A general model completeness result for expansions of the real ordered field

ANNALS C PURE ANL APPLIED LOGIC

ELSEVIER Annals of Pure and Applied Logic 95 (1998) 185-227

A general model completeness result for expansions of the real ordered field

Steve Maxwell Mathematical Institute, University qf Oxjiird, 24-29 St. Giles. O,xford OX1 3LB, C’K

Received May 20, 1997; received in revised form January 29, 199X Communicated by I. Moerdijk

Abstract

We approach the subject of o-minimality from the point of view of tame systems, following the work of Charbonnel and Wilkie. This gives some general sufficient conditions for a system to be model complete and o-minimal. We are then able to obtain the following generalisation of a recent result of Gabrielov (which in his case applied only to analytic functions): A polynomially bounded o-minimal expansion of the real ordered field by a collection of restricted C” functions, which is closed under partial differentiation, is model complete. @ 1998 Elsevier Science B.V. All rights reserved.

AMS cluss~fication: Primary 03ClO; Secondary 03C80; I251 5

Keywords; Model complete; O-minimal structures; Tame systems

0. Introduction

A tame system will usually be some collection of definable subsets in a structure

expanding R = (@; +, - , .; 0,l; <), where the key part of the definition of tameness is

that for each set A in the tame system, 3N E N such that all sets of the form A n X (where X is affine) have at most N connected components. The main general result

concerns an expansion of E to a structure ([w, {f})fE~ with R = (JnE.\+ R,, where

for each n E Nf, R, is a collection of C’ functions defined on all of Iw”. We show

that if the quantifier-free definable subsets of the structure form a tame system and

the topological closure of an existentially definable set is also existentially definable,

then the structure (a, {f}) .fE~ is model complete. From here it is easy to show that

the collection of all definable sets is tame, and hence obtain the o-minimality of the

structure. (Recall that (E, {f})ft~ is o-minimal if each definable subset of [w has a

finite boundary.)

0168-0072198/$19.00 @ 1998 Elsevier Science B.V. All rights reserved. PII SOl68-0072(98)00012-S

186 S. Maxwell I Annals of Pure and Applied Logic 95 (1998) 185-227

The main ideas used in the construction of suitable tame systems originally appeared

in a paper by Charbonnel [l]. However, many of his proofs contain errors, some of

which were rectified by Wilkie [9]. This material is covered in Sections 1 and 2.

The third section contains a Cell Decomposition Theorem for certain tame systems,

under an assumption regarding the boundary of sets. In the fourth section we explore

further some properties relating to functions in tame systems, including the selection

of a CP function and other results which are used in a more recent paper by Wilkie

[lo], in which he proves the o-minimality of iw expanded by unrestricted Pfaffian

functions. We then work towards weakening the boundary assumption to establish

the model completeness and o-minimality results already mentioned (Theorem 6.5 and

Corollary 6.6). In the final section, we note that the definable sets in uny reduction

(expanding E) of an o-minimal structure form a tame system. This mostly reduces

the issue of model completeness of such a structure to the verification that if A is

existentially definable, then 2 is existentially definable. By considering only bounded

sets, we are able to use a generalisation of a result of Gabrielov [3], concerning

bounded semianalytic sets, to deduce the model completeness of certain expansions

of R by restricted C” functions, We also obtain Gabrielov’s results on the model

completeness of certain reductions of R,, as a corollary.

1. Tame systems

The whole of this first section is essentially the account given by Wilkie [9] of

original work by Charbonnel [I]. We will be working throughout with the set R of

real numbers, with its usual topology, and where R” will always be assumed to have

the product topology. We begin by fixing some notation.

Notation 1.1.

1. If A C_ R”, then 2 will denote the topological closure of A in R”.

2. If A C R” and m dn, then we denote the projection of A onto its first m coordinates

by JJ,,,L4, i.e.

JLwJAl= {h,...,xm): 3x,+1 . ..3x.(xt,...,x,,x,+l,...,x,)~A}.

3. Unless otherwise stated, Z will always denote an n-tuple (xi,. . .,x,,) and 9 will

always denote an m-tuple (vi,. . . , y,).

4. We call a set X C R” an afine set iff it is a translation of a subspace of R”, i.e.

X = {X : A4X = if} for some n x n matrix A4 and n-tuple Z.

5. For f: R” + W”, let the graph of f be denoted by

r(f) = ((2, ?) : .fG> = B 1.

6. If A G W+‘, and X E R”, then we denote the fibre above X in A by

A,-={~ER:(T,~)EA}.

IX7 S. MaxwellI Annals of Pure and Applied Logic 95 11998) 185-227

Following the usual convention we let

rw=([w;+,-;;O,l; <).

We call a subset of BP semi-algebraic iff it is a quantifier-free definable set (where

we allow parameters) in the structure @.

We say that a set is 3i-definable (in a given structure) if it is existentially definable,

i.e. definable by a formula of the form 3 $$(X, _? ) where 4 is a formula without

quantifiers.

Definition 1.2. An K&system is a sequence

+=(.Pn:nEN+)

(where N+ = N\(O)), such that for each n E PQi+, -yn is a collection of subsets of Iw”.

If 2 and 9’ are iW-systems, we will write 9 C: 9 to mean that ,% C .yi (Vn E Ni ).

We also write A E S to mean that A E :Pn for some n E !+J+.

Definition 1.3. An [W-system 9 is called tame if conditions (Tl)-(T7) are satisfied

(tin E N’ ):

(Tl) ~WE.‘P,;

(T2) AE&P,c(A)E~~ where n-:iw” ---f R” is any permutation of coordinates;

(T3) A E & =+ (A x 58) E ,Pj+,,,;

(T4) A,BE;~~P,(A~B)E:Y~;

(T5) Every semi-algebraic subset of Iw” is in J+;

(T6) A E Yn =F 3m an and closed BE .Ym such that A =II,,,[B];

(T7) A E :‘p, =S 3 N E N such that for all affine sets X & aB”, the set A nX has at most

N connected components. We will refer to N as a (T7) connected components

number for A.

Remark 1.4.

It is trivial that (T5) + (T 1 ), however, we retain condition (Tl ) because of

Remark 1.10.

It is condition (T7) which leads us to use the word tame. We also note that it is the

most important condition in the sense that there are clear connections between (T7 )

and the definition of o-minimal&y. In fact, it is not difficult to prove that every set

definable in an o-minimal expansion of R has a (T7) connected components number

(Lemma 7.1).

It is an easy exercise to verify that if ,P is a tame R-system, and A E .‘p,. BE -Pm,

where n,mE M+, then AxBEY~+,. (For proofs of this and other easy results in

this section see either Wilkie [9] or Maxwell [7].)

4. We will often let B C [w” denote a ‘box’, i.e. a set of the form

188 S. Maxwelll Annals of Pure and Applied Logic 95 (1998) 185~-227

where Z E R” and E > 0. All such B are easily seen to be semi-algebraic and hence

lie in all tame systems. (If we do not specify whether a box is open or closed, it is

because it makes no difference to the argument.)

Suppose 9 is any R-system. Then we define a new system 9” as follows:

9;=

I ijAi:&J+;A ,,..., A, E% ) i=I

.9”=(.qnu:nEN+).

We observe that 9 C P”.

Lemma 1.5. 9 is tame * LP” is tame.

Proof. It is easy exercise to verify that conditions (Tl)-(T7) hold. 0

Suppose 9) is any R-system. Then we define a new system Pu as follows:

@= {rI,,,[C]:mEkl+; m>n; CEPm},

9”=(~~:nfNf).

We observe that 9 C .C?‘u (since m = n is allowed).

Lemma 1.6. 9 is tame + 9” is tame.

Proof. Again, most of the proof is straightforward. I give proofs of conditions (T4)

and (T7) for 9’“.

Let A,B E 9”, with A =II,,,[C], B=II,,[D]. We will assume wlog that p3m. Then

=II m+p-n,n[C x [Wp-l nql+m-n,,[~ x R-7

= &+p--n,n[4C x ~p-n)l nKn+p--n,m x IWrn-“l,

where 0 : (Wm+p-” + [W”‘+p-” i s the permutation of coordinates which swaps the

(n + i)th with the (p + i)th coordinate for each i E { 1,. . . , m - n}, and fixes all other

coordinates. Now by considering the coordinates in 3 groups - the first n; the next

(p - n); the next (m - n) - we see that, in fact,

A n B =IIm+p--n,n[o(C x lR-*) n (D x R”-“)I.

Now (T3), induction, (T2) and (T4) for 9 imply that A nBEP’“, and hence (T4)

holds for 9’.

S. Maxwrlll Annals of’ Pure and Applied Logic 95 ilW8i 185-227 189

For (T7), suppose that A =II,,,[C], iV is a connected components number for C and

X C: R” is affine. Then A nx =ll,,,[C n (X x KYn)]. Now since X x KY’-” is alline.

C n (X x IR”-‘) has at most N connected components. Hence N is also a connected

components number for A since the number of connected components of a set does

not increase under projection. 0

Suppose Y is any R-system. Then we define a new system @ as follows:

Remark 1.7.

1. When r = 1, we interpret the expression AI n n;=, z simply as Ai. Hence it is clear

that 9 C -Yn.

2. The rea&r for the intersection in the definition of .q” is to ensure that (T4) holds

for YE.

3. If R”EY~, then AEP~P,JEP~~, since A = R” n2. It is not important for our

purposes whether 9”” IS closed under topological closure (i.e. whether A E 9” =+

2 E IV”), and in fact in general it is not.

Lemma 1.8. Y is tame * Yp” is tame.

Proof. Again I omit the straightforward proofs of (Tl)-(T5). For the proof of (T6)

for Yfi, we note that by (T6) for .“P we may let AI =l&,,,[C] where m >,a, C E Y,,,

and C is closed. We also let B= R” n OF=, AT. Then B is closed and by (T5) for

./p ‘~ fl, B E 2”. Now,

A = A, n n Ai =II,,,[C] n B =&,[C fl (B x ET-“)] i=2

But, being the intersection of 2 closed sets, C n (B x Wpn) is closed, and using some

of the other conditions it is easy to get [C n (B x Rm-n)] E 2”. The proof of (T7) is

somewhat more complicated:

Again, by (T6) for 9 we may let A, =II,,,[C] where m>n, C E .‘p, and C is

closed. Let Z & R” be affine. Let B = A2 x . . x A, E .Pn,_,.

We now define A c R” as follows:

A={(x I,., x,,x ,,., x, ,..., x I,., xn):x ,,..., x,ER}.

Let Y=Z’nA. Then

= n,,-,. [(B XndC]) n Y]

190 S. Maxwell1 Annals of Pure and Applied Logic 95 (1998) 185-227

= Ilnr,n [K-n+m,nr[(B x 01 n y]

= &,?I 0 KIr--n+m,nr [(BxC)rl(Y x urn)]

= &--n+m,n[(B x C) n (Yx ~m-n)l

since C is closed. Now B, C E 9’ + (B x C) E P. Also Z affine + Y x lP-” affine. Now

since the number of connected components of a set does not increase under projection,

we are done if we can prove the following claim:

Claim 1.9. Let n E N+, DE 9$. Then 3J E N such that for all afine sets X 2 t!P, the set DnX has at most N connected components, (i.e. D has a (T7) connected components number).

Proof. Since the proof contains a large number of small steps, we label them to make

it easier to follow:

Recall that the affine sets of Iw” are exactly those which are of the form {X E [w” :

MX = i?} where M is an n x n matrix over [w and a E Iw”. It follows, by multiplying

out the equation A4z - a = 0, that there exists a polynomial p(X, y’) E Z[?i, y’] (where

y’ is an m-tuple with m = n2 + n) such that p has the property that given any affine

set Z & [w”, there exist parameters bl, . . . , b, E [w such that Z = {z : ~(2, g) = 0).

Let EC W+m+2 be the semi-algebraic set (E E 9?,+,,,+2) defined by -

(~,~,R,E):~P(~,~)~<&A &%R2 . i=l

Note that E fl (D x KF2) E Pnntrn+2 (by the fact that D, E E 9 and by (T3), induction

and (T4)) and hence has a (T7) connected components number, N say. We will

show that N is also a (T7) connected components number for 0.

(The reason we include f throughout the proof is to ensure that N works uniformly

for all affine sets X C [w”. We note also that for the remaining steps, we need not

be concerned with whether the sets we are considering lie in 9.)

Assume, for a contradiction, that X & aB” is affine such that DnX has more than N

connected components. Then, by point set topology, 3 pairwise disjoint, open sets

Ur,...,U,v+r such that

(DnX)n U,#f!I (for i= 1 ,...,N+ 1) and @r-M)& U Ui. i=l

We let z=(DnX)n Ui (for i= l,...,N + 1).

