Decomposing functions of Baire class 2 on Polish spaces
aa r X i v : . [ m a t h . L O ] A ug DECOMPOSING FUNCTIONS OF BAIRE CLASS ONPOLISH SPACES
LONGYUN DING, TAKAYUKI KIHARA, BRIAN SEMMES, AND JIAFEI ZHAO
Abstract.
We prove the Decomposability Conjecture for functions ofBaire class 2 on a Polish space to a separable metrizable space. Thispartially answers an important open problem in descriptive set theory. Introduction
In descriptive set theory, the study of decomposability of Borel functionsoriginated by a famous question asked by Luzin around a century ago: Is ev-ery Borel function decomposable into countably many continuous functions?This question was answered negatively. Many counterexamples appeared inthe literature (cf. [8, 10]) show that, even a function of Baire class 1 is notnecessarily decomposable. Among these counterexamples, the Pawlikowskifunction P : ( ω + 1) ω → ω ω stands in an important position. Indeed, Solecki[15] proved that:Let X, Y be separable metrizable spaces with X analytic,and let f : X → Y be of Baire class 1. Then f is notdecomposable into countably many continuous functions iff P ⊑ f , i.e., there exists embeddings φ : ( ω + 1) ω → X and ψ : ω ω → Y such that ψ ◦ P = f ◦ φ .Later, Pawlikowski and Sabok [13] generalized this theorem onto all Borelfunctions from an analytic space to a separable metrizable space. MottoRos [11, Lemma 5.6] also gave an elegant proof for all functions of Baireclass n with n < ω .A natural generalization of Luzin’s question is to replace continuous func-tions with Σ γ -measurable functions. We write f ∈ dec( Σ γ ) if there existsa partition ( X k ) of X with each f ↾ X k is Σ γ -measurable; and also write f ∈ dec( Σ γ , ∆ δ ) if such a partition can be a sequence of ∆ δ subsets of Mathematics Subject Classification.
Primary 03E15, 54H05, 26A21.The Research of the first author was partially supported by the National NaturalScience Foundation of China (Grant No. 11725103). The second author was partiallysupported by JSPS KAKENHI Grant 17H06738 and 15H03634, and also thanks JSPSCore-to-Core Program (A. Advanced Research Networks) for supporting the research. X . It is trivial to see that, for δ ≥ γ , f ∈ dec( Σ γ , ∆ δ ) implies the Σ δ -measurability of f . It is also well known that, for any Σ δ -measurable func-tion f with δ > γ , we have f ∈ dec( Σ γ ) ⇐⇒ f ∈ dec( Σ γ , ∆ δ +1 ) (cf. [11,Proposition 4.5]).A slightly more finer notion of Baire hierarchy was essentially introducedby Jayne [3] for studying the Banach space of functions of Baire class α . Afunction f : X → Y is called a Σ α,β function (or more precisely denoted by f − Σ β ⊆ Σ α ) if the preimage f − ( A ) is Σ α in X for every Σ β subset A of Y . The following theorem discovers a deep connection between this notionand decomposability: Theorem 1.1 (Jayne-Rogers [4]) . Let
X, Y be separable metrizable spaceswith X analytic, and let f : X → Y . Then f − Σ ⊆ Σ ⇐⇒ f ∈ dec( Σ , ∆ ) . This theorem was generalized in [6] to the case that X is an absoluteSouslin- F set and Y is an arbitrary regular topological space.It is conjectured that the Jayne-Rogers Theorem can be extended to allfinite Borel ranks as follows: The Decomposability Conjecture (cf. [1, 11, 13]). Let
X, Y be separablemetrizable spaces with X analytic, and let f : X → Y . Then for n ≥ f − Σ n ⊆ Σ n ⇐⇒ f ∈ dec( Σ , ∆ n ) . Furthermore, for 2 ≤ m ≤ n we have f − Σ m ⊆ Σ n ⇐⇒ f ∈ dec( Σ n − m +1 , ∆ n ) . This conjecture was further generalized to The Full Decomposability Con-jecture (see [2, Section 4]) which covers all infinite Borel ranks. Motto Rospresented an equivalent condition of the decomposability conjecture (see [11,Conjecture 6.1]). Another interesting equivalent condition with some extrarestrictions on spaces and on relation between m, n , concerning computabil-ity on Borel codes from A to f − ( A ), was given by Kihara in [9]. Mostrecently, Gregoriades-Kihara-Ng [2] proved f − Σ m ⊆ Σ n = ⇒ f ∈ dec( Σ n − m +1 , ∆ n +1 ) . It is clear that the case m = n = 2 in the decomposability conjecture isjust the Jayne-Rogers Theorem. Remarkable progress is due to Semmes, thethird author of this article. In his Ph.D. thesis [14], Semmes proved the case m ≤ n = 3 for functions f : ω ω → ω ω . In his proof, many kinds of gamesfor characterizing Borel functions were widely applied. From the viewpointof Jayne’s work [3] in functional analysis, the zero-dimensionality constrainton Semmes’ theorem was strongly desired to be removed. In this article, wegeneralize Semmes’ theorem to arbitrary Polish spaces: Theorem 1.2.
The decomposability conjecture is true for the case that X is Polish space and m ≤ n = 3 . ECOMPOSING FUNCTIONS OF BAIRE CLASS 2 ON POLISH SPACES 3
It is worth noting that in our proof, no game for Borel functions areinvolved. This is the key point that this proof can be extended to all Ploishspaces. This theorem consists of two cases: (a) m = 2 , n = 3, and (b) m = n = 3. We will prove them in sections 3 and 4 respectively.Following the outline of Semmes’ proof, the proof appearing in this arti-cle was developed by the first and the forth authors. Almost at the sametime, the second author independently gave a detailed exposition of Semmes’strategy. He also asserted that the use of games for Borel functions is mis-leading, and emphasized the use of finite injury priority argument instead.Soon after reading it, Motto Ros pointed out that the same argument in thesecond author’s proof also works well, with some minor modifications, forarbitrary Polish spaces.The authors would like to thank Rapha¨el Carroy and Luca Motto Rosfor helpful suggestions. The first author is grateful to Slawomir Solecki forhis suggestions and encouragement. The second author also would like tothank Vassilios Gregoriades, Tadatoshi Miyamoto, and Yasuo Yoshinobu forvaluable discussions. 2. Preliminaries
All topological spaces considered in this article are separable metrizable.
