Omega-powers and descriptive set theory

ω -po wers and descriptiv e set theory . Dominique LECOMTE J . Symboli c Lo gic 70, 4 (2005), 1210-1232 Abstract. W e study the sets of the infinite sentenc es constructible with a dictionary ov er a finite alp habet, from the vie w- point of descriptive set theory . Amo ng other things, this giv es some true co-an alytic sets. The case where the dictionary is finite is studied and giv es a natural e xample of a set at the lev el ω of the W adge hierarchy . 1 Introdu ction. W e consider the fi nite alphabet n = { 0 , . . . , n − 1 } , where n ≥ 2 is an inte ger , and a dictio nary ov er this alphabet , i.e., a subset A of the set n <ω of finite words with let ters in n . Definition 1 The ω - pow er associated to A is the set A ∞ of the infini te sentence s cons tructible with A by concate nation. So we have A ∞ := { a 0 a 1 . . . ∈ n ω / ∀ i ∈ ω a i ∈ A } . The ω -po wers play a crucial role in the character ization of subsets of n ω accept ed by finite au- tomata (see Theorem 2.2 in [St1]). W e will stud y these ob jects from the v iewpoi nt of descripti ve set theory . The reader should see [K1] for the c lassical res ults of th is theo ry; we will also use the no tation of this book. The ques tions we stu dy are the followin g: (1) Wha t are the possib le lev els of topo logical comple xity for the ω -powers? This questio n was asked by P . Simonn et in [S], and st udied in [St2]. O. Finke l (in [F1]) and A. Louvea u prove d indepen dently that Σ 1 1 -complet e ω -po wers exi st. O. Finkel p rov ed in [F2] the exis tence of a Π 0 m -complet e ω -po wer for each inte ger m ≥ 1 . (2) What is the topolo gical comple xity of the set of dictionari es whose associ ated ω -power is of a gi ven le vel of comple xity? T his question arises naturally when we look at the character izations of Π 0 1 , Π 0 2 and Σ 0 1 ω -power s obtained in [St2] (see C orollar y 14 and Lemmas 25 , 26 ). (3) W e w ill recall that an ω -po wer is an analyt ic subse t of n ω . W hat is the topological complexity o f the set of cod es for analyt ic sets which are ω -po wers? This question was ask ed by A. L ouv eau. T his questi on also makes sense for the set of codes for Σ 0 ξ (resp., Π 0 ξ ) sets w hich are ω -po wers. And also for the set of codes for Borel sets which are ω -powers . As usual with descript iv e set theory , the point is not only the computat ion of topolog ical com- ple xities, bu t also the hope that these co mputations will lead to a bette r understa nding of the studi ed object s. Many sets in this paper won’ t be clope n, in particu lar won’ t be recursi ve. This giv es unde- cidabi lity results . 1 • W e giv e the answer to Q uestion (2) for the very first le vels ( {∅} , i ts dual class and ∆ 0 1 ). This contains a study of the case w here the dictiona ry is finite. In particu lar , we sho w that the set of dictionarie s whose associa ted ω -po wer is generated by a dictiona ry w ith two words is a ˇ D ω ( Σ 0 1 ) -comple te set. This is a surpris ing result beca use this comple xity is not clear at all on the defini tion of th e set. • W e giv e two proof s of the fact that the relation “ α ∈ A ∞ ” is Σ 1 1 -complet e. One of these proofs is used later to giv e a partial answer to Q uestion (2). T o understand this answer , the reader should see [M] for the basic not ions of ef fecti ve descri ptiv e set theo ry . Roughly speak ing, a set is ef fectiv ely Borel (resp ., effect iv ely Borel in A ) if i ts construc tion ba sed on basic clopen sets can be code d with a recurs iv e (resp., recurs iv e in A ) sequence of integers . T his answer is the Theor em. The fo llowing sets ar e true co-analyti c sets: - { A ∈ 2 n <ω / A ∞ ∈ ∆ 1 1 ( A ) } . - { A ∈ 2 n <ω / A ∞ ∈ Σ 0 ξ ∩ ∆ 1 1 ( A ) } , for 1 ≤ ξ < ω 1 . - { A ∈ 2 n <ω / A ∞ ∈ Π 0 ξ ∩ ∆ 1 1 ( A ) } , for 2 ≤ ξ < ω 1 . This result also comes from an analysis of Borel ω -po wers: A ∞ is B orel if and only if we can choos e in a Borel way the decompo sition of any sentenc e of A ∞ into words of A (see L emma 13). This analysis is also related to Question (3) and to some Borel unifor mization r esult for G δ sets locally with Borel projectio ns. W e will specify these relations . • A natural ordinal rank can be defined on the complement of any ω -po wer , and w e study it; its kno w ledge giv es an upper bound of the complex ity of the ω -po wer . • W e study the link between Question (1) and the exten sion or dering on finite sequences of integers. • Finally , we giv e some exampl es of ω -powers complete for the classes ∆ 0 1 , Σ 0 1 ⊕ Π 0 1 , D 2 ( Σ 0 1 ) , ˇ D 2 ( Σ 0 1 ) , ˇ D 3 ( Σ 0 1 ) and ˇ D 2 ( Σ 0 2 ) . 2 Finitely generated ω - powers. Notation. In order to answer to Question (2), we set Σ 0 := { A ⊆ n <ω / A ∞ = ∅} , Π 0 := { A ⊆ n <ω / A ∞ = n ω } , ∆ 1 := { A ⊆ n <ω / A ∞ ∈ ∆ 0 1 } , Σ ξ := { A ⊆ n <ω / A ∞ ∈ Σ 0 ξ } , Π ξ := { A ⊆ n <ω / A ∞ ∈ Π 0 ξ } ( ξ ≥ 1) , ∆ := { A ⊆ n <ω / A ∞ ∈ ∆ 1 1 } . • If A ⊆ n <ω , then we set A − := A \ {∅} . • W e define, for s ∈ n <ω and α ∈ n ω , α − s := ( α ( | s | ) , α ( | s | + 1) , ... ) . • If S ⊆ ( n <ω ) <ω , then we set S ∗ := { S ∗ := S (0) . . . S ( | s | − 1) /S ∈ S } . 2 • W e define a recursi ve map π : n ω × ω ω × ω → n <ω by π ( α, β , q ) :=  ( α (0) , . . . , α ( β [0])) if q = 0 , ( α (1 + Σ j 0 β ( m ) > 0) and ( ∀ q ∈ ω π ( α, β , q ) ∈ A ) ] . Pro position 2 ([S]) A ∞ ∈ Σ 1 1 for all A ⊆ n <ω . If A is finite, then A ∞ ∈ Π 0 1 . Pro of. W e define a contin uous map c : ( A − ) ω → n ω by the formula c (( a i )) := a 0 a 1 . . . W e hav e A ∞ = c [( A − ) ω ] , and ( A − ) ω is a Polish space (compac t if A is finite).  Pro position 3 If A ∞ ∈ ∆ 0 1 , then ther e exists a finite subs et B of A suc h that A ∞ = B ∞ . Pro of. Set E k := { α ∈ n ω /α ⌈ k ∈ A and α − α ⌈ k ∈ A ∞ } . It is an op en subset of n ω since A ∞ is open, and A ∞ ⊆ S k > 0 E k . W e can fi nd an integer p su ch that A ∞ ⊆ S 0 0 such that A ∞ = ( A ∩ n ≤ p ) ∞ , by Proposition 3. S o let s 1 , . . . , s k , t 1 , . . . , t l ∈ 2 <ω be such that A ∞ = S 1 ≤ i ≤ k N s i = n ω \ ( S 1 ≤ j ≤ l N t j ) . For each 1 ≤ j ≤ l , and for each sequence s ∈ [( A − ) <ω ] ∗ \ {∅} , t j 6≺ s . So we ha ve A ∞ ∈ ∆ 0 1 ⇔    ∃ p > 0 ∃ k , l ∈ ω ∃ s 1 , . . . , s k , t 1 , . . . , t l ∈ 2 <ω S 1 ≤ i ≤ k N s i = n ω \ ( S 1 ≤ j ≤ l N t j ) and ∀ 1 ≤ j ≤ l ∀ s ∈ [( A − ) <ω ] ∗ \ {∅} t j 6≺ s and ∀ α ∈ n ω { α / ∈ S 1 ≤ i ≤ k N s i or ∃ β ∈ p ω [( ∀ m > 0 β ( m ) > 0) and ( ∀ q ∈ ω π ( α, β , q ) ∈ A ) ] } . This sho ws that ∆ 1 is a K σ subset of 2 n <ω . 3 T o sho w tha t it is n ot Π 0 2 , it is enoug h to see that its inters ection with the clos ed se t { A ⊆ n <ω / A ∞ 6 = n ω } is dense and co-dense in this closed set (see (b)), by B aire’ s theorem. S o let O be a basic clopen subset of 2 n <ω meeting this clos ed se t. W e may assume that it is of the form { A ⊆ n <ω / ∀ i ≤ k s i ∈ A and ∀ j ≤ l t j / ∈ A } , where s 0 , . . . , s k , t 0 , . . . , t l ∈ n <ω and | s 0 | > 0 . Let A := { s i /i ≤ k } . Then A ∈ O and A ∞ is in Π 0 1 \ {∅ , n ω } . There are two cases. If A ∞ ∈ ∆ 0 1 , then we hav e to find B ∈ O w ith B ∞ / ∈ ∆ 0 1 . Let u 0 , . . . , u m ∈ n <ω with S p ≤ m N u p = n ω \ A ∞ . Let r ∈ n \ { u 0 ( | u 0 | − 1) } , s := u 0 r | u 0 | + max j ≤ l | t j | and B := A ∪ { s } . Then B ∈ O and s ∞ ∈ B ∞ . L et us sho w tha t s ∞ is no t in the int erior of B ∞ . Otherwise, we could find an inte ger q such that N s q ⊆ B ∞ . W e w ould ha ve α := s q u 0 u 0 ( | u 0 | − 1) r ∞ ∈ B ∞ . As N u 0 ∩ A ∞ = ∅ , the decomposition of α into no nempty words of B would start with q times s . If this decompositi on could go on, then w e would hav e u 0 = ( u 0 ( | u 0 | − 1)) | u 0 | . Let v ∈ n <ω be such that N v ⊆ A ∞ . W e ha ve v ( u 0 ( | u 0 | − 1)) ∞ ∈ A ∞ , so ( u 0 ( | u 0 | − 1)) ∞ ∈ N u 0 ∩ A ∞ . But th is is absu rd. Therefore B ∞ / ∈ ∆ 0 1 . If A ∞ / ∈ ∆ 0 1 , then we ha ve to find B ∈ O such that B ∞ ∈ ∆ 0 1 \ { n ω } . Notice th at n ω 6 = S i ≤ k N s i . So let v ∈ n <ω be non cons tant suc h tha t N v ∩ S i ≤ k N s i = ∅ . W e set D := A ∪ [ r ∈ n \{ v (0) } { ( r ) } ∪ { v (0) | v | } , B := A ∪ { s ∈ n <ω / | s | > max j ≤ l | t j | and ∃ t ∈ D t ≺ s } . W e get B ∞ = S t ∈ D N t ∈ ∆ 0 1 and N v ∩ B ∞ = ∅ , so B ∞ 6 = n ω .  