We now choose Ro E Iw large enough so that for each i = 1,. . . , N + 1, 3~ E Ti with

cF=, xi’ CR;. We let

S. Maxwelli Annals of’ Pure and Applied Logic 95 (1998) 185-227 191

6. Define C Cr R” as follows:

Then since the U,‘s are open, C is closed. By definition of BR(), C is also bounded.

It follows that C is compact.

7. Choose h’ E R”’ such that X = {X : p(X, 6) = 0} (where p is the polynomial in 1).

8. We show that CflX=0. Suppose ZECnx. Then ZEDnx, hence FEUD’ Ui

(by 4). But now X $! C, by definition of C in 6.

9. By 7 and 8, we see that the continuous function 2 H 1 p(X, &)I is non-zero throughout

C, and hence (since C is compact by 6) obtains a minimum value on C, say EO >O.

10. Now let us define the affine set Y = R” x {(~,Ro,BO)}. We will show that if S = (E 13

(D x [Wm.+*)) n Y, then S has more than N connected components, contradicting the

choice of N in 3. We remark that, putting together the definitions for E and Y, we

obtain S = SO x ((6, Ra, ~0)) where

so= x:(xED)AIp(x,ij(<EoA i ( i;, x%&g}. c

11. We show SC (U,“=’ Ui) x lW+*.

Suppose Z E SO. Then (by 9) X $ C. But since % E D and X E BRA, it follows from 6

that 1E U;“_:’ U,.

12. We show that SfI(Ui x FP+*)#(d for each i= l,...,N + 1. (t)

Let iE{I,...,N+l}, then by 5, 3FETi such that ~:=,x,?<R~. We can now find

a suitably small neighbourhood W of X such that:

(a) W C U; (since X E U,, Ui open);

(b) / p(W_ g)l <CO (‘v’i~ E W) (since x E X implies that p(X, $) = 0, and p is contin-

uous );

(c) W C B,Q (since C:=, x’ CR;).

Nowsincex~TiCIDnX~X~D, 3~c~WsuchthatK~~D.Butnow(b)and

(c) imply wo E SO. Hence (tio,g,Ro,~g) E S, but also by (a) (?&J,~,Ro,Eo) E (U, x

[w m+2), which is what we require to prove (t).

13. Now since the Ui’s are disjoint and open, 11 and 12 imply that S has more than

N connected components, which is the required contradiction to the choice of N

in 3. C1

This concludes the proof of Lemma 1.8.

Remark 1.10. By checking the proofs used, we observe that if we change our basic

domain to any non-empty set instead of R in the definitions, then

.P satisfies (Tl)-(T4)+PU and @ satisfy (Tl)-(T4).

192 S. MaxweNIAnnals of Pure and Applied Logic 95 (1998) 185-227

Also, if our basic domain is any topological space (with Cartesian powers of the

domain still equipped with the product topology), then

9 satisfies (Tl )-(T4) + PR satisfies (Tl )-(T4).

Systems satisfying (Tl)-(T4) are called pure in Wilkie [9].

2. The Charbonnel closure of a tame system

The results in this section are again those of Charbonnel [l], and also appear in

the form presented here in Wilkie [9]. The proofs given for Lemmas 2.6 and 2.7 are

my own. From now on, we will be assuming that Y is a tame system. Following

the notation used in Karpinski-Macintyre [4], we define the Charbonnel closure of Y,

C/z(Y), as follows:

P=Y,

Y m+’ = (((Ym)“)=)R (Vm E N),

Ch(Y)= u Y”.

mEN

Corollary 2.1. Ch(Y) is a tame system, which is closed under the operations ofjnite

intersection, jinite union, projection and topological closure.

Proof. An easy induction shows that Y” is a tame system (v’m E N). Conditions (Tl)-

(T3) and (T5)-(T7) now follow immediately for Ch(Y). It is easy to see that

mdk+YmCY’k (*)

from which (T4) for Ch(9) follows easily. Hence Ch(9) is tame.

By (T4) and induction, Ch(Y’) is closed under finite intersection.

IfA 1,. . . ,Ak E Ch(Y), then let m = max{mi : Aj E YmS}. By (*), Al,. . . ,Ak E 9”’ and

hence U,“=, Ai E (Ym)“. Then lJ,“=, Ai E (((Yym)U)IT)K and it follows that l_l%, Ai E

Ch(9).

If AE Ch(Y),, (i.e. A E Ch(9) and A&W) and k<n, then let m E N such that

A E 9”. Then I&[A] E (9”“)“. Hence &k[A] E (((ym)U)n)n and it follows that

%,k[A] E CA(Y). If A E Ch(Y),, then let m E N such that A E 9”‘. Then A E ((9”“)“)‘. Hence

(R” n 2) E (((Y”“)U)n)fi (by definition of “) and it follows that AE Ch(9). q

Definition 2.2. We define a Ch-formula (in a given language expanding Z(E)) as

follows:

(i) All atomic formulae are Ch-formulae.

(ii) If 4, $ are Ch-formulae and x is a variable, then (4 A rc/), (4 V $) and 3x4 are

Ch-formulae.

S. Maxwell/ Annals of Pure and Applied Logic 95 ll998) 185-227 193

(iii) If 4 is a C/z-formula, then 6 is a C/z-formula, where

(iv) Nothing else is a C&formula.

Remark 2.3. By the definition of 3, all C/z-formulae are formulae. The converse is,

of course, not generally true, but a major part of this work will be to find conditions

under which it is true (up to equivalence in suitable structures).

Although the proof of the following corollary is straightforward, it is a key obser-

vation as it will be used implicitly throughout the rest of this work (in particular the

implication 3h * 1).

Corollary 2.4. The following are equit’alent:

1. A E Ch(.y).

2a. A is equal to a$nite expression consisting of sets in .Y acted on by the operations

of finite intersection, finite union, projection and topological closure.

2b. A can be dejined by a Ch-formula in the structure (@,{X})X~.~.

3a. A is equal to a finite expression consisting of sets in Ch(Y) acted on by the

operations of jinite intersection, finite union, projection und topological closure.

3b. A can be defined by a Ch-formula in the structure (a, {Y})rEch(.~ 1.

Proof. The definition of Ch(Y) makes 1 + 2a trivial and 2a + 3a is immediate from

the fact that .Y & Ch(Y). Also 3a+ 1 is just a reformulation of the second half of

Corollary 2.1. We now complete the proof by showing that 2a ++ 2b and observing

that the same proof gives 3a@ 3b. This depends on an almost trivial correspondence

between our operations and logical symbols. To make the point clearly, we assume

that if X E :/:,, then $x(X) is the n-place predicate belonging to our language such that

(R{X})XE.u b:x(x) @ xEX.

We also write d(X) - I@) to mean

(R {X})XE v I== VZ(G) * $(X)).

The following are then all trivial to verify:

em&) - 4,463 A h&O

~.4ubG) N 4Acf) v hm;

Sn,,,.,,[~]~~~“~Y,+l,...,~Y,~~~~,Y,+l,...,Y,);

4#> cy #A(3

This concludes the proof. q

194 S. MaxweNlAnnals of Pure and Applied Logic 95 (1998) 185-227

We now look at some of the measure theoretic properties of the system C/z(Y). This

was an approach first introduced by Charbonnel [l], and it gives rise to some new and

powerful methods stemming from the use of Fubini’s Theorem and Corollary 2.8.

We begin with an easy lemma.

Lemma 2.5. All sets in Ch(9) are Lebesgue measurable.

Proof. Let A E Ch(Y), (n E fW+), then by (T6), 3m E N+ with m >n and a closed

BE Ch(Y),, such that A = II,,,[B]. But now

A=fiII,,,[Bfl[-i,i]m] i=l

and so A is the countable union of compact, hence measurable, sets. It follows that A is measurable. 0

We will let ,un denote Lebesgue measure on R”; e.g. if A & R” is null then we write

pn(A) = 0. The main motivation behind a measure-theoretic approach is that we will

be able to utilise the following consequence of Fubini’s Theorem.

Suppose A C IV+’ is measurable and let B = {X E R” : p 1 (A:) > 0). Then p,,+i (A) > 0

*Pn(B)>O. This result will prove very useful in the following lemmas.

Lemma 2.6. Let A E Ch(Y),, (n E IV+). Suppose A contains no interior points. Then

A is p,,-null.

Proof. Since we have already noted that A is the countable union of compact, hence

closed, sets which lie in Ch(Y), it will suffice to prove the lemma for A closed (since

a countable union of null sets is null).

The proof is by induction on n. The case n = 1 is easy since by (T7), if A E Ch(Y)l with no interior points, then A is a finite set of points.

Suppose the result is true for n. Now suppose for a contradiction that A E Ch(Y),+I is closed and contains no interior points, but is not pn+t-null. Let N E N be a (T7 1 connected components number for A. Now for each s E /Vi, I E Z, we define the

sets:

We note that by Corollary 2.4, Cs,l E Ch(9’) for each s E N+, 1 E Z.

Now let

c= u u Cs,r. sEN+ IEL

S. MaxwellI Annals of Pure and Applied Logic 95 (1998) 185-227 195

Then we note that (by definition of N)

X E C ++ A, contains an interval,

X E R”\C % A,- contains at most N points.

Now by Fubini’s Theorem, since A is not null, C is not null, which implies that

3r E N+, k E Z such that Cr,k is not null. But Cr,k E CZZ(Y)~ and is a closed set

(because A is closed and the mapping t H t + 2i/r is continuous). Our inductive hy-

pothesis now implies that Cr,k has an interior point, hence 3 an open box Bo C C,.k.

Suppose X E Bo. Then by the definitions of N and Cr,k, we have

k k+l 3E ;,--

[ 1 such that

2i - 2 t + ---, t+2’ CA,-,

r r r I

for some i E {i, . , N}. It follows that E E Bc, 3iE{l,...,N} such that (*i) holds:

[

k+2i- l,k+2i cA,.

I r 1 (*i) Our strategy will be to construct open boxes BN c BN_~ C . . C Bo such that (*i) fails

VX E B;. The result then follows since we have BN n Bo = 0, which is a contradiction

since BN C Bo.

Consider B; x Zi where Bi & KY” is an open box and Z; C R is an open interval. Since

A has no interior points, 3(&a) E Bi x Z, such that @,a) @A. Furthermore, since A is

closed, this is an interior point of Rnfl \A and hence 3 an open box Bi+, C: B, such

that ‘d’x E B,+I (X, a) @A. We are now done by iterating this argument for i = 1,. , N,

where we let Zj =((k + 2i - l)/r,(k + 2i)/r). 0

Lemma 2.7. Suppose A E Ch(Y), (n E N+) contains no interior points. Then 2 con-

tains no interior points.

Proof. By induction on n. The case n = 1 is again easy. Suppose the result is true

for n. Suppose, for a contradiction, that A E Ch(9’),+1, with no interior points, such

that Bo x I s 2 where Bo C R” is an open box and Z C R is an interval.

Let N be a (T7) connected components number for A, and let I,, . . ,IP,T+~ be any

pairwise disjoint intervals such that Z, C Z for i = 1,. . , N + 1.

Now since Bo has an interior point (and lies in Ch(Y)), the following argument

holds for i = 0: Let

Suppose C,+l has no interior points. Then since Ci+i E Ch(Y) (by Corollary 2.4), the

inductive hypothesis implies that Ci+ 1 has no interior points. Hence the set B,\C,+, is

non-empty and open, hence contains an open box, Di+l say. But then the open box

(Di+l x Ii+,) Cl but does not meet A, which is a contradiction. Hence Ci+l contains

an open box, Bi+l say.

196 S. Maxwelli Annals of Pure and Applied Logic 95 (1998) 185-227

But now observe that we may iterate the argument for i = 0, 1, . . . , N. Then since

BN+I CBN C...CBO,

Ntl

‘v’ZEBN+~ 31 . ..3t~+l /\(tiEZi /\(%,ti)~A). i=I

Because the ii’s are pairwise disjoint, by definition of N, A,- contains an interval

(VT E BN+~). But now by Fubini A is not null. Hence A has an interior point (by

Lemma 2.6), which contradicts the hypothesis on A. 0

Corollary 2.8. Let A E Ch(Y). Then the following are equivalent:

1. A has no interior points.

2. 2 has no interior points.

3. A is null.

4. A is null.

Proof. (1 +2) and (2 =+3) are the results of Lemmas 2.7 and 2.6.

(3 =+4) and (4 + 1) are trivial. 0

We also get the following easy consequence:

Corollary 2.9. Suppose A, B C Iw” such that A has an interior point, B does not have

an interior point, and B E Ch(Y). Then A n (W\B) h as an interior point. (Note that

A n (W\B) need not be in Ch(Y).)

