Wadge Hierarchy and Veblen Hierarchy Part I: Borel Sets of Finite Rank Author(s): J. Duparc Source: The Journal of Symbolic Logic, Vol. 66, No. 1 (Mar., 2001), pp. 56-86 Published by: Association for Symbolic Logic Stable URL: http://www.jstor.org/stable/2694911 Accessed: 04/01/2010 12:38
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http://www.jstor.org THE JOURNAL OF SYMBOLICLOGIC Voluime 66. Number 1, March 2001
WADGE HIERARCHY AND VEBLEN HIERARCHY PART I: BOREL SETS OF FINITE RANK
J. DUPARC
Abstract. We consider Borel sets of finite rank A C A'" where cardinality of A is less than some uncountable regular cardinal K5. We obtain a "normal form" of A, by finding a Borel set Q, such that A and Q continuously reduce to each other. In more technical terms: we define simple Borel operations which are homomorphic to ordinal sum. to multiplication by a countable ordinal, and to ordinal exponentiation of base K;*under the map which sends every Borel set A of finite rank to its Wadge degree.
?1. Introduction. We recall a tree T on A is a subset of ACIDthat is closed under
subsequences i.e., satisfies Vu E T Vv (v C u - v E T).When T is some tree, we write [T] to denote the set of its infinite branches: {a E AW: Vn E co (a [ n) E T}. By a Borel set we mean a Borel subset A of A'A,where AA (which contains at least two elements to avoid trivialities) is some well-ordered alphabet of cardinality less than some uncountable regular cardinal K, and the topology on A' is the product topology of the discrete topology on AA: A is open if and only if there exists some U C A"<-"such that A= {ua oAC, : u U}. As soon as it was introduced, the class of Borel sets (the closure by comple- mentation and countable union of the class of open sets) was set up in a hi- erarchy by Baire: V is the class of open sets, and by induction for 4 > 1, Lo; = {UnEOA,1 -An E U(<;V} where -A,] stands for the complement of the set An. The Baire hierarchy of height co, was later subdivided by Hausdorff and Kuratowski into K wil levels (they worked with K = wl) by introducing the sepa- rated difference operation[. But an ultimate refinementwas to come with Wadge's work based on the notion of inverse images of continuous functions. A continuousreduction of A to B is a continuous function f: AC-0 ACBsuch that A - f' B. It is denoted A < w B; intuitively it means that A is less complicated than B -strictly if there is no reduction of B to A; which is denoted Received January 30, 1997; revised May 20, 1999. ? 2001, Association for Symbolic Logic 0022-4812/01/6601 -0002/$4. I0 56 WADGEHIERARCHY AND VEBLENHIERARCHY 57 determinacy, states that two Borel sets have same Wadge degree if and only if they are Wadge equivalent or one is Wadge equivalent to the complement of the other [1]. We recall the definition of a self-dual set: a set A that satisfies A =jw -A. Willing to concentrate on non self dual sets we need to consider another definition of the Wadge degree that put apart the self dual ones. DEFINITION 1. o dj(0) = d,(- 0) 1 o dA = sup{dw(B) + 1: B non self dual A B sup V0 o ... o Vo(2) n Coa n (denoted 'CEosince it is the first fixpoint of the exponentiation of base K) while the Borel rank only runs through integers. And things get even much worse for Borel sets of infinite ranks: the Wadge degree of sets of Borel rank co climbs up to V1(2) (characterized also as the K"th fixpoint of the exponentiation of base K, therefore denoted fej. And when dealing with all wol Borel ranks the Wadge hierarchy involves sup,,(,, V, (2) distinct degrees. As a consequence, it is quite a problem to provide for each Wadge degree af a description of a Borel set Q(ca) of degree precisely a. A solution for K = col was given essentially by Wadge [9], and confirmed after ten years of work culminating with Louveau and Saint-Raymond [2], [3]. We have given a new solution in the case af < V1(2), that is for AO sets (and we are writing an extension to all Borel ranks). This solution is canonical because its characteristic feature is to give a Borel definition of Q(a) whose is isomorphic to the canonical description of the ordinal Oaitself. Let us quote Wadge in the conclusion of his Ph. D. Thesis: One goes from the standard definition of the Wadge degree to ours by: o if A is non self dual, only three possibilities occur: (a) df A = 2.n with n integer, then d'cA = n + 1 (b) d' A = A + 2n with n integer and A limit (necessarily cof(A) = IKholds), then d'A = A + n (c) d WA = A+2n +I1 with n integer and A limit (necessarilycof(A) < IKholds), then dfA = A+n o if A is self dual, also there is only three cases: (a) d' A = 2.n + 1 with n integer, then d,?UA= n + 1 (b) diA = A + 2n with n integer and A limit (necessarily cof(X) < I%holds), then d?A = A + n (c) divA = A + 2n + 1 with n integer andA limit (necessarily cof(A)K = %holds), thenOA d A = X+n+ 1 58 J. DUPARC "It would also be very interesting to know of some "inductive" definition of the degrees, which allows a given degree to be constructed from those below it. This might, for example, allow properties of subsets of 0%oto be proved by induction on the degree of the set involved" ([9], p. 323) Wadge worked in the Baire space co;w he defined set theoretic counterparts of the sum and < wol-multiplicationso that on the basis of the empty set (which degree is 1) he was able to generate sets of degree al for any al < c<' but couldn't go any further. In particular, with