Algebra in the Stone-Čech compactification: theory and applications

Remark 9.24. Theorem 9.21 and 9.23 are valid with (N, +) in place of G. This is an imediare consequence ofthe fact that W* is a lefr ideal of (BZ, +) by Exercise 4.3.5.

Weconclude this section by listing some open questions which are tantalizingly easy to formulate, but very hard to answer.

202

9

racrnpolcnu

We shall see, as a consequence of Thrmrems 12.29 and 12.45, that it is consischat there are idempotents p in @N such that, if p = q + r , then bxh q and r arc in Z p. In particular, such idempotcnts are both sr-maximal and sL-maximal.

+

Quatlon 9 3 . Out it be shawn in ZFC that Z*contains lq?dmal

idempotents?

As a eoasequence of Exercise 9.1.4 we know mat no minimal idempotent in Z* can be right maximal.

Quation 9.26. Are there any idmtpoents in Z* which are both minirnalondmaxid?

If (p& isaseyemeof idempotentsin/3Zsuchthatpn a . )

9.4 'Identitiesin /JS If S is any discr*e sernigmup. then an identity of S is also an identity of /3S. This follows from the fact that the equations se = s and es = s, which arc valid for every s E S, extend to @Sbecause the maps p, and L are continuous on /3S. In this section. we shall see that this is the only way in which an identity can arise in @S. We shall show, in fact, that @ Scannot have a unique right identity which is a member of S*. If S is a weakly leftcancellative semigroup. S is aleft ideal in @S(by Theorem4.31) and @Scan have no right identities in S*. However, a simple example of a semigroup S in which S* has rightidentities. although S has none, is provided by putting S = (N,A). In this case, every element of S' is a right identity for @S.(SeeExercise 4.1.11.) In coneast, we shall see that, if S is commutative. then no element of S* can be a left identity for @S. Theorem 9.28. Let S be a discrete semigmup. IfBS has a right identity q E S*, then /3S has at leasf 2*' right identities in S', where K = Ilq 8. PmoJ For each s E S. let I, = [t E S :rt = s). Since sq = s, we have q E A;I[(s)]. This is an open subset of BS and so I, = S n ~ ; ~ [ [ sE) lq. Choose A E q with IAl = ~ K . Then ( A n I, : s E A] is a collection of K sets with the K-uniformfinite

B

a,,,

intersdon pmpnty. It follows from Theorem 3.62 that contains 2' r uniform uldlters. We shall show that hzry ulnafilter p in this set is a right identity

for as.

To see this, lets E S. We can choose t E A n I,. because A and I, srr: both in q. Then I, E p. Now I, g I,. because u E I, implies that su = (st)u = s ( t u ) = st = s. SO I, E p. Hence s p E cC{su :u E I,) = (s)and t h m f m s p = r. It follows from the continuity of pp that x p = x for every x E pS. 0

CoroUy 939. Ln S be any discn?te semigmup. I f q E pS is a unique right identity for PS, then q E S.

Corollary 930. Let S be any discrete semigroup. I f q E pS is an ideniityfor pS. then q E S.

Theorem 931. Let S be a left cancellative discme semigmup. I f p is a right identity for PS, then p e S lrnd p is an identityfor S and hencefor BS. Pmof: Lets E S. Since s p = r, we must have p E S, because S* is a left ideal of pS @y Theorem 4.31). This implies that p is a left identity for S because, for every t E S, we have st = spt and sot = pt. So p is an identity for S, and we have noted that this implies that p is an identity for pS. 0

Theorem 932. Lct S be a commumMtrve discmte semigroup and let q E S. Then q cannot be a left identify for PS. lim st = sq. we have sq = s for every s

#A" , .,

E S. By the continuity of pq, this imPli& that

xq = x for evety x E pS. So q is an identity for flS, wnaadicting Corollary 9.30.

We obtain a partial analogue to l h e w m 9.28.

Theorem 933. Let S be a discrete semigmup. I f ,9S has a I@ idenfity q E S with 11q11 = K. then S* has at least Z2' elements p with the propeny that ps = s for every s E S.

P&

The proof of Theorem 9.28 may be copied. with left-right switches, except for 0 the last sentence. We now see that, if S is countable, any right identity on S* is close to being a right identity on /3S.

Theorem 934. Let S be any discrete countable semigmup. IfS has a right idcntiry q, then S has a left ideal V such that q is a right identity on V and S\V ishire.

Pmof: Let U = S*\ cC(Sq fl S ) .For each s E S, ( s q )n S is either a singleton or empty. Thus. by Corollary 3.37. S q n S* is nowhere dense so that U is a dense open subset of S*.

204

9 ldcmpotents

Let T = {t E S : tq E SJ. We claim that S\T is finite. To see this, suppose the conhary. Thenthere is an element x E U n(S\T)*, and we can choose X E x satisfyin* ' X S U and X S S\T. We have x E X and xq E Since x = xq, it follows fmm

z.

~ h m r c m 3 . 4 0 t h a t s = y q f o r s o m e s € ~ a n d s o w y E ~ . ~ y = s ~ f o r s o mP e and s o w y E X'. The second possibility cannot hold because X* S U . So we may assume the first. Then s E and so, slnce s is isolated In PS, we must have s = tq for some t E X . Thls contradicts the assumption that X E S\T. So S\T is finite & therefore T* = S*. Since q is idempotent, T q S T. So pq maps T to T. It foUows from Theorem 3.35 that. if V = { t E T : tq = 1 ) . then V E x for every x E T*. So T\V is finite and therefore S\V is finite. Furthermore, it is clear that V is a left ideal of S.

0

Corollary 9.35. Ler S be any discrete countable semigroup. VS' has a right identity. then S* has at least 2' right identities.

P m f Let q E S* be a right identity for S* and let V be the left ideal of S guaranteed by Theorem 9.34. By Theorem 9.28, with V in place of S. there are 2<elements of S* which are right identities on 7. If p is one of these elements and if x E S*, then the fact that up = u for evety v E V implies that xp = x, k a u s e x E 7. Exe-

9.4.1. Let S be a right cancellative discrete semigroup and let p be a left identity for,% Rove that p E Sand that p is an identity f a S and hence for&% (This is Theorem 9.3 1 with the wads "left" and "right" interchanged.)

Notes Theorem 9.10 was proved by I. Protasov (in a personal communication). Topologiesof the kind discussed in Theorem 9.13 have becn studied by T. Paparyao 11911. Theorem 9.15, Theorem 9.16, Corollary 9.17, and Theorem 9.19 are due to I. Rotasov (in personal communications). Corollary 9.17 answers an old unpublished question of E. van Douwen, who wanted to know if there were any homogeneous regular topologies that are maximal among topologies with no isolated points. The semigroups X,,defined in Exercise 9.2.3, were intsoduced by E. Zelenuk in [246]. Theorem 9.28 is due to I. Baker. A. Lau, and I. F'ym, in [9].

. .

Chapter 10

Homomorphisms

In this chapter, we shall consider continuous homomorphismsdefined on subsemigroup of pS. where S denotes a countable discrete semigroup which is commutative and cancellative. Given two semigmups which are also topological spaces. we shall say that they are algebraically and topologically equivalent or that one is a copy of the other. if there is a mapping from one to the other which is both a homeomorphism and an algebraic isomorphism. We have seen that copies of MI occur everywhere. because of the remarkable facility with which homomorphisms can be defined on W.(See Theorem 6.32. for example). In contrasl if S and T are countably infinite discrete semigroups which are commutative and cancellative, there are surprisingly few continuous injective homomorphisms mapping @ T into p S or T * into S*. and none at all mapping pT into S*. In particular. there are no topological and algebraic copies of BN in W*. Of coune, any injective homomorphism from T into S d w s determine a continuous injective homomorphism from pT into pS, as well as a continuous injective homomorphism from T' into S' (by Exercise 3.4.1 and Theorem 4.8). We shall see in Theorems 10.30 and 10.31 that it is true that every continuous injective homomorphism from BT into @ S ,and almost hue that every continuous injective homomorphism from 7' into S*, must arise in this way as the extension of an injective homomorphism from T into S. If S and T aregroups andif L and M are principal left ideals in Sand T*respectively, defined by nonminimal idempotents. we shall see that any mapping between L and M which is both an isomorphism and a homeomorphism must arise from an isomorphism between S and T . We shall also see that any two distinct principal left ideals of pN defined by nonminimal idempotcnts cannot be copies of each other. Whether this statement holds for minimal left ideals is an open question. We remind the reader that, if S is a commutative discrete semigmup. then S is contained in the center of p S (by Theorem 4.23). We shall frequently use this fact without giving any further reference. We shall use additive notation throughout this chapter, except on one m i o n where we present two results about (N..),h e c a w (W, +) is the most important semigroup to which our discussion applies and because this notation in convenient for some of the proofs.

206

I0 Hnnomorphism

10.1 Homomorphisms to the Circle Group In this section we develop some algebraicinformationrelated to nahual homomorphisms froma subsemigroup of (R, +). the primary example being N, to the circle p u p T. This infomation is of interest for its own s a k and will be useful in Section 10.2 and

3

again in Section 16.4 where we pmdua a large class of examples of sets with special mmbinatorial propdes. We take the circle group T to be R/Z and shall use @ for the group optation in T. We define n to be the natural projection of R onto T. If 11.1~ E T we shall write tl < t2 ifthereexistsul.u2 E R suchthatn(ul) =rl.x(uz) = tz andul s u2 < U I f.

+

Lemma 10.1. Let S bc an infinite discretesonigmup and lei h : S + T be an injective homomorphism, with h denoting its continuous u t m i o n . t c i U = [X E S* : (S E S :z(x)
Pmof: Let x E S*, let TI 3 [S E S : j;(X) < h(s)] and Tz = [s E S : h(s) < b ) ) . Then TI n Tz = 0. Since IS\(Tl U Tdl 5 1. TI E x or TZ E x. Thus x P U n D and XEUUD. We now show that U n Z and D n Z an non-empty. Let p be an idempotcot in S. Since z i s a homomorphism @y Corollary 4.22). p E 2. Suppose that p E U . Thm we can choose a sequence (r.)zl in h[S] for which n(0) < t. f a every n and .t + ~ ( 0 ) . Now %s] is a compact subsemigroupof T and is t h e m f o ~agroup (by Exmise2.2.3). Hence,fore.chn E W.wecanchoose y. E BSforwhiihh(y.) = -1. Lety E S* bea limit point of the sequence (y.)gl. Then y € Z and h(yJ = -t. < n(0) forevny n. Foreachn E W, y, E cC([s E S : h(s) < n(0))) andso y E cC((s E S :h(s) < ~ ( 0 ) ) ) . Thus [s E S : h(s) < n(0)) E y and y E D. This shows that p E U implies that D n Z # 0. Similarly, p E D implies that U n Z # 0. So U n Z and D n Z are non-empty. We now show that U is a right ideal of BS. Suppose thaix E U and that y E BS. p o s e u and v in R such that n(u) = h(x) and n(v) = h(y). Let X = (s E S : h(x) < h(s)J. Then X E x. Givena E X, let 7i E (u. u $) satisfy n(ii) = h(a). Put .Y = (S E S : h(s) E n[(v+u-ii,v+u+f -ii)]). ~own[(v+u-~,v+u+f-alis a neighborhood of n(v) =z(y) in T, and so Y. E y. By Theorem 4.15. (a b :a E X andb~Y.)~x+yW . eclaimthat(a+b:a~Xandb~Y.]~[s~S z(x y) < h(s)) so that x y E U as required. To this end let a E X and b E .Y Pick tu E (v + u - ii, v + u f 3 such that n(w) = h(b). Since h and n are homomorphisms we have z(x y) =-Nu v) and h(a b) = n(7i w). Since u + v e Z + w c~+v+f,wehaveh(x+~)
+

+

+

+

+ +

+

+

+

10.1 Hanomorpbisms to the Circle Group

201

'Ibis establishesthat U is aright ideal of BS. The proof thM D is a right ideal ofpS is essentially the same. 0 'Ibe following notation docs not reRect its depcndmcc on the c h a i a of the semi-

P

P s.

m t i w 103. Let S be asubaemigroup of (nt, +)and lctu be apoaitive real number. We define functions g, : S -P Z, f, : S -P I-;, f), and h. : S -r T as follows for x E S:

z,

We shall use &, and & to denote the continuous extensions of these functions $1, and T respectively, where S has the discrete topology. which map BS to P Z , 'Ihcse exist because the spaces BZ. [-i,i], and T an compact. Note that g,(x) is the nearest integer toxu if xu $! Z+ f .We shall use the fact that. for any p E BS and any neighborhood U of z ( p ) , the fact that is continuous implies that A-'[u] is a neighborhood of p and hence that f,-'[u] E p (by 'Ib-m 3.22).

[-I.

z

Lemm 103. Let S be a subsemigroup of -@, +) and let a r 0. The map h. is a homomorphismfmm S to T.Consequently h, :BS -r T is also a homomorphism Proof: We note that h&) = %(ax) and so h. is a homomorphism. 'Ihe second 0 conclusion then follows by Comllary 4.22. Definition 10.4. Let S be a subsemigroup of (R, +) and let u > 0.

Observe that U. and D, an the set of points of P S for which value from above and below respectively.

5 approaches its

Ddlaitlon 105. Let Sbe a subsemigroup of @. +) and l a a E R Then a is irrational with respect to S if and only if

208

10 Homomorphisms

Saying that a is irational with respect to S means that h. is one-to-one on S and that 0 p ha[S\M1. Notice that "inational with respect to W is simply "irrational". Notice also that if IS1 c c, then the set of numbers irrational with respect to S is dense in R. Remark 10.6. Suppose that S is a subsemigmup of 09, +)and fhai a > 0 is irmtioml with rrspect to S. k t U. D and Z be the sets defined in Lemma 10.1 with h = ha. Then U = Ua. D = Da and Z = Z,\(O). Lcmma 10.7. Let S be a subsemigmup of respect to S. Then Ud U Dm = S*.

(R,+) and let a

7

0 be irmtioml with

P m f :This is immediate from Lemma 10.1 and Remark 10.6.

0

The following rrsult is of interest because, given the cohtinuity of pp, it is usually easier to describe left ideals of BS. (See for example Theorem 5.20.) Theorem 10.8. Let S be a'subsemigmup of (R.+) and let a z 0 be irrational with respect 10 S. Then We and Dm are right ideals of BS.

Pmof:f i s is immediate fmm Lemma 10.1 and Remark 10.6.

0

In the following lemma, all we care about is that cCflS(Na)n U, # 0 and that d p s ( N a ) n D, # 0. We get the smngerconclusion for free, however. Lemma 10.9. Let S be a subsemigroup of (W,+) and let a > 0 be irrational with respect to S. Thenfor each a E S\(01, cCgs(Wo) ilXu # 0 and ctps(Na)n Y, # 0.

P m f : We apply Lemma 10.1 with h replaced by h, and S replaced by No. Then c f p ~ ( N arl) X, = U rl Z # 0 and cCps(Na) n Y, = D n Z # 0. For the remainder of this section we resnict our attention to PI. Notice that, while h. is a homomorphism on (flW. it is not a homomorphismon (BN. .). However. close to0 it is better behaved.

+),

Theorem 10.10. Lcr S = N and let a be a positive irrational number: Then Xu and Ymare left ideals of (BPI, .).

PmoJ We establish the statement for X,, the proof for Y, being nearly identical.

&,

Letp E Xa andletq E BN. We€irstobservethat,foranym. n E N.if 1 f,(n)l < then f.(mn) = mf,(n). To see that qp E X . let 6 > 0 be given (with 4 5 For each m E N, let B, = ( n E W : 0 < f,(n) < ): and note that 9, E p. Thus by Theorem 4.15. (mn : rn E N a n d n ~ B , ) ~ q p . s i n c e ( m n : m ~ N a n d n ~ B , ) ~ { k ~f ,W( k:) O < s<) , i t follows that qp E X,. 0

i).

Ofcourse, as the kernel of a homomorphism. Z, is a subsemigroup of (BN. +) for any a z 0 . Consequently, by Theorem 10.8, if a is irrational. then X. and Y, are subsemigroups of (BW,+).

.. A

,

10.1 Homomorphisms to Ihe Circle Group

209

CornMary 10.11. Lct S = N and let a be a positive irrational numbez Then 2, is a ~ubsemigmupof (BN. .I. pmof: Sirmaisirrational. (n E N : f,(n) = 0) = OsobyLemma10.7. Z = X.uY, so the result follows from Theorem 10.10. 0 The followingresult is our main tool to be used in Section 16.4 in deriving nontrivial explicit examples of IP* sets and central* sets.

Tbeomn 10.12 Let S = N and let a r 0. Thm & is an isomorphism and a homeomorphismfmm Z, onto ZI,, with inverse 87,. If a is i m r i o ~ l then , & takes X, onto Y I ~ , andtakes . Y onto XI/,. Proof: We show that

Tben using the fact that the same assertions are valid with I / a replacing a proves the t h w ~ m Notice . that the hypothesesof statements(4) and (5) are nonvacuous only when a is irrational. TO verify (1). let p,q E Z. and let 0 < c < Let A = (m E N : Ig,(m) am1 c r). If n and m are in A, then Ig,(m) g,(n) - a(m n)l c 2c < so g a b ) + g d n ) = g a b n). Also, A E p and A E q so

+

+

4.

4

+

To verify (2) and (3). k t p E Z. and let q = z ( p ) . La r E (0 ') and let B = '7 (n E N : Ifa(n)l c €1. Foranyn E N,wchave I fa(n)l c (ifandonly ~f Im-nal < E w h m m = g,(n). Now Im nal c c if and only if lm; - nl < The second inequality establishes that n is the integer closcst to m and hence that n = g L(m). m Thus we have seen that n E B implies both I fi (g.(n))l c 5 and n = gl(g,(n)). So " "

-

.

.

2.

..

To verify (4). let p E X., let r r 0 be given with s c 1/2, and let B = (n E W : -6 < fil,(n) < 0). We need to show B E g,(p). Pick 8 > 0 with 8 < E a. Let C = ( n E N : O < fa@)
.