Proof. Since A has an interior point, so does A” (the interior of A). Also, since

BE Ch(Y), i? does not have an interior point (by Corollary 2.8). It follows that

3X E A” n (lR”\B), but since it is open, X must be an interior point of this set. Hence

X is also an interior point of the set A n (W\B) as required. 0

3. The Cell Decomposition Theorem for Ch(Y)

In this section we will prove the Cell Decomposition Theorem (CDT) for sets in

Ch(sP) modulo an assumption which says roughly that if a set is in Ch(Y), then so

is its boundary. Our main motivation lies in the fact that if we have cell decomposi-

tion, then it follows easily that Ch(9’) is closed under complement, since given a set

A E Ch(Y),, and a decomposition of R” partitioning A, the complement of A is simply

the (finite) union of all the cells which have empty intersection with A. We begin with

the following lemma.

Lemma 3.1. Suppose C E Ch(Y)n+l and has no interior points and let Bo CR” be a

box. Then there exists p E N and a box B C Bo such that for each X E B, the Jibre C’

contains exactly p points.

S. Maxwell1 Annals of Pure and Applied Logic 9S ilYY8) 185-227 197

Proof. Let N be a connected components number for C, and for k = 1,. . , N + 1, let

BI;=

i rEBo:3r,~r:>0...3rr>O~(x,r, +...+v,)EC

I=1 i

We show first that BN+I cannot contain an interior point, since otherwise, by def-

inition of N, if X E BN+I, Cy contains an interval. But then since pn(B~+ I )>O, by

Fubini, p,,+,(C) > 0 and by Corollary 2.8, C has an interior point, which contradicts

the hypothesis on C.

We can now conclude that 3p E (0, 1, . . . , N} such that BP contains an interior point.

but B,+I does not. But BP+, is easily seen to be in C/r(Y), and so, by Corollary 2.9,

BP\BP+l has an interior point. Hence we may choose a box B C: BP\BP+,, and it follows

that V’:u E B, /Cyl = p as required. 0

We now prepare for the proof of the CDT for Ch(.Y’), by showing that if the graph

of a function lies in Ch(.Y), then the function has desirable continuity properties.

Lemma 3.2. Let f :A ---f R such that TEE Ch(.Y),+,. Let D, =D= {XEA: ,f’ is

discontinuous at x}. Then

(a) DE Ch(Y),;

(b) D contains no interior points.

Proof. (a) We first define the function g : A ---) R (which we think of as l/f) as follows:

T(g)= {(X,y):3z((x,z)Ez(f) Azy= 1) v ((%O)ET(f)A y= l)}.

By Corollary 2.4, T(g) E Ch(Y),,+, . Let us now define the set E as follows:

E= {x~A:3y3r>O((x,y)~T(f)r\(x,y+r)~T(f))~(x,O)~T(g)}

Note that r(f) E Ch(Y),+i =+ A E Ch(Y),, and hence, by Corollary 2.4, E E Ch(.V),,+,

We will show that E = D.

Suppose X E D. Then 3s > 0 such that for all neighbourhoods U of 2, 36 E C’ n A

such that If(u) - S(X)1 >F. It follows that 3 a sequence of points {?7i} converging to

X such that

If(ii,) - f(Y)1 >E (Vi E N).

There are now 2 cases to consider:

(*I

(i) {f(&)) is unbounded, which implies that (X,0) E T(g), and hence %E E.

(ii) {f(i&)} is bounded and h ence, by BolzanoWeierstrass, contains a subsequence

convergent, say, to a E 52. It follows that @,a) E r(f), but (*) ensures that a # f(x).

Hence, since (X, f(x)) E r(f), we have X E E.

Conversely, if X E E, we easily get a sequence {iii} converging to X, for which

{f(@)} does not converge to f(Z), which gives discontinuity of f‘ at X. Hence D = E

and so D E Ch(.Y),. This completes the proof of (a).


(b) We show that D contains no interior points. Suppose D does contain an interior

point. Then let B CD be an open box. For each m E N, we let

It is clear that F,,, E Ch(Y) (Qm E AJ). Also UrnEN F, = B, and hence pCLn(UmEN F,) >O.

Now since a countable union of null sets is itself null, ZlA4 E IV such that ,~,,(FM) > 0,

and hence (by Corollary 2.8) FM has an interior point. We may therefore assume that

B has been shrunk so that

Q? E B If(Z)l <AL (t)

It follows from (t) and the fact that B is open that QZ E B(i, 0) @T(g), and hence,

since B CD, that Q? E B, the fibre r(f),- contains at least 2 points.

Now obviously r(f) has no interior points, and hence by Corollary 2.8, the same is

true of r(f). But now applying Lemma 3.1, we may assume that B has been shrunk

so that QZ E B, r(f),- contains exactly p points for some p E N\{O, 1).

We now consider the following sets:

Since Qi U (22 = B, one of them must have strictly positive measure and hence an

interior point (by Corollary 2.8, since Qi, QZ E Ch(9)). We may therefore assume

wlog that B C Ql.

Let us now define a function h : B -+ R by:

~EBA(%~)ET(~)A

h(Z)=y * P--l

Note that:

(i) This is indeed a function since If(j&l = p (QZ E B);

(ii) h(n) >f(Z) (QF E B) since B C Ql ;

(iii) I’(h) E Ch(Y), by Corollary 2.4.

Now for each k E N+, let

Tk = {X E B : h(x) - f(x) > l/k}.

Now because T(h),I’(f) E Ch(Y), we have Tk E Ch(Y) (Qk E Nf). Also,

U kEN+Tk = B, since h(x) >f(x) (QZ E B). Hence pn(UkEN+ Tk) PO, implying that

I, >O for some K E N+. It now follows by Corollary 2.8 that T, has an interior

point, so again we may assume that B & T K. We now use the fact that (5, h(x)) E r(f)

(and B is open, B & TK) to obtain

QZ E B 32 E B such that f(G) > f(Y) + i. ($)

S. MaxwellI Annals of Pure and Applied Logic 95 (1998) l&--227 199

Fix Uo EB. Let L E N be such that f(Za) + L/K >A4 (recall that, by (t), IS(X)1 <M

(tin E B)). Now using ($) at each step, we may construct a sequence {Go,. . . ,UL} 5 B

such that for each i = 0,. . . , L - 1, we have f(?ii+t ) > f(&) + l/K. Hence iiL E B and

,f(U~)>f(Ua) + L/K >,M, which contradicts (t). Therefore, D contains no interior

points, as required. 0

We now introduce some definitions which are essentially the same as those used

when working in an o-minimal structure (see, for example, vdDries [2]), but where we

replace ‘definable’ by ‘in C/r(Y)‘.

Notation 3.3. 1. For each A E Ch(,Y),, let

C(A)={f:A+RIf 1s continuous and r(f) E Ch(.Y),, I},

C,(A)=C(A)U {-W(X)>,

where dms are thought of as constant fimctions on A.

2. For f, g E C,(A), we write f <g sfs f (X) <g(X) (EEA).

3. For f ,g E C,(A), with f <g, let

(f>g)A =(f,g)={(%y)EA x R:f(G<y<g@)}.

Definition 3.4. Let (il,. . , i,) be a sequence of zeros and ones of length n. An (il, . ,

&,)-cell is a set in Ch(Y),, obtained by induction on n as follows:

(i) A (0)-cell is a one element set (1.) 2 R (i.e. a point).

A (1 )-cell is an open interval (a, 6) 2 R where a, b E R U {+x}, a < 6.

(ii) For n31, an (it,...,i,,O)-cell is the graph r(f) of a function ~EC(A) where A

is an (iI,. . , &)-cell.

An (iI,. , i,, 1)-cell is a set of the form (f, g).d where A is an (iI,. . , i,)-cell and

f,gEC,(A) with f <g.

We say that A C_ iw” is a cell @ A is an (iI,. . , &)-cell for some tuple (ii,. . , i, )

C_{O, I}“, where n E fV)+.

Remark 3.5.

1. All cells lie in Ch(Y). This is by (T5) and the fact that for a cell A, by definition

of C(A), f > g E C(A) * Qf ), Ug) E Ch(.Y),+I 2. A cell is open iff it is a (1, 1, . . , 1 )-cell.

3. Each cell is homeomorphic under a coordinate projection to an open cell:

Let A be an (il,. . . , i,)-cell. Let A( 1) < . <i(m) be the indices i. E { 1,. ,n} for

which i;=l, so that m=il +...+i,. Define p:R”--,Rm by

It is easy to show (by induction on n) that p maps A homeomorphically onto an

open cell p(A) C R”.

200 S. Maxwelll Annals of Pure and Applied Logic 95 (1998) 185-227

4. Each cell is pathwise connected, where the path may always be chosen in Ch(Y).

(The proof is by induction on n, considering the projection of the cell. The fact that

the path may be chosen in Ch(Y) follows from the fact that, because + is in the

language, the midpoint of the line between two points may be defined by a suitable

formula.)

Definition 3.6. 1. A decomposition of L%” is a finite set {Cl,. . , Ck} of pairwise disjoint

cells which cover Iw” such that if n > 1, the following condition also holds: The set

{IIn,n--l [C,], . . . ,IIIn,n--l [Ck]} is a decomposition of EP-‘.

2. A decomposition B of Iw” is said to partition a set A & R” if for each cell C E 8,

either CLA or CnA=Q).

Remark 3.7.

1. We will also use the notion of decomposition of a cell C, which is defined in the

same way as a decomposition of Iw”, except with C replacing [w” and II,,,_l[C]

replacing P-l.

2. We will be using the fact, which is easy to prove, that the decompositions of

a cell C G iw” are exactly the restrictions to C of the decompositions of [w” which

partition C.

3. We observe that if the decomposition 6 partitions A, then we may write & = {Cl,. . . ,

ck,ck+l,..., Cl} such that

A=C, u.. .uCk and (&!“\A)=&+, u...uC/.

Definition 3.8. The boundary of a set A (denoted 6(A)) is defined to be A\A’.

Note that if X E 6(A), then any neighbourhood of X contains both a point in A and

a point not in A.

We are now ready to state and prove the CDT for Ch(Y), modulo an assumption

which will later be discharged in certain cases. Recall that our main interest in the

CDT is that it implies that Ch(Y) is closed under taking complements.

Assumption 3.9. Let A E Ch(Y)n (n E IV+). Then 3T E Ch(Y’)n such that:

(i) &A) 2 T; (ii) T has no interior points.

Theorem 3.10. Suppose Assumption 3.9 holds. Then the following conditions hold

Vn E N+:

(In) Given any AI,..., Ak E Ch(Y),, there is a decomposition of UP partitioning

each of AI,...,Ak.

(IIn) For each function f:A--t [w (A C KY) such that f(f) E Ch(Y),+l, there is

a decomposition 6 of R” partitioning A, such that the restriction f JC : C -+ IF! to each

cell C E 6, with C 2 A, is continuous.

S. Maxwell1 Annals of Pure and Applid Logic 95 (1998) 185-227 201

Proof. By induction on n. The cases n = 1 are fairly straightforward:

(11) Since A 1,. . , Ak E C%(Y), they each consist of a finite number of intervals and

points, hence there is a jinite set of points and intervals, pairwise disjoint and covering

R, which partitions each of the intervals and points making up Al,. , Ax. Since all

semi-algebraic sets are in Ch(,Y), this is the required decomposition.

(111) Given f : A + R (A C R), recall Df(= D) denotes the set of points at which

,f‘ is discontinuous. By Lemma 3.2, DE Ch(.Y’)r and contains no interior points. It

follows that D is a finite set of points. Now using (Ii ) we can decompose [w such that

we partition both A and 0. This gives a suitable decomposition 6.

We now suppose that (Ii ), . . . , (In-l ) and (II, ), , (II,_1 ) hold and we derive first

(I,,) and then (II,).

(I,) Let Al,. . ,Ak E Ch(y)n with corresponding 7’1,. . . Tk as in Assumption 3.9.

We begin by proving the following lemma:

Lemma 3.11. Let c” he a decomposition of 58” ttshich partitions T,, . , Tk nnd

(Al r: TI ), . . (Aa n Tk). Then 6 also partitions AI,. , An.

Proof. Suppose not. Then 3 a cell C E 8, and points Z,~E C, such that for some

i E { 1,. ,k}, ~ZE Ai, 6$ Ai. Then since cells are pathwise connected (by part 4 of

Remark 3.5) we can join Z and b by a path in C. Clearly there is a point X on

the path such that X E &A;) (X may be 5 or 7;). Then X E T, (by assumption on T, ).

which implies that C C 7; (since Q partitions T, ), and hence Z E (A; n Ti). But now

C C (A; n T, ) (because (5” also partitions Ai n T, ) and so 7; E A,, a contradiction. : i

We now continue with the proof of (In). Let Tk+i = A; n I; (for i = 1,. . . , k). By the

lemma it will suffice to find a decomposition of RR” which partitions T,, , TZX. For

reasons which may not yet be clear, we also let T” = lJ:L, T,. For each i E { 0, 1, ,2k},

wedefinethesetsT/CiW”-‘,forj=l,...,N,+I, where N, is a (T7) connected com-

ponents number for I;, as follows:

7’/= x~R”-‘:~r~~r?>O...~ri>Ojj ((X,/y +..-+r,,)eT,) r , m=I

Note that T,’ E Ch(Y),_i, and that

X E T;’ e 3 at least j points in the fibre above X in T,.