his method, he couldn't reach a p-complete set (which degree is precisely <' ). We hope this work is an attempt to respond to the wish formulated by Wadge. ?2. Away with self-dual sets, the conciliating Wadge hierarchy. We say a set A is conciliating if it is a subset of A<', where AA is of cardinality less than K. And given two conciliating sets A, B, we define the conciliatory version C(A, B) of the Wadge game W(A, B), which is this game when the sequences a, ,8 produced by the players may be finite. Remember that the Wadge game W(A, B) is a two players game, where I begins and alternatively player I chooses a letter from AA, player H a letter from AB. Thus after co moves, I determines cl E AA and H determines /1 e A`; and H wins if and only if al E A < ,8 E B otherwise it is I who wins. These rules are designed so that A A WC -A, because the strategy that consists in skipping at first move and then copying H's run is winning for player I in C (-A, A). (b) The conciliatory Wadge hierarchy joins the previous one when sets are crammed with a specific letter "b" (for blanks) so as to turn them into sets of infinite sequences. That is, given a conciliating set A C A Given A, B two conciliating sets, it is easy to see that a player wins C(A, B) if and only if the same player wins W(Ab,Bh): so, A < B a Ah < Bh. and in particular, Ah #w -A . In fact, there's an obvious isomorphism between strategies a for C(A, B) and Uh for W(Ah, Bh): a passes if and only if a,, chooses a "blank". (c) But we also get much more in: THEOREM3. For every Borel set offinite rank A C AWO,A is non self-dual if and only if there is an alphabet AB and a conciliatory set B C A<' such that A - Bh. Furthermore,one can get AB=AA and B n AA A (so that B is just A extended by addingfinite sequences). The proof of theorem 3 will be indicated at the end of the study, for it needs the further operations to be achieved. Now there's a trivial procedure which derives the structure of the self-dual classes from the non self-dual ones - see Proposition 12. So we concentrate on non self-dual sets. We define the Wadge degree of a conciliating set A, denoted dcA, simply as the ordinal dWA". It is a consequence of theorem 3 that for all B C A<' if B" is Borel of finite rank, then d'cB = sup{(d?C) + 1: C conciliatory 'eo = sup VOo ... o Vo(2), n Ew( n ca has an "iterated Cantor normal form of base K" which describes a, using < K parameters and only "arithmetical" operations on ordinals. The sum: (f8,y) | ) ,8 + y, the product: (,f, v) |-) ,B v (for each v, 0 < v < A) and the exponentiation: | - Kf. To each of these operations we associate an operation defined on conciliating sets. Sum: (A, B) | ) A + B, multiplication: (A, v) | ) A * v (for each v, 0 < v < K) and exponentiation: A |-* A-. Remember that in our previous definition d'0 is 1, we will always write 0 for 0 regarded as a subset of A`w as well as a subset of A'W. So starting with d??C= 1, we show: THEOREM 4. GivenA C A<', B C A (a) dc?(A + B) = (dcoA) + (dc B) (b) dc?(A * v) = (dc A).v (c) dcA Vo(d?A) - (here proved only for d ?A < VI (2)) i.e., dcA where -1 if cof(sup{A dc?A:< A=Oor A is limit})= 0 E = d0 if cof(sup{f < dA: A = 0 or A is limit}) = K 1+1 ifco < cof(sup{,{ < d0A: A = Oor A is limit}) < K. PROOF. (a) is Proposition 17, (b) is Lemma 21 and (c) is Lemma 37. -1 60 J. DUPARC In particular, for every cO < 'eo, this yields the following way to define a Borel set Q(ca)b of degree a: Oahas an iterated Cantor normal form: n Of = E: Kdi.vj, with Too > do > . .. > Xn and Vi < n 0 < vi < X i=0 where Xj itself is written in normal form, and so on. Then, in order to get Q(ca), evaluate this "iterated polynomial of ordinals" when the ordinal 1 is replaced by the empty set, and the ordinal operations are replaced by their above set theoretic counterparts. The above results (a), (b) and (c) guarantee that the resulting Borel set Q(ca)b is of degree a, provided that the correction introduced in (c) by the variable value E E { -1, 0, 1} is duly taken into account. This will be precisely defined and proved later in the paper see Definition 32 and Theorem 33. REMARKS 5. (a) (i) Given A and A C ACIDit is very easy to build A' C (AAU A)w0such that A' w A. (ii) Same remark holds with B C A<', B' C (AB U A)?w, B' =c. B. In case of conciliatory sets, every letter E A \ AB can be regarded as a skip i.e., B' can be defined as follows: for /3E (AB U A)?<0, let /3E B' X /3' E B, where ,/' = /( /A -. AB). That is, /3' is /3,except all letters not in AB are removed. So in the sequel we assume every alphabet is as enriched as desired. (b) We always take b for "blanks" and s for "skips" as new letters not in the considered alphabets; and of course s 7&b. We also write co* for co \ {0}; so A<'* denotes the set of nonempty finite sequences of elements of A. (c) Consider the version of the Wadge game W (A, B) where players play finite sequences instead of single letters, so that I has produced a = ao Ia I ^ . . . and Hp=/3 bKIb1^ .... Obviously, from a winning strategy for a player in this new version, one gets a winning strategy for the same player in the usual Wadge game. So we may here and there allow player H to play finite sequences. In fact, all notions we define are extrinsic. i.e., they are properties of (A, AC7)or even (A, AC4,AWO), rather than A. We hope what we mean is clear from the context. We start defining basic operations on both conciliatory sets