210

10 Homomorphisrm

EsereLsc 10.11. It is a fact that if F and C are disjoint finite nonempty s u b of R and F u C is hncarly independent over Q then U. n D, # 0. Use this

nu,F naeG

fact, together with Theorem 10.8 to show that (BN. +) has a collection of 2' p a h v i x disjoint nght ideals. (lXs fact also follows frmn Theorem 6.44.) Exerdsc 10.13 Let a z 0 and let p,q E BN. Rovc that & ( p 1.&(p) + G ( q ) + 11. &(q)*&(p)

-

+ q ) E ( c ( p )+

ExereLsc10.1.3. L e t a > Obehtional.1ctk E Nwithk > a . a n d I c t d = k - a . Pmve that (i) (ii) (iii) (iv)

for all p E BN. h?.(p) = -%(PI, Z = Za, X, = Ys,and .Y = Xs.

10.2 Homomorphisms from BTinto S* We h o w that topological copies of j3N arc abundant in N*. Indeed, every infinite compact subset of N* contains a copy of this bind, by Theorem 3.59. However, as we shall show in this section. there are no copies of the right topological semigroup BN in W*. In fact, if @ : gN + N* is a continuous homomorphism, then @[BNis finite and

19WIl = 1.

We shall show that, ifS and T are countable, commutative, cancellative semigrovps and if S can be embedded in the unit cimle, then we can often asscrt that @ [ T wis] a finite gmup whenever 41 : BT -t S* is a continuous homomorphism.

Lemma 10.13. Let S be an infinite dircmie semigmup which can be algebmically embeaUed in T and kt V be an infinite subsemigmup of S. Then,for every x and p in y E V*for which x y p # y p x. S*, there

+ +

+ +

P m . Let h : S + T be an injective homomorphism, with : BS + T denoting its continuous extension. Let U and D be defined as in L e y 10.1. By this lemma, applied to V instead of S. if U'- ( y E V* : (s E V : h(y) -c h ( s ) ) E y ) and D' = ( y E V* : {S E V :h ( s ) < h ( y ) )E y), then U' and D' are non-empty. Let x . p E S*. I f x E U. we can choosey E D'. Since D' E D and U and D are rightidealsofgS.x+y+pe U a n d y + p + x e D a n d s o x + y + p # y + p + x . Similarly, i f x E D, we can choose y E U' and deduce that x y + p # y p x.

+

+ +

Lemma 10.14. Suppose that S is a countably injnirc discnte semigmup which can be algebmicaNy embedded in T and that T is a counrab& infinite commutative we* leji cnncellative semigmup. Let @ : BT -+ S be a conti~uu)ushomomorphism Then K(@[T*]) is afinite group. Furthemon, for every c E T. ( b E T : 9 ( b c ) E K($[T*]))is infinire.

+

10.2 Homomaphisms from ,9T into S

p m j O w e that by 'Ibmrem 4.23, i f a E S, then a commutes with every member of PS. Also, if t E T, then r commutes with mry member of PT and hence #(I) commutes with evny member of #[)TI. We shall show that if p is any minimal idempatent in #[To] and q E T* with #(q) = P. then

+

(1) there exists r E T such that @(r) E #[T*] p and (2) given any c E T and any B E q. Ihmexists t E T such that #(b) K(#[T9I).

+ #(c)

E

To see this, let such p and q be given, let c E T. and let B E q. Let V = (a E S : a + #(q) E #[BT]) We claim that V is fmite. Suppose instcad that V is infinite, and noticethatVisasuhsemigroupofS. (Givena.b E V.a+b+#(q) = a + b + @ ( q ) + #(q) = a #(q) b @(q).)Forevery y E V*.we have Y + 9(q) E b[PTl, by the continuityofprr.,. Thenforewryy E V*, wehavrb(c)+y+@(q) = y+$(q)+#(c) which contradic& Lemma 10.13. Choose A E #(c) for which A n V = O. Now #(c) #(q) = #(q) #(c). #(c) #(q) E cC(A d(q)). and @(q) #(c) E cC(O[Bl+ W ) ) . It follows fmm Theorem 3.40 that one of the two following possibilities must hold:

+

+

+ +

+

+

+

+

(i) a + # ( q ) = # ( v ) + # ( c ) f o r s o m e a E A a a d s o m e v € P T ; (ii) u #(q) = #(b) #(c) for some b E B and some u E PS.

+

+

The h t possibility is ruled out by the assumption that A n V = 8. So we may suppose that (ii) holds. Lett I b c. Then #(t) E ,9S p and so #(r) p = @(r). Since#(:) = #(r+q)+#(q)andt+q E T* (by Theorem4.31).#4) E #[Tw1+p E K(#[T*]). Thus (1) d ( 2 ) hold. Now let p be a minimal idempoteat in #IT*] and pick by (1) t E T such that $(t) E #[TL]+p. Let F = #[T*]+$(t). Then, since t E T, F = #(r)+#[T*]so F is both a minimal left ideal and a minimal right ideal of #[T*]. and is therefore a gmup (by Theorem 1.61). Since F is compact. it must be f i ~ t ebaause . an infinite compact subset of )S cannot be homogeneous by 6.38. If p' is any other minimal idempotent m @[T*l.there exists t' E T such that $I(:') E #[T'] p'. Now @(t t') = #(I' t). and so the minimal left ideals $[T*] + p' = #IT*] # ( t l ) and #[T*] + p = @[T*]+ &(t)of #[,9T] intersea They are k f o ~ equal. This showsthat F is theunique minimal left ideal of#[Tg1 and so F = K (#IT*]) (by Theorem 1.64). The fact that (b E T : &(b c) E K(#[TW])]is infinite for each c E T follows immediately h m (2). 0

+

+ +

+

+

+

+

+

10.15. Suppose :hut S is a countably infinite discrete semigroup which can be algebraically embedded in the unit circle. Let T be a countably infurire commutative weakly left cancefhtive semigmup with the property that any idea! of T hpr furite complement and fer @ : OT + S* be a continuous homomorphism. Then @[)TI is finite and @[Ta]is afinite gmup.

Th-

212

10 Homomorphisms

p w j BYLemma 10.14, if F = K(@[T*]),then F is a finite group. Let I = #-'[Q. We claim that I r l T is an ideal of T. It follows h m Lemma 10.14 that I fl T # 0. c E I n T, then @(c)= @(c)+ p for some minimal idempotent p of @IT*].Thereexis@ q E T* for which @(q)= p. For every b E T. we have @(b c) = @(b) @(c)a

+

+

and we have shown that I n T is an ideal of T. Thus T\I is finite. Given q E T*. I E q and so @[T*l E F. Since #[BT] @[T\IJ U F, @ v T ]is finite. S i c e #[+*I is a finite cancellative semigroup. it is 0 group by Exercise 1.3.1. Notice that the hypotheses of Thmrem 10.15 an satisfied in each of the following three cases.

(i) T is a countably infinite commutative group; (i) T = (W. +); or (iii) T = (W, V). The following lemma is simple and well known.

Lemma 10.16. k t G be a commwnfr've gmup wirh identity 0 which contains no elemenu offurite order apart fmm 0. If IGI 5 r then G can be embedded in (R. +). Pmqf Let IGI = n. Suppose that G\[O) has bem arranged as a K-sequence(k),,, and that, for each fi c x , Gp denotes the s u b p u p of G generated by (a, : a 8). Z. We then We define fo : Go -t R by putting jo(na0) = n far each n make the inductive assumption that 0 c fl c K and that we have defined an injective homomorphism f, : G. + R for every a < 0. We also assume that fa g fat whenever a c a' < 8. We shall show that we can define fg so that these properties hold with fi replaced by @ 1. Let H = Uaca G, and f = UaCBfa. Then f : H + R is an injective homomorphism. Notice that Gg = H &#. Assume first that Zag n H = {O). Then we can choose s E R\(Q f [HI), because IQf [Hll < c. We define fg : H %g + 1R by stating that fg(b ma) = f (b) + ms for every b E H and every m € Z. Then fg is well defined because %a8 n H = (0) and is easily seen to be an injective homomorphism. Thus we may assume that kag = c for some k E Z\[O) and some c E H. We define f) : H +Zag + R by stating that fS(b mag) = f (b) f (c) for every b e H and every m E Z. To see that fg is well defined. suppose that b +map = b' m'ag. where b, b' E H andm.ml E Z. Thenk(bl-b) = k(m-mf)ag = (m-mr)candsok(f (bl)- f (b)) = (m - mf)f (c) and f (b) Tf(c) = f(b') $f(c). It is easy to see that fg is a homomorphism. To see that it is injective, suppose that fg(b + map) = O for some b E H and some m E Z. Then f (b) + f (c) = O and so f (kb mc) = 0. Since f is injective, kb mc = 0. That is to say, k(b + mag) = 0 and so b + mag is of finite order in G and hence b + mag = 0.

+

+ +

+

+

+

+

+

+

+

+

,

102 Homomorphism from 87 into S

213

Thus we can define an injective homomorphism fp : Gp + B for every p c K, sothat fp C f p whenever p < +9' < K. Then Up...fp : G + R is an injective homomorphi~m.

meorein 10.17. Let G be a countably injnite c o m t a t i v e gmup which conrainr no ,ton-trivialfinitesubgmups, and let T be a countably infinite commutative weakly Iefi cancellative semigmup with the pmprrry that any ideal of T hasfinite complement. If 6 :p T + G* is a continuous homomorphism, then O[BT] isfinite and 1q5[T8]1 = 1. P m 4 By Lemma 10.16, Gcanbeembeddcdin W andthereforeinl. (If Sisacountable subsemigroup of W.we can choose a E R such that a is irrational with respect to S. The mapping ha then embeds S in T). Thus it follows from Theorem 10.15 that q$[pT] is finite and 6[T*l is a group. By Theorem 7.17. G* contains no non-trivial finite subgroups and so 16[T8]I = 1. 0

Theorem 10.18. If@ : BN + W is a continuous homomorphism. $[pN isfinite and I61W*ll = 1. P m f This follows fmm M e m 10.17 with T = Wand G = Z.

Question 10.19. Is there any continuous homomorphism 6 : BN + N* for which 16[BNl # 17 We note that, in view of Theorem 10.18. there are two simple equivalent versions of Question 10.19

Corollary 1030. Thefollowing statements are equivalent. (a) There is a nontrivial continuous homomorphismfmntBN to N*. (b) T h e r e u i s t p # q i n N ' n r c h t h a t p + p = q + p = p + q = q + q = q .

(c) There is afinite subsemigmup of W whose elements are not all idempotent.

Pmof (a) implies (b). Let 6 : pW + N* be a nontrivial continuous homomorphism. Then by Theorem 10.18.q5[@~is finiteandthereissomeq E W* suchthatq4[N8] = [q). There is a largest x E W such that 6 ( x ) # q. (Since 6 is nontrivial, there is some such x. and if infinitely many x have 4(x) # q and r E cL(x E N : #(x) # q ) n W*. then 6 h ) # 4.) Let P = W ) . (b) implies (c). The set (p, q] is such a subsemigroup. (c) implies (a). Let S be a finite subsemigroup of W* whose elements arc not all idempotent and pick p E S which is not an idempotent. Define a homomorphism f from W to S by stating that f (n)is the sum of p with itself n times and let # be the continuous extension o f f to BN. Then +is a homomorphism by Comllary 4.22 and 6 is nontrivial because ( p ) is not a semigroup. 0 Question 10.21. Which discrete semigmups S have the property that any continuour homomorphismfromBS to S' murr have afinite image?

10 Hommaphiuns

103 Homomorphisms from T* into S* ' A good deal of specialized notation will apply throughout this section. S and T denote countably infinite discrete semigroups which are commutative and cancellpti~. G will denote the group generated by S and 0 will denote the identity of G . (So G , is a wuntably infinite commutative group in which every element can be expressed a the difference of two elements of S). We shall regard S as embedded in G and BS a embedded in pG. We shall mume that T has an identity. CIhis is no real mhiction, because rn identity can be adjoined to any semigroup). We shall assume that we have chosen sequential orderings for G and T . For each rn E N , G , will denote the set whose elements are the fitst m elements of G, and T, will denote the set whose elements are the first m elements of T . We shall suppose that # : T* -, S* is a continuous injective homomorphism. We choose q to be a nonminimal idempotent in 7' and we put p = #(q). (The existence of a nonminimal idempotent in T* follows fmm Corollary 6.34.)

P w J The mapping # defines an isomorphism from T* onto #[To]. Lemma 1023. For every s x E K(pS).

E

S and every x E 6 3 , s

+x

0

E K ( p S ) implies that

P m J Suppose that s + x E K(BS). Since K ( p S ) is a union of groups (by Theorem 1.64). there is an idempotent u € K ( B S ) for which s + x u = s x . By Lemma 8.1 this implies that x + u = x and hence that x E K(pS). 0

+

+

Lemma 10.24. There is a subset Y = (t, : n E N)of T with thefollowingpropm'cs: ( i ) F o r e v e ~ y € Y * M d e v e r y b , c E T . # ( b + q ) fB G + $ ( c + y ) + p . (ii) For every m and n in N, every a E Gm, and every b , c E T,, ifm c n, then #(b+t,+q)#a+#(c+t.+q). Pmqf We can choose a minimal ide-nt u E T* satisfying u = q + u = u + q (by Theorem 1.60). We note that # ( u ) is in the smallen ideal of #[T*]by Lemma 10.22. We claim that. for every b E T . # ( b q ) 6 pG $(u). To see this, we note that # ( b q ) E pG # ( u ) implies that # ( b q ) = # ( b q ) # ( u ) E K ( # [ T * ] ) .By Lemma 10.22. this would imply that b q E K ( T a )and hence that q E K ( T * ) (by Lemma 10.23). This conhadicts our choice of q. So for each b E T . # ( b + q ) has a c l o p neighborhood Wb in pG disjoint from the compact subset pG # ( u ) of pG. For each a E G and each b, c E T , let

+

+ + +

+

+

+ +

+

+ + + p, E.,b., is a clopen subset of T*. + + + p = a +# ( c + u + u ) +6 ( q ) =

By the continuity of the map x H a # ( c x ) We note that u p E..b.r, because a 9 ( c u )

.

10.3 Homomorphisms h m T*into S

215

a + $ ( ~ + ~ ) + @ ( ~ + qSinceu+q ). = u , a + @ ( c + u ) + p E f l G + @ ( u ) EBG\Wb. n u s UaEG U%TUcETEa.bVc# T*, becnuse u does not belong to this union. It follows from Exercise 3.4.5 that there is a non-empty open subset U of T*disjoint from UoeG UbETUccTE..b+-. Wecanchoax an i n f ~ ~subset t e V of T for which V* S U . Weclaimthat,ify E V'andb, c E T,then@(b+q) f flG+@(c+y)+p. Suppose. on the conwry. that @(b q) E BG @(c y) p. Since Wb is a neighborhood of @(b q) in BG and we are assuming that @(b q) E cL(G @(c y) p), we must have a @(c y) p E Wb for some a E G. This implies that y E E.,bsc and contradicts our assumptions that y E U and U r l E..b.c = 0. To obtain property (ii). we shall enumerate V as (u. : n E N) and then replace (v.)E, by a subsequence ( t n ) E l defined inductively. We put 11 = ul. We then assume that we have chosen elements tl. t2,. .,tk in h : n ~N)withthefollowingpmperty:i f i . j ~ ( l . ,..., 2 k),aeGi.b.c~6,and i <j,then@(b+ti+q)#a+@(c+tj+q). . S u p p o s e t h a t i ~ { 1 , 2 , k).Ifa€Gandb,c~T,weclaimtherrarratmoata finite number of values of n far which $(b + ti +q) = a @(c v, 9). To see this, suppose that this set of values of n is infinite. Then the corresponding elements v. have a.limit point y in V* satisfying @(b ti q) = a +@(c+ y + q ) = a @(c y) p. We have seen that this cannot occur. It follows thatwecan choose v, withthepropenythat@(b+ti+q) # a+@(c+v.+q) whenever i E (1.2,. . . k). a E Gk and b. c E Tk. We put q + l = u,,. rhis shows that we can define a sequence ( t . ) ~ , with the property specified in (ii). We then put Y = (1. : n E N). The fact that 'Y c V* implies that pmpeny (i) holds as well.

+

+

+

+

+ +

+ +

+

+

+ +

.

...,

+

+ +

+ + + + +

.

Lemma 1025. Let Y denote the set guar&teed by Lemma 10.24. let y E Y* and let t E T. Then @(t y) p is right cancelable in PG.

+ +

Pmof: Ifwesuppoxtheconwry,then@(t+y)+p=x+@(t+y)+pforsome x E G* (by Theorem8.18). Now @(t y ) + p E cl((@(t tn q) :n E N)). because @(r y) p = @(r y q) and t y q E c W t +I. q :n EN)). We also have x @(t y ) p E et((G\(O)) $0 y) p). It follows fromThaorem 3.40 that@(t+t.+q)=x'+$(t+y)+pforsomen ~ N a n d s o m e x 'E BG,orelse

+ + + + + + +

+ + + + + +

+ + +

@(t+y'+q)=a+@(t+y)+pfors~mey'~Y*andsomca#OinG. The lint possibility contradicts condition (i) of Lemma 10.24, and so we shall assume the second. Let m E N satisfy a , -a E G, and r E T,. Since @(t y' q ) E c L ( ( @ ( t + t n + q ) : n > m ) ) a n d a + @ ( t + y ) + p ~ c t ( ( a + @ ( t + t . + q ) : n >mJ). a second application of Theorem 3.40 allows us to deduce that @(t tm q) = a @(t+yV+q)or@(t+t.+q)=-a+@(t+y"+q)forsomcn r minNand some y" E cl((f. : n > m)). Now, by condition (i) of Lemma 10.24. neither of these equationscan hold if y" E Y*. Soy" = r, forsome r > m. By condition (ii) of Lemma 10.24, neither of these equations can hold if r # n. However, they cannot hold if r = n. by Lemma 6.28. Thus @(t y) p is right cancelable in BG.