We now apply the inductive hypothesis (I,_, ) to obtain a decomposition .F of PX’-’

which partitions the sets T/ (i = 0,. . ,2k; j = 1,. . . , Ni i- I ). Then for each cell c‘ E .F

and for each i E (0, 1, . . ,2k}, there is a unique j,” E (0, 1, . . . , N,, m} such that ‘dX E C.

the fibre above Y in Ti contains exactly j,” points (where ‘j,’ = x8’ means that the fibre

above X in Tj contains an interval).

We now show for each cell C E ,P (note there are only finitely many such C) that

we can find a decomposition of the cell C x iw C: [w” which partitions TI, , T?k. By

202 S. Maxw~elllAnnals of Pure and Applied Logic 95 (1998) 185-227

then forming the union of all such decompositions, we have the required decomposition

of LQ”.

Claim 3.12. Suppose C E 9 and C has no interior points. Then we can find a decomposition of the cell C x R! C KY’ which partitions T,, . . . , Txk.

Proof. As in part 3 of Remark 3.5, we can project C homeomorphically onto the open

cell p(C) C [Wm (for some m E N+, m <n - 1). For convenience of notation we will

assume the projection is onto the first m coordinates. Note that p(C) E C/z(Y) since it

is a cell. Now for i = 1 , . . . ,2k, let T: E Ch(Y),+I be given by

Ti* = {(q ,...,%,y):%7+1...~~*-1(~i,...,~,,~,+I,...,~,-I,Y)E7;}.

Now by ([,,,+I ) (note that m + 1 <n), we have a decomposition, 6* say, of P’+‘, such

that 8” partmons T;“, . . . , TT* and p(C) x R.

For each cell G* E b* such that G’ C p(C) x [w, we form the set G & [w” by

It is now easy to check that G is a cell and hence that the collection of sets

{G:G*E~*A(G*C~(C)XR)}

forms a decomposition of the cell C x Iw, which partitions TI, . . . , TZk as required. I?

Claim 3.13. Suppose C E F such that C has an interior point. Then we can find a decomposition of the cell C x R C R” which partitions Tl , . . . , T&.

Proof. First, note that j,” # 0~) for i E (0, 1, . . . ,2k}, since otherwise by applying

Fubini and Corollary 2.8, z has an interior point. Hence for each i E (0, 1,. . . ,2k}, there is ji(= j:) E (0, 1, . , Ni} such that ‘d? E C the fibre in 7; above X contains ex-

actly ji points. Now for each i E (0, 1,. . ,2k} we can define the function J;! : C ---) 58

(for I= I,...,ji) by

3 at least (I- 1) z’s with [(%,z)~T~Az<y]

’ A 3 at least (j; - I) z’s with [(Z,z) E 8 AZ> y]

It should be clear from previous proofs that we can express this in a formula of the

kind which ensures r(f;J) E Ch(Y). Note that the definition is not sound if ji = 0, but

this is a trivial case, since then any decomposition of C x [w will partition 7;:. Since Cc iw”-’ (and by part 2 of Remark 3.7) we may apply the inductive hypoth-

esis (&-I ) separately to each of the f/‘s, on each occasion obtaining, in particular,

a decomposition H/ of the cell C, such that the restriction of the function f/ to each

cell in 2; is continuous. Now we may apply the inductive hypothesis (In-t) to obtain,

in particular, a decomposition 8 of the cell C, which partitions all the cells appearing

in any of the X/‘s

S. Maxwell/ Annals of Pure and Applied Locqic 95 (199Sj 1X.5-227 203

Let GE 8. Then we complete the proof of our claim by constructing a decompo-

sition of G x [w which partitions T,, . , T2k. By definition of 8, we have functions

91 < . < gj,, : G --$ iw which are continuous and whose graphs are in Ch( 9) (where

gY = fi 1~; for r = 1,. . ,jo). Hence the following is a decomposition of the cell G x iw:

Now if i E { 1,. ,2k}, then the number of points in the fibre above .U in r, is con-

stant throughout G, f/ is continuous on G for I = 1,. . . ,j; and r, C TO. Hence for each

1E{l,..., jl}, 3rE{l,..., jo} such that ,I;’ 1~ = ~1~. It follows that the given decompo-

sition of G x [w partitions TI, . . . , T2k. E

Now (I,,) follows immediately from our 2 claims.

We next assume that (Il),...,(l,) and (II1),...,(ll,,_~) hold, and prove that (II,) fol-

lows. Let ,f : A + R (A & W) such that r(f) E Ch(Y),,+ 1. Recall

D,, = D = {X E A : f is discontinuous at X}.

By Lemma 3.2, D E Ch(Y), and has no interior points. Applying (I,) we obtain a de-

composition 8 of 1w”, partitioning both A and D. It will now suffice to show that if C

is a cell in d such that CC A, then we can obtain a decomposition .Y? of the cell C

such that the restriction of J’ to each cell in .@ is continuous.

(i ) Suppose C E 8, C C A and C has an interior point. Suppose further, for a con-

tradiction, that fit is discontinuous. Then Dn C # 0, implying that C C_ D (since d

partitions D). But then D has an interior point, which is a contradiction. Hence ,/‘I(- is

continuous as required.

(ii) Suppose C E 8, C CA and C has no interior points. Then we can project C

homeomorphically onto p(C) C OB”’ for some m < n. (For convenience of notation we

again assume that the projection is onto the first m coordinates.) Now define the func-

tion ,f* : p(C) + R by

r(f*>= { (x,,.,xm,y>: ( h~~Grn)Ep(C)A &n+, .-.~Z,(,~l,.,xn,zm+l,.,Zn,_Y)E~(.f)

)) Clearly r(j‘*) E Ch(Y),+1, hence by inductive hypothesis (II,, ) we have, in particular,

a decomposition X’* of the cell p(C) such that the restriction of f * to each cell in

.@* is continuous. For each cell G” E ,X*, let G = (G* x EJY”‘) fl C. Then it is easy

to check that G is a cell and that the restriction of ,f to G is continuous. Now the

collection of all such G forms the required decomposition of the cell C. This proves

(IIn) and so completes the proof of the CDT for Ch(.Y). Cl

4. Further properties of sets in Ch(9)

The aim of this section is to investigate further what general restrictions apply to

sets (particularly the graphs of functions) which lie in Ch(#Y). We will first show that

204 S. MaxweNIAnnals of’ Pure and Applied Logic 95 (1998) 185-227

given A E Ch(Y’)a),+,, we can find a continuous function whose graph lies in Ch(9),+m

and is contained in A, and whose domain is a box in R”, provided that we have the

obvious necessary condition that IIn+m,n [A] contains an interior point, (Lemma 4.1).

By the end of this section we will have shown that we can strengthen this to the

selection of a CP function (for each p E FV). The latter result (Theorem 4.19) is used

in Wilkie [IO], and also in Claim 5.5.

Lemma 4.1 (Selection of a continuous function). Let n, m E N+. Suppose A E Ch(Y),+,

and B C R” is a box such that

Then 3 a box B’ c B and a continuous function f : B’ + W’ such that r(f) E

Ch(Y),+, and r(f)cA.

Proof. The main step in the proof is the following claim:

Claim 4.2. Under the given hypotheses, 3 a box B &B and a function (not necessarily

continuous) f : B -+ Rm such that r(f) E Ch(Y),+,,, and r(f) CA.

Proof. We proceed by induction on m, assuming first that m = 1. Suppose that A n (B x IF?) has an interior point. Then there is a box B C B and a point b E 08 such that

B x {b} &A. But this is the graph of a function f : B + R which also lies in Ch(9’)

since it is semi-algebraic.

Suppose then that A n (B x R) has no interior points. Then by Lemma 3.1, there

exists a box fi C B and 3k E N such that VT E 8, IA,-/ = k. Note that the hypothesis on

A ensures that k # 0. Now the following set is the graph of a function f :B + R:

{

(T,rl):rtBA3r:>0--.3r~>O~((r,r, +...+q)EA) .

i=l i

Furthermore, r(f) CA and, by Corollary 2.4, r(f) E Ch(9). Hence we have proved

the claim for the case m = 1.

Now suppose m > 1 and the result is true for positive integers less than m. Then

by inductive hypothesis for (m - I), 3 a box B” LB and a function f * : B* ---f tV-’

such that r(f *) E Ch(Y) and r(f *) 2 IIn+m,n+m_-l [A]. Consider the relation S C (W’+’

given by

By inductive hypothesis (for m = l), 3 a box B ‘_B* and a function f”: g -+ R such

that r(f”) C S and r(j) E Ch(.Y). Now define f : B ---f R by

r(f)={(~,yI,...,y,):xEBAf*(x)=(yl,...,y,-,)AS(~)=y,}.

Then it is easy to check that f satisfies the required condition. q

S. Maxwell1 Annals of Pure and Applied Logic 95 (1998) 185-227 205

To complete the proof of the lemma, it will suffice to show that there is a box

B’ c_j such that f le, is continuous. To do this we define, for j = 1,. . . , m, the functions

fi : B 4 R! as follows:

fjW=v, @ fw=(Yl,...,Ym)~

Recall that Dj; = {X E j : fj is discontinuous at X}. We define the set D by D = Dh U .

U Df,, Then D E Ch(Y) (since by Lemma 3.2, DA E Ch(Y) for j = 1,. . . , m). Also, by

Lemma 3.2, Dh has no interior points for j = 1,. . . , m, and hence by Corollary 2.8 is

null. It follows that D is null and hence has no interior points. Now by Corollary 2.9,

!?\D has an interior point. Hence we may choose a box B’ C (B\D). Since each f, is

continuous on B’, it follows easily that flak is also continuous as required. 0

Lemma 4.3. Let A E Ch(Y),+, be such that A 2 KY” x (0, m) and A has no interior

points. Let ‘40 = (7: (X,0) ~1). Then A0 has no interior points.

Proof. We first note that A0 E Ch(9). Suppose for a contradiction, that A0 has an inte-

rior point. Then we can find an open box Bo C Ao. Then since A has no interior points,

by Lemma 3.1, 3 E N and a box B C Bo such that for each X E B, the fibre Ax contains

exactly k points. Now as in the proof of Lemma 4.2, there is a function f : B + R such

that r(f) E Ch(Y), (X, f(E)) E A and such that (X, y) E A =+ f(F) < y (VT E j). More-

over, by Lemma 4.1, there exists a box B’ & j such that fl~l is continuous. Now we

may take B’ to be a closed, hence compact, box and thereby obtain that fl~r has a

minimum value, say E > 0. Hence, E E B’(T, 0) 62, and therefore X $Z Ao, contradicting

the fact that B’ C: Ao. 0

The remainder of this section consists of lemmas to establish Theorem 4.19, the

selection of a CJ’ function, which will be used in the proof of Claim 5.5.

Notation 4.4. If A C: W+‘, then we say that (5, 00) E 1 if

(?&O)E{(Z,z):3y((x,y)~A~yz== 1 AZ>O)}.

Similarly for (5, -00) ~2 where z >O is replaced by z ~0.

Throughout the rest of this section I will often leave it as an exercise to the reader

to verify that certain sets are in C/z(Y). The method required will always involve the

same ‘tricks’ that I have already been using.

Lemma 4.5. Let h : B x (0, 3.) --+ R (where B C R” is an open box and 2 > 0) such that

T(h) E Ch(.sP),+2 and h is a continuous function. Let & = S be dejined as follows:

where R+, = iw U {ho}. Then S E Ch(9) and S has no interior points.


Proof. The verification that S E Ch(Y) is routine, where we notice that yi # y2 can

also be expressed as yi < y2 V y2 < yl. To show that S has no interior points, we begin

with the following claim:

Claim 4.6. Suppose (5, 0, b), (a, 0, c) E T(h) (where 6, c E LQm; b cc). Then (a, 0,~) E

T(h) (Vz E [hcl).

Proof. Let d E (6, c). Let V C B be a box around Z, and let r] E (0,J.). It will suffice

to show that 3(x, y) E V x (0, n) such that (2, y,d) E T(h).

Since (a, 0, b) E T(h) and b <d,

3(x’, y’,z’) E V x (0, y) x (-cqd) such that (X’, y’,z’) E T(h).

Also since (a, 0, c) E T(h) and d CC,

3(~“, y”,z”) E V x (0, yl) x (d, m) such that (_Z”, y”,z”) E T(h).