and sets of infinite sequences. DEFINITION 6. (a) Let A C A', B C A'B, andA, C ABwith AB \AA 0 A B a{ A'A :a A} " Uu((t)' \t : u E A`w, t E AB \ AA, PEE B} (b) Let A C A A )B = a c A<(: Oa E Al U {u-(t)f A: u E A` The next operation allows a player to choose at first move the set he wants to be in charge of. In particular, when A is non self dual, \, is self dual and is the -A <,,-least above A (and -A). DEFINITION 7. Let A, B C A', assume {AA,AB} is a partition of A in two nonemptysets, /A : \,B{(ao) aoEAA,oaEA}U{(bo)'73: boEAB ,EEB} DEFINITION 8. Assume AA C AB and JAB\ AA I> 2 and {A+. A_ } is a partition of AB \ AA in two nonempty sets, (a) Let A C AW,B C AC, B + A = A U{u^(a)'7i: u c A`w, (a e A+ and/i E B) or (a e A and/i 0 B)} (b) Let A C AA, B C AB, B + A = A U {u^(a)<73: u E A`W, (a E A+ and/i E B) or (a E A_ and/i B)} The previous operation can be regarded as if it allows a player in charge of B + A to begin the game being in charge of A and at any moment of the run, to restart, being in charge this time of B or of -B depending on whether the first letter not in AA that he plays is in A+ or in A_. In particular when A C A', B C AB, the sum can be expressed by operations SO~~~~~B and it is immediate that B + A w A X -B We generalize definition 7 by allowing a player to choose among < K many sets the one he wants to be in charge of. DEFINITION 9. Let I C A<'@ such that {Ki'A: i E I} be a partition of At i.e., Vi E I Vu E A<( i\u 0 I andVu E A") wi E I (i C u or u Ci) Let Vi EI Ai C A, Zis1Ai = Ufi : a E Ail} iCI The next operation is very close to the previous one, except that, defined on conciliatory sets, it allows a player to choose no set at all (e.g., skipping indefinitely). These two operations are designed so to give (when used properly) sets whose Wadge degree is a limit ordinal of cofinality < K. But while the previous operation will be used to generate sets in spaces of the form AWthat are self dual, the next one only allows to create conciliatory sets therefore non self dual ones. DEFINITION 10. Let I C AA, Vi E I Ai C AA , sup Ai {(i)%or: i E I, E Ai} iEI 62 J. DUPARC REMARKS 11. (a) Every operation defined above preserves the Wadge ordering, e.g., A (A B) C ,, A-(BC) and (A + B) + C w A + (B + C) (d) Given A C A', B C A', as well as A C A<', B C A -(B+A) -w B+-A where - X denotes the complement of X. PROPOSITION 12. Let A C AWand A be self dual, then there exists (Ai)iE,, I C A<'*, each Ai C AA7 non self dual and Ai t. A such that A W EiEI Ai /-Ai Moreover,if A, isfinite then A _, for some i E I . Ai The proof is an easy consequence of THEOREM 13 (Martin, Wadge). Let A C AW, A Borel, A is non self dual ]az o EEA'O, Vn E co, A[a [n]--, A whereVu E A<(, A[u] a{ E A4: u% e A}. PROOF OF PROPOSITION 12. The proof goes by induction on dwA. Set Io { a n,a: a E AA & n, is the least integer such that A[&[an] Zw A}. Clearly, jIol < wt if IAA1 < wo and IIoI < IAAIotherwise. Furthermore, for u E Io, A[u] t, A hence A = EZEo A[u]; by induction hypothesis each A[u] _= EZE1, A[u]i with A[u]i C AA, non self dual and A[u]i Z A[u]. Thus A -tU EuEo (ZEjI A[u]i) = E A [u]i where I ={u"i : u c Io, i I,}I verifies: o if AqIA > co, II < IAJA.jI A= IJA o if JAAI< wo, each Il, is finite so is I too; and there exists u, i such that A[u]i is oA[t]i of maximal Wadge degree; thus clearly A _-w \\, 12-1 WADGE HIERARCHY AND VEBLEN HIERARCHY 63 PROOF OF THEOREM 13. (See also [6].) (=#) The fact A[a [ n] A is self dual z=o Va e A', 3n E co, A[a [ n] w A. Towards a contradiction, assume that there exists a that satisfies Vn E co, A [a n] >,, A. For each integer n let a? be a winning strategy for 11 in the 11 imposed game W(A, A) where player il's n first moves must be a [ n and a' a winning strategy for 11 in the 11 imposed game W (A, -A) where also player il's n first moves must be a n. Given e E 2', consider the following sequence of co Wadge games. 6fE(?) -f ( 1) Uf6(2) -fE(3) II I II I II I II I 1I 2 a0 a0 a30 a0a4 a00 I- a01 ~2 - a0 a03 a I a2 a3 a4 a10 a1I a - 3 1 I 2 a3 a2 a2 0 I- a. a2 a2 a3 a3 0I a3 a3 a4 0 a4 where nk is the smallest integer greater than nk- 1 such that player 11 having already played the nk first values of a in game number k, player 11 in the first game has 64 J. DUPARC answered with at least k letters i.e., (0.(0) o o< (1) o ... O p(k)) * (a n k) has length at least k. (We need to proceed this way because in a Wadge game player 11 is allowed to skip, hence the above picture is not accurate). We then build a sequence (oG '+')nC,, where oa'["+' * /J is the sequence played by 11 in the first ganie when I plays /3 in the game number n: pono 11 nk~)O .Os~h)(i Hence, by construction 1 V/ EEAV Vm > n (au"'Z */3) rn = (G-I;F1 *pi) n which shows that the sequence of continuous functions (at Ln-l )co converges uni- formly in e. Therefore its limit f 2' F-* A' is a continuous function, so that B f - A is Borel hence satisfies the Baire property: B is either meager or comeager on a basic open set N = { e E 20/v C e} for some v E 2<`. We define X : N, X ) N, by +(e)(n) = e(n) if n :& lh(v) + 1 and X(e))(n) = 1-e(n) if n = lh(v) + 1. This implies V8eENv((e)(EB B eB) which contradicts the fact B has the Baire property. 