fθ ∩ (U × V ) 6= ∅.

Notice that Fθ ∩A1 is ∆2X -dense in Fθ, so there is (x, y) ∈ (Fθ ∩A1)∩ (U × V ).

Since Fθ ∩ A0 is ∆2X -dense in Fθ, there are (xk, yk) ∈ A0 ∩ Fθ converging to

(x, y) (in both ∆2X and the product topology, since the former is a refinement of

the latter). For each k, there is nk such that (xk, yk) ∈ C2nk . Since each C2nkis closed and (x, y) /∈ C2nk , we can suppose that the sequence (nk) is strictlyincreasing. This shows that (xk, yk) ∈ I ∩ (U × V ), for k big.

To show this for Σ2X , notice first that I is Σ1

1, since Fθ and “(x, y) ∈ C2nk”are. Two applications of the separation theorem for Σ1

1 sets show that I∆2X =

IΣ2X , but then Fθ ⊆ I

∆2X = I

Σ2X , which is what we are looking for.

We set for each ~u = (u0, u1) ∈ T\~∅,

~t(~u) := (sq0, tq1) if ~u = (sq0w, tq1w)

We are ready to construct our function. We do this by constructing thefollowing objects:

- a sequence (xs)s∈2<ω\∅ of points of X,- a sequence (Xs)s∈2<ω\∅ of Σ1

1 subsets of X,

- a map Φ : ~t(~u)|~u ∈ T\∅ → ω.

We will require these objects to satisfy the following, for each s ∈ 2<ω\∅and each (u0, u1) ∈ T\∅:

1. xs ∈ Xs,2. Xsε ⊆ Xs ⊆ ΩX ∩Ks(0),3. diamGH(Xs) ≤ 2−|s|,4. (xu0 , xu1) ∈ Fθ ∩ CΦ(~t(u0,u1)),5. Xs0 ∩Xs1 = ∅,6. the parities of Φ(sq0, tq1) and (q)0 coincide.


Suppose these objects have been constructed. For α ∈ 2ω, the sequence(Xα|n)n>0 is a decreasing sequence of clopen subsets in ΩX with vanishingdiameter, so this defines f(α) ∈ ∩n>0Xα|n. Notice that this defines a continuousfunction (since the Gandy-Harrington topology is finer that the usual topology).Also, this function is injective in each Nε by 5. Notice that by 2, f maps Nε toKε, so that it is injective.

Note that if (α, β) ∈ dT e∩Sε, then Φ(~t((α, β)|n)) = 2N+ε for n big enough.Since (xα|n, xβ|n) → (f(α), f(β)) and C2N+ε is closed, then (f(α), f(β)) ∈C2N+ε ⊆ Aε.

We now show that the construction is possible by induction on the lengthof s. By (*) we obtain an even natural number Φ(0, 1) ≥ 1 and (x0, x1) ∈ Fθ ∩CΦ(0,1)∩ΩX2 . As ΩX2 ⊆ Ω2

X and CΦ(0,1) ⊆ K0×K1, we can find Σ11 sets X0, X1

of diameter at most 2−1 such that (x0, x1) ∈ (X0 × X1) ⊆ (ΩX)2 ∩ CΦ(0,1) ⊆(ΩX)2 ∩ (K0 ×K1).

Let l ≥ 1, we will construct all our objects for s ∈ 2l+1. First, let m(s, t) ∈ ωsuch that (xs, xt) ∈ Rm(s,t). Note that (xsl , xtl) ∈ Fθ∩(Usl×Utl), where Usl , Utlare defined by:

Usl :=x ∈ X|∀s ∈ 2l∃x′s ∈ Xs

(x = x′sl ∧ ∀~u ∈ T ∩ (2l × 2l)

(x′u0, x′u1

) ∈ Fθ ∩ CΦ(~t(u0,u1)) ∩Rm(u0,u1)),

Utl :=x ∈ X|∀s ∈ 2l∃x′s ∈ Xs

(x = x′tl ∧ ∀~u ∈ T ∩ (2l × 2l)

(x′u0, x′u1

) ∈ Fθ ∩ CΦ(~t(u0,u1)) ∩Rm(u0,u1)).

Observe that these sets are Σ11, so (*) gives

Φ(sl0, tl1) > maxq<l

Φ(sq0, tq1)

of correct parity, and

(xsl0, ytl1) ∈ Fθ ∩ CΦ(sl0,tl1) ∩ (Usl × Utl).

As xsl0 ∈ Usl , there are witnesses (xs0)s∈2l . Similarly, by ytl1 ∈ Utl , we obtain(xs1)s∈2l .

Note that xsl0 6= xsl1, since (xsl0, xsl1) ∈ CΦ(sl0,sl1) and (xsl1, xtl1) ∈CΦ(~t(sl1,tl1)), and Φ(sl0, sl1) > Φ(~t(sl1, tl1)). Similarly, xtl0 6= xtl1.

We can also see that xs0 6= xs1. Suppose without loss of generality thats(0) = 0. Consider an s(T )-path ps from sl to s, say (u0, u1, . . . , uk) whichexists since s(T ) ∩ (2l, 2l) is a connected graph in 2l. By the definition ofUsl and Utl , there are two q.a. equivalent s(A0 ∪ A1)-paths, one from xs0 toxsl0 and the other from xs1 to xsl1. Also (xsl0, xtl1) ∈ A0 ∪ A1 ⊆ R and(xsl1, xtl1) ∈ Rm(sl,tl) ⊆ R. We obtain the following graph.


xsl0 xu10 xuk−10 xs0

xtl1

xsl1 xu11 xuk−11 xs1

Rε1m(u0,u1) Rεkm(uk−1,uk)

6=

Rε1m(u0,u1) Rεkm(uk−1,uk)

R

R

Therefore, we can apply quasi-acyclicity. Thus xs0 6= xs1. The case s(0) = 1is handled in a similar way.

Finally, we take small enough Σ11 neighborhoods for each xsε and ysε, which

completes the construction.

Corollary 5.5. Let A0, A1 be disjoint analytic relations in X such that A0∪A1is quasi-acyclic. Exactly one of the following holds:


2. There is f : 2ω → X injective continuous such that for ε ∈ 2, Sε ∩ dT e ⊆(f × f)−1(Aε).

Proof. This corollary comes from Proposition 3.17.

Corollary 5.6. Let X be a Polish space and A0, A1 be disjoint analytic relationson X. The following are equivalent:

1. there is a s-acyclic relation R such that A0 ∩ R is not separable fromA1 ∩R by a pot(∆0

2) set.2. there is a locally countable relation R such that A0 ∩ R is not separable

from A1 ∩R by a pot(∆02) set.

3. there is a quasi-acyclic relation R such that A0 ∩R is not separable fromA1 ∩R by a pot(∆0

2) set.4. there is f : 2ω → X injective continuous such that Sε ∩ dT e ⊆ (f ×

f)−1(Aε) for all ε ∈ 2.

Proof. 1 implies 3 and 2 implies 3 are a consequence of Proposition 3.17. 3implies 4 is a direct consequence of the previous theorem. For the other direc-tions, observe that R := (f × f)[dT e ∩ (S0 ∪ S1)] is a locally countable s-acyclicrelation.

Corollary 5.7. Let X be a Polish space, A0, A1 disjoint analytic relations onX such that A0 is contained in a pot(∆0

2) quasi-acyclic relation. Exactly one ofthe following holds:


2. There is f : 2ω → X injective continuous, such that (Sε ∩ dT e) ⊆ (f ×f)−1(Aε).

Proof. Let R be a quasi-acyclic pot(∆02) relation such that A0 ⊆ R. It is enough

to show that A0∩R = A0 is not separable from A1∩R by a pot(∆02) set, by the

previous lemma. But if it were separable by P ∈ pot(∆02), then P ∩ R would

be a pot(∆02) set that separates A0 from A1.


Corollary 5.8. Let X,Y be Polish spaces, and A0, A1 disjoint analytic subsetsof X × Y , such that A0 is contained in a pot(∆0

2) locally countable relation orA0 ∪A1 is a locally countable relation. Exactly one of the following holds:


2. There are f : 2ω → X and g : 2ω → Y injective continuous, such that(Sε ∩ dT e) ⊆ (f × g)−1(Aε).

Proof. The exactly part is clear. Suppose that A0 is not separable from A1 by apot(∆0

2) set. Note that X,Y are not countable, so that we can find ∆02 disjoint

non-countable sets X0, X1 ⊂ X and Y0, Y1 ⊆ Y such that X0 ∪ X1 = X andY0 ∪ Y1 = Y .

Notice then that there are ε, ε′ ∈ 2 such that A0∩ (Xε×Yε′) is not separablefrom A1 ∩ (Xε×Yε′) by a pot(∆0

2) set. Without loss of generality, suppose it isε = ε′ = 0. We also fix continuous injections iY : 2ω → X1 and iY 2ω → Y1.

Define A′ε ⊆ X0 ⊕ Y0 by A′ε := (x, y)|(x, y) ∈ Aε ∩ (X0 × Y0). Note thatif A′0 is separable from A′1 by a pot(∆0

2) set, then A0 ∩ (X0 × Y0) is separablefrom A1 ∩ (X0 × Y0) by a pot(∆0

2) set.So let f ′ : 2ω → X0 ⊕ Y0 be the injective continuous function obtained

by applying Theorem 5.4. Let α ∈ 2ω. Notice that (0α, 1α) ∈ S0 ∩ dT e. Inparticular, (f ′(0α), f ′(1α)) ∈ A′0 ⊆ (0 ×X0)× (1 × Y0).

