Baire-class $\xi$ colorings: the first three levels

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Baire-class ξ colorings: the first three levels

Dominique Lecomte, Miroslav Zeleny

To cite this version:

Dominique Lecomte, Miroslav Zeleny. Baire-class ξ colorings: the first three levels. 2011. �hal-00586487�

Baire-class ξ colorings: the first three levels

April 26, 2011

• Universit´e Paris 6, Institut de Math´ematiques de Jussieu, Projet Analyse Fonctionnelle Couloir 16-26, 4`eme ´etage, Case 247, 4, place Jussieu, 75 252 Paris Cedex 05, France

dominique.lecomte@upmc.fr

• Universit´e de Picardie, I.U.T. de l’Oise, site de Creil, 13, all´ee de la fa¨ıencerie, 60 107 Creil, France

•1_{Charles University, Faculty of Mathematics and Physics, Department of Mathematical Analysis}

Sokolovsk´a 83, 186 75 Prague, Czech Republic zeleny@karlin.mff.cuni.cz

Abstract. TheG0-dichotomy due to Kechris, Solecki and Todorˇcevi´c characterizes the analytic

re-lations having a Borel-measurable countable coloring. We give a version of theG0-dichotomy for

Σ0ξ-measurable countable colorings when ξ ≤ 3. A Σ0ξ-measurable countable coloring gives a

cov-ering of the diagonal consisting of countably manyΣ0ξ squares. This leads to the study of countable

unions ofΣ0ξ rectangles. We also give a Hurewicz-like dichotomy for such countable unions when

ξ ≤ 2.

2010 Mathematics Subject Classification. Primary: 03E15, Secondary: 54H05

Keywords and phrases. Borel chromatic number, Borel class, coloring, dichotomy, Hurewicz, partition, product

2_{The work is a part of the research project MSM 0021620839 financed by MSMT and partly supported by the grant}

GA ˇCR 201/09/0067.

Acknowledgements. This work started in summer 2009, and was continued during spring 2010, when the first author

was invited at the University of Prague by the second one. The first author is very grateful to the second one for that. We would also like to thank Alain Louveau for some nice remarks made in the Descriptive Set Theory Seminar of the University Paris 6. They were useful to simplify our presentation.

1 Introduction

The reader should see [K] for the standard descriptive set theoretic notation used in this paper. We study a definable coloring problem. We will need some more notation:

Notation. The lettersX, Y will refer to some sets. We set ∆(X) := {(x0, x1) ∈ X2| x0= x1}.

Definition 1.1 (1) LetA ⊆ X2. We say thatA is a digraph if A does not meet ∆(X).

(2) LetA be a digraph. A countable coloring of (X, A) is a map c : X → ω such that A does not

meet(c×c)−1 ∆(ω)
.

In [K-S-T], the authors characterize the analytic digraphs of having a Borel countable coloring. The characterization is given in terms of the following notion of comparison between relations.

Notation. LetX, Y be Polish spaces, A (resp., B) be a relation on X (resp., Y ), and Γ be a class of sets. We set

(X, A) _Γ(Y, B) ⇔ ∃f : X → Y Γ-measurable with A⊆ (f ×f )−1(B).

In this case, we say thatf is a Γ-measurable homomorphism from (X, A) into (Y, B). This notion essentially makes sense for digraphs (we can takef to be constant if B is not a digraph).

We also have to introduce a minimum digraph without Borel countable coloring:

• Let ψ : ω → 2<ω _{be a natural bijection. More specifically, ψ(0) := ∅ is the sequence of length 0,}

ψ(1) := 0, ψ(2) := 1 are the sequences of length 1, and so on. Note that |ψ(n)| ≤ n if n ∈ ω. Let n ∈ ω. As|ψ(n)| ≤ n, we can define sn:= ψ(n)0n−|ψ(n)|. The crucial properties of the sequence(sn)n∈ω

are the following:

-(sn)n∈ωisdense in 2<ω. This means that for eachs ∈ 2<ω, there isn ∈ ω such that snextends

s (denoted s ⊆ sn).

-|sn| = n.

• We putG0 := {(sn0γ, sn1γ) | n ∈ ω and γ ∈ 2ω} ⊆ 2ω× 2ω. Note thatG0 is analytic (in fact

difference of two closed sets) since the map(n, γ) 7→ (sn0γ, sn1γ) is continuous.

The previous definitions were given, whenΓ = ∆11, in [K-S-T], where the following is proved:

Theorem 1.2 (Kechris, Solecki, Todorˇcevi´c) LetX be a Polish space, and A be an analytic relation

onX. Then exactly one of the following holds:

(a) There is a Borel countable coloring of(X, A), i.e., (X, A) _∆1

1 ω, ¬∆(ω)
,

(b)(2ω_,G 0) _Σ0

1 (X, A).

This result had several developments during the last decade. Here is a non-exhaustive list: - We can characterize the potentially open sets via a Hurewicz-like test, and in finite dimension it is a consequence of the previous result. Let us specify this. The following definition can be found in [Lo2] (see Definition 3.3).

Definition 1.3 (Louveau) LetX, Y be Polish spaces, A be a Borel subset of X×Y , and Γ be a Borel

class. We say thatA is potentially in Γ denoted A ∈ pot(Γ)
if we can find a finer Polish topology σ (resp., τ ) on X (resp., Y ) such that A ∈ Γ (X, σ)×(Y, τ )
.

The pot(Σ0

1) sets are the countable unions of Borel rectangles. A consequence of this is that

the Borel hierarchy built on the Borel rectangles is exactly the hierarchy of the classes of the sets potentially in some Borel class. The good notion of comparison to study the pot(Γ) sets is as follows (see [L3]). LetX0, X1, Y0, Y1be Polish spaces, andAε0, Aε1⊆ Xε×Yεbe disjoint. We set

(X0, Y0, A00, A01) ≤ (X1, Y1, A10, A11) ⇔

∃f : X0→ X1 ∃g : Y0→ Y1continuous withA0ε⊆ (f ×g)−1(A1ε) for each ε ∈ 2.

The following theorem is proved in [L1], and is a consequence of Theorem 1.2:

Theorem 1.4 LetX, Y be Polish spaces, and A0, A1 be disjoint analytic subsets ofX × Y . Then

exactly one of the following holds:

(a) The setA0can be separated fromA1 by a pot(Σ01) = (∆11×∆11)σ set (i.e., there isS ∈ pot(Σ01)

withA0⊆ S ⊆ ¬A1),

(b) 2ω, 2ω, ∆(2ω),G0
≤ (X, Y, A0, A1).

In [L1], it is also proved that we cannot havef one-to-one in Theorem 1.2.(b) in general. It is easy to check that Theorem 1.2 is also an easy consequence of Theorem 1.4. This means that the study of the Borel countable colorings is highly related to the study of countable unions of Borel rectangles. - We can extend Theorem 1.2 to any finite dimension, and also in infinite dimension if we change the space in which lives the infinite dimensional version ofG0(see [L2]).

- B. Miller recently developped some techniques to recover many dichotomy results of descriptive set theory, but without using effective descriptive set theory. He replaces it with some versions of Theorem 1.2. In particular, he can prove Theorem 1.2 without effective descriptive set theory.

WhenA is Borel, it is natural to ask about the relation between the Borel class of A and that of the coloringf when Theorem 1.2.(a) holds. This leads to consider ∆0

ξ-measurable countable colorings

(or equivalentlyΣ0ξ-measurable countable colorings). We have the following conjecture:

Conjecture 1 Let1 ≤ ξ < ω1. Then there are

- a0-dimensional Polish spaceXξ,

- an analytic relation_AξonXξ

such that for any0-dimensional Polish space X, and for any analytic relation A on X, exactly one of

the following holds: (a)(X, A) _∆0

ξ ω, ¬∆(ω)
,

(b)(Xξ,Aξ) _Σ0

1 (X, A).

We will prove it when1 ≤ ξ ≤ 2, and in these cases we do not have to assume that A is analytic. We will also prove it whenξ = 3, which is much more difficult. We should not have to assume that X is0-dimensional when ξ ≥ 2, but we have to do it when ξ = 1.

We saw that the study of the Borel countable colorings is highly related to the study of count-able unions of Borel rectangles, and gave some motivation for studying Σ0ξ-measurable countable

colorings. This motivates the study of countable unions ofΣ0ξrectangles. Another motivation is that

(X, A) _∆0

ξ ω, ¬∆(ω)
is equivalent to the fact that ∆(X) can be separated from A by a (Σ 0 ξ×Σ0ξ)σ

set, by the generalized reduction property for the classΣ0ξ(see 22.16 in [K]).

Conjecture 2 Let1 ≤ ξ < ω1. Then there are0-dimensional Polish spacesX0ξ,X1ξand disjoint analytic

subsets_A0_ξ,_A1

ξofX0ξ×X1ξsuch that for any Polish spacesX, Y , and for any pair A0, A1of disjoint

analytic subsets ofX ×Y , exactly one of the following holds:

(a) The setA0can be separated fromA1 by a(Σ0ξ×Σ0ξ)σ set,

(b)(X0

ξ,X1ξ,A0ξ,A1ξ) ≤ (X, Y, A0, A1).

It is trivial to prove this whenξ = 1. We will prove that Conjecture 2 holds when ξ ≤ 2, which is significantly more and more difficult whenξ increases. We use effective descriptive set theory, and give effective strengthenings of our results. The reader should see [M] for basic notions of effective descriptive set theory. In particular, we will see that to test whether an analytic relation has aΣ0ξ

-measurable countable coloring, it is enough to test countably many partitions instead of continuum many. We will use the topologyTξ generated by the Σ11 ∩ Π0<ξ subsets of a recursively presented

Polish space (introduced in [Lo1]) whenξ is 2 or 3 (T1 is just the basic topology). The last result can

be strengthened as follows (see [L3]).

Theorem 1.5 Let1 ≤ ξ ≤ 2. Then there are 0-dimensional Polish spaces_X0

ξ,X1ξ and disjoint analytic

subsetsA0

ξ,A1ξ ofX0ξ×X1ξ such that for any recursively presented Polish spaces X, Y , and for any

pairA0, A1of disjointΣ11subsets ofX ×Y , the following are equivalent:

(a) The setA0cannot be separated fromA1 by a(Σ0ξ×Σ0ξ)σ set.

