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
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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
v E T).When T is some tree, we
subsequences i.e., satisfies Vu E T Vv (v C u
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 complementation and countable union of the class of open sets) was set up in a hierarchy by Baire: V is the class of open sets, and by induction for 4 > 1,
where -A,] stands for the complement of the
Lo; = {UnEOA,1
-An E U(<;V}
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 separated difference operation[. But an ultimate refinementwas to come with Wadge's
work based on the notion of inverse images of continuous functions.
A continuousreductionof A to B is a continuous function f: AC-0
such that
ACB
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 <w. If
A < w B < w A, in other words, A and B continuously reduce to each other, then
A is said Wadge-equivalentto B, denoted _ w.
it is a consequence of Borel determinacy
Given two Borel sets A C A(", B C AWB,
[5] that either A <? B or -B <? A. In case K = wal, Martin proved that <? is
wellfounded on Borel sets [1]. This result still holds in case K > col and allows
to define the Wadge degree of a Borel set by induction: dw(0) = dw(-0) = 0 and
dt?A = sup{dw(B) + 1: B <w A}. Wadge's lemma, as a consequence of Borel
-
-
Received January 30, 1997; revised May 20, 1999.
? 2001, Association
for Symbolic
0022-4812/01/6601
56
Logic
-0002/$4.
I0
WADGEHIERARCHYAND 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.
1
DEFINITION 1.
o dj(0) = d,(- 0)
o dA = sup{dw(B) + 1: B non self dual A B <w A} (for either A self dual or
not, A >a 0)1.
Thus the Wadge degree of A is a measure of the complexity of A, just as its Borel
rank or its level in the Hausdorff-Kuratowski hierarchy. But the Wadge degree is
a much finer measure. Computing the height of the Wadge hierarchy requires a
hierarchyof fast increasing ordinal functions: the Veblen hierarchy [7].
DEFINITION 2. The Veblenhierarchy(V: )<,,, of continuousfunctions from i+ \
{O} - iii+ -. {O}isdefinedby: Vois thefunction that enumeratesordinalsof cofinality
X or 1 that are closed under ordinal addition, namely Vo(1)
1, Vo(a + 1)
VoJ(a) X, Vo((A) = iA when cof((A)= X, Vo((A)= &A+'whenco < cof((A)< K.
For > 0, V, is thefunction that enumeratesordinalsof cofinality K or 1 that are
closed undereachfunction V(for any 4 < I.
The Wadge degree of Borel sets of finite rank ranges from 1 to
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 characteristicfeature 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
K = %holds), then
OA d A =
A limit (necessarily cof(A)
(c) divA = A + 2n + 1 with n integer and
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
to be
below it. This might, for example, allow properties of subsets of 0%o
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 <? B if and only if H has a winning
strategy in the game W(A, B) where I can't skip while H is allowed to skip, even
co times, provided he also plays (a letter in AB) co times. Assuming co stands for
co {O} and s (for skip) is a letter not in AB, a strategy for H is some function
A<( I)*
Ua:
AB U {s} such that (a*a)( /s) is infinite for any a E AA (au*a
denotes the sequence (a (a(n))/n < lh(a)) and (a*a) ( /s) stands for what remains
from a *a once every occurrence of the letter s has been removed).
In the conciliatory form C(A, B), both players are allowed to skip, even indefinitely if they want so. Thus their final choices a, ,8may be finite: af E AAC ,8 E AB<'.
8 E B. We write
In any case, we say that H wins C(A, B) if and only if a E A <
A <c B if H has a winning strategy in the conciliatory Wadge game C(A, B). We
define -C and <c the same way we defined =, and <a on the basis of <?. That is
A C B if A <c B <? A, and A <C B if A <c B butB Ha A.
Now, for the purpose of considering only non self-dual sets, it is enough to replace
sets of co-sequencesby sets of < co-sequences. This result is based on three facts:
(a) In the conciliatory Wadge hierarchy,every conciliating set A is non self-dual:
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<C',let "b" denote a letter
not in AA, and set
A" = {c
c (AAU {b})cw/a( /b) E A}
where al ( /b) is the sequence obtained when removing every occurrence of the letter
b.
59
WADGE HIERARCHY AND VEBLEN HIERARCHY
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):
A < B
so,
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:
A is non self-dual if and
THEOREM
3. For every Borel set offinite rank A C AWO,
only if there is an alphabetAB and a conciliatoryset B C A<' such that A - Bh.
A (so that B is just A extended
Furthermore,one can get AB=AA and B n AA
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'sa trivial procedurewhich 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 <c B}.
Now if af is less than
'eo = sup VOo ... o Vo(2),
(
n Ew
n
cahas 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<W,
A", B" both Borel,
(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
E
= d0
1+1
<
if cof(sup{A dc?A:
< dA:
if cof(sup{f
A=Oor
A = 0 or A is limit})
ifco < cof(sup{,{ < d0A:
PROOF. (a) is Proposition
17, (b) is Lemma
0
A is limit})=
=
K
A = Oor A is limit})
21 and (c) is Lemma
< K.
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.
it is very easy to build A' C (AAU A)w0such that
(a) (i) Given A and A C ACID
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 :aA}
(b) Let A C A<wO,B C A<',
A
)B =
a c A<(:
Oa E
"U
u((t)' \t
andAA
: u
E A`w,
t
E AB \ AA, PEE B}
C ABwith AB\ AA #4 0
Al U {u-(t)f
A: u
E
A`<O' t EEAB
\ AA,
t EEBI
A player in charge of A B is like a player in charge of A endowed with an extra
move that erases everything played before and turns A into B. But this extra move
may be used only once and is materialized by playing a letter in AA \ AB for the first
time in the game.