+ + + + +

+ +

+

E cW[b Y +ql)and@(b+y+q) G c~(G+#(c+y+q)), it follows fromTheomn WOthat@(b+t. +q) E BG+@(c+ y + q ) forsomen E N,

Pmof: Sinced(b+y+q)

aelse)(b+z+q)=a+@(c+y+q)fasomez~Y'andsomea~G.Thefvst possibility is ruled out by condition (i) of Lemma 10.24, and so we assume the ssecod We choose m E N such that a. -a E G, and b, c E T,. Now @(b z + q) e

+

+ +

+ +

+

c!(()(b 1. q) :n s ml) andn+)(c+y +q) E c!(a I@@ r, q) :n > m)). So a second application of Theorem 3.40 shows that there exists n z m in N and u ~cC({t,: n s m ) ) s a t i s f y m g ) ( b + t . + q ) = a + ) ( c + u + q ) a @ ( b + u + q ) = a @(c I,, q). By condition (i) of Lemma 10.24, neither of these equations can hold if v E Y*. So u = r, for some r > m. However, by condition (ii) of Lemma 10.24, neither of these equations can hold if r # n. Thus r = n. E In. -a). Byadding@(y+q)onthe So@(b+t.+q) = a'+)(c+r.+q),wherea' rightof thisequation, wesee that @(b+q)+@(b+y+q) = a'+)(c+q)+@(t,+y+q). By Lemma 10.25. Q(r, +y +q) is right cancelablein PC. So@(b+q) = al+@(c+q) and)(c+q) E G + b ( b + q ) .

+

+ +

+

Lemma 10.27. For every c E T. @(c q) E G

+ p.

Pmof:Let Y be the set guaranteed by Lemma 10.24 and let y E Y*. F a every b E T. wehave@(c+q)+@(b+y +q) = @(b+q)+@(c+y+q). ItfollowsfromComllary 620that@(b+y+q) E B G + @ ( c + y + q ) o r @ ( c + y + q ) ~ P G + @ ( b + y + q ) . In either case, Lemma 10.26 implies that @(c q ) E G @ (b q). By choosing b to be the ~dentityof T,we deduce that @(c q) E G p. 0

+

+

+

+

+

In the statement of the following t h e o m , we remind the reader of the standing hypotheses that have been apply~ngthroughout this section.

Theorem 10.28.

b t S and T be countably infinite. comrmrrative, and cancellative semigmupsmdasswne that T has an identify. Let G be the gmup generated by S and assume that ) : T ' -+ S* is a continuous injecfive homomorphism Then there is an injective homomorphim f : T + G such that f (x) = )(x) for every x E T*.

Pmof: By Lemma 10.27, foreacht E T, themexistsa E G forwhich@(t+q) = a + p . This element of G is unique. by Lemma 6.28. We define f : T + G by stating that @t q) = f (r) p. It 1s easy to check that f is an injective homomnphism, and so f is also an injective homomorphism (by Exercise 3.4.1 and Corollary 4.22). Since )(t q) = f (1) + p forevery r E T, it follows by continuity that@(x+q) = 7 ~ x 0 p forevery x E BT. Let 'I be the set guaranteed by Lemma 10.2%and let y E Y*.-Fw e v y x E T*, w _ e h a v e ~ ( x ) + @ ( y + q ) = 4 ( x + y + q ) = f ( x + ~ ) + ~ = f ( x ) f+( y ) + p = t ( x ) )(y q). Now )(y q ) is right cancelable in BG, by Lemma 10.25. So f (XI= @(XI. 0

+

+

+

+

+

+

+

10.3 Homomorphisms from T ' into S

217

C o d q 10.29. Let f denote the mapping d @ d in Theorem 10.28 and let T'

=

f-'[S]. Then T' is a subsemigmup of T forwhich T\T1 isfnite. PRX$ It is immediate that T' is a subsemigroup of T . _By T h m m 10.28, T ( x ) = $ ( x ) E S* for every x E T'. Thus, if x E T'. then S E f ( x ) so T' = f - ' [ S ] E x by Lemma 3.30. Consequently T\Tr is finite. 0

Thmrem 10.30. Let S and T A countably infinite, cornmufafive, and cancellative semigmups andassume that T has an idemMfyand that $ : T* -+ S' is a continuous injective homomorphism. There is a subsemigmup T' of T for which T\Tf isfinite, and an injective homomorphism h : T' + S for which h ( x ) = $ ( x )for every x E T'. Proof. Let f denote the mapping defined in Theorem 10.28. We put T' = f -'IS] and let h = fir.Our claim then follows from Corollary 10.29 and Theorem 10.28. o The necessity of introducing T' inTheorem 10.30. is illushated by choosing T = o and S = N. Then T*and S* are isomorphic, but there are no injective homomorphisms from T to S. In this example. T' = T\[O). In the case in which S is a group we have S = G and h e k e T' = fm'[S] .= f e l [ G ]= T .

Theorem 1031. Let S and T be countably infinite, cornmutafive, and camellarive semigmups and assume thar T has an identify. Suppose that 0 : pT + pS is a continuous injective ho~ornotphism Then there is an injective homomorphism f : T -r Sforwhich0 = f . Pmof Let $ = O I p . We observe that $ [ T W g ] S,because x E T * implies that $ ( x ) is not isolated in pS and hence that $ ( x ) 6 S. B_yTheorem 10.28 there is an injective homomorphism f : T + G such that @ ( x )= f ( x )for evety x E T*. Let Y denote the set guaranteed by Lemma 10.24 and let y E Y*. For each t E T, we have@@ y q ) = 0-W 0 ( y + q ) = @(r) $ ( y q ) and also 0 0 y + q ) = 70 Y + = f 0 )+ f (Y + q ) = f (0 + $(Y + q ) . BYLemma 10.25. $(Y q ) is right cancelable in pG. So f ( t ) = @(t).Now f ( t ) E G and 6 0 ) E pS so f ( t ) e S.

+

+ +

+

+

+

+

+

Theorem 1032. Lct S and T be countably infinite, commutative, and cancellative semigroups andassume that T has an identity. Then S* doesnot contain any topological and algebraic copies of pT.

Pmof This is an immediate consequence of Theorem 10.31.

0

We donot know theanswer tothefoIlowingquestion. (Although, ofcourse, algebraic copies of Z are plentiful in N*. as are topological ones.)

Question 1033. Does N* confain an algebraic and ropological copy of Z? E x e ~ . l O 3 . 1 .Show that the only topological and algebraic copies of N* in N* are the sets of the form (kN)*,where k E W.

I

10 Ho-

218

Exerdse 103.2. Show that the only topological and algebraic copies of Z* in Z ' the set9 of the fonn (kZ)*, where k E Z\(O).

a

Exercise 1033. Show that the only topological and algebraic copies of (N*. +) ia (N*, .)arc those induced by mappings of the fonn n H k" from N to itself, whae k E N\{lJ.

Exercise 103.4. Let S be an a t e d i m t e canccllative semigroup. Show that S* contains an algebraic and topological copy of N. (flmt: By Theormi 8.10. there is a. element p E S* which is right cancelable in BS. Define p, for each n E N by stating that pl = p and p . + ~= p. p for every n E W. Then (p. : n 6 NJ is an algebraic and topological copy of N.)

+

Exerrise 10.35. Show that N* contains an algebraic and topological copy of o. (Hint: Let p be an idempotent in N* for which FS((3")zl)E p and let q E ~ ( ( ( 2 . 3 ": n E N)) n N*. Definex. for every n E o by stating that y, = p, X I = p q p and &+I = x. + X I for every n E N. Then (x. : n E w ) is an algebraic and topological copy of o . To see that (x, :n E o ) is discrete. consider the alterations of 1's and 2's in the ternary expansion of integen. Forexample, (&F, 3' C,,F~2 . 3 ' XrEF33[ : max Fl < min F2 and max F2 < min F,) E X I . )

+ +

+

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10.4 Isomorphisms Defined on F'rincipal Left and Right Ideals In this seftion, we shall discuss continuous isomorphisms between principal left ideals or principal right ideals defined by idempotents. Throughout section, S and T will denote countably inhnite discrete semigroups. which are commutative and canceUative. G will denote the group generated by S and H will denote the group generated by T.

Lemma 1034. Suppose that p and q an nanminimal idempotem in S and T* nspective3.. and that f :q B T q + p BS p is a continuous isomorphism Ir!

+ +

+ +

S ' = ( ~ E G : ~ + ~ E ~ ~ S ) ~ M ~ T ' = ( ~ E 7ken@(q)=pand H : ~ + ~ E B T ) . then is an isomorphism f : T' + S' m h that +(I q) = f (I) p for every I E T'.

+

+ +

+

+ +

PmoJ Since 9 is in the center of q BT q. @(q) is in the center of p BS p. So @(q) = a p for some a E C (by Theorem 6.63). Since @(q) is idempotenk it follows from Lemma 6.28 that a is the identity of G and hence that p = f (9). Foreveryt E T1.t+q = q + ( t + q ) + q isinq+pT+qandisinthecenterofthis semigroup. So g ( r q) is in the center of p ,l?S p. It follows from Theorem 6.63 that *(r q) = s p for some s E G. We note that s E S' and that the element s is unique (by Lemma 6.28). So we can define a mapping f : T' -+ S' such that $0 + q ) = f ( I ) + p for every r E T'.

+

+

+ +

+ +

It is easy to see that f is injective and a homomorphism. To see that f is smjectivc, 1 e t s ~ S ' S. m ~ s + p i s h t h e c m t a o f p + B S f p.s+p=@(x)forsomexinthe center of q pT q. We must have x = r q for some r E T' by Th6.63, and a this implies that f ( t ) = s.

+ +

+

em^ 1035. L a pond q be nmininaal identpofentc in S and T* rrspcaivcly. Lrt Sf = (a E G :a+p E BS)andlerT' = ( b E H : b+q E BT). I f @ :pT+q + pS+p is a continuous isomorphism, then $(q) is o ~nminimalidempotent, BS + p = BS + @(q),and there is an isomorphism f : T' -+ S'for which @(t q ) = f (t) + $(q) for every t E T'.

+

+

+

+

p = BS $(q). On the one hand. $ ( q ) E BS p and so BS + $ ( q ) E j3S + p. Oa the other hand @-'(PI= q and so p = $($-'(p)+q) = p+@(q) andthenforeBS+p E BS+ $(q). It follows that @(q)is a nonminimal idempotent (byTheorem 1.59). Wealsoobse~ethat(aE G : a + p E BS) = (a E G :a+$r(q) E BS). Tosee.tor example,that[a E G : a+p E 0s) g (a E G : a+$(q) E BSl,leta E Gandassume

p m J We first show that pS

+

thata+p~BS.Thena+@(q)=a+$(q)+p=$(q)+a+pEpS+BSEBS. The result ~ n follows v from Lemma 10.34 and the obsuvation that $ defines a

+ +

wnrinuous isomorphism from q + pT + q onto $(q) BS $(q). (To see that $(q)+BS+$(q) E $(q+BT+q),letr E $(q)+BS+@(q). T h e w = r+$(q)so r = $(v)forsornev E pT+q. T h e n q f v E q+BT+qand$(q+v) = $(q)+r = r.)

a We omit the proof o f the following lemma, since it is essentially similar to the preceding proof.

Lemma 10.36 Let p andq be ~ M l i n i m aidcmpotents l in S* and T* respectively. Let St = (a E G : a+p E pS)andletT' = ( b E H :b+q E BT). I f @ :q+BT -+ p+BS

+

is a continuous isomorphism then @(q)is a nonminimal idempotent, p pS = $(q) + gS, and them is an isomorphism f :T' -r S' such that $0 q ) = f ( t ) @(q)for every t E T'.

+

+

Lemma 1037. Suppose that y E G9\K(/3G). Lcr f : H -+ G be an isomorphism Then them is an element r E T' for which j ( x ) + y is right cancelable in BG. :BH -+ BG is an isomorphism and so & K _ c ~ H ) ]= K(pG). By Lemma 6.65. T n K(BH) # 0. and so f[Tln K(f3G) = f [T flK(BH)] f 0. It follows from 'lheonm 6.56 (with S = G ) that there is an element v E-G* such that f [TI E u and v y is right cancelable in BG. Since v E TIT].v = f ( x ) for some x E T*.

Prwf: We note that

+

Theorem 1038. Suppose that G and H are counrabk discrete commutative gmups and that L and M am principal lefi ideals in BG and pH respectively, defined by nonminimal idempotents. I f @ : M -t 4 is a continwus isomorphism there is an isomorphism f : H + G for which $ = f i ~ .

no

10 ~om~morphism~

PPZR$ LetL = pG+pandM = BH+q,whempandq arenonminimalidcmpolm*lin

G* d H * respectively. By applyingLemma 10.35 with G = S = S a n d H = T = TI, we see that $(q) is nonminimal and t h m is an isomorphism f : H + G such tha~$(t + q) = f(t) $(q) for every t E H. By continuity, this implies that $(x+q) = T(x) +$(q)forweryx E BE. Forx E BH,wehave$(q+x+q) = f(q+_x)+!b(q) = f"(a)+T~)+$(q)and also$(q+x+q) = $(q)+*(x+q) = $(q)+f (x)+$(q). Sof (q)+f(x)+$(q) = $(q) + f i x ) + $(q) for every x E OH. By Lemma 10.37. we can choose x E BH such that f ( x p $(q) is right cancelable in BG. Tbus $(q) = f (q). This implies that $(x q) = f (X q) for every x E pH. 0

-

+

+

+

Theorem 1039. Suppose that L and M a n principal IejY ideals in ON defined by nunminimal idempotents. If$ : M + L is a continuous Lromorphisrn then M = L and $ is the identie map. Pmof Let p and q be nonminimal idempotents in BN for which L = BN + p and M = p N + q . TbenL=pZ+pandM=pZ+q.since,givenr E B Z , r + p = r + p p E BW p because N* is a left ideal in BZ (by Exercise 4.3.5). So L and M are principal left ideals in BZ defined by nonminimal idempotents in Z*. It follows fium Theorem 10.38 that there is an isomorphism f : Z + Z for which $ = f i ~ Now there are precisely two isomorphisms from Z to itself: the identity map r H t and the map r H -I. We can ~ l e _ o uthe t possibility that f is the second of these maps. because this would imply that f (q) $ BN. Hence f is the identity map and so is Jr.

+

+

Theorem 10.40. Let G and H be countable commutativc groups and let L and M be principal right idealr in G* and H* respectively, dqined by nonminimal idempotents. Suppose that there is a continuous Gomorphism $ : M + L. Then there is an isomorphism f : H -+ G for which f [MI = L. Pmof Let p and q be nonminimal idempotents in G* and H* respectively for which L s p + BG and M = q + p H . By applying Lemma 10.36 with G = S = S' and H = T = T'. we see that $(q) is nonminimal and them is an isomorphism : H + G suchthat$(t+q) = f (t)+$(q). Itfollows. bycontin$y,that$(x+q) = f (x)+$(q) for every x E M. Putting x = q shows that M q ) = f (q) $(q). On the other haid, we also have, for every x E H, that $(q x qj = T(q) .h~+$(q)_and$(q+x+q) = $(qJ+W_+x+q) = $(q)+f(q)+f(x)+$(q). for wery * E BH. We can So f (4) f (x) + $(q) = y ( q ) f (4) f ( x )) 4 $ choose x E H* for which f (x) $(q) is right cancelable in BG (by Lemma 10.37) and so f"(q) = S ( q ) R q ) . Itfollowseasily that $(q)+B_G = f"(q)+flG. We alsoknow that $(q)+BG = L 0 (by Lemma 10.36). So = f (q) BG = *(q) BG = L.

t

+

+

+

+

+

+

T[M]

+

+ +

+

+

+

Tbe preceding results in this section tell us nothing about principal left or right ideals defined by minimal idempotents.

.

If S is any discrete semigroup, we know that any Wo minimal left ideals in BS are both homeomorphic and isomorphic by Theorems 1.64 and 2.11. However. the maps which usually define homeomorphisms between minimal let3 ideals are different from those which define isomorphisms. It may be the case that one minimal left ideal cannot be mapped onto another by a continuous isomorphism. It is tantalizing that we do not know the answer to either of the following questions.

Question 10.41. Are there two minimal le~?ideals in BRI with the property that one cannot be mapped onto the other by a continuous homomorphism? Question 10.42. Are there w o distinct minimal Iej3 ideals in PRI with the property that one can be mapped onto the other by a continuous homomorphism?

Notes Most oftheresult of Section 10.1 arc from [331, resultsofcollaboration withV. Bergelson and B. Kra Theorem 10.8 is due to J. Baker and P.Milnes in [lo]. Theorem 10.18 is from [228] and answers a question of E. van Douwen in [XO]. Mosl of the results in Section 10.3 were proved in collaboration with A. Maleki in [180].

'Chapter 1I

The Rudin-Keisler Order

Recall that if S is a discrete space. then the Rudin-Keislcr order SRKon BS is dehned bystatingeafforanyp,q ~ B S . p i ~ ~ q i f a n d ~ n l y i f t h e r e i s a f u n cft i:oSn+ S for which f (q) = p. Alw dthat if p. q E BS, we write p C R K q if p SRK9 and ~ F R ~K ~ ~ n d ~ ~ ~ t e p a ~ ~ 9 i P.f ~ 5 ~ ~ 4 a n d q 5 As a consequence of Thkrem 8.17 one notes that while the relation SRKis clearly reflexive and transitive. it is not anti-symmetric. The relation SRK does induce an antisymmetric partial order on the set of equivalence classes of the equivalence relation ~ R K .

The RudiiKeisler order has proved to be an important tool in analyzing spaas of ulhahlters. It shows how one ultrafilter can be essentially differcat from another. There are deep and difficult theorems about the Rudin-Keisler order which we shall not attempt to prove in this chapter. Our aim is to indicate some of the connections between the Rudin-Keisler order and the algebraic structure of BS. If S is a discrete semigroup. one would expect that the Rudin-Keisler order on BS would have little relation to the algebra of BS,because any bijective mapping from one subset of S to another induces a Rudin-Keisler equivalence between ulaafilters, and a bijstion may have no respect for algebraic structure. Nevertheless, there are algebraic properties which are related to the RudiiKeisler order. For example, if S is countable and cancellative. an element q E S' is right cancelable in @S if and only if p < R K pq and q C R K pq for every p E S*as we shall see in Theorem 11.8.