(b) The setA0cannot be separated fromA1 by a∆11∩ (Σ0ξ×Σ0ξ)σset.

(c) The setA0cannot be separated fromA1 by aΣ01(Tξ×Tξ) set.

(d)A0∩ A1 Tξ×Tξ

6= ∅.

(e)(_X0

ξ,X1ξ,A0ξ,A1ξ) ≤ (X, Y, A0, A1).

2 Some general effective facts

One can hope for an effective strengthening of Conjecture 1:

Effective conjecture 1 Let1 ≤ ξ < ω1. We can find a0-dimensional Polish spaceXξand an analytic

relationAξonXξsuch that(Xξ,Aξ) 6_∆0

ξ ω, ¬∆(ω)
, and for any α ∈ ω

ω_{with1 ≤ ξ < ω}α

1, for any

0-dimensional recursively in α presented Polish space X, and for any Σ1

1(α) relation A on X, one

of the following holds: (a)(X, A) _∆1

1(α)∩∆0ξ ω, ¬∆(ω)
,

(b)(Xξ,Aξ) _Σ0

We will see that this effective conjecture is true when 1 ≤ ξ ≤ 3. The following statement is a corollary of this effective conjecture, and is in fact a theorem:

Theorem 2.1 Let1 ≤ ξ < ωCK₁ ,X be a 0-dimensional recursively presented Polish space, and A be

aΣ11relation onX. We assume that (X, A) ∆ξ0 ω, ¬∆(ω)
. Then (X, A) ∆11∩∆0ξ ω, ¬∆(ω)
.

A consequence of this is that to test whether an analytic relation has aΣ0ξ-measurable countable

coloring, it is enough to test countably many partitions instead of continuum many. Another con-sequence is the equivalence between Conjecture 1 and the Effective conjecture 1. We have in fact preliminary results that will help us to prove also the equivalence between (a)-(d) in Theorem 1.5, in the general case.

Lemma 2.2 Let 1 ≤ ξ < ωCK₁ , X, Y be recursively presented Polish spaces, A ∈ Σ11(X) ∩ Σ0ξ,

B ∈ Σ11(Y ) ∩ Σ0ξ, andC ∈ Σ11(X ×Y ) disjoint from A×B. Then there are A′, B′∈ ∆11∩ Σ0ξ such

thatA′_×B′_{separatesA×B from C. This also holds for Π}0

ξinstead ofΣ0ξ.

Proof. Note thatA and {x ∈ X | ∃y ∈ B (x, y) ∈ C} are disjoint Σ11 sets, separable by aΣ0ξsubset

ofX. By Theorems 1.A and 1.B in [Lo1], there is A′_{∈ ∆}1

1∩ Σ0ξseparating these two sets. Similarly,

B and {y ∈ Y | ∃x ∈ A′ (x, y) ∈ C} are disjoint Σ11sets, and there isB′∈ ∆11∩ Σ0ξ separating these

two sets. The proof forΠ0ξis identical to the one forΣ0ξ. 

Theorem 2.3 Let1 ≤ ξ < ωCK₁ ,X, Y be recursively presented Polish spaces, and A0, A1be disjoint

Σ11 subsets of X × Y . We assume that A0 is separable from A1 by a Σ0ξ× Σ0ξ

σ set. ThenA0 is

separable fromA1by a∆11∩ (∆11∩ Σ0ξ)×(∆11∩ Σ0ξ)

σ set.

Proof. By Example 2 of Chapter 3 in [Lo2], the family N (n, X)

n∈ω is regular without parameter.

By Corollary 2.10 in [Lo2], Π0ξ(X), as well as Σ0ξ(X) =

S

η<ξ Π0η(X)

σ, are regular without

parameter. By Theorem 2.12 in [Lo2],Σ0ξ(X)×Σ0ξ(Y ) is also regular without parameter. By Theorem

2.8 in [Lo2], the family Φ := Σ0ξ(X) × Σ0ξ(Y )

σ is separating, which implies the existence of

S ∈ ∆1

1∩ Φ separating A0fromA1.

With the notation of [Lo2], letn be an integer with (0∞, n) ∈ W and C0∞_,n= S. Then (0∞, n) is

inWΦ, which by Theorem 2.8.(ii) in [Lo2] is

( (α, n) ∈ W | ∃β ∈ ∆11(α) ∀m ∈ ω α, β(m)
∈ WΣ0 ξ(X)×Σ0ξ(Y )andCα,n= [ m∈ω C_α,β(m) ) .

This implies thatS ∈ ∆11∩ ∆11∩ (Σ0ξ×Σ0ξ)

σ. It remains to check that

∆11∩ (Σ0ξ×Σ0ξ) = (∆11∩ Σ0ξ)×(∆11∩ Σ0ξ).

The second set is clearly a subset of the first one. So assume thatR = A×B ∈ ∆11 ∩ (Σ0ξ×Σ0ξ). We

may assume thatR is not empty. Then the projections A, B are Σ11 sinceR ∈ ∆11. Lemma 2.2 gives

A′, B′∈ ∆1

Recall that ifA is a relation on X and D ⊆ X, then D is A-discrete if A ∩ D2= ∅.

Proof of Theorem 2.1. We apply Theorem 2.3 toY := X, A0:= ∆(X) and A1:= A. As

(X, A) _∆0

ξ ω, ¬∆(ω)
,

∆(X) is separable from A by a (Σ0ξ× Σ0ξ)σ set. Theorem 2.3 gives Cn, Dn∈ ∆11 ∩ Σ0ξ such that

S :=S

n∈ω Cn×Dn∈ ∆11 separates∆(X) from A. As the set of codes for ∆11∩ Σ0ξ subsets ofX

isΠ₁1 (see Proposition 1.4 and Theorem 1.A in [Lo1]), the∆1₁-selection theorem and the separation theorem imply that we may assume that the sequences(Cn) and (Dn) are ∆1₁. Note that(Cn∩ Dn) is

a∆11covering ofX into A-discrete ∆11∩ Σ0ξsets. AsX is 0-dimensional we can reduce this covering

into a∆11 covering ofX consisting of ∆11 ∩ Σ0ξ sets, which are in fact∆0ξ. This gives the desired

partition. 

Notation. Following [Lo1], we define the following topologies on a0-dimensional recursively in α presented Polish space X, for any α ∈ ωω_{. LetT}

1(α) be the topology of X, and, for 2 ≤ ξ < ω1,

Tξ(α) be the topology generated by the Σ11(α) ∩ Π0<ξ subsets ofX. The next proposition gives a

reformulation of the inequality(X, A) _∆1

1(α)∩∆0ξ ω, ¬∆(ω)
of the Effective conjecture 1.

Proposition 2.4 Let1 ≤ ξ < ωCK₁ ,X be a 0-dimensional recursively presented Polish space, and A

be aΣ₁1relation onX. Then (X, A) _∆1

1∩∆0ξ ω, ¬∆(ω)
is equivalent to ∆(X) ∩ A Tξ×Tξ

= ∅.

Proof. Assume first that (X, A) _∆1

1∩∆0ξ ω, ¬∆(ω)
. Then there is a partition (Bn) of X into

A-discrete ∆1

1 ∩ ∆0ξ sets. In particular, Theorem 1.A in [Lo1] implies thatBn is a countable union

of∆11 ∩ Π0<ξ sets ifξ ≥ 2. In particular, BnisTξ-open and ∆(X) is disjoint from A Tξ×Tξ

(even if ξ = 1).

Conversely, assume that∆(X) ∩ ATξ×Tξ = ∅. Then each element x of X is contained in a A-discreteΣ11∩ Π0<ξ set (basic clopen set ifξ = 1). Lemma 2.2 implies that each element x of X is in

fact contained in aA-discrete ∆11∩ Π0<ξ set ifξ ≥ 2. It remains to apply Proposition 1.4 in [Lo1] and

the∆11-selection theorem to get the desired partition. 

One can also hope for an effective strengthening of Conjecture 2 generalizing Theorem 1.5:

Effective conjecture 2 Let1 ≤ ξ < ω1. Then there are

-0-dimensional Polish spaces_X0_ξ,_X1_ξ, - disjoint analytic subsetsA0

ξ,A1ξof the spaceX0ξ×X1ξ, not separable by a(Σ0ξ× Σ0ξ)σ set,

such that for anyα ∈ ωω_{such that1 ≤ ξ < ω}α

1, for any recursively inα presented Polish spaces X, Y ,

and for any pairA0, A1 of disjointΣ11(α) subsets of X ×Y , the following are equivalent:

(a) The setA0cannot be separated fromA1 by a(Σ0ξ×Σ0ξ)σ set.

(b) The setA0cannot be separated fromA1 by a∆11(α) ∩ (Σ0ξ×Σ0ξ)σset.

(c) The setA0cannot be separated fromA1 by aΣ01 Tξ(α)×Tξ(α)
set.

(d)A0∩ A1

Tξ(α)×Tξ(α)

6= ∅.

In fact, the statements (a)-(d) are indeed equivalent:

Theorem 2.5 Let1 ≤ ξ < ωCK₁ ,X, Y be recursively presented Polish spaces, and A0, A1be disjoint

Σ11 subsets ofX ×Y . The following are equivalent:

(a) The setA0cannot be separated fromA1 by a(Σ0ξ×Σ0ξ)σ set.

(b) The setA0cannot be separated fromA1 by a∆11∩ (Σ0ξ×Σ0ξ)σset.

(c) The setA0cannot be separated fromA1 by aΣ01(Tξ×Tξ) set.

(d)A0∩ A1 Tξ×Tξ

6= ∅.

Proof. Theorem 2.3 implies that (a) is indeed equivalent to (b). It also implies, using the proof of

Proposition 2.4, that (c) implies (a), and the converse is clear. It is also clear that (c) and (d) are

equivalent. 

A consequence of this is that Conjecture 2 and the Effective conjecture 2 are equivalent.