Now consider the line segment joining (X’, y’) to (X”, y”) in Rn+‘. This line is clearly

contained in the (convex) set V x (0, r). Now since h is continuous, we may apply the

Intermediate Value Theorem along this line segment to obtain (2, y) E V x (0, ty) such

that (2, y,d) E T(h) as required to prove the claim. 0

It follows immediately from the claim that the fibre above (Z,O) in T(h) has non-

zero measure (/&ES). Now, for a contradiction, suppose that S has an interior point. ,

Then {(%y):@,O,y)EOh)} h as non-zero measure (by Fubini) and hence an interior

point (by Corollary 2.8). But then by Lemma 4.3, T(h) has an interior point, which

is clearly absurd since h is a function. 0

For the next few lemmas, we will need to work with sets which are ‘almost func-

tions’. We therefore introduce a new name for sets of this kind:

Definition 4.7. Let A & U x R” (UC KY) be such that

(i) K? E U3y’E EP(?,y’) EA;

(ii) {XE U: 3~13j&(jl #& A(X,y’l)EA A(X,~2)EA)} is p,-null.

Then we say that A is the graph of a pseudofunction.

Remark 4.8. 1. We will denote pseudofunctions by f, 6, etc.

2. All functions are obviously also pseudofunctions.

3. We borrow the notation of functions, writing f: U---f Rm and J’(f). Also , if ii~ 7.J

and there is a unique y’ E R” such that (a, y’) E r(f), then we will write J(Z) = y. A

4. If f : U + Rm is a pseudofunction such that UC [w” is open, and T(j‘) E G(Y),

then the set in part (ii) of the definition also lies in Ch(Y) and hence by Corol-

lary 2.9 3 an open box B C U such that the restriction of f to B is a function.

Notation 4.9. Let E E R, n E N+ and i E { 1,. . , n}. Then we let Ej denote the n-tuple

(0,. . . ,O,&,O,. . .) 0), where E appears as the ith coordinate.


From here until the end of Claim 4.15, we fix a pseudofimction f^: U -+ R (where

U 2 iw” is open) such that r(f) E CJz(,Y),+i We note that for F # 0 and i E { 1,. . . , n},

if Z+E~ and x lie in U and f takes unique values at both points, then the expression

[.f?(? + E;) - j(X)]/c takes a unique real value. This motivates the definition below of

more pseudofunctions i,, . , @,,, which will be useful in trying to investigate the points

at which J’ is actually a C’ function. For i = I,. .,n, let

r( Lj, ) = i

(X, E, y) : x E u A (X + Ei) E u A 32,3z2[(X,z, ) E r(f) 1 A(x+Ej,Z~)Er(f’)ACy=z2-Z,]A(E#o)

Although it is difficult to write down the domain of i,, it is easy to check that if X E U,

then Gi is always defined (as a pseudofunction) on V x ((-q, q)\(O)) for some PI >0

and box V (c R” around X (since U is open). We define also for i = 1,. . . , n, the sets

G, = {(F. y) : (X, 0, y) E r(&)}

Remark 4.10.

1. r(Lji),Gj~Ch(Y) for i=l,..., n.

2. If ZE U and (df/&;)(Z) exists, then (Z,(8,?/&,)(5)) E G;.

Lemma 4.11. If f is a function which is difirentiahle and has continuous derivative,

then I‘(Z~/&,) = Gi, .for i = 1,. ,n, and hence r(8f/8xx,) E Ch(-Y).

Proof. We first fix i E { 1, . , n}. Then by part 2 of Remark 4.10, we see that r( 8f/cSxi)

2 Cf. To show the converse, it is sufficient to verify that if Z E U and 6 > 0, then there

exists an open box B around (Z,O) such that the following holds:

(We note that differentiability of f at Z trivially gives us (only) that there ex-

ists an interval I C Iw, about zero, such that (t) holds with V(X, E) E {z} x I replacing

Y(T,E)EB.)

First, we find a box W & KY’ around Li such that W C U and

n A

Jg(“)- $5) <a (b%E W) 2

This is possible by continuity of a!/&, at 5. It is easy to see that we may now choose

an open box B 2 KY+’ such that

YfE, E W ‘@,E)EB.

Suppose (X,6) E B (wlog~>O). Then 3!y such that (x,E, y) E r(ii), namely .v =

[_?(X+Ei)- .~(X)]/E. Now, let $ : I + R denote the function

f( )=.ftXl,.~~,Xi-l,Xi+t,Xi+l,...,Xn), “I ^


where I is some open interval containing [0, E]. Then f is a differentiable function, and

so the Mean Value Theorem holds for f. Hence 31 E (0, E) such that (df/dt)(q) = y.

But now,

”

y++g 2 ,( +%I.

Since (Y, q) E B, we have that (5 + vi) E W. Hence,

i.e.

This proves that (t) holds V(?, a) E B from which the result follows. 0

We will now define a set C which is easily seen to lie in C/z(Y) and we aim to

show in what follows that C has no interior points and includes all the points at which

1 is not a C’ function.

3Yl3Y2CYlf Y2 A 6 Yl> E m A 6 Y2y2) E m)

Ci= X: V(X,m)EGiV(T,-W)EGiV 2

3yl3y2(yl # ~2 A (FYI) E Gi A (%y~) E Gi)

C&f. i=l

Lemma 4.12. Let f * denote the restriction off to the open set U\C. Then f + is a

function, and the partials af * /a x, exist and are continuous for i = 1,. . _ , n.

Proof. Fix i E { 1,. . . , n). We note first that f * is a function, since otherwise 32 E U\C

3y,3y~(yt # yz A (a, yl) E r(f”) A (5, y2) E r(f)). But then b E Ci, which implies 5 E C,

a contradiction.

Now, we show that the partial af */axi exists. Suppose, for a contradiction, that it

does not exist at ZE U\C. Then &(z,E) (which does indeed take a unique value for

small E, since U\C is open) does not tend to a limit as E -+ 0, and hence either (i) or

(ii) must hold:

(i) 3~13~2 E [w (with yl # ~2) such that (Z, yr ),(7i, ~2) E Gi. But then ifs Ci, which

implies Z E C, a contradiction.

(ii) Either (Z, co) E Gi or (5, -00) E G,. But then ZE C;, which implies Zf C, a

contradiction.


Therefore df */aXi : U\C -+ R is a function defined on an open set. By part 2 of

Remark 4.10 and the fact that Gi is closed, we have that

(*)

Now suppose, for a contradiction, that i?f */ax< is discontinuous at a E U\C. Then

either (iii) or (iv) holds:

(iii) 3y # (af */axi) such that (a, y) E r(df */hi). But then applying (*), we have

that (5, y), (a, (df */8xi)(Z)) E Gi which implies that Z E Ci, leading to a contradiction.

(iv) There exists a sequence {Zj} + si such that (df */axi) is unbounded. But then

by (*) there exists a sequence {yj} 4 fm such that (?ij, yj) E Gi (Vj E F+J). It follows

that (Z, &tco) E G; and so Z E C,, which again leads to a contradiction. 0

Lemma 4.13. C has no interior points.

Proof. Suppose, for a contradiction, that C has an interior point. Then IJk, C, has non-

zero measure (by Corollary 2.8), hence Ci has non-zero measure for some i E { 1,. . , n},

which implies that Ci has an interior point (by Corollary 2.8). We may let B C C, be an

open box (where we note that B C Ii, since C, C_n). Now by applying Lemma 4.1, we

may assume that B has been shrunk such that f is a continuous function (which we will

now denote simply by f) on some open box which strictly contains B. In particular,

we may assume that the restriction of [f(x + Ei) - f (X)1/& = i, to B x (--I, i,)\{O} is a

continuous function (where A>O). We now denote g^ilsXt-i,;.)\fa) simply by g. Note

that T(g) E Ch(Y),,+z.

We now let g+ denote g(Bx(s,j,) and g- : B x (0, I.) t R be defined by g-(X, F) =

~(2, -6). Also, we let G,G+ and G- be defined in the obvious way corresponding

to how G, was defined from ii. Now, by Lemma 4.5, both S,+ and S,- (as defined in

that lemma) have no interior points, but since they both lie in Ch(Y), we can apply

Corollary 2.9 so that we may assume that B has been shrunk so that B n (Sg- U S,- ) = 0).

Hence \JX E B, (* 1) and (*2) hold:

3!y~R~{ico} such that (x,y)~G+, say y=y+(X), (*1)

3!y~R’U{fm} such that (X,~)EG-, say y=y-(X). (*2)

(Of course the existence of some such y is trivial since g is defined on the set

B x (-i.,R)\{O}.) Now, let

U, = (X E B : y+(x) = +m = y-(X)},

U, = {X E B : y+(x) = --03 = y-(Y)},

U, = {x E B : y+(x) # y-(x)}.

It is easy to verify that G = G+ U G-, and hence that B = U1 U UZ U Us (since also

B C C,). Moreover, given that (*l) and (*2) hold, we can easily show that U1, Ul, iJ3 E

Ch(Y). Hence by Corollary 2.9 it will suffice for a contradiction to show that UI, UZ, U,

210 S. Muxwelll Annals of Pure and Applied Logic 95 (1998) 185-227

do not contain interior points. In fact we only need the result for Ur and U3 since the

proof for U, is clearly symmetrical to the proof for Ut.

Claim 4.14. VI contains no interior points.

-- Proof. We suppose, for a contradiction, that Bi C UI is a box and let a,a +qi E B1 (where q >O). Let L C R ‘+I denote the line segment joining the points (Z,f(ii)) and

(a + vi, f(E + vi)). Suppose that there exists a segment I CL with one endpoint 5 such

that I C_ r(f). Then 3z E [w (Z,z) E G+, which is a contradiction since z # y+(Z) = $00.

Hence there is no such interval. Now we note that condition (T7) for r(f) implies that

LnT(f) has a finite number of connected components, and hence we may assume,

by reducing q if necessary, that there are no points in r(f) n L lying strictly between

the two endpoints of L. (t)

For X E I&+t,,, [L], let Z(Z) E R’ denote the point such that (Z, Z(Y)) EL. Now, we use

the fact that (Z, v) E G+ =+ y = +cq to obtain

36 E (0, q) with f(ir + Si) > I(5 + Zi).

Similarly, since (a + vi, JJ) E G- j Y = +oc,

3~ E (6, q) with f(Z + ;jji) < l(iT + yi).

But now by continuity off, we may apply the IVT to get E E (6, y) such that f(Z+Ei) =

E(a + Ei). This contradicts (t). Cl

Claim 4.15. U3 contains no interior points.

Proof. Suppose, for a contradiction, that B3 2 Us is an open box. Let ?i E B3 and let

b=y+(Z); c=v-(Z) (wlogbtc). Let d=(b+c)/2. Now by (*l) and (*2), 3q>O

and Bi C B3, a box of length 21 around 5, such that

YW~B; x (O,rl), g+(E,~)<d and g-(Y,y)>d.

Let y E (0, q). Then

= f((a+yi)-7i)-f(a+7i)

-Y

- - since (Z + yi, yi) E B$ x (0, r]). But this is clearly a contradiction. 0

These two claims conclude the proof that C contains no interior points. 0

S. MaxwellIAnnals of’ Pure and Applied Logic 95 (1998) 185-227 211

Definition 4.16. We recall the inductive definition of a CP function (where p E N):

Suppose J‘: CT ---f Iw” (where U C iw” is open). Then we say f is C’ if all the partials

af /axi (for i = 1,. . , n) are defined and continuous. Let p > 1. Then we say ,f is CP

if .f is C’ and the differential df : U --t Rm” is CP-‘. By convention f is Co iff ,f is

continuous.

Remark 4.17. Let f : U ---f R” (U C RF). Then the following are standard results and,

in particular, we will be using them for all cases of verifying whether a function is CP.

1. f is CP @ af /i3xi exists and is CP-’ (for i = 1,. . ,n).

2. Let f be written (f,,...,fm):U+Rm. Then f is CP w fi:U-iw is CP (for

j=l ,...,m).

Theorem 4.18 (Almost everywhere smoothness of pseudofunctions). Let n,m E Nit.

p E N, f^ : U + Rm (U 2 ET’ is open) be a pseudofunction such that I’(f) E Ch(.Y),+,.

Then 3 a closed set C E Ch(Y),, containing no interior points. such that the restric-

tion of ,j to iJ\C is a CP function.

Proof. We first consider the case m = 1, and proceed by induction on p. Let p = 1.

(Note this also covers the case p = 0.) Then if we take C to be defined as it was

before Lemma 4.12, the result follows immediately from Lemmas 4.12 and 4.13 and

part 1 of Remark 4.17.

Letm=l,p>l.LetC~I)ECh(Y),bethesetwhichwitnessesCwhenm=l,p=l

(using inductive hypothesis). Now with Gi defined as after Remark 4.8, we define for

i= I,...,n:

Then since C(t ) has no interior points, hi is a pseudofunction. Moreover, ii lc,\c, ,/ =

Z.f*/axi (where f * is as in Lemma 4.12) and r(h,) is easily shown to be in Ch(Y)

for i=l,...,n.