No w we will study F := { A ⊆ n <ω / ∃ B ⊆ n <ω finite A ∞ = B ∞ } . Pro position 5 F is a co -nowher e dens e Σ 0 2 -har d subset of 2 n <ω . Pro of. By Propositio n 3, i f A ∞ = n ω , then th ere exis ts an inte ger p such that A ∞ = ( A ∩ n ≤ p ) ∞ , so Π 0 ⊆ F and, b y Theore m 4, F is co-no where den se. W e define a continuou s map φ : 2 ω → 2 n <ω by the formula φ ( γ ) := { 0 k 1 /γ ( k ) = 1 } . If γ ∈ P f := { α ∈ 2 ω / ∃ p ∀ m ≥ p α ( m ) = 0 } , then φ ( γ ) ∈ F . If γ / ∈ P f , then the co ncatenatio n m ap is an homeomorphism from φ ( γ ) ω onto φ ( γ ) ∞ , thus φ ( γ ) ∞ is not K σ . S o φ ( γ ) / ∈ F , b y P roposit ion 2. Thus the preimage of F by φ is P f , and F is Σ 0 2 -hard.  Let G p := { A ⊆ n <ω / ∃ s 1 , . . . , s p ∈ n <ω A ∞ = { s 1 , . . . , s p } ∞ } , so th at F = S p G p . W e hav e G 0 = Σ 0 , so G 0 is Π 0 1 \ Σ 0 1 . 4 Pro position 6 G 1 is Π 0 1 \ Σ 0 1 . In particula r , G 1 is Π 0 1 -complet e. Pro of. If p ∈ ω \ { 0 } , then { 0 , 1 p } / ∈ G 1 since B ∞ = { s ∞ } if B = { s } . Thus { 0 } is not an interior point of G 1 since the seque nce ( { 0 , 1 p } ) p> 0 tends to { 0 } . So G 1 / ∈ Σ 0 1 . • Let ( A m ) ⊆ G 1 tendin g to A ⊆ n <ω . If A ⊆ {∅} , then A ∞ = ∅ = {∅} ∞ , so A ∈ G 1 . If A 6⊆ {∅ } , then let t ∈ A − and α 0 := t ∞ . There e xists an intege r m 0 such that t ∈ A m for m ≥ m 0 . Thus w e may assume that t ∈ A m and A ∞ m 6 = ∅ . S o let s m ∈ n <ω \ {∅} be such that A ∞ m = { s m } ∞ = { s ∞ m } . W e ha ve s ∞ m = α 0 . L et b := min { a ∈ ω \ { 0 } / ( α 0 ⌈ a ) ∞ = α 0 } . • W e will sho w that A m ⊆ { ( α 0 ⌈ b ) q /q ∈ ω } . Let s ∈ A m \ {∅} . As s ∞ = α 0 , we can find an i nteger a > 0 such th at s = α 0 ⌈ a , and b ≤ a . Let r < b and q be integ ers so tha t a = q .b + r . W e hav e, if r > 0 , α 0 = ( α 0 ⌈ a ) ∞ = ( α 0 ⌈ b ) ∞ = ( α 0 ⌈ q .b )( α 0 ⌈ a − α 0 ⌈ q .b ) α 0 = ( α 0 ⌈ b ) q ( α 0 ⌈ a − α 0 ⌈ q .b ) α 0 = ( α 0 ⌈ a − α 0 ⌈ q .b ) α 0 = ( α 0 ⌈ r ) α 0 = ( α 0 ⌈ r ) ∞ . Thus, by minimality of b , r = 0 and we are done. • Let u ∈ A . W e can find an inte ger m u such that u ∈ A m for m ≥ m u . S o the re exists an inte ger q u such that u = ( α 0 ⌈ b ) q u . T herefo re A ∞ = { ( α 0 ⌈ b ) ∞ } = { α 0 ⌈ b } ∞ and A ∈ G 1 .  Remark. N otice that this shows that we can find w ∈ n <ω \ {∅} such that A ⊆ { w q /q ∈ ω } if A ∈ G 1 . Now we stud y G 2 . T he ne xt lemma is just Corollary 6.2.5 in [Lo]. Lemma 7 T wo finite sequenc es which commute ar e powers of the same finite seque nce. Pro of. Let x and y be finite sequence s w ith xy = y x . Then the subgroup of the free group on n genera tors generated by x and y is abelian, hence isomorph ic to Z . One generato r of this subgroup must be a finite sequen ce u such that x and y are bo th po wers of u .  Lemma 8 Let A ∈ G 2 . T hen the r e exists a finite subset F of A suc h that A ∞ = F ∞ . Pro of. W e will sho w more. Let A / ∈ G 1 satisfy ing A ∞ = { s 1 , s 2 } ∞ , with | s 1 | ≤ | s 2 | . Then (a) The decompositi on of α in to w ords of { s 1 , s 2 } is unique for ea ch α ∈ A ∞ (this is a conseque nce of Corollarie s 6.2.5 and 6.2.6 in [Lo]). (b) s 2 s 1 ⊥ s 1 q s 2 for each inte ger q > 0 , and s 2 s 1 ∧ s 1 q s 2 = s 1 s 2 ∧ s 2 s 1 . (c) A ⊆ [ { s 1 , s 2 } <ω ] ∗ . • W e prov e the first two point s. W e split into cases. 2.1. s 1 ⊥ s 2 . The result is clear . 5 2.2. s 1 ≺ 6 = s 2 6≺ s ∞ 1 . Here also, the result is clear (cut α into words of len gth | s 1 | ). 2.3. s 1 ≺ 6 = s 2 ≺ s ∞ 1 . W e can w rite s 2 = s m 1 s , where m > 0 and s ≺ 6 = s 1 . Thus s 2 s 1 = s m 1 ss 1 and s m +1 1 s ≺ s q 1 s 2 if q > 0 . B ut s m 1 ss 1 ⊥ s m +1 1 s oth erwise ss 1 = s 1 s , and s , s 1 s 2 would be po w ers of some sequence, which contra dicts A / ∈ G 1 . • W e p rov e (c). Let t ∈ A , so that ts ∞ 1 , ts 2 s ∞ 1 ∈ A ∞ . T hese sequenc es sp lit after t ( s 1 s 2 ∧ s 2 s 1 ) , a nd the decomposition of ts ∞ 1 (resp., ts 2 s ∞ 1 ) into words of { s 1 , s 2 } starts with us i (resp., us 3 − i ), whe re u ∈ [ { s 1 , s 2 } <ω ] ∗ . So ts ∞ 1 and ts 2 s ∞ 1 split after u ( s 1 s 2 ∧ s 2 s 1 ) by (b). But w e m ust ha ve t = u becaus e of the posit ion of the spl itting po int. • W e prove Lemma 8. If A ∈ G 0 , then F := ∅ works. If A ∈ G 1 \ G 0 , th en let w ∈ n <ω \ {∅} such that A ⊆ { w q /q ∈ ω } , and q > 0 such that w q ∈ A . Then F := { w q } works . S o we may assu me that A / ∈ G 1 , an d A ∞ = { s 1 , s 2 } ∞ . As A ∞ ⊆ S t ∈ A − { α ∈ N t /s 1 s 2 ∧ s 2 s 1 ≺ α − t } is compact, we get a fi nite subset F of A − such that A ∞ ⊆ S t ∈ F { α ∈ N t /s 1 s 2 ∧ s 2 s 1 ≺ α − t } . W e hav e F ∞ ⊆ A ∞ . If α ∈ A ∞ , then let t ∈ F such that t ≺ α . By (c), we hav e t ∈ [ { s 1 , s 2 } <ω ] ∗ . The sequen ce t is the beginni ng of the decompos ition of α in to words of { s 1 , s 2 } . Thus α − t ∈ A ∞ and we can go on like this. This sho ws that α ∈ F ∞ .  Remark. The inclusion o f A ∞ = { s 1 , s 2 } ∞ into { t 1 , t 2 } ∞ does not imply { s 1 , s 2 } ⊆ [ { t 1 , t 2 } <ω ] ∗ , e ven if A / ∈ G 1 . Indeed, take s 1 := 01 , s 2 := t 1 := 0 and t 2 := 10 . But we hav e | t 1 | + | t 2 | ≤ | s 1 | + | s 2 | , which is the case in general : Lemma 9 Let A , B / ∈ G 1 satisfy ing A ∞ = { s 1 , s 2 } ∞ ⊆ B ∞ = { t 1 , t 2 } ∞ . Then ther e is j ∈ 2 suc h that | t 1+ i | ≤ | s 1+[ i + j mod 2] | for each i ∈ 2 . In particula r , | t 1 | + | t 2 | ≤ | s 1 | + | s 2 | . Pro of. W e may assume that | s 1 | ≤ | s 2 | . Let, for i = 1 , 2 , ( w i m ) m ⊆ { t 1 , t 2 } be sequences such that s ∞ 1 = w 1 0 w 1 1 . . . (resp., s 2 s ∞ 1 = w 2 0 w 2 1 . . . ). B y the proof of Lemma 8, there is a minimal inte ger m 0 satisfy ing w 1 m 0 6 = w 2 m 0 . W e let u := w 1 0 . . . w 1 m 0 − 1 . The sequenc es s ∞ 1 and s 2 s ∞ 1 split after s 1 s 2 ∧ s 2 s 1 = u ( t 1 t 2 ∧ t 2 t 1 ) . Similarly , s ∞ 1 and s 1 s 2 s ∞ 1 split after s 1 ( s 1 s 2 ∧ s 2 s 1 ) = v ( t 1 t 2 ∧ t 2 t 1 ) , where v ∈ [ { t 1 , t 2 } <ω ] ∗ \ {∅} . S o we get s 1 u = v . Similarly , with the sequen ces s 2 s ∞ 1 and s 2 2 s ∞ 1 , we see that s 2 u ∈ [ { t 1 , t 2 } <ω ] ∗ \ {∅} . So we m ay assume that u 6 = ∅ since { s 1 , s 2 } / ∈ G 1 . If t 1 6⊥ t 2 , then we may assume that ∅ 6 = t 1 ≺ 6 = t 2 . So we m ay ass ume that we are not in the case t 2 ≺ t ∞ 1 . Indeed, otherwis e t 2 = t m 1 t , where ∅ ≺ 6 = t ≺ 6 = t 1 (see the proof of Lemma 8). Moreov er , t 1 doesn ’t finish t 2 , otherwis e we w ould ha ve t 1 = t ( t 1 − t ) = ( t 1 − t ) t and t , t 1 − t , t 1 , t 2 would be po wers of the same sequen ce, which contrad icts { t 1 , t 2 } / ∈ G 1 . As s i u ∈ [ { t 1 , t 2 } <ω ] ∗ , this sh ows that s i ∈ [ { t 1 , t 2 } <ω ] ∗ . So we are done since { s 1 , s 2 } / ∈ G 1 as before. 6 Assume for ex ample that t 2 = w 1 m 0 . L et m ′ be maximal with t m ′ 1 ≺ t 2 . Notice that ut m ′ 1 ≺ s 1 s 2 ≺ s 1 s 2 s ∞ 1 . W e hav e ut 2 ≺ s 1 s 2 s ∞ 1 , otherwise we would obtain ut m ′ +1 1 ≺ s 1 s 2 s ∞ 1 ∧ s 2 s ∞ 1 = s 1 s 2 ∧ s 2 s 1 ≺ s ∞ 1 , which is absurd. S o w e get | t 2 | ≤ | s 1 | since | u | + | t 2 | + | t 1 t 2 ∧ t 2 t 1 | ≤ | s 1 | + | s 1 s 2 ∧ s 2 s 1 | . Similarly , | t 1 | ≤ | s 2 | since ut m ′ +1 1 ≺ s 2 2 s ∞ 1 . T he ar gument is similar if t 2 = w 2 m 0 (we get | t i | ≤ | s i | in this case for i = 1 , 2 ).  