For any subset A of a topological space X , we denote by A the closure of A in X and denote A c = X \ A for brevity.We recall some basic notations. A topological space is called a Polishspace if it is separable and completely metrizable, and is called an analyticspace if it is homeomorphic to an analytic subset of a Polish space. Givena separable metrizable space X , Borel sets of X can be analyzed into Borelhierarchy, consisting of Σ ξ , Π ξ subsets for 1 ≤ ξ < ω . As usual, we denote ∆ ξ = Σ ξ ∩ Π ξ .Let X, Y be two separable metrizable spaces, and f : X → Y . We say f is Σ α -measurable if f − ( U ) ∈ Σ α for every open set U ⊆ Y . For thedefinition of the Baire classes of functions, one can see [7, (24.1)]. It is wellknown that a function is of Baire class ξ iff it is Σ ξ +1 -measurable (cf. [7,(24.3)]).In the section of introduction, we already presented notion of Σ α,β func-tions, f − Σ β ⊆ Σ α and dec( Σ γ , ∆ δ ). The following proposition give somewell known properties which will be used again and again in the rest of thisarticle. Proposition 2.1 (folklore) . Let
X, Y be two separable metrizable spaces,and let f : X → Y . Then the following are equivalent: (i) f ∈ dec( Σ γ , ∆ δ ) . (ii) There exists a sequence ( A n ) of Σ δ subsets with X = S n A n suchthat every f ↾ A n is Σ γ -measurable. LONGYUN DING, TAKAYUKI KIHARA, BRIAN SEMMES, AND JIAFEI ZHAO (iii)
There exists a sequence ( A n ) of Σ δ subsets with X = S n A n suchthat every f ↾ A n ∈ dec( Σ γ , ∆ δ ) .Proof. (i) ⇒ (ii) and (ii) ⇒ (iii) are trivial. We only prove (iii) ⇒ (i).For every n < ω , since A n ∈ Σ δ , we can choose a sequence ( B mn ) m<ω of ∆ δ sets such that A n = S m B mn . Moreover, since f ↾ A n ∈ dec( Σ γ , ∆ δ ) , there exist two sequences ( C kn ) k<ω , ( D kn ) k<ω of Σ δ sets with A n ⊆ [ k C kn , A n ∩ C kn = A n \ D kn such that each f ↾ ( C kn ∩ A n ) is Σ γ -measurable. Note that B mn ∩ C kn = B mn \ D kn ∈ ∆ δ , and f ↾ ( B mn ∩ C kn ) is Σ γ -measurable for all n, k, m < ω .Let ( K l ) l<ω be an enumeration of all B mn ∩ C kn , n, k, m < ω . Then [ l K l = [ n,k,m ( B mn ∩ C kn ) = X. For each l < ω , put K ′ l = K l \ ( S i Let X, P be two separable metrizable spaces, and let D ⊆ X , h : D → P a function of Baire class . Let B P be a countable topologicalbasis of P , and for each V ∈ B P , let G V be a countable class of subsets of D such that h − ( V ) = [ G V . If h / ∈ dec( Σ , ∆ ) , then there exist V ∈ B P , G ∈ G V , and a closed set F ⊆ D satisfying: (a) For any open set U with F ∩ U = ∅ , h ↾ ( h − ( V c ) ∩ F ∩ U ) / ∈ dec( Σ , ∆ );(b) G ∩ F is dense in F ; (c) F ∩ D = ∅ .Proof. Let { U k : k < ω } be a topological basis of X . For any V ∈ B P andany closed subset F ⊆ X , we denoteΓ V ( F ) = { k < ω : h ↾ ( h − ( V c ) ∩ F ∩ U k ) ∈ dec( Σ , ∆ ) } , Θ V ( F ) = { x ∈ F : ∀ k < ω ( x ∈ U k ⇒ k / ∈ Γ V ( F )) } . It is trivial to see that Θ V ( F ) ⊆ D is closed. ECOMPOSING FUNCTIONS OF BAIRE CLASS 2 ON POLISH SPACES 5 For any G ∈ G V , we define closed set F αV,G for α < ω as follows: F V,G = X,F α +1 V,G = G ∩ Θ V ( F αV,G ) ,F λV,G = \ α<λ F αV,G , for limit ordinal λ. Since X is second countable, there exists a ξ < ω such that F αV,G = F ξV,G for each V, G and α ≥ ξ .If there exist V ∈ B P , G ∈ G V such that F ξV,G = ∅ , then V, G , and F = F ξV,G fulfil clauses (a) and (b). Set U = X in (a), we can see (c) is alsofulfilled.Assume for contradiction that, for any V ∈ B P , G ∈ G V , we have F ξV,G = ∅ . For α < ξ and k ∈ Γ V ( F αV,G ), put H αV,G,k = F αV,G ∩ U k . Note that h ↾ ( h − ( V c ) ∩ H αV,G,k ) ∈ dec( Σ , ∆ ) . Now define a subset H of all x satisfying that, there exist V , V ∈ B P with V ∩ V = ∅ , and for i = 1 , 2, there exist G i ∈ G V i , α i < ξ , and k i ∈ Γ V i ( F α i V i ,G i ) such that x ∈ H α V ,G ,k ∩ H α V ,G ,k . Since h ↾ ( h − ( V ic ) ∩ H α V ,G ,k ∩ H α V ,G ,k ) ∈ dec( Σ , ∆ ) ( i = 1 , , and h − ( V ic ) is Σ in D for i = 1 , 2, by Proposition 2.1, we have h ↾ ( D ∩ H α V ,G ,k ∩ H α V ,G ,k ) ∈ dec( Σ , ∆ ) . Therefore, h ↾ ( D ∩ H ) ∈ dec( Σ , ∆ ).For any x ∈ D , V ∈ B P , and G ∈ G V with x ∈ G , we claim that there exist α < ξ and k ∈ Γ V ( F αV,G ) such that x ∈ H αV,G,k . There is unique α < ξ suchthat x ∈ ( F αV,G \ F α +1 V,G ). Note that x / ∈ ( F \ G ∩ F ) for any closed set F ⊆ X ,so x / ∈ (Θ V ( F αV,G ) \ F α +1 