61
WADGEHIERARCHYAND VEBLENHIERARCHY
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:
DEFINITION 8. Assume AA C AB and JAB\
of AB \ AA in two nonempty sets,
boEAB
AA I>
2 and {A+.
,EEB}
A_
} is a partition
(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
X
and it is immediate that B + A w A
-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,
: a E Ail}
Zis1Ai = Ufi
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
iEI
{(i)%or: i E I,
E Ai}
62
J. DUPARC
REMARKS 11.
(a) Every operation defined above preserves the Wadge ordering, e.g.,
===
A<w A'&B<wB'
(b) Given A C Ai',
A+B<wA'+B'
B C A<(O,
= Ah
(AB)b
Bb and (A + B)h = Ab + Bh
(c) Given A C Aw, B C AB, C C Aw or A C A{w, B C A'w, C C A<', the
operations
and + are associative:
C ,, A-(BC)
and (A + B) + C w A + (B + C)
(A B)
(d) Given A C A', B C A', as well as A C A<', B C A<O,
-w B+-A
-(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
that
A
Moreover,if A, isfinite then A
W EiEI
/-Ai
_,
t.
A such
Ai
for some i
E
I.
Ai
The proof is an easy consequence of
THEOREM 13 (Martin, Wadge).
A is non self dual
whereVu E A<(, A[u]
PROOF OF PROPOSITION
Let A C AW, A Borel,
]az
o EEA'O, Vn E co, A[a [n]--,
a{ E A4:
A
u% e A}.
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]
Clearly, jIol < wt if
IAA1
Zw A}.
< wo and IIoI < IAAIotherwise.
A hence A = EZEoA[u]; by induction
Furthermore, for u E Io, A[u] t,
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 [u]i
A -tU EuEo (ZEjI A[u]i) = E
where I ={u"i
: u c Io, i
I verifies:
I,}
> co, II < IAJA.jI
o if AqI
A
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
63
WADGE HIERARCHY AND VEBLEN HIERARCHY
PROOF OF THEOREM 13.
(See also [6].)
(=#) The fact A[a [ n] <w A is immediate. To show that A[a [ n] >w A holds
it is enough to consider the game W (A, -A) in which player I applies a winning
strategy a and 11 skips indefinitely. The resulting sequence a = U*s"' satisfies
Vn E co, A[a [ n] >? A.
(<=) The proof is very close to Martin's proof that Borel Wadge degrees are
wellfounded. We prove
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(?)
II
-f
I
II
1I
a1I
1I
I-
a3
0I
a3
I
a4
a0
a3
a4
-
a
a2
II
a03
a0
2
a2
a2
0
-
a2
a10
I
a30
~2
a01
aI
a.
II
I
a0
I-
-fE(3)
Uf6(2)
2
a0
a00
( 1)
3
a3
a2
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:
11 nk~)O .Os~h)(i
pono
Hence, by construction
rn =
V/ EEAV Vm > n (au"'Z */3)
(G-I;F1
1 *pi)
n
which shows that the sequence of continuous functions (at Ln-l )co converges uni2' F-* A' is a continuous function, so
formly in e. Therefore its limit f
B
f
A
is
Borel
hence
satisfies
the
Baire property: B is either meager or
that
comeager on a basic open set N = { e E 20/v C e} for some v E 2<`. We define
if
X : N, X ) N, by +(e)(n) = e(n) if n :& lh(v) + 1 and X(e))(n) = 1-e(n)
n = lh(v) + 1. This implies
B eB)
V8eENv((e)(EB
which contradicts the fact B has the Baire property.
We say A is initializable when A >a A
13-1
'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
{
B, C <w A
B B <w B
C >?, B
A-w
B
AC such
that
'C
PROOF OF PROPOSITION 14. Set
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
and every node in T has a proper extension in T). By Dependent Choice [T] 7
holds. We define B by:
(a) If [T]> A 74 0 then B = A n [T]
(b) otherwise B = AA
0
0
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 <? A holds and B <? A does not hold otherwise A
would be initializable. Hence we have B <w A.
We define U as the set of positions that exit T for the first time:
u E T, a E AA, u^(a)
U = {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 <a A.
We show that A =, BLC holds by proving both A <? B C and A >?
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
(b) In the Wadge game W(B
(a) In both cases
(i) [T]
(ii) [T]
A
A
' C.
' C, A) the following strategy is winning for 11.
A n [T]
0 and B is defined by B
0 then B is defined by B =AA
as long as Player I stays in charge of B 11 has a winning strategy against I that
remains in T. At any move if I decides to become in charge of C, then 11 still
being in T is in a position that is equivalent to the initial one and therefore
can easily beat I.
It just remains to prove that C >? -B holds. Towards a contradiction
assume C <? B, then
A-w
B
which contradicts B <w A.
C <w B
B-w
B
14H
?3. The Borel counterpartof ordinalsum. 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 <? B +(D
A <? B orA ?-
B
PROOFOFLEMMA15. 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 EA <
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] <? B or
A[a r n] <? -B so A <w B or A < B.
151
LEMMA16. Let A C A`), B C A(', C C AO, all non self dual and C >S (?, If
-A,A <?, B+C,thenA ?w B+DforsomeD
CAc -D D <? CandDnon self
dual.
PROOFOFLEMMA16. By hypothesis there exists winning strategies for players H:
r+and z in respectively W (A, B + C), W (A, B + -C). Set
F ={ae E AA :
*aa andi
*a contain no letter in AB\ AC}
a
B + -C?
i.e., a E F if and only if both questions r+* a E B + C? and * ae
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 <? B or A <? -B holds. Thus A <? B + D for any
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 <? C with the help of r+
and D <? -C with the help of z-. So A <? B + D; and if D is non self dual
the work is done. If D is self dual then D _S EicY Di, each Di C AC?non self
dual and as A <w B + D with A non self dual there necessarily exists Di such that
A <w B + Di-cf. Lemma 18.