Cor ollary 10 G 2 is a ˇ D ω ( Σ 0 1 ) \ D ω ( Σ 0 1 ) set. In particula r , G 2 is ˇ D ω ( Σ 0 1 ) -comple te. Pro of. W e will app ly the Hausdorf f deri vat ion to G ⊆ 2 n <ω . This mean s tha t we define a decreasi ng sequen ce ( F ξ ) ξ <ω 1 of closed subsets of 2 n <ω as follo ws: F ξ :=   \ η<ξ F η   ∩ G if ξ is ev en ,   \ η<ξ F η   \ G if ξ is odd. Recall that if ξ is e ven, then F ξ = ∅ is equi va lent to G ∈ D ξ ( Σ 0 1 ) . Ind eed, we set U ξ := ˇ F ξ . W e ha ve U ξ +1 \ U ξ = F ξ \ F ξ +1 ⊆ G if ξ is ev en and U ξ +1 \ U ξ ⊆ ˇ G if ξ is odd. Similarly , U ξ \ ( S η<ξ U η ) ⊆ ˇ G if ξ is limit. If F ξ = ∅ , then let η be minimal such that F η = ∅ . W e hav e G = S θ ≤ η, θ odd U θ \ ( S ρ<θ U ρ ) . If η is odd, th en ˇ G = S θ < η, θ ev en U θ \ ( S ρ<θ U ρ ) ∈ D η ( Σ 0 1 ) , thus G ∈ ˇ D η ( Σ 0 1 ) ⊆ D ξ ( Σ 0 1 ) . If η is ev en, then G = S θ < η, θ odd U θ \ ( S ρ<θ U ρ ) ∈ D η ( Σ 0 1 ) and the same co nclusion is true. Con vers ely , if G ∈ D ξ ( Σ 0 1 ) , then let ( V η ) η<ξ be an inc reasing seq uence of open sets with G = S η<ξ , η od d V η \ ( S θ < η V θ ) . By induction, we check that F η ⊆ ˇ V η if η < ξ . This clearly implies that F ξ = ∅ because ξ is eve n. • W e will sho w that if A / ∈ G 1 satisfies A ∞ = { s 1 , s 2 } ∞ , then A / ∈ F M := F M ( G 2 ) , where M is the smallest odd inte ger g reater than or equal to f ( s 1 , s 2 ) := 2Σ l ≤| s 1 | + | s 2 |− 2 n 2( | s 1 | + | s 2 |− l ) . W e argue by contrad iction: A is the limit of ( A q ) , where A q ∈ F M − 1 \ G 2 . Lemma 8 gi ves a finite subset F of A , and w e may assume that F ⊆ A q for each q . Thus we ha ve A ∞ ⊆ A ∞ q , and the inclus ion is strict. T hus we can fi nd s q ∈ [ A <ω q ] ∗ such that N s q ∩ A ∞ = ∅ . Let s q 0 , . . . , s q m q ∈ A q be such that s q = s q 0 . . . s q m q . No w A q is the limit of ( A q ,r ) r , where A q ,r ∈ F M − 2 ∩ G 2 , and we may assume that { s q 0 , . . . , s q m q } ∪ F ⊆ A q ,r for ea ch r , and that A q ,r / ∈ G 1 becaus e A q / ∈ G 1 ⊆ G 2 . Let s q ,r 1 , s q ,r 2 such that A ∞ q ,r = { s q ,r 1 , s q ,r 2 } ∞ . By Lemma 9 we ha ve | s q ,r 1 | + | s q ,r 2 | ≤ | s 1 | + | s 2 | . Now we let B 0 := A 0 , 0 and s 0 i := s 0 , 0 i for i = 1 , 2 . W e ha ve B 0 ∈ F M − 2 ∩ G 2 \ G 1 , A ∞ ⊆ 6 = B ∞ 0 = { s 0 1 , s 0 2 } ∞ , and | s 0 1 | + | s 0 2 | ≤ | s 1 | + | s 2 | . 7 No w we iterate th is: for each 0 < k < n 2( | s 1 | + | s 2 | ) , we get B k ∈ F M − 2( k +1) ∩ G 2 \ G 1 such that B ∞ k − 1 ⊆ 6 = B ∞ k = { s k 1 , s k 2 } ∞ and | s k 1 | + | s k 2 | ≤ | s k − 1 1 | + | s k − 1 2 | . W e can find k 0 < n 2( | s 1 | + | s 2 | ) such that | s k 0 1 | + | s k 0 2 | < | s k 0 − 1 1 | + | s k 0 − 1 2 | (with the con vention s − 1 i := s i ). W e set C 0 := B k 0 , t 0 i := s k 0 i . So we hav e C 0 ∈ F M − 2( k 0 +1) ∩ G 2 \ G 1 , C ∞ 0 = { t 0 1 , t 0 2 } ∞ and | t 0 1 | + | t 0 2 | < | s 1 | + | s 2 | . No w we iterate this: for each l ≤ | s 1 | + | s 2 | − 2 , we get t l 1 , t l 2 , k l < n 2( | t l − 1 1 | + | t l − 1 2 | ) and C l ∈ F M − 2Σ m ≤ l ( k m +1) ∩ G 2 \ G 1 satisfy ing C ∞ l = { t l 1 , t l 2 } ∞ and | t l 1 | + | t l 2 | < | t l − 1 1 | + | t l − 1 2 | (with the co n vention t − 1 i := s i ). W e h av e | t l 1 | + | t l 2 | ≤ | s 1 | + | s 2 | − 1 − l , thus 2Σ l ≤| s 1 | + | s 2 |− 2 ( k l + 1) ≤ 2Σ l ≤| s 1 | + | s 2 |− 2 n 2( | t l − 1 1 | + | t l − 1 2 | ) ≤ f ( s 1 , s 2 ) and this constr uction is possible . But we hav e | t | s 1 | + | s 2 |− 2 1 | + | t | s 1 | + | s 2 |− 2 2 | ≤ 1 , thus C | s 1 | + | s 2 |− 2 ∈ G 1 , which is absurd. Let A / ∈ G 2 . As A / ∈ G 1 , we can find s, t ∈ A which are not po wers of the same sequenc e. Indeed, let s ∈ A − and u with minimal length such that s is a power of u . T hen any t ∈ A \ { u q /q ∈ ω } works , because if s and t are po wers of w , then w has to be a power of u . Ind eed, as u ≺ w , w = u k v with v ≺ u , an d v has to be a po w er of u by minimality of | u | and Lemm a 7. Assume that more ove r A ∈ F 2 k + 2 . No w A is the limit of ( A k ,r ) r ⊆ F 2 k + 1 ∩ G 2 for each integer k , and we may assume that s, t ∈ A k ,r / ∈ G 1 . Let s k ,r 1 , s k ,r 2 be such that A k ,r ∞ = { s k ,r 1 , s k ,r 2 } ∞ . By Lemm a 9 we ha ve | s k ,r 1 | + | s k ,r 2 | ≤ | s | + | t | and f ( s k ,r 1 , s k ,r 2 ) ≤ f ( s, t ) . By the precedi ng poin t, we must ha ve 2 k + 1 < f ( s , t ) . Thus T m F m ⊆ G 2 . Notice that F m +1 ( ˇ G 2 ) ⊆ F m , so that F ω ( ˇ G 2 ) = ∅ and G 2 ∈ ˇ D ω ( Σ 0 1 ) . • Now le t us show tha t { 0 } ∈ F ω ( G 2 ) ( this will imply G 2 / ∈ D ω ( Σ 0 1 )) . It is enough to see that { 0 } ∈ \ m F m . Let E ( x ) be the bigge st inte ger less than or equal to x , p k ,s := 2 k +1 − E ( | s | / 2) and k ∈ ω . W e de fine A ∅ := { 0 } and, for s ∈ ( ω \ { 0 , 1 } ) ≤ 2 k + 1 and m > 1 , A sm := A s ∪ { (01 p k,s ) m ; (0 2 1 p k,s ) m } if | s | is ev en, A s ∪ { s ∈ [ { 0 , 1 p k,s } <ω ] ∗ /m ≤ | s | ≤ m + p k ,s } if | s | is odd. Let us sho w that A s ∈ G 2 (resp., ˇ G 2 ) if | s | is e ven (resp., odd). First by inducti on w e get A sm ⊆ { 0 , 1 p k,s } <ω . Therefore A ∞ sm = { 0 , 1 p k,s } ∞ if | s | is odd, because if α is in { 0 , 1 p k,s } ∞ and t ∈ [ { 0 , 1 p k,s } <ω ] ∗ with minimal length ≥ m beg ins α , then t ∈ A sm . Now if | s | is e ven and A ∞ sm = { s 1 , s 2 } ∞ , th en 0 ∞ ∈ { s 1 , s 2 } ∞ , thus for example s 1 = 0 k +1 . ( 01 p k,s ) ∞ ∈ { s 1 , s 2 } ∞ , thus s 2 ≺ (01 p k,s ) ∞ and | s 2 | ≥ | (01 p k,s ) m | since s 2 0 ∞ ∈ { s 1 , s 2 } ∞ . But then (0 2 1 p k,s ) ∞ / ∈ { s 1 , s 2 } ∞ since m > 1 . Thus A sm / ∈ G 2 . As ( A sm ) m tends to A s and ( A s ) | s | =2 k +2 ⊆ G 2 , we ded uce from this that A s is in F 2 k + 1 −| s | \ G 2 if | s | ≤ 2 k + 1 is odd, and that A s ∈ F 2 k + 1 −| s | ∩ G 2 if | s | ≤ 2 k + 1 is ev en. Therefor e { 0 } is in T k F 2 k + 1 = T m F m .  8 Remarks. (1) The en d of this proo f also shows that G p / ∈ D ω ( Σ 0 1 ) if p ≥ 2 . Indeed, { 0 } ∈ F ω ( G p ) . The only thing to change is the definition of A sm if | s | is ev en: we set A sm := A s ∪ { (0 j +1 1 p k,s ) m /j < p } . (2) If { s 1 , s 2 } / ∈ G 1 and { s 1 , s 2 } ∞ = { t 1 , t 2 } ∞ , then { s 1 , s 2 } = { t 1 , t 2 } . Indeed, { t 1 , t 2 } / ∈ G 1 , thus by Lemma 9 we get | s 1 | + | s 2 | = | t 1 | + | t 2 | . By (c) in the pro of of Lemma 8 and the previo us fact, s i = t a i ε i , where a i > 0 , ε i , i ∈ { 1 , 2 } . As { s 1 , s 2 } / ∈ G 1 , ε 1 6 = ε 2 . T hus a i = 1 . Conjectur e 1. Let A ∈ F . Then there exis ts a finite s ubset F of A such that A ∞ = F ∞ . Conjectur e 2. L et p ≥ 1 , A , B / ∈ G p with A ∞ = { s 1 , . . . , s q } ∞ ⊆ B ∞ = { t 1 , . . . , t p +1 } ∞ . Then Σ 1 ≤ i ≤ p +1 | t i | ≤ Σ 1 ≤ i ≤ q | s i | . Conjectur e 3. W e hav e G p +1 \ G p ∈ D ω ( Σ 0 1 ) for each p ≥ 1 . In particu lar , F ∈ K σ \ Π 0 2 . Notice that Conjectures 1 and 2 imply Conjecture 3. Indeed, F = G 1 ∪ S p ≥ 1 G p +1 \ G p , so F ∈ K σ if G p +1 \ G p ∈ D ω ( Σ 0 1 ) ⊆ ∆ 0 2 , by Proposit ion 6. By Proposi tion 5 we hav e F / ∈ Π 0 2 . It is enoug h to see that F ω := F ω ( G p +1 \ G p ) = ∅ . W e ar gue as in the proof of Coro llary 10. This time, f ( s 1 , . . . , s q ) := 2Σ l ≤ Σ 1 ≤ i ≤ q | s i |− 2 n q (Σ 1 ≤ i ≤ q | s i |− l ) for s 1 , . . . , s q ∈ n <ω . The fact to notice is that A / ∈ F M ( G p +1 \ G p ) if A / ∈ G p satisfies A ∞ = { s 1 , . . . , s p +1 } ∞ and M is the minimal o dd integ er greate r than or equal to f ( s 1 , . . . , s p +1 ) . So if A ∈ F 2 k + 2 ∩ F \ G p , th en Conject ure 1 gi ves a finite subset F := { s 1 , . . . , s q } of A . The set A is the limit of ( A k ,r ) r ⊆ F 2 k + 1 ∩ G p +1 \ G p for each integ er k , and we may assume that F ⊆ A k ,r . Conjecture 2 implies th at f ( s k ,r 1 , . . . , s k ,r p +1 ) ≤ f ( s 1 , . . . , s q ) and 2 k + 1 < f ( s 1 , . . . , s q ) . Thus T m F m ⊆ ˇ F ∪ G p . S o F ω ⊆ ( ˇ F ∪ G p ) ∩ G p +1 \ G p = ∅ . 