13-1 We say A is initializable when A >a A ' A (which implies n+1 A-W A 'A .. A for any integer n). We remark that an initializable set is necessarily non self dual because, as a consequence of Proposition 12, every self dual set is obviously not initializable. PROPOSITION 14. Let A C AAW i-A #w A and A 'A :aw A then there exists B C AO)WC C AC such that { B, C T = {u E A`A: A[u] >? A} and [T] = a{ E AA : Vn E co a [ n E T}. Since A is non self dual we remark that T is a nonempty pruned tree (that is T 7 0 and every node in T has a proper extension in T). By Dependent Choice [T] 7 0 holds. We define B by: (a) If [T]> A 74 0 then B = A n [T] (b) otherwise B = AA The reader should notice that B is designed so that player 11 has a winning strategy in W(B, B) that always remains in T. Obviously B is initializable in case (b). To show that B is initializable in case (a) it is enough to remark that for any u E T player 11 has a winning strategy in WADGE HIERARCHY AND VEBLEN HIERARCHY 65 W (A, A [u]). In particular, when player I remains in T, this winning strategy requires 11 to stay in T. Therefore this winning strategy is also a winning strategy for 11 in W(B, B[u]) as long as I remains in T, But if I exits T, since at that time 11 is still in a position that is in T and [T] -. A :&0 holds, this position can be extended to an infinite sequence of [T] which is not in A. Obviously in both cases B U = {u^(a): u E T, a E AA, u^(a) ? T} Set AC U U AA. Choose uo E U ( U :&0 since A is not initializable) and ao E AA. Assume for any a E AC?a ( U/ao) denotes a after every occurrence of a letter in U has been replaced by a0. Set C = EIEAc 1/ "Al where Al is defined by o if lE u then A, = f{ EAC:A 1(a(U/ao)) E A} o if I , u then Al = {a E AC0: uo'(a(U/ao)) E A} = Au, By definition of U, for any u E U, A :w All holds. Therefore, since A is non self dual one gets A :w C, thus C ? B ' C are verified: (a) In the Wadge game W (A, B C) the following strategy is winning for 11. Player 11 stays in charge of B as long as I stays in T and in that case 11 copies I's run. If and when I exits T i.e., comes to a position u^(a) with u E T and u' (a) E U then 11 decides to be in charge of C and even more precisely to be in charge of Au. This shows A ? -B holds. Towards a contradiction assume C A-w B C ?3. The Borel counterpartof ordinal sum. We come now to the study of non self dual but non initializable sets of the form B + C; at the start when C is minimal, and then when dwCC> 1. 66 J. DUPARC LEMMA15. Let A C A", A non self dual and B C AW -A,A PROOFOF LEMMA15. The hypothesis means 1 has winning strategies in both games W (A, B + (0) and W (-A, B + (0). As A is non self dual there exists a E A , such that Vn < co A[a [ n] >w A. Assume in both games I plays a, then, in at least one game, 11 plays at some point to be in charge of B or of -B instead of (3. Because otherwise both 11 players would remain in charge of (D which would give a E A < a E -A. So, depending on whether 11 chooses B or -B in the first Wadge game or in the other one, this gives for some n E co: A[a [ n] LEMMA16. Let A C A`), B C A(', C C AO, all non self dual and C >S (?, If -A,A F ={ae E AA : *aa andi *a contain no letter in AB\ AC} i.e., a E F if and only if both questions r+ * a E B + C? and * aea B + -C? reduce to r+ * a E C? and - * a E -C?. F is closed. If F 0, since A is non self dual there exists a E A, such that for all integer n A [a n] A. For F is empty, 'r+* a or Z * a contains a letter in AB \ AC which shows that either A D < C as desired. If F 0, set for each a f F: ma, =min{n: Va' E A, (a n)^a'/ F} and G = {a AeA \ F: 3/# E AQs.t. (a I (Ma -1)) E F n C}. Finally set D = (C n F) U G so D C AC; clearly D The previous lemmas are just what is required to compute dw(A + B): PROPOSITION17 (Wadge). Let A C AW, B C AO, both Borel and non self dual, du,(A + B) = (dwA) + (dwB) PROOFOF PROPOSITION 17. By induction on dwB: o dwB = 1, hence B=-,, (D or -B, (D, thus by lemma 15: d,(A + B) =sup{dw$C+I C I C, C o dWB> 1, hence B >w (O, thus by Lemma 16: - d,,(A + B) = supf d,',C + I1: C :Sw C, C <,, A + B} - = supf dw C + I: C :Sw C, C (dWA) + (dWB) 17 H A similar study for E and sup: LEMMA 18. Let Ai C A', A C A,, A, Ait Borel, BAC A' (a) -A :Sw A & A PROPOSITION 19. For all i E I, let Ai C A<;, Bi C A() non self dual, A C A' and Ai, Bi, A all Borel, (a) (i) In the case 3i+, i-(Bi+B W -Bj_ & Vj Bi+, Bi z Bj) (that is the case Eic, Bi -t \K) Bi+ dw Eici Bi = sup{dw Bi : i E I} + 1. (ii) Otherwise dw Eici Bi = sup{dwBi: i E I} 68 J. DUPARC (b) Assume Vi E I 3j E I Ai <, Aj, dw(sup Ai)h = sup{dwA6: i E I}. iCI PROOFOF PROPOSITION 19. (a) (i) is clear. - (ii) Otherwise means Vi+, i_ (Bs -W Bi z==* 3j B1, Bi_ <, B1). Hence there are only two possibilities: o 3jVi Bj >w Bi, thus dw Ei Bi = dw0B = sup{dwBi :i E I} o Vi3j Bi )b (b) dw(sup Ai = sup{dwC + 1: -C :w C, 3i E I C or -C ?4. The Borel counterpart of ordinal < n-multiplication. With the help of the previous operations, we inductively define the Borel counterpart of ordinal multi- plication less than K. DEFINITION 20. Let A C A<', o Ao1=A o A*(v+1)=(A*v)+A o A.*=supA*O ,forAlimit. We remark that if B C A = sup{dwB: B w B, B = sup{dwB B w B, 30 < A B =sup{dw(A*0)b: 0 = sup{(dc?A).0: 0 < Al = (dN?A).i 21 H On the basis of the empty set, using only previously defined operations, one can't reach the Kth level of the Wadge hierarchy (the one that corresponds to Lo-complete sets). Even when defining (as Wadge did in case I= co,1)the Borel counterpart of i-multiplication as the limit of: A '-A A ...... 