16H
The previous lemmas are just what is required to compute dw(A + B):
PROPOSITION
17 (Wadge). Let A C AW, B C AO, both Borel and non self dual,
du,(A + B) = (dwA) + (dwB)
PROOFOFPROPOSITION
17. By induction on dwB:
o dwB = 1, hence B=-,, (D or -B,
d,(A
+ B) =sup{dw$C+I
(D, thus by lemma 15:
C
I C, C <wA
+ B}
= sup{dwC + 1: -C Sw C, C <w A or -C <, A}
=sup{d.C + 1 dw0C< d$A}
= duoA + 1
67
WADGEHIERARCHYAND VEBLENHIERARCHY
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 <w (A +D),
D <w B}
= sup{dw(A + D) + 1: D Sw D, D <w B}
sup{(dwA) + (dwD) + 1: dwD < dwB}
(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'
Bs*3i E I A <w Bi.
(a) -A :Sw A & A <W EiCii
(b) Assume Vi e I 3j e I -Ai <(. Aj,
(i) -A :Sw A & -AA <w (supic, Ai)6 =* 3i (E I A <w Ai,
(ii) -A <w A & A <w (supicz Ai)6 z=# A <w Z(a)cA['] A6
whereVa C A \ I A(,) = Ai forsome i E I.
PROOF OF LEMMA 18.
(a) Given a winning strategy for player 11 in W (A, EicI By), let a E AO'satisfy
Vn E coA[a [ n] >w A and assume I plays a, necessarily,11 chooses to be in charge
of some Bi instead of Eic, B
(b)
(i) As A is non self dual, there exists a sequence a E A,, such that Vn E
I A[a [ n] >w A. Assume I plays a in both games W (-A, (supiEI Ai)")
and W (A, (supiCi Ai)") while 11 applies a winning strategy in each of them.
Necessarily at some point, in at least one game, 11reaches a position (i) which
is equivalent to being back to the initial position (()) but this time being in
in charge of Ai instead of Eic, Ai. If this happens in the first game it gives
-A <w Ai', and in the second one A <w Ar'. If A <w A?, the work is done; if
-A <w At then A <w -Ab thus A <w Ab for some j E I.
(ii) As A is self dual, A -w ,Ej
A', where for all j E J. A' C Al and A'
non self dual. So, as both A <w (supic, Ai)" and -A <w (supic, Ai)" hold;
we see -A', A' <W (supiEI Ai)6 holds for every j E J; hence by (b)(i)
Vj E J 3i E I A' <w Ail; thus
Z,jz
A' <W ZiEI Ai
18H
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) AssumeVi E I 3j E I Ai <, Aj,
dw(sup Ai)h = sup{dwA6: i E I}.
iCI
PROOFOFPROPOSITION
19. (a) (i) is clear.
(ii) Otherwise means
i_ (Bs -W - Bi
Vi+,
are only two possibilities:
o 3jVi Bj >w Bi, thus dw Ei
o Vi3j Bi <w Bj, hence
=sup{dwC?1:
3j B1, Bi_ <, B1). Hence there
Bi = dw0B = sup{dwBi :i E I}
Bi = supf dw C + 1:
dwoEi
z==*
C :Sw C, C <w EicI
-C
w C, 3ieI
i
C<wBi}
= sup{dwBi + 1: i e I}
= sup{dwBi: i e I}
(b)
dw(sup Ai
)b
= sup{dwC + 1: -C :w C, 3i E I C or -C <w A6}
iCI
sup{dwA6 + 1: i E I}
= sup{dwAb: i e I}
19 H
?4. The Borel counterpart of ordinal < n-multiplication. With the help of the
previous operations, we inductively define the Borel counterpart of ordinal multiplication 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<Cis initializable and A, A <? B then A * v <? B holds
for any v < K. So in particular;if A is initializable and B is the least (up to =w and
complementation) initializable set above A; then B reduces A * v for each v < .
In fact the degree of B is exactly (d?A).K as will be clear at the end of the whole
study.
LEMMA 21. Let A C A<', v < K6,A" Borel,
d? (A * v) = (dc?A).v
PROOFOFLEMMA 21. This is by induction on v:
o by Proposition 17: d?A * (v + 1) = dw(A * v + A)h = d(A
(dcA).v + dc0A= (dcA).(v + 1)
o for A limit, by Proposition 19(b):
d0 (A * A) = dw(supA * 0)"
= sup{dwB:
B
w B, B <w (supA * 0)6}
= sup{dwB
B
w B, 30 < A B <w (A * 0)6}
* v)6 + dwA=
69
WADGE HIERARCHY AND VEBLEN HIERARCHY
0<A}
=sup{dw(A*0)b:
= 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.
Let A C A<', and ff f AA (a symbolfor "Back Space"),
a++ A} wherea&' is inductivelydefinedby:
A- = {a E (AAU {f < })<o
DEFINITION 22.
O
V)U = ()
with lh(<&) = k:
+
= ^a(a)
if a
(a (a))
- =
(a (0<-))
aF (k - 1) if k> O
()
if k =O
(at (f ) )
o for a infinite: (a) P = lim(a C n) , where,given A/, E A<w,
o for finite
=-
nEco
for u E A`
u Climfln
"3nVp > n/rlh(u)
nEco
u
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:
PROPOSITION23. A C As
0
B C A<W,
A <?c B
#=
A - <?c By
It is an immediate consequence of next lem(Iz4):
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 <e A; thus - (Bj
<c? A-', hence -(By)
<c B- which contradicts the non self dual
(-B)
character of conciliatory sets.