Define f : 2ω → X by f ′(0α) = (0, f(0α)) and f(1α) = iX(1α) and g : 2ω →Y by f ′(1α) = (1, g(1α)) and g(0α) = iy(0α).

Separation by a Dξ(Σ01) set

Let b : ω → 2<ω be the surjection given by ∅, ∅, 0, 0, 1, 1, 00, 00, 01, 01,10, 10, 11, 11 . . . We are taking every sequence twice for reasons that will beclear in the following section. Note that 2|b(q)| ≤ q. Also, let I : ω<ω → ω bethe injection defined by I(∅) = 1 and I(s) = p

s(0)+10 p

s(1)+11 . . . p

s(|s|−1)+1|s|−1 , where

(pk)k∈ω is the increasing sequence of prime numbers.Fix 1 ≤ ξ < ω1. For every limit ordinal λ < ξ, we inductively fix a strictly

increasing sequence (αλn)n∈ω of odd ordinals, cofinal in λ. We define the functionϕξ : ω<ω → (ξ + 1) ∪ −1 by induction on the length of s by ϕξ(∅) = ξ and,

ϕξ(sn) =

θ if ϕξ(s) = θ + 1,−1 if ϕξ(s) = −1 ∨ ϕξ(s) = 0αλn if λ is limit

Let Tξ = s ∈ ω<ω|ϕξ(s) 6= −1 and If Seqξ = I(Tξ), let iξ : Seqξ → ω bethe increasing bijection. We define the bijection 〈〉ξ : Tξ → ω by 〈s〉ξ = iξ I(s).

Observe that for all u ∈ ω<ω and all q ∈ ω, if uq ∈ Tξ, then 〈uq〉ξ ≥〈u〉ξ + 1 + |b(q)|. In fact, note that

I(u) < I(u0) < . . . < I(uq).

Theniξ(I(u)) < iξ(I(u0)) < . . . < iξ(I(uq)).

Thus, 〈u〉ξ + q + 1 ≤ 〈uq〉ξ.

Separation by a Dξ(Σ01) set 63

Fix ξ, for each u ∈ Tξ we define (t0u, t1u) ∈ (2 × 2)<ω by tε∅ = ∅, andtεuq = tεub(q)0〈uq〉ξ−〈u〉ξ−|b(q)|−1ε. Notice that by the previous remark, this iswell defined and |tεu| = 〈u〉ξ.

Finally, we define the following relations on 2ω, for ε ∈ 2.

Aξε = (t0uγ, t1uγ)|γ ∈ 2ω ∧ (|u| ≡ ε mod 2).

Notice that |u| is even if and only if ϕξ(u) ≡ ξ mod 2. By Proposition 3.15,A0 ∪ A1 is s-acyclic.

Lemma 5.9. Let 0 < ξ < ω1. Then Aξ0 is not separable from Aξ1 by apot(Dξ(Σ0

1)) set.

Proof. We proceed by contradiction, so suppose that A0ξ is separable from A1

ξ

by C ∈ pot(Dξ(Σ01)). Let τ ′ be a topology where C ∈ Dξ(Σ0

1). By Lemma 3.3there is a dense Π0

2 subset H of 2ω such that C ∩H2 is Dξ(Σ01) in H2. So there

is (Oθ)θ<ξ an increasing sequence of open sets such that

C ∩H2 =⋃

η<ξ,η 6≡ξ mod 2(Oη\ ∪θ<η Oθ) ∩H2.

Fix (t0uγ, t1uγ) ∈ H2. We first show, by induction on ν < ξ, that if ϕξ(u) = ν,then (t0uγ, t1uγ) /∈ Oν .

Suppose that ν = 0. Then, ξ ≡ ε mod 2, where (t0uγ, t1uγ) ∈ Aε. Observethat if ξ is even, then A0 ∩H2 ∩O0 ⊆ C ∩H2 ∩O0 = ∅. On the other hand, ifξ is odd, O0 ∩H2 ⊆ C ∩H2, so that A1 ∩H2 ∩ O0 = ∅. So in both cases, weobtain that (t0uγ, t1uγ) /∈ O0.

Suppose now that it is true for any η < ν. Suppose by contradiction thatthere is u ∈ Tξ and γ ∈ 2ω such that ϕξ(u) = ν, and (t0uγ, t1uγ) ∈ Oν .

We claim that for any k ∈ ω there is pk ∈ ω and γk ∈ 2ω such that(t0upkγk, t

1upk

γk) ∈ H2 and tεuγ|k v tεupkγk. Else, fix p ∈ ω. Then, the func-tion f : Nt0up → Nt1up defined by f(t0upγ) = t1upγ is an homeomorphism but(H ∩Nt0up)∩f−1(H ∩Nt0up) = ∅, which contradicts the fact that H is comeager.Thus, we obtain a sequence in H2 such that tεupkγk → tεuγ.

First case: ν ≡ ξ mod 2. In this case, (t0uγ, t1uγ) ∈ A0 ∩ H2. Then,(t0uγ, t1uγ) ∈ ∪η<ξ,η 6≡ξ mod 2(Oθ\ ∪θ′<θ Oθ′). Notice that, because of the parity,(t0uγ, t1uγ) ∈ Oη for some η < ν. As Oη is open, (t0upkγ, t

1upk

γ) ∈ Oη for k largeenough. If ν = ν′ + 1, then ϕξ(upk) = ν′ ≥ η and so we obtain a contradictionwith our induction hypothesis. If ν is limit ϕξ(upk) is eventually increasing, soϕξ(upk) > η for k large, and again we obtain a contradiction.

Second case: ν 6≡ ξ mod 2. Then, (t0uγ, t1uγ) ∈ A1 ∩H2. If (t0uγ, t1uγ) /∈ Oηfor η < ν, then (t0uγ, t1uγ) ∈ A1 ∩Oν\ ∪η<ν Oη, which contradicts the definitionof the Oθ. Then, we obtain (t0uγ, t1uγ) ∈ Oη for some η < ν. We obtain the samecontradiction as in the previous case.

Now, consider α ∈ H, so that (α, α) ∈ A0 ∩H2 ⊆ ∪η<ξ,η 6≡ξ mod 2(Oθ\∪θ′<θOθ′). Then there is η < ξ such that (α, α) ∈ Oη. By the same argument asbefore, we can find (t0(pk)γk, t

1(pk)γk) ∈ H2 such that (t0(pk)γk, t

1(pk)γk) → (α, α).

Therefore, (t0(pk)γk, t1(pk)γk) ∈ Oη, which contradicts what we just proved.

Let us define A = (A0,A1), (B(A0),B(A1)).


Theorem 5.10. Let 0 < ξ < ω1, and A0, A1 be disjoint analytic relations in Xsuch that A0 ∪A1 is a quasi-acyclic relation. The following are equivalent:

1. A0 is not separable from A1 by a pot(Dξ(Σ01)) set.

2. There are f : 2ω → X an injective continuous function and (A,B) ∈ Asuch that A ⊆ (f × f)−1(A0) and B ⊆ (f × f)−1(A1).

3. There are f, g : 2ω → X injective continuous functions such that A0 ⊆(f × g)−1(A0) and A1 ⊆ (f × g)−1(A1).

Proof. (1.⇒ 2.) Let (Rn)n∈ω be witnesses for A0 ∪ A1 being quasi-acyclic.Suppose that A0 is not separable from A1 by a pot(Dξ(Σ0

1)) set.Suppose that A0 cannot be separated from A1 by a pot(Dξ(Σ0

1)) set. Inparticular,X is non-empty. As such, there is a continuous surjection h : ωω → Xand a closed subset Y of ωω such that h|Y is injective and h[Y ] = X. LetA′ε = (h × h)−1(Aε) ∩ Y 2. Notice that A′0 is not separable from A′1 by apot(Dξ(Σ0

1)) set.Suppose that there are an injective continuous function f ′ : 2ω → ωω and

(A,B) ∈ A such that A ⊆ (f ′ × f ′)−1(A′0) and B ⊆ (f ′ × f ′)−1(A′1). There aretwo cases. Either

∀α ∈ 2ω(f ′(α), f ′(α)) ∈ A′0or

∀α ∈ 2ω(f ′(0α), f ′(1α)) ∈ A′0.

In both cases, f ′[2ω] ⊆ Y . Let f : 2ω → X be defined by f = h f ′. Thenf is the desired injective continous reduction. Thus, it is enough to prove thetheorem for ωω. Also, without loss of generality, we can suppose that A0, A1, Rnare Σ1

1 subsets of (ωω)2.For each ν ≤ ξ, if ξ ≡ ε mod 2 and ν + ε ≡ ε′ mod 2, then let Fν :=

Aε′ ∩⋂θ<ν F

εν . So Fν

∆2X = F εν . In particular, by Lemma 5.1, Fξ is a non-empty

Σ11 subset of (ωω)2.We also define Dν := (t0uw, t1uw)|w ∈ 2<ω∧ϕξ(u) = ν, so that for example

Dξ = (s, t) ∈ 2<ω × 2<ω|s = t. Also, observe that the tree generated by thev-predecesors of the (t0u, t1u) is ∪ν<ξDν . In particular, s(∪ν<ξDν ∩ (2× 2)n) isan acyclic connected graph.

Let S := x ∈ X|(x, x) ∈ Fξ.• Case 1: S = ∅. Then there is (x, y) ∈ Fξ such that x 6= y. Let O0, O1 be

disjoint Σ01 neighborhoods of x and y, respectively.