'AA )AA A '-A (where every sequence that "crosses all arrows" is rejected) one can't get up to the NE th level (which corresponds to E -complete sets). ?5. The Borel counterpart of quasi-exponentiationof base x'. We introduce an operation A X-* A" which is simply the priority, because a player in charge of a set A - is like a player in charge of A that may as often as he wants to, get back to an earlier position and restart from it. In more practical terms, he is allowed to use the "Back Space" key, the one familiar to your computer. DEFINITION 22. Let A C A<', and ff f AA (a symbolfor "Back Space"), + A- = {a E (AAU {f < }) O V)U = () o for finite with lh(<&) = k: + (a (a)) = ^a(a) if a - (a (0<-)) = aF (k - 1) if k> O (at (f ) ) =- () if k =O o for a infinite: (a) P = lim(a C n) , where,given A/, E A We remark that (-A)V = (Aj-) and that the set (A -)b is obviously initializable since for a player in charge of such a set every position is equivalent to the initial one because the "backspace"symbol allows to erase what has already being played. This operation is the Borel counterpart of the first Veblen function V0 since dc. (A') = Vo(d.0A) holds for any Borel conciliatory set A (proved here only when d?A < V1(2) in lemma 37). This operation preserves the Wadge ordering: 0 PROPOSITION23. A C As B C A PROOF OF PROPOSITION23. (Iz4): It is an immediate consequence of next lem- ma, because given T a winning strategy for 11 in C (A, B), the strategy u1 is winning for 11 in C (A-, By). - (is ): By contradiction and determinacy, A As B -B LEMMA 24. Recall s denotes a letterfor skips, s is not in A, nor in AB and a( /s) standsfor the sequence a when every occurrenceof s has been erased. Given r any strategyfor H: -: (AA U{s}) ar : (A, {Uu s}) (T*a')( /s) = ((or * a)( /s)) PROOF OF LEMMA 24. First, we view this lemma as asserting that player 11 has a winning strategy a, in the two player game G? (AA, AB); where players play each at a time, a letter in AAU{f-} for I and a letter in ABU{*(-} for if. It is I who begins, both are allowed to skip (materialized by the play of "s"). So, in co steps, I builds a and if replies with ,6 = a, *a. The strategy a, is winning if and only if it satisfies the requisites of the lemma. So we describe the strategy o, by explaining how player if responds in this convenient game. In fact we allow 11 to play finite sequences, so as is the strategy easily derived from the one explained below. Secondly, after n moves, I has played (al n + 1) with a E (AA U {S ((})wo and if has played (f [ n + 1) with ,6 E (AB U {S, ((- })'. We describe CT by explaining if's nth move. For that purpose we build, for each n, finite sequences uI, , uI, utn, with the property that (Vm > n)[(u, C um) = (zn aA)]u ^I I =_ un satisfies un un( /s) = (a n + 1)( /s)A zn'^II satisfies unII^II = rC*uIand uII UII ( /s). At the first move: o if I plays a E AA, iresponds with r ((a)); so u (a); a[ = (a). o if I plays s,i7 responds with r((s)); so u ( u1 = (s). o if I plays ff,if responds with r((s)): uf (; z^fis not ((-) but zIf (s). At move n + 1 (n > 0): I has played some u E (AA U { , ((})n+1 so that u( /s)< = UXIE AA UII (I/5) with U = un- 11 o If I plays a E AA responds with r(ua'(a)). Now u$, = u/j'(a) and UnI~ = n (a). o If I plays s, 11 responds with r(uaI4'(s)). Now u/+= U" and ui U o If I plays (-, different cases appear: o If uI =() then, by construction, u = P with 0 < p < n, and 11 responds with c(U^I^(s)). Now u/+, () and UnI+l = 5P+l. O If UI = u^'(a) with a E AA and u E A"W, then, by construction, u;I u'^(a)"sP with 0 < p < n and u'( /s) = u. WADGE HIERARCHY AND VEBLEN HIERARCHY 71 (a) If u' 7 (; by construction, at move n, one has ui w where v' =r*u' and v'w' = r*(u'%a"sP). 11 erases wl - i.e., he plays the letter "*-<-" k times where k = lh(w'( /s)), and then plays -r(u"'(s)). All this means that 11 answers with the sequence .. . -, r(u' (s))). Now u$Il u and uI+I = u' 7s). k (b) If U' = (), then u = . Player 11 erases everything he played before and plays r ((s)). So that now u/+ ()and uI+, = (s) This strategy is winning for 11 because: o if I played a such that lim(a [ n( /s)>)< = lim u a"// E Al", then by n n construction limuaI = ao' E U {s }) with a/" = and 11 ,6 n (A, a'( /s), played such that ,6( /s)+ = -coe' o if I played a such that lim ui = u E AWO,we have two possibilities to explain n the finiteness of u: (a) u is finite because after a certain move n, I always skipped until the end of the game. In that case, Vm > n, ui- u and by construction u- a,{'u's( n-); hence, Uz =*(u s(' -)) and a' = zn s'' works because a'( /s) =u( /s) = u (I oa I/s)-. (b) u is finite because after a certain move n, I erased his moves except the first n an infinite number of times. i.e., there exists a strictly increasing sequence of integers (Pi)iz, such that up = UI and Vq > po q , {pi :1 }E c U.I By construction ^I~ = Usi and VqVi pi < q < pi+I U p U so that lim UjI =s and Vi E co Vm > pi U = s') C U I, thus 11 played ,6 with IJ( /s) A = (r * (aus))( / s). The sequence a= uzI sW works because a'( /s) = U^I(/s) = u = a/a( Is)P. 24-1 We introduce an other operation which is, in a way, the inverse of the "priority". ( is defined on conciliatory sets while the following ' is defined on subsets of AW). The operations + and sup create non initializable sets, so that proofs may proceed by induction on the Wadge degree since any non initializable set can be broken into sets of lesser degree. On the contrary, - gives only initializable sets. So if we also want to