23
PROOF OF PROPOSITION23.
70
J. DUPARC
LEMMA 24. Recall s denotes a letterfor skips, s is not in A, nor in AB and a( /s)
standsfor the sequence a whenevery occurrenceof s has been erased.
Given r any strategyfor H:
(AA U{s})<WD
-:
I
AB U {S}
there exists a strategyfor II:
u
ar : (A, {U
such that Va E (A, U {s, (
-* AB U {S, ((-}
s})<w, I
E (A, U {s})WO, a'( /s) = a ( /s) P and
J})w 3a'
(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
, uI,
if's nth move. For that purpose we build, for each n, finite sequences uI,
utn, with the property that
(Vm > n)[(u, C um)
^I
un
satisfies
I
=
=
(zn
_
un
un(
u
aA)]
/s) = (a
^II
^II
= rC*uIand uII
satisfies unII
zn'
n + 1)( /s)A
I
UI ( /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<Swith 0 < lh(u,) < n and u^Ic (AA U s )<C with 0 < lh(u I) < n. By
u/f
such that v( /s)=
construction, 11 has played some v E (AB U {S5,*-})<(
UII (I/5)
with
U
=
o If I plays a E
UnI~
=
n
unAA
11 responds
with
r(ua'(a)).
Now
u$,
=
u/j'(a)
and
(a).
o If I plays s, 11 responds with r(uaI4'(s)). Now u/+=
U
U" and ui
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/+,
O If UI = u^'(a) with a E
u'^(a)"sP
AA
() and UnI+l = 5P+l.
and u E A"W, then, by construction,
with 0 < p < n and u'( /s) = u.
u;I
71
WADGE HIERARCHY AND VEBLEN HIERARCHY
(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
u and uI+I = u' 7s).
.. . -, r(u'
(s))). Now u$Il
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)>)<
n
= lim u
a"// E Al", then by
n
construction limnuaI = ao' E (A, U {s }) with a/" = a'( /s), and 11 played ,6
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
u and by construction
end of the game. In that case, Vm > n, uin-);
a,{'u's(
us(' -))
and a' = zn s'' works
hence, Uz =*(u
because a'( /s) =u( /s) = u (I ao 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
c
}E
VqVi pi < q < pi+I
Vi E co Vm > pi
U
By construction
U.I
U
p
=
U
^I~
so that lim
s') C U I,
/ s). The sequence a=
IJ( /s) A = (r * (aus))(
a'( /s) = U^I(
/s) = u = a/
a( Is)P.
UjI
=
Usi
=s
and
and
thus 11 played ,6 with
uzI sW
works because
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.
(
DEFINITION
25.
GivenA C 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<CW}and A(F{)a subset of
infinitebranchesof this tree. Infact, TA(F,)representsthe allowedpositions in a Wadge
gamefor a player in chargeof the set A (F,); so we describe the tree by explaining how
72
J. DUPARC
a run goes. In this play, odd moves are auxiliary moves where the player answers
questions about whetheror not he will "stay" all along in the closed sets F1's. Even
moves are lettersfrom A,. So assuming the run is a E [TA(F)], for all integer n, it
must satisfy:
o a (2n) E A,
o a(2n + 1) = (0) or a (2n + 1) = (1)'v
with v E Aw0; dependingon 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 compelledto 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.
Otherwisewhateverhis oddmoveswill 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 willprecisely 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 chargeof 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
r
V
aeEA(FI)
a~~
A
1
REMARKS 26.
a e [TA(FU)] A
aE
A
/
a f [TA(F)]A 3a' (w'a' E [TA(F,)]A 2 E A)
wherew is the longest initial segment of a that belongs to TA(F,).
(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.
73
WADGE HIERARCHY AND VEBLEN HIERARCHY
(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) <w A.
(c) For any A C A', the set of all A(F), where (FULcA<' ranges over all possible
sequences of closed subsets of AA, is <,-well-founded. And there cannot be two
minimal non self dual elements with one Wadge equivalent to the complement of the
other. Because if (FtI)UCA<' and (Gl1)ucA<`* are two sequences of closed subsets of
A
AA such that A(F)
A
then one can find (H1),uCA`* such that:
A(G")
A
A(H") <w A(F")
and A(H")
<w A(G);
thus A(Hz)
<w A(F),
A(G)
set
o
Hu = F(ao.ak.) if u
o HU = G(ao.... -ak) ifu
(ao0..
a2k)
(ao,. , a2k+1)
Hence all minimal element are Wadge equivalent and only two cases might appear:
a minimal element for <w A (F,) is self dual or it is non self dual and its complement
is not of the form A(G).
We let A' denote a <w-minimal element of the form A(F") where (F11)uCA<'*
ranges over all sequences of closed subsets of A'4 indexed by nonempty finite sequences of elements in AA. The operation ' preserves the Wadge ordering:
PROPOSITION27. Let A C AO, B C AO A, B both Borel,
A <w B =>
A- <w Bs
But in contrast with - the converse is rather false: A' <w B- =# A <w B.
PROOF OF PROPOSITION27. Assume
: A<O*
AB U { s} is a winning
For each u E A<`* set HU
-
o-
BB(GL) <
A' = A ) A<(V*; B
and
strategy for player f in the game W (A, B).