V,G ). From the definition of Θ V ( F αV,G ), we can find a k ∈ Γ V ( F αV,G ) such that x ∈ U k . Then we have x ∈ F αV,G ∩ U k = H αV,G,k .In the end, we consider h ↾ ( D \ H ). First, for any x ∈ ( D \ H ), if x ∈ H αV,G,k for some V ∈ B P , G ∈ G V , α < ξ , and k ∈ Γ V ( F αV,G ), weclaim that h ( x ) ∈ V . If not, we can find a V ′ ∈ B P such that h ( x ) ∈ V ′ and V ∩ V ′ = ∅ . Since h − ( V ′ ) = S G V ′ , we can find an G ′ ∈ G V ′ suchthat x ∈ G ′ . Hence x ∈ H α ′ V ′ ,G ′ ,k ′ for some α ′ < ξ and k ′ ∈ Γ V ′ ( F α ′ V ′ ,G ′ ),contradicting x / ∈ H . Secondly, let d be a compatible metric on P . For any n < ω , let ( V nm , G nm , k nm , α nm ) m<ω LONGYUN DING, TAKAYUKI KIHARA, BRIAN SEMMES, AND JIAFEI ZHAO be an enumeration of all ( V, G, k, α ) with diam( V ) ≤ /n , G ∈ G V , α < ξ ,and k ∈ Γ V ( F αV,G ). Denote H nm = H α nm V nm ,G nm ,k nm . For any x ∈ D , we can find a V ∈ B P with diam( V ) ≤ /n such that h ( x ) ∈ V and a G ∈ G V with x ∈ G . Hence x ∈ H αV,G,k for some α < ξ and k ∈ Γ V ( F αV,G ). It follows that D ⊆ S m H nm . Put K nm = H nm \ S k Let F = h F , · · · , F k i be a finite sequence of closed sets of X with F ⊇ · · · ⊇ F k , U an open subset of X , and let P ⊆ Y .(i) If k = 0, i.e., F = h F i , then we say F is P - sharp in U if U ∩ F = ∅ ,and for any open set U ′ ⊆ U with U ′ ∩ F = ∅ , we have f ↾ ( f − ( P ) ∩ F ∩ U ′ ) / ∈ dec( Σ , ∆ ) . We also say F itself is P -sharp in U for brevity.(ii) If k > 0, then we say F is P - sharp in U if F k is P -sharp in U , andfor any open set U ′ ⊆ U with U ′ ∩ F k = ∅ , F ↾ k is P -sharp in someopen set U ′′ ⊆ U ′ .A similar notion named δ - σ -good was presented in [14]. The followingpropositions are trivial, we omit the proofs. Proposition 3.3. Suppose F = h F , · · · , F k i is P -sharp in U . Then forany open set U ′ ⊆ U with U ′ ∩ F k = ∅ , we have F is P -sharp in U ′ . Proposition 3.4. Suppose F is P -sharp in U . Then for any m < lh( F ) , F ↾ m is P -sharp in some open set U ′ ⊆ U . The following lemma is modified from [14, Lemma 4.3.6]. Lemma 3.5. Suppose F = h F , · · · , F k i is P -sharp in U . Let ( C l ) l Assume for contradiction that there exists l = 0 such that F is not P \ C l -sharp in U , then there is an open set U l ⊆ U with U l ∩ F = ∅ suchthat f ↾ ( f − ( P \ C l ) ∩ F ∩ U l ) ∈ dec( Σ , ∆ ) . Since C and C l are disjoint closed subsets of P , Proposition 2.1 gives f ↾ ( f − ( P ) ∩ F ∩ U l ) ∈ dec( Σ , ∆ ) , contradicting that F is P -sharp in U .For k > 0, assume that we have proved for all k ′ < k . Since F k is P -sharpin U , from the arguments for k = 0 above, we may assume that there is anopen set U ⊆ U with U ∩ F k = ∅ such that F k is P \ C l -sharp in U forany l = 0. Assume for contradiction that there are k + 1 many l = 0, say, l = 1 , · · · , k, k + 1, such that F is not P \ C l -sharp in any open set U ′ ⊆ U .Particularly, F is not P \ C -sharp in U , so there exists an open set U ⊆ U with U ∩ F k = ∅ such that F ↾ k is not P \ C -sharp in any open set U ′ ⊆ U .Similarly, we can find a sequence of open sets U k +1 ⊆ U k ⊆ · · · ⊆ U ⊆ U such that U l ∩ F k = ∅ and F ↾ k is not P \ C l -sharp in any U ′ ⊆ U l for0 < l ≤ k + 1. By Definition of P -sharp, there is a open set U ∗ ⊆ U k +1 suchthat F ↾ k is P -sharp in U ∗ , contradicting the induction hypothesis. (cid:3) Let p · , · q be the bijection: ω × ω → ω as following: p , q = 0 , p , j + 1 q = p j, q + 1 , p i + 1 , j − q = p i, j q + 1 . Denote Ω = { z ∈ ω : ∃ i ∃ ∞ j ( z ( p i, j q ) = 1) } . It is well known that Ω is Σ -complete subset of 2 ω .For any z ∈ ω and l < ω , we call sequence z ↾ ( p , l q + 1) , z ↾ ( p , l − q + 1) , · · · , z ↾ ( p l, q + 1)the l -th diagonal of z , and call z ↾ ( p l, q + 1) the end of l -th diagonal. For s ∈ <ω , we denote lh( s ) = i the length of s . If s ⊆ z and lh( s ) = p i, j q + 1,then s is in ( i + j )-th diagonal. Moreover, the l -th diagonal of z is alsonamed the l -th diagonal of s when p l, q < lh( s ).For s = ∅ , let lh( s ) = p i, j q + 1. We denote row( s ) = i, col( s ) = j . If i + j > 0, we call s ↾ ( p i + j − , q + 1) the end of the last diagonal of s , denoted by u ( s ). If j > 0, we call s ↾ ( p i, j − q + 1) the left neighbor of s , denoted by v ( s ).For proving the following theorem, we need an order (cid:22) on 2 <ω define by t (cid:22) s ⇐⇒ lh( t ) < lh( s ) or (lh( t ) = lh( s ) , t ≤ lex s ) , where ≤ lex is the usual lexicographical order. We also denote t ≺ s when t (cid:22) s but not t = s . Denote s M = max (cid:22) { t : t ≺ s a } . LONGYUN DING, TAKAYUKI KIHARA, BRIAN SEMMES, AND JIAFEI ZHAO Theorem 3.6. Let X be a Polish space, Y a separable metrizable space,and let f : X → Y . Then f − Σ ⊆ Σ ⇐⇒ f ∈ dec( Σ , ∆ ) . Proof. The “ ⇐ ” part is trivial, we only prove the “ ⇒ ” part.Assume for contradiction that f / ∈ dec( Σ , ∆ ). We will define a contin-uous embedding ψ : 2 ω → X and an open set O ⊆ Y such that ψ − ( f − ( O )) = Ω . Thus f − ( Y \ O ) is Π -complete subset of X , contradicting f − Σ ⊆ Σ .For any open set V ⊆ Y , since f − ( V ) is Σ , we can fix a system of openset W mn ( V ) ⊆ X ( m, n < ω ) with f − ( V ) = [ m \ n W mn ( V ) . Denote G m ( V ) = T n W mn ( V ). For G = G m ( V ), we also denote W n ( G ) = W mn ( V ).For s ∈ <ω with s = ∅ , let lh( s ) = p i, j q + 1. We say s is an inheritor if j > s ( p k, i + j − k q ) = 0 for any k < i , otherwise we say s is an innovator . Note that s is always an inheritor if i = 0, j > 0, and is alwaysan innovator if j = 0.Fix a compatible metric d on X with d ≤ 1. We will inductively construct,for each s ∈ <ω , an open set V s of Y , a G δ set G s of X , a closed set F s of X , an open set U s of X , and a sequence of open sets ( U ws ) s (cid:22) w ≺ s a of X satisfying the following:(0) diam( U s ) ≤ − lh( s ) , U s a ∩ U s a = ∅ , U s a ⊆ U ws , U s a ⊆ U ws ;(1) V s a = V s a , G s a = G s a , F s a = F s a ;(2) for s, t ∈ <ω , we have V s = V t or V s ∩ V t = ∅ ;(3) there exist m < ω such that G s = G m ( V s );(4) if col( s ) > 0, then F s a = F s a ⊆ F s ;(5) F s ∩ U ws = ∅ for each w ;(6) U s = U ss , and U w s ⊇ U w s for w (cid:22) w ;(7) G s ∩ F s is dense in F s ;(8) if s is an inheritor, then we have V s = V v ( s ) , G s = G v ( s ) , F s = F v ( s ) ;furthermore, (a) if s a U ws ∩ F u ( s ) = ∅ for each w ; (b) if s a U s ⊆ W n ( G s ) for all n < lh( s );(9) if s is an innovator, then U s ∩ F s ∩ G m ( V t ) = ∅ for all m < lh( s ) andall lh( t ) < lh( s );(10) by letting P s = Y \ S t (cid:22) s V t , F s = h F s ↾ (lh( s ) − row( s )) , · · · , F s ↾ (lh( s ) − , F s i , ECOMPOSING FUNCTIONS OF BAIRE CLASS 2 ON POLISH SPACES 9 for t (cid:22) s ≺ t a 0, we have(a) if t a F t is P s -sharp in U st ;(b) if t a F u ( t a is P s -sharp in U st .When we complete the construction, for any z ∈ ω , we set ψ ( z ) to be theunique element of T k U z ↾ k . From (0) and (6), we see that ψ is a continuousembedding from 2 ω to X . Put O = [ t ∈ <ω V t . If z ∈ Ω, let i be the least i such that there are infinitely many j with z ( p i, j q ) = 1. Then there is J < ω such that z ( p i, j q ) = 0 for all i < i andall j > J . Hence for any j > J , z ↾ ( p i , j q + 1) is an inheritor. Denote V = V z ↾ ( p i ,J q +1) , G = G z ↾ ( p i ,J q +1) . By (8), we have V = V z ↾ ( p i ,j q +1) and G = G z ↾ ( p i ,j q +1) for all j ≥ J . If j > J with z ( p i , j q ) = 1, by (8)(b), we have ψ ( z ) ∈ W n ( G ) for all n ≤ p i , j q . Since there is infinitely many such j , we have ψ ( z ) ∈ G ⊆ f − ( V ).Therefore, f ( ψ ( z )) ∈ V ⊆ O .If z / ∈ Ω, we show that f ( ψ ( z )) / ∈ O . If not, there exits t and m suchthat ψ ( z ) ∈ G m ( V t ). Fix an i > max { m , lh( t ) } . Since z / ∈ Ω, for largeenough j , we have z ( p i, j q ) = 0 for all i < i , so z ↾ ( p i , j q + 1) is aninheritor. Let J be the largest j such that z ↾ ( p i , j q + 1) is an innovator.Denote F = F z ↾ ( p i ,J q +1) . By (8), we have F = F z ↾ ( p i ,j q +1) for all j ≥ J . From (5) and (6), we seethat F ∩ U z ↾ ( p i ,j q +1)) = ∅ for any j > J , and hence ψ ( z ) ∈ F . It followsfrom (9) that F ∩ U z ↾ ( p i ,J q +1) ∩ G m ( V t ) = ∅ . Thus ψ ( z ) / ∈ G m ( V t ). Acontradiction!Now we turn to the construction.First, set D, P, h, B P , G V as follows:(i) P = Y , D = X , and h = f ;(ii) B P is a countable basis of Y ;(iii) for each V ∈ B P , let G V = { G m ( V ) : m < ω } .Applying Lemma 3.1 with these D, P, h, B P , G V , we get an open set V ⊆ Y ,a G δ set G ∈ G V , and a non-empty closed set F ⊆ X . Then put V ∅ = V, G ∅ = G, F ∅ = F, U ∅ = X. Secondly, assume that we have constructed V t , G t , F t , U t , and U wt for t, w ≺ s a 0. We will define for s