3 Is A ∞ Borel? No w w e will see tha t the maximal complexit y is po ssible. W e essentially gi ve O. Fink el’ s exam- ple, in a lightly simpler vers ion. Pro position 11 Let Γ := Σ 1 1 or a Bair e class. The ex istence of n ∈ ω \ 2 and A ⊆ n <ω suc h that A ∞ is Γ -complet e is equiv alent to the e xistence of B ⊆ 2 <ω suc h that B ∞ is Γ -complete . Pro of. Let p n := min { p ∈ ω /n ≤ 2 p } ≥ 1 . W e define φ : n ֒ → 2 p n := { σ 0 , . . . , σ 2 p n − 1 } by the formula φ ( m ) := σ m , Φ : n <ω ֒ → 2 <ω by the formula Φ( t ) := φ ( t (0)) . . . φ ( t ( | t | − 1)) and f : n ω ֒ → 2 ω by the formula f ( γ ) := φ ( γ (0)) φ ( γ (1)) . . . Then f is an homeomorphis m from n ω onto its range and reduce s A ∞ to B ∞ , w here B := Φ[ A ] . The in vers e function of f reduces B ∞ to A ∞ . So we are done if Γ is stable under intersection with closed sets. Otherwise, Γ = ∆ 0 1 or Σ 0 1 . If A = { s ∈ 2 <ω / 0 ≺ s or 1 2 ≺ s } , then A ∞ = N 0 ∪ N 1 2 , which is ∆ 0 1 -complet e. If A = { s ∈ 2 <ω / 0 ≺ s } ∪ { 10 k 1 l +1 /k , l ∈ ω } , then A ∞ = 2 ω \ { 10 ∞ } , which is Σ 0 1 -complet e.  Theor em 12 The set I := { ( α, A ) ∈ n ω × 2 n <ω /α ∈ A ∞ } is Σ 1 1 -complet e. In fact, (a) (O. F inkel , see [F1]) T her e exist s A 0 ⊆ 2 <ω suc h that A ∞ 0 is Σ 1 1 -complet e. (b) Ther e e xists α 0 ∈ 2 ω suc h that I α 0 is Σ 1 1 -complet e. 9 Pro of. (a) W e set L := { 2 , 3 } and T := { τ ⊆ 2 <ω × L/ ∀ ( u, ν ) ∈ 2 <ω × L [( u, ν ) / ∈ τ ] or [( ∀ v ≺ u ∃ µ ∈ L ( v , µ ) ∈ τ ) and (( u, 5 − ν ) / ∈ τ ) and ( ∃ ( ε, π ) ∈ 2 × L ( uε, π ) ∈ τ )] } . The set T i s the set of pruned trees over 2 with label s in L . It is a closed subset of 2 2 <ω × L , thus a Polish space. T hen we set σ := { τ ∈ T / ∃ ( u , ν ) ∈ 2 ω × L ω [ ∀ m ( u ⌈ m, ν ( m )) ∈ τ ] and [ ∀ p ∃ m ≥ p ν ( m ) = 3] } . • Then σ ∈ Σ 1 1 ( T ) . Let us show that it is complete. W e set T := { T ∈ 2 ω <ω /T is a tree } and I F := { T ∈ T /T is ill-fo unded } . It is a well-kno w n fact that T is a Polish space (it is a closed subset of 2 ω <ω ), and that I F is Σ 1 1 -complet e (see [K1]). It is enough to find a Borel reduc tion of I F to σ (see [K2]). W e define ψ : ω <ω ֒ → 2 <ω by the formula ψ ( t ) := 0 t (0) 10 t (1) 1 . . . 0 t ( | t |− 1) 1 , and Ψ : T → T by Ψ( T ) := { ( u, ν ) ∈ 2 <ω × L/ ∃ t ∈ T u ≺ ψ ( t ) and ν = 3 if u = ∅ , 2 + u ( | u | − 1) otherwise } ∪ { ( ψ ( t )0 k +1 , 2) /t ∈ T and ∀ q ∈ ω tq / ∈ T , k ∈ ω } . The map Ψ is Baire class one. Let us sho w that it is a reduction . If T ∈ I F , then let γ ∈ ω ω be such that γ ⌈ m ∈ T for each inte ger m . W e hav e ( ψ ( γ ⌈ m ) , 3) ∈ Ψ( T ) . L et u be the limit of ψ ( γ ⌈ m ) and ν ( m ) := 2 + u ( m − 1) (r esp., 3 ) if m > 0 (resp ., m = 0 ). These objects sho w that Ψ( T ) ∈ σ . Con versel y , we hav e T ∈ I F if Ψ( T ) ∈ σ . • If τ ∈ T and m ∈ ω , then w e enu merate τ ∩ (2 m × L ) := { ( u m,τ 1 , ν m,τ 1 ) , . . . , ( u m,τ q m,τ , ν m,τ q m,τ ) } in the lexi cographic ordering. W e define ϕ : T ֒ → 5 ω by the formula ϕ ( τ ) := ( u 0 ,τ 1 ν 0 ,τ 1 . . . u 0 ,τ q 0 ,τ ν 0 ,τ q 0 ,τ 4)( u 1 ,τ 1 ν 1 ,τ 1 . . . u 1 ,τ q 1 ,τ ν 1 ,τ q 1 ,τ 4) . . . The set A 0 will be made of finite subs equences of sen tences in ϕ [ T ] . W e set A 0 := { u m,τ q +1 ν m,τ q +1 . . . u p,τ r ν p,τ r /τ ∈ T , m + 1 < p, 0 ≤ q ≤ q m,τ , 1 ≤ r ≤ q p,τ , [( m = 0 and q = 0) or ( q > 0 and ν m,τ q = 3 and u m,τ q ≺ u p,τ r )] , ν p,τ r = 3 } (with th e con vention u m,τ q m,τ +1 ν m,τ q m,τ +1 = 4 ). It is cle ar that ϕ is continu ous, and it is enou gh to se e that it reduces σ to A ∞ 0 . So let us assume that τ ∈ σ . This means the existe nce of an infinite branch in the tree with infinitely many 3 labels . W e cut ϕ ( τ ) after the first 3 label of the branch corr espondin g to a sequ ence of length m > 1 . Then we cut after the first 3 label corresp onding to a sequence of length at least m + 2 of the branch. And so on. T his clea rly gi ves a decompo sition of ϕ ( τ ) into words in A 0 . If such a decompositi on exis ts, then the first word is u 0 ,τ 1 ν 0 ,τ 1 . . . u p 0 ,τ r 0 ν p 0 ,τ r 0 , and the second is u p 0 ,τ r 0 +1 ν p 0 ,τ r 0 +1 . . . u p 1 ,τ r 1 ν p 1 ,τ r 1 . So we ha ve u p 0 ,τ r 0 ≺ 6 = u p 1 ,τ r 1 . And so on. T his gi ves an in finite branch w ith infinitely many 3 labels. 10 • By Proposit ion 11, we can also ha ve A 0 ⊆ 2 <ω . (b) Let α 0 := 10 10 2 10 3 . . . , ( q l ) be the sequence of prime numbe rs: q 0 := 2 , q 1 := 3 , M : ω <ω → ω defined by M s := q s (0)+1 0 . . . q s ( | s |− 1)+1 | s |− 1 + 1 , φ : ω <ω → 2 <ω \ {∅} defined by the formulas φ ( ∅ ) := 1010 2 = 10 10 2 M ∅ and φ ( sm ) := 10 2 M s +1 10 2 M s +2 . . . 10 2 M sm , and Φ : 2 ω <ω → 2 n <ω defined by Φ( T ) := φ [ T ] . • It is clear that M sm > M s , and that M and φ are well defined and one-to-one . S o Φ is con tinuous: s ∈ φ [ T ] ⇔ ∃ t ( t ∈ T and φ ( t ) = s ) ⇔ s ∈ φ [ ω <ω ] and ∀ t ( t ∈ T or φ ( t ) 6 = s ) . If T ∈ I F , then we can find β ∈ ω ω such that φ ( β ⌈ l ) ∈ Φ( T ) for ea ch integer l . Thus α 0 = (10 10 2 M β ⌈ 0 )(10 2 M β ⌈ 0 +1 . . . 10 2 M β ⌈ 1 ) . . . ∈ (Φ( T )) ∞ . Con versel y , if α 0 ∈ ( Φ( T )) ∞ , then ther e exist t i ∈ T such that α 0 = φ ( t 0 ) φ ( t 1 ) . . . W e ha ve t 0 = ∅ , and, if i > 0 , then M t i ⌈| t i |− 1 = M t i − 1 ; from this we deduce that t i ⌈| t i | − 1 = t i − 1 , because M is one-to -one. So le t β be the limit of the t i ’ s. W e hav e β ⌈ i = t i , thus β ∈ [ T ] and T ∈ I F . Thus Φ ⌈T reduce s I F to I α 0 . T herefo re thi s last set is Σ 1 1 -complet e. Indeed, it is clear that I is Σ 1 1 : α ∈ A ∞ ⇔ ∃ β ∈ ω ω [( ∀ m > 0 β ( m ) > 0) and ( ∀ q ∈ ω π ( α, β , q ) ∈ A ) ] . Finally , the map fro m T into n ω × 2 n <ω , which ass ociates ( α 0 , Φ ( T )) to T clearly reduc es I F to I . So I is Σ 1 1 -complet e.  Remark. This proof sho ws that if α = s 0 s 1 . . . and ( s i ) is an antichai n for the extension ordering, then I α is Σ 1 1 -complet e (here we ha ve s i = 10 2 i +1 10 2 i +2 ) . T o see it, it is enough to notice that φ ( ∅ ) = s 0 and φ ( sm ) = s M s . . . s M sm − 1 . S o I α is Σ 1 1 -complet e for a dens e set of α ’ s. W e w ill deduce from this some true co-analy tic sets. But we need a lemma, which has its own interes t. Lemma 13 (a) The set A ∞ is Bor el if and on ly if ther e exis t a Bor el function f : n ω → ω ω suc h that α ∈ A ∞ ⇔ ( ∀ m > 0 f ( α )( m ) > 0) and ( ∀ q ∈ ω π ( α, f ( α ) , q ) ∈ A ) . (b) Let γ ∈ ω ω and A ⊆ n <ω . T hen A ∞ ∈ ∆ 1 1 ( A, γ ) if and only if, for α ∈ n ω , we have α ∈ A ∞ ⇔ ∃ β ∈ ∆ 1 1 ( A, γ , α ) [( ∀ m > 0 β ( m ) > 0) and ( ∀ q ∈ ω π ( α, β , q ) ∈ A )] . Pro of. The “if” directio ns in (a) and (b) are clear . W e hav e seen in the proof of Proposition 4 the “if” directi on of the equiv alences (the exi stence of an arbitrar y β is necessary and suf ficient). So let us sho w the “only if” directions . 