We construct the following:1. xs ∈ X, ys ∈ X for each s ∈ 2<ω,2. Xs, Ys Σ1

1 subsets of X for each s ∈ 2<ω,3. V(s,t) a Σ1

1 subset of X2, for each (s, t) ∈ ∪ν≤ξDν ,And we want these objects to satisfy the following:1. (xs, yt) ∈ V(s,t) ⊆ (Xs × Yt) ∩ ΩX2 ,2. Xsε ⊆ Xs ⊆ ΩX ∩O0, Ysε ⊆ Ys ⊆ ΩX ∩O1, and V(sε,tε) ⊆ V(s,t),3. diamGH(Xs),diamGH(Ys),diamGH(V(s,t)) ≤ 2−|s|,4. Xs0 ∩Xs1 = ∅ = Ys0 ∩ Ys1,5. V(s,t) ⊆ Fν if (s, t) ∈ Dν .


Suppose first that we have constructed these objects. For each α ∈ 2ω, definef0 : 2ω → X by f0(α) = ∩n∈ωXα|n and f1 : 2ω → X by f1(α) = ∩n∈ωYα|n.Notice that these functions are continuous for the Gandy-Harrington topologyof X, so they are also continuous for the usual topology. Notice also that thesefunctions are injective, thanks to condition 4. Finally, define f : 2ω → X byf(εα) := fε(α). It is continuous and, since fε[2ω] ⊆ Oε, it is injective.

Now, suppose that (0α, 1β) ∈ B(Aξε). Then, (0α, 1β) = (0t0uγ, 1t1uγ), withu ∈ Tξ such that |u| ≡ ε mod 2. Note that we can find F (α, β) ∈ X2 such thatF (α, β) = ∩i>nV(α|i,β|i) ⊆ Aε, by condition 5. Thus (xα|i, yβ|i)→ F (α, β) inthe Gandy-Harrington topology ofX2. Since this topology is finer, this sequencealso converges in the square of the Gandy-Harrington topology of X, so that(f(0α), f(1β)) = (f0(α), f1(β)) = F (α, β), which shows that f is as required.

Let us show that the construction is possible. We do this by induction onthe length of s.

Notice first that Fξ∩(O0×O1) is a non-empty Σ11 subset ofX2, so that we can

find (x∅, y∅) ∈ Fξ∩(O0×O1)∩ΩX2 . As ΩX2 ⊆ Ω2X , we can find X∅, Y∅ ∈ Σ1

1(X)of diameter at most 1 such that (x∅, y∅) ∈ X∅ × Y∅ ⊆ Ω2

X ∩ (O0 × O1), andV(∅,∅) ∈ Σ1

1(X2) with diameter at most 1, such that V(∅,∅) ⊆ Fξ∩(X∅×Y∅)∩ΩX2 .These sets satisfy all of the conditions.

Suppose we have constructed our objects for all s ∈ 2m. Notice that(xs, yt) ∈ V(s,t) ⊆ A0 ∪ A1, so that for each (s, t) ∈ (2 × 2)m we can fix k(s, t)such that (xs, yt) ∈ Rk(s,t). Let u ∈ ω<ω and q ∈ ω such that m + 1 = 〈uq〉ξ.In particular, there is w ∈ 2<ω, such that (t0uw0, t1uw1) = (t0m+1, t

1m+1).

Let

O0 =x ∈ X|∀s ∈ 2m∃x′s, y′s ∈ X

(x′t0uw

= x∧∀(s, t) ∈ (2× 2)m ∩

⋃θDθ

((x′s, y′t) ∈ V(s,t) ∩Rk(s,t)

)),

and

O1 =y ∈ X|∀s ∈ 2m∃x′s, y′s ∈ X

(y′t1uw

= y∧∀(s, t) ∈ (2× 2)m ∩

⋃θDθ

((x′s, y′t) ∈ V(s,t) ∩Rk(s,t)

)).

Notice that these sets are Σ11 and that (xt0uw, yt1uw) ∈ (O0 × O1) ∩ Fν ⊆

(O0 × O1) ∩ ∩η<νFη∆2X , where ϕξ(u) = ν. Let η := ϕξ(uq) < ν, so we can

find (xt0uw0, yt1uw1) ∈ Fη ∩ (O0 × O1). For s ∈ 2m, let xs0, ys0, xs1, ys1 be thewitnesses for the facts that xt0uw0 ∈ O0 and yt1uw1 ∈ O1 respectively.

First observe that |u| 6≡ |uq| mod 2. Thus, for some ε ∈ 2, (xt0uw1, yt1uw1) ∈Ut0uw,t1uw ⊆ Aε and (xt0uw0, yt1uw1) ∈ A1−ε. Since A0 and A1 are disjoint, thisimplies that xt0uw0 6= xt0uw1. A similar argument proves that yt1uw0 6= yt1uw1

We need to show that xs0 6= xs1 (similarly for ys0 and ys1). But first, observethat if s 6= t ∈ 2m, then xsε ∈ Xs and xtε′ ∈ Xt, so that xsε 6= xtε′ by condition4. Similarly ysε 6= ytε′ . Also, since xsε ∈ O0 and ytε′ ∈ O1, no “x” is equal tono “y”.

The fact that G := s(∪θ<ξ Dθ ∩ (2 × 2)m

)is a connected acyclic graph

provides a single G-path from s to t0uw. This path gives us two q.a.-equivalents(R)-paths by the definition of Oε, one from xs0 to xt0pw0 and another from xs1to xt0pw1.

Observe that (xt0pw0, yt1pw1) ∈ s(R), (xt0pw1, yt1pw1) ∈ s(R), for any (u, v) inthe G-path (xuε, yvε) ∈ Rk(u,v) or R−1

k(u,v) accordingly, and xtε′ 6= yt1pw1 6= ytε′


for everyone in the s(R)-paths. Thus, by quasi-acyclicity, xs0 6= xs1. Similarly,one can prove that ys0 6= ys1.

Thus, it is easy to find Xs0, Xs1 ⊆ Xs (respectively Ys0, Ys1 ⊆ Ys) disjointΣ1

1 sets with the required properties, as well as V(sε,sε) and V(s,t) accordingly.• Case 2: S 6= ∅. Then, ∅ 6= S2 ∩ Fξ ⊆ S2 ∩ A0. On the other hand, note

that S2 ∩A1 is irreflexive. Else, there is (x, x) ∈ Fξ ∩A1 ⊆ A0 ∩A1.The rest of the proof is similar, so we mention the parts were they differ. We

will in fact obtain here that f0 = f1. So we construct only xs ∈ X,Xs ∈ Σ11(X)

and V(s,t) ∈ Σ11(X2). Any condition on ys will then be unnecessary. 1. will

include (xs, xt) ∈ V(s,t) instead.Let us show that Aε ⊆ (f ×f)−1(Aε) for ε ∈ 2. So suppose that (t0uγ, t1uγ) ∈

Aε. Let F (α, β) ∈ ∩kV(t0uγ|k,t1uγ|k) ⊆ Aε, since the V(s,t) are ΣX2-closed non-empty with vanishing diameter. Also, (xt0uγ|k, xt1uγ|k)→ F (α, β) in ΣX2 , so alsoin τ1(X). This shows that (f(α), f(β)) = F (α, β) ∈ Aε.

In the construction, we start by picking x∅ ∈ S. We must replace Oε by

U0 =x ∈ X|∀s ∈ 2m∃x′s ∈ X

(x′t0uw

= x∧∀(s, t) ∈ (2× 2)m ∩

⋃θDθ

((x′s, x′t) ∈ V(s,t) ∩Rk(s,t)

)),

and

U1 =x ∈ X|∀s ∈ 2m∃x′s ∈ X

(x′t1uw

= x∧∀(s, t) ∈ (2× 2)m ∩

⋃θDθ

((x′s, x′t) ∈ V(s,t) ∩Rk(s,t)

)).

As before, we obtain that (U0 × U1) ∩ Fν 6= ∅, which provides (xt0um0, xt1um1),and the witnesses xs0, xs1.

Let us show that xt0u 6= xt1u for u 6= ∅. If |u| = 1, then (xt0u , xt1u) ∈ A1. Butthis set is irreflexive so they must be different. For u = u′p, we obtain thatxteu′p

= xteu′w ∈ Xte

u′, so by Condition 4, they are different (since t0u′ 6= t1u′). The

rest of the proof follows in a similar manner.(2.⇒ 3.) Let h0(α) = 0α and h1(α) = 1α. Then (h0(α), h1(β)) ∈ B(R) if

and only if (α, β) ∈ R for any relation R on 2ω. Composing the function from2 with these ones, we obtain the required reduction.

(3. ⇒ 1.) This is Lemma 5.9

We notice that A is in fact an antichain for square reductions.

Proposition 5.11. There is no continuous injective function f : 2ω → 2ω suchthat either Aε ⊆ (f × f)−1(B(Aε)) or B(Aε) ⊆ (f × f)−1(Aε) for all ε ∈ 2.

Proof. Note that if f : 2ω → 2ω is injective then Diag(2ω) = (f×f)−1[Diag(2ω)].Also, Diag(2ω) ⊆ A0. If Aε ⊆ (f × f)−1(B(Aε)), then Diag(2ω) ⊆ B(A0), acontradiction.