proceed by induction on the Wadge degree of By we need an operation that makes the Wadge degree of an initializable set decrease. In fact, -/ is an operation only if we make it functional which requires AC. But that functional character is not needed in the proofs. C DEFINITION 25. GivenA AA and (F,,)uEA`, a sequence of closed subsets of ACA abbreviatedby (Fu)5 we define A(F, ) C [TA(F4J, where TA(F,)is a nonempty pruned (n.e.p.) tree on the alphabet AA U {(O)} U {(1)"v: v E A o a (2n) E A, o a(2n + 1) = (0) or a (2n + 1) = (1)'v with v E Aw0; depending on which option he takes. Set u = (a(2i)/i < n): a a(2n + 1) = (0): the player guarantees the sequence (a(2i)/i E co) will be in F.,. The player can choose this option only if there exists a sequence /3 E AA satisfying u c /3 E F1,. Otherwisehe is compelled to take the other option. a(2n + 1) = (1)'v: the player chooses the option to get out of Fu; and indicates with v E A'w the way he wants to exit: a will satisfy u'v is a an initial segment of that verifies u'v'^/ f F, for any /3 E A,. Here also, the player can only take this option if there exists such a sequence v. Otherwisewhatever his oddmoves will be they willform a sequencebelonging to F., thereforehe is compelled to take option (0). From now on, the next odd moves are determined by the promise made. By promise we mean: c in case a(2n + 1) - (0) then the sequence (a(2k)/k E co) must belong to Fu o in case a (2n + 1) - (1)^v then the next lh(v) odd moves will precisely be V. So at any even move a new promise is made that must be coherent with all the previous ones and the player is committed to every one of them. A move that does not respect a promise is illegal. Finally, TA(FI,)is simply the tree of legal positions for a player in charge of A (F,) in this game. To match with our generalpurpose of workingwith spaces of theform AW we define A (F,) as a subset of A'0 where A ,A4 U{(0)} U {(1)'v: v E As } by a r a e [TA(FU)] A E A aeEA(FI) a~~A V / a f [TA(F)]A 3a' (w'a' E [TA(F,)]A 2 E A) 1 wherew is the longest initial segment of a that belongs to TA(F,). REMARKS 26. (a) It is clear from the definition of A (F,) that a player J in charge of A(F, ) in a Wadge game has a winning strategy if and only if he has a winning strategy that always remains inside TA(F,,).Therefore in the sequel we will always consider that any strategy involving a set of the form A (F,) will remain in the underlying tree; in other words it restricts its moves to the legal ones. WADGE HIERARCHY AND VEBLEN HIERARCHY 73 (b) For any sequence (Fl)UCA<(&* of closed subsets of A' and any A C AW,the following obviously holds: (i) A(F") -w (-A)(Fu) (ii) A(F) A(H") o HU = G(ao.... -ak) ifu (ao,. , a2k+1) Hence all minimal element are Wadge equivalent and only two cases might appear: a minimal element for We let A' denote a But in contrast with - the converse is rather false: A' PROOF OF PROPOSITION27. Assume A' = A ) A<(V*; B BB(GL) < and : A We are done, since AS1 = A(FI) <. A(H",) by minimallity of A'; thus A' Let us have a look at what kind of set is A'. It is non self dual if and only if A is initializable. PROPOSITION28. Let A C A', A Borel, (a) if A is self dual, then A' is self dual: A - A == A =-W A (b) if A is non self dual but not initializable, then A' is self dual: - (A #w -A & A #w A A) A --W A (c) if A is initializable, then A' is non self dual: A- A 'A )AS X Al PROOF OF PROPOSITION28. (a) If A w -A, then by Proposition 27 ASw ( A) -W (A*) (b) The proof is by induction on dwA. By Proposition 14 there exists B C AA initializable and C C A' such that B, C o if C At B then, by determinacy, C -w -B thus \, -W which is C,* C,* self dual. o if C >by B then =w C'. But in that case, since both C,* C,* C ?m B and C >y -B hold C cannot be initializable otherwise it would lead to B ' C On = (a (0), a (2), .. , a (2n)) Vn AA WADGE HIERARCHY AND VEBLEN HIERARCHY 75 where: F,1 n F(a(0)a(l).I....(2i)) i On = (a (0), a (2),. a (2n))"vn1 A 4 =-n (a(O),a(2),... a(2i))"wj'A'. kns. at(2i+1)=(1) wi (By convention vo = , = F0 = A"). From that position legal moves are now restricted to the closed set F,1n On- Assume that by construction the following holds: [S] n On C Fn and On n [S] 54 0 (i.e., (a (0), a (2),. a (2n))'v,1 E S) We set (i) a (2n + 1) = (1) 'wn if and only if there existsWne A<10 such thatVn C- w and (a (0), a(2),. a (2n))'wn E S. One has On+1 = (a(0), a(2), a (2n))'WnjA, and Fn1l-Fn. (ii) a (2n + 1) = (0) otherwise. Notice that necessarily [S] n 0n C F(a(o)a(1).....o(2n)) Then On+1 = On and Fn+l = Fn n F(a(Q)a(1)Q.(2n)) In both cases the requirement On+, n [S] C Fn+l and On+1n [S] : 0 holds. o On even moves: a(2n + 2) = anything such that (a(0), a(2), . a (2n), a(2n + 2)) W( A, A[(a (O),a (2),. , a (2n))'Vn This is true because A W( A 'A, A[(a&(O),a(2), . , a (2n))Vn]). But in that game, as long as I does not "cross the arrow" (that is plays a letter that is not in AA) the winning strategy requires if's run to remain inside S (otherwise I 76 J. DUPARC would have a winning strategy then). From that winning strategy we easily derive a strategy that remains in S and is winning for ff in the game W( A, A[(a(O), a(2),.