3 The case ξ = 1

(A) Continuous colorings

As in [L3], we can separate Conjecture 1 in two parts. We introduce the following notion, that will help us to characterize the relationsA for which there is a continuous homomorphism from A into any relation without countable continuous coloring:

Definition 3.1 Letξ be a countable ordinal, Π0

0:= ∆01, andX be a 0-dimensional Polish space. A

familyF of subsets ofX is ξ-disjoint if the elements of F are Π0

ξ and pairwise disjoint.

The first part ensures the existence of complicated examples.

Lemma 3.2 (a) Assume that(Cε

i)(ε,i)∈2×ω is a0-disjoint family of subsets of the spaceX such that

X\(S

(ε,i)∈2×ω Ciε) 6= ∅ and no clopen set meetingX\(

S (ε,i)∈2×ω Ciε) is ( S i∈ω Ci0×Ci1)-discrete. Then(X,S i∈ω Ci0×Ci1) 6∆0 1 ω, ¬∆(ω)
.

(b) There is a0-disjoint family (Cε

i)(ε,i)∈2×ωof subsets of2ωsatisfying the assumption (and thus the

conclusion) of (a).

Proof. (a) We argue by contradiction, which givesf :X → ω continuous such that f(x) 6= f(y) if (x, y) ∈S

i∈ω Ci0× Ci1. We set Dk:= f−1({k}), so that (Dk)k∈ω is a partition ofX into clopen

sets discrete forS

i∈ω Ci0× Ci1. Choosez ∈X\(

S

(ε,i)∈2×ω Ciε), and k with z ∈ Dk. This gives

(x, y) ∈ (S

i∈ω Ci0×Ci1) ∩ Dk2, which is absurd.

(b) We setCε

i := N02i+ε₁, so thatS_i∈ω C_i0×C_i1= {(02i1α, 02i+11β) | i ∈ ω and α, β ∈ 2ω}. Note that

{0∞_{} =X\(}S

(ε,i)∈2×ω Ciε). If C is a clopen neighborhood of 0∞, thenN0i⊆ C if i is big enough.

This gives an integeri with (02i1∞, 02i+11∞) ∈ (S

i∈ω Ci0×Ci1) ∩ C2. 

Lemma 3.3 LetX be a 0-dimensional Polish space, (Cε

i)(ε,i)∈2×ω be a0-disjoint family of subsets

ofX, X be a 0-dimensional Polish space, and A be a relation on X. Then one of the following holds:

(a)(X, A) _∆0

1 ω, ¬∆(ω)
,

(b)(X,S

i∈ω Ci0×Ci1) Σ01 (X, A).

Proof. Assume that (a) does not hold. Let us fix a compatible complete metric onX. In the sequel, the diameter will refer to this metric (this will also be the case in all the proofs where diameters are involved to come). We enumerate a basis N (p, X)

p∈ω for the topology ofX made of clopen sets.

• We build

- an increasing sequence of integers(pi)i∈ω,

- a sequence(xk)k∈ω of points ofX.

We want these objects to satisfy the following conditions: (1) (x2i, x2i+1) ∈ A ∩ N (pi, X)2

(2) N (pi+1, X) ⊆ N (pi, X)

(3) diam N (pi, X)
≤ 2−i

(4) There is no covering of N (pi, X) consisting of A-discrete clopen subsets of X

• Assume that this is done. Then we can define a point x of X by {x} =T

i∈ω N (pi, X). Note that

(xk)k∈ω tends tox. We define f :X → X by f(z) := x if z /∈S(ε,i)∈2×ω Ciε,f (z) := x2i+ε ifz ∈ Ciε.

Note thatf is continuous. Moreover, f (y), f (z)
= (x2i, x2i+1) ∈ A if (y, z) ∈ Ci0×Ci1, so that (b)

holds.

• Let us prove that the construction is possible. We set N (p−1, X) := X. Assume that (pi)i<l and

(x2i, x2i+1)i<l satisfying (1)-(4) have been constructed, which is the case for l = 0. We choose a

covering ofN (pl−1, X) with basic clopen sets of diameter at most 2−l, contained in N (pl−1, X).

Then one of these basic sets, sayN (pl, X), satisfies (4). It remains to choose (x2l, x2l+1) in the set

A ∩ N (pl, X)2. 

We setX1:= 2ωandA1:= {(02i1α, 02i+11β) | i ∈ ω and α, β ∈ 2ω} =S_i∈ω Ci0×Ci1, so thatA1

is aΣ01relation onX1.

Corollary 3.4 LetX be a 0-dimensional Polish space, and A be a relation on X. Then exactly one

of the following holds: (a)(X, A) _∆0

1 ω, ¬∆(ω)
,

(b)(X1,A1) _Σ0

1 (X, A).

Moreover, there are a non0-dimensional Polish space X, and a closed relation A on X, for which

neither (a), nor (b) holds (with this couple(X1,A1) or any other). There are also a 0-dimensional

Polish spaceX, and a relation A on X (a difference of two closed sets), for which it is not possible

Proof. Note first that (a) and (b) cannot hold simultaneously, by Lemma 3.2. Lemma 3.3 implies that

(a) or (b) holds.

• Consider now X :=R and A := {(0, 1)}. Then (a) does not hold since R is connected. If (b) holds, then we must havef (02i1α) = 0 and f (02i+11β) = 1. By continuity of f , we get f (0∞) = 0 = 1.

This would be the same with any (X1,A1). Indeed, as (X1,A1) 6_∆0

1 ω, ¬∆(ω)
, we have

Π0[A1] ∩ Π1[A1] 6= ∅, since otherwise there would be a clopen subset C of the 0-dimensional space

X1 separatingΠ0[A1] from Π1[A1], and we would have ∆(X1) ⊆ C2 ∪ (¬C)2⊆ ¬A1. So we can

choosex ∈ Π0[A1] ∩ Π1[A1], x2i∈ Π0[A1] such that (x2i) tends to x, y2i+1∈ Π1[A1] such that (y2i+1)

tends tox, y2iwith(x2i, y2i) ∈A1, andx2i+1with(x2i+1, y2i+1) ∈A1. Thenf (x2i) = 0, f (y2i+1) = 1

and we conclude as before.

• Consider X := 2ω_{andA := {0}∞_}×(2ω_\{0∞_{}). Then (a) does not hold since if a clopen subset C}

of2ω _contains0∞_{, then it contains also someα 6= 0}∞_{, so that(0}∞_{, α) ∈ A ∩ C}2_{. If (b) holds, then}

f (02i_{1α) = 0}∞_{for each integeri and f is not one-to-one.}

This argument works as soon asΠ0[A1] has at least two elements. If we argue in the other factor,

then we see that an example(X1,A1) with injectivity must satisfy thatA1 is a singleton {(α, β)}.

As(X1,A1) _Σ0 1 (2

ω_,G

0), α 6= β. So take a clopen subset C ofX1 containing α but not β. Then

∆(_X1) ⊆ C2∪ (¬C)2⊆ ¬A1. 

The notion of a0-disjoint family is essential in the following sense:

Proposition 3.5 LetX be a 0-dimensional Polish space, and A be a relation on X. The following are

equivalent:

(a) For any0-dimensional Polish space X, and any relation A on X, (X, A) 6_∆0

1 ω, ¬∆(ω)
⇒ (X, A) Σ01 (X, A).

(b) There is a0-disjoint family (Cε

i)(ε,i)∈2×ωof subsets ofX such that A⊆Si∈ω Ci0×Ci1.

Proof. (a) ⇒ (b) We set X := X1 and A :=A1. By Lemma 3.2, we get f :X → 2ω such that

A⊆(f ×f)−1_(A

1). We set Ciε:= f−1(N02i+ε₁).

(b)⇒ (a) By Lemma 3.3 we get (X,S

i∈ω Ci0×Ci1) Σ0

1 (X, A), so that (X, A) Σ01 (X, A). 

(B) Countable unions of open rectangles (i.e., open sets)

The content here is completely trivial. It is just the fact that a subset of a metric space is not open exactly when it contains a point that we can approximate by a countable sequence contained in its complement. We give some statements since the situation will be more complicated in the caseξ = 2. As in (A) we can characterize the tuples(X0_,X1_,A0_,A1_{) ≤-below any tuple (X, Y, A}

0, A1) with A0

Lemma 3.6 (a) Assume that (C_iε)i∈ω is a 0-disjoint family of subsets of the space Xε such that

X0_\(S

i∈ω Ci0)
× X1\(

S

i∈ω Ci1)
6= ∅ and no open set meeting X0\(

S i∈ω Ci0)
× X1\( S i∈ω Ci1)
is disjoint fromS i∈ω Ci0× Ci1. Then X0\( S i∈ω Ci0)
× X1\( S

i∈ω Ci1)
is not separable from

S

i∈ω Ci0×Ci1by a(Σ01×Σ01)σ set.

(b) There are0-disjoint families of subsets of 2ω _{satisfying the assumption (and thus the conclusion)}

of (a).

Proof. (a) is obvious.

(b) We setCε

i := N0i₁, so that S_i∈ω C_i0× C_i1= {(0i1α, 0i1β) | i ∈ ω and α, β ∈ 2ω}. Note that

{0∞_{} =X}ε_\(S

i∈ω Ciε). If O is an open neighborhood of (0∞, 0∞), then N02i⊆ O if i is big enough.

This gives an integeri with (0i₁∞_{, 0}i₁∞_{) ∈ (}S

i∈ω Ci0×Ci1) ∩ O. 

Lemma 3.7 Let_X0,_X1 _{be0-dimensional Polish spaces, (C}ε

i)i∈ωbe a0-disjoint family of subsets of

Xε_{,X, Y be Polish spaces, and A}

0, A1be disjoint subsets ofX×Y . Then one of the following holds:

(a)A0is separable fromA1by a(Σ01×Σ01)σ set,

(b) _X0,_X1_{, X}0_\(S

i∈ω Ci0)
× X1\(

S

i∈ω Ci1)
, Si∈ω Ci0×Ci1
≤ (X, Y, A0, A1).