So suppose that B(Aε) ⊆ (f × f)−1(Aε) holds. Let (0t0uγ, 1t1uγ) ∈ B(Aε),so that (f(0t0uγ), f(1t1uγ)) = (t0vγ′, t1vγ′) ∈ Aε. We claim that ϕξ(u) ≤ ϕξ(v).We proceed by induction on ϕξ(u). Notice that is is obvious for ϕξ(u) = 0.

Suppose that it holds for all ν < ϕξ(u). Notice that we can find pk ∈ ω andγk ∈ 2ω such that (t0upkγk, t

1upk

γk) ∈ A1−ε and (t0upkγk, t1upk

γk) → (t0uγ, t1uγ).Then, by continuity, (t0vkγ

′, t1vkγ′) := (f(t0upkγ), f(t1upkγ)) → (t0vγ′, t1vγ′). In

particular, for k large, (t0v, t1v) v (t0vk , t1vk

). Thus the sequence vk is a strictextension of v. Therefore ϕξ(vk) < ϕξ(v). But, by the induction hypothesis,ϕξ(upk) ≤ ϕξ(vk) < ϕξ(v). If ϕξ(u) = ν + 1, then ν = ϕξ(upk) < ϕξ(v), so we


obtain our result. If ϕξ(u) is a limit ordinal, then (ϕξ(upk))k∈ω is cofinal in it,so ϕξ(v) = ϕξ(u).

Finally, let α ∈ 2ω, so that (0α, 1α) = (0t0∅α, 1t1∅α) ∈ B(A0). Therefore,(f(0α), f(1α)) = (t0vγ′, t1vγ′) , so that ϕξ(v) = ξ and thus v = ∅ which contra-dicts the injectivity of f .

We would like to point out the special case ξ = 1, i.e., separability by apot(Σ0

1) set. In this case, we can take A0 = Diag(2ω) and A1 = H0, as onewould expect from Theorem 2.3.

Corollary 5.12. Let 0 < ξ < ω1 and A0, A1 be disjoint analytic relations in Xsuch that A0∪A1 is s-acyclic or locally countable. The following are equivalent:

1. A0 is not separable from A1 by a pot(Dξ(Σ01)) set.

2. There are f : 2ω → X an injective continuous function and (A,B) ∈ Asuch that A ⊆ (f × f)−1(A0) and B ⊆ (f × f)−1(A1).

3. There are f : 2ω → X and g : 2ω → Y injective continuous such thatAξε ⊆ (f × g)−1(Aε) for all ε ∈ 2.

Proof. This holds by Proposition 3.17.

Corollary 5.13. Let 0 < ξ < ω and let X be a Polish space and A0, A1 bedisjoint analytic relations on X. The following are equivalent:

1. There is a quasi-acyclic relation R such that A0∩R is not separable fromA1 ∩R by a pot(Dξ(Σ0

1)) set.2. There is a locally countable relation R such that A0 ∩R is not separable

from A1 ∩R by a pot(Dξ(Σ01)) set.

3. there is an s-acyclic relation R such that A0 ∩ R is not separable fromA1 ∩R by a pot(Dξ(Σ0

1)) set.4. There is f : 2ω → X an injective continuous function such that Aξε ⊆

(f × f)−1(Aε) or B(Aξε) ⊆ (f × f)−1(Aε) for all ε ∈ 2.

Proof. 1 implies 3 and 2 implies 3 are direct consequences of Proposition 3.17,and 3 implies 4 is a consequence of Theorem 5.10. For the other directions,observe that R := (f × f)[Aξ0 ∪ Aξ1] and R′ := (f × f)[B(Aξ0) ∪ B(Aξ1)] are s-acyclic locally countable relation. Indeed, the injectivity of f guarantees thats-acyclicity is preserved, and as pointed out after the definition of Aξ0 ∪A

ξ1, this

union is s-acyclic.Let P = Aξ0 ∪ Aξ1. Notice that P satisfies that P ∩ P−1 ⊆ Diag(2ω) (i.e.,

it is antisymmetric). To show R′ is s-acyclic, it is enough to show B(P ) iss-acyclic. Suppose not, and take a s(B(P ))-path. It needs to be of the form(ε0γ0, ε1γ1, ε0γ2 . . . , ε1γ2k+1) with (ε0γ0, ε1γ2k+1) ∈ B(P ) for some 2k + 1 > 2.

Consider the path (γ0, γ1, γ2, . . . , γ2k+1) which is in s(P ), and that satisfiesthat (γ0, γ2k+1) ∈ s(P ). Thus, the path cannot be injective.•Case 1: 2k+1 = 3. Then the path is (γ0, γ1, γ2, γ3). Notice that γi 6= γi+2

for all i ∈ 2, otherwise εiγi = εiγi+2, which contradicts the injectivity of thes(B(P ))-cycle. Then, we can suppose without loss of generality that γ0 = γ1.In this case (γ0, γ2, γ3) is still a s(P )-path. We must have γ2 = γ3 or γ0 = γ3.In either case, we obtain (γ0, γ2) ∈ s(P ), so by antisymmetry they are equal,which we already established is not possible.


•Case 2: 2k + 1 > 3. Let γi, γi + 1, . . . , γj be a path such that γi = γj ,and which is minimal with this property. If i + 3 ≤ j, then we would have ans(P )-cycle. So that j < i + 3. If j = i + 2, then γi = γi+2 which we alreadyproved cannot be.

It rests the case where j = i + 1. We can remove γi+1 to obtain a news(P )-path (γ′0, . . . , γ′2k), with 2k > 2. Again, by acyclicity, this path is notinjective. Notice that this path satisfies that there is some l such that γ′l =γi 6= γi + 2 = γ′l+1. We can then find a s(P )-path γ′n, γ

′n+1, . . . , γ

′m such that

it contains γ′l, γ′l+1, γ′n = γ′m and is minimal with this property. The acyclicityof s(P ) guarantees that this path has exactly 3 elements. We can assume it isγ′l, γ

′l+1, γ

′l+2. We obtain (γ′l, γ′l+1) ∈ s(P ), so by antisymmetry they are equal,

a contradiction.

Corollary 5.14. Let A0, A1 be disjoint analytic relations in X such that A1is contained in a quasi-acyclic pot(Π0

1) relation. Exactly one of the followingholds:

1. A0 is separable from A1 by a pot(Σ01) set.

2. There is f : 2ω → X injective continuous such that either— Diag(2ω) ⊆ (f × f)−1(A0) and H0 ⊆ (f × f)−1(A1), or— B(Diag(2ω)) ⊆ (f × f)−1(A0) and B(H0) ⊆ (f × f)−1(A1).

Proof. Let R be a quasi-acyclic pot(Π01) relation containing A1. Suppose that

there is P a pot(Σ01) set that separates A0∩R from A1∩R. Consider S = R∩¬P .

Then, A1 is contained in S and S ∩ A0 6= ∅. Notice that P ∈ pot(Σ01), so that

S is a pot(Π01) set. In particular, A0 is separable from A1 by ¬S, which is a

pot(Σ01).

So, if A0 is not separable from A1 by a pot(Σ01), neither is A0 ∩ R from

A1 ∩R. Aplying the previous corollary, we obtain 2.

Corollary 5.15. Let 0 < ξ < ω1, X,Y be Polish spaces, and A0, A1 be disjointanalytic subsets of X ×Y such that A0 ∪A1 is locally countable. Exactly one ofthe following holds:

1. A0 is separable from A1 by a pot(Dξ(Σ01)) set.

2. There are f : 2ω → X and g : 2ω → Y injective continuous, such thatAε ⊆ (f × g)−1(Aε) for all ε ∈ 2.

Proof. The exactly part is clear. Suppose that A0 is not separable from A1 bya pot(Dξ(Σ0

1)) set.Define A′ε ⊆ (X ⊕ Y )2 by A′ε := (x, y)|(x, y) ∈ Aε. Note that if A′0 is

separable from A′1 by a pot(Dξ(Σ01)) set, then A0 is separable from A1 by a

pot(Dξ(Σ01)) set.

So let f ′, g′ : 2ω → X ⊕ Y be the injective continuous functions obtained byapplying Theorem 5.10. Let α ∈ 2ω. Then (f ′(α), g′(α)) ∈ A′0. In particularf ′(α) ∈ 0 ×X and g′(α) ∈ 1 × Y . Define f : 2ω → X by f ′(α) = (0, f(α))and g : 2ω → X by g′(α) = (1, g(α)).

Separation by a pot(∆(Dξ(Σ01))) set

In this section, we keep the previous definition of ϕξ, 〈〉ξ, as well as that ofthe sequence (t0u, t1u). We define for ε = 0, 1 the following subtrees T εξ := s ∈Tε|∀k < |s|s(0) ≡ ε mod 2.

Separation by a pot(∆(Dξ(Σ01))) set 69

We then set:

Dξ0 := (t0uγ, t1uγ)|γ ∈ 2ω ∧ u ∈ Tξ\∅ ∧ (ϕξ(u) ≡ ξ mod 2⇔ u ∈ T 0ξ ),

Dξ1 := (t0uγ, t1uγ)|γ ∈ 2ω ∧ u ∈ Tξ\∅ ∧ (ϕξ(u) ≡ ξ mod 2⇔ u ∈ T 1ξ ).

Lemma 5.16. Let 0 < ξ < ω1. Then Dξ0 is not separable from Dξ1 by apot(∆(Dξ(Σ0

1))) set.