Ex-

11.0.1. Let S be a discrete space and let p q E BS if and only if p E S.

€

BS. Show that p 5 q for every

Exercise 11.02. Let S be a countable discrete space and let p E BS. show that (q E BS : q SRKp} has cardinality at most c. Exercise 11.03. Let S be a discrete semigroup and let a E Sand p a p SRKp and pa I R p. KShow that a p pa Z R Kp if S is right cancellative.

%RK

E BS. Show that p if S is left cancellative and that

Exercise 11.0.4. Let S be a countable discrete space and let p be a P-point in S*. If q E S* satisfies q ~ R p. K show that q is also a P-point in S*.

11.1 Conndons with Right Cancelability

11.1 Connections with Right Cancelabiity In this section we shall explore the connections between the relation R K and right cancelability in BS. T h w arise from the fact that both concepts arc related to the topological property of smmg discreteness.

Definition 11.1. Let S be a discretespace and let p, q E BS. The temorpmducf p Q q of p and q is defined by ~Q~=(A~SXS:(S:(~:(S,~)EA)E~]E~].

The reader is asked to show in Exercise 11.1.1 that p- Q q- is an ulaafilter on S x S. I f A S SxS.thenA e p@qifandonlyifAcontainsasetofthefom( ( s ,t ) : s E P and t E Q,), where P E p and Q, E q for each s E P. Note that. if s. r E S, we have s 8 t = (s. t ) where. as usual, we identify the point (s,1) with the principal ulaPfilter it generates. If we choose any bijectio56 : S x S -+ S. can rrgard pOq as being an ultra6lur on S by identifying it with # ( p @ q), where # : p(S x S ) + BS is the continuous extension of 6. In panicylar. if s. t E S, then s Q t can be identified with the element #(s, t ) of S. In this section, we shall sometimes think of Q as being a binary operation on S, assuming that a bijection 6 has been chosen. Of course, different bijections will result in p @ q being identified with different elements of pS. These will, however, all be Rudin-Keisler equivalent. So it does not matter which one we choose for sNdying properties of the relation S R K .

we

Lemma 113. Let S be a discrete space and let p, q E S*. Then p 4

p Q q and

Proof: If nl and nz are the projection maps from S x S onto S. it is easy to see that % ( p ~ q ) = ~ B n d l r ? ( ~ @ q ) S= O q .P ~ Q~KQ q a n d qSRK p Q q . Now tbere is no member of p @ q on which nl or n2 is injective, and so p +RK p @ q and q FRKp Q q by Theorem 8.17. 0 Lemma 113. Let S be a discrete space and let p, q 6 BS. Then

p Q q = lim lim(s,t) s-.pt-q

where s and t denote elements of S ond the limits are taken to be in B(S x S). Pmofi This is Exercise 11.1.2. Recall that in Section 4.1. we extended an arbitrary binary operation on S to BS. Lrmma 11.4. Let S be a discrete spoce and let * be a binary opemtion defined on S. L e r p , q , x , y ~ p S . I f x ~ ~ ~ p a n d ~ S ~ ~ q , t hP Q e qn . x * y 5 ~ ~

!

1

! i

224

11 Tbe Rudin-Kcislcr Order

~ m o f k: t f : s + ~ a n d g : ~~+b e f n n ~ t i m s f o t w h i c h f ( ~ ) = x a n d ~ ( q ) ~ Wedefine h : S x S +. S by h(s, t ) = f ( s ) * g ( t ) . Thcn x ( p O q ) = s-pt-q lim lim h(s. t ) =

f(s) * 8(t) =

fcP) * z ( q ) = x * y.

o

ComUary 115. Let S be a discrete space und let * be a binary opemtion defttcd on S. For every p.q E PS, we hav~p * q ~ R pK 8 q. Funhnnore, i f h : S x S + S is defined by h(s, I ) = s * t, then h ( p O q ) = p * q.

-2

.:

'

.

.,

Pmof: The proof is the same as the proof of Lemma 11.4. with f and g taken to be the identity maps. a Corollary 11.5 shows that for any p, q E BS, p C3 q is-a I R Kuppe bound of the set of alI elements of the form p * q , where * denotes a binary operation on S. We now investigate the possibility that p * q * R K p 63 q . When we say that an arbitrary binary operation t o n S is left cancellative. we mean exactly the same thing as when the operation is assumed to be associative. That is for each s E S. the map t H s * I is injective.

Theorem 11.6. Lets beadiscretespaceandlet* bea lefrcollcellativeb i ~ r y o p e m t i o n definedon S. men, for every p, q

E BS,

thefollowing are equivalent

(1) P * ~ * R K ~ 8 % (2) There is a set D E p for which D * q is strongly discrete and s whenever s und t are distinct members 0f.D.

*q

# t *q

P m f : ( I ) implies (2). Ifp*q *=a pQq, there is aser A e p 8 q on which the mapping (s, I ) H s * I from S x S to S is injective (by Corollary 11.5 and Theorem 8.17.) We may supposethat A = U,,D((sl x E,). where D E pand E, E q foreachs E D. Then for each s E S,ctgs(s 8 El)is a neighborhood of s * q in pS. If s and s' are distinct elements of D. s * E, and s' * E,, are disjoint, and so ctps(s * Es) and cEbs(sl * E,,) are disjoint. (2) implies (1). Suppose that for some D E p. D * q is strongly discrete and s * q # t * q whenever s and t are distinct members of D. Then, for each s E D, there is a neighborhood Usof s * q in BS such that Usn Us,= 0 whenever s and s' are distinct elements of D. For each s E D. we can choose E, E q for which s 8 E, c U,. I f A = U,,,([sl x E,). then A E p 8 q and, since s * E, IIs' * E,! = 0 when s # s' and s * r # s * I' when t # t', the mapping (s. t ) t+ s * t is injective on A. So P * ~ * R K~ 8 4 0 Corollary 11.7. Let S be a countable lefr cmcellarive discrete semigroup and let q be a right cance&ble element of IS. Then pq % R K p 8 q f o r every p E BS.

Proof: B y Theorem 8.1 1. Sq is strongly discrete in BS and sq # t q whene s # 1 in S. and so our claim follows from Theorem 11.6. a

''

-.

11.1 Conmctions wit3 Right Cancelability

..

'22s

We are now able to establishthe equivalenceof statements( I ) , (4)and (5) of Theorem 8.18 under weaker hypotheses.

meorem 11.8. Suppose that S is a countable c~cellarivediscrete semigroup Md that q E S*. Then the following statements are equivalent.

Pmof: (1) implies (2). If q is right cancelable in @S,it follows fromLemma 11.2 and Corollary 11.7 that p
PmoJ We may suppose that S is a group. because there are groups of cardinality K. (The direct sum of K copies of '& provides an example.) It then follows from Theorern6.30 that there is asubset P of S' such that [PI = Z2<, up # bp when a # b in S and Sp is strongly discrete in pS for every p E P. and (fiS)p fl (fiS)q = 0 whenever p and q are distinct elements of P. ByLemma11.2andTheorem11.6.x < R K xpforevetyxf S a n d e v e s y p e P. Since (fiS)p n (fiS)q = 0, x p xq if p and q are distinct elements of P.

+

Theorem 11.10. Let S be an injnire discme cancellative semigraup. There is no element of S whose Rudin-Keisler predecessors form a subsemigmup of fiS. Pmof:htpES*and~etL={~€fiS:x5~~p).htK=llp~~and~etREphave cardinality K. We can find a subset V of S with cardinality K such that aq # bq when a f b in R and Rq is strongly discrete in @Sfor every uniform ultrafilter q on V by Theorem 6.30. Since R can be mapped bijectively onto V. we can choose a uniform ultrafilter q on V for which p 'JRR q. By Theorem 11.6 and Lemma 11.2. we have p < R K pq. Since p, q E L and pq 4 L. L is not a subsemigmup of fiS. 0

Theorem 11.11. Let S be an inj'inite discme I@ cancellafive semigmup. Let D 2 S and let q E S*. Suppose thar Dq is strongly discrete in fiS and thar sq # t q whenever s andt are distinct elementsof D. Suppose also thatx. y E S*, p € 5 n S*, x . ~ R Kp, and Y ~ R 4.K Then XY ~ R pq. K

Corollnry 11.1%Let S be a discme countable lrfr cancellatiw semigmup and lei q be a nght cancelable element of&% Then.for every x, p E S*. x SRKp q l i e s t h a ~ xq ~ R ~K 9 . Pmof: By Thmrem 8.1 l(1). lheorcm 11.11 applies with y replaced by q.

X

0

Theorem 11.13. Let S be a countable cmcellative semigroup. Let q be a right cancelable element of fiS and let p E fiS sdsfy p SRKq. Ifx. y E S* d x q SRKyp, then x SRKy.

Pro05 Let f : S + ~ b e a h m c t i m f o r w h i c h ~ @ ~ ) =Ifq x ~~. S , t h e n p ~ S a n d ~ R xq K J R K yp =,RK y by Exmisc 11.0.3, s o w may assume that q E S*. Let? = (b E S : f (bp) E Sq). For each b E B, there is a unique c E S for which f (bp) = cq, by Lcmma6.28.So we can define a function g : S + S by putting g(b) = c if b € B, and extending g arbiaarily to S\B. We may suppose that a(y) # x, sox

since othrrwise x ~ R yK as we wish to prove. So there is a set V E y for which g[V] qf x. We choose V g B in the case in which B E y and we chaose V G S\B if B f Y. We now claim that (*) for each X E x and each Y E y, if X forsomea E ~ a n d s o m e z€7.

E S\g[Vl

and Y

S V, then a q = 7(zp)

To see this, suppose instead that_(*) fails and pick X E x and Y e y witnessing this failure. Now xq E c t ( X q ) f ~cC f [Yp] so by Theorem 3.40 and the assumption that (*) fails. we must have wq = f (bp) for some w E j7 and some b E Y. Then wq ~ R bp. K bp -RK p by Exercise 11.0.3, and p SRK q by assumption. If w E S*. then by Theorem 11.8, q < R K wq, a contradiction. Thus w E S and s o b E B and w = g(b). Also b E Y S V so that w E g[V] n X, conwdicting the fact that g[V] n X = O. Thus (*) is established. ~ e At = [a E s : aq E A ( B S ) ~ I I .~f A f x, then leaing x = S\(A u ~ [ v I ) dso A E x. For each a E A, let and Y = V one obtains contradiction to (*). C. = [z E BS : aq = f (zp)). Then C. = (f o p,)-'[[aq]] so C. is closed. We claim that for each a E A, C. flcf (Ubs,,,lnl Cb) = 0. Indeed, by Theorem 8.1 l.aq g cl(bq : b E A\(a)) so pick aneighborhood W of aq which misses (bq :b E ~ \ ( a ) ] . Cb = 0. Given z E C., pick D E z such that 7[&] G W. Then En UbeA,lal Thus one can pick for each o E A a clopen set V, such that C. G V. and V, fl (UbEA,lal Cb) = 0. Let A be sequentially ordered by c.For the first element a0 of A. let U, = V., For later elements. let U. = V\. Ubso Vb. Then for each a E A, U, is clopen. Ca c Ua. and Uo n Ub = 0 if. # b. Define h : 5 + S by h(s) = a if s E U,, defining h arbiaarily on S\UaEAU. We claim that h(y) = x. Suppose~n~ead that h(y) # x and pick X E x such that h(y) f and pick Y E y such that h[Y] r l X = 0. We may assume that X c_ S\g[Vl

a

-

x

11.1 Connsrio~with Right Cancelability

227

Corollary 11.14. k t S be a countable cancellarive semignxq and k t p E BS. The following statements a n equivalent. (I) (2) (3) (4)

p is r i g h t c ~ ~ e l a binl eBS. Forweryx, y E BS, xp R K yp implies thatx SRK y. Foreveryx. y E BS, xp -nr yp impliesthatx ~ R y.K For every x E~S, p
Pmof If p E S, fhen (I) is true by Lemma 8.1. (2) and (3) arc m e by Exercise 11.0.3. 11.0.1 and 11.0.3. Thus we may assume that p E S. and (4) is true by E& The equivalence of (1) and (4) follows from Theorem 11.8. To see that (I) implies (2),letx.y ~ p S w i t h x p yp. S ~ Ifx.y ~ €S*,thenx SRK ybyThearem11.13and if x E S, them x ~ R yKby Exercise 11.0.1. Thus it puffices to show that one cannot have x E S and y E S. Indeed in this case p ~ R xp K by Theorem 11.8 and yp -RK p by Exercise 11.0.3 so one would conclude that p < R Kp. It is hivial that (2) implies (3). To see that (3) implies (1). assume that (3) holds and suppose that p is not right cancelable in BS. Pickx y in BS such that xp = yp and p i c k ~ ~ x a en yds~u c h t h a t x n ~= ~ . ~ o w x ~ ~ ~ a n d y p ~ ~ s o j t f o l l o w s hmTheorern3.40thatsp=zp.whms ~ ~ a n EdL zo r s G ~ a n d X.z ~~y ( 3 ) . this implies that s *RK z and hence. by Exercise 11.0.1, that z E S. This contradicts 0 Lemma 6.28. because s # z.

+

Tbeore~i~ 11.15. Let S be a countable discrete space and let * be a left cancellnrive binary operation dejaed on S. Suppose that p, q E S and that A € p has thepmpeny that A *q isdiscme in BS and that a *q # b *q whenevera and b are distinct elements of A. Then p * q ~ R qK* p implies that p and q are Rudin-Keister compamble.

*

We suppose that p * q SRK q p, and note that we then have p O q S R K q + p by Theorem 11.6 and the fact that "discnte" and ''strongly discrete" are the same in this case. La n, and nz denote the projection maps of S x S onto S. Then 6(x @ q) = x

P&

and~(xOq)=qforeveryx€~S. Let f : S -+ S x S be a h c t i o n for which 7(q * p)_= p O q. If P E p and Q E q. p @ q belongs toeach of the sets cC(P O q ) and cC(f [Q c pl). It follows from

Theorem 3.40 that either (i) ~ ( b * p ) = x @ q f o r s o m e b~~a n d s o m e x ~ P o r (ii) ~ @ q = f ( z * p ) f o r s o m e a E ~ a n d s o m e z€ 3 .

70

Now if (i) holds for some P and Q, then 6 o Ab is a continuous extension of the function s H r2(f (b * s)) taking p to q so thac q 5.q~ p. We may therefore suppose that (ii) holds for every P E p and every Q E q.

228

11 l ix Rudin-Keister Orda

Let B = ( b E S : % ( j ( b t p)) E SJ and de6ne.g : S -+ S by stating g(b) = 6 ( 7 ( b * p)) if b E B and defining g a r b i W y on S\B. We claim that z ( q ) = p, so suppose instead that z ( q ) # p. Rck P E p and Q E q e c h nB = 0. By (ii) pick %me a E P and some z E Q such that a Q q = f ( z * P), Thus a = % (a Q q ) = f~ (f( z * p)) and a is imlated so them is some R E z 6 t h , ~ [ f i ~ * P ] ] = ( a ] . P i c k b ~ R n Q . ~ h e n % ( ~ ( b * p ) ) = a s o b ~ ~ a n d'"t g(b) = a, a contradiction. 0

z[Q]

Corollary 11.16. Let S be a counrablc discretespaceandlet p, q q Q p, then p and q are Rudin-Keisler comparable.

E

S*. I f p a q

snK

P m f The hypotheses of T h e o m 11.15 are clearly satisfied if Q is regarded as a binary operation on S via a fixed bijection between S x S md S. 0 Corolhrg 11.17. Let S be a discrete countable cancellative semigroup and let q be a right cancelable element of S*. I f p E S' and pq C R K qp, then p and q a n Rudin-Keisler comparable.

P m f Theorem 11.15 applies by Theorem 8.1 l(1).

0

If S is an infinite discrete semigroup, we know that 6.9 is rarely commutative. In the following corollary. we show something stronger than nonsommutativity in the case in which S is countable and cancellative. In this case, there are elements p and q of S* for which pq and qp arc not Rudin-Keisler comparable. Corollary 11.18. Let S be a countable infinite discrete cancellative semigroup. There are elements p, q E S*for which pq and qp are not Rudin-Keisler comparable.

P m f By Theorem 6.36, there are elements p and q of S* which are not Rudin-Kcislcr comparable. We may suppose that p and q are right cancelable in BS, because there is an infinitesubset T of S with the propeqthat every element of T* is right cancelablein pS (by Theorem 8.10). Since Scan be mapped bijectively onto T, p and q are RudinKeisler equivalent to elements in T*. Our claim then follows from Corollary 11.17. 0

Exercise 11.1.1. Let S be a discrete space and let p, q E BS. Show that p Q q E B(S x S).

Exercise 11.12. Rove Lemma 11.3. Exercise 11.13. We cannot normally use the most obvious function to establish a relation of the form p Q q CRK q O p. Let f- : N x N -+ N x N be defined by f (s. t ) = ( I . s). Show that, for any P. q E N*, f (q Q P ) # P Q q. Exercise 11.1.4. Let S be an infinite cancellative semigroup. Show that there is no element p in S* for which ( q € BS : q ~ R pKJ is a subsemigroup of BS. (Hint: Consider the pmof of Theorem 11.10.)