Proof. Assume that (a) does not hold. Pick(x, y) ∈ A0∩ A1, and(xi, yi) in A1tending to(x, y). We

definef :X0_{→ X by f (z) := x if z /∈}S

i∈ω Ci0,xiifz ∈ Ci0. Note thatf is continuous. Similarly, we

defineg :X1_{→ Y , so that (b) holds. }

We define _Xε₁ := 2ω_{, A}0

1 := {(0∞, 0∞)} = X01\ (

S

i∈ω Ci0)
× X11\ (

S

i∈ω Ci1)
and also

A1

1:= {(0i1α, 0i1β) | i ∈ ω and α, β ∈ 2ω} =

S

i∈ω Ci0× Ci1. As in (A) we get the two following

consequences:

Corollary 3.8 LetX, Y be Polish spaces, and A0, A1 be disjoint subsets ofX×Y . Then exactly one

of the following holds:

(a)A0is separable fromA1by a(Σ01×Σ01)σ set,

(b)(_X0

1,X11,A01,A11) ≤ (X, Y, A0, A1).

Proposition 3.9 Let_X0,_X1_{be0-dimensional Polish spaces, andA}0_,A1_⊆X0_×X1_{be disjoint. The}

following are equivalent:

(a) For any Polish spacesX, Y , and any A0, A1⊆ X ×Y disjoint,

A0 is not separable fromA1 by a(Σ01×Σ01)σ set ⇒ (X0,X1,A0,A1) ≤ (X, Y, A0, A1).

(b) There is a0-disjoint family (Cε

i)i∈ω of subsets ofXεsuch that the inclusionsA1⊆Si∈ω Ci0×Ci1

and_A0⊆ X0_\(S

i∈ω Ci0)
× X1\(

S

4 The case ξ = 2

(A) Baire class one colorings Lemma 4.1 (a) Assume that (Cε

i)(ε,i)∈2×ω is a1-disjoint family of subsets ofX such that no

non-empty clopen subset ofX is (S

i∈ω Ci0×Ci1)-discrete. Then (X,

S

i∈ω Ci0×Ci1) ∆02 ω, ¬∆(ω)
.

(b) There is a1-disjoint family (Cε

i)(ε,i)∈2×ωof subsets ofωωsatisfying the assumption (and thus the

conclusion) of (a).

Proof. (a) We argue by contradiction, which gives a∆02-measurable mapf :X→ω with f(x)6=f(y)

if(x, y) ∈S

i∈ω Ci0×Ci1. We setDk:= f−1({k}), so that (Dk)k∈ω is a partition ofX into ∆02sets

discrete forS

i∈ω Ci0×Ci1. By Baire’s theorem, there are an integerk and a nonempty clopen subset

C ofX such that DkcontainsC. This gives (x, y) ∈ (S_i∈ω Ci0×Ci1) ∩ C2⊆ (

S

i∈ω Ci0×Ci1) ∩ Dk2,

which is absurd.

(b) Letb : ω → ω<ω_{be a bijection. We setC}ε

i :=b(i) 2|b(i)|+ε
∞ , so that [ i∈ω C_i0×Ci1=  u(2|u|)∞, u(2|u|+1)∞
| u ∈ ω<ω _. If∅ 6= C ∈ ∆0

1(ωω), then C contains some basic clopen set Nu, and u(2|u|)∞, u(2|u|+1)∞
is in

(S

i∈ω Ci0×Ci1) ∩ C2. 

Remark. There are a1-disjoint family (Cε

i)(ε,i)∈2×ω of subsets ofωω, and a relationA on ωω such

that(ωω, A) 6_∆0

2 ω, ¬∆(ω)
and (ω ω_,S

i∈ω Ci0×Ci1) 6Σ01 (ω

ω_{, A), so that Lemma 3.3 cannot be}

extended toΣ02-measurable countable colorings.

Indeed, we setCε

i := {u(2i+ ε)∞ | u ∈ ωi} and A :=



u(2|u|)∞, u(2|u|+ 1)∞
| u ∈ ω<ω _.

Then(Cε

i)(ε,i)∈2×ωis clearly a1-disjoint family. Lemma 4.1 gives the first assertion. For the second

assertion, assume, towards a contradiction, thatf : ωω_{→ ω}ω _{is continuous and satisfies the inclusion}

S

i∈ω Ci0×Ci1⊆ (f ×f )−1(A). If i ∈ ω, then there is ui∈ ω<ωwith

f [C_i0]×f [C_i1] ⊆

ui(2|ui|)∞, ui(2|ui|+1)∞
.

In particular, for anyα, β ∈ ωω _{we get}

f (α), f (β)
= limi→∞



f (α|i)(2i)∞
_{, f (β|i)(2i+1)}∞
 = limi→∞ ui(2|ui|)∞, ui(2|ui|+1)∞
.

But this implies thatf is constant, which is absurd. To fix this, we refine the notion of a ξ-disjoint family.

Definition 4.2 Let1 ≤ ξ < ω1. Aξ-disjoint family (Ciε)(ε,i)∈2×ω of subsets of a0-dimensional Polish

space_{X is said to be comparing if for each integer q there is a partition (O}qp)p∈ωofX into ∆0ξsets

such that, for eachi ∈ ω,

(a) ifq < i, then there is piq∈ ω such that Ci0∪ Ci1⊆ O pi

q q ,

(b) ifq ≥ i and ε ∈ 2, then Cε

Lemma 4.3 There is a comparing 1-disjoint family (C_iε)(ε,i)∈2×ω of subsets of ωω satisfying the

assumption (and thus the conclusion) of Lemma 4.1.(a).

Proof. Letb : ω → ω<ω_{be a bijection satisfyingb}−1_{(s) ≤ b}−1_{(t) if s ⊆ t. It can be built as follows. Let}

(pq)q∈ωbe the sequence of prime numbers, andI : ω<ω→ ω defined by I(s) := ps(0)+10 ...p

s(|s|−1)+1 |s|−1 if

s 6= ∅, and I(∅) := 1. Note that I is one-to-one, so that there is an increasing bijection ϕ : I[ω<ω] → ω. We setb := (ϕ ◦ I)−1: ω → ω<ω_{. We define(C}ε

i)(ε,i)∈2×ω as in the proof of Lemma 4.1.(b), so that

(Cε

i)(ε,i)∈2×ω is a1-disjoint family. It remains to see that (Ciε)(ε,i)∈2×ωis comparing. We set

O_qp:=              N

b(i)(2|b(i)|+ε)maxl≤q (|b(l)|+1)−|b(i)| ifp = 2i+ε ≤ 2q+1,

ωω_\(S

p′_≤2q+1 Op ′

q ) if p = 2q+2,

∅ if p ≥ 2q+3,

so that(Oqp)p∈ω is a partition ofωω into∆01 sets. Note that (b) is fulfilled. Ifq < i, then there is at

most one couple(j, ε) ∈ (q+1)×2 such that b(j)(2|b(j)| + ε)maxl≤q(|b(l)|+1)−|b(j)|_{is compatible with}

b(i). If it exists and if |b(i)| ≥ maxl≤q (|b(l)| + 1), then Ci0∪ Ci1⊆ O 2j+ε

q and we setpiq:= 2j +ε.

Otherwise,C_i0∪ Ci1⊆ O 2q+2

q and we setpiq:= 2q+2. 

We have a stronger result than Conjecture 1, in the sense that we do not need any regularity assumption onA, neither that X is 0-dimensional.

Lemma 4.4 LetX be a 0-dimensional Polish space, (Cε

i)(ε,i)∈2×ω be a comparing1-disjoint family

of subsets ofX, X be a Polish space, and A be a relation on X. Then one of the following holds:

(a)(X, A) _∆0

2 ω, ¬∆(ω)
,

(b)(X,S

i∈ω Ci0×Ci1) Σ01 (X, A).

Proof. IfA is not a digraph, then choose x0with(x0, x0) ∈ A, and put f (x) := x0. So we may assume

thatA is a digraph. We set

U :=[ nV ∈ Σ01(X) | ∃D ∈ Π01(ω×X) V ⊆ [ p∈ω Dpand∀p ∈ ω A ∩ D2p= ∅ o . Case 1.U = X.

There is a countable covering ofX into A-discrete Σ02sets. We just have to reduce them to get a

partition showing that (a) holds.

Case 2.U 6= X.

ThenY := X \U is a nonempty closed subset of X.

Claim IfV ∈ Σ01(X) meets Y , then V ∩ Y is not A-discrete.

We argue by contradiction. AsV ∩ U can be covered with some S

p∈ω Dp’s, so is V . Thus

• We construct a family (xu)u∈ω<ω of points ofY , and a family (X_u)_u∈ω<ω of open subsets of Y .

We want these objects to satisfy the following conditions: (1) xu∈ Xu

(2) Xup⊆ Xu

(3) diam(Xu) ≤ 2−|u|

(4) (x_u(2n), x_u(2n+1)) ∈ A if u ∈ ωn

(5) xu(2n+ε)= xuifu /∈ ωnandε ∈ 2

• Assume that this is done. We define f :X→Y ⊆X by {f (x)} :=\ q∈ω Xp0...pq−1= \ q∈ω Xp0...pq−1,

wherepi satisfiesx ∈ O_ipi witnessing comparability, so thatf is continuous. Note that f (x) is the

limit ofxp0...pq−1, and thatxu(2|u|+ε)= xu(2|u|+ε)2= ... = x_u(2|u|+ε)q+1for each(u, ε) ∈ ω<ω×2. Thus

f (x) = limq→∞xu(2|u|+ε)q+1= x_u(2|u|+ε)ifx ∈ C_iεandu := pi₀...pi_i−1, and

f (x), f (y)
= (x_u(2|u|), x_u(2|u|+1)) ∈ A if(x, y) ∈ C_i0×C_i1. So (b) holds.

• Let us prove that the construction is possible. We choose x∅∈ Y and an open neighborhood X∅of

x∅ inY , of diameter at most 1. Assume that (xu)u∈ω≤l and (Xu)u∈ω≤l satisfying (1)-(5) have been

constructed, which is the case forl = 0.

An application of the claim gives(x_u(2l), x_u(2l+1)) ∈ A ∩ X_u2ifu ∈ ωl_{. We satisfy (5), so that the}

definition of thexu’s is complete. Note thatxu∈ Xu|l ifu ∈ ωl+1.