Proof. We proceed by contradiction, so suppose that Dξ0 is separable from Dξ1 byC ∈ pot(∆(Dξ(Σ0

1))). Let τ ′ be a topology where C is ∆(Dξ(Σ01)). By Lemma

3.3 there is a dense Π02 subset H of 2ω such that H2 ∩ C is ∆(Dξ(Σ0

1)) in H2.So there is, for ε ∈ 2, an increasing sequence (Oεθ)θ<ξ of open sets such that

H2 ∩ C = H2 ∩⋃

η<ξ,η 6≡ξ mod 2(O0

θ\ ∪θ′<θ O0θ′).

H2\C = H2 ∩⋃

η<ξ,η 6≡ξ mod 2(O1

θ\ ∪θ′<θ O1θ′).

As in the proof of Lemma 5.9, if (t0uγ, t1uγ) ∈ H2 such that ϕξ(u) 6= 0,we can find a sequence (t0upkγk, t

1upk

γk)k∈ω in H2 such that (t0upkγk, t1upk

γk) →(t0uγ, t1uγ).

To simplify notation, we suppose that ξ is even, but the proof for ξ odd issimilar.•Let (t0uγ, t1uγ) ∈ H2. We claim that if u ∈ T εξ \∅ and ν ≤ ϕξ(u), then

(t0uγ, t1uγ) /∈ Oεu. We show it by induction on ϕξ(u). Let us suppose u ∈ T 0, theproof is similar in the other case.

–First case: ϕξ(u) = 0. Then (t0uγ, t1uγ) ∈ D0 ∩H2, and by the definition ofthe Oεν ’s, (t0uγ, t1uγ) is not in O0

0.–Second case: ϕξ(u) = θ + 1 ≡ 0 mod 2. Then (t0uγ, t1uγ) ∈ D0 ∩ H2. In

particular, there is ν odd such that (t0uγ, t1uγ) ∈ O0ν\⋃ν′<ν O

0ν′ . Suppose by

contradiction that there is ν ≤ ϕξ(u) such that (t0uγ, t1uγ) ∈ O0ν . Then, we can

suppose that ν = θWe can find a sequence (t0upkγk, t

1upk

γk)k in H2 such that (t0upkγk, t1upk

γk)→(t0uγ, t1uγ). Since O0

θ is open (t0upkγk, t1upk

γk) ∈ O0θ for k big enough. Observe

that ϕξ(upk) = θ, which contradicts the induction hypothesis.–Third case: ϕξ(u) = θ + 1 ≡ 1 mod 2. Then (t0uγ, t1uγ) /∈ D0 ∩ H2.

If (t0uγ, t1uγ) ∈ O0ϕξ(u) then it is in O0

θ , else, it is in C, which contradictsits definition. Suppose by contradiction that there is ν ≤ ϕξ(u) such that(t0uγ, t1uγ) ∈ O0

ϕξ(u). Then, we can suppose that ν = θ

We can find a sequence (t0upkγk, t1upk

γk)k in H2 such that (t0upkγk, t1upk

γk)→(t0uγ, t1uγ). Since O0

θ is open (t0upkγk, t1upk

γk) ∈ O0θ for k big enough. Observe

that ϕξ(upk) = θ, which contradicts the induction hypothesis.–Fourth case: ϕξ(u) is limit. Then (t0uγ, t1uγ) ∈ D0∩H2. In particular, there

is ν odd such that (t0uγ, t1uγ) ∈ O0ν\⋃ν′<ν O

′ν . Suppose by contradiction that

there is ν ≤ ϕξ(u) such that (t0uγ, t1uγ) ∈ O0ν . Then, we can suppose that ν < ϕ

As explained before, we can find a sequence (t0upkγk, t1upk

γk)k∈ω in H2 suchthat (t0upkγk, t

1upk

γk) → (t0uγ, t1uγ). Since Oν is open (t0upkγk, t1upk

γk) ∈ Oν


for k big enough. Observe that ϕξ(upk) is cofinal in ϕ, which contradicts theinduction hypothesis.•Let α ∈ H. If (α, α) ∈ H2 ∩ C, then there is ν < ξ such that (α, α) ∈

O0ν . Again, we can find a sequence (t0(pk)γk, t

0(pk))k∈ω in H2 ∩ Dξ0 such that

(t0(pk)γk, t0(pk)) → (α, α). Therefore (t0(pk)γk, t

0(pk)) ∈ O0

ν . Notice that whetherξ is successor or limit, ϕξ(u) ≥ ν, which contradicts our claim. Thus (α, α) /∈H2∩C. A similar argument shows that (α, α) /∈ H2\C. This is a contradiction,so that D0 cannot be separated from D1 by a pot(∆(Dξ(Σ0

1))) set.

Let A := (Aξ0,Aξ1), (Aξ1,A

ξ0), (Dξ0,D

ξ1). Notice that Aξε is not separable

from Aξ1−ε by a pot(∆(Dξ(Σ01))) set, as otherwise Aξ0 is separable from Aξ1 by a

pot(Dξ(Σ01)) set.

Theorem 5.17. Let 0 < ξ < ω1, and A0, A1 be disjoint analytic relations in Xsuch that A0 ∪A1 is contained in a quasi-acyclic pot(Π0

1) relation. Exactly oneof the following holds:

1. A0 is separable from A1 by a pot(∆(Dξ(Σ01))) set.

2. There are f : 2ω → X and g : 2ω → Y injective continuous and (A,B) ∈A such that A ⊆ (f × g)−1(A0) and B ⊆ (f × g)−1(A1).

Proof. The exactly part is a consequence of Lemmas 5.9 and 5.16.Let R be a quasi-acyclic relation, with witnesses (Rn)n∈ω such that A0∪A1 ⊆

R. Suppose that A0 is not separable from A1 by a pot(∆(Dξ(Σ01))) set. In

particular, X is non-empty. As such, there is a surjection h : ωω → X anda closed set Y of ωω such that h is injective in Y and h[Y ] is injective. LetA′ε = (h × h)−1(Aε) ∩ Y 2. Notice that A′0 is not separable from A′1 by apot(∆(Dξ(Σ0

1)) set.If there is f ′ : 2ω → ωω injective continuous and (A,B) ∈ A such that A ⊆

(f × g)−1(A′0) and B ⊆ (f ′× g′)−1(A′1). We want to show that f ′(α), g′(α) ∈ Yfor all α ∈ 2ω. Notice that there are (αk, βk) in A or B (regardless of what theyare), such that (αk, βk) → (α, α). By continuity and the fact that Y is closed,we obtain that f ′(α), g′(α) are in Y .

Let f, g : 2ω → X be defined by f = h f ′ and g = h g′. Then, f and g areinjective continous functions whose product is the reduction we need. Thus, itis enough to prove the theorem for ωω. Also, without loss of generality, we cansuppose that A0, A1, R,Rn are Σ1

1.By applying Lemma 5.2 we obtain that Gξ is a non-empty Σ1

1 subset of X2.Notice also that

Aε ∩Gξ ⊆F εξ if ξ is evenF 1−εξ if ξ is odd.

,We show this by induction on ξ. If ξ = 1, Aε ∩G1 ⊆ Aε ∩ A1−ε = F 1−ε

1 . Ifξ is limit,

Aε ∩Gξ ⊆⊆⊆ Aε ∩⋂ν<ξ

(Aε ∩Gν)⊆⊆⊆ Aε ∩⋂ν<ξ

(F εν )⊆⊆⊆ Aε ∩⋂

ν<ξ,ν pair(F εν )⊆⊆⊆ F εξ .

Finally, if ξ = ξ′ + 1, without loss of generality, suppose ξ′ is even, so that ξ isodd.

Aε ∩Gξ ⊆⊆⊆ Aε ∩A1−ε ∩Gξ′ ⊆⊆⊆ Aε ∩ F 1−εξ′ .


Notice that this last set is contained in F 1−εξ , as required.

So, if Aε ∩ Gξ 6= ∅, if e has the correct parity then F eξ 6= ∅. Therefore, byfollowing the proof of Theorem 5.10, we obtain injective continuous functionsf, g : 2ω → X such that A0 ⊆ (f × g)−1(Ae) and A1 ⊆ (f × g)−1(A1−e). Thus,in the following, we suppose that Aε ∩ Gξ = ∅, and we will prove that we canobtain a reduction to the other example.

Define Dεν := (t0uw, t1uw)|w ∈ 2<ω∧ϕξ(u) = ν∧∃e(ξ+e ≡ ν+ε mod 2∧u ∈

T e) for ν < ξ, and Dξ = (s, t) ∈ 2<ω×2<ω|s = t. Note also they are pairwisedisjoint. Observe that s(∪ν≤ξ,ε∈2D

εν ∩ (2× 2)n) is an acyclic connected graph.

Let S := x ∈ X|(x, x) ∈ Gξ.• If S = ∅. Then there is (x, y) ∈ Gξ, such that x 6= y. Let O0, O1 be disjoint

Σ01 neighborhoods of x and y, respectively.For each s ∈ 2<ω and each couple (s, t) ∈ ∪ν≤ξDε

ν . We construct thefollowing:

1. xs ∈ X, ys ∈ X,2. Xs, Ys Σ1

1 subsets of X,3. V(s,t) a Σ1

1 subset of X2,And we will ask for these objects to satisfy the following:1. (xs, yt) ∈ V(s,t) ⊆ (Xs × Ys) ∩ ΩX2 ,2. Xsε ⊆ Xs ⊆ ΩX ∩O0, Ysε ⊆ Ys ⊆ ΩX ∩O1, and V(sε,tε) ⊆ V(s,t),3. diamGH(Xs),diamGH(Ys),diamGH(V(s,t)) ≤ 2−|s|,4. Xs0 ∩Xs1 = ∅ = Ys0 ∩ Ys1,

5.V(s,t) ⊆ Gξ if (s, t) ∈ Dξ,V(s,t) ⊆ Aε ∩Gν if ν < ξ ∧ (s, t) ∈ Dε

ν .