11 (a) W e define f : n ω → ω ω by the formula f ( α ) := 0 ∞ if α / ∈ A ∞ , and, otherwis e, f ( α )(0) := min { p ∈ ω /α ⌈ ( p + 1) ∈ A and α − α ⌈ ( p + 1) ∈ A ∞ } , f ( α )( r + 1) := min { k > 0 / [ α − α ⌈ (1 + Σ j ≤ r f ( α )( j ))] ⌈ k ∈ A and α − α ⌈ ( k + 1 + Σ j ≤ r f ( α )( j )) ∈ A ∞ } . W e get π ( α, f ( α ) , 0) = α ⌈ f ( α )(0) + 1 ∈ A and, if q > 0 , π ( α, f ( α ) , q ) = ( α (1 + Σ j 0 β ( m ) > 0) and ( ∀ q ∈ ω π ( α, β , q ) ∈ A ) } , which is a closed subse t of X × Y . The projecti on Π X [ F ∩ ( X × N s )] is Borel if A ∞ is Borel , since it is { S ∗ γ /S ∈ ( A ∩ n s (0)+1 ) × Π 0 0 β ( m ) > 0) an d ( ∀ q ∈ ω π ( α, β , q ) ∈ A )]) . This sh ows that Π is Π 1 1 . The same a rgu ment works with Σ . F rom this we can ded uce that Σ 1 is Π 1 1 , if we forg et γ and tak e the section of Σ at θ ∈ W O ∩ ∆ 1 1 such that | θ | = 1 . Similarly , Σ ξ and Π ξ are co-ana lytic if ξ ≥ 1 . For getting θ , we see that the relation “ A ∞ ∈ ∆ 1 1 ( A, γ ) ” is Π 1 1 . • Let us look at the proo f of T heorem 12. W e will sho w that if ξ ≥ 1 (resp., ξ ≥ 2 ), then Σ ξ \ I α 0 (resp., Π ξ \ I α 0 ) is a true co-anal ytic set. T o do this, we will re duce W F to Σ ξ \ I α 0 (resp., Π ξ \ I α 0 ) in a Borel way . W e change the definition of Φ . W e set t ⊆ α 0 ⇔ ∃ k t ≺ α 0 − α 0 ⌈ k , E := { ( α 0 ⌈ p ) r /p ∈ ω \ { 2 } , r ∈ n \ { α 0 ( p ) }} , F := { U ∗ 6⊆ α 0 /U ∈ φ [ T ] <ω } , Φ ′ ( T ) := φ [ T ] ∪ { s ∈ n <ω / ∃ t ∈ E ∪ F t ≺ s } . This time, Φ ′ is Baire class one, since s ∈ Φ ′ ( T ) ⇔ s ∈ φ [ T ] or ∃ t ∈ E t ≺ s or ∃ U ∈ (2 <ω ) <ω ( ∀ j < | u | U ( j ) ∈ φ [ T ]) and U ∗ 6⊆ α 0 and U ∗ ≺ s. The proof of Theorem 12 remains va lid, since if α 0 ∈ (Φ ′ ( T )) ∞ , then the decompos itions of α 0 into words of Φ ′ ( T ) are actuall y dec omposition s into words of φ [ T ] . 13 • Let us sho w that (Φ ′ ( T )) ∞ ∈ Σ 0 1 ∩ ∆ 1 1 (Φ ′ ( T )) if T ∈ W F . T he set (Φ ′ ( T )) ∞ is [ S ∈ φ [ T ] <ω ,l ∈ n \{ 1 } ,m ∈ n \{ 0 } [( [ s/ ∃ t ∈ F t ≺ s N S ∗ s ) ∪ N S ∗ l ∪ N S ∗ 1 m ∪ ( N S ∗ 101 \ { S ∗ α 0 } )] . If α ∈ n ω , then α co ntains infinitely man y l ∈ n \ { 1 } or finish es with 1 ∞ . As 1 2 and the sequences beg inning with l are in Φ ′ ( T ) , the cl open sets are subse ts of (Φ ′ ( T )) ∞ since φ [ T ] and th e seq uences beg inning with t ∈ F , l or 1 m are in Φ ′ ( T ) . If α ∈ N S ∗ 101 \ { S ∗ α 0 } , t hen let p ≥ 3 be maximal such that α ⌈ ( | S ∗ | + p ) = S ∗ ( α 0 ⌈ p ) . W e hav e α ∈ (Φ ′ ( T )) ∞ since the sequenc es be ginning with ( α 0 ⌈ p ) r are in Φ ′ ( T ) . Thus we get the inclusion into (Φ ′ ( T )) ∞ . If α ∈ (Φ ′ ( T )) ∞ , then α = a 0 a 1 . . . , w here a i ∈ Φ ′ ( T ) . Either for all i we hav e a i ∈ φ [ T ] . In this case, there is i such that a 0 . . . a i 6⊆ α 0 , otherwise we could find k with α 0 − α 0 ⌈ k ∈ (Φ( T )) ∞ . But this contradi cts th e fac t that T ∈ W F , as i n the proof of Theo rem 12. So we hav e α ∈ S ∃ t ∈ F t ≺ s N s . Or there exists i m inimal such that a i / ∈ φ [ T ] . In this case, - Either ∃ t ∈ E t ≺ a i and α ∈ S S ∈ φ [ T ] <ω ,l ∈ n \{ 1 } ,m ∈ n \{ 0 } [ N S ∗ l ∪ N S ∗ 1 m ∪ ( N S ∗ 101 \ { S ∗ α 0 } )] , - Or ∃ t ∈ F t ≺ a i and α ∈ S S ∈ φ [ T ] <ω S s/ ∃ t ∈ F t ≺ s N S ∗ s . From this we dedu ce th at ( Φ ′ ( T )) ∞ is Σ 0 1 . Finally , we ha ve α ∈ (Φ ′ ( T )) ∞ ⇔  ∃ t ∈ n <ω ∃ b ∈ ω <ω [( | t | = 1 + Σ j < | b | b ( j )) and ( ∀ 0 < m < | b | b ( m ) > 0) and ( ∀ q < | b | π ( t 0 ∞ , b 0 ∞ , q ) ∈ Φ ′ ( T ))] and [ ∃ l ∈ n \ { 1 } tl ≺ α or t 1 2 ≺ α ] . This sho ws that (Φ ′ ( T )) ∞ is ∆ 1 1 (Φ ′ ( T )) . Therefore , Φ ′ ⌈T reduce s W F to Σ ξ \ I α 0 if ξ ≥ 1 , an d to Π ξ \ I α 0 if ξ ≥ 2 . So these sets are true co-ana lytic set s. But Σ 1 ∩ I α 0 is Π 1 1 , by Lemma 13 . As Σ 1 \ I α 0 = Σ 1 \ ( Σ 1 ∩ I α 0 ) , Σ 1 is not Borel . Thus Σ is not Borel, as before. T he ar gument is similar for Σ ξ , Π ξ ( ξ ≥ 2 ) and Π . And for ∆ too.  Question. Does A ∞ ∈ ∆ 1 1 imply A ∞ ∈ ∆ 1 1 ( A ) ? Probably not. If the answer is positi ve, ∆ , and more general ly Σ ξ (for ξ ≥ 1 ) and Π ξ (for ξ ≥ 2 ) are tru e co-analytic sets. Remark. In any c ase, ∆ is Σ 1 2 becaus e “ A ∞ ∈ ∆ 1 1 ” is equi valen t to “ ∃ γ ∈ ω ω A ∞ ∈ ∆ 1 1 ( A, γ ) ”. This ar gument sho ws that Σ ξ and Π ξ are Σ 1 2 ( θ ) , where θ ∈ W O satisfies | θ | = ξ . W e can s ay more abo ut Π 1 : it is ∆ 1 2 . Indeed, in [St2] we hav e the follo wing characteriza tion: A ∞ ∈ Π 0 1 ⇔ ∀ α ∈ n ω [ ∀ s ∈ n <ω ( s ≺ α ⇒ ∃ S ∈ A <ω s ≺ S ∗ )] ⇒ α ∈ A ∞ . This gi ves a Π 1 2 definitio n of Π 1 . T he same fac t is true for Σ 1 : Pro position 16 Σ 1 and Π 1 ar e co-no wher e dense ∆ 1 2 \ D 2 ( Σ 0 1 ) subsets of 2 n <ω . If ξ ≥ 2 , then Σ ξ and Π ξ ar e co-nowher e dense Σ 1 2 \ D 2 ( Σ 0 1 ) subsets of 2 n <ω . ∆ is a co-nowher e den se Σ 1 2 \ D 2 ( Σ 0 1 ) subset of 2 n <ω . 14 Pro of. W e hav e seen that Σ 1 is Σ 1 2 ; it is also Π 1 2 becaus e A ∞ ∈ Σ 0 1 ⇔ ∀ α ∈ n ω α / ∈ A ∞ or ∃ s ∈ n <ω [ s ≺ α and ∀ β ∈ n ω ( s 6≺ β or β ∈ A ∞ )] . By Propositi on 4, Π 0 is co-no where d ense, and it is a su bset of Σ ξ ∩ Π ξ ∩ ∆ . So Σ ξ , Π ξ and ∆ are co-no where dense, and it remain s to se e that they are not open. It is enough to notice tha t ∅ is not in their interi or . Look at the pro of of Theorem 12; i t sho ws that for ea ch integ er m , there is a su bset A m of { s ∈ 5 <ω / | s | ≥ m } such that A ∞ m / ∈ ∆ 1 1 . B ut the ar gument in the pro of of Proposition 11 sho ws that we can ha ve th e same thing in n <ω for each n ≥ 2 . This gi ves the resul t becau se the seq uence ( A m ) tends to ∅ .  W e can say a bit more abou t Π 1 and Σ 2 : Pro position 17 Π 1 , Π 1 and Σ 2 ar e Σ 0 2 -har d (so the y ar e not Π 0 2 ). Pro of. Conside r the map φ define d in the proof of Propo sition 5. B y Proposit ion 2, if γ ∈ P f , then φ ( γ ) ∞ is Π 0 1 . Moreo ver , as φ ( γ ) is an antichain for the extensio n ordering, the decomposition into words of φ ( γ ) is unique . This sho ws that φ ( γ ) ∞ is ∆ 1 1 , because α ∈ φ ( γ ) ∞ ⇔ ∃ β ∈ ∆ 1 1 ( α ) [( ∀ m > 0 β ( m ) > 0) an d ( ∀ q ∈ ω π ( α, β , q ) ∈ φ ( γ )) ] . So φ ( γ ) ∈ Π 1 if γ ∈ P f . So the preimage o f an y of the se ts in the s tatement by φ is P f , an d the r esult follo ws.  4 Which sets are ω -powers? No w w e come to Quest ion (3). Let u s spec ify what we m ean by “codes for Γ -sets”, where Γ is a gi ven class, and fix some notati on. • For the B orel classe s, we will es sentially co nsider the 2 ω -uni versal sets used in [K1] (see Theor em 22.3). For ξ ≥ 1 , U ξ , A (resp. U ξ , M ) is 2 ω -uni versal for Σ 0 ξ ( n ω ) (resp. Π 0 ξ ( n ω ) ). So we hav e - U 1 , A = { ( γ , α ) ∈ 2 ω × n ω / ∃ p ∈ ω γ ( p ) = 0 and s n p ≺ α } , where ( s n p ) p enumerat es n <ω . - U ξ , M = ¬ U ξ , A , for each ξ ≥ 1 . - U ξ , A = { ( γ , α ) ∈ 2 ω × n ω / ∃ p ∈ ω (( γ ) p , α ) ∈ U η, M } if ξ = η + 1 . - U ξ , A = { ( γ , α ) ∈ 2 ω × n ω / ∃ p ∈ ω (( γ ) p , α ) ∈ U η p , M } if ξ is the limit of the strictly increasing sequen ce of odd ordina ls ( η p ) . • For the cla ss Σ 1 1 , we fix some bijectio n p 7→ (( p ) 0 , ( p ) 1 ) between ω and ω 2 . W e set ( γ , α ) ∈ U ⇔ ∃ β ∈ 2 ω ( ∀ m ∃ p ≥ m β ( p ) = 1) and ( ∀ p [ γ ( p ) = 1 or s 2 ( p ) 0 6≺ β or s n ( p ) 1 6≺ α ]) . It is n ot hard to see that U is 2 ω -uni versal for Σ 1 1 ( n ω ) , and we us e it here beca use of the compactn ess of 2 ω × n ω , rathe r than the ω ω -uni versal set for Σ 1 1 ( n ω ) gi ven in [K1] (see Theorem 14.2). 