We choose an open neighborhoodXu ofxu inY , of diameter at most 2−l−1, ensuring the

inclu-sionXu⊆ Xu|l. 

We setX2:= ωωandA2:= u(2|u|)∞, u(2|u|+1)∞
| u ∈ ω<ω = Si∈ω Ci0×Ci1, so thatA2is

aΣ02relation onX2. As in Section 3.(A) we get the two following consequences:

Corollary 4.5 LetX be a Polish space, and A be a relation on X. Then exactly one of the following

holds:

(a)(X, A) _∆0

2 ω, ¬∆(ω)
,

(b)(X2,A2) _Σ0

1 (X, A).

Proposition 4.6 Let_{X be a 0-dimensional Polish space, and A be a relation on X. The following are} equivalent:

(a) For any Polish spaceX, and any relation A on X, (X, A) 6_∆0

2 ω, ¬∆(ω)
⇒ (X, A) Σ01 (X, A).

(b) There is a comparing1-disjoint family (Cε

(B) Countable unions ofΣ02rectangles

Lemma 4.7 (a) Assume that(Cε

i)(ε,i)∈2×ω is a1-disjoint family of meager subsets of X such that

no nonempty clopen subset ofX is (S

i∈ω Ci0× Ci1)-discrete. Then ∆ X\(

S (ε,i)∈2×ω Ciε)
is not separable fromS i∈ω Ci0×Ci1by a  Σ02 X\( S i∈ω Ci1)
×Σ02 X\( S i∈ω Ci0)
 σ set.

(b) There is a comparing1-disjoint family (Cε

i)(ε,i)∈2×ω of subsets of ωω satisfying the assumption

(and thus the conclusion) of (a).

Proof. (a) We argue by contradiction, which givesCn∈ Π0₁ X\(S_i∈ω Ci1)
on one side, and also

Dn∈ Π0₁ X\(S_i∈ω C_i0)
with ∆ X\(S_{(ε,i)∈2×ω} C_iε)
⊆ S_n∈ω (Cn×Dn) ⊆ ¬(S_i∈ω C_i0×C_i1).

In particular, X\(S

(ε,i)∈2×ω Ciε) =

S

n∈ω Cn∩ Dn, and Baire’s theorem givesn and a nonempty

clopen subsetC ofX such that C \(S

(ε,i)∈2×ω Ciε) ⊆ Cn∩ Dn. Note thatC \(Si∈ω Ci1) ⊆ Cnand

C \(S

i∈ω Ci0) ⊆ Dnsince theCiε’s are meager andX\(

S

(ε,i)∈2×ω Ciε) is dense inX\(

S

i∈ω Ciε).

The assumption gives(x, y) ∈ (S

i∈ω Ci0×Ci1) ∩ C2. Then(x, y) ∈ (

S

i∈ω Ci0×Ci1) ∩ (Cn×Dn),

which is absurd. (b) Let(Cε

i)(ε,i)∈2×ωbe the family given by Lemmas 4.1.(b) and 4.3. As theCiε’s are singletons, they

are meager. 

Remark. Note that∆ X\(S

(ε,i)∈2×ω Ciε)
= ∆(X) ∩  X\(S i∈ω Ci1)
× X\(Si∈ω Ci0)
 is a closed subset of X\(S

i∈ω Ci1)
× X\(Si∈ω Ci0)
. This shows that the spaces X02,X12of Conjecture

2 cannot be both compact, which is quite unusual in this kind of dichotomy (even if it was already the case in [L2]). Indeed, our example shows thatA0

2,A12must be separable by a closed setC, and C,A12

must have disjoint projections. IfX0

2,X12 are compact, thenC and its projections are compact too.

The product of these compact projections is a(Σ0

2×Σ02)σ set separating A0₂fromA1₂, which cannot

be. We will meet an example whereX = 3ω_{. This fact implies that we cannot extend the continuous}

maps of Theorem 1.5.(e) to3ω_{in general.}

To ensure the possibility of the reduction, we introduce the following notion:

Definition 4.8 Let1 ≤ ξ < ω1. Aξ-disjoint family (Ciε)(ε,i)∈2×ω of subsets of a0-dimensional Polish

spaceX is said to be very comparing if for each integer q there is a partition (Opq)p∈ωofX into ∆0_ξ

sets such that, for eachi ∈ ω,

(a) ifq < i, then there is pi_q∈ ω such that C0

i ∪ Ci1⊆ O pi q q , (b) ifq ≥ i and ε ∈ 2, then Cε i ⊆ O2i+εq , (c) if(ε, i) ∈ 2×ω, thenS r≥i T q≥r O2i+εq = Ciε.

Lemma 4.9 There is a very comparing1-disjoint family (Cε

i)(ε,i)∈2×ωof subsets ofωωsatisfying the

assumptions (and thus the conclusion) of Lemma 4.7.(a).

Proof. Let (Cε

i)(ε,i)∈2×ω be the family given by Lemmas 4.1.(b), 4.3, and 4.7.(b). It remains to

check Condition (c). Note first that the inclusionS

r≥i

T

q≥r Oq2i+ε⊇ Ciεholds for any comparing

Conversely, S r≥i T q≥r Oq2i+ε= S r≥i T

q≥r N_{b(i)(2|b(i)|+ε)}maxl≤q (|b(l)|+1)−|b(i)|

= {b(i)(2|b(i)| + ε)∞} = Cε i.

This finishes the proof. 

Notation. We now recall some facts about the Gandy-Harrington topology (see [L2]). LetZ be a recursively presented Polish space. TheGandy-Harrington topology GH on Z is generated by the Σ11subsets ofZ. We set ΩZ := {z ∈ Z | ω1z= ωCK1 }. Then ΩZisΣ11, dense in(Z, GH), and W ∩ ΩZ

is a clopen subset of(ΩZ, GH) for each W ∈ Σ₁1(Z). Moreover, (ΩZ, GH) is a 0-dimensional Polish

space. So we fix a complete compatible metric_{dGH on (Ω}Z, GH).

The following notion is important for the next proof.

Definition 4.10 Leta be a countable set, ξ < ω1, and F := (Siε)(ε,i)∈2×ω be aξ-disjoint family of

subsets ofaω_{. We say thats ∈ a}<ω_{isF-suitable if there is (α, β) ∈}S

i∈ω Si0×Si1such thats = α ∧ β

is the longest common initial segment ofα and β.

Example. In the next proof, we will take a := 3, ξ := 1, and s will be suitable when s is empty or finishes with2. If θ : ω → {s ∈ 3<ω _{| s is suitable} is a bijection such that |θ(i)|}

i∈ω is

non-decreasing, then we can define a1-disjoint family F of subsets of 3ωbyS_iε:= {θ(i)εα | α ∈ 2ω}, and s is suitable exactly when s is F-suitable.

In particular, a non suitable sequence is of the formsεt, where s is suitable, ε ∈ 2 and t ∈ 2<ω_(we

will use this notation in the next proof). If∅ 6= s is suitable, then we set s−:= s|max{l < |s| | s|l is suitable}.

Lemma 4.11 LetX be a 0-dimensional Polish space, (Cε

i)(ε,i)∈2×ω be a very comparing1-disjoint

family of subsets ofX, X, Y be Polish spaces, and A0, A1be disjoint analytic subsets ofX×Y . Then

one of the following holds:

(a)A0is separable fromA1by a(Σ02×Σ02)σ set,

(b) X\(S i∈ω Ci1),X\( S i∈ω Ci0), ∆ X\( S (ε,i)∈2×ω Ciε)
, Si∈ω Ci0×Ci1
≤ (X, Y, A0, A1).

Proof. We may assume thatX, Y are recursively presented and that A0, A1 areΣ11. Assume that (a)

does not hold. By Theorem 2.5 we getN := A0∩ A1T2×T26= ∅. Lemma 2.2 implies that

(x, y) /∈ A1 T2×T2

⇔ ∃C, D ∈ Σ11∩ Π01 (x, y) ∈ C ×D ⊆ ¬A1

⇔ ∃C, D ∈ ∆1

1∩ Π01 (x, y) ∈ C ×D ⊆ ¬A1.

This and Proposition 1.4 in [Lo1] show thatN is Σ11. We construct

- A sequence(xu)u∈3<ωof points ofX,

- A sequence(yu)u∈3<ω of points ofY ,

- A sequence(Xu)u∈3<ωofΣ₁0subsets ofX,

- A sequence(Yu)u∈3<ωofΣ₁0subsets ofY ,

We want these objects to satisfy the following conditions: (1) (xu, yu) ∈ Xu×Yu

(2) (xs, ys) ∈ Vs⊆ N ∩ ΩX×Y ifs is suitable

(3) Xuε⊆ Xuifu is suitable or u = s0t, and Xs1t2⊆ Xs

(4) Yuε⊆ Yuifu is suitable or u = s1t, and Ys0t2⊆ Ys

(5) Vs⊆ Vs−if∅ 6= s is suitable

(6) diam(Xu), diam(Yu) ≤ 2−|u|

(7) diamGH(Vs) ≤ 2−|s|ifs is suitable

(8) (xs0, ys1) ∈ Π0[(Xs×Ys) ∩ Vs]×Π1[(Xs×Ys) ∩ Vs]
∩ A1ifs is suitable

(9) (xs0t, ys1t) = (xs0, ys1) if s is suitable and t ∈ 2<ω

• Assume that this is done. We define φ : ω<ω_{→ 3}<ω_{byφ(∅) := ∅ and}

φ(sn) :=( φ(s)2ε if n = 2|s|+ε or s 6= ∅ and n 6= s(|s|−1)
,

φ(s)ε if (n = 2q+ε and q 6= |s|) ands = ∅ or s 6= ∅ and n = s(|s|−1)
 . This map allows us to defineΦ : ωω_{→ 3}ω_{byΦ(γ)(p) := φ γ|(p+1)
(p), and Φ is continuous.}

As (Cε

i)(ε,i)∈2×ω is very comparing, there are some witnesses(Opq)p∈ω. Letx ∈X. As in the

proof of Lemma 4.4, we associate the sequence(pq)q∈ω∈ ωωdefined byx ∈ Oqpq. As(Ciε)(ε,i)∈2×ω

is very comparing, (pq)q∈ω is not eventually constant ifx /∈S_{(ε,i)∈2×ω} Ciε. Thus Φ (pq)q∈ω
has

infinitely many2’s in this case. If x ∈ C_iε, then

Φ (pq)q∈ω
= Φ pi0...pii−1(2i+ε)∞
= φ(pi0...pii−1)2ε∞.