Suppose first that we have constructed these objects. For each α ∈ 2ω, definef : 2ω → X by f(α) = ∩n∈ωXα|n and g : 2ω → X by g(α) = ∩n∈ωYα|n.Notice that these functions are continuous in the Gandy-Harrington topologyof X, so they are also continuous in the usual topology. Notice also that thesefunctions are injective, thanks to condition 4.

Now, suppose that (α, β) ∈ Dξε. Then, there is N such that (x, y) =(t0uγ, t1uγ), with ϕξ(u) = ν and u ∈ T e, for e with the correct parity. Sothat there is n such that (α|i, β|i) ∈ Dε

ν for each i > n. Note that we canfind F (α, β) ∈ X2 such that F (α, β) = ∩i>nV(α|i,β|i) ⊆ Aε, by condition 5.Thus (xα|n, yβ|n) → F (α, β) in the Gandy-Harrington topology of X2. Sincethis topology is finer, this sequence also converges in the product topology ofthe Gandy-Harrington of X, so that (f(α), g(β)) = F (α, β) ∈ Aε, which showsthat f × g satisfies what we need.

Let us show that the construction is possible. We do this by induction onthe length of s.

Notice first thatGξ∩(O0×O1) is a non-empty Σ11 subset ofX2, so that we can

find (x∅, y∅) ∈ Gξ∩(O0×O1)∩ΩX2 . As ΩX2 ⊆ Ω2X , we can find X∅, Y∅ ∈ Σ1

1(X)of diameter at most 1 such that (x∅, y∅) ∈ X∅ × Y∅ ⊆ Ω2

X ∩ (O0 × O1), andV(∅,∅) ∈ Σ1

1(X2) with diameter at most 1, such that V(∅,∅) ⊆ Fξ∩(X∅×Y∅)∩ΩX2 .Suppose we have constructed our objects for all s ∈ 2m. Notice that

(xs, yt) ∈ V(s,t) ⊆ A0 ∪ A1 ⊆ R, so that for each (s, t) we can fix a k(s,t)such that (xs, yt) ∈ Rk(s,t). Let u ∈ T 0∪T 1 and q ∈ ω such that m+ 1 = 〈uq〉ξ.In particular there is w ∈ 2<ω, such that (t0uw0, t1uw1) = (t0up, t1up).


Let O0 =x ∈ X|∀s ∈ 2m∃x′s, y′s ∈ X

(x′t0uw

= x ∧ ∀(s, t) ∈ (2 × 2)m ∩∪θDθ

((x′s, y′t) ∈ V(s,t) ∩ Rk(s,t)

)), and O1 =

y ∈ O1|∀s ∈ 22n+ε∃x′s, y′s ∈

X(y′t1uw

= y ∧ ∀(s, t) ∈ (2× 2)m ∩ ∪θDθ

((x′s, y′t) ∈ V(s,t) ∩Rk(s,t)

)).

Notice that these sets are Σ11 and that (xt0uw, yt1uw) ∈ (O0×O1)∩Gν , where

ϕξ(u) = ν. Observe that ϕξ(up) = η < ν. But, (O0 × O1) ∩ Gν ⊆ (O0 ×O1) ∩ Gη+1 ⊆ (O0 × O1) ∩ Aε ∩Gη

∆2X , for ε such that (t0up, s1

up) ∈ Dεη. In

particular, the last one is non empty. By Proposition 3.19, there is (xt0up , yt1up) ∈(O0 × O1) ∩ Aε ∩ Gη. For s ∈ 2m, let xs0, ys0 (xs1, ys1 respectively) be thewitnesses for xt0up ∈ O

0 (yt1up ∈ O1 respectively).

If ν = ξ, then (t0uw1, t1uw1) ∈ D0ξ and (t0uw0, t1uw1) ∈ D0

η. Therefore,(xt0pw1, yt1pw1) ∈ V(t0pw,t1pw) ⊆ Gξ. However, (xt0pw0, yt1pw1) ∈ Aε. Since we sup-posed that Aε and Gξ are disjoint, this implies that xt0pw0 6= xt0pw1. A similarargument proves yt0pw0 6= yt0pw1

If ν 6= ξ, then (xt0pw1, yt1pw1) ∈ V(t0pw,t1pw) ⊆ Ae, for some e. Observe thatν 6≡ η mod 2, since for ν limit, η is odd by definition, and for ν successor,ν = η + 1. Thus, e must differ from ε, by definition of De

ν .We need to show that xs0 6= xs1 (similarly for ys0 and ys1). But first, observe

that if s 6= t ∈ 2m, then xsε ∈ Xs and xtε′ ∈ Xt, so that xsε 6= xtε′ by condition4. Similarly ysε 6= ytε′ . Also, since xsε ∈ O0 and ytε′ ∈ O1, no “x” is equal tono “y”.

Since G := s(∪θ<ξ,e∈2 D

eθ ∩ (2 × 2)m

)is a connected acyclic graph. This

provides us with a single G-path from s to t0upw. This path provides two q.a.-equivalent s(R)-paths by the definition of Oε, one from xs0 to xt0uw0 and anotherfrom xs1 to xt0uw1. Since xt0uw0Ryt0uw1Rxt1uw1, by quasi-acyclicity xs0 6= xs1.Similarly, one can prove ys0 6= ys1.

Thus, it is easy to find Xs0, Xs1 ⊆ Xs (respectively Ys0, Ys1 ⊆ Ys) disjointΣ1

1 sets with the required properties, as well as V(sε,sε) and V(s,t), accordingly.• Suppose now that S 6= ∅. The proof is similar, so we mention the parts

were they differ. We will in fact obtain here that f = g. So we will onlyconstruct xs ∈ X,Xs ∈ Σ1

1(X) and V(s,t) ∈ Σ11(X2). Any condition on ys will

then be unnecessary. 1. will include (xs, xt) ∈ V(s,t) instead. Everything elsestays the same. The same proof show that if we suppose these objects found,we can construct f which satisfy what is need (with g := f).

In the construction, we start by picking x∅ ∈ S. We must change Oε byO :=

x ∈ X|∀s ∈ 2m∃x′s ∈ X

(x′sm = x ∧ (x′s, x′t) ∈ V(s,t) ∩ Rk(s,t)

). The rest

follows in a similar manner.

As we mentioned on the introduction, there is a fundamental difference be-tween the succesor and limit cases. In the succesor case, A is an antichain,however, in the limit case, we can reduce its size.

Proposition 5.18. If ξ = ξ′ + 1, then A is an antichain

Proof. We begin by showing that there are no functions f, g : 2ω → 2ω suchthat Aε ⊆ (f × g)−1(Dε). Let Aν =

(t0uα, t1uβ)|(u ∈ T 1

ξ ∧ ϕξ(u) ≤ ν) ∨ (u ∈T 0ξ ∧ ϕξ(u) < ν). It is clearly open.Suppose that u ∈ T 1

ξ . Then, for all γ ∈ 2ω, (t0uγ, t1uγ) ∈ Aϕξ(u). On the otherhand, (t0uγ, t1uγ) /∈ Aν , for any ν < ϕξ(u). Indeed, if it where, then there arev ∈ T 0

ξ ∪ T 1ξ , and α, β ∈ 2ω such that (t0uγ, t1uγ) = (t0vα, t1vβ). But then, v v u,


and in particular ϕξ(u) ≤ ϕξ(v). We obtain that v ∈ T 1ξ and thus ϕξ(v) ≤ ν <

ϕξ(u), a contradiction. We obtain that (t0uγ, t1uγ) ∈ Aϕξ(u)\ ∪ν<ϕξ(u) AνSuppose that u ∈ T 0

ξ . Then, for all γ ∈ 2ω, (t0uγ, t1uγ) ∈ Aϕξ(u)+1. Onthe other hand, (t0uγ, t1uγ) /∈ Aϕξ(u) (so neither in /∈ Aν , for any ν < ϕξ(u)).Indeed, if it where, then there are v ∈ T 0

ξ ∪ T 1ξ , and α, β ∈ 2ω such that

(t0uγ, t1uγ) = (t0vα, t1vβ). But then, v v u, and in particular ϕξ(u) ≤ ϕξ(v). Weobtain that v ∈ T 0

ξ and thus ϕξ(v) < ϕξ(u), a contradiction. Again, we obtainthat (t0uγ, t1uγ) ∈ Aϕξ(u)+1\ ∪ν<ϕξ(u)+1 Aν

This allows us to show that D0 is separated from D1 by Dξ((Aν)ν<ξ). In-deed, if (t0uγ, t1uγ) ∈ D0 and u ∈ T 1

ξ , then ϕξ(u) 6≡ ξ mod 2. In particular,Aϕξ(u)\ ∪ν<ϕξ(u) Aν ⊆ Dξ((Aν)ν<ξ. If u ∈ T 0

ξ , then ϕξ(u) ≡ ξ mod 2, whichmeans ϕξ(u) + 1 6≡ ξ mod 2. Again Aϕξ(u)+1\ ∪ν<ϕξ(u)+1 Aν ⊆ Dξ((Aν)ν<ξ.