(G(o*-1)( /S)). Since o* is continuous, each H1 is
closed in AO).Furthermore, A
EAA <W B
EAB because a winning strategy
for player nf of the underlying Wadge game consists in:
o applying a on even moves and,
o on odd moves, answering the questions whether the sequence he is playing on
even moves will be in G(,f* )( /s) or not with:
o (0) if I answers (0) when questioned about Hi,,
o (1)'v'
if I answers (1)'u'
when questioned about HU, with'
that verifies
VPSE A' (a *U)( /S )'v/ /S
G(0T*I)/ )
The sequence v' is given by the fact from now on, I will produce on even
moves a sequence a E AO) such that u' C a, u'a 0 H2,; so for some
integer n:
VPSE AB
(u
'
(a^
[
n))^/)
Is)
=
((a *u) ( Is))^'v'^/
0 G(a *U)( /S)-
As player if can skip, he just has to wait until this event happens. This
necessarily makes him reach the rewarded v'.
74
J. DUPARC
We are done, since
AS1 = A(FI) <.
A(H",)
by minimallity of A'; thus A' <? B'.
27+
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 =-W A
A -
(b) if A is non self dual but not initializable,then A' is self dual:
A --W
A
(A #w -A & A #w A 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 <w A, -B <w C and A -w B 'C.
Easily one can see
(B
wB\
)C)'
-we,
Cs
o if C At B then, by determinacy, C -w -B thus \,
self dual.
o if C >by B then
=w C'.
which is
-W
C,*
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
C --w C which gives A =-, C contradicting
to B ' C <W C therefore B
the fact A is not initializable.
Granted with C not being initializable and satisfying C <w A we apply the
induction hypothesis and get C' is self dual.
(c) Assume A' = A(F). It is sufficient to show there exists a E [T(F,)] that
verifies A(F") <? A(F')[a [ n] for all integer n. Set
S
{w E A`:
A <w A[w]}.
We remark that S is a n.e.p. tree because A is non self dual. We inductively define
a by:
o a (O)is anything so that (a (O)) E S.
o assume (a (O),a (1), . . . , a (2n)) has been defined and the promise made states
that the run will be in the closed set Fn and the next necessary moves are going
to be vn so that the run will also take place inside the basic open set
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<n. a(2i+1)=(0)
and
=
On
(a (0), a (2),.
a (2n))"vn1 A 4
=-n
a(2i))"wj'A'.
(a(O),a(2),...
kns. at(2i+1)=(1)
wi
= F0 = A"). From that position legal moves are
(By convention vo = ,
now restricted to the closed set F,1n OnAssume 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<10such 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
n [S] C Fn+l and On+1n [S] : 0
On+,
holds.
o On even moves: a(2n + 2) = anything such that
(a(0), a(2), .
a (2n), a(2n + 2))<A'A C [S]
n 0n+1
We now show that a works i.e., Vk E co A' <w A'[a [ k]. It is enough to show
that the property holds for k even. Set k = 2n.
First we remark that A being initializable there exists a strategy T that always
remainsinside S and is winning for 11 in the Wadge game
W( A, A[(a (O),a (2),.
, a (2n))'Vn
This is true because A <w A[(a&(O),
a(2), . &.
., a(2n))'vn] holds since the sequence
(a(0), a(2), . &., a(2n))<vn belongs to S. Elsewhere, ALA <w A holds because
A is initializable. Therefore there exists a winning strategy for 11 in the game
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].
of
A'
it
is enough to show ff has a winning strategy in the game
By minimallity
WJ(A (Gv)
(1).
(2n))])
for some specific distribution of closed sets (Gv)VEA<CV* that will be described in a
moment.
In that game let 11 apply the strategy z on even moves. These moves are legal
and ff is sure to win provided he gives the right answer on odd moves. But to ensure
this it is enough than whenever 11 needs to know if its run will be in say F, that
he "asks" I to answer the question for him. This is simply done by imposing G,
where v is a sequence such that (a (O),a(2), . .&, a(2n))'((z*v)( /s)) = u to be the
inverse image by the continuous function induced by the strategy of the set Flu.
So have shown
, A(Fly)[(a(O)a,
A
A` <w
A(GV)
<? A(F{)[(a(O), (1,..
., a(2n))] = A"[a [ k + 1]
28 d
Since - is an exponentiation on sets, the next proposition shows it is the inverse
of - - so a logarithm on sets.
A" Borel,
PROPOSITION 29. Let A C A<',
(a) (A )b* <? Ahb
(b) assumefor any B C A', B <w A there exists some conciliatory set C C A<',
C"=w B. Then
(A)h
-
Ah.
We do not need case (b) in the forthcoming proofs of the exponential character
of -; we may have only noticed case (a). Anyway the hypothesis will be proven for
___ At holds for
sets of degree less than VI (2) (in fact A'), so it is true that (Aj"
any At E A' (and even for any Borel).
PROOFOFPROPOSITION
29. (a) (A )b
n E co
F(11)
getamap
{
(AA U b)t
- {aE
(AAU {,b})<t
v
?I
b
-*
X
Vm > n, u C (a( /b)
v=(u,
if u C v( /b)P
VE
(AA U{<,b})'
Then the sequence
VVnE
(F<v,)vE(AAU{((
enough to allow player
11
Set for each u E A`0* and
Ab.
x
in the game W
that satisfies:
then
mE co
b})<-*
n)
[m)P}
-<
(va)
[m >-= (u,n).
of closed subsets of (AAU { b, ((
((Ajb(F
<),
Ab)
})t
is
to know in advance
what will be the first moves of the resulting sequence a&(/b)K - i.e., if u is some
finite sequence that will satisfy u C a( /b)<, then, sooner of later but surely during
the run, I will inform 11 of this fact. So, since he can skip by playing b, his strategy
77
WADGE HIERARCHY AND VEBLEN HIERARCHY
will be to wait as long as necessary (possibly until the game is over) to be sure of
a( l/b)P (nn)before playing exactly it.