Ifx ∈X\(S

i∈ω Ci1), then the increasing sequence (n0k)k∈ωof integers such thatΦ (pq)q∈ω
|n0kis

suit-able or of the forms0t is infinite. Condition (3) implies that (X_{Φ((pq)q∈ω)|n}0

k)k∈ω is non-increasing.

Moreover,(X_{Φ((pq)q∈ω)|n}0

k)k∈ωis a sequence of nonempty closed subsets ofX whose diameters tend

to0, so that we can define {f (x)} :=T

k∈ω XΦ((pq)q∈ω)|n0 k = T k∈ω XΦ((pq)q∈ω)|n0 k. This defines a continuous mapf :X\(S

i∈ω Ci1) → X with f (x) = limk→∞ xΦ((pq)q∈ω)|n0

k. Similarly, we define

g :_X\(S

i∈ω Ci0) → Y continuous with g(x) = limk→∞yΦ((pq)q∈ω)|n1 k.

If x /∈S

(ε,i)∈2×ω Ciε, then the sequence (kj)j of integers such that Φ (pq)q∈ω
|kj is suitable

is an infinite subsequence of both (n0

k)k∈ω and (n1k)k∈ω. Note that (VΦ((pq)q∈ω)|kj)j∈ω is a

non-increasing sequence of nonempty closed subsets ofΩX×Y whose GH-diameters tend to0, so that we

can defineF (x) by {F (x)} :=T

j∈ω VΦ((pq)q∈ω)|kj⊆ N ⊆ A0. AsF (x) is the limit (in (X ×Y, GH),

and thus in X × Y ) of (x_{Φ((pq)q∈ω)|kj}, y_{Φ((pq)q∈ω)|kj})j∈ω, we get F (x) = f (x), g(x)
. Therefore

∆ X\(S

(ε,i)∈2×ω Ciε)
⊆ (f ×g)−1(A0).

Note thatxs0= xs02= ... = x_s0q+1 for eachs suitable. Thus

f (x) = limq→∞ xφ(pi 0...pii−1)20q+1= xφ(pi0...pii−1)20 ifx ∈ C_i0. Similarly,g(y) = y_φ(pi 0...pii−1)21ify ∈ C 1 i and S i∈ω Ci0×Ci1⊆ (f ×g)−1(A1).

• Let us prove that the construction is possible. As N 6= ∅, we can choose (x∅, y∅) ∈ N ∩ ΩX×Y, aΣ11

subsetV∅ofX×Y with (x∅, y∅) ∈ V∅⊆ N ∩ ΩX×Y of GH-diameter at most1, and a Σ10neighborhood

X_∅ (resp.,Y_∅) ofx_∅(resp.,y_∅) of diameter at most1. Assume that (xu)u∈3≤l,(yu)u∈3≤l,(Xu)u∈3≤l,

(Yu)u∈3≤l and(Vs)_s∈3≤l_{suitable satisfying (1)-(9) have been constructed, which is the case for l = 0.}

Lets ∈ 3<ω_{be suitable. Note that(x}

s, ys) ∈ (Xs×Ys) ∩ Vs⊆ A1 T2×T2

. We chooseX′, Y′∈ Σ0 1

with(xs, ys) ∈ X′× Y′⊆ X′× Y′⊆ Xs× Ys. AsΠε[(X′× Y′) ∩ Vs] is Σ11,Πε[(X′×Y′) ∩ Vs] is

Σ11∩ Π01. In particular,Πε[(X′×Y′) ∩ Vs] is T2-open. This shows the existence of

(xs0, ys1) ∈ Π0[(X′×Y′) ∩ Vs]×Π1[(X′×Y′) ∩ Vs]
∩ A1.

Note that(xs0, ys1) ∈ X′×Y′⊆ Xs×Ys. We setxs1:= xs,ys0:= ys. We definedxu, yuwhenu ∈ 3l+1

is not suitable butu|l is suitable.

Assume now thatu ∈ 3l+1is suitable, but notu|l. This gives (s, ε, t) such that u = sεt2. Assume first thatε = 0. Note that xs0t= xs0∈ Xs0t∩ Π0[(Xs×Ys) ∩ Vs]. This gives

xu∈ Xs0t∩ Π0[(Xs×Ys) ∩ Vs],

and alsoyuwith(xu, yu) ∈ (Xs0t∩ Xs)×Ys
∩ Vs= (Xs0t×Ys) ∩ Vs. Ifε = 1, then similarly we get

(xu, yu) ∈ (Xs×Ys1t) ∩ Vs.

Ifu and u|l are both suitable, or both non suitable, then we set (xu, yu) := (xu|l, yu|l). So we

definedxu, yu in any case. Note that Conditions (8) and (9) are fulfilled, and that(xs, ys) ∈ Vs− ifs

is suitable. Moreover,xu∈ Xu|lifu|l is suitable or u|l = s0t, and xu∈ Xsifu = s1t2, and similarly

inY . We choose Σ10setsXu, Yuof diameter at most2−l−1with

(xu, yu) ∈ Xu×Yu⊆ Xu×Yu⊆

  

Xu|l×Yu|lifu is not suitable or u|l is suitable,

X_u|l×Ysifu = s0t2,

Xs×Yu|lifu = s1t2.

It remains to choose, whens is suitable, Vs∈ Σ₁1(X × Y ) of GH-diameter at most 2−l−1 such that

(xs, ys) ∈ Vs⊆ Vs−.  We set_Xε₂:= ωω_{\{u(2|u|+1−ε)}∞_{| u ∈ ω}<ω_{} =X\(}S i∈ω Ci1−ε), A02:= ∆(ωω\{u(2|u|+ε)∞ | (u, ε) ∈ ω<ω×2}) = ∆ X\( [ (ε,i)∈2×ω Ciε)
, andA1 2:= 

u(2|u|)∞_{, u(2|u|+1)}∞
_{| u∈ ω}<ω _{= S}

i∈ω Ci0×Ci1.

Corollary 4.12 LetX, Y be Polish spaces, and A0, A1 be disjoint analytic subsets ofX ×Y . Then

exactly one of the following holds:

(a)A0is separable fromA1by a(Σ02×Σ02)σ set,

(b)(X0

Remark. In the remark after Lemma 4.7 we announced an example withX:=3ω_{. In fact, we already}

met it after Definition 4.10. Recall that the formulaSε

i := {θ(i)εα | α ∈ 2ω} defines a 1-disjoint

family of subsets of3ω_{, which are clearly meager. It is also clear that no nonempty clopen subset of}

3ωis(S

i∈ω Si0×S1i)-discrete. One can check that the formula

Op_q:=              S

t∈2|θ(q)|−|θ(i)| N_θ(i)εtifp = 2i+ε ≤ 2q+1,

3ω_\(S

p′_≤2q+1 Op ′

q ) if p = 2q+2,

∅ if p ≥ 2q+3, defines witnesses for the fact that(Sε

i)(ε,i)∈2×ωis very comparing.

To close this section, we notice that the notion of a very comparing1-disjoint family gives only a sufficient condition, and not a characterization like in 3.5, 3.9 or 4.6:

Proposition 4.13 Let _{X be a 0-dimensional Polish space, (Ci}ε)i∈ω be a very comparing1-disjoint

family of subsets ofX, Xε_⊆X\(S i∈ω Ci1−ε),A0⊆ ∆ X\( S (ε,i)∈2×ω Ciε)
, and A1⊆ S i∈ω Ci0×Ci1

be as in the definition of≤. Then for any Polish spaces X, Y , and any disjoint analytic subsets A0, A1

ofX ×Y ,

A0 is not separable fromA1 by a(Σ01×Σ01)σ set ⇒ (X0,X1,A0,A1) ≤ (X, Y, A0, A1).

5 The case ξ = 3: Baire class two colorings

Remark. Unlike whenξ ∈ {1, 2}, we cannot haveA3of the formSn∈ω Cn0×Cn1, where(Cnε)(ε,n)∈2×ω

is a 2-disjoint family. Indeed, we will see that there is a Borel graph _{G ⊆ 2}ω_{× 2}ω _{of a partial}

injection such that (2ω,G) 6_∆0

3 ω, ¬∆(ω)
. We would get f : X3 → 2

ω _{continuous such that}

A3⊆ (f ×f )−1(G), and f[Cn0]×f [Cn1] would be a singleton. The set (f ×f )[A3] would be countable,

and (2ω_{, (f × f )[A}

3]) _∆0

3 ω, ¬∆(ω)
, (X3,A3) ∆03 ω, ¬∆(ω)
would hold, which is absurd.

However, the following result holds.

Theorem 5.1 There are a0-dimensional Polish space X3 and an analytic relationA3 on X3 such

that for any Polish spaceX, and for any analytic relation A on X, exactly one of the following holds:

(a)(X, A) _∆0

3 ω, ¬∆(ω)
,

(b)(_X3,A3) _Σ0

1 (X, A).

We can take X3 = ωω, but this is not the most natural thing to do. Note that we can replace

X3 with any copy of it. Our spaceX3 will be a denseGδ subset of2ω, in fact a copy ofωω. This

Gδsubset is not necessary to see that(X3,A3) satisfies the “exactly” part of Theorem 5.1 (i.e., that

(_X3,A3) 6_∆0

3 ω, ¬∆(ω)
), but it is useful to build and ensure the continuity of the homomorphism

of Statement (b). The definition of X3 and A3 is based on the construction of the following basic

Recall the sequence(sn)n∈ω defined in the introduction. InG0, we putsn, i.e., a finite sequence

of elements of 2, before the changed coordinate. InA3, we will put a finite sequence of elements

of2<ω_{, together with a way to recover them after concatenation, before the changed coordinate. In}

order to do that, we identifyω with ω2.

Notation. Let< ., . >: ω2→ ω be a natural bijection. More precisely, < n, p >:= (Σk≤n+p k)+p.