On the other hand, suppose that (t0uγ, t1uγ) ∈ D1 ∩ Dξ((Aν)ν<ξ) 6= ∅. Bythe definition of D1, if u ∈ T 1

ξ , then ϕξ(u) ≡ ξ mod 2. In particular Aϕξ(u) ∩Dξ((Aν)ν<ξ = ∅, a contradiction. A similar argument gives us a contradictionif u ∈ T 0

ξ .So D0 is separable from D1 by a pot(Dξ(Σ0

1)) set, so that there cannot besuch functions. The case Aε ⊆ (f × g)−1(D1−ε) is handled in a similar way.

Now, suppose that there are continuous injective functions f, g : 2ω → 2ω,such that Dε ⊆ (f × g)−1(Aε). We extend our definition of ϕ to ϕt : A0 ∪A1 →ξ + 1 by ϕt(t0uγ, t1uγ) = ϕξ(u) and to ϕs : D0 ∪ D1 ∪ Diag(2ω) → ξ + 1 byϕs(t0uγ, t1uγ) = ϕξ(u).

We claim that if (αn, βn)n∈ω is a sequence in Dε that converges to (α, β) ∈D1−ε ∪ Diag(2ω), then ϕs(αn, βn) < ϕs(α, β). In fact, let u ∈ T 0 ∪ T 1 suchthat (α, β) = (t0uγ, t1uγ). Then, given k ∈ ω, (t0uγ|k, t1uγ|k) v (αn, βn) =(t0unγn, t

1unγn) for n large enough. So that un us a strict extension of u, i.e.,

ϕξ(un) < ϕξ(u). A similar argument shows the analogous claim for ϕt.We claim that for all (α, β) ∈ Dε, ϕt(f(α), g(β)) ≥ ϕs(α, β). We show this

claim by induction on ϕs(α, β).If ϕs(α, β) = 0, it is obvious. If ϕs(α, β) = ν+1, let (αn, βn) be a sequence in

D1−ε that converges to (α, β) ∈ D0∪D1∪Diag(2ω) and such that ϕs(αn, βn) = ν.Thus, ν ≤ ϕt(f(αn), g(βn)) < ϕt(f(α), g(β)).

If ϕs(α, β) = ν is limit, we take our sequence (αn, βn) as before, except thatwe ask that ϕs(αn, βn) = νn to be cofinal in ν. In particular, we obtain thatνn ≤ ϕt(f(αn, βn)) < ϕt(f(α), g(β)), so this is at least ν.

The last claims allows us to show that for all α ∈ 2ω, ϕt(f(α), g(α)) ≥ϕξ(α, α), so that f(α) = g(α).

To finish, let (α, β) ∈ D0 such that ϕξ(α, β) = ξ. Therefore ϕt(f(α), f(β)) ≥ξ′, so it is equal to ξ′, otherwise f(α) = f(β), which contradicts injectivity off . But ξ ≡ ξ′ + 1 mod 2, so that (f(α), f(β)) ∈ A1, which is a contradiction.

For the last case, suppose that we have injective continuous functions f, g :2ω → 2ω such that Aε ⊆ (f × g)−1(A1−ε). A similar argument as above showsthat for all (α, β) ∈ A0 ∪ A1, ϕt(α, β) ≤ ϕt(f(α), g(β)). In particular f =g, as Diag(2ω) ⊆ A0. Recall that if f is injective, then Diag(2ω) ⊆ (f ×f)−1(Diag(2ω)). But then, Diag(2ω) ∩A1 6= ∅, a contradiction.

Proposition 5.19. If ξ is a limit ordinal, then A is not an antichain.

Proof. We show that there is an injective continuous function f : 2ω → 2ω suchthat Dξε ⊆ (f×f)−1(Aξε) for all ε ∈ 2. By an analogous argument, we can obtain


g : 2ω → 2ω such that Dξε ⊆ (g × g)−1(Aξ1−ε) for all ε ∈ 2.To obtain f let us construct a function φ : 2<ω → 2<ω such that, for every

s, t ∈ 2<ω\∅,

1. If s v t then φ(s) v φ(t),2. there is n < min|φ(s0)|, |φ(s1)| such that φ(s0)(n) 6= φ(s1)(n),3. if there are u ∈ ω<ω and w ∈ 2<ω such that (s, t) = (t0uw, t1uw) then

there are v ∈ ω<ω and z ∈ 2<ω such that (φ(s), φ(t)) = (t0vz, t1vz),4. if u, v are as before, (u ∈ T 0

ξ ⇐⇒ ϕξ(u) ≡ ξ mod 2) if and only if|v| ≡ ξ mod 2,

5. if u, v are as before, then ϕξ(u) ≤ ϕξ(v).

Suppose that we already had constructed the function. By 1. there is isa continuous function f : 2ω → 2ω defined by f(α) = ∪φ(α|n). By 2. it isinjective. If (α, β) ∈ Dε, then there is u such that (u ∈ T 0

ξ ⇐⇒ ϕξ(u) = ε) and(α, β) = (t0uγ, t1uγ). Thus (f(α), f(β)) = (t0vγ′, t1vγ′), with ϕξ(v) ≡ ξ+ε mod 2.

To construct the function, we proceed by induction on the length of s.Suppose that |s| = 1, so we only have two options (0) and (1). Notice that(0, 1) = (t0(0), t

1(0)), so that (0) ∈ T 0 and ϕξ(0) 6≡ ξ mod 2. To satisfy 3, 4, 5,

we must then choose v with ϕξ(v) 6≡ ξ and ϕξ(0) ≤ ϕξ(v). Clearly (0) works,and so φ(0) = (0) and φ(1) = (1).

We want to construct φ(s) for s ∈ 2n+1, for n > 0, so suppose that we haveconstructed φ(s) for |s| < n + 1. Notice there is one and only one u such that(t0u, t1u) ∈ 2n+1. We begin by defining simultaneously φ for these sequences, andextending the definition to the rest of the sequences in 2n+1.• |u| > 1. Then there are u0 ∈ ω<ω and w ∈ 2<ω such that tεu = tεu0

wε. Bycondition 3, (φ(t0u0

w), φ(t0u0w)) = (t0vz, t1vz) for some v ∈ ω<ω and z ∈ 2<ω. Let

q ∈ ω such that z v b(q) and φ(u) ≤ φ(vq). We can find such a q because ifφ(v) = ν + 1, then φ(vq) = ν, but φ(u) < φ(u0) ≤ ν + 1 so that φ(u) ≤ ν. Ifφ(v) is limit, then (φ(vq))q∈ω is cofinal in v and φ(u) < φ(u0) ≤ φ(v). Defineφ(tεu0

wε) := tεvq.By definition, there is N ∈ ω such that tεvq = tεvz0Nε. Finally, let φ(sε) :=

φ(s)0Nε, for any s ∈ 2n. Conditions 1 and 2 clearly hold. Let us verify theother conditions.

First notice that (φ(t0u), φ(t1u)) = (t0vq, t1vq) by definition, so that 3 is verified.To verify 4, suppose that u0 ∈ T 0

ξ and φ(u0) ≡ ξ+ε mod 2. Then φ(u0) ≡ φ(v)mod 2. But φ(u0) 6≡ φ(u) mod 2 and φ(v) 6≡ φ(vq) mod 2, so that u ∈ T 0

ξ andφ(u) ≡ φ(vq) mod 2 which verifies 4. Also, in this case, Condition 5 followsfrom the choice of vq.

Suppose there are u1 ∈ ω<ω, w′ ∈ 2<ω and e ∈ 2ω such that (s, t) =(t0u0

we, t1u0we). Therefore, by the induction hypothesis, (φ(t0u1

w′e), φ(t1uqw′e)) =

(φ(t0u1w′)0Ne, φ(t1u1

w′)0Ne) = (t0vz0Ne, t1vz0Ne). Thus, conditions 3,4,5 are ver-ified.• Case u = (p) for p ∈ ω\∅. Let w = (t0u)|n. Notice there are infinitely

many q’s such that φ(w) v b(q). Since ξ is a limit ordinal, ϕξ(q) is strictlyincreasing. Thus q can be chosen so that ϕξ(p) ≤ ϕξ(q).

If p is even, then, (p) ∈ T 0 and ϕξ(p) 6≡ ξ. Notice that ϕξ(q) 6≡ ξ. Letφ(tε(p)) := tε(q). This contains φ(w), and thus, verifies condition 1 and 2. If p isodd, φ(tε(p)) := tε(q0).


Let w0 and w1 be the sequences such that φ(T ε(p)) = φ(w)wεε. Notice theyare different if p is odd. As in the previous case, we define φ(sε) := φ(s)wεε, forany s ∈ 2n. Notice how the choice of wε only depends on the last coordinate ofsε.

Conditions 3, 4, 5 are verified as before for (φ(t0u), φ(t1u)). For the othercases, (φ(t0u0

we), φ(t1u0we)) = (φ(t0u0

w)wee, φ(t1u0w)wee) = (t0vzwee, t1vzwee), by

the induction hypothesis. This verifies the conditions.

In the sequel, we fix ξ. If ξ is successor, then let E := A, and if ξ is limit,then let E := (D0,D1). So, by Theorem 5.17 and Propositions 5.18 and 5.19, Eis an antichain basis for the property of non separability by a pot(∆(Dξ(Σ0

1))).Let us set C := E ∪ (B(A),B(B))|(A,B) ∈ E.