.~~

11.2 Connections with Left Cancelability in N*

112 Connections with Left Cancelabiity in N* Because BS is a right topological semigmup. left cancelability in BS is more difficult to handle than right cancelability. As we have seen, it is easy to derive connections between h e Rudin-Keisler order and right cancelability. However, whether the analogousresults hold for left cancelability remains a difficult open question. even in BN. For example, we do not know whetherq C R K p + q forevery q E N', whenever p is aleft cancelable element of N*. However. we shall see that there is a rich set of left cancelable elements p in N* for which this statement is true. Recall that, for each k E N, the natural homomorphism hk : Z + 4 has a continuous extension he : BZ -t Zk which is also a homomorphism-(by Theorem 4.8). If x . y E BN and k E N, we may writex = y (mod k) if hk(x) = hk(y). Theorem 11.19. Suppose that p E W and that there exists A E p with the pmpeny that x = p (mod k) for every x E A* and every k E N. m e n p < R K p + q and q < R K p+qforeveryq E PI-. Pmo$ Letq E N*. Weobse~ethatthercisatmostoneaE Aforwhicha+q E A*+q. q and b q E A* q. Then To see this. suppose that a , b E A satisfy a q E A' hk(a) = hk(b) for every k E N and so a = b. We may replace A by A\(a), if such an element a exists. So we may suppose that A q is disjoint from A* + q. We claim that a q # b q whenever a # b in A and A q is strongly discrete in BN. Otherwise we should have b q = x q for some b E A and some x E A\{b), by Theorem 8.11. Now this cannot hold if x E A*; neither can it hold if x E A (by Lemma 6.28). Thus our theorem follows from Tbeorem 11.6 and Lemma 11.2. . ,

+

+

+

+

+

+

+

+

+

+

-

-

Comment 1120. The set ofultrafilters p in N* which satisfy the hypolhesesof Theorem 11.19 contains a dense open subset of W*. To see this. let U be any non-empty clopen q (mod k)). subset of N* and let q E U . For each k E N, let Vk = ( x E BN : x Then Vk is a clopen subset of BN. Since q E U r l Vk, this set is a non-empty Gs-subset of N* and therefore has a non-empty interior in N* (by Theorem 3.36). Any ultrafilter p in the interior of U n Vk satisfies the hypotheses of Thwrem 11.19.

n&

nla_,

Theorem 1121. Lct (a.)zl be an infinite increasing sequence in N. Suppose that either ofrhe rwo following conditionr is satirfied: (i) For every n E N. an+!is a multiple of a,,. (ii) For every m, n E N with m # n, a,, is not a multiple ofa,. I f A = { a . : n E N ) a n d p E A * , t h e n p C R K p + q a n d q < R K p+qforevery

q E A*.

nEla.

Pmof: We first consider the case in which q E For each n E M, let so that B. E q. B. = If condition (i) is satisfied, we define f : N -+ N by stating that

n;:

f (n) = max(a, : a,ln)

if amlnfor m e m,

230

1 I Tbe Rudin-Kdskr otdu

+

dcfining f arbitrarily otherwise. For mry n E N and every b E B., f (a* b) a,. Sou,,,+B. anda.+B,medisjointifm #n. Sincea,+B, Ea,+q,thesuA+p is strongly discrete in BPI and our claim follows from Theorem 11.6 and Lmuna 1 1 2 If mndition (ii) is sahsfied, we define g :N + N by stating that * g(n) = min[a,,, :a, 1 n} if a,

1 n for some m,

delininggarbiharilyotkwk. Foreveryn € N a n d e v e r y b € B . . g ( a . + b ) = a , , . Sowecanagaindeducethat A + q isdiscreteandthatTheomn 11.6andLnnma112 apply. Now let q be an arbiuaty dement of W. For each n E N, we can choose b. E N satisfyingb.+q = O ( m o d k ) f o r m r y k E {1,2, ...,n]. Tosathis,letn E Nbe given. T h e n ( c ~ N : c ~ q ( m o d k ) f o r m r y kE (1,2,.:.,n)] ~ q s o p i c k c i n t h i s set Pickm ENsuchthatn!m w candletb. =n!m-c. LetB={b. : n E N ) a n d letr E B*.Thenq+r ~ n ~ , ~ s i n c e f o r e a c h k , ( b : b + q r a ~ ( r n o d k ) ) ~ r . By what we have already proved. with q +r in place of q, we know that a +q +r # b + q + r when a and b m distinct members of A (by Lemma 6.28) and A q + r is discrete in DM. This is equivalent to stating that a q r # x q r if a 6 A and x E X\(a). by ~heormn8.1 1. This implies that a q # x q for every a E A and every x E X\(a) and hence that A q is strongly discrete. So Th11.6 and Lemma 11.2 apply once again. 0

+ +

+

+

+

+ +

+

Comment 1122. By Ramsey's Theorem Clbnwn 5.6). evey infinite sequence in N contains an infinite subsequence satisfying the hypotheses of Theorem 11.21. So the set of ulaafiltm p E PI* to which this theorem applies also contains a dem open subw of N*.

Thearun 1133. k t S be a discrere countable cancellarive semigroup and let p be a P-point in S. men p CRK pq and q
ies 8.36.8.37, and 8.38 and Theorem 8.39. Qncstion 1135. I f p is a I@ cancelable ulnafiter in No,mun ir be hue that p C R K p + q a n d q e ~ ~ p + q f o r ~ e rEM*? yq

231

113 FortherC~ooswitb~heAlgebnofBS

113 Further Con~~ections with the Algebra of BS we remind the rrada that a family F of subsets of a given set is said to be almost disjoint if the intcneaioo of any two distinct members of 9 is finite.

~mmo 1136. Let A be a subset of W. IfIAl successor in H.

c, the elements ofA have a common

Prwf: We index A as (p,),,nc, allowing repetitions if IAI < c. Let (Ex),en be an almost disjoint family of infinite subsets of {2" : n E N). (Such a family exists by Theorem 3.56.) For each x E R we choose q, E E n N* such that q, =RK p,. For each F E 3 where Pf(W) denotes the set of non-empty finite subsetsof R,we put sF = &F qz, with the terms in the sum occurring in the order of inmasing indices. We order Pj(B) by set inclusion and choose q to any limit point of the net (sp)p.pfCR) in BN, noting that q E H. Recallthatsupp(n)istheb'marysappmofn. Forcachx E lR,wedefine f.:N + N by stating that

m),

fx(n) = min(E, n (2"' :m E supp(n)))

if E,

n (2"' :m E supp(n)J # 0

and defining f,(n) d i e y otherwise. We shall show that f, (q) = q, for each x E W.We suppose that this equation does not hold. Then we can chB E q, such that f,-'[B] (e q. and we can choose Q E such that f,-'[B] !lQ = 0. there exists F E Pjm) such that Since q is a limit point of the net ~ of Nsatisfyig x E F and Q E SF. Wecanchmse adisjoint family ( B y ) y Eofsubsets B , G E , n B a n d . f o r e v e r y y ~ F.By G E y , B y E q y . U L d B y n E x = 0 i f y # x . Let M denote the set of all integers m of the form m = E y c ~ , n ywhen . n,, E By for each y E F. We obsmve that this expression for m is unique. smce each n, is a power of 2, and that f,(m) = n, E B. Suppose that F is arranged in increasing order. By Exercise 4.1.6. M E XteF qy = SF. Thus we can choose n E M n Q. Since fI(n) € B. it follows that n € f; [B] n Q. mntradicting our assumption that this set is empty. So f,(q) = q, and we have 0 PI ~ R G K ~ R qK for every x E W. ComUay 1127. Let A be a subset of N* with IAl 5 r The elements of A have a common SRK-successorin my givm minimal lrft ideal of PN,and they also have a common 5.x-successor in any givm minimal right ideal of pN. Furthermore, there is a I& ideal L of pM and a right ideal R of pN such that x < R Ky for every x E A and everyy~LUR. P m f By Theorem 11.26, the elements of A have a common S R K - S U C C ~ S Sq~ ~in BN. We may suppose that x
+

+

232

11 The Rndin-Kcisler Order

lhcorem 11.21. We canchoose u to lie in any given left ideal or in any given right i of BN. LetL = BN+pand R =p+BN. Then* < R K q -RK P ~ R YKf a e v a y x E A andeveryy E L U R . Corollary 11.m.Every minimal lefl idealof BN c o n t a h an increasing snr-ehain of onirr rype cf, and so does every minimal righr ideal of BN.

Pro@ Chains of this kind can be conskucted by an obvious transfinite induction. usiw Corollary 11.27 and Theorem 11.9. D

Theorem 11.29. There are at nmst c elmvnrs qfV whose ~RK-successors form subsemigroup of SN.

0

Pmof Suppose that p 6 W* has the property that its ~RK-successors from a subsemigroup of BN. We shall show that q E K ( B W implies that p S R Kq. To see this, let L be the minimal left ideal of @ N containing q. By Corollary 1127, there is an element u E L for which p S R ~u. There is also a minimal left ideal M of BN such that p SRK v forevery u E hi. So p ~ R yK for every y E M + u since the successors of p form a semigroup. However. M u is a left ideal contained in L and q. thus M u = L. Consequently p SRK The result now follows from the fact that any given element of N' can have at most c ~RK-predecessors (by Exercise 11.0.2). 0

+

+

11.4 The Rudin-Frohi Order There are other useful topological order relations on spaces of ulwfiltm. one of chew. being the Rudin-Frolik order.

DeGnition 11.30. Let S be a countable dismte space. The Rdin-Fmlfi order E on BS is defined by stating that, if p, q E BS. p L-q if there is an injective function f : S + j3.S for which f [Sl is discrete in BS and f ( p ) = q . Ifp,q€flS,wemaywritepCqifpEqandqgp.

It is immediate that the relation E is reflexive, and we shall see in the next lemma that it is transitive. It is not however, anti-symmetric. Lemma 11.31. Let S be a countable discrete space. The xlation

c on BS is transitive.

Proof Let g : S -+ B S be an injective function for which g[S] is discrete in flS, and let X be a countable discrete subset of BS. Then 3x1 is also discrete. because it follows from Exercise 11.4.1 that defines a homeomorphism from BS onto clgs g [ S ] . Suppose now that p, 9 , r E BS and that p q and g & r. So there injective functions f and g from S to BS for which f [Sl and g [ S ] are discrete, f (p) = q.

r

are

a A

P 11.4 Thc Rudh-Fmlik Order

and z(q) = r. Then

ao f

is injective by Exmise 11.4.1. a f [s]] is discrete, and

~ T f ) ( p=) z('(r(p)) = r. So P p

c r.

We now obsewe that the assertion that p E q is stronger than the assertion that 4.

5RK

Lemma 11.32. k t S be a countable discrete space and suppose that p, q E BS satisfy p c 9 . T ~ ~ P q. ~ R K

Pmof Thii is an immediate consequence of Lemma 6.37 and the fact that this is a purely set theoretic statement. 0

We now see that there are elements 9 E S* which are not right canalable, but have the property that p C pq for every p E S*. We remind the reader that an idempotent q E S* is strongly right maximal if the equation xq = q has the unique solution x = q in S*. (And strongly right maximal idempotents exist by Theorem 9.10.) Lemma 1133. Let G be a discrete countabk gmup and let q be a right maximal idempotent in G*. Ln C = l x E G* : xq = 9). Then, for every p E BG, dther p E (BG)C ar then exists P E p such that Pq is discrete in BG.

Pmof: Suppose that p $ (BG)C. Since C is finite (by Theorem9.4). (J3G)Ciscompact. So we can choose P E p for which P n (J3G)C = 0. We claim that Pq is discrete. If we assume the contrary, then a9 E c!((P\(a])q) for some a E P. This implies that a9 = xq for some x E F\i\(a). Then a-'x E C so x E aC. conbadiciing our choice of P. Theorem 11.34. Let G be a discrete countable gmup and let q be a strongly right maximal idempotenr in G*. Then p G pq for every p E G'. In fact, for every p E G*. either p C pq andq
Pmof: Suppose that p E G* and that p # pq. Let C = l x E G* : x q = q ) . Then C = [ q )and so we are assuming that p $ (BG)C. because p E (BG)q implies that p = pq. It follows from Lemma 11.33 that we can choose P E p such that Pq is discrete (and hence, being countable, strongly discrete) in BS. Further aq # bq when a and bare distinct elements of P by Lemma 6.28. S o p c pq andq < R Kpq, by Exercise 11.4.5.Lemma 11.2 andTheorem 11.6.

Exercise 11.4.1. Let S be a countable discrete space and let f : S -c pS be an igectivefunction for which f [S] is discrete in BS. Show that f is also injective, where f : BS -+ BS denotes the continuous extension o f f . (Hint: let x and y be distinct elements of J3S and let X and Y be disjoint subsets of S for which X E x and Y E y. Observe that f [XI n f[Yl = f[Xln ~ [ Y=I 0 and apply Theorem 3.40.) Exercise 11.42. Let S be a countable discrete space and let p E BS. Show that 119 E BS: 9 C PII 5 c.

234

I1 'Ihe RndibKeiila Older

Excrrhc11.43. L e t p , q E P N . I f p c q r a d q ~ p , w e s h p l l s a y t h a t p a n d q m Rudin-FmM equivalent. Show that p and q are Rudin-Fmlik equivdent if and only if they are Rudin-Keisler equivalent. (Hint Use Th8.17 and Lemma 11.32)

Exercise 11.4.4. L* p be a weak P-point in N o . Show that, for any q E No.q 5 ~ R q. K Deduce that p is minimal in N* for Ihc Rudin-FmUk adcr.

implies that p

E x e d 11.4.5. Let S be a discrete countable s e m i p u p and let p, q E S . Suppose that there is a set P E p such that Pq is discrete in /3S and 4q # bq whenever and b are distinct elements of P. Show that p C pq. (Hint Once you have show. that p & pq. apply Lemma 11.2. Theorem 11.6, and Lemma 11.32 to deduce that one cannot have pq E p.)

Exercise lld.6. Let S be a discrete countable semigroup and let q e S* be right cancelable in P S . Show that p C pq for w r y p E S*. (Hint Use Exercise 11.45 and Theorem 8.11.)

Exe&

11.4.7. Show that every element of W has 2' distinct Rudin-FmM successors. (Hint: use Exercise 11.4.6, Theorem 6.30. and Theorem 8.1 1.)

Exerdse 11.4.S. Let S be a discmt countable cancellative wmigmup and let p 6 S. Show that p is right cancelable in P S if and only if-there is an injective function f : S + /3S for which f[Sl is discrete in P S and f (q) = q p for every q S'. (Hint: Use Theorem 8.11. For thc sufficiency consider the proof of Lemma 6.37. For the necessity consider 4.)

Notes S o w of the d b of Sections 11.1 and 11.2 are from [103], obtained in collaboration with S. Garcia-Femim. T h e o m 11.10 contrasts strongly with the situation that holds for the Comfort order of ulhdiltcrs. Given an ultrafilter p on a set S , a space X is said to be p-compeer if and only if whenever f : S -+ X . p- lim f (s) exists in X. One then sss

defims p ic q for p and q in flS if and only if every q-compact space is pcompact. It is easy to see that p S R Kq implies that p Sc q. It is shown in I1031 that given any semigroup Sand any p E P S . (q E /3S :q 5~ p ) is a subsemigroup of /3S. Corollary 11.27 and Exercise 11.4.8 are based on resullr, in [46], a result of collaboration with A. Blass. Exercise 11.4.4 shows that the existence of elements of RP which are RudiFmlik minimal can be pmved in ZFC, because the existence of weak P-points in N*can be proved in ZFC. (See [170].) This c o n m u with the Rudin-Keisler order. The existence of elements of W* which are minimal in the Rudin-Keisler order does follow from the Continuum Hypothesis. but cannot be demonstrated in ZFC.In f a any ultrafilter in N' which is minimal in the Rudin-Keisler order has to be a P-point in W*. (See [69].)

'3

TBe Rudin-Keisler oPdcrwas introduced andstudied by M. Rudin in [211] and [212] and independently by H.Keisler in kchuts at U W in 1%7. The RudiiFrolP order was implicitly intmduced by 2 Frdllt in 1951 and some of its basic properties w a e established by M. Rudin in [212]. There are sevrral very interestingproperties of the Rudin-Keisler and Rudin-FroHl ordm that we have not presented because they do no4 relate to the algebra of BS. Included among these is the equivalence of the following three statements for p E BS: (a) p is minimal in the Rudin-Keisler order. @) p is a selective ultrafilter. F a t is, given any f : S + S, there is some A E p such that the reshiction o f f to A is either one-to-one or constant.) (c) p is a Rmnscy ultrafilter. (That is, for every n E N and every partition (Po,PI) of [Sl", t h a exist some A E p and some i E {O. 1) such that [A]"c Pi.) Another interesting result is that the Rudin-Keisler equivalmce classes (which are the same as the RudiiFmllL equivalence classes by Exercise 11.4.3) of Rudin-Frolik pedccessors of a given element of BN are linearly ordered by Sna. A systematic presentation of these and other results can be found in Sections 9 and 16of [69].

Chapter 12

Ultrafilters Generated by Finite Sums i

We have seen in Theorem 5.8 that if p is an idempotent in (PA,+) and A E p, then A contains FS((x,,)zl) for some sequence (&)El in W. However, it is not necessarily m e that FS((x,)r=,) E p. We investigate in this chapter the existence and properties of ulmfilters satisfying the requirement that each A E p contains some FS((x&) with FS((x,)Z,) E p.

12.1 Martin's Axiom We shall have several pmofs in this chapter for which Martin's Axiom is a hypothesis. Since we expect that Martin's Axiom may be unfamiliar to many of our readers, we include in this section an elementary introduction. . We use the terminology which is standard when dealing with Martin's ~ x i o k although some of it may seem strange at first glance.

Definition 12.1. Apartidly ordered set is a pair (P. 5 )where P is a nonempty set and 5 is a transitive and reflexive relation. that is apanial order. Notice that a partial order is not required here to be anti-symmetric. (However, the partial orden that we will use are anti-symme~c.)

Definition123. Let (P, 5)be a panidly ordered s e t Afirrer in P is a subset G of P such that (1) G # 0. (2) for every a , b E G, there is some c E G such that c 5 a and c 5 b. and (3) f o r e v e r y a s G a n d b ~ P,ifa Sb,thenb EG. Thus, if S is a set, a filter on S is a filter in the partially ordered set (P(S)\(O], G).

Definition 12.3. Let (P. 5 ) be a partially ordered set. A subset D of P is dense in P if and only if for every a E P there is some b s D such that b 5 a.