Note that the inverse bijectionq 7→ (q)0, (q)1
is build as follows. We set, for q ∈ ω,

M (q) := max{m ∈ ω | Σk≤mk ≤ q}.

Then we define (q)0, (q)1
:= M(q)−q+(Σk≤M (q)k), q−(Σk≤M (q)k)
. More concretely,

ω = {< 0, 0 >, < 1, 0 >, < 0, 1 >, . . . , < M (q), 0 >, < M (q)−1, 1 >, ..., < 0, M (q) >, ...}. Ifu ∈ 2≤ω and n ∈ ω, then we define (u)n∈ 2≤ω by (u)n(p) := u(< n, p >) if < n, p >< |u|.

Here also we define< α0, α1, ... >∈ 2ω by< α0, α1, ... > (< n, p >) := αn(p), for any sequence

(αn)n∈ω of elements of2ω. In particular, α 7→ (α)n

n∈ω and(αn)n∈ω7→< α0, α1, ... > are inverse

bijections.

Lemma 5.2 Letu, v ∈ 2<ω_.

(a)u ⊆ v implies that (u)n⊆ (v)nfor eachn ∈ ω.

(b)|(u)0| ≤ |u|.

(c)|(u)n| ≤ |u|+1−n if n ≤ |u|+1.

Proof. (a) If< n, p >< |u|, then (u)n(p) = u(< n, p >) = v(< n, p >) = (v)n(p) because of the

inequality< n, p >< |v|, so that (u)n⊆ (v)n.

(b) We set, for n, q ∈ ω, cn

q := Card({p ∈ ω |< n, p >< q}). As < ., . > is a bijection, we get

cn_q+1≤ cn

q+1. As cn0= 0, cnq≤ q. We are done since |(u)n| = cn_|u|.

(c) Note first that< n, p >= (Σk≤n+pk)+p < (Σk≤n′_+p′ k)+p′=< n′, p′ > if n+p < n′+p′, and

that(q)0+(q)1= M (q) ≤ q < q+1. This implies that q =< (q)0, (q)1 ><< n, q+1−n > if n ≤ q+1.

It remains to apply this toq := |u| since |(u)n| = cn_|u|. 

We can viewG0 as the countable unionSn∈ω Gr(ϕn), where ϕnis the homeomorphism defined

on the basic clopen setNsn0onto the clopen setNsn1 defined byϕn(sn0γ) := sn1γ. The setA3 will

also be the countable union of the graphs of some homeomorphisms, indexed byω<ω_{instead ofω.}

Their domain and range will beGδsubsets of2ωinstead of clopen sets. We first define the closures

of theseGδ’s. They will be copies of2ω. In fact, our homeomorphims will also be defined on the

closure of these final domains. We will fix the coordinates whose number is in one of the verticals before that of the number of the changed coordinate. This leads to the following notation.

Notation. Ift ∈ ω<ω_{andk ≤ |t|, then we set Σ}t

k:= Σj<k t(j)+2
, and

Σt:=< Σj<|t| t(j)+2
, 0 >=< Σt|t|, 0 >

(Σtwill be the number of the unique changed coordinate). We setwn:= sn0, so that |wn| = n+1 and

We define the following objects fort ∈ ω<ω. • We first define a copy Ktof2ω by

Kt:=α∈ 2ω| ∀k < |t| (α)Σt k= (wt(k))00 t(k)+1−|(wt(k))0|₁₀∞_and ∀0 < i < t(k)+2 (α)_Σt k+i= wt(k)
i0 ∞ _.

This is well defined since|wt(k)| = t(k)+ 1, so that we can apply Lemma 5.2.(b) to u := wt(k)and

t(k)+1−|(w_t(k))0| ≥ 0. In particular, the last 1 in (α)Σt

k is at the positiont(k)+1. Here is the picture

ofKtwhent = (4, 2):

• We define a non-trivial partition (K0

t, Kt1) of Ktinto clopen sets byKtε:= {α ∈ Kt| α(Σt) = ε}.

• We define a homeomorphism ϕt: Kt0→ Kt1byϕt(α)(m) := 1 if m = Σt,α(m) otherwise.

We can view the construction ofKt, Ktεand ϕtinductively. Indeed, K∅= 2ω,K_∅ε is the basic

clopen setNε, andϕ∅(α)(m) is 1 when m = 0, α(m) otherwise. Then

Ktn:= {α ∈ Ktwn(0)| (α)Σt |t|= (wn)00 n+1−|(wn)0|₁₀∞_{and∀0 < i < n+2 (α)} Σt |t|+i= (wn)i0 ∞_}, Kε

tn:= {α ∈ Ktn| α(< Σt_|t|+n+2, 0 >) = ε}, and ϕtn(α)(m) is 1 when m is equal to < Σt_|t|+n+2, 0 >,

α(m) otherwise.

The set_S3:= {α ∈ 2ω | ∃m ∈ ω ∀n ∈ ω ∃p ≥ n (p)0= m and α(p) = 1} is a standard Σ03

-complete set (see 23.A in [K]). We will more or less recover this example, but the1’s have to be well placed. This leads to the following technical but crucial notion.

Definition 5.3 We say thatu ∈ 2<ω _{isplaced if u 6= ∅ and there is t ∈ ω}<ω _{such thatN}

u∩ Kt6= ∅,

(|u|−1)0= Σt|t|, andu(|u|−1) = 1 if (|u|−1)1> 0. We also say that t is a witness for the fact that u

is placed.

This means that the last coordinate ofu has a number on the vertical Σt

|t|, on which the

coordi-nates of the elements ofKt are left free byt, and which is the first vertical with this property. The

coordinates ofu whose number is on one of the verticals before the previous one are determined by t. Finally, the last coordinate of u is 1, except maybe if this coordinate has the number Σt, which is

Examples. Letα ∈ Knj= K(n,j). Then

• α|1, α|(< 0, n+1 > +1) are placed with witness ∅.

• α|(< n+2, 0 > +1), α|(< n+2, j +1 > +1) are placed with witness (n).

• α|(< n + 2 + j + 2, 0 > +1) is placed with witness (n, j). If α(< n + 2 + j + 2, q >) = 1, then α|(< n+2+j +2, q > +1) is placed with witness (n, j).

We are now ready to define_X3andA3.

Notation. We setX3:=α∈ 2ω | ∀n ∈ ω ∃p ≥ n α|p is placed . Let t ∈ ω<ω. We set

Ht:= {α ∈ Kt0| ∀n ∈ ω ∃p ≥ n (p)0= Σt|t|andα(p) = 1},

and_A3:=St∈ω<ω Gr(ϕ_t|Ht). In this sense, we recoverS3. More concretely,

A3=

[

t∈ω<ω

n

(u0γ, u1γ) | |u| = Σtandu0γ ∈ Ktand∀n ∈ ω ∃p ≥ n (u0γ)(< Σt_|t|, p >) = 1

o .

Lemma 5.4 Lett ∈ ω<ω_{andε ∈ 2.}

(a)(Ktn)n∈ω,wn(0)=εis a sequence of pairwise disjoint meager subsets ofKtε.

(b) Any nonempty open subset ofKε

t contains one of theKtn’s.

Proof. (a) This comes from the fact that the last1 in (α)_Σt

|t| is at the positionn+1 if α ∈ Ktn.

(b) A nonempty open subset ofKtεcontains a basic clopen setC of the form

{α ∈ Ktε| εu ⊆< (α)Σt |t|, (α)Σ

t

|t|+1, ... >},

where u ∈ 2<ω_{. We choose n ∈ ω such that εu ⊆ w}

n. It remains to see that Ktn ⊆ C. So let

m =< i, p >≤ |u|. Note first that M (q) ≤ min q, M (q+1)
. Thus, as |u|≤ |wn| = n+1,

i = (m)0≤ (m)0+(m)1= M (m) ≤ M (|u|) ≤ M (n+1) ≤ n+1 < n+2.

Lemma 5.2.(a) allows us to write

(εu)(m) = (εu)(< i, p >) = (εu)i(p) = (wn)i(p) = (α)Σt |t|+i(p) = (< (α)_Σt |t|, (α)Σ t |t|+1, ... >)i(p) = (< (α)Σ t |t|, (α)Σ t |t|+1, ... >)(m).

This finishes the proof. 

We now start to prove the required properties of_X3andA3.

Lemma 5.5 (a) The setX3 is a denseΠ02 subset of2ω. In particular, X3 is a0-dimensional Polish

space.

(b) Lett ∈ ω<ω_{. The setH}

tis a denseΠ0₂subset ofKt0.

(c) The set_A3is aΣ03subset ofX23. In particular,A3is an analytic relation onX3.

(d) Letβ ∈ ωω_{. Then}T

n∈ω Kβ|(n+1)⊆X3.

(e)(_X3,A3) 6_∆0

Proof. (a)X3 is clearly aΠ02 subset of2ω. So let us prove its density. We just have to prove that

α ∈ 2ω _{| ∃p ≥ n α|p is placed is dense in 2}ω _{for each integer n. So let ∅ 6= w ∈ 2}<ω_{. Note that}

α := w1∞∈ N_w(0)= K_∅w(0). Letp ≥ max(n, |w|) with (p)0= 0. Then α|(p+1) is placed with witness

t := ∅.

(b)Htis clearly aΠ02 subset of Kt0. So let us prove its density. We just have to prove the density

inK_t0 of the set α ∈ K_t0 | ∃p ≥ n (p)0 = Σt_|t|andα(p) = 1 , for each integer n. If |t| = 0,

then Kt0 = K∅0. As in the proof of (a), we see (with w(0) := 0) that α := w1∞ ∈ Nw ∩ Kt0 and

p ≥ max(n, |w|) with (p)0= 0 are suitable. If |t| ≥ 1, then we argue similarly. We put again w1∞, in

the coordinates not determined byt.

(c) By (b),A3is aΣ03 subset of2ω×2ω. So we just have to see thatA3⊆X23, which is clear.