By minimallity of (A-)b:
(A- )b
(b) Ab
Assume
?n
<? At
<w (A)b(F<v)
by induction on dwAb is quick, using Borel Determinacy.
(A-)b
At :w
, thus (A)b*
(A)>
<w -Ab.
<w Ab; so there exists a conciliatory set B C A<',
This leads to (Aj"
B <a A and (A')b
such that
Bb <w At
-w
by induction hypothesis,
--w B -w
(B)*
(A-
hence
-B <? A
(bt
< w(A-)
b
(-B
)
b
-
< w(-)b
(B
)bl
<w (A-)bl
thus
<w (A
(A -)
is non self dual for it is initializable.
but this contradicts the fact (A)b*
29-d
On the other hand:
30. Let A C AX, B C A<W, A Borel,
PROPOSITION
(a) A* <w Bb'
. A <w (B )b
Bb
A*
<w A
<w
(b)
<
- (Bjb
30.
PROOFOFPROPOSITION
(a) ('#=-) is immediate since by Propositions 27 and 29(a)
A <W (B)b
As A
w ((B)-)
<w B".
(az~) is just a priority argument. Assume A' - A(F, ) and z is a winning strategy
for 1 in W (A*, Bb).
*
A<(
We describe a
(AB U {*, b})<w a winning strategy for if in
W (A, (Bj)).
Of course a is a strategy that lets 11 play finite sequences instead
of single letters, but anyway it is easy from a to derive a (classical) strategy a'
U {,b}).
I-(AB
AAw
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
0
C F(ao al):
((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
with O*u = v such that v+PI =
-*(ao,
(ao,... Ia,+,) and nf has replied
=
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,
and a 0 {b,
? (})<w
-}:
(u^-"(a))'Ph = uePh (a)
(U -(b)) Ph = 2ePh (b)
*< )) Ph =
(u^'(a,
(u'a(b, )*-))"P
(u
)( *)
-(*-
so in particular, v( /b)P
o All = (0)(
E A'
Vi < n (A
and
3a E A'
U
(*-
)
_PP
())
an, An)( /b). Assume also:
a
al an)< EF((\
,.I
Iai+J) with i < i + j < n
(ao, ...
...
We define a (u) = a ((ao,.
(i) if
((
EA (ao,
3?
o A = (l)'(ai+1,...
( zzVa
=
A8, al, A7...,
Z*(ao,
=
=(2^(*-))Pb
P'
)
-ePh
=
.
I ai+1) F(.))
a,,+1)) by:
(0)
(ao,...
zz4
:0 E A' (ao,...
a,,+1)%a
E
F(ai.
a,,1+1)%a E
?1)
then
Aj1+l= (0) and Vi < n
An+1 = All
Oa(u) = (b,1+1,Bn+l)
where
bn+I
((ao, A8 +..
A
B,1+I =((ao,
Vi < n (A =(O)
(ii) if
==>
a
A.
.
a
3a EA" (ao*,
but
Va E A' (ao,...
A
I an+l)%a
F(aai)
a,,+
))
a1,
A"+'))
an+l)
'a E F(ao.....ai))
WADGE
HIERARCHY
AND
VEBLEN
79
HIERARCHY
then
) and Vi < n A"+l = A"
A"+l = l)(a,+
+(u)
(b,1+1 5 Bn+
)
where
bi+I = T((ao, A+l .
Bn+I =
((ao, A
an, A+lan+,))
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
A" a ,A.
A",
V-Pb =-r*(ao,
AA)
there exists an integer k such that
>)
,((,,,,..0
(v(
A
=*(ao,
. Iajl,A"-l)
ifj>0
k
0
=(ifj
then set
for i < j:
A"+
fori >jand 30a E A'
fori > jandVae
u' '
A:+=
E F(a,,......at)
E A' ua
And nf plays a(u)
An
;)
lF(a
A+
A"+= (l)-(aj+
.
an+,)
+ where
Bj, bj++,BBj+l,...,I bn+l B.+,)
b(B.,)^(b,
k
givenO < p < n + 1 -j:
((ao, AO8 ..aj+p-1
bj+p
((ao, A+1
Bj+p-T
.
Aj+P
(0(), (O)...
) or
(ii) ( (1)^w, (1)(w,
(iii) ((0), (0) I .
k
A j+))
aj+p,
m < co) is ultimately constant:
aj+p-1,, Any+lj1,
Note that for each n E co the sequence (a "'I/n <
it takes in fact one of the three following forms:
(i)
aj))
.
or
(\)^w,
I (0), (1)"w,
...
wO
Inductively define the sequence (Mn )nE, by
o mO is the least integer such that Vl > mOAl= Ai;
o for n > 0, mn is the least integer such that
Vn Vl > mnVn' < n (A' =
& Anmtl
AMnm
AM1)
80
J. DUPARC
Since I played (at/i E co
>=
a E AO, 11 played 6 = U*a which satisfies
i < n}Vq > Pn (
supfmM
VflPn
=Z*(ao,A ...
[
Pn)
an, A >nC(>
[q)
Hence
' = lim(f i
*(ao, A0...
q)P" =r l
no
q<w
an, An)
A? .. I,.
=rt*(aO,
By construction every position (ao, A10,.
ak- 1, Ank
because:
Vn < co A
(0)
a E F(ao).....ll)
AIn,,
j
An -(1)"w
(ao.
Va'
an)'W
E AO(ao,.
-
I)
, an, An"...)
is legal for any integer k
C a
, an)-w-'a
F(ao,...,a,)
Hence
U*a
6
f( /b)l
E (B)b
(=
filbP(
/b)) E B
Which proves oais a winning strategy for if in W (A, (Bjb).
(d)(z=):
is by contradiction. Assume (Bjb
-A ?w B~b .~
(-A)
?