* , a(2n))'v"]). Hence there exists a strategy z winning for ff in WJ( A , A[(a&(O),a(2),. a (2n)) such that VP6E A'7 *e on n [s]. By minimallity of A' it is enough to show ff has a winning strategy in the game WJ(A (Gv) , A(Fly)[(a(O)a, (1). (2n))]) for some specific distribution of closed sets (Gv)VEA A` if u C v( /b)P then VE (AA U{<,b})' VVnE mE co -< (va) [m >-= (u,n). Then the sequence (F (A- )b (b) Ab ?n (A-)b by induction on dwAb is quick, using Borel Determinacy. Assume At :w (A)> , thus (A)b* B (B)* --w B -w (A- hence -B On the other hand: PROPOSITION30. Let A C AX, B C A A (az~) is just a priority argument. Assume A' - A(F, ) and z is a winning strategy for 1 in W (A*, Bb). We describe a A<( * (AB U {*, b}) AAw I-(AB U {,b}). We assume I plays (ai/i E co >E A()A (a) At the first move: a ((ao)) = 'r ((ao)) - (b) At the second move: (i) if ]a E AO) such that (ao, al)'a C F(ao al): 0 ((ao, al)) = (z ((ao, (0))), z ((ao, (0), al))) and set AO (0) 78 J. DUPARC (ii) otherwise: a ((ao, al)) = Kz (Kao,(1)(a) ()(a), a, and set Ao = (1)' (a ). (c) At move n + 2: assume I has played u = (ao,... I a,+,) and nf has replied with O*u = v such that v+PI =-*(ao, A' a,, An, I an, A) where 'P" is defined exactly as -P except that it does not take care of the letter b, it just erases it as if there were no letter at all: ()= () for u e (A. U {s, ? (}) (u^-"(a))'Ph = uePh (a) (U -(b)) Ph = 2ePh (b) *< Ph (u^'(a, )) = -ePh (u'a(b, )*-))"P =(2^(*-))Pb (u -(*- )( *) ) P' = (( U (*- ) _PP ()) so in particular, v( /b)P = Z*(ao, A8, al, A7..., an, An)( /b). Assume also: o All = (0)( 3? EA (ao, ,.Ial an)< EF((\a . o A = (l)'(ai+1,... I ai+J) with i < i + j < n ( zzVa E A' (ao, ...... I ai+1) F(.)) We define a (u) = a ((ao,. a,,+1)) by: < = :0 E E Vi n (A (0) zz4 A' (ao,... a,,1+1)%a (i) if and 3a E A' (ao,... a,,+1)%a E F(ai. ?1) then Aj1+l= (0) and Vi < n An+1 = All Oa(u) = (b,1+1,Bn+l) where ((ao, A8 +.. a A )) bn+I a,,+ B,1+I =((ao, A A.. a a1, A"+')) Vi < n (A =(O) ==> 3a EA" (ao*, an+l) 'a E F(ao.....ai)) (ii) if but Va E A' (ao,... I an+l)%a F(aai) WADGE HIERARCHY AND VEBLEN HIERARCHY 79 then A"+l = l)(a,+ ) and Vi < n A"+l = A" +(u) (b,1+1 5 Bn+ ) where bi+I = T((ao, A+l . an, A+lan+,)) Bn+I = ((ao, A a. Al+a ,,+,, A"+')) (iii) if 3i < n A = (0) and Va E A' u''"a F(.aOf,.), then get j =the least such i. As v (=what has been played until then by 1) satisfies V-Pb =-r*(ao, A" a , A.A", AA) there exists an integer k such that (v( ,((,,,,..0 >) =*(ao, A . Iajl,A"-l) ifj>0 k =(ifj 0 then set for i < j: A"+ An ' fori >jand 30a E A' u' E F(a,,...... at)A:+= fori ;) > jandVae E A' ua lF(a A+A"+= (l)-(aj+ . an+,) And nf plays a(u) b(B.,)^(b, Bj, bj++, BBj+l,...,I bn+l B.+,)+ where k given O < p < n + 1 -j: bj+p ((ao, AO8 ..aj+p-1 Aj+P aj)) ((ao, A+1 . aj+p-1,, Any+lj1, A j+)) Bj+p-T aj+p, Note that for each n E co the sequence (a "'I/n < m < co) is ultimately constant: it takes in fact one of the three following forms: (i) (0(), (O)... ) or (ii) ( (1)^w, (1)(w, . or (iii) ((0), (0) I . I (0), (1)"w,(\)^w, . . . k wO Inductively define the sequence (Mn )nE, by o mO is the least integer such that Vl > mOAl= A i; o for n > 0, mn is the least integer such that Vn Vl > mnVn' < n (A' = AMnm& Anmtl AM1) 80 J. DUPARC Since I played (at/i E co >= a E AO, 11 played 6 = U*a which satisfies VflPn supfmM i < n}Vq > Pn ( [ Pn) =Z*(ao,A ... an, A >nC(> [q) Hence ' = lim(f i q)P" =r l *(ao, A0... an, An) q - By construction every position (ao, A10, . ak- 1, Ank I) is legal for any integer k because: Vn < co A (0) a E F(ao).....ll) (ao. C AIn,, an)'W a An -(1)"w j Va' E AO(ao,. , an)-w-'a F(ao,...,a,) Hence U*a 6 E (B)b f( /b)l (= filbP( /b)) E B Which proves oa is a winning strategy for if in W (A, (Bjb). (d)(z=): is by contradiction. Assume (Bjb AwA, by determinacy this means -A ?w B~b .~ (-A) ? (B~bja A (A*) n, BA .. >B lb Bb which contradicts the non self dual character of B". (~z): is immediate. 30+ ?6. A normal form for Borel sets of finite rank. Everything is ready for definition of canonical Borel sets of finite rank and even A,-sets. But first, let us have a look at canonical Sn+icomplete sets. Set: A, - A and Vn e , A E+l (AB.- LEMMA 31. Given 0 C A PROOF OF LEMMA 31. First, remark that given A C AA E,A" e + defining for each integer n the map f: (A A w - eachfA(AUb})t since is continuous f limfn is a Baire class re1 ffunction. (Ai) Since = fJ (o) e Ah, one obtains (Aj" E S"?+2.This proves (o"n)b e n+ ((d)bWe prove reduces every ? set by induction(Abt)s on n. Assume reduces any frset, weoBe show for any A e k+2 AE y (ng+ i)e by induction on do A . WADGE HIERARCHY AND VEBLEN HIERARCHY 81 Since A E V)1+2 there clearly exists a sequence (F1)LEAC'Aof closed subsets of A' such that A(F')En+l A ) =u n u A(k,, k,,_ .k,_ ....ko) kil E co k,1,-IE co k,,2 _ w where A(k,, k,, - i._0) is closed or open depending on the parity of n. For exam- ple set F1, = A(k,,kk,0_j...kO) if A(kSk __._ko) is closed and F, = -A(k k.__ ko) if A(k,,Sn_1-.ko) is open, with (kn,.kn-1. ,ko) q$(lh(u)) for some bijection q$be- tween co and ,n+l. Hence by A By the forthcoming lemma 37 the Wadge degree of any Lo+, complete set is 2 when n = 0 and it is i& when n > 0; so that Borel sets of finite rank are all those n of Wadge degree less than supnE, = KEO. n DEFINITION 32. Let 0 < a < NE, VI(2), a admits a Cantor normalform of base K: = lk a -Vk +- + l.VO with 'e, > a > ak > ... > ao > ? andVi < k, O < vi < K. Define: Q(a) = Q(KC'k) * Vk + ...... + Q(K'o) * VO where Q(t&') with 0 < /3 < 'E, is defined by: IfB = 0: Q(t'Cf) = Q(1) =? If, = n E co andn 54 0: Q(Kfl) Q( + 0) If , = A + n, cof(A) = K and n E co: Q*tC) Q(P) If = A + n, cco ofQ(A) < K and 0 54 n E co: Q(Kfl) =Q(fi-1)' If, = A, co < cof(A) < K: Q(') sup n()). Where /3 - Op is a fixed map for limit /3 < 'e, of cofinality < K, and where card (0r) < K and is cofinal in P. For example: (a) if P&= K then K ('+1) K" n Oor < =fl Eco,6= =K P/3} n 82 J. DUPARC (b) if P < K#then [l A 1,1-8 + ' . + 0/?Aj with fi >/n> ..