15 • For the class ∆ 1 1 , it i s dif ferent be cause ther e is no uni versal s et. But we c an us e the Π 1 1 set of codes D ⊆ 2 ω for the Borel se ts in [K1] (se e Theorem 3 5.5). W e may assume t hat D , S and P are ef fectiv e, by [M]. • The sets we are interes ted in are the follo wing: A ξ := { γ ∈ 2 ω / U ξ , A γ is an ω -po wer } , M ξ := { γ ∈ 2 ω / U ξ , M γ is an ω -power } B := { d ∈ D /D d is an ω -power } , A := { γ ∈ 2 ω / U γ is an ω -power } . As we mention ned in the int roduction , Lemma 13 is also relat ed to Question (3). A rough ans wer to this question is Σ 1 3 . Indeed, w e ha ve, for γ ∈ 2 ω , γ ∈ A ⇔ ∃ A ∈ 2 n <ω ∀ α ∈ n ω ([( γ , α ) / ∈ U or α ∈ A ∞ ] and [ α / ∈ A ∞ or ( γ , α ) ∈ U ]) . W ith L emma 13, we ha ve a better estimatio n of the comple xity of B : it is Σ 1 2 . Indeed, for d ∈ D , D d is an ω -power ⇔ ∃ A ∈ 2 n <ω ∀ α ∈ n ω ([( d, α ) / ∈ S or ∃ β ∈ ∆ 1 1 ( A, d, α ) [( ∀ m > 0 β ( m ) > 0) and ( ∀ q ∈ ω π ( α, β , q ) ∈ A ) ]] and [ α / ∈ A ∞ or ( d, α ) ∈ P ]) . This arg ument also shows that A ξ and M ξ are Σ 1 2 . W e can say more about thes e tw o sets. Pro position 18 If 1 ≤ ξ < ω 1 , then A ξ and M ξ ar e Σ 1 2 \ D 2 ( Σ 0 1 ) co-mea ger subset s of 2 ω . If mor eover ξ = 1 , then the y ar e co-nowher e den se. Pro of. W e set E 1 := { γ ∈ 2 ω / U 1 , A γ = n ω } , E η +1 := { γ ∈ 2 ω / ∀ p ( γ ) p ∈ E η } if η ≥ 1 , and E ξ := { γ ∈ 2 ω / ∀ p ( γ ) p ∈ E η p } (wher e ( η p ) is a strictly incre asing seque nce of odd ordinals cofinal in the limit ordin al ξ ). If s ∈ 2 <ω , then we set γ ( p ) = s ( p ) if p < | s | , 0 otherwis e. Then s ≺ γ and U 1 , A γ = n ω , so E 1 is dense . If γ 0 ∈ E 1 , then for all α ∈ n ω we can find an integer p such that γ 0 ( p ) = 0 and s n p ≺ α . By compactness of n ω we can find a finite subset F of { p ∈ ω /γ 0 ( p ) = 0 } such that for each α ∈ n ω , s n p ≺ α for some p ∈ F . Now { γ ∈ 2 ω / ∀ p ∈ F γ ( p ) = 0 } is an open neighb orhood of γ 0 and a subset of E 1 . So E 1 is an open subset of 2 ω . Now the map γ 7→ ( γ ) p is contin uous an d open, so E η +1 and E ξ are dens e G δ subset s of 2 ω . Then we noti ce tha t E ξ is a su bset of { γ ∈ 2 ω / U ξ , A γ = n ω } (res p., { γ ∈ 2 ω / U 1 , A γ = ∅} ) if ξ is odd (re sp., ev en). Indeed, this is clear fo r ξ = 1 . Then we use the formulas U η +1 , A γ = S p ¬ U η, A ( γ ) p and U ξ , A γ = S p ¬ U η p , A ( γ ) p , and by ind uction we are done. As ∅ an d n ω are ω -powers, we get the resul ts about Baire ca tegory . N o w it re mains to see that A ξ and M ξ are not open. But by induction again 1 ∞ ∈ A ξ ∩ M ξ , so it is enou gh to see that 1 ∞ is not in the interior of these sets. • Let us sho w that, for O ∈ ∆ 0 1 ( n ω ) \ {∅ , n ω } and for each integer m , we can find γ , γ ′ ∈ 2 ω such that γ ( j ) = γ ′ ( j ) = 1 for j < m , U ξ , A γ = O and U ξ , M γ ′ = O . For ξ = 1 , w rite O = S p N s n q k , where q k ≥ m . Let γ ( q ) := 0 if there exis ts k su ch th at q = q k , γ ( q ) := 1 otherwis e. The same argumen t appl ied to ˇ O gi ves the complete result for ξ = 1 . 16 No w w e ar gue by induct ion. Let γ p ∈ 2 ω be such th at γ p ( q ) = 1 for < p, q >< m and U η, M ( γ ) p = O . Then de fine γ by γ ( < p , q > ) := γ p ( q ) ; we hav e γ ( j ) = 1 if j < m and U η +1 , A γ = S p U η, M ( γ ) p = O . The arg ument w ith ˇ O stil l works. The ar gument is similar for limit ordinals. • Now we apply this fact to O := N (0) . This giv es γ p , γ ′ p ∈ N 1 p such that U ξ , A γ p = N (0) and U ξ , M γ ′ p = N (0) . But ( γ p ) , ( γ ′ p ) tend to 1 ∞ , γ p / ∈ A ξ and γ ′ p / ∈ M ξ .  Cor ollary 19 A 1 is ˇ D 2 ( Σ 0 1 ) \ D 2 ( Σ 0 1 ) . In particular , A 1 is ˇ D 2 ( Σ 0 1 ) -comple te. Pro of. By the precedi ng proof, it is enough to see that A 1 \ { 1 ∞ } is open . S o let γ 0 ∈ A 1 \ { 1 ∞ } , p 0 in ω with γ 0 ( p 0 ) = 0 , and A 0 ⊆ n <ω with U 1 , A γ 0 = A ∞ 0 . If α ∈ n ω , then s n p 0 α ∈ U 1 , A γ 0 , so we can find m > 0 such that α − α ⌈ m ∈ A ∞ 0 ; thus there exists an inte ger p such that γ 0 ( p ) = 0 and s n p ≺ α − α ⌈ m . By compact ness of n ω , th ere are finite s ets F ⊆ ω \ { 0 } and G ⊆ { p ∈ ω /γ 0 ( p ) = 0 } such that n ω = S m ∈ F, p ∈ G { α ∈ n ω /s n p ≺ α − α ⌈ m } . W e set A γ := { s ∈ n <ω / ∃ p γ ( p ) = 0 and s n p ≺ s } for γ ∈ 2 ω , so tha t A ∞ γ ⊆ U 1 , A γ . Assume that γ ( p ) = 0 for each p ∈ G and let α ∈ U 1 , A γ . Let p 0 ∈ ω be suc h that γ ( p 0 ) = 0 and s n p 0 ≺ α . W e can find m 0 > 0 and p 1 ∈ G suc h that s n p 1 ≺ α − α ⌈ ( | s n p 0 | + m 0 ) , and α ⌈ ( | s n p 0 | + m 0 ) ∈ A γ . Then we can find m 1 > 0 and p 2 ∈ G su ch that s n p 2 ≺ α − α ⌈ ( | s n p 0 | + m 0 + | s n p 1 | + m 1 ) , and α ⌈ ( | s n p 0 | + m 0 + | s n p 1 | + m 1 ) − α ⌈ ( | s n p 0 | + m 0 ) ∈ A γ . And so on. Thus α ∈ A ∞ γ and { γ ∈ 2 ω / ∀ p ∈ G γ ( p ) = 0 } is a clo pen neighborho od of γ 0 and a subset of A 1 .  Pro position 20 A is Σ 1 3 \ D 2 ( Σ 0 1 ) and is co-nowher e dense. Pro of. L et U := { γ ∈ 2 ω / ∀ β ∈ 2 ω ∀ α ∈ n ω ∃ p [ γ ( p ) = 0 and s 2 ( p ) 0 ≺ β and s n ( p ) 1 ≺ α ] } . By compactn ess of 2 ω × n ω , U is a dense open subset of 2 ω . Moreo ver , if γ ∈ U , then U γ = ∅ , so U ⊆ A and A is co -no where dense. It re mains to see tha t A is not open, as in the proof of Prop osition 18. As U 1 ∞ = n ω , 1 ∞ ∈ A . Let p be an inte ger satisfying s 2 ( p ) 0 = ∅ and s n ( p ) 1 = 0 q . W e set γ p ( m ) := 0 if and only if m = p , an d also P ∞ := { α ∈ 2 ω / ∀ r ∃ m ≥ r α ( m ) = 1 } . Then ( γ p ) tends to 1 ∞ and we ha ve U γ p = { α ∈ n ω / ∃ β ∈ P ∞ ∀ m m 6 = p or s 2 ( m ) 0 6≺ β or s n ( m ) 1 6≺ α } = { α ∈ n ω / ∃ β ∈ P ∞ ( β , α ) / ∈ 2 ω × N 0 q } = ¬ N 0 q . So γ p / ∈ A .  5 Ordinal ranks and ω -powers. Notation. T he fact that the ω -po wers are Σ 1 1 implies the exi stence of a co -analytic rank on th e com- plement of A ∞ (see 34.4 in [K1]). W e w ill consider a natural one, defined as follo ws. W e set, for α ∈ n ω , T A ( α ) := { S ∈ ( A − ) <ω /S ∗ ≺ α } . T his is a tre e on A − , which is well fou nded if an d only if α / ∈ A ∞ . 17 The rank of this tree is the announc ed rank R A : ¬ A ∞ → ω 1 (see page 10 in [K1]): we hav e R A ( α ) := ρ ( T A ( α )) . L et φ : A − → ω be one-to-one , and ˜ φ ( S ) := ( φ [ S (0)] , . . . , φ [ S ( | s | − 1)]) for S ∈ ( A − ) <ω . This allows us to de fine the map Φ fro m the set of trees on A − into the set of trees on ω , w hich assoc iates { ˜ φ ( S ) /S ∈ T } to T . As ˜ φ is one-t o-one, Φ is con tinuous: t ∈ Φ( T ) ⇔ t ∈ ˜ φ [( A − ) <ω ] and ˜ φ − 1 ( t ) ∈ T . Moreo ver , T is well-fou nded if and only if Φ ( T ) is w ell-fou nded. Thus, if α / ∈ A ∞ , then we hav e ρ ( T A ( α )) = ρ ( Φ[ T A ( α )]) becau se ˜ φ is strictly monoto ne (see page 10 in [K1]). Thus R A is a co- analyt ic rank because the functi on fro m n ω into the set o f trees on ω <ω which a ssociates Φ[ T A ( α )] to α is continuous , and beca use the rank of the w ell-fou nded trees on ω defines a co-analyti c rank (see 34.6 in [K1]). W e set R ( A ) := sup { R A ( α ) /α / ∈ A ∞ } . By the boundedne ss theorem, A ∞ is Borel if and only if R ( A ) < ω 1 (see 34.5 and 35.23 in [K1]). W e can ask the questio n of the link between the complexit y of A ∞ and the o rdinal R ( A ) when A ∞ is Borel. Pro position 21 If ξ < ω 1 , r ∈ ω and R ( A ) = ω .ξ + r , then A ∞ ∈ Σ 0 2 .ξ +1 . Pro of. The reader should see [L] for operati ons on ordin als. • If 0 < λ < ω 1 is a limit ordinal, then let ( λ q ) be a strictly increas ing co-final se quence in λ , with λ q = ω .θ + q if λ = ω . ( θ + 1) , and λ q = ω .ξ q if λ = ω .ξ , w here ( ξ q ) is a stric tly incre asing co -final sequen ce in the limit ordin al ξ othe rwise. By induction , we define E 0 := { α ∈ n ω / ∀ s ∈ A − s 6≺ α } , E θ + 1 := { α ∈ n ω / ∀ s ∈ A − s 6≺ α or α − s ∈ E θ } , E λ := { α ∈ n ω / ∀ s ∈ A − s 6≺ α or ∃ q ∈ ω α − s ∈ E λ q } . • L et us sho w that E ω .ξ + r ∈ Π 0 2 .ξ +1 . W e may assume that ξ 6 = 0 and that r = 0 . If ξ = θ + 1 , th en E λ q ∈ Π 0 2 .θ +1 by induction h ypothesis , thus E ω .ξ + r ∈ Π 0 2 .θ +3 = Π 0 2 .ξ +1 . Otherwise, E λ q ∈ Π 0 2 .ξ q +1 by inducti on hypo thesis, th us E ω .ξ + r ∈ Π 0 ξ +1 = Π 0 2 .ξ +1 . • Let us show that if α ∈ A ∞ , then α / ∈ E ω .ξ + r . If ξ = r = 0 , it is clear . If r = m + 1 and s ∈ A − satisfies s ≺ α and α − s ∈ A ∞ , then we hav e α − s / ∈ E ω .ξ + m by indu ction hypot hesis, thus α / ∈ E ω .ξ + r . If r = 0 and s ∈ A − satisfies s ≺ α and α − s ∈ A ∞ , then we ha ve α − s / ∈ E λ q for each inte ger q , by induction hypothe sis, thus α / ∈ E ω .