(d) Letα ∈T

n∈ω Kβ|(n+1). Note that the sequence Σβ|(n+1)n+1

n∈ωis strictly increasing. In particular,

p :=< Σβ|(n+1)_n+1 , β(n+1)+1 > +1 ≥< Σβ|(n+1)_n+1 , 0 > +1 ≥ Σβ|(n+1)_n+1 +1≥ n and t := β|(n + 1) are witnesses for the fact factα ∈X3.

(e) We argue by contradiction, which gives a partition(Cn)n∈ω ofX3intoA3-discreteΠ02 sets. Fix

n ∈ ω and t ∈ ω<ω. Let us prove that there isi ∈ ω such that Cn∩ Kti= ∅. We argue by contradiction.

By Lemma 5.4,Cn∩ Ktεis dense inKtεfor eachε ∈ 2. As Cn∩ KtεisΠ02, it is comeager inKtε. By

(b),Htis also comeager inKt0, so that this is also the case ofCn∩ Ht. In particular,ϕt[Cn∩ Ht]

is comeager inKt1, andCn∩ ϕt[Cn∩ Ht] too. In particular, Cn∩ ϕt[Cn∩ Ht] is not empty, which

contradicts theA3-discreteness ofCn.

Applying this inductively, we constructβ ∈ ωω _{such thatC}

n∩ Kβ|(n+1)= ∅ for each n ∈ ω. By

compactness, there isα ∈T

n∈ω Kβ|(n+1), andα /∈Sn∈ω Cn=X3. But this contradicts (d). 

The following uniqueness properties will be important in the sequel.

Lemma 5.6 Lett ∈ ω<ω_{andα ∈ K} t.

(a) Assume thatu ∈ 2<ω _{is placed with witnesst. Then Σ}

t< |u|, the last 1 in u strictly before the

position< Σt

k+1, 0 > is at the position < Σtk, t(k)+1 > for each k < |t|, and t is unique.

(b) Letp > Σtbe such thatα|p is placed with witness t′. Thent ⊆ t′.

Proof. (a) As(|u|−1)0= Σt|t|, we may assume that|t| ≥ 1. Let α ∈ Nu∩ Kt. Then

(α)_Σt

k= (wt(k))00

t(k)+1−|(wt(k))0|₁₀∞

for eachk < |t|.

In particular,α(< Σt

k, t(k)+1 >) = 1. As < n+p+1, 0 >>< n, p > and (|u|−1)0= Σt|t|, we get

|u| > |u|−1 ≥< Σt

|t|, 0 >= Σt≥< Σtk+1, 0 >>< Σtk, t(k)+1 >. Thus u(< Σtk, t(k)+1 >) = 1.

• Let us prove that the last 1 in u strictly before the position < Σt

k+1, 0 > is at the position

< Σt

k, t(k)+1 >. The consecutive integers between the values < Σtk, t(k)+1 > and < Σtk+1, 0 >

are< Σt

k, t(k)+1 >, < Σtk−1, t(k)+2 >, ..., < 0, Σtk+t(k)+1 > and < Σtk+1, 0 >. So we have to

see thatu(< Σt

There arek′< k and i < t(k′)+2 such that Σt_k−j = Σt

k′+i. In particular,

t(k)+1+j = Σt_k+1−1−Σtk′−i ≥ t(k′)+2−i.

Lemma 5.2.(c) applied tow_t(k′₎implies that|(w_t(k′₎)_i| ≤ t(k′)+2−i. Thus

α(< Σtk′+i, t(k)+1+j >) = 0,

and we are done.

• As the last 1 in u strictly before the position < Σt

|t|, 0 > is at the position < Σt|t|−1, t(|t|−1)+1 >,

t(|t|−1) is determined. It remains to iterate this argument to see the uniqueness of t.

(b) We argue by induction onl := |t|, and we may assume that our property is proved for l. So let t ∈ ωl+1,α ∈ Kt, andp > Σtbe such thatα|p is placed with witness t′. Note thatα ∈ Kt|landp > Σt|l.

By the induction assumption, we gett|l ⊆ t′.

Let us prove that t|l 6= t′. We argue by contradiction, so that (p − 1)0 = Σt ′

l = Σtl. Note that

< Σt

l, t(l)+1 >< Σt< p. Thus (p−1)1> t(l)+1 > 0 and α(p−1) = 1 = α(< Σtl, (p−1)1 >). As

α ∈ Kt, the last1 of (α)Σt

l is at the positiont(l)+1, which is absurd.

This shows the existence oft′(l). Let β ∈ Nα|p∩ Kt′. The last1 of (β)_Σt

lis at the positiont ′_(l)+1.

As< Σt

l, t′(l)+1 >< p−1 =< Σt ′

|t′_|, (p−1)1 >, it is also the last 1 of (β|p)Σt

l= (α|p)Σtl. But the last

1 of (α|p)_Σt

l is at the positiont(l)+1, so that t

′_{(l) = t(l) and t ⊆ t}′_{. }

Definition 5.7 Letu ∈ 2<ω_{andl ∈ ω.}

(a) Ifu is placed, then let t ∈ ω<ω_{be the unique witness given by Lemma 5.6.(b). We will consider}

• the length l(u) := |t|

• the sequence ul(u)_{∈ 2}|u|_{\{u} defined by u}l(u)_{(m) := 1−u(m) if m = Σ}

t,u(m) otherwise. Note that

ul(u)is placed with witnesst, so that l(ul(u)) = l(u) and (ul(u))l(u)= u • the digit ε(u) := u(Σt). Note that ε(ul(u)) = 1−ε(u).

(b) We say that u is l-placed if u is placed and l(u) = l. We say that u is (≤ l)-placed (resp., (< l)-placed, (> l)-placed) if there is l′≤ l (resp., l′_{< l, l}′_{> l) such that u is l}′_-placed.

The following lemma will be crucial in the construction of the homomorphism. We construct some finite approximations of the homomorphism. The lemma says that these finite approximations can be constructed independently.

Lemma 5.8 Letu 6= v ∈ 2<ω_{be placed withε(u) = ε(v). Then {u, u}l(u)_{} ∩ {v, v}l(v)_{} = ∅.}

Proof. Note first thatε(ul(u)) = 1−ε(u) = 1−ε(v) = ε(vl(v)). Thus u /∈ {v, vl(v)} and ul(u)6= v. If ul(u)= vl(v), thenu = (ul(u))l(ul(u))= (vl(v))l(vl(v))= v, which is absurd. 

When we consider the finite approximations of an element of A3, we have to guess the finite

sequence t. We usually make some mistakes. In this case, we have to be able to come back to an earlier position. This is the role of the following predecessors.

Notation. Letu ∈ 2<ω. Note that< ε > is 0-placed. This allows us to define u−:= ∅ if |u|≤ 1,

u|max{n < |u| | u|n is placed} if |u| ≥ 2. and, forl ∈ ω,

u−l:= ∅ if |u|≤ 1,

u|max{n < |u| | u|n is (≤ l)-placed} if |u| ≥ 2,

Before proving our main theorem, we study the relation between these predecessors and the placed sequences.

Lemma 5.9 Letl ∈ ω and u ∈ 2<ω _{bel-placed with |u| ≥ 2.}

(a) Assume thatu−isl-placed. Then ε(u−) = ε(u). If moreover (ul)−isl-placed, then (ul)−= (u−)l. (b)u−lisl-placed if and only if (ul₎−l_{isl-placed. In this case, ε(u}−l_{) = ε(u) and (u}l₎−l_{= (u}−l₎l_.

(c) Assume thatu−or(ul)−is(< l)-placed. Then u−= u−l= (ul)−= (ul)−lis(l−1)-placed.

(d) Assume that u− or (ul₎− _{is (> placed. Then exactly one of those two sequences is (>}

l)-placed, and the other one isl-placed. If u−(resp.,(ul₎−_{) is(> l)-placed, then u}−l_{= (u}l₎−
l

(resp.,

u−l= u−) andε(u−l) = ε(u) (resp., ε (ul)−l
= ε(ul_)).

(e)l(u−l) ∈ {l−1, l}.

Proof. Lett ∈ ωl_{(resp. t}′_{∈ ω}<ω_{) be a witness for the fact thatu (resp., u}−_{) is placed, andα ∈ N} u∩Kt.

Claim. Assume that(|u|−1)1= 0. Then u−= u−l= (ul)−= (ul)−lis(l−1)-placed.

Proof. Note thatl ≥ 1 since |u| ≥ 2. The consecutive integers between the values < Σt

l−1, t(l−1)+1 >

andΣtare< Σt_l−1, t(l−1)+1 >, < Σt_l−1−1, t(l −1)+2 >, ..., < 0, Σt_l−1+t(l −1)+1 > and Σt.

By Lemma 5.6.(b), Σt < |u| and the last 1 in u strictly before the position Σt is at the position

< Σt

l−1, t(l−1)+1 >. This shows that u|(Σt+1) and u|(< Σt_l−1, t(l−1)+1 > +1) are placed and

u|(Σt+1)
−= u|(< Σt_l−1, t(l−1)+1 > +1) since t(l−1)+j > 0 if 1 ≤ j ≤ Σt_l−1+1. As (|u|−1)1= 0,

|u| = Σt+1 and the sequence u−= u|(< Σt_l−1, t(l−1)+1 > +1) is (l−1)-placed, which implies that

u−= u−l= (ul)−= (ul)−l. ⋄

(a) By the claim,(|u|−1)1> 0. Thus u|(Σt+1)$ u is l-placed, u|(Σt+1) ⊆ u−andΣt< |u−|. As

u−⊆ α, we can apply Lemma 5.6.(c) and t ⊆ t′. Thust = t′ since |t| = |t′| = l, and the equalities ε(u−_{) = (u}−_)(Σ

t′) = u(Σ_t) = ε(u) hold.

Assume now that(ul₎−_{isl-placed. As u}l_{isl-placed with witness t, there is some β ∈ N}

ul∩ K_t.

Asu|(Σt+1) ⊆ u−$ u, we get u|(Σt+1)

l

⊆ (ul₎−_{⊂ β. Thus |(u}l₎−_{| > Σ}

t. Lemma 5.6.(c) implies

thatt is the witness for the fact that (ul₎−_{isl-placed. If u}−_{= u|(< Σ}t

l, j0 > +1), then there is no