a s a 1. We consider the followingtwo cases: Case 1. Assume s a v = v ( s a i ) , u = u ( s a i ). For i = 0 , 1, put V s a i = V v , G s a i = G v , F s a i = F v . Note that s = u or s is also an inheritor with u ( s ) = u . By (5) and (8)(a), U s M s ∩ F u = ∅ . Subcase 1.1. If col( s a a > 0, then s a a U s a such that U s a ⊆ U s M s , U s a ∩ F u = ∅ , diam( U s a ) ≤ − (lh( s )+1) . By (10)(b), we see F u is P s M -sharp in U s M s . By Proposition 3.4, F v is P s M -sharp in some open set U ⊆ U s M s , and hence U ∩ F v = ∅ . Denote W = T n ≤ lh( s ) W n ( G v ). Since W ⊇ G v , by (7) we have W ∩ F v is open densein F v , and hence W ∩ U ∩ F v = ∅ . We can find U s a such that U s a ⊆ U ∩ W, U s a ∩ F v = ∅ , diam( U s a ) ≤ − (lh( s )+1) . By shrinking we may assume that U s a ∩ U s a = ∅ . For i = 0 , 1, we set U s a is a = U s a , U s a s a = U s a , and for other t , set U s a it = U s M t . Subcase 1.2. If col( s a a 0) = 0, then s a a U s a i for i = 0 , U s a in Subcase 1.1 with one more requirement U s a ∩ U s a = ∅ .It is trivial to check clauses (0)–(9). Note that P s a = P s a = P s M and F s a = F s a = F v . Since (10) holds for s M , it also holds for s a s a Case 2. Assume s a V l , G l , F l and U l for each l < ω as the following:Since s (cid:22) s M ≺ s a 0, by (10)(a), we have F s is P s M -sharp in U s M s . So f ↾ ( f − ( P s M ) ∩ F s ∩ U s M s ) / ∈ dec( Σ , ∆ ) . Denote F − = F s , U − = U s M s . Assume that we have defined V k , G k , F k and p k for k < l . Set D, P, h, B P , G V as follows:(i) P = P s M \ S k 0, i.e., for s ≺ t (cid:22) s M .In the end, denote A = [ lh( t ) ≤ lh( s ) [ m ≤ lh( s ) G m ( V t ) . Then A is G δ set. By (3) and V L ⊆ P s M , we have G L ∩ A = ∅ . It followsfrom Lemma 3.1 that G L ∩ F L is dense in F L , so A ∩ F L is nowhere densein F L . We can find two open sets U s a and U s a such that U s a ∩ U s a = ∅ ,and for i = 0 , 1, we have U s a i ⊆ U L , U s a i ∩ F L = ∅ , U s a i ∩ F L ∩ A = ∅ , diam( U s a i ) ≤ − (lh( s )+1) . Now put V s a i = V L , G s a i = G L , F s a i = F L , and U s a is a = U s a , U s a s a = U s a . (cid:3) Corollary 3.7. Let X be a Polish space, Y a separable metrizable space,and let f : X → Y . If f / ∈ dec( Σ , ∆ ) , then there exists a Cantor set C ⊆ X such that f ↾ C / ∈ dec( Σ , ∆ ) .Proof. Let ψ be the continuous embedding defined in Theorem 3.6. Put C = ψ (2 ω ). (cid:3) The decomposability conjecture for m = n = 3Before proving Theorem 1.2 for m = n = 3, we prove a known result first:for functions of Baire class 1, f − Σ ⊆ Σ ⇒ f ∈ dec( Σ , ∆ ) . This is an easy corollary of Solecki’s theorem (see [15, Theorem 4.1]), since f ∈ dec( Σ , ∆ ) ⇐⇒ f ∈ dec( Σ ) and P − Σ Σ . Furthermore, thisresult is also a special case of [13, Corollary 1.2], [11, Corollary 5.11], or [2,Theorem 1.1]. In order to show a completely different method of proof, wepresent a direct proof which follows the same idea as in the previous section.The readers can skip directly to Theorem 4.7. Lemma 4.1. Let X, P be two separable metrizable spaces, and let D ⊆ X ,and h : D → P a function of Baire class . Let B P be a countable topologicalbasis of P . If h / ∈ dec( Σ , ∆ ) , then there exist a V ∈ B P and two closedsets E ⊆ F ⊆ D satisfying: (a) for any open set U with E ∩ U = ∅ , we have h ↾ ( h − ( V ) ∩ U ∩ E ) / ∈ dec( Σ , ∆ ); (b) for any open set U with F ∩ U = ∅ , we have h ↾ ( h − ( V c ) ∩ F ∩ U ) / ∈ dec( Σ , ∆ );(c) E ∩ D = ∅ .Proof. Let { U k : k < ω } be a topological basis of X . For any B ∈ B P , wedenote F B = { x ∈ X : ∀ k ( x ∈ U k ⇒ h ↾ ( h − ( B c ) ∩ U k ) / ∈ dec( Σ , ∆ )) } . It is trivial to see that(i) F B is closed,(ii) h ↾ ( h − ( B c ) \ F B ) ∈ dec( Σ , ∆ ), and(iii) for any open set U with F B ∩ U = ∅ , h ↾ ( h − ( B c ) ∩ F B ∩ U ) / ∈ dec( Σ , ∆ ) . Assume for contradiction that, h ↾ ( h − ( B ) ∩ F B ) ∈ dec( Σ , ∆ ) for any B ∈ B P . We denote H = [ B ∈B P ( h − ( B ) ∩ F B ) ,H = [ B ∈B P ( h − ( B c ) \ F B ) . It is straightforward to check that, h ↾ H i ∈ dec( Σ , ∆ ) for i = 1 , H = D \ ( H ∪ H ). For any x ∈ H and any B ∈ B P , we have h ( x ) ∈ B ⇒ x ∈ h − ( B ) ⇒ x / ∈ F B ,h ( x ) / ∈ B ⇒ x ∈ h − ( B c ) ⇒ x ∈ F B . So h ↾ H is continuous.Let ˜ Y ⊇ Y be a Polish space. By Kuratowski’s theorem (cf. [7, (3.8)]),there is a G δ set G ⊇ H and a continuous function