237

121 Martin's Axiom

Thus, in more usual terminology, a subset of P is dense if and only if it is cofinal

down& &4idtiaa 12.4. Let (P, 5 )be a partially ordered set (a) Elements a and b of P are compatible if and only if t h m is some c E P such

that c

j

a and c 5 b. (Otherwise they are incompatible.)

(b) (P. 5 ) is a C.C.C. porrially orderedset if and only if whenever D is a subset of

p consisting of pairwise incompatible elements, one has that D is countable. In the above definition. "c.c.c." stands for "countable chain condition". However, it a c ~ d l asserts y that all antichains are countable.

Definition 12.5. (a) Let M be an-infinite cardinal. Then MA(K) is the assertion that whenever ( P , j ) is a C.C.C.partially ordered set and 4) is a collection of at most w dense subsets of P, there is some filter G in P such that for all D € 4). G flD # 0. (b)Manin's Axiom is the assenion that for all cardinals K. if w 5 K < c. then MA(%). We first note that M d n ' s Axiom follows fmm the continuum hypothesis.

Theorem 12.6. MA(w) is true. PmoJ Let (P, j)be a partially ordered set and let {A. : n E NJ be a set of dense subsets of P. Choose al AI. Inductively, having chosena. E A", pick a,+l E A,+l such that am+,j a,. Let G = {b E P : there is some n E W with a. 5 b]. 0 . .

Notice that, for the truth of MA(@)one does not need that (P, 5 ) is a C.C.C. partial order. However, the next theorem (together with Exercise 12.1.1) shows that the C.C.C. requirement cannot be deleted.

Theorem 12.7. There exist a partially ordcmd set (P. 9 ) and a collection D of wl dense subsets of P such that no filter in P meets every member of a. P m J Let P = { f : there exists A E Pf(N) such that f : A + wl). For f, g E P apethatf jgifandonlyifgs f.Foreacha<w~,letD,=[f ~ P : a i s i n t h e range o f f ] and k t 4) = (D, :a < wl]. Trivially each D. h dense in P. Suppose now that we have a filter G such that G n D. # 0 for each a < wl. Let g = UG. We claim that g is a function. To this end, let a. b. c be given with (a, b) E g and (a.c) E g. Pick f, h e G such that (a, b) E f and (a. c) E h and pick k E G such t h a t k c f andk~h.Thm(a,b)~kand(a,c)€ksob=c.Butnowgisafunction whose domain is contained in Nand whose range is all of ol. which is impossible. 0 We now illustrate a typical use of Martin's Axiom The partially ordered set used is similar to those used in succeeding sections. Recall that a set A of subsets of a set S is an almost disjoint family if and only if whenever A and B are distinct members of A, / A n BI < w.

12 Ulmfiltm Ocoaatd by Fik Sums

238

&tldtioa 1WI. La S be a set. A set A of subsets of S is a maaimal a h o s t disjoint fmniy if and only if it is dd almost disjoint family which is not properly contained in any 0th- almost disjoint family.

Tacomu lZ9. Let A A a a ' m a l almost d i j o h if m i @ ofinfmite subsers of a set S. ? Then A is not countably infinite.

PmoJ This is Exerrise 12.1.2.

0

As a consequence of Theorem 12.9, one sees immediately that the continuum hypothesis implies that any infinite maximal almost disjoint family of subsetp of N mnst have cardinality c. We shall see that the same sameresutl follows from Martin's Axiom.

Lemma 1210. Irt A A a family of inginire subsets of fi with the pmpeny that, f m every 3 E Pf(A).N\ U 3 is infmite. Let IAl = K. If MA(K),then there is an infmitc

subset B ofN such thaf A n B isfinitefor every A E A.

Prwf:Let Q = [ ( H ,3):H E Pf(N)and 3 E *(A)). H . 3 j ( H , 7 )if and only if

Order Q by agreeing that

E H', (2) 3 E P ' , and (1) H

(3) (H'\H)

n u 3 = 0.

It is mutinc to verify that 5 is a partial order on Q (which is in fact antisymmetric). Next notice that if ( H , 3)and (H'. 3')are incompatible elements of Q, then H # H' (since otherwise ( H , 3 U 3 ' ) 5 ( H , 3)and ( H , 3 U 3')j (H', 3')). Thus, since IPf(N)l= o.it follows that ( Q , 5 ) is a C.C.C. partially ordered set Now, for each A E A. let D ( A ) = [ ( H ,3)E Q : A 6 3).Then givm any ( H , 3)E Q , one has ( H , 7 U [ A ) )€ D ( A ) and ( H , 3 U [ A ) )5 ( H . 3).So D ( A ) is dense. For each n E N, let E ( n ) = {(H. 3)E Q : H\(1.2,. ,n ) # 0). Given any ( H , 3)E Q, we can choose m z n such that m $ U3. Then ( H U [m).3)E E ( n ) and ( H U [ m ) . 7 )5 ( H , 3).So E ( n ) isdense in Q By M A ( K ) ,there is a filter G in Q such that G # D ( A )l # 0 for each A E A and G n E ( n ) # 0 for each n E N. Let

..

B = U [ H :there is some 7 with ( H , 3)E GI. Since G n E ( n ) # 0 for each n E N, it follows that B is infinite. We shall show that A n B isfiniteforeveryA € A. Tolhisen4letA E Aandpick(H, 3)E G n D ( A ) . We claim that A n B E H. Let x E An B and pick some (H', 3')E G such that x e H'. Since ( H , 3)E G and (H', 3')E G , pick (H". 3")E G such that (H", P") j ( H . 3)and (H", 3")5 (H'. 3'). Then x E H' E H", (H"\H) n IJ 3 = 0. and A E 3. SOXE H as required.

12.1 Mmcin'sAxiom

239

P m f L a x E W\ U,, ..F For each a -z K. we can choose a clopen subset Dm of N* such that Fa E D. and x f Dm.We have Dm r A; for some & c N by nKorem 323. Now U,, D. # Pi*. Hence, if A = [A. :a < K), them A satisfies the hyporheses of Lemma 12.10. So there is an infinite subset B of A such that A. n B is finite for every a < K. Our claim then follows from the fact that B* is a non-empty clopen subset of N* disjoint from each D. and therefom disjoint From each .F Corollary 12.12. Let K < c and let ( G . ) , be a family ofopcn subsets of N'. Assuming M W ) , G, 0. h i n t ~ * ( n , . Ga) 0.

ifn,,,

+

+

P m $ This is the dual of Cnollary 12.11.

0

Theorem l2.U. Let A be an infinite almondkjoint family of inmite subsets of N with IAl = x. If MA(x), then A is not a nuuimal almost disjoint family.

Proof:This follows immediately from Lemma 12.10.

0

Corollary 12.14. Let A A an i n t i t e m ' m a l almost disjoint family of subsets of N andassume Martin'sAxian. 7%mIAl= c.

P&

By Theorem 12.13 one cannothave IAl e c and, since A E

Pm.(A15 c.

We also seeas a c o n q u e n a o f the next c d a r y t h a t Martin's axiomis the strongest assmion of irs kind which is not simply false. Cornltarg 12.15. MA([) isfolse.

Pnn$ Let B be an infinite. set of pairwise disjoint infinite subsets of N. A routine application of Zom's Lemma yields a maximal almost disjoint family A with B A. ~ & e s s a r i IAl l ~ 5 r so Theorkn 12.13 together with the assumption of MA(() yield a contradiction.

Exercise 12.1.1. k

0

c that the @ally

ordered set in Theorem 12.7 is not a C.C.C.

partially ordered set. Exercise E1.2. Rove Theomn 12.9.

Exercise 12.13. Recall that a P-point in a topological space is a point x with the property that any countable intnsection of neighborhoods of x is again a neighborhood of x. It is a consequence of Theorems 12.29. 12.35. and 12.36. that Martin's Axiom implies that there are P-points in A*. Rovide a direct pmof of this fact. (Hinc Let 'A Use (Fa),,. be an enumeration of all countable families of clopen subsets of . of proper clopen subsets of Corollary 12.1 1 to consrmct an increasing family (D.).,, A* with the property that. for every a < c, either 5% E D. or W\ 3, 5 Dm.)

n

n

12 UltrafiltersGenerated by Finite Sums

240

Exerdse 12.1.4; Show that Martin's Axiom implies that N* satisfies the foUowiag sumg form of the Baire Category Theorem: If 3 is a family of nowhere dense subsctp of N ' with 1 3 1c c, then IJ3 is nowhere dense in N*. (Hint: Let K < c and let (F,),,, be a family of nowhere dense subsets of N*. Use Corollary 12.11 toconsmct an increasing family (Dm).,, of proper clopen subsets of N* for which Fa c D. for every u < K . )

12.2 Strongly Summable Ultrafilters -Existence We deal in this section with the relationships among three, classes of ultrafilters on N. One of these classes, namely the idempotents. we have already investigated extensively.

Dellnition 12.16. Let p E PW. (a) The ultrafilter p is stmngly surnmnbk if and only if for every A E p. there is a sequence (x.)zl in PI such that FS((x,)zl) 2 A a d FS((x.)zl) E p . (b) The ultrafilter p is weakly summble if and only if for every A E p, there is a sequence (x.)z, in N such that FS((x.)E,) Recall that r = c t ( p E

:p

5 A.

+ p = PI.

Theo~m 12.17. The set r = [ p E P N : p is weakly summable). hpalticulag every idempotent is weakly summablc.

Proof:'Ibis is an immediate consequence of T h m m 5.12. We set out to show that every strongly summable ultrafilter is an id em pot en^

Lemma 12.18. Let p be a stmngly summabk ulnofilter: For each A E p there is an increasing sequence ( x . , ) ~ ~ in N such that FS((X,,)~=~) g A and for each m E W, F%(xn),m,) E P.

Proof:Fori E lI,21,letCi = (3"(3k+i) :n, k E a]. (Thatis,Ci isthesetofpositive integers whose rightmost nonzero ternary digit is i.) Pick i E (1.2) such that Ci E p. Define f : N + a b y f ( 3 " ( 3 k + j ) ) = n f o r a l l n , k ~ o a n d j ~(1,2).Noticethat if x, y E Ci and f (x) = f(y). then x y $ Ci. Consequently. if {x, y,x y) Ci, then f (x Y) = minlf (x), f (y)J. Let A E p and pick a sequence ( x , ) z l such that FS((x.)zl) E p and

+

+

+

By the observation above, the sequence ( x , ) z l is one to one, so we may presume that it is increasing. Let m E N and suppose that FS((x,)&) $ p. Then m > I. Now

i

E, 12.2 Strongly Summable Ulka6ltm

-Existence

241

whee x n ~ u x n= 0. Since FS((X.);';) is finite and, by assumption, CnEUxn+ FS((xn)%) f P, pick F with 0 # F E 11.2,. ,m - 1) such that Z.,FX. FS((X.)&) E p. Given any y E &PX. FS((xn)zA). one has that for some G ~.?'j(Im,m+ I.m+2, ...I).y= & ~ x ~ + & ~ x ~ . w h e r e

+

..

so, as observed above, f (y) 5 f ( Z . G F X ~ . Pick a sequence (y.)El with FS((yn)El) E & F X ~ F S ( ( ~ . ) K ~E) p. Then for each k E W. f (yk) 5 f (&Fx.) such that f (yk) = f (yc). Then yk +ye f Ci, a contradiction.

+

+ FS((xn)&)

and so pick some k # t

Theorem 12.19. Let p be a strongly sununabfe ultmfilur. Then p = p + p. E p and pick by Lemma 12.18 a sequence (x.)zl in W such that G p. It suffices to show that FS((x,)gl) E A and for each m E W, FS((x.)F-) FS((x,,)ZJ C {y E N : -y A E p ) , so let y E F S ( ( x n ) z l )and pick F E f'j(W) such that y = ZnEpx.. Let m = max F 1. Then FS((x,)z4) E p and E -Y +A. FS((x.)&J

Pmof: Let A

+

+

Lemma 12.20. Let p be a strongly swnmable ulnafilter: For every A E p there is a sequence ( x n ) z lin N such thut

Pmoj LetA e p . Fori

E {0.1.2}let

and pick i E (0.1.2) such that B; E p. Pick by Lemma 12.18 an increasing sequence (x,)rsl in W such c h a t F S ( ( ~ . ) ~2~A ) n B, and for each m E N,FS((x.)&,) E p. We claim that ( x , ) z l is as required, so suppose not and pick the first n such that 5 x. < z~'+'+'. Then in fact x.+l 5 4 Z:=l XI. Pick k E o such that Z"+' x, < 2"+'+'. (If n = 1 this is immediate. Otherwise, x, s 4 st so ;C :: x, c 23X+i-1so C:,l x, < Pk+'+' + 23k+i-1 c 2U+i+2and. since ZYxlX, E Bi. Z:=' x, < 2)'+'+'.) Now x.+, 5 4Z:=I xr c 23k+i+3and *,,+I E Bi SO *,,+I c Since also x, c x,,+l we have that 23k+i+1< x, x.+l c sox. X"+I Bi. a contradiction. 0

z:=,

+

+

z"+'+~,

The following easy theorem shows that not nearly all idempotents are strongly summable.

Theorem 12.21. Let p E ct K ( p N . +). 7hen p is not stmngly swnmable.

242

12 Urhafiltetll Gcnanca by Fdte Sums

Pmof. Snppoge that p is shungly summable and pick by Lermna 12.20 a ( x , , ) ~such , that F S ( ( x J z l ) E p, and f a each n E N.X.+I =- 4 X:,l 1,. But FS((x,,)r-,) is not p i m e syndetic, mhadicting Corollary 4.41.

&,, %

Theorem UZt. m e n is a weakly summable ulmr@er which is not an idempotem Pmof:lhis was shown in Corollary 8.24.

0

.'

Theorem 1223. The set is not a semigroup. Infaet. thek exist idempotems p andq inpNsuchthatp+q $ r.

Pmof.This was shown in Exercise 6.1.4.

D

We now introduce a special kind of smngly summabk ulhaflter. Its definition is somewhat complicatad. but we shall prove in Section 12.4 that these idempoteats can only be written in a trivial way as sums of other ulaafilters. Recall that for y E N we define the binary support of y by y = Cntlu& 2". In a ) define the factorial soppon of y. similar fashion, for elements y of F S ( ( t ! ) E lwe

Lkfidtion 1224. (a) For y E F S ( ( t ! ) z , ) ,fsuppb) is defined by y = 6t~Iw,) t!. (b)The ultrafilter p E is a special stmngly sununable uhmJilter if and only if (1) fmeachm E N.FS((n!)zO=,) E p.

We shall show that M d n ' s Axiom implies that s p s i a l strongly sununable ulhaf~lters exist. Indeed, we shall show that Martin's Axiom implies that any family of subsets of FS((n!)r=,) which is contained in an idempotent and has cardinality less than c, is

contained in a special stmngly summable idempotent.

Lkfid11on 1225. (a) 8 denotes the set of infinite subsets S of N with the property &at m divides n whenever m , n e S and rn < n. 8fdenotes the set of non-empty finite subsets of N which have the same property. (b) If X is a subset of N which can be anangcd as an increasing sequence ( x . ) z l , we put FS,(X) = F S ( ( X . ) ~ for ~ ) each m E N. We put FS,(X) = n m e N c(p~(Fs,(X)).

Lanna 1226. Let 9 denote a fmily of subsets of N with thehcfnite intersectionp m p e q , such that mN E 9 for every m e N. Suppose that B E P rmd that Zi contains on idemporenr p. Suppose also rhar B* = ( b E B : -b + B E p] E 3 and that - b + B ' E 3 f o r e v e r y b E Be. L e t m o ~< c a n d a n d ( ~ ~ ~ ~ ~ r hsw. a r I PThenit I

p m j We may suppose that F is closed under finite intersectia Let Q = (F E df : FS(F) B'). We d e h e a partial order on Q by stating that

F'S F i f F E F'. Since Q is countable, it is hivial that it satisfies t h C.C.C. condition. ForeachA E 3 , l e t D(A) = ( F E Q : F n A # 0). Toseethat D(A)isdcnsein Q. let F E Qandlctm = ll F. Wecanchoosea E AnB.narR(F)(-b+B*)nmW. Then F U (01 5 F and F E D(A). Thusit follows~mMartin'sAxiomIhat~isafi1 in Q t h a t ( A ) # Bforevery A E ~ L.e t X = U a . We shall show that X has the properties required.To see that Xis infinin, let m E N and pick F E n D(mN. l%en X n mN # 0. Since d is a filter. any two elements x and y of X belong to a set F E cD. So, if x -z y, x divides y . 'lhus X E d. Furthermore. if H is any finite subset of X, H E F for some F E and so FS(H) s 8. Thus FS(X) E B. Finally, for any A E 3,there exists F E 10 n D(A) and so X n A # 0.

Lemma 12.27. Lei p be an idempoienr in N*, lef 9 C p, and lei o 5 1 3 15 K. then MA(c) implies rluu there exists X E d such that FS,(X)

g nAeFA. K

< r

P~~~:L~~$=~U(~N:~EN)U(B*:BEP)U(-~+B~:BE~ and note that by Lemmas 4.14 and 6.6, $ p. Let X be the set of finite intersections of members of and note that X E p and IXI 5 K. Well order 3as (A&, (with repetition if (31< K). Pick by Lunma 12.26 (with X in place of 3 and Ao in place of 8 ) some Xo E 6 for which FS(Xo) E Ao, and X o n A f 0foreveryA E X. We then make the inductive assumption that 0 4 0 5 K and that we have defied X, E d for every a < p so that the following conditions are satisfied:

a

(a) F S ( X e ) ~ A . a n d X . n A # O f o r e v e r y A ~ R a n d (b) if 8 < a,then IX.\Xal < o. Using condition (b)and the facts that X is closed under finite intersections and that mN E X for every m E W one sees that (*) for any finite nonempty 3J E

X U (Xu :a < 0). there exist A 6 X and a s 0

~ithAnx,Gno. In particular, by condition (a), (A n X. : A E H and a -z B ] has the finite intersection ProPcrtY. We apply Lemma 12.26. with (A n X, : A E X and a < 01 in place of 3 and Ag in place of B. Pick Wg E d such that FS(Wg) s Ag and Wg n A n X. # 0 for every A E X and every a -z B . By the 0bSe~ation(*), n ( W g n A n X. : A E JC and a -z f l ) # 0 so by Corollary 12.12. there exists an infinite subset Xg of M such that Xg* Wg n A n X, for every A E X and every a < p. We may suppose that

12 U l u a Generated ~ ~ by Finite Sums

2A4

Xg C W8.We may also suppose that Xp 6 6.because Xg n m N # 0 for every m E N, and so Xg contains a set in 8. It is clear that conditions (a) and (b) are satisfied with p in place of a. We put X = X,. If a 5 K. X\X, i s a n d so there is some n E N such that FS.(X) c FS(X,) and thus FS,(X) g FS(X,) c 0

x.