Now, we apply the inductive hypothesis for p- 1, (m = 1) to each hi, to obtain closed

sets CL,, . . , Ci$! E Ch(Y’)n containing no interior points and such that the restriction of

i, to U\C&, is a CP-’ function (for i = 1,. . . ,n). Now, let

Then C is a closed set in C%(Y) containing no interior points. Moreover, if X is in

the open set U\C, af/axj exists and is CP-’ (for i= l,...,~). Hence fluic is a CJ’

function (by part 1 of Remark 4.17). Hence, the result holds for m= 1 (VIE IV).

Suppose m>l, pEF+J. Write f=(f 1 ,..., f^,):U+FP.Then~t ,,.., fm:U-+IWare

pseudofunctions. Now, applying the result with m = 1, we have closed sets Cr;, . . . ,

Cf E Ch(S;P), with no interior points such that the restriction of fj to U\Cf is a CP ,n ,


function (for j = 1,. . . , m). Now let C= Uy=i Cl. Then (by part 2 of Remark 4.17),

f^ is a CP function on U\C. Hence C satisfies all the required conditions. 0

Theorem 4.19 (Selection of a CJ’ function). Let n, m E N+; p E N. Suppose A E

Ch(Y)n+m and B & Iw” is a box such that ‘G E B 3j E R” (??, j) E A. Then 3 an open

box B’ 5 B and a CP function f : B’ -+ Rm such that r(f) E Ch(S),+, and r(f) CA.

Proof. The result has already been proved in Lemma 4.1 in the case p = 0. Hence, we

may assume that we have an open box B* c B and a continuous function f * : B* + IR”

such that r( f * ) E Ch(9’),+m and r(f *) 2 A. Now by Theorem 4.18, 3 a closed

set C E Ch(Y), with no interior points, such that f * is CP on B*\C. But then by

Corollary 2.9, B’\C has an interior point, hence we may choose an open box B’ C B*\C and let f = f * 1~‘. Then since the restriction of f + to B*\C is CP, we have the

result. 0

5. Defining the boundary

In this section we will prove that if A E Ch(Y)n is of a certain form then 3T E

Ch(5Q such that 6(A) C T and T has no interior points (Theorem 5.3). Recall that

A will then satisfy Assumption 3.9, which if it is satisfied by all the sets in Ch(Y),

implies the CDT for Ch(9’). We begin with the following lemma, in which we denote

a k-tuple (xi,. . . ,xk) by 2, and an I-tuple (vi,. . . ,_yr) by j.

Lemma 5.1. Let k, 1 E N+. Let g: Rk+’ --+ [0, co) be a C’ function. Dejne the sets F (=F,) and E (=E,) as follows:

,

Let a”~ R“ be such that a”~ b(&+l,k[Z(g)]), where Z(g) denotes Suppose further that {j : g(& 9) = 0) has a finite non-zero number ponents and that at least one of these is bounded. Then GEE.

the zero set of g. of connected com-

Remark 5.2.

1. This lemma is general in the sense that we do not require that T(g). F and E are

in Ch(Y). 2. It is easy to verify that the result still holds when the range of g is the whole of Iw

(and where we allow a<0 as well in the definition of F), but we will only need

to use the lemma as stated.

Proof. Let P G R’ be a bounded connected component of {p : g(& j) = 0). Let H C [w’ be closed and bounded, with the property that each point of P is an interior point of

S. Maxwelll Annals of Pure and Applied Logic 95 (1998) 185-227 213

H and such that H does not meet any other connected component of {p : g(a”, j) = O}.

(This is possible since by hypothesis there are a finite number of such connected

components and 52’ has the topological property of being a normal space.) Since

~5 E IIk+l,k[Z(g)] and C E &IIk+,,,[Z(g)]), we have a sequence {zj} --f G such that for

each j E N, g[{Gj} x R’] C_ (0, co). Now it follows, since H is compact and g is con-

tinuous, that we can define Sj E R (where Sj > 0) by

6j = min{g(csJ, j) : 5 E H} (for each j E kJi>.

Since 3@ EH such that g(2, @) = 0, it also follows by continuity of g around (6, ,i?)

that {Sj} + 0. Our aim will be to show that for all arbitrarily large j, (cl/-, ~3, ) E F.

For each j E N, let 6j E H be such that g(zj, 6j) = 6,. Since H is compact, we may

assume, by choosing a subsequence if necessary, that 3 6 E H such that { 6j} + &.

Then {C;,} + a” and {g} --t 0, and so g(& 6) =0 (by continuity of g), which im-

plies that 8 E P. But b is therefore an interior point of H (by definition of H) and

hence 33 E N such that Vj >J, 6j E H”. We now show that (Jg/ayi) (fij, 6j) = 0 for

i= l,...,m (Vj>J):

Let j>J, iE{l,. . . , m}. For convenience of notation, we will assume that i = m.

NOW, let I C R be any open interval such that bj?,, E I and {(bj,, . , bj,,,_, } x I C H.

(This is possible since 6j E H”.) Then by definition of Sj,

Since bj,,( is in the interior of I, this implies that (zg/ay,) (Cj, &j) = 0, as claimed.

It now follows that (Gj, S,) E F (Vj> J) since g(cT,, 6j) = Sj and S, > 0. But then 2 E E

as required (since {((zj, Sj)> --f (G, 0)). 0

We are now aiming for the following:

Theorem 5.3. Let n, m E N+. Let A E Ch(Q be such that we can write

A={x:Sy’f(x,y’)=O},

where f : iR”+“’ ---f R is C’ and r(f) E Ch(Y),+,+I.

Then 3 T E Ch(Y), such that T has no interior points and 6(A) C T.

We will begin by defining a set T 5 R” which we verify is in Ch(Y),,, and then

prove the claims that T has no interior points and 6(A) C T. First, let g : lFP’+2m -+ iw

be defined by

g(% 9, y’2) = (f(X v’))2 + 2 (q2 - cy: + z; >J2. i=l

Remark 5.4.

1. Q) E Ch~~P)n+l+2m+l (by Corollary 2.4); 2. rang 2 [0, cc);

214 S. MaxweNIAnnals of Pure and Applied Logic 95 (1998) 185-227

3. g is a C’ function (since it is the sum of products of C’ functions), and hence

by Lemma 4.11 the graphs of each of the first order partial derivatives of g are in

CA(y);

4. 32g(E,q,~,Z)=O~~(~,v’)=O~ A~JlvMsl); 5. Let (si, c) E [w” x [0, co). Then {(j,.?) : g(Z, c, y”,Z) = 0} is a subset of [-c, ~1’~.

Now, recalling the definitions of F and E from Lemma 5.1, we let

i

&>OA3jz g(x,q,$,t)=&A (

F= (X,q,E): m

&&q?y’>z’)=O)A ~(&%q&i)=O) i=l I=1

E={(%q):(%q,O)d},

Q={(fr):3q((~,q)tEAq>OAq=~)},

T = {X : (X, 0) E a}.

Note that F E Ch(Y),+z (by Corollary 2.4 and parts 1 and 3 of Remark 5.4). It is

now easy to deduce that E, Q E Ch(Y), and, in particular, that T E Ch(Y)n.

Claim 5.5. T has no interior points.

Proof. Suppose, for a contradiction, that T has an interior point. Then Q has an interior

point (by Lemma 4.3) so E has an interior point (by definitions of Q,E). Hence F has an interior point (by Lemma 4.3) and so 3(Z, q) E [Wn+’ and an interval I C_ [w such

that { (5, q)} x I & F. Let g* : R?“’ + 52 denote the map given by g*(j,z’) = g(Z, q, 3,s).

Now, since Vt EZ 3y’ 32 such that g*(y’,Z)=s, and T(g*)E Ch(Y), we can apply

Theorem 4.19 to select a Cl function (which we think of as a path) y :I’-+ [W2m (with

Z’LI) such that r(y) E Ch(Y) and such that y(t)=(y’,Z) implies g*(y’,Z)= t. Now consider the map g* o y : I’ -+ I’. It is easily seen to be the identity. However, if t E I’, using the chain rule, (g* o y)‘(t) = (g*)‘(y(t)) . y’(t) = 0 (since tit E I’ (g*)‘(y(t)) = 0, by definition of F). Hence, since g* o y is a function of one variable, it is constant,

which contradicts the fact that it is the identity. 0

Claim 5.6. 6(A) C T.

Proof. We first introduce some notation for the proof. Let 9 E P”, then let

]]jl] =IWiX{Iyij: i= I,.. .,rB}.

Let Z E 6(A). We will now consider the cases ZE E A and 7i $A separately.

(i) Suppose Z E A. Then 3g~ IF such that f(Z, 6) = 0. Let llg]] = qo. Let q E [qo, co). Then 3z’g(Z, q, 6,Z) = 0 (by part 4 of Remark 5.4), so

6% 4) E rI n+l+2m,n+lmI)l vq E [40, ~1). (t)

S. Maxwelll Annals of Pure und Applied Loyic 95 (1998) 185-227

However, since si E 6(A), 3 a sequence (5,) -?i such that 7ij @A

J’(iz,, j) # 0 (V,y’ E iw”, ‘dj E FV) and, hence,

@i>4, _$Z) # 0 (Yq, $2) E R ‘+2m, Vj E FV), by part 4 of Remark

Therefore (Gj,q) 4 %+1+2~,~+1 [Z(g)] (Vj E N, Vq E R), and so

(%q) E mL+l+2m,n+l [-%)I) Wq E [qo, m(;))> by (t).

215

(VjEE). so

5.4.

Let c E [qo, cm). Then {(y”,z’) : g@, c, y,z’) = 0) has only finitely many connected com-

ponents, since T(g) E C/r(Y), and all of them are bounded (by part 5 of Remark 5.4).

Now by part 3 of Remark 5.4, all the hypotheses for Lemma 5.1 hold (with k = n + 1

and 1 = 2~). Hence (a, c) E E. Thus (5, q) E E (Vq E [qo, x)), hence,

G%r)EQ @YE (0,;) (=(O,,x) if yo-0)).

Therefore E E T (by definition of T) as required.

(ii) Suppose ZiGA. Then since ZE 6(A), 3 sequences {zj} +E and {$} such that

,/‘(a,,$) = 0 (Vj E N). Wlog let us assume that for each j E W, gj has been chosen

such that if S(?ij,C;)=O then \\F,11 >\jE;,]]. Letjj = \lb;l/(VjE N).

Now suppose that {qj} is bounded. Then {b,} would contain a subsequence which

converges, say to b’. But since f(‘Fij, gj) = 0 (Vj E N), it follows by continuity of J‘ that

j(a, 6) = 0, which contradicts the hypothesis that a @A. Hence {q,} is unbounded, and

we may also assume wlog that qj >0 (Vj E N). Let J E N. Then ZIZg(&,q.l, bJ,z’) =0

(by part 4 of Remark 5.4), so

(aJ,SJ) E K+,+2,,,+1Mg>1. (*I

Let qE[O,qJ), then g(&,q,y’,Z)#O (V(y’,z)~Ft~~) by minimality of ilgJ\l and

part 4 of Remark 5.4. Therefore,

Hence by (*), since we assume that qJ #O, (aJ,qJ)rG(~n+l+2m,n+l[Z(g)]) and (by

parts 3 and 5 of Remark 5.4) we may apply Lemma 5.1 to obtain (ZJ, qJ ) E E.

This gives (Zj,q,i) E E (Vj E N) and, recalling that {qj} is unbounded, we have

(a,,rj)E Q (Vj E N) where y, = l/qj and SO {rj} --) 0. Thus Fi E T as required (since

{?j} + a). D

Now since we have already remarked that T E Ch(.Y), Theorem 5.3 follows imme-

diately from the two claims. 0

6. Some sufficient conditions for a structure to be model complete and o-minimal

In this section we shall impose extra conditions on Y, chosen so that we can dis-

charge Assumption 3.9.

216 S. Maxwelll Annals of Pure and Applied Logic 95 (1998) IM-227

Notation 6.1. Throughout this section, for each n, R, will denote a collection of

functions f: OF’+ R, where we denote UnEN+ R, by R. We will be interested in the

definable sets in the structure (@, {f})fE~, where we mean dejnable by a usual formula, not a Ch-formula. We will use the notation q@) N I+@) to mean

Lemma 6.2. Let n E N+ and AC Iw” be dejned in @,{f})fE~ by an 31_formula,

4(X) say. Then there exists a polynomial p E [w[zl, . . . ,zl+,,+,,,] and functions ft, . . . ,

ft E R such that 4(F) N Zly’p(f, (Z, j), . . . , ft(%, 7),x, y’) = 0. (By this we do not mean

that every J;: must lie in R,+,, but that for each J;, there is some subset of {xl,. . .,x,,

yi, . . . , y,,,} to which it is applied )

Proof. This is a standard transformation of a formula, the method for which I will

outline below: We use first the fact that if ri, ~2 are terms, then ri <z2 N (3y3z(zi +

y2) = ~2 A yz = 1). Then by ‘unnesting’ terms, we can obtain 4(X) - I&Z), where I,@)

consists of formulae of the form s(fr (2, y’), . . . , f/(X, j),X, y) = 0 (where fi, . . . , fl E R and q E [w[zl, . . . ,zI+~+,J), joined only by connectives from the set {A, V}. The proof

is now completed by noting the following tricks: (41 = 0 A q2 = 0) - (qf + qi = 0) and

(41 =ovq2 =O)-(q1,qz=O). q

Definition 6.3. We define 3 conditions on the collection of functions R: (RI) Each R, consists of C’ functions.