g : G → ˜ Y such that g ↾ H = h ↾ H . Put H = { x ∈ D ∩ G : h ( x ) = g ( x ) } . Since h is of Baireclass 1, we see H is G δ subset of D and H ∪ H ∪ H = D . Note that H is F σ subset of D and H is Σ subset of D . It follows that h ∈ dec( Σ , ∆ ).A contradiction!Therefore, there exists a B ∈ B P such that h ↾ ( h − ( B ) ∩ F B ) / ∈ dec( Σ , ∆ ) . Since B = S { V ∈ B P : V ⊆ B } , we can find a V ∈ B P with V ⊆ B suchthat h ↾ ( h − ( V ) ∩ F B ) / ∈ dec( Σ , ∆ ) . In the end, define E = { x ∈ F B : ∀ k ( x ∈ U k ⇒ h ↾ ( h − ( V ) ∩ U k ) / ∈ dec( Σ , ∆ )) } . Then we have E ∩ D = ∅ , and h ↾ ( h − ( V ) ∩ U ∩ E ) / ∈ dec( Σ , ∆ ) ECOMPOSING FUNCTIONS OF BAIRE CLASS 2 ON POLISH SPACES 13 for any open set with U ∩ E = ∅ . Note that V c ⊇ B c . So V, E and F B satisfy clauses (a)–(c) as desired. (cid:3) In the rest of this section, we fix X be a Polish space, Y a separablemetrizable space, and f : X → Y a Σ -measurable function. Definition 4.2. Let F = h F , · · · , F k i be a finite sequence of closed sets of X with F ⊇ · · · ⊇ F k , U an open subset of X , and let P = h P , · · · , P k i bea sequence of pairwise disjoint subsets of Y .(i) If k = 0, i.e., F = h F i , P = h P i , then we say F is P - sharp in U if U ∩ F = ∅ , and for any open set U ′ ⊆ U with U ′ ∩ F = ∅ , we have f ↾ ( f − ( P ) ∩ F ∩ U ′ ) / ∈ dec( Σ , ∆ ) . We also say F itself is P -sharp in U for brevity.(ii) If k > 0, then we say F is P - sharp in U if F k is P k -sharp in U , andfor any open set U ′ ⊆ U with U ′ ∩ F k , F ↾ k is P ↾ k -sharp in someopen set U ′′ ⊆ U ′ . Proposition 4.3. Suppose F = h F , · · · , F k i is P -sharp in U . Then forany U ′ ⊆ U with U ′ ∩ F k = ∅ , we have F is P -sharp in U ′ . Proposition 4.4. Suppose F is P -sharp in U . Then for any m < lh( F ) , F ↾ m is P ↾ m -sharp in some open set U ′ ⊆ U . Let P = h P , · · · , P k i , 0 ≤ j ≤ l , and let C ⊆ P j . We denote P \ C = h P , · · · , P j \ C, · · · , P k i . Lemma 4.5. Let F = h F , · · · , F k i , P = h P , · · · , P k i . Suppose F is P -sharp in U . Let ≤ j ≤ k and ( C l ) l 0, assume that we have proved for all k ′ < k . Case 1. If j = k , since F k is P k -sharp in U , from the arguments for k = 0above, we may assume that there is an open set U ⊆ U with U ∩ F k = ∅ such that F k is P k \ C l -sharp in U for any l = 0. It follows that F is P \ C l -sharp U for any l = 0. Case 2. If j < k , assume for contradiction that there are more than one l , say, l = 0 , 1, such that F is not P \ C l -sharp in any open set U ′ ⊆ U .Particularly, F is not P \ C -sharp in U . Note that F k is P k \ C l -sharp in U for any l < m , so there exists an U ⊆ U with U ∩ F k = ∅ such that F ↾ k is not ( P ↾ k ) \ C -sharp in any open set U ′ ⊆ U . Similarly, wecan find an open set U ⊆ U with U ∩ F k = ∅ such that F ↾ k is not( P ↾ k ) \ C -sharp in any U ′ ⊆ U . By Propositions 4.3 and 4.4, there isan open set U ∗ ⊆ U such that F ↾ k is P -sharp in U ∗ , contradicting theinduction hypothesis. (cid:3) Theorem 4.6. Let X be a Polish space, Y a separable metrizable space, andlet f : X → Y be of Baire class . If f − Σ ⊆ Σ , then f ∈ dec( Σ , ∆ ) .Proof. Assume for contradiction that f / ∈ dec( Σ , ∆ ). We will define acontinuous embedding ψ : 2 ω → X and an G δ set G ⊆ Y such that ψ − ( f − ( Y \ G )) = Ω. Thus f − ( G ) is Π -complete subset of X , con-tradicting f − Σ ⊆ Σ .It it well known that Y is homeomorphic to a subspace of R ω . Withoutloss of generality, we may assume Y = R ω . Granting this assumption, we canfix a sequence of continuous functions f n : X → Y pointwisely convergingto f . Fix a compatible metric d on X with d ≤ s = ∅ , let lh( s ) = p i, j q + 1. Now we redefine inheritors and innova-tors. We say s is an inheritor if j > s ( p k, i + j − k q ) = 0 for any k ≤ i (note: it was for any k < i in the definition of inheritor in Theorem 3.6),otherwise we say s is an innovator . Note that s is always an innovator if j = 0 or s ( p i, j q ) = 1.We will inductively construct for each s ∈ <ω an open set V s of Y , twoclosed sets E s , F s of X , an open set U s of X , and a sequence of open sets( U ws ) s (cid:22) w ≺ s a of X satisfying the following:(0) diam( U s ) ≤ − lh( s ) , U s a ∩ U s a = ∅ , U s a , U s a ⊆ U ws ;(1) F s a ⊆ F s a ;(2) V s ⊆ V ∅ and F s ⊆ E ∅ for any s = ∅ ;(3) for any s, t = ∅ with row( s ) = row( t ), we have V s = V t or V s ∩ V t = ∅ ;(4) if col( s ) > 0, then V s a , V s a ⊆ V s and F s a ⊆ F s ;(5) E s ⊆ F s ;(6) E s ∩ U ws = ∅ for each w ;(7) U s = U ss , and U w s ⊇ U w s for w (cid:22) w ;(8) if s is an inheritor, then we have V s = V v ( s ) , F s = F v ( s ) , E s = E u ( s ) ;(9) if s is an innovator, then V s ∩ V t = ∅ for any t with t ≺ s and row( t ) =row( s ); furthermore, there exists n ≥ lh( s ) such that f n ( U s ) ⊆ V s ; ECOMPOSING FUNCTIONS OF BAIRE CLASS 2 ON POLISH SPACES 15 (10) if s = ∅ , by letting V − s = (cid:26) V s ↾ (lh( s ) − , row( s ) > ,V ∅ , row( s ) = 0 ,P rs = V − s \ [ { V t : t (cid:22) r, row( t ) = row( s ) } , P rs = h P rs ↾ (lh( s ) − row( s )) , · · · , P rs , V s i , F s = h F s ↾ (lh( s ) − row( s )) , · · · , F s , E s i , then for any t (cid:22) s ≺ t a 0, we have(a) if t a F t is P st -sharp in U st ;(b) if t a F u ( t a is P su ( t a -sharp in U st .When we complete the construction, for any z ∈ ω , we set ψ ( z ) to be theunique element of T k U z ↾ k . From (0) and (7), ψ is continuous embeddingfrom 2 ω to X . Put G m = [ row( t )= m V t , G = \ m<ω G m . If z ∈ Ω, there exist i < ω and a strictly increasing sequence j k > z ( p i , j k q ) = 1 for any k < ω . Since z ↾ ( p i , j k q + 1) is an innovator, by (9),there is n k > p i , j k q such that f n k ( ψ ( z )) ∈ V z ↾ ( p i ,j k q +1) . It follows from (9)that f ( ψ ( z )) / ∈ V t whenever row( t ) = i . Thus f ( ψ ( z )) / ∈ G i ⊇ G. If z / ∈ Ω, we show that f ( ψ ( z )) ∈ G . For any m < ω , there exists J m < ω such that z ( p i, j q ) = 0 for any i ≤ m and any j > J m . So z ↾ ( p m, j q + 1)is an inheritor for any j > J m . Denote V m = V z ↾ ( p m,J m q +1) , u mj = u ( z ↾ ( p m, j q + 1)) . By (8), we have V m = V z ↾ ( p m,j q +1) for all j > J m . Since all u mj are innovators,by (9) we can find an n j ≥ lh( u mj ) such that f n j ( ψ ( z )) ∈ V u mj . By (4) and(8) we have f n j ( ψ ( z )) ∈ V m for all j > J m . So f ( ψ ( z )) ∈ V m for each m .Again by (4) we have V m +1 ⊆ V m for any m < ω . So f ( ψ ( z )) ∈ V m +1 ⊆ V m ⊆ G m for all m < ω . It follows that f ( ψ ( z )) ∈ G .Now we turn to the construction.First, set D, P, h, B P as follows:(i) P = Y, D = X, h = f ;(ii) B P is a countable basis of Y .Applying Lemma 4.1 with these D, P, h, B P , we get an open set V of Y andtwo closed sets E ⊆ F of X . Then put V ∅ = V, F ∅ = F, E ∅ = E, U ∅ = X. Secondly, assume that we have constructed V t , E t , F t , U t , and U wt for t, w ≺ s a 0. We will define for s a s a 1. We consider the followingtwo cases: Case 1. Assume s a v = v ( s a , u = u ( s a V s a = V v , F s a = F v , E s a = E u . Note that either s = u , or s is also an inheritor with u ( s ) = u , so E s = E u .By (7), E u ∩ U s M s = ∅ , so we can define an open set U s a such that U s a ⊆ U s M s , U s a ∩ E u = ∅ , diam( U s a ) ≤ − (lh( s )+1) . We set U s a s a = U s a and U s a t = U s M t for other t .To check (0)–(10), the only nontrivial one is (10)(a) with t = s a 0. Notethat, if s a a s a a 0) = 0, i.e., u = v , so P s a s a = P s M u and F s a = F u . Since (10) holds for s M , it holds for s a E u ∩ ( U s M s \ U s a ) = ∅ . By (10)(b) andProposition 4.3, we see F u is P s M u -sharp in U s M s \ U s a . By Proposition 4.4, F u ↾ (row( v ) + 1) is P s M u ↾ (row( v ) + 1)-sharp in some open set U ⊆ ( U s M s \ U s a ). Thus F v is P s M v -sharp in U , and hence f ↾ ( f − ( P s M s ) ∩ F v ∩ U ) / ∈ dec( Σ , ∆ ) . We inductively define V l , E l , and F l for each l < ω . Denote F − = F v .Assume that we have defined V k , E k , and F k for k < l . Set D, P, h, B P asfollows:(i) P = P s M s \ S k 0, if t a F t is P s a t -sharp in U s a t . By Lemma 4.5, we can find an natural number L t suchthat, for any l ≥ L t , F t is P s a t \ V l -sharp in some U lt ⊆ U s a t . If t a L t such that, for any l ≥ L t , F u ( t a is P s a t \ V l -sharp in some open set U lt ⊆ U s M t . Then we set L = max { L t : s ≺ t (cid:22) s a } and U s a t = U Lt for t (cid:22) s a ≺ t a 0, i.e., for s ≺ t (cid:22) s a F L ⊆ E s , we can see that ( F s ↾ row( s a a F L a E L is ( P s M s \ V L ) a V L -sharp in U .Pick an x ∈ ( f − ( V L ) ∩ E L ∩ U ). Since f ( x ) ∈ V L , there is an n > lh( s )such that f n ( x ) ∈ V L . Then we can define an open set U s a such that U s a ⊆ U, f n ( U s a ) ⊆ V L , x ∈ U s a , diam( U s a ) ≤ − (lh( s )+1) . Then put V s a = V L , E s a = E L , F s a = F L , and U s a s a = U s a . ECOMPOSING FUNCTIONS OF BAIRE CLASS 2 ON POLISH SPACES 17 Case 2. Assume s a s (cid:22) s M ≺ s a 0, by (10)(a), wehave F s is P s M s -sharp in U s M s . Thus E s is V s -sharp in U s M s , and hence f ↾ ( f − ( V s ) ∩ E s ∩ U s M s ) / ∈ dec( Σ , ∆ ) . Set D, P, h, B P as follows:(i) P = V s , D = E s ∩ U s M s ∩ f − ( P ), h = f ↾ D ;(ii) B P is a countable basis of P such that V ⊆ P for each V ∈ B P .Applying Lemma 4.1 with these D, P, h, B P , we get an open set V of Y and two closed sets E ⊆ F ⊆ D ⊆ E s with E ∩ U s M s ⊇ E ∩ D = ∅ .From Lemma 4.1 and F ⊆ E s , we can see that ( F s ↾ row( s a a F a E is( P s M s \ V ) a V -sharp in U s M s .Pick an x ∈ ( f − ( V ) ∩ E ∩ U s M s ). Since f ( x ) ∈ V , there is an n > lh( s )such that f n ( x ) ∈ V . Then we can an open set U s a such that U s a ⊆ U s M s , f n ( U s a ) ⊆ V, x ∈ U s a , diam( U s a ) ≤ − (lh( s )+1) . Then put V s a = V, E s a = E, F s a = F, and U s a s a = U s a , and U s a t = U s M t for other t .To check (0)–(10), it is trivial for s = ∅ . For s = ∅ , the only nontrivialclauses are (3), (9), and (10). Note that row( s a > 0, so V − s a = V s . Notealso that either s is also an innovator, or col( s a 0) = 0, i.e., u ( s ) = v ( s ). Inboth cases, (4) and (9) imply that there is no t ≺ s a t ) =row( s a 0) and V − t = V s . So (3) and (9) hold. Therefore, P s a s a = V s \ V ,thus P s a s a = ( P s M s \ V ) a V . Similarly, P s a t = P s M t and P s a u ( t a = P s M u ( t a for t ≺ s a ≺ t a 0. Since (10) holds for s M , it holds for s a F ∩ ( U s M s \ U s a ) = ∅ . By Lemma 4.1, f ↾ ( f − ( V s \ V ) ∩ F ∩ ( U s M s \ U s a )) / ∈ dec( Σ , ∆ ) . Now we define for s a (cid:3) Theorem 4.7. Let X be a Polish space, Y a separable metrizable space,and let f : X → Y . Then f − Σ ⊆ Σ ⇐⇒ f ∈ dec( Σ , ∆ ) . Proof. The “ ⇐ ” part is trivial, we only prove the “ ⇒ ” part.Since f − Σ ⊆ Σ implies f − Σ ⊆ Σ , it follows from Theorem 3.6that f ∈ dec( Σ , ∆ ), i.e., there exists a sequence of G δ set X n such that S n X n = X and each f ↾ X n is of Baire class 1. Then Theorem 4.6 gives f ↾ X n ∈ dec( Σ , ∆ ). Therefore, we have f ∈ dec( Σ , ∆ ). (cid:3) Corollary 4.8. Let X be a Polish space, Y a separable metrizable space,and let f : X → Y . If f / ∈ dec( Σ , ∆ ) , then there exists a Cantor set C ⊆ X such that f ↾ C / ∈ dec( Σ , ∆ ) . Proof. If f / ∈ dec( Σ , ∆ ), by Corollary 3.7, there exists a Cantor set C ⊆ X such that f ↾ C / ∈ dec( Σ , ∆ ). It is clear that f ↾ C / ∈ dec( Σ , ∆ ).If f ∈ dec( Σ , ∆ ), i.e., there exists a sequence of G δ set X n such that S n X n = X and each f ↾ X n is of Baire class 1, then there is some X n suchthat f ↾ X n / ∈ dec( Σ , ∆ ). Let ψ be the continuous embedding defined inTheorem 4.6. Put C = ψ (2 ω ). (cid:3) References [1] A. Andretta, The SLO principle and the Wadge hierarchy , in: Fundations of theformal sciences V, Studies in Logic, vol. 11, pp. 1–38. Coll. Publ., London, 2007.[2] V. Gregoriades, T. Kihara, and K. M. Ng, Turing degrees in Polish spaces and de-composability of Borel functions , submitted, 2016.[3] J. E. Jayne, The space of class α Baire functions , Bull. Amer. Math. 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Motto Ros and B. Semmes, A new proof of a theorem of Jayne and Rogers , RealAnal. Exchange 35 (2009/2010), 195–204.[13] J. Pawlikowski and M. Sabok, Decomposing Borel functions and structure at finitelevels of the Baire hierarchy , Ann. pure Appl. Logic 163 (2012), 1748–1764.[14] B. Semmes, A game for the Borel functions, Ph.D. thesis, ILLC, University of Ams-terdam, Amsterdam, Holland, 2009.[15] S. Solecki, Decomposing Borel sets and functions and the structure of Baire class 1functions , J. Amer. Math. Soc. 11 (1998), 521–550. (Longyun Ding and Jiafei Zhao) School of Mathematical Sciences and LPMC,Nankai University, Tianjin, 300071, P.R.China E-mail address : [email protected] (Longyun Ding), [email protected] (JiafeiZhao) (Takayuki Kihara) Graduate School of Informatics, Nagoya University,Nagoya, 464-8601, Japan E-mail address : [email protected] (Brian Semmes) StudyLab Language School, Moscow, Nikolskaya St. 10,Nikolskaya Plaza Office Centre, Russian Federation E-mail address ::