ξ + r . • Let s ∈ A − such that s ≺ α / ∈ A ∞ . W e ha ve ρ ( T A ( α − s )) = sup { ρ T A ( α − s ) ( t ) + 1 / t ∈ T A ( α − s ) } ≤ su p { ρ T A ( α ) (( s ) t ) + 1 / ( s ) t ∈ T A ( α ) } ≤ ρ T A ( α ) (( s )) + 1 ≤ ρ T A ( α ) ( ∅ ) < ρ ( T A ( α )) . 18 The first inequ ality comes from the fact that the m ap from T A ( α − s ) into T A ( α ) , which associates ( s ) t to t is strictly monotone (see page 10 in [K1]). W e ha ve ρ ( T A ( α )) ≥ [ sup { ρ ( T A ( α − s )) / s ∈ A − , s ≺ α } ] + 1 . Let us sho w that we actually hav e equality . W e ha ve ρ ( T A ( α )) = ρ T A ( α ) ( ∅ ) + 1 = sup { ρ T A ( α ) (( s )) + 1 / s ∈ A − , s ≺ α } + 1 . Therefore , it is enough to notice that if s ∈ A − and s ≺ α , then ρ T A ( α ) (( s )) ≤ ρ T A ( α − s ) ( ∅ ) . But this comes from the fac t that the m ap from { S ∈ T A ( α ) / S (0) = s } into T A ( α − s ) , which associates S − ( s ) to S , preserv es the exten sion ord ering (see p age 352 in [K1]). • L et us sho w that, if α / ∈ A ∞ , th en “ ρ ( T A ( α )) ≤ ω .ξ + r + 1 ” i s equ iv alent to “ α ∈ E ω .ξ + r ”. W e do it by i nduction on ω .ξ + r . If ξ = r = 0 , then i t is clear . If r = m + 1 , the n “ ρ ( T A ( α )) ≤ ω .ξ + r + 1 ” is equiv alent to “ ∀ s ∈ A − , s 6≺ α o r ρ ( T A ( α − s )) ≤ ω .ξ + m + 1 ”, by the preceding po int. This is equi valent to “ ∀ s ∈ A − , s 6≺ α or α − s ∈ E ω .ξ + m ”, which is equiv alent to “ α ∈ E ω .ξ + r ”. If r = 0 , then “ ρ ( T A ( α )) ≤ ω .ξ + r + 1 ” is equi valen t to “ ∀ s ∈ A − , s 6≺ α or there ex ists an inte ger q such that ρ ( T A ( α − s )) ≤ λ q + 1 ”. T his is equiv alent to “ ∀ s ∈ A − , s 6≺ α or there e xists an int eger q such that α − s ∈ E λ q ”, which is equi valen t to “ α ∈ E ω .ξ + r ”. • If α / ∈ A ∞ , then ρ ( T A ( α )) ≤ ω .ξ + r + 1 . By the preceding point, α ∈ E ω .ξ + r . Thus we hav e A ∞ = ¬ E ω .ξ + r ∈ Σ 0 2 .ξ +1 .  W e can find an upper bound for the rank R , for some Borel classes: Pro position 22 (a) A ∞ = n ω if and only if R ( A ) = 0 . (b) If A ∞ = ∅ , then R ( A ) = 1 . (c) If A ∞ ∈ ∆ 0 1 , then R ( A ) < ω , and ther e exists A p ⊆ 2 <ω suc h that A ∞ p ∈ ∆ 0 1 and R ( A p ) = p for each int e ger p . (d) If A ∞ ∈ Π 0 1 , then R ( A ) ≤ ω , and ( A ∞ / ∈ Σ 0 1 ⇔ R ( A ) = ω ) . Pro of. (a) If α / ∈ A ∞ , then ∅ ∈ T A ( α ) and ρ ( T A ( α )) ≥ ρ T A ( α ) ( ∅ ) + 1 ≥ 1 . (b) W e hav e T A ( α ) = {∅} for each α , and ρ ( T A ( α )) = ρ T A ( α ) ( ∅ ) + 1 = 1 . (c) By compactne ss, there exist s s 1 , . . . , s p ∈ n <ω such that A ∞ = S 1 ≤ m ≤ p N s m ∈ ∆ 0 1 . If α / ∈ A ∞ , then we ha ve N α ⌈ max 1 ≤ m ≤ p | s m | ⊆ ¬ A ∞ , th us ρ ( T A ( α )) ≤ m ax 1 ≤ m ≤ p | s m | + 1 < ω . So we get the first point. T o see the second one, we set A 0 := 2 <ω . If p > 0 , the n w e set A p := { 0 2 } ∪ [ q ≤ p { s ∈ 2 <ω / 0 2 q 1 ≺ s } ∪ { s ∈ 2 <ω / 0 2 p +1 ≺ s } . Then A ∞ p = S q ≤ p N 0 2 q 1 ∪ N 0 2 p +1 ∈ ∆ 0 1 . If α p := 0 2 p − 1 1 ∞ , then ρ ( T A p ( α p )) = p . If α / ∈ A ∞ p , then ρ ( T A p ( α )) ≤ p . 19 (d) If A ∞ ∈ Π 0 1 and α / ∈ A ∞ , then le t s ∈ n <ω with α ∈ N s ⊆ ¬ A ∞ . Then ρ ( T A ( α )) ≤ | s | + 1 . Thus R ( A ) ≤ ω . If A ∞ / ∈ Σ 0 1 , then w e hav e R ( A ) ≥ ω , by Propositi on 21. Thus R ( A ) = ω . Con versel y , we apply (c ).  Remark. Notice that it is not true that if the W adge cla ss < A ∞ > , havin g A ∞ as a co mplete set, is a subclass of < B ∞ > , then R ( A ) ≤ R ( B ) . Indeed, fo r A w e tak e the e xample A 2 in (c), and for B we tak e the example for Σ 0 1 that we met in the proof of P roposit ion 11. If we e xchange the roles of A and B , then w e see that th e con verse is also false. This exa mple A for Σ 0 1 sho ws that Proposition 21 is optimal fo r ξ = 0 sinc e R ( A ) = 1 and A ∞ ∈ Σ 0 1 \ Π 0 1 . W e can say more : it is not tru e that if A ∞ = B ∞ , th en R ( A ) ≤ R ( B ) . W e use again (c): we tak e A := A 2 and B := A \ { 0 2 } . W e hav e A ∞ = B ∞ = A ∞ 2 , R ( A ) = 2 and R ( B ) = 1 . Pro position 23 F or ea ch ξ < ω 1 , ther e e xists A ξ ⊆ 2 <ω with A ∞ ξ ∈ Σ 0 1 and R ( A ξ ) ≥ ξ . Pro of. W e use th e not ation in the proof of Theorem 15. Let T ∈ T , an d ϕ : T → T Φ ′ ( T ) ( α 0 ) defined by the formula ϕ ( s ) := ( φ ( s ⌈ 0) , . . . , φ ( s ⌈| s | − 1)) . Then ϕ is strictly m onoton e. If T ∈ W F , then α 0 / ∈ (Φ ′ ( T )) ∞ and T Φ ′ ( T ) ( α 0 ) ∈ W F . In this case , ρ ( T ) ≤ ρ ( T Φ ′ ( T ) ( α 0 )) = R Φ ′ ( T ) ( α 0 ) (see page 10 in [K1]). Let T ξ ∈ W F be a tree w ith rank at least ξ (see 34.5 and 34.6 in [K1]). W e set A ξ := Φ ′ ( T ξ ) . It is clear that A ξ is what we were lookin g for .  Remark. Let ψ : 2 n <ω → { T rees on n <ω } define d by ψ ( A ) := T A ( α 0 ) , and r : ¬ I α 0 → ω 1 defined by r ( A ) := ρ ( T A ( α 0 )) . Then ψ is co ntinuous, th us r is a Π 1 1 -rank on ψ − 1 ( { W ell-fou nded trees on n <ω } ) = ¬ I α 0 . By the boundedne ss theorem, the rank r and R are not bound ed on ¬ I α 0 . Propositio n 23 specifies this result. It sh ows th at R is no t bounded o n Σ 1 \ I α 0 . 6 The extension ordering . Pro position 24 W e equip A with the exte nsion ord ering. (a) If A ⊆ n <ω is an antichain , then A ∞ is in {∅} ∪ { n ω } ∪ [ Π 0 1 \ Σ 0 1 ] ∪ [ Π 0 2 ( A ) \ Σ 0 2 ] , and any of these cases is possibl e. (b) If A ⊆ n <ω has finite antic hains, th en A ∞ ∈ Π 0 2 (and is not Σ 0 2 in gene ral). Pro of. L et G := { α ∈ n ω / ∀ r ∃ m ∃ p ≥ r α ⌈ m ∈ [( A − ) p ] ∗ } . Then G ∈ Π 0 2 ( A ) and contains A ∞ . Con versel y , if α ∈ G , then we ha ve T A ( α ) ∩ ( A − ) p 6 = ∅ for each integer p , thus T A ( α ) is infinite. (a) If A is an anti chain, then each sequenc e in T A ( α ) has at most one extens ion in this tr ee addin g one to the length. Thus T A ( α ) is finite splitting. This implies that T A ( α ) has an infinite branch if α ∈ G , by K ¨ onig’ s lemma. Therefore A ∞ = G ∈ Π 0 2 ( A ) . 20 - If we tak e A := ∅ , then A is an antic hain and A ∞ = ∅ . - If we tak e A := { (0) , . . . , ( n − 1) } , then A is an antichain and A ∞ = n ω . - If A ∞ / ∈ {∅ , n ω } , then A ∞ / ∈ Σ 0 1 . Indeed , let α 0 / ∈ A ∞ and s 0 ∈ A − . By uniqueness of the decompo sition into words of A − , the sequenc e ( s n 0 α 0 ) n ⊆ n ω \ A ∞ tends to s ∞ 0 ∈ A ∞ . - If we tak e A := { (0) } , then A is an antic hain and A ∞ = { 0 ∞ } ∈ Π 0 1 \ Σ 0 1 . - If A is finite, then A ∞ is Π 0 1 \ Σ 0 1 or is in {∅ , n ω } , by the fac ts ab ove and Propo sition 2 . - If A is infini te, the n A ∞ / ∈ Σ 0 2 becaus e the map c in the proof o f P roposit ion 2 is an homeomorphis m and ( A − ) ω is not K σ . - If A := { 0 k 1 /k ∈ ω } , then A is an antichain and A ∞ = P ∞ , which is Π 0 2 \ Σ 0 2 . (b) The intersectio n of P ∞ with N 1 can be made with the chain { 10 k /k ∈ ω } . So let us assume that A has finite anticha ins. • Let us sho w that A is the u nion of a finite set and of a finite uni on of infinite subsets of sets of the form A α m := { s ∈ n <ω /s ≺ α m } . L et us enumera te A := { s r /r ∈ ω } . W e construct a sequence ( A m ) , finite or not, of subsets of A . W e do it by inductio n on r , to decide in which set A m the sequen ce s r is. First, s 0 ∈ A 0 . A ssume that s 0 , . . . , s r ha ve been put into A 0 , . . . , A p r , with p r ≤ r and A m ∩ { s 0 , . . . , s r } 6 = ∅ if m ≤ p r . W e choose m ≤ p r minimal such that s r +1 is compatible with all the sequ ences in A m ∩ { s 0 , . . . , s r } , we put s r +1 into A m and we set p r +1 := p r if possi ble. Otherwise, we put s r +1 into A p r +1 and we set p r +1 := p r + 1 . Let us sho w that there are only finitely many infinite A m ’ s. If A m is infinite, then there exists a unique sequence α m ∈ n