(R2) Each quantifier-free definable set of (E, { f})fER has a (T7) connected compo-

nents number.

(R3) If A is 31 -definable in (E, {P})J.~R, then so is 2.

For the following results, we will let Y denote the quantifier-free definable sets of

((W,{f})fE~. We note that it is trivial to check that Y satisfies conditions (Tl)-(T5)

and that whenever (RI) holds, Lemma 6.2 implies that each set in 9’ is the projection

of a closed set (the zero set of a continuous function) and, hence, (T6) is satisfied.

Therefore Y is tame if (Rl) and (R2) hold.

Lemma 6.4. Suppose that condition (R3) holds. Then Ch(9’) =Y’, or equivalently Ch(Y) corresponds to the 31-definable sets in (@, {f })J~R.

Proof. Recall that Ch(Y) = UrnEM Y”, hence it is sufficient to show that 9” = Yn

(‘dm E N+). Noting that Y” = Y, we see that by an induction, we need only show

that ((Yn)‘)” = 9’“. If A E (z%‘~)~, then we may write A = A1 n nl=, & where Al,. . . ,

A, E Y”. But by condition (R3), we have 22,. . . ,A,. E Yn. Hence A E Y”, since Y”

is closed under intersection, and so we have (Yn)n = 9’“. The result now follows

easily from the fact that U and II commute. 0

Theorem 6.5. Suppose that conditions (Rl)-(R3) are satisfied. Then the theory of

0% If )) /ER is model complete.

S. MaxwellI Annals of Pure and Applied Log? 95 (1998) 185-227 217

Proof. By Lemma 6.2, we may deduce that all 3r-definable sets can be expressed

in the form {X: 3_?f(F, j) = 0} where ,f is a C’ function and r(f) E ,Y. Hence by

Lemma 6.4, this is true for all sets in C%(Y). Now since (Rl) and (R2) hold, .Y’ is

tame and so by Theorem 5.3, we have that for each A E Ch(.Y’), there exists T E Ch(.Y)

such that 6(A) C T and T has no interior points, i.e. Assumption 3.9 is satisfied. Hence

the CDT for CIr(.Y) holds, and in particular Ch(.Y) is closed under complement. But

then by Lemma 6.4, the 3t-definable sets are closed under complement from which a

standard argument gives model completeness of the theory of (R, { .f }) , Ed. II

Corollary 6.6. Suppose that conditions (Rl)-(R3) are satisfied. Then the structure

@8, {f)) fER is o-minimal.

Proof. Let 4 define a set A C Iw. Now, by model completeness, 4 - $ where $ is

some 3, -formula. Hence A E Yn, which implies that A satisfies condition (T7) and is

therefore a finite union of intervals and points of [w as required. 0

7. A generalisation of a result of Gabrielov

Having obtained Theorem 6.5, we now aim to explore in this section how easy it is

to find structures satisfying conditions (Rl)-(R3). We begin by looking at (RT).

Lemma 7.1. Every set dejinable in an o-minimal expansion of iw hus u (T7) connected

components number.

Proof. Let 4(X) be a formula defining a set A C_ [w” in an o-minimal structure. Let

p(X, y’) E Z[Y, y’] be a polynomial with the property that if Z C_ [w” is affine, then there

exist parameters b,, . , b, E R such that Z = {X : p(X, 6) = 0) (see proof of Claim I .9).

Then it suffices to show that 3N E N such that for any b’ E IF, the set

(3 : &c) A p(X, 6) = O}

has at most N connected components. This is Theorem 0.3 in Knight-Pillay-Steinhom

[51. 0

Lemma 7.1 and Theorem 6.5 together imply that any o-minimal structure

(E, {f })fER, satisfying (Rl) and (R3) is model complete. At first sight, this may not

seem to be leading us towards any interesting results, since almost all o-minimality

results are actually deduced from model completeness of the structure. However, while

reductions of o-minimal structures are trivially o-minimal, reductions of model com-

plete structures are not in general model complete. Therefore, in the hope that we may

be able to apply Theorem 6.5 to reductions of o-minimal structures, we demonstrate

in the remainder of this section how some further conditions can be added, enabling

us to weaken condition (R3) so that it refers only to the closure of quantijer-free de-

finable sets. These are much easier to analyse than 31-definable sets and, in particular,

218 S. MaxwellI Annals of Pure and Applied Logic 95 (1998) 185-227

we can then make use of a theorem of Gabrielov [3] concerning the definability of

the closure of quantifier-free definable sets in certain structures. This leads us finally

to Theorem 7.8, which is a generalisation of the model completeness result obtained

by Gabrielov in [3]. We now introduce some notation, so that condition (R5) may be

more easily stated below.

Notation 7.2. Suppose f E R, and D : { 1,. . . , n} -+ {-cqO,+oo}. Let Z, C r-1,1]” be

dejned by

I-, = [-l,O),

1, =&(I) x . . x LT(,), where Zo = [-l,l],

z+m = (0,11.

Then for each such o we define fn : Z, ---f iw by

h(X) = f& 1,

where

xj o(i) =o, X, = (01 (x), . . . , O&C)) and ai =

$ C$i)#O.

All the conditions on R in which we will subsequently be interested are now listed

together here for ease of reference.

(RI) Each R, consists of C’ functions;

(R2) Each quantifier-free definable set in (E, {f}),fER has a (T7) connected com-

ponents number;

(R3) If A is 31-definable in (R,{f})fE~, then so is 2;

(R4) Suppose that W C [- 1, I]” is a set of the form

W= .FE[-lJ]“:f(Z)=OA

{

A (Xj>O)A A (Xi<O) a(i)=+m 0(i)=-cc 1

where f(F) is of the form p(J (X), . . . ,f&f),~) with p E R[zl,. . . ,zj+,J and

f, ,..., fi~R and where o:(l)..., FZ} + { -00, 0, +CXI}. Then w is 3, -definable

in (E, {f}) PER; (R5) LetfER,anda:{l,...,n}-t{-~,O,+co}.Then3h~R,suchthathlr~=f,;

(R6a) All functions in R are C”;

(R6b) (9 {f})j ER is polynomially bounded, i.e. for each definable function f : [w---f

53, there exist a E Iw and p E N such that

If &>I <xp Vx E (a, 00);

(R6c) R is closed under partial derivation, i.e. if f E R, and i E { 1,. . . , n} then g E R,.

S. Ma.wellI Annals of‘ Pure and Applied Logic 95 (1998) 185-227 71’)

Remark 7.3. 1. We note that (R4) is easily implied by the statement that if A is

quantifier-free definable in (a, {f}).fE~, then so is 2. We give the condition in this

more complicated form however, in an attempt to state our final results in the strongest

possible form.

2. We will only ever be interested in the case when (RI) holds. Therefore, when

condition (RS ) holds, it implies that for each .f E R,, f0 must always be extendab!e to

a function with domain Iw” which is C’. This can be seen to restrict the behaviour at

infinity of functions in R, and has the same effect as allowing only restricted functions.

3. By condition (R6) we mean the conjunction of conditions (R6a), (R6b) and (R6c).

In what follows, we will always be assuming that (RI) and (R2) hold, and we will

show first that (R3) may be weakened to (R4) whenever condition (R5) holds. We go

on to show that if (R6) holds then (R4) holds. This leads us to some further model

completeness results.

Theorem 7.4. Suppose (Rl), (R2), (R4) und (R5) hold. Then (RR,{j‘})fGK is model

complete.

By Theorem 6.5, it is enough to show that (R3) holds. Recall that to verify (R3) we

must show that the closure of an It-definable set is itself 31-definable. (Note that unless

stated otherwise, definability will always be with respect to the structure (R, {.f}) fen.)

Before giving the proof we illustrate the general method by applying it to a simple case

in the following worked example. Note that in general (and in the example below), if

A = rI,,,n[B] then II,,,[B] &A, but II,,,,[B] #I. Consequently, the method is designed

to pick up the points in A\II,,,[B].

Example 7.5. Suppose that

,4=(x, :hz(n,,x2)~B}=(0,~).

Here A will be our 3r-definable set and we will show how our method can obtain 2

by considering the closure of quantifier-free definable subsets of [- 1, l]*. Let

J-,=(-x,-l], Jo=[-l,l] and J+ccl=[l,;xj).

We now deal separately with each of the regions

1. B, =BnJo xJo;

2. B2=B~JoxJ+,;

3. Bx=BnJ+, XJO.

Note that for this example, we do not deal with the other 6 similar regions since they

are all empty. We aim to define corresponding A,, A’, A3 C [w such that A, U A: U A? = 2.

1. Let A1 =TI~,J[B~]=[;,~].


2. Consider the set

Let Az=II2,3[B;]=[O,i].

3. Consider the set

B; = {(x1,x2) E (0, l] x [-1, l] :2&x2 = 1 Ax1 >0}

={(n,,X~)E[-1,1]2:2X2=X~ AX] >O}.

Let AT = II2,r [z;] = [0, 11, and let

A3={x:3y(xy=1AyEA;)}=[l,co).

Now .~~UA~UA~=[~,~]U[O,~]U[~,~~)=[O,KI)=~ as required.

Remark 7.6. This example illustrates the idea of inverting variables so that we may ob-

tain information about unbounded sets by considering quantifier-free definable bounded

sets. Note that it is precisely condition (R5) which will make this possible when we

consider sets definable in (E, {f})fE~ which are not simply semi-algebraic as they

have been in this example.

We now proceed with the proof of Theorem 7.4, which is based on the method in

the example above.

Proof. Let A C iWm be an 3r-definable set. Then we may use Lemma 6.2 to deduce

that A may be written in the form A = II,,[B], where

B = {X : p(f, (X), . . , f+),X) = 0)

for fl , . . . , fl E R. We recall also the following notation:

I- w =[-190) J- cc =(--m,--11, IO =[-1, l] and Jo =[-l,ll,

~+co =(Q 11 J+m = [Lo).

Fix cr:{l,...,n}+ {-co, 0, +oo}. Then define B, as follows:

Ba=BnJ,(l,x.,.xJ,(,,.

Next, we define B;f G I,, as follows:

B,:={x~Z,:p(f,(x,),...,f~&)&)=O}.

Now by (R5), we may let hr , . . . , hr E R be such that for i = 1,. . . ,I,

h,(x)=A&)=fi(x,) EEZ,,.

S. Maxwell/ Annals of’ Pure and Applied Logic 95 (1998) 185-227 221

This allows us to rewrite Bs as

B; = {x E I, : p(h,(x), . . . ) h@)&) = O}.

But now by multiplying up by the denominators coming from the remaining X,, we

can find a polynomial q E [w[zl, . . ,z[+,J such that

B;: = {x E 1, : q(h, (X), . . . , hl(lf),F) = 0).

It clearly follows by the definition of I,, that B; can be written in the form W as in

(R4), hence applying this condition gives us that @ is It-definable and clearly then

so is &,[@I. Now let

A,= x’:3j 1 t A (xiyj=l)A /\ (Xi=yi)Ay'EIL,m[%l . ~(i)SO a(i)=0 11

Now with C denoting the set of all maps CJ : { 1,. . _, PI) + { -CG, 0, -too}, I leave it to

the reader (with reference to Example 7.5) to verily that J= UISEZAd. Since each A,

is 3r-definable, we have that 2 is also 31 -definable as required. 0

For each .f’ E R, let the restricted function f : R” + R be given by

J’(X) = j(X) X E [- 1, l]“,

0 otherwise.

Let g,, = {.f : J’ E R} and E = UnEN+ &. What follows is a generalisation of a result

of Gabrielov in [3], for which he assumes that l? is a collection of restricted analytic

functions, which is closed under partial derivation. (He also assumes that the collection

contains the coordinate functions and the constant functions 0 and 1 and forms a ring.

In our case these conditions are suppressed because we state the conclusion in terms of

definability in a structure expanding R, rather than geometrically.) The proof given here

is virtually identical to that given by Gabrielov, but we include it for completeness and

to verify that it works under weaker hypotheses than those considered by Gabrielov.

Although this result is stronger than it needs to be for our purposes (to imply (R4)),

we include it in this form since it may be of some independent interest.

Theorem 7.7. Let R satisfy conditions (R2) and (R6). Let X & [-1, 11” he a set of the form

where Jf is of the form p(f,(E), . . . ,&),X)=0 with PE [w[zl, . . . ,zlfn] and f,, . . . ,

J EI?, and where each ii is either a polynomial or is in 8. Then ?? is quantifkfree

definable in (IX, { f}},fEj.