Theorem 5.20. Let A0, A1 be disjoint analytic relations in X such that A0∪A1is contained in a pot(Π0

1) quasi-acyclic relation. Exactly one of the followingholds:


2. There is f : 2ω → X an injective continuous function and (A0,A1) ∈ Csuch that Aε ⊆ (f × f)−1(Aε).

Proof. The exactly part is a corollary of Theorem 5.17. Suppose A0 is notseparable from A1 by a pot(∆(Dξ(Σ0

1))) set. Let R be a pot(Π01) quasi-acyclic

relation with witnesses Rn. Without loss of generality, we can suppose that Xis recursively presented and that A0, A1 and Rn are all Σ1

1 sets.Let us consider Gξ. So, if Aε ∩ Gξ 6= ∅ then Fξ 6= ∅ and by following the

proof of Theorem 5.10, we obtain f : 2ω → X such that one of the followingholds, for all ε ∈ 2,

— Aε ⊆ (f × f)−1(Aε),— B(Aε) ⊆ (f × f)−1(Aε),— Aε ⊆ (f × f)−1(A1−ε),— B(Aε) ⊆ (f × f)−1(A1−ε).In particular, on the case that ξ is a limit ordinal, the proof of 5.19 provides

(A,B) ∈ C which reduces to one of the previous cases with a single function f ′.By composition, we obtain 2.

So we can suppose that Aε ∩ Gξ = ∅. Let S = x ∈ X|(x, x) ∈ Gξ. Wehave two cases.•If S 6= ∅. We showed in the proof of Theorem 5.17 that one can find

f : 2ω → X such that Dε ⊆ (f × f)−1(Aε).•If S = ∅, then there are (x, y) ∈ Gξ non-equal, and O0, O1 disjoint Σ0

1subsets of X. By looking at the proof of theorem 5.17, one can find injectivecontinuous functions f0, f1 : 2ω → X such that Dε ⊆ (f0 × f1)−1(Aε). Noticethat by the construction there, fε(α) ∈ Oε. Let us define f : 2ω → X by f(εα) =fε(α). Notice it is continuous, injective, and that B(Dε) ⊆ (f × f)−1(Aε).

Proposition 5.21. C is an antichain basis on Theorem 5.20.

Proof. Note that E and B(E) := (B(A),B(B))|(A,B) ∈ E are antichains. Soit is enough to show that we cannot have a reduction between one couple in Eand another in B(E).


Let us see that there is no injective continuous function f : 2ω → 2ω such thatfor some (A,B), (A′,B′) ∈ E , A ⊆ (f × f)−1(B(A′)) and B ⊆ (f × f)−1(B(B′)).So, by contradiction, suppose that such f,A,B,A′ and B′ exist.

As we have previously pointed out, Diag(2ω) ⊆ (f × f)−1Diag(2ω). ButDiag(2ω) ⊆ A0, so this rules out the cases where A is A0 or A1. Suppose that(A,B) = (D0,D1). Let α ∈ 2ω. There are (αk, βk) ∈ D0 such that (αk, βk) →(α, α). Therefore (f(αk), f(βk))→ (f(α), f(α)), by continuity. However, for allk ∈ ω, (f(αk), f(βk)) ∈ N0×N1 which is closed, so that (f(α), f(α)) ∈ N0×N1,a clear contradiction.

It rests to show that there is no injective continuous function f : 2ω → 2ωsuch that for some (A,B), (A′,B′) ∈ E , B(A) ⊆ (f × f)−1(A′) and B(B) ⊆(f × f)−1(B′). Proposition 5.11 already ruled out the cases where (A,B) =(A′,B′) = (Aε,A1−ε). If (A,B) 6= (A′,B′) and ε ∈ 2, define fε : 2ω → 2ω byfε(α) = f(εα), which are clearly continuous and injective. But f0 × f1 is areduction from (A,B) to (A′,B′), which by Proposition 5.18 cannot be the case.

So it only rests the case (A,B) = (A′,B′) = (D0,D1). This is essentially theproof of Theorem 5.18. That is to say, we first extend the definition of ϕξ totwo new definitions ϕdξ : D0 ∪ D1 ∪ Diag(2ω) → ξ + 1, and ϕbξ : B(D0 ∪ D1 ∪Diag(2ω)) → ξ + 1. In a similar way, we show that these functions force that(f(0α), f(1α)) ∈ Diag(2ω), which is clearly a contradiction.

Corollary 5.22. Let 0 < ξ < ω1 and A0, A1 be disjoint analytic relations inX such that A0 ∪ A1 is contained in a pot(Π0

1) s-acyclic or locally countablerelation. Exactly one of the following holds:

1. A0 is not separable from A1 by a pot(∆(Dξ(Σ01))) set.

2. There are f : 2ω → X and g : 2ω → Y injective continuous and (A,B) ∈E such that A ⊆ (f × g)−1(A0) and B ⊆ (f × g)−1(A1).

Proof. This holds by Proposition 3.17.

Corollary 5.23. Let 0 < ξ < ω and let X be a Polish space and A0, A1 bedisjoint analytic relations on X. The following are equivalent:

1. There is an s-acyclic relation R such that A0 ∩ R is not separable fromA1 ∩R by a pot(Dξ(Σ0

1)) set.2. There is a locally countable relation R such that A0 ∩R is not separable

from A1 ∩R by a pot(Dξ(Σ01)) set.

3. There is a quasi-acyclic relation R such that A0∩R is not separable fromA1 ∩R by a pot(Dξ(Σ0

1)) set.4. There are f : 2ω → X injective continuous and (A,B) ∈ C such that

A ⊆ (f × f)−1(A0) and B ⊆ (f × f)−1(A1).

Proof. 1 implies 3 and 2 implies 3 are direct consequences of Proposition 3.17,and 3 implies 4 comes from Theorem 5.20. For the other directions, observethat by Proposition 3.15 and the definition of our examples, A ∪ B is a s-acycliclocally countable relation, for any (A,B) ∈ C. Then R := (f × g)[A ∪ B] is therelation required for 1. and 2.

Corollary 5.24. Let 0 < ξ < ω1, X,Y be Polish spaces, and A0, A1 disjointanalytic subsets of X × Y such that A0 ∪ A1 is contained in a pot(Π0

1) locallycountable relation. Exactly one of the following holds:

Separability of oriented graphs by Γ sets 77


2. There are f : 2ω → X and g : 2ω → Y injective continuous and (A,B) ∈A, such that A ⊆ (f × g)−1(A0) and B ⊆ (f × g)−1(A1) .

Proof. The proof is similar as the proof in the case of Corollary 5.15.

Separability of oriented graphs by Γ setsLet G be an oriented graph. We want to characterize when it is separable

from its inverse graph G−1 by a set in pot(Γ), for Γ a Wadge class of Borelsets. So fix Γ. Let S0 and S1 the relations on 2ω given by Theorem 3.14. Wedefine G = S0 ∪ S−1

1 . As pointed out after Theorem 3.14, S0 and S1 are in thebranches of the tree generated by a frame. In particular, S0 ∪ S1 ⊆ N0 ×N1.

Lemma 5.25. Let Γ be a Wadge class of Borel sets or Γ = ∆0ξ for ξ countable

non zero. Then G is not separable from G−1 by a pot(Γ) set.

Proof. Suppose there is C ∈ pot(Γ) such that G ⊆ C and such that C∩G−1 = ∅.In particular, S0 ⊆ C and C ∩ S1 = ∅ a contradiction with Theorem 3.14.

We obtain our dichotomy.

Theorem 5.26. Let Γ be a Wadge class of Borel sets or Γ = ∆0ξ for ξ countable

non zero, X be a Polish space, and G be an analytic oriented graph on X.Exactly one of the following holds:

1. G is separable from G−1 by a pot(Γ) set.2. (2ω,G) c (X,G).

Proof. The exactly part holds by the preceding Lemma. Indeed, if C separatesG from G−1 then f−1(C) separates G from G−1.

Suppose that G is not separable from G−1 by a pot(Γ) set. Then there arefunctions f0 : 2ω → X and f1 : 2ω → X such that S0 ⊆ (f0 × f1)−1(G) andS1 ⊆ (f0 × f1)−1(G−1).

We define f : 2ω → X by f(εα) = fε(εα), so that f is continuous. If(α, β) ∈ S0 ⊆ N0×N1 then (f(α), f(β)) = (f0(α), f1(β)) ∈ G. If (α, β) ∈ S−1

1 ⊆(N1 × N0), then (f(β), f(α)) = (f0(β), f1(α)) ∈ G−1, so that (f(α), f(β)) ∈G.

We can obtain an injective version for the non self-dual Wadge classes ofBorel sets contained in ∆0

2. Let ξ be a countable non zero ordinal. DefineGξi := B(A0) ∪ (B(A1))−1.

Theorem 5.27. Let ξ be a countable non-zero ordinal. For any Polish spaceX and any analytic quasi-acyclic graph G on X. Exactly one of the followingholds:

1. G is separable from G−1 by a pot(Dξ(Σ01)) set.

2. There is an injective continuous function f : 2ω → X such that Gξi ⊆(f × f)−1(G).


Proof. Lemma 5.25 shows that both statements cannot hold simultaneously.Suppose that G is not separable from G−1 by a pot(Dξ(Σ0

1)) set. ByTheorem 5.10, there exists f : 2ω → X injective continuous such that eitherA0 ⊆ (f × f)−1(G) and A1 ⊆ (f × f)−1(G−1) or B(A0) ⊆ (f × f)−1(G) andB(A1) ⊆ (f × f)−1(G−1). However, note that the former option is not possible.Indeed, if it were the case, we would have Diag(2ω) ⊆ (f × f)−1(G), whichcontradicts the fact that G is not reflexive.

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