ω such that A m ⊆ A α m . Let us ar gue by contrad iction: there ex ists an infinite sequence ( m q ) q such that A m q is infinite. Let t 0 be the common beginnin g of the α m q ’ s. There exis ts ε 0 ∈ n such that N t 0 ε 0 ∩ { α m q /q ∈ ω } is infinite. W e choose a sequence u 0 in A ext ending t 0 µ 0 , w here µ 0 6 = ε 0 . Then we do it again: let t 0 ε 0 t 1 be the common beginnin g of the elements of N t 0 ε 0 ∩ { α m q /q ∈ ω } . There exists ε 1 ∈ n su ch that N t 0 ε 0 t 1 ε 1 ∩ { α m q /q ∈ ω } is infinite . W e ch oose a seq uence u 1 in A exten ding t 0 ε 0 t 1 µ 1 , wher e µ 1 6 = ε 1 . The sequ ence ( u l ) is an infinite antich ain in A . B ut this is absurd . Now let us choose the longes t sequence in each none mpty finite A m ; this gi ves an antich ain in A and the result. • No w let α ∈ G . There are tw o cases. Either for ea ch m and for ea ch integ er k , α ⌈ k / ∈ [ A <ω ] ∗ or α − α ⌈ k 6 = α m . In this case, T A ( α ) is fi nite splitting. As T A ( α ) is infinite, T A ( α ) has an infinite branch witnessing that α ∈ A ∞ , by K ¨ o nig’ s lemma. O therwise, α ∈ S s ∈ [ A <ω ] ∗ ,m { sα m } , which is counta ble. Thus G \ A ∞ ∈ Σ 0 2 and A ∞ = G \ ( G \ A ∞ ) ∈ Π 0 2 .  7 Examples. • W e ha ve seen exampl es of subse ts A of 2 <ω such that A ∞ is complet e for the classes {∅} , { n ω } , ∆ 0 1 , Σ 0 1 , Π 0 1 , Π 0 2 and Σ 1 1 . W e will giv e some more e xamples, fo r some classes of Borel sets. Notice that to sho w that a set in such a no n self-dua l class is comple te, it is e nough to sh ow that it is true (see 21.E, 22.10 and 22.26 in [K1]). 21 • For the class Σ 0 1 ⊕ Π 0 1 := { ( U ∩ O ) ∪ ( F \ O ) / U ∈ Σ 0 1 , O ∈ ∆ 0 1 , F ∈ Π 0 1 } , we can take A := { s ∈ 2 <ω / 0 2 1 ≺ s or s = 0 2 or ∃ p ∈ ω 10 p 1 ≺ s } , since A ∞ = { 0 ∞ } ∪ S q N 0 2 q +2 1 ∪ N 1 \ { 10 ∞ } . • For the class ˇ D 2 ( Σ 0 1 ) := { U ∪ F / U ∈ Σ 0 1 , F ∈ Π 0 1 } , we can take Example 9 in [St2]: A := { s ∈ 2 <ω / 0 ≺ s or ∃ q ∈ ω (101) q 1 3 ≺ s or s = 10 2 } . W e hav e A ∞ = [ p ∈ ω [ N (10 2 ) p 0 ∪ ( [ q ∈ ω N (10 2 ) p (101) q 1 3 )] ∪ { (10 2 ) ∞ } , which is a ¬ D 2 ( Σ 0 1 ) set. T o wards a contrad iction, assu me th at A ∞ is D 2 ( Σ 0 1 ) : A ∞ = U 1 ∩ F = U ∪ F 2 , where the U ’ s are open and the F ’ s are close d. Let O be a clop en set separatin g ¬ U 1 from F 2 (see 22.C in [K1]). Then A ∞ = ( U ∩ O ) ∪ ( F \ O ) would be i n Σ 0 1 ⊕ Π 0 1 . If (10 2 ) ∞ ∈ O , then we would ha ve N (10 2 ) p 0 ⊆ O for some inte ger p 0 . But the seque nce ((10 2 ) p (1 2 0) ∞ ) p ≥ p 0 ⊆ O \ U and tends to (10 2 ) ∞ , which is absurd. If (10 2 ) ∞ / ∈ O , then w e would ha ve N (10 2 ) q 0 ⊆ ¬ O for so me inte ger q 0 . But the sequen ce ((10 2 ) q 0 (101) q 1 ∞ ) q ≥ q 0 ⊆ F \ O and tends to (10 2 ) q 0 (101) ∞ , which is absurd . • For the cla ss D 2 ( Σ 0 1 ) , we can take A := [ A <ω 1 ] ∗ \ [ A <ω 0 ] ∗ , where A 0 := { 0 10 , 01 2 } and A 1 := { 0 10 , 01 2 , 0 2 , 0 3 , 10 2 , 1 2 0 , 10 3 , 1 2 0 2 } . W e hav e A ∞ = A ∞ 1 \ A ∞ 0 . Indeed , as A ⊆ [ A <ω 1 ] ∗ , w e hav e A ∞ ⊆ A ∞ 1 . If α ∈ A ∞ 0 , then its decompo sition into words of A 1 is unique and made of words in A 0 . T hus α / ∈ A ∞ and A ∞ ⊆ A ∞ 1 \ A ∞ 0 . Con versel y , if α = a 0 a 1 . . . ∈ A ∞ 1 \ A ∞ 0 , with a i ∈ A − 1 , then there are two cases. Either there are infinitely many index es i (say i 0 , i 1 , . . . ) such that a i / ∈ A 0 . I n this case, the words a 0 . . . a i 0 , a i 0 +1 . . . a i 1 , . . . , ar e in A and α ∈ A ∞ . O r there exists a maximal index i such that a i / ∈ A 0 . In th is case, a 0 . . . a i 0 , 10 2 , 1 2 0 ∈ A , thus α ∈ A ∞ = A ∞ 1 \ A ∞ 0 . P ropos ition 2 sh ows that A ∈ D 2 ( Σ 0 1 ) . If A ∞ = U ∪ F , with U ∈ Σ 0 1 and F ∈ Π 0 1 , then we ha ve U = ∅ beca use A ∞ 1 is no where dense (e very sequen ce in A 1 contai ns 0 , thus the sequences in A ∞ 1 ha ve infinit ely many 0 ’ s). Thus A ∞ would be closed . But this cont radicts the fact that ((01 2 ) n 0 ∞ ) n ⊆ A ∞ and tends to (01 2 ) ∞ / ∈ A ∞ . T hus A ∞ is a true D 2 ( Σ 0 1 ) set. • For the class ˇ D 3 ( Σ 0 1 ) , we can tak e A := ([ A <ω 2 ] ∗ \ [ A <ω 1 ] ∗ ) ∪ [ A <ω 0 ] ∗ , where A 0 := { 0 2 } , A 1 := { 0 2 , 01 } , and A 2 := { 0 2 , 01 , 10 , 10 2 } . W e hav e A ∞ = ( A ∞ 2 \ A ∞ 1 ) ∪ A ∞ 0 . Indeed, as A ⊆ [ A <ω 2 ] ∗ , we ha ve A ∞ ⊆ A ∞ 2 . If α ∈ A ∞ 1 , then its decomposi tion into words of A − 2 is unique and made of words in A 1 . If moreov er α / ∈ A ∞ 0 , then it is clear that α / ∈ A ∞ and A ∞ ⊆ ( A ∞ 2 \ A ∞ 1 ) ∪ A ∞ 0 . Con versel y , it is clear that A ∞ 0 ⊆ A ∞ . If α = a 0 a 1 . . . ∈ A ∞ 2 \ A ∞ 1 , then the ar gument abov e still works . W e h av e to check that s := a 0 . . . a i 0 / ∈ [ A <ω 1 ] ∗ . It is clear if a i 0 = 10 . Otherwise, a i 0 = 10 2 and we ar gue by co ntradicti on. 22 The length of s is ev en and the decompos ition of s into words of A 1 is un ique. It finishes with 0 2 , and t he e ven coordi nates of the sequen ce s are 0 . Therefore, a i 0 − 1 = 0 2 or 10 ; we ha ve th e same th ing with a i 0 − 2 , a i 0 − 3 , . . . Because of the parity , some 0 remains at the beginni ng. But this is absurd. No w we hav e to check that a 0 . . . a i 0 / ∈ [ A <ω 1 ] ∗ . It is clea r if a i = 10 2 . Otherwise, a i = 10 and the ar gument above wor ks. Finally , we hav e to check that if γ ∈ A ∞ 1 , then γ − (0) ∈ A ∞ . There is a sequence p 0 , p 1 , . . . , finite or not, such that γ = (0 2 p 0 )(01 )(0 2 p 1 )(01 ) . . . 0 ∞ . T herefo re γ − (0) = (0 2 p 0 10)(0 2 p 1 10) . . . (0 2 ) ∞ ∈ A ∞ . If we set U i := ¬ A ∞ 2 − i , then we see that A ∞ ∈ ˇ D 3 ( Σ 0 1 ) . If α finishes with 1 ∞ , then α / ∈ A ∞ 2 ; thus A ∞ 2 is no where d ense, just like A ∞ . Thus if A ∞ = ( U 2 \ U 1 ) ∪ U 0 with U i open, then U 0 = ∅ . By uniqu eness of the decompo sition of a sentenc e in A ∞ i into wo rds of A i +1 , we see that A ∞ i is no where dense in A ∞ i +1 . So let x ∅ ∈ A ∞ 0 , ( x n ) ⊆ A ∞ 1 \ A ∞ 0 con ver ging to x ∅ , and ( x n,m ) m ⊆ A ∞ 2 \ A ∞ 1 con ver ging to x n . T hen x n,m ∈ U 1 , which is absurd . Thus A ∞ / ∈ D 3 ( Σ 0 1 ) . • For the cla ss ˇ D 2 ( Σ 0 2 ) , we can take A := { s ∈ 2 <ω / 1 2 ≺ s or s = (0) } . W e can write A ∞ = ( { 0 ∞ } ∪ [ p N 0 p 1 2 ) ∩ ( P f ∪ { α ∈ 2 ω / ∀ n ∃ m ≥ n α ( m ) = α ( m + 1) = 1 } ) . Then A ∞ / ∈ D 2 ( Σ 0 2 ) , otherwise A ∞ ∩ N 1 2 ∈ D 2 ( Σ 0 2 ) and would be a comeager subset of N 1 2 . W e could fi nd s ∈ 2 <ω with ev en length such that A ∞ ∩ N 1 2 s ∈ Π 0 2 . W e define a continuous functi on f : 2 ω → 2 ω by formulas f ( α )(2 n ) := α ( n ) if n > | s | +1 2 , (1 2 s )(2 n ) otherwise, and f ( α )(2 n + 1) := 0 if n > | s | 2 , (1 2 s )(2 n + 1) otherwise. It reduces P f to A ∞ ∩ N 1 2 s , whic h is absurd. 23 Summary of the complexity r esults in this paper: Baire categ ory comple xity | ξ = 1 ξ = 2 ξ ≥ 3 Σ 0 no where dense Π 0 1 \ Σ 0 1 Π 0 co-no where de nse Σ 0 1 \ Π 0 1 ∆ 1 co-no where de nse K σ \ Π 0 2 Σ ξ co-no where de nse Π 1 1 \ ∆ 1 1 Π 1 1 \ ∆ 1 1 Π ξ co-no where de nse Π 1 1 \ Π 0 2 Π 1 1 \ ∆ 1 1 Π 1 1 \ ∆ 1 1 ∆ co-n owhe re den se Π 1 1 \ ∆ 1 1 Σ ξ co-no where de nse ∆ 1 2 \ D 2 ( Σ 0 1 ) Σ 1 2 \ Π 0 2 Σ 1 2 \ D 2 ( Σ 0 1 ) Π ξ co-no where de nse ∆ 1 2 \ Π 0 2 Σ 1 2 \ D 2 ( Σ 0 1 ) Σ 1 2 \ D 2 ( Σ 0 1 ) ∆ co-no where de nse Σ 1 2 \ D 2 ( Σ 0 1 ) G ξ ( ξ ∈ ω ) Π 0 1 \ Σ 0 1 no where dense ˇ D ω ( Σ 0 1 ) \ D ω ( Σ 0 1 ) Π 1 1 \ D ω ( Σ 0 1 ) F Π 1 1 \ Π 0 2 A ξ co-meage r ˇ D 2 ( Σ 0 1 ) \ D 2 ( Σ 0 1 ) co-no where de nse Σ 1 2 \ D 2 ( Σ 0 1 ) Σ 1 2 \ D 2 ( Σ 0 1 ) M ξ co-meage r Σ 1 2 \ D 2 ( Σ 0 1 ) co-no where de nse Σ 1 2 \ D 2 ( Σ 0 1 ) Σ 1 2 \ D 2 ( Σ 0 1 ) B Σ 1 2 A co-no where de nse Σ 1 3 \ D 2 ( Σ 0 1 ) 8 Refere nces. 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