(B~bja
A, by determinacy this means
Aw
A
(A*) n, BA ..
>B lb
Bb
which contradicts the non self dual character of B".
is immediate.
(~z):
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<W, so that 0t is nl-complete (e.g., 0 =?(
-(
or
O =O.2),
Vn <wc, (Q" )b is S?1 -complete
(it is a S"+ set, and every S"+ set continuouslyreduces to it).
PROOF OF LEMMA 31. First, remark that given A C AA
E,A" e
each integer n the map f:
is continuous
f
reduces any
do A .
w
A
-
eachfA
(AUb})t
1 ffunction.
(Ai)
Since
limfn is a Baire class re
Ah, one obtains (Aj"
We prove
((d)b
(A
E S"?+2.This proves (o"n)b
reduces every
?
+ defining for
since
=
fJ (o)
e
e n+
on n. Assume
set by induction
(Abt)s
frset, we
oBeshow for any A e
E
k+2 A
y (ng+ i)e
by induction on
81
WADGE HIERARCHY AND VEBLEN HIERARCHY
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
kil E co k,1,-IE co k,,2_
where A(k,, k,,
ple set F1, =
-
A(k,,Sn_1-.ko)
A(k,, k,,_ .k,_
....ko)
w
is closed or open depending on the parity of n. For examif A(kSk __._ko) is closed and F, = -A(k k.__ ko) if
A(k,,kk,0_j...kO)
is open, with (kn,.kn-1.
,ko) q$(lh(u)) for some bijection q$bei._0)
tween co and ,n+l.
Hence by <w-minimality A' E Lo+, thus A'
tion 30(a):
A <W
(?0-1)
_ (-n+l
<w
thus by Proposi-
(Q-n)h
31
)
By the forthcoming lemma 37 the Wadge degree of any Lo+, complete set is 2
when n = 0 and it is
when n
i&
>
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 +-
with 'e, > a > ak > ... > ao > ? andVi
Define:
Q(a)
=
Q(KC'k)
where Q(t&') with 0 < /3 < 'E,
+
l.VO
< k, O < vi < K.
* Vk + ......
+ Q(K'o)
is defined by:
IfB = 0:
If
Q(t'Cf)
= n E co andn
If,
= A + n,
,
54 0:
cof(A) = K and
= A + n, cco
If,
= A, co < cof(A)
Where /3
n E co:
ofQ(A) < K and 0 54 n E co:
If
card (0r)
< K:
Op is a fixed map for limit /3 <
< K and is cofinal in P. For example:
'e,
Q(
Q*tC)
Q(P)
sup n()).
of cofinality
K ('+1)
=fl
n
n Eco,6=
Oor
=K<
0)
=Q(fi-1)'
then
K"
+
Q(Kfl)
Q(Kfl)
=?
= Q(1)
Q(')
-
(a) if P&= K
* VO
P/3}
< K, and where
82
J. DUPARC
(b) if P < K#then
A
1,1-8
+ '.
[l
0/?Aj
+
with fi >/n> ..>. > Po > O and Vi < n, O<ui <.
(i) if uo is limit
(0# =
K
/t10.,u
. .
tt'.,,+
: 'U < 'U }
(ii) if io is not limit
1#
THEOREM 33.
+
=S f K,)1,
Let 0 <a<
Kfl.
***+
-
(,u0
1) +
KO
C
0
: }#
E
d,'?Q(aE) = aE
PROOFOFTHEOREM
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:
a
with '6, > ae >- ?k >
Recall:
f
. ..
>
=
K
Clk
.Vk
+
. . . +
K;?g?.VO
> Oand Vi < k, O < vi < K.
ao
-(a) = Q('f'k)
*
Vk +
.
* VO0
?Q(1'0)
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) =
where sup0c,0 = ao so
sup Q('&O)
d'Q (a) = sup{dt,,A + 1:
A gw A,
= sup{dwA
+ 1:
-A
=sup{dA
+ 1:
dwA <
wA, ,A
KtJ
A, A ?wQ(a)
<w Q K)
for someO
}
0 E c0}
for some O
< ao}
a.
(b) Otherwise: the result relies on
LEMMA34. Let A C A'
-A, A
<w Q(a)h
A non self dual,
=>, A <W Q()b
for some
f/
< a.
because
do?Q(a) = sup{dt,,A + 1: -A :9, A, -A, A <w Q (a)h}
=sup{dwA
+1:
-A
wA,
A<wQ(fl)b
= sup{dwA + 1: dwA < fl for some fl < a}
: a.
forsomefl<a}
83
WADGE HIERARCHY AND VEBLEN HIERARCHY
PROOFOFLEMMA34. 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) < (and
i
n > 0)
= ao + I if A=0(andn>O)
i <. Note that < a.
so that in any case Q (ao) = (Q
CLAIM35. Let A C AO, A initializable,
or A <w Q(4) hforsome
<
-A, A <WQ~()
[b
,-A
with 6 E {-1, 0, 1} so that 4 + E < ao holds
Notice Q (j n('+ Q
in any case: either there exists an integer m 74 0 such that 4 + m = 4 in
K
and Q ([) =
(S(+1)
K
(for the same 6)
which case Q (4) '= Q (S(+0
or there is no such integer and then 4 + E < 4 < ao. Moreover, from
-
b
--A or A <W Q (0Y) we get A <w (Q (G +
(?
o =?
some ,8 < a since Q (4j +
~~~~~~~~~~b
) so that A <w Q (fl) for
+ i).
PROOFOFCLAIM35. It comes from Proposition 29(a) and Proposition 30(b).