Corollarg 1228. Let F be a family ofsubsets of N contained in an idempotent pfor which FS((n!)zl) E p. Let L be any infinite subset of N and let w 5 K c c. I/ MA(K) and 1 3 15 K, then there exisfs X E 8 with FS(X) c FS((n!)z,) such fhai FSk(X) E Tiforevery A E 3and L\ UxEFS(X) fsupp(x) is inbite. Pmof: We may suppose that 3is closed under finite intersections. By Exercise 12.2.1. we may suppose that FS((n!)Zm) E 3 forevuy m E N.. By Lemma 12.27, there exists Y E 6 such that FS,(Y) s Ti for every A E F. We claim that Y n FS((n!)z_) # 0 for every m E A. To see this. let m E N. Since FS,(Y) = nk,,FSt(Y) s (FS(n!)z-), there exists k E N for which FSt(Y) FS((n!)E,). Thus we can inductively chowe a sequencc (x,)z, in Y n FS((n!)zl) with the property that, foreach n E W. max fsupp(x.)c t. cmin fsupp(x.+l) for somet. E L. If X = {x,, : n E N), then X has the properties required. 0 Notice that, while the previous results only use MA(w) for a fixed following theorem uses the full strength of Martin's Axiom.

K

< c, the

Theom l2.29. L e t 3 beaf a m i l y o f s u b s e r s ~ c o n t a i m d i n ~ Et qN* for which FS((n!)zI) E q. If 1 3 1c c, Martin's Axiom implies that then is a specid strongly summable ulnofilter p for which 3 E p. Pmof: We assume Martin's Axiom. Let (Y.)., be an enumeration of P(W and let (L,),,, be an enumeration of the infinite subsets of N. We can choose Zo E (Yo. N\Yo) such that Zo E q. By Corollary 12.28 applied to

F U (Zo), we can choose Xo 6 8 with FS(Xo) C FS((n!)z,) such that FS,(Xo) -

s

A n Zo for every A E 3 and Lo\ U,,,(xo,fsupp(x) is infinite. We then make the inductive assumption that 0 c fl c c and that X, E 8 with FS(X,) 2 FS((n!)z,) has been defined for every u c fl so that the following conditions hold:

is a compact subsemigroup of N* and By Lemma 5.1 1. for each a c fl, FS,(X,) hence so is FS,(X,) which thereforecontains an idempotent r. Since FS(Xo) c FS((n!)zl), FS((n!)z,) E r. Wecan choose Zg E (Ye. N\Yp) satisfying Zg E r. By Corollary 12.28 (applied to (FS,(X,) :a ifi and m E N) U (Zg] in place of 3)we

no,g

'

can choose Xp E d with FS(Xd S FS((n!)z,) such that Lg\ U,,Fstx.,fsupp(x) -. ?.

is

infinite. F S d X p ) G % . P ~ ~ F s , ( x ~ ) C FS,(X.) forevery a c fi andevny m E N. Thus F S d X g ) S FSdX,) for every a < fi. So conditions ( a H c ) are satisfied with p in place of a. This shows that we can define X, E 6 for every a < c so that conditions (a)-(c) an satisfied. We put p = (B C N : FS,(X,) E B for some a c c]. It is clear that p is a filter. For every Y M. Y E p or W\Y E p, and so p is an ulnafilter. Since FS,(Xo) G A, we have 3 c p. Since FS(Xo) G FS((n!)zl). we have by Exercise 12.2.1 that for each m E N, c Ti and thus FS((n!)&) E p. Given A E p. A = Y, for some a and so FS,(X,) for some m. FS,(X,) A. Condition (3) of the definition holds directly, and so we have that p is a special strongly summable ulnafilter.

nA,,

c

Exerdse 1221. Show that if p is an idempotent in N* and FS((n!)zl) E p. thm for all m E N, FS((n!)&) E p. (Hint: Use Lemma 6.6 and consider the proof of Lemma 12.18.)

12.3 Strongly Summable Ultrafilters -Independence In the last section we saw that Martin's Axiom implies that smngly summableultrafilters exist In this section we show that their existence cannot be deduced in ZFC. Definition 1230. P1 is a union ultmfilter if and only if 9.L is an ultrafilter on 9 5 . 0 and for every A E 9.L there is a sequence (F.)Z1 of pairwise disjoint members of +(W) suchthatFU((FdzI) E P1 andFU((F.)ZI) G A. It would appear that union ultrafilters and strongly s u m a b l e ultrafilters are essentially equivalent notions. Thisis indeedhue. as weshallshortly see. Thecomspondenm in one direction is in fact immediate.

Theorem 12.31. Lc: 9.L be a union u11mpIter: Let Then p is a smngly summable ulfmfilter:

Pmof: The function f : 4 0 -r M defined by f (F) = CraF 2'-I is a bijection, so p is trivially an ultrafilter. To see that p is strongly sumable, let A E p. Let A = IF E P,(N) : &F 2'-I E A ] . Then A E e( so pick a sequenm (F.)zl of painvise E 9.L and FU((F,)zl) disjoint members of .Pf(W) such that FU((F,Jr=I) each n. letx. = C,.F. 2'-I. Then FS((x,)Z,) E p and F S ( ( x , ) ~ , )G A.

E A.

For 0

To establish the comspondence in the reverse direction, we need some purely atithmetic lemmas. They are somewhat stronger than needed bere, but they will be used in their full strength in the next section.

246

12

WtraEIta~Geaaaad by Fmite Sums

,

Lemma 12.32. Lrr (&,)ED=, be a sequence in N such that, for each n € N,x.+, 4 Zy=, x,.' Then each member of W can be written in at nost one way cu a l i m r combinmion of xn 's with c@efUsfmm 11.2.3.4).

Pmoj Suppose not and let z be the smallest eooaterexamplc. Then (taking as u s w &B Xr = 0) OW. has

where {FI, F2.F3, F4) and (GI,G2. G3, G4J me each pairwise disjoint and not each Fi = Gi. Let n = maxu=,(F'i U Gi) and assume without loss of generality that n E l-f-l Fi. Then n $ u = l Gi. since if it were we would have z x. as a s&Um counhrexample. l%us

-

Lemma 12.33. Lcr (x.)El be a sequence in N such that, for each n E N,*.+I > 4 x,. 7?meach member of Z CM be written in at most one way as a linear combination of x. 's with coe&kiemfmm (-2. 1 , 1.2).

xy,,

Pmof: Assume that we have

l%us, by Lemma 1232, om has

-

so by Lemma 12.32. F

n G = 0.

Theorem 12.35. Let p be a amngly sununable ultmflter and pick by L c m 12.20 sequence (x,)El in W such that FS((xn)EI) E p and for each n E N,. x.+l > 4 Zy,l

XI. Define yp : FS((xn)zl)+ Pf(W by ( p ( z t E ~ x O = [XI : t E F). Let 91 = [A E Pf(N) :yp-'[Al E p). 7h-n 91 is o union ulnafitez

Notice first that by Lemma 12.32, the function (p is well defined. One has then immediately that 91 is an oltratiltcr. To see that 1L is n union ultnrfilter, let A € 1L. 'Let A = (p-l[Al. Then A E p so pick a sequence (y,)gl such thatFS(CV,):,) GA and FS((y,Jz,) E p. For n E N, let F,, = yp(y,,). By Lemma 12.34. one has for each H E Pf(N) that yo(&^ y,,) = UIEHFn Consequently, one has that

p,&

o

FU((F.)Z,) 2 A andFU((Fdz,) E 91.

Theorem 12.36. Lcr U be a mion &mfiIfer and let q = m%(U), whrm m = : p(pf (N))-t ON is the confi~ousatmsion of max : 3'f(W) -r N. Then q is a P-point of W. Plwf: Notice that q is a noaprincipal u l ~ l t cbecause r 91 is and the function max is finiteto one on ,F"(N). To see that q is a P-poiat, let (A,)E, be a sequence of membess of q. We need to show that them is some B E q such that B* G A.'. We may presume that Al = N and that A.+I G A. and n # A.+I for all n. @I particular A,, E B.) Dcfim f :N + M by f (I) = maxln E N :x E A"). Let

n?=,

,n:

and and pick i E 10.1) such that 3% E U. Pick a pairwise disjoint sequence in such that W((F,,)Z1) G Bi and W((F,)Z,) E 91. We may pmume that min F,, min F,,+l for all n. Since FU((F,)?=l) E 91. we have that (maxG : G E FU((F.)zl)) 6 q. Since

e(W)

12 UIwfilters Gcrrnted by Finite Sums

248

We show first that i # 0. so suppose instcad that i = 0. Pick k E N suchthat fa n g k,maxF, s m a x F 1 . Sinceqisnonprincipal[maxF. : n E N a n d n z k ) e q . Let f = min FI. Now Ac+l E q so pick n 2 k such that max F. E A ~ + IThen .

a conwdiction. Thusi = 1. LetB=[maxF, : n EN). N o w , i f n , k ~ N a n d m a x F . E B\At, t h a n 5 min F. c f (max F,,)< kso IB\AkI < k. ThusB* A.*asrequimj.

nz,

0

We have need of the following fact which it would take us too far afield to prove.

Theorem 12.37. The existence of P-points in M' c m m t be established in ZFC. P&

r222.

VI, 941.

0

Corollary 1238. The existence ofshongly sunvnable ulnafilterscan notbe established

in ZFC. n it is consistent relative to ZFC that there are no P-points Pmof: By l l ~ e ~ r n12.37 in N*. Theorem 12.35 shows that if strongly summable ultrafilters exist, so do union ultrafilters, while Theorem 12.36 shows that if union ultmilters exist so do P-points in

W*.

0

F.xercise 12.3.1. As in T h e o m 12.31. let U be a union ulaafilter and let p = : F E A ) : A E 91). Define f : P f ( N T) N by f ( ~ = ) ZIEF21-I. Rove that f (U) = p.

SF^'-:

Exercise 12.33. As in Theomm 12.35. let p be a strongly summable ultra6ltcr and pick a sequence (x.)zl in N such that FS((x.)zl) E p and for each n E N, x.+l r 4x:,,xt. Define (p : FS((x.)z,) -t P,(N) by (P(C~FX,)= [x, : I E F ) and extend yp arbih;lrily lo the rest of N. Let U = [A S Pf(W) :cp-'[A] E p). Prove that G(P) = U.

12.4 Algebraic Properties of Strongly Summable Ultrafilters Recall that for any discrete semigroup (S, +), an idempotent p in BS is strongly right m i m l in f3S if and only if (q E BS : q p = pl = (p]. We saw in Theorem 9.10 that strongly right maximal idempotents in f3Pd exist.

+

Theorem 1239. Let p E BN be a strongly surnnrable ultrajlte,: Then p is a srmngly right ll~~n'mal idempotent.

249

12.4 Algebraic Roperties

Proof. By Theorem 12.19, p is an idempotent. Let q f p be given. We show that p # p. Pick A e p\q and pick by Lemma 12.20 a sequence (X,);=~_N such that FS((xJK,) G A, for each m E N. FS((x,)~,,,) E p. and X.+I 2 4 xl for all n. It suffices to show that

4

+

(In fact equality holds, but we do not care about that,) Indeed, one has then that p andthus p # q p. ( Y EN: -Y +FS((xn)El) E PI q s o F s ( ( + ) z I ) To this end, let a E N such that -a + FS((x.)zl) E p. Then

e +

e

+

Pick F andG in 9if(N) such that yl = & ~ x ~ - a n d y = z ~ I , G X , . By L e m 12.34. F n G = 0. Also, a yl and a n are in FS((x,)zl). So pick H and K in P f ( W XI and a + A = EIGK XI. Then such that a yl =

+

+

+

+

+

%X%Vf XI = &PAK XI Z ~ E P2x1 ~K Thus, by h I l l l M 12.32 SO & G A H X ~ GAH = FAK and G flH = F n K. Since F n G = 0. one concludes from these equations that F E H. S o a = &H\FXI and hencea € FS((x,,)zI) as required.

We saw in Corollary 7.36 that maximal group6 in K(BN) are as large as possible (and highly noncommutative). That is, they all contain a copy of the free group on 2' generators. We set out to show now that if p is strongly summable, then H(p) is as small (and wmmutative) as possible, namely a wpy of %.

Lemma 12.40. Let (x,)zl be a sequence in N such thatfor each n, x.+l 4 Z:, x,. Ler F, GI, andG2 befinire subsets ofN such that G I n G2 = 0 and F U GI u GZ# 0. Lcra=mio(FUG~UGz)andlet

Pmof: Assume that t # 0. Let F' = F\(GI U G2). let GI' = ( G I \ F ) U (G2 n F), and let GZ' = G2\F. Then F', GI'. and G2' are pairwise disjoint and

' = E ~ S F '-X ~

-EnEG2'k.

& E ~ , l ~ n

12 Ultratiltus GmcmAby Fite Sums

250

S i n e e t f O , F ' U G ~ ' u G ~ ' 0. f Letb=max(FiUGI'U-G2')and.notiamotbza. Then if b s F', we have

Otherwise b E (GI' U G 2 9 so that

The foUowing lemma is exceedinglytechnical and we apologize in advance for that fact It will be invoked twicc in the p m f of the main result of this section. b4 seqllCnCeS in N a n d a s s m that Lemma 12.41. Ln (x,b:,. (y.):,. and for each n, x.+l > 4 x,. Let L E N and assume that whenever k, j > L, one has SOPIW H ~ E ,g ~ ( N such thor zk yj = Z n f ~ t .xn. J L C ~b = max Ht,t curd (USthat wheneverminlk, jl r L one has min Hk.j > b. L e t t = ( & H ~ . ~ \ H ~ , ~x.) + , yt

x,,I

+

-

P 4 Wehtshowthatforanyk > f.Hk,t\Hk,e+~ = Hc,c\He,t+~ andHt,k\Hc+~.k = Ht.c\Hf+l+c. Indeed, consider the eqoatiws

and Zk

+ Yt+I = CneHt,c+l X n ,

From the first two equations we have

while the last two equations yield

Thus by Lemma 12.33, one has that Ht,c\Hk,t+l = He,c\He,e+l. Likewise. working with z t + ~ zt. one gets that He.k\Hf+l.k = He.t\He+l.c.

-

124 Algebraic Ropnties

Xb+l

Xb+l

xb+l

sndsoltl < -.Similarly. 1st c -andthusIs+rl < -. 4 4 2 Now let k > f be given. Then zk yc = &H,,, x,, so

+

Tocompletethe proof, it thmfmsuffices toshow thats = -1. Fmm thederivatioos above we have that z k yk = C n € ~xn k & E ~ k xn r s and we also know that Zk Yk = & ~ t , Xn. ~ Thus L E H ~ ,Xn - &A Xn %€GI Xn = S 1. k t Kt = (Fk\Gk) U (Gk\Fk) and let K2 = Fk n Gk. Thm

+

+

+

++ -

+

As amnsequence of the next cheorrm (andTheorern 12.29) it is consistent with ZFC that Ume are maximal groups of (BN, +) that are just copies of Z. We do not know whether the existence of such small maximal groups can be established in ZFC.

Theorem 12.42. k t p be a stmngly sumnuable ultrafilter: Ifq E W and them exists r ~ N * s u c h t h o t q + r = r + q = p , t h e n q ~ Z + p. InpaniculazH(p)=Z+p.

Prwf: Let us notice first that since, by Theorem 4.23, Z is contained in the center of (075, +), Z p is a subgroup of (BZ. An4 by Exercise 4.3.5, Z p 5 BN. Consequently Z p H(p).The "in particular" conclusion then follows from the

+

+

+).

+

12 Ultrafitera Oemtcd by Fimte Sum

252

k t ssscrtian immediately, because any member of H(p) is in W and is imrntible wim respect top. So assume we have q and r in N* such that q r = r q = p and s u p p that (l+q,q.-l+q),pickAl i epsuch q p Z + p . 'Ihmlikewise,p$! Z+q. Sincep $ Al q, A1 q, and 1 A1 # q. By Lemma 12.20 wemay presume (by that -1 passing to a subset) that AI = F S ( ( X I , ~ where ) ~ ~ )for each n, XI,,,+I > 4 qj. and for each m, FS((xl,,):-) E p. Let BI = (x E N : -x AI E q ] . Since p = r +q. Bl E r. tick yl E BI. Let

+

e

+ <

+

+

+

C; = (-yl + Al) n (X E N : -X

xy=l

+ AI E r ) n n:=-,(N\(-i + A ~ ) ) .