Proof. We first introduce some notation. Whenever, a property is true of all the 2ji’s,

we write it as true of g, e.g. X = {Z :f(Z) = 0 A g(X) >O}. The function I( . )/ : R” -+ R will denote the Euclidean norm. We now proceed with the proof in a number of stages.

222 S. Maxwelll Annals of Pure and Applied Logic 95 (1998) 185-227

Stage A: Let

A= (&,6):3? d&X)<; A_

i i ytX,~~~~/,~~~min(Y)=6 ’ II

where d denotes the distance function and &,,(y) = min{g”, (y), . . . , ijk(y)}. Note that

A C (O,OO)~. We show that ?IIC E N+, PI >O such that

[(E, 6) E A and E E (0, pl)] + ?-’ < 6. (t)

The reason for using K - 1 rather than K will become clear in Stages B and C.

We first suppose that (E,O) ~2, and show that E = 0. Suppose not, then 3 sequences

{pi} -+ E>O, {Si} -+ 0 such that (Ei, Si) E A (Vii E N). Also, 3 a corresponding sequence

of witnesses for X, say {Zi}, which, being bounded, has an accumulation point, z say.

Now, by definition of A, 37&Y with /jZ - 711 GE/~. However, by the properties of

the sequences, sul+*, Ilz_Y,,‘li~,2 coin GO. Hence, by definition of X, $7 EX with

)(Z - j7lj d&/2, which is a contradiction.

Now note that since X is definable in ($ {f})fE~, so is A. But by (R6b), (D,

{f})f,~ is polynomially bounded and the above argument showed that if (E, S) l 2

and E > 0, then 6 > 0. Hence we can find K E N+ such that the graph of the polynomial

a?’ ‘lies underneath’ the set 2 for small values of x, i.e. 3~ E N’, pl >0 such that

(t) holds as required. From now on we fix K and ~1.

Stage B: Suppose that we are given a function II/ : [0, co) + [w with the property that

t,b(t) = o(P) as t--f 0. Then we can find p >O such that

([E,&EA and EE(~,P)]+E~+$(E)<&

By the given property of 1+4, we may let p2 >O have the property that

W(t>l < 1

\JtE(o,pZ), yy-

Now let p = min{pl, ~2, l/2}. Then if E E (0, p),

EK + l+b(&)<2EK<EK-’ <6

as required.

Stage C: Let gK,X,i denote the sum of the terms which have order at most K and

appear in the Taylor expansion of gi about the point X. Again, whenever a property is

true of all the g”K,j(,i’s, we write it as true of kjK,:. Let

I$@, 7) !Ef &(y) 3 IIX - 711” A (g”(X) > 0 v I/x - 711 > 0).

The point behind this definition is that the variable 7 occurs only semi-algebraically

in the formula 4(X,7), a fact which we shall be using in Stage G. We now let

G = {y :&) = 0 A f#@, jq}

and show that Z E x ($ ?i E ??,.

S. Maxwell1 Annals of‘ Pure and Applied Logic 95 (1998) 185-227 223

First note that it is trivial that Z EX % ZE 6. It therefore suffices to show be-

low that ZE??~+EE~, and then that ZE~\X +?i E&. For both of these, we let

& : [0,03) + [w denote the function given by

$q(t)= suP{/BK,u.j(Y) - ii(V)1 : IlY-~ll dt; i= l,...,k}.

Now, since by (R6a) the ii’s are Cm (on [-1, I]“), they are certainly C“ which implies

that r&,(t) = o(t’) as t --f 0, and so we may let ~2, p >O be as defined in Stage B.

Let 3 E Gz and ;’ > 0. We will find c E X such that (]a - ~11 <y. Let C E & such that

(15 - ??]I< min{y, ~2). Then

where we are using first the fact that C E & and then the definitions of p2 and $q. It

follows that G(T) > 0, and because T E G;r *f(C) = 0, we have that C EX as required.

Now, we let z E~\X and y > 0, and show that we can find C E C, such that I]z-z\\ < 7.

Choose b such that VEX and E< min{;l,p} where c = I/Z - $11. Let

Then clearly (E, 6) E A, since 5 EX. But now by Stage B, E” + I+&(E) t6. Hence by

definition of 6, we can find C EX such that 11~ - C/( da and such that

@,i”(c) - $q(c)>a’.

Now by the definition of $g, we get that i,,JC) > E” 3 11~ - ?(lK, for i = 1,. . . k, from

which T E Gz as required.

Sfagr D: Let

(Note that if E is such that 36 with (a, 6) E B, then the definition of B implies that

VP E (0, a], the set over which the minimum is taken is non-empty.) We note also that

Bc (0,~)~. We show that 32~ t+J+, qt>O such that

([E,~)EB and FE(~,~~)]+EI-‘<~.

As before, we show first that if (a,O) E 3, then e = 0. Suppose, for a contradiction,

that 3 sequences {si} + F, {Sj} -+ 0 and {Zj} with an accumulation point Z say, such

that for each j, d(Zj,X)>aj. Then note that d(2,X) 26. Now choose fl E (0,~) and let

J E N be large enough such that ‘v’j>J, p lies in the interval (0,&j). Then since for this

p, the minimum is taken over a non-empty set, there exists a sequence of witnesses

{Gj};>J such that for each j aJ,

(i) 115, - Uill = /3.

224 S. Maxwelli Annals of Pure and Applied Logic 95 (1998) 185-227

(ii) 4_<zi,Zj) holds.

(iii) Jf(iij)I <Sj.

Let V be an accumulation point of {u/}j>J. Then If(zS)I =0 (using (iii) and the

facts that T is continuous and {Sj} ---) 0). Also ]]Z - till =/I by (i). And (ii) implies

g(Uj)> ]]Zj - Ej]lK =p” = (1~ - UIJK. H ence g”(u) 2 11~ - fllK, by continuity of @. There-

fore &2,25) holds, and it follows that u EX. But d@,X) 2:~ >/I = (1~ - u]], which is a

contradiction. Stage D now follows easily by applying the polynomial boundedness as

before. Again, we fix 1 and yli from now on.

Stage E: Suppose that we are given a function $: [O,oo) -+ Iw with the property that

t&t) = o(tL) as t ---f 0. Then we can find r] > 0 such that

[(E,~)EB and EE(O,~I)] + si.++(&)<2j.

The proof is as for Stage B.

Stage F: Let _& denote the sum of the terms which have order at most 2 and

appear in the Taylor expansion of 7 about the point X. Let

Then we will show that Z EX ($ Z E Fz.

Suppose that ZgX and that pi >O. We show that 3b~Fz with J/Z - pII <pi. We

first note that since, by (R6a), f is C”, it is certainly of class C’, and hence

]!i,,z(y)l = If( + o(/]z - 711)” as J/Z - 711 -+ 0. Therefore $2 >O such that we have

If,,,<~>l G lh>I + lb - AI’ w h enever 11~ - 711 <pz. Now we note that since z E ?&,

we can find b E & such that ](z - b]] < min{pi,pz}. Now @,b) holds and f(b) =0,

and so I~~,,(b)l < ](a - bl12. Th’ is implies that b E Fz as required.

Conversely, suppose that 5 EFZ, but Z $x. We show first that we can find yi , ~2, ya >O

such that the following hold:

(a) Q’p E (0, vi] 37 E FZ such that 115 - uJ( =/I;

(b) 4W)a~2;

(c) If /I E (0, ys] and (p, S) E B, then p’ + @f(p) t6, where & : R --f R is the function

given by

if+(t) = sup ,,u_7i,,<t 1fm -&ml.

The existence of y2 follows from the fact that Z @x. The fact that f is (at least) C’

implies that $/r(t) = o(t’) as t -+ 0, and hence Stage E ensures the existence of ~3. For

yi, we note that the set

{P : 30 E Fii A IF - VII = PII

is 3i-definable in (E, {I}),-,R (the functions appearing in fl,,; and JK,s are semi-

algebraic or in r? since it is closed under partial derivation by (R~c)), and hence is the

projection of a set which has finitely many connected components, by (R2). Therefore,

the set itself has a finite number of connected components, which, together with the

fact that ZE& implies that we can find a suitable yi.

S. Maxwell I Annals of Pure and Applied Logic 95 fl998) 185-227 225

NOW let y = min{yt, yz, ys}. Then (a) ensures the existence of 6, where

B = sup PE(O.:~l

(min{l_%? : 4477) A 112 - Al= PI) .

Then by (b) and the definition of B, (y, S) E B and hence by (c), 7’. + $f(y) < S. Now

by definition of 6, we may find j30 E (0, y] such that

min{lj(j3l:4@,Y) A IlE - Yll =Po) >Y^ + $jclf(:l).

But

It follows that

min{ljj&,&9 : W,T) A Ila - 31) = PO} >yi >B”,

i.e. $7 with l]iz - 711 = ,!3 0 such that 7 E F;, which contradicts (a).

Stage G: x is quantifier-free definable in (R, {fl})fEi.

Re-expressing Stage F gives us that

_- where x(x,~,E) is the formula

~,.,O) 3 Ilk - yll” A (g”(X) > 0 v I/z - jql > 0)

A l&(Y)1 d 11x - 711” A 112 - VII <E.

Now we note that x may be viewed as a semi-algebraic formula in the variables y

and E with coefficients being either partials of f or 5 at X of order between 0 and

p, where p = max{rc, A}, or polynomials in X, or elements of [w. Now, by Tarski [8],

we may obtain a quantifier-free formula x’(X) which defines x, where x’(X) is in

the language of rings together with function symbols for the partial derivatives of r

and g of order at most ~1. However, since f ,, . . ,fI, Lj,, . , ,ijk are in I? or polynomial,

and (R6c) implies that k is closed under partial differentiation, this implies that x is

quantifier-free definable in (E, {.j))jEd, as required. C

Theorem 7.8. Let R satisfy (R2),(R5) and (R6). Then (E,{f})f.E~ is model

complete.

Proof. Immediate from Theorems 7.4 and 7.7, since (R6a) clearly implies (Rl ). C

Corollary 7.9. Suppose that R satisjies (R5) and (R6c) and that each Jimction in R

is unalytic. Then (@, {f })fE~ is model complete.

226 S. MaxwelllAnnals of Pure and Applied Logic 95 (1998) 185-227

Proof. Condition (R6a) clearly holds. For the conditions (R2) and (R6b), we turn our

attention to [w,, the structure obtained by adding all restricted analytic functions to iw.

This structure may be presented in many (equivalent) ways, one of which is to consider

for each 12 E Nf, a function symbol for each analytic function f : [w” ---f U&! which can

be continued analytically at infinity. All of these function symbols are then added to

@ to give the structure [w,,,. The fact that R satisfies (R5) and contains only analytic

functions implies that the structure ($ {f}) Jew is a substructure of Iw,,. However, it is

known that [w,, is both o-minimal and polynomially bounded, and hence (9 {f})fE~

satisfies (R2) (by Lemma 7.1) and (R6b) as required. q

Remark 7.10. 1. Corollary 7.9 is a result first obtained by Gabrielov in [3]. An un-

published version of Theorem 7.8, where the condition (R2) is replaced by the a priori

stronger assumption of o-minimality of the structure, has been proved by van den Dries,

using a method which is a generalisation of Gabrielov’s argument. Both these results

are stated in a slightly different form, i.e. as referring to functions which are restricted

in the sense of being set to zero outside the unit cube. This is a condition which is

essentially the same as our condition (R5).

2. We note that Theorems 6.6, 7.4 and 7.7 also imply that (R, {f})fE~ is o-minimal

whenever R satisfies (IQ), (R5) and (R6). However, the o-minimality result is immedi-

ately implied by the main theorem in Wilkie [lo], where he only needs the hypotheses

that R consists of functions satisfying conditions (R2) and (R6a).

Acknowledgements

I would like to thank my supervisor Alex Wilkie, who has helped enormously with

many of the ideas used in this work, and with its presentation. I also express my

gratitude to the EPSRC and to Brasenose College, Oxford for financial support, and to

the referee for many helpful suggestions.

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Math. 125 (1) (1996) I-12.

[4] M. Karpinski, A. Macintyre, o-minimal expansions of the real field: a characterisation, and an application

to Pfaffian closure, submitted, University of Bonn and University of Oxford, 1997.

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dissertation as part of D.Phil Thesis, University of Oxford, 1996.

[8] A. Tarski, A decision method for elementary algebra and geometry, 2nd ed. revised, Berkeley and

Los Angeles, 1951.

[9] A.J. Wilkie, Some remarks on a paper of Charbonnel, preprint, University of Oxford, 1994.

[lo] A.J. Wilkie, A general theorem of the complement and some new o-minimal structures. preprint,

University of Oxford, 1996.

A general model completeness result for expansions of the real ordered field

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