>. > Po > O and Vi < n, O (0# = tt'.,,+ . . /t10.,uK : 'U < 'U } (ii) if io is not limit 1# =S f K,)1, + ***+ Kfl. (,u0 - 1 ) + KO 0 C : }# THEOREM 33. Let 0 d,'?Q(aE) = aE PROOFOF THEOREM 33. We prove dc7Q (a) = a by induction on a. The case a= 1 is simply definition. Assume a > 1, a admits a Cantor normal form of base i: Clk a = K .Vk + . . . + K;?g?.VO with '6, > ae >- ?k > . .. > ao > Oand Vi < k, O < vi < K. Recall: f -(a) = Q('f'k) * Vk + . ?Q(1'0) * VO0 Assume k = 0 and vo = 1:: the result comes directly from Proposition 17 and Lemma 2 1. Assume k > 0 or vo > 1 so that a = i'i: (a) if co < cof(ao) < K: the result derives from Proposition 19(b) because, Q(a) = sup Q('&O) where sup0c,0 = ao so d'Q (a) = sup{dt,,A + 1: A gw A, A, A ?wQ(a) } = sup{dwA + 1: -A wA, ,A =sup{dA + 1: dwA < KtJ for someO < ao} a. (b) Otherwise: the result relies on LEMMA34. Let A C A' A non self dual, -A, A because do?Q(a) = sup{dt,,A + 1: -A :9, A, -A, A =sup{dwA +1: -A wA, A PROOFOF LEMMA34. The proof distinguishes between A initializable and not initializable. In case A is not initializable (which is claim 36) one can "break" A in smaller pieces so to apply induction. In case A is initializable (which is claim 35) the inverse of ,- operation is required. Assume ao = A + n with A limit, n E co and set: 4= ato if cof(A) = K = ao - I if o < cof(A) < (andi n > 0) = ao + I if A=0(andn>O) so that in any case Q (ao) = (Qi <. Note that < a. CLAIM35. Let A C AO, A initializable, -A, A which case Q (4) '= Q (S(+0K and Q ([) =- (S(+1)K (for the same 6) or there is no such integer and then 4 + E < 4 < ao. Moreover, from b ~~~~~~~~~~b --A or A < W Q (0Y) we get A -A or A A,A (o (Q (4 - 1)~"* v)-=Q (K -) V) and (,jo-l).V < a. (ii) If c < cofG4) < ii then A ((sup Q (o)") * v =- (c(ao- )v) and K (ceo-l).v < a. (iii) If cof(d) =i then A < w (< ))b for some 4 < 4 notice Q ()- = n K((+6) with 6 E {-1, 0, 1} and K < K < a. PROOFOF CLAIM 36. The proof goes by induction on dwA. By Proposi- tion 14, there exists B. C C Al, B, C C >Cl where II < i, Ci is non self dual and Ci <, C <, A. So, for each i E I: (i) 4 is successor -=Ci C (ii) co < cof(G) < K - Ci Set vc = sup{vi : i E I}. Clearly C ?w ((supoco, LI(0)) * vc) and 0 < VC < K. (iii) cofG4) = z- Ci IB C * vc)b for some vc < n co < cof)W< K (B cof()=/ JB To get the result, it is sufficient to set 4 max{fB, Sc} in case cofG4) and v= VC + VB in the other cases. 36H34H33 LEMMA 37. GivenA C A dc A- = Vo(d?A) (d?A+e) {-1 if cof(sup{f. < dcoAA = 0 or A is limit}) = 0 where e = if cof(sup{). < dcoAA = 0 or A is limit}) = K t+Iif co < cof(sup{). < dcA : A = 0 or A is limit}) < ;. PROOF OF LEMMA 37. Assume dc0A = a. The previous proposition showed A -c - Q (a) or -Q (a). So A-v-c Q (a)<' or -Q (a) '; and since dcoA = dcof(a)>' dc?Q (a)' we just have to refer to Definition 32 and compute dc?Q(a)< ao 1: 0 (a) =(D)-( (D d Q (a)' =I ; (-l 1 << OC: Q (a)= - Q-? ( . dc7Q (a)[ ' _ (0-) WADGE HIERARCHY AND VEBLEN HIERARCHY 85 a = A+ n, cof(A) = an, E c: Q ()=Q (K`l) d~ol (a),- Kas~ a =A + n, co < cof(A) < A, n E co: Q?(x)' = Q : d (a)'v - <,a+1) - Since ((0 * 2) ")b is L +I-complete, the Wadge degree of any Lo+ -complete set is 2 if n = 0 and ii if n > 0. So Borel sets of finite rank are those of degree less n than supneir =eo. This remark leads to: n THEOREM 38. Let A C A', A non self-dual. A E #U En -a, 0 < a <'EO A orA - n o A U s(A) >? B is obvious and does not depend on s(A); o to show A U s(A) and set for each u E A^w u E s(A) - (z*u)'bo E Bb, and ()E s (A) 4=* ' ) E B. The strategy that consists in skipping when I skips and applying z (turn- ing "b" into "s") when I plays a letter in AA is clearly winning for 11 in C (A U s(A), B). 3 REFERENCES [1] A.S. KEcHRis,Classical descriptiveset theory, Graduate texts in mathematics, vol. 156, Springer Verlag, 1994. [2] A. LOUVEAU,Some results in the Wadge hierarchyof Borel sets, Cabal seminar 79-81, Lecture Notes in Mathematics, vol. 1019, Springer Verlag, 1983, pp. 28-55. [3] A. LOUVEAUand J. ST.RAYMOND, The strength of Borel Wadgedeterminacy, Cabal seminar81-85, Lecture Notes in Mathematics, vol. 1333, Springer Verlag, 1985, pp. 1-30. 86 J. DUPARC [4] , Les propriftts de reductionet de normepour les classes de Bornliens,Fundamenta Mathe- maticae, vol. 131 (1988), pp. 223-243. [5] D. A. MARTIN, Borel determinacy,Annals of Mathematics, vol. 102 (1975), pp. 363-37 1. [6] R. VAN WESEP, Wadge degrees and descriptive set theory, Cabal seminar 76-77, (Proceedings of the Caltech-UCLA logic seminar, 1976-77), Lecture Notes in Mathematics, vol. 689, Springer, Berlin, 1978, pp. 151-170. [7] 0. VEBLEN, Continuous increasingfunctions of finite and transfiniteordinals, Transactionsof the American Mathematical Society, vol. 9 (1908), pp. 280-292. [8] W W WADGE, Degrees of complexity of subsets of the Baire space, Notices of the American Mathematical Society, (I1972),pp. A-714. [9] , Reducibilityand determinatenesson the Baire space., Ph.D. thesis, Berkeley, 1984. EQUIPE DE LOGIQUE MATHEMATIQUE CNRS URA 753 ET UNIVERSITE PARIS VII U.F.R. DE MATHEMATIQUES 2 PLACE JUSSIEU, 75251 PARIS CEDEX 05. FRANCE E-mail: duparc~logique.jussieu.fr