A, A <
so
b
Q(
= dwA' <
* -A*', A` <
(Q g)-b)*
<
<w Q ()
< a, and by induction -As or A'
-A or A
<w?Q
Q
(4)b
b
thus
35 -1
(UYb
CLAIM36. Let A C Al, A non self dual but not initializable,
b
<WQ (
A,A
A <w Q (If) for some I3< K0- = a
,
whichrelies on:
(i) if 4 is successor then A <w ( (d- 1)
(Q(4 -
1)~"* v)-=Q
(ii) If c < cofG4) <
notice
((sup
Q ()-
= n
* v
Q (o)")
(iii) If cof(d)
=i
ii
(o
(K
for some v < i notice
and
V)
< a.
(,jo-l).V
then A <w ((supoc?
') * v) for some v < t
oe Q(O>)
=-
(c(ao-
then A < w
K((+6)
-)
* v)b
with
)v)
(< ))b
6
K (ceo-l).v < a.
and
for some 4 <
E {-1, 0, 1}
4
and K
notice
< K
< a.
PROOFOFCLAIM36. The proof goes by induction on dwA. By Proposition 14, there exists B. C C Al, B, C <w A with B initializable; so that
A _w B
C. By induction:
if C is self dual then
84
J. DUPARC
C
where II < i, Ci is non self dual and Ci <,
>Cl
A.
C <,
So, for each i E I:
(i)
4
-=Ci<c
is successor
(Ql
(
* v,)b for some vi < K.
- 1)
Set vc = sup{vi: i E I}. Clearly0 < vc < K and
C <w (Ql
(ii)
Set vc = sup{vi : i
0
1)Y * vc)b.
- Ci <? ((supoc<? Q (0))
< cof(G) < K
co
-
E
.
for some vi <
* vi
* vc)
I}. Clearly C ?w ((supoco, LI(0))
and
< VC < K.
(iii) cofG4) =
for some Si < .
Ci <?, Q (gi)b
z-
Set Sc sup{ji: i E I}. Clearly 0 < Sc < K and C LwI (Syb
So that, for both C non self dual and C self dual, the following holds:
4 is successor
IB <?W(Q
co < cof)W<
*
('-1)
C <?w (Ql
for some VB < K
for some vc < n
VB)
- 1)> * vc)b
K
(B <w ((suPOCO,
IC <w
for someVB
* VB)
fl (0)
<
K
b
* vc)
((supoo, fi (0))
for some vc < K
cof()=/
JB
Q (YB)'
<W
C <W?Q
for some
for some
(c)b
To get the result, it is sufficient to set 4
and v=
LEMMA 37.
VC
+
VB
SB <
c<d
max{fB,
Sc} in case cofG4)
in the other cases.
36H34H33
we have
GivenA C A<w with d(0A <
dc A- = Vo(d?A)
(d?A+e)
if cof(sup{f. < dcoAA = 0 or A is limit}) = 0
if cof(sup{). < dcoAA = 0 or A is limit}) = K
co < cof(sup{). < dcA : A = 0 or A is limit}) <
t+Iif
{-1
wheree =
;.
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:
(a)
1 <<
=(D)-(
(D
0
d Q (a)'
=I
; (-l
OC:
Q (a)=
-
Q-? (
. dc7Q (a)[ ' _
(0-)
85
WADGE HIERARCHY AND VEBLEN HIERARCHY
a = A+ n, cof(A) = an,
E c:
Q ()=Q
Kas~
d~ol (a),-
(K`l)
a =A + n, co < cof(A) < A, n E co:
Q?(x)'
Since ((0 * 2)
is 2 if n = 0 and
")b
is
d
:
= Q
(a)'v
-
-
<,a+1)
+I-complete, the Wadge degree of any Lo+ -complete set
L
if n > 0. So Borel sets of finite rank are those of degree less
ii
n
=eo.
than supneir
This remark leads to:
n
A non self-dual.
Let A C A',
THEOREM 38.
A E
#U
A orA
0 < a <'EO
-a,
En
-
n<co
Recall that if A is self-dual, there exists I C A<('
A _W Eic
Ai
where each Ai is non self dual and verifies Ai <w A.
Thus theorem 38 provides for every Borel set of finite rank A a normal form Q
Wadge equivalent to A.
At last we give a proof of the first theorem saying every non self dual A of finite
Borel rank is Wadge equivalent to Bb for some conciliatory B, and moreover, B
can be obtained from A by only adding finite sequences.
3. First part of Theorem 3 directly comes from previous ThePROOFOFTHEOREM
orem. It just remains to prove last sentence of Theorem 3. It simply relies on the
such that A -w
fact, given A C A' and B C A<'
that verifies A U s (A) -c B.
o A U s(A)
>? B is obvious
o to show A U s(A)
<I
B", one can define
and does not depend on s(A);
z is a winning strategy
B, assume
and set for each u E A^w
u E s(A)
()E
s (A)
-<I>
4=* '
(z*u)'bo
s(A)
C A`
for 1 in W (A, Bb)
E Bb, and
) E B.
The strategy that consists in skipping when I skips and applying z (turning "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.
The strengthof Borel Wadgedeterminacy,Cabal seminar81-85,
[3] A. LOUVEAUand J. ST.RAYMOND,
Lecture Notes in Mathematics, vol. 1333, Springer Verlag, 1985, pp. 1-30.
86
J. DUPARC
, Les propriftts de reductionet de normepour les classes de Bornliens,FundamentaMathe[4]
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, Wadgedegrees and descriptiveset 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, Continuousincreasingfunctions 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.
, Reducibilityand determinatenesson the Baire space., Ph.D. thesis, Berkeley, 1984.
[9]
EQUIPE DE LOGIQUE
MATHEMATIQUE
CNRS URA 753 ET UNIVERSITE
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