Sinceyl~B1andp=q+r,9~q.Pickzlrylwithz1~C1.Sinceyl+zl~A~. pick JI,I,I E 9,dW such that YI zl = ~ n E ~ I , ,XI,". ,I Let bl = max J1.1.1 andlet XI,,,. al = bl 1. Note that a1 is the smallestmember of N with XI.., s Cn~j,,,,, Assume now thatk > 1 and we have chosen Ak-I = FS((xk-lc)zl) E p. Bt-1 E r.Ck-1 E q , y k - ~~ B k - 1 ~ z k - ECk-1 1 E ( x E N : - x + A k - 1 ~r).bk-1.aodak-1 such thatforeachn, xk-l..+l r 4 xk-1.j. and fneach m.FS((Xk-l..)&) E p. ~ )Also. P C I t i q : E Z T h e i F S ( ( ~ k - ~ , n ) z ~ ~E_ P, +a n~d) F S ( ( ~ ~ . r n ) ~ = ~ , -E, +P. and Irl 5 k]. Thus we may choose by Lemma 12.20, At = F S ( ( X ~ , ~E )p~where ~) foreachn.xk,.+~ > 4Cyel xk,j, foreachm. FS((xt,.):-O,) E p, At $! r + q forany r E Z with 111 5 k. and

+

+

x;,I

m s FS((X~-I..),=~~ _,,I

~F~((xI..)E,-,+~).

c

Note that At At-1. Now Ak E p = r + q . Let Bk = (-zk-1

+ At-I) n (X E N :-x + Ak E q) nBt-I.

m e n Bk E r. Pick yk E Bk with yk > zk-I. Let Ck = Ct-1

n (-yk + A t ) n (X E N : -x + A t

E

r ) nn f = - , ( ~ \ ( - i

+ At)).

Then Ct E q. Pick zk E Ck with zk > yt. Let a k be the smallest member of N with XI., w yk zk. Pick Jk,r,t E Pj(Pr) with yk zt = x , E ~ , , k , r ~ k . n and let h = max Jk.k.t. The inductive construction being complete, let t ,k, j E N be given with e 5 min(j , kl.

+

+

I f k I jwehavezk ECk G C j g - y j + A j s o y j + z k €A,. I f k < jwehavcyj E B GBk+l ~ c-zk+Aksoyj+zk~At.

+

Consequently yj zk E At. Further, if i > l , then Ai E At+! E F S ( ( X ~ . . ) ~ ~ ~ + ~ ) Consequently, we may choose Jr.k,j E P,(N) such that yj ZI = Z,EJtl,j XC." and, if L 4 min(j, k]. then min Jc,t.j z bc. Note that, if C = k = j. then the definition of Jr,k.j agrees with the one given earlier.

+

253

12.4 Algebraic Ropatiu1

N o w l e t f = ( C n E ~ , . l . , \ ~ l . l . 2 * ~ . . ) - ~Fork ~ . > 1.letFk = J ~ . t , ~ n J ~ . ~ . ~ a n d 1 Gk = h , ~ . t ~ J ~ , zByLemma12.41 ,k. wehavcthat Irl < -xl.bl+l andforeachk > I, 4 Zk = XncFb X1.n f yC 3 &ck X1.n f. Next let L = 111 (or t = 2 if It1 5 1). Lett' = ( ~ n e ~ ~ xt.") , ~ -, yt. ~ For \ ~ ~ , ~ ~ ~ ~ k 7 l , k t Fk' = Jt.r,c n Jt.r.t+l and Gk' = Jt.t.k n Jt.c+l,t. Then again by Lemma t' and yk = EnEGkv xt." 1'. 12.41 we have that for each k r C, zk = CnEFt,xt.., To complete the +of we show that t = r'. Indeed, once we have done this we will At)), a ha* It+] = ~ I E ~ t + I , +x I~ .Enf At while zt+l E Ct+l g Ct g (W\(t contradiction. Now, given n, xt, E At G FS((XI,.)~=~)so pick T . such that x e , = EmETm XI.^ Also, if n # k, then XI.. E FS((XI,,)~=~) so by Lemma 12.34, T. fl fi = 0. Let k = L 1. We show that ifn E Je,k,t, then min T . 2 a t 2. Indeed, zt yk E A t 5 W(XI.,),&+~) so pick H with zt yk = E ~ ~ Hand x min I , H ~ 3 at 2. On the other hand. usingat the second equality the fact that the T.s' an pairwise disjoint ~ ,= ~C , ~n E ~ ~where ~ . Ln = UmEJ <,,, T,,. Thus L = H wehavezk +yk = C ~ G Jxt.n and in particular each min T . z_ min H 2 at 2. Now let Ft" =,Unefi, T . andlet GI" = UnSGtrT,,.S i c e Fk'E Jt.k,c+l =J ~ , ~ , ~ and Gk' S Jt.t+l,t = Jc,k,k, we have min Fk" ? at 2 and minGk" 2 a t +2. Note also that zk = Ensfi" X I + r' and yk = EnEGkXI," rr. Now let

+

-

+

-

+

+

+

+

+

+

+

+

+ L I = (n € F k : n 2 a t + Z),

+

L2 = ( n € F k : n < a t + 2 ] , MI = ( n € G k : n ~ a ~ + 2 ] .

and Mz=(nEGt:n
+

m e n zk = L ~ L X1.n I C n s XI." ~ ~+ t and equating both expressions for z r , we have Cns~,"XI,#

- %ELI

X1.n

+ EneAd2XI.,,

= Crn,~,XI."

= EneL2X1.n

-I.

SO,

+ - t'. f

Likewise. from the expressions for yk we have

- C n c ~XI,"I = C n s ~XI."2 + I ' -I. Since min(Fkn U L I ) 2 a t + 2 and min(GkUU MI) 2 a t + 2, we have by Lemma & E ~ t n ~ ~ . n

12.40 that

(1) C.,L~XI..+ r - f ' = O o r I

E.f~,x~.,+t-t'l

(2) Z n ~ ~ 2 +I' ~ ~- .t = n Om I CnfYXI.^

+

f'

2 f x ~ . ~ and <+~

- 11 z fx1.,,~+2.

We show in the remainder of this section that it is consistent with the usual axioms of set theory that there exist idcmpotmts in (PN, which can only be written as a sum of elements of PN in a trivial fashion.

+)

Lanmo 12d3. Lct (x.)FEl be an increasing sequrncc in N with the pmpeny thaf for each n. E N, x,, divides x.+l. I f a , b, m G 4 a 5 x,. x,+l divides b, and a + b E F S ( ( X , ) ~ ~thena ) , E F S ( ( ~ , J ~ ~ ) a mEfFS((xm)zI). b

...,

Pmof: P i c k F ~ ~ ~ ~ s u c h ~ a ~ + eb t =~ = ~ ~~ n~[ ~1 ,x2m~,] &. let H = F \ ( l , 2, ....ml. Then a E,,GX,= &Hx, b (where &x, = 0). r lx,+I. SO a CeeGx, = Then X,+I dividts C t E ~ x , b while la - E t s ~ ~ < 0 C l r xt~ b = 0.

-

-

-

-

-

P m 4 By Theonm 12.39 it suffices to show that r = p. So suppost wt and pick A p\r. Pickan increasing sequence (x,,)zl such that FS((x.)zl) G A, FS((x.)zl) € p. and for each n E N, x. dividesx.+l. Then FS((x,,)zl) € q + r so picka E W such that -a FS((x.)zl) E r. Pick m E N such that x, Z. a. Pick

+

+

Then a b E FS((x,)Zl). a 5 x,. FS((x,)zl) g A, a contradiction.

and X.+I divides b so by Lemma 12.43, b E 0

- + +

Of course, any idempotent p in ,9N can be written as ( p n) ( p n) for any n E Z. We see now that if p is a special strongly summable ultrafilter. then this is the only way that p can be written as a sum.

Theomn 12.6. Let p be a special strongly summable ultmftlrr and let q, r E PN. I f p=q+r.thenqandrareinZ+p.

+n ,:

Pmof: We claim that it suffices to show that r E Z have k E Z such that -k r E nElce(Nn). Then (k Lemma12.44.k+q=-k+r=p.

+

cC(Nn). For assume we

+ q) + (-k + r ) = p so by

Thus we supposc that r f? Z Define

+ n'R,,cL(Nn),so that also q 4 Z +

c.t(Nn).

a : N + ~ ~ ~ ,..., ( 0 t.) 1

xzl

.

by the equation x = u ( x ) ( f ) t! (It is easy to see tbM any positive integer has a unique suchfacrorial expansion.) Let Z be the mntinuous extension of u to BN. Let M = [t E W : B(q)(t)# tj and let L = (t E W : b?(q)(t)# 01. We claim hat both M and L are infinite. To pee that M is infinite. suppose instead we have some k k E N such tbat for all r > k. I(q)(r)= r. L e r = (k I)! - E,,, I ( q ) ( f ). t!. We claim that q z E ce(Nn),a contradiction. To see this, let n E W be given with n > k. We show that -z + N(n + I)! E q. Now. by the continuity of Z,

+

nzl

+

To see this. let w

E N(n

+ I)! + E:,,

.

Z(q)(t).t ! Then

as required. Similarly, if L is finite so that for some k, one has Z(q)(t)= 0 for all r z k. and k z = C,=, Z(q)(t).t! ,then q z E ce(Nn). Since p is a special strongly swnmable ulhafiltcr, pick a sequence ( z n ) z ,such that FS((zn)Zl) S FS((r!JPO_,). W ( z n ) Z , ) P and M\U(fsup~Cy): Y EFS((tm)El)I is infinite. Let K = U(fsupp(y): y E F S ( ( Z , ) ~ ~ ) ) . Now FS((z,,)E,) E q r so pick a E N such that -a FS((z.),O,,) E r. Pick m E Nsuchthala < m!andpickt s s > msuchthate E M\K ands E L. Now WV!)Et+j)E P so

-

+

so pick b E N(C

n?=,

+

+ I)! + E:=,B(q)(r).t! such that -b + FS((t!)PO,(+,)E r. Pick

12 Ultrafilters Gcncrated by Fimte Sums

256

S i c e s E L, Z(q)(s) # 0 so

x:=l Z(q)(r) .r! - a so xisli;(q)(t)

.t! - a

=

x:=,q . t ! w h m each c, E 10.1.. ..,t) and not all cl*s

are 0. Now b + x e FS((z,)zl)

12

2 s! - a > 0

+ (b - a ) so for some H e 9f(N),

Pick I E Pf(N)such that &,,H zn = = 1\{1,2.. ...e). Then

t! and let 11 = I r l f1,2,.

.

.. el and

+

Now (t I)! divides the left hand side of this equation. On the other hand, C $ K so e~Itande~MsoZ(q)(e)ie-l.Thus

Notes The presentation of Martin's Axiom is based on that by K. Kunen in [171]. See the book by D. Fremlim 1941 for substantial information about Martin's Axiom. It was shown in 1972 in [I 171that the (then unproved) Finite Sums Theorem togetha with the continuum hypothesis implied the existence of an ulhafilter p on N such that for every A E p. ( x E N : -x A E p ) E p. And A. Taylor (unpublished) showed that the continuum hypothesis could be weakened to Martin's Axiom. Of course. we now know that such an ultrafilter is simply an idempotent in (pN, +). (See the notes to Chapter 5 for the history of the discovery of this fact.) Wtth this knowledge it appeared that nothing interesting remained in [117]. Then. in 1985, E. van Dwwen noted in

+

Notes

257

comenation that in fact the ultrafilter produced there had the stronger pmperty that it was generated by sets of the fonn FS((xm)z,).and wondered if the existence of such an ultrafilter could bc proved in ZFC. This question led to the investigation of shangly sununable ultrafilters, beginning with [1311. Union ultrafiltersw e n introduced by A. Blass in (441. T k m 12.37, om o f h nooelemmtary results used in this book that we do not prove, is due to S. Shelah. Caollary 12.38 is due to P. Matet [181]. The proof of Corollary 12.38 presented here is from [471. a result of collaboration with A. Blass. Theorem 12.39 is from (461,a result of collaboration with A. Blass. Theorem 12.42 is from 11361. Some of the results of this chapter have been gennalised in [I511 (a m u l t of collaboration with I. Protasov) to countably infinite abelian groups. If G is a p u p of this kind, the existence of strongly summable nonprincipal ulmfilters on G follows fmm Manin's Axiom,but cannot be demonstrated in ZFC.If G can be embedded in the unit circle, then any strongly summable ultrafilter p E G* has the property that the quation p x = p has the unique solution x = p in G ', and x, does the equation x p = p. It also follows from Martin's Axiom that there are nonprincipal strongly summable ultrafilters p E G* with the remarkable algebraic property that, for any x . y E G*. x + y =pimpliesthatxandy a r e b o t h i n G + p . If G is a countably infinite Boolean group, the existence of a nonprincipal strongly summable ultrafilter p on G has an interesting consequence. The ultrafilter p can be used to define a non-discrete extremally disconnected topology on G for which G is a topological group. It is not known whether the existence of a topological group of this kind can be demonstrated in ZFC.

+

+

Chapter 13

Multiple Structures in @S

We deal in this chapter with the relationships among different operations on the same set S.as for example the semigroups (N. +) and (W. .). We also include a section considering the relationships between the lefi continuous and right continuous extensions of an operation. on S.

13.1 Sums Equal to Products in BZ The earliest applicationsof the thtory of compact right topological semigroups to Ramsey Theory involved the semigroups (N,+) and (N. .). (See Chapters 5 and 17 for several examples.) In investigating these applications. the question arose whethcr the equation q + p = s .r has any solutions in W'. or indeed in Z*. These questions remain open, but we priscnt several partial results in this section. We begin with the only nontrivial instance of the distributive law known to hold in pZ. (For more information about the distributive law see Section 13.2.) Since we are dealing with two operations, the notations Ap and pp are now ambiguous. We adopt the convention in Sections 13.1 and 13.2 that pp(q) = q p and Ap(q) = p + q and introduce the notations rp(q) = 4 . p and ep(q) = p .q. If p E pZ, - p denotes (-1). p = (-A : A E p )

+

PmoJ Since. by 'Iheor+m 4.23. Z is contained in the center of (pZ. .), the second assertion follows from the first. Now n ( p q ) = (t. o pq)(p) and n p n q = (p., o L.)(p) so it sufficesto show that the functions 4. o pq and p.., o f . agree on Z . So let m E Z be given. Then

. +

. + .

siorr the disaibutivelaw hclds in 2,the functions L, o A. and A.,

o I.

agree on 2.0

We have - i d l y used the cancmical mapping from Z m Z . befm. In this chaptm it will be used in several proofs, so we fix some notation for i t

Ddinitim l3.Z Let n E N. lben h. :Z N Z denotes the canonical mapping.

6

Notice that the continucus extension :P Z -r h, of h, is a homomnphism both horn (BZ. to (Z.. +) and from (BZ. 9 to (Z. .) (by Lemma 2.14).

+)

DeBnltion 13.3. Let b be a 14 n u m k satisfying b > 1. We define #b : N N o by putting &An) = Llogb(n)j. As usual @b :BN + B o denotes the continuous extension of @b.

+

Pmof: (i) Choose r r 0 with the property that I logb(l t)1 < I if r E (-r. r). If m E Z, n E N, and Iml ien, we have logb(m + n ) = logb n logb(l and hence %(m + n ) E M n ) + {-LO. 1). For e a ~ hi E (-1,0,1), we define Mi E Z by putting m E Mi if [n E N : qb(m+n) =#b(n)+i] E p. SithesetsMipartitionZwecrmchoogej E [-1.0, I) suchthat Mi E q. Foreuchm E Mj.let N, = (n E N : k ( m + n ) =@b(n)+ j) E p. We have: & ( q + ~ )= q - , ~ ~ ~ - / $ ~ @ b ( m + n )

+ z)

+

= q- lim p- Lim (@b(n)+ j) mcMl ncN. = &(P) + j. Ci) 'Ihis follows in the saw way h m the observation that, for every m, n E A. @b(mn)E M m ) + M I ) (0.1). o

+

PIT@ We may assume that p E N*. since otherwise, by Lemma 13.1. we could replace p by -p and q by -9. Now by Exercise 4.3.5, q p E N*.And trivially 0 . p = 0. while n E -N implies that n p E -N8. Thus n E N. Suppose that n > 1. We can then choose_b E 1 sattfying 1 b2 < n. So $b(n) 2 2. It follows from, Lemma 13.4-that &(n. p) = &(p) k for some t 2 2 (-LO, I]. By Lemma 6.28, this is a in W. On the other hand, k ( q p) E &(PI contradiction. 0

+

.

+

+

+

+

.

It is easy to find solutions to the equation q p = n r w h a n E Z. The next two results are concerned with multiple solutions to such an equation for fixed p and r. Such multiple solutions arc used in Theorem 13.9 to characterize the existence of q and sinZ*withq+p=s.r.

13 Multiple S m ~ l u r r ain PS

260

Ltmma13.6. Lctp.r e Z ' , l e t s , t ~ p ' b , a n d f e t r n , n ~IZf s + p = m . r w r + p = n r, then m = n.

.

P m J We may supposethat p € N ', since otherwise. by Lemma 13.1, we could r e p l w p, s, 1, andr by -p, -s, -I, and -r respectively. This implies that s + p and t + p are in N*,by Exercise 4.3.5, and hence that m # 0 and n # 0. Therefore r, m, and n all in BN a all in -BN. We may suppose that r, m, n E BN, since otherwise we could replacer, m, and n by -r. -m, and -n respectively. If n # m. we may suppose without loss of generality that n > m. We can choose b E R satisfying b > 2 and mb4 c_ n. So @b(n)? @b(m) 4. By Lemma 13.4. @b(n.r ) = @b(m.r) k for some k ?_3 in N. We all0 know that & ( ~ + ~ ) a n d @ b ( t + ~ ) ~ n b o t h i n @ b ( ~ ) + [ - 1 , 0 . 1 1 . S O @ ~ V + P=)@ b ( s + p ) + e for same e E 1-2, -1.0, I, 2). Since C f k, this contradicts Lemma 6.28. This establishes that n = m. 0

+

+

Lemma 13.7. Lctp. r E Z*andletm, n, m', n' E Z. Ifm+p = n.randm'+p = n'.r, then (m, n) = (m',n').

Pmof By Lemma 13.6, n = n'. It then follows from Leks 6.28 that m = m'. As we remarked at the beginning of this section, we are primarily concerned with two questions. namely whether there exist solutions to the equation q + p = s r in N' and whether there exist such solutions in Z*. Now. of course, an affirmative answer to the fint question implies an affirmative answer to the second. Further, the reader is asked to show in Exercise 13.1.1 that if p, r, s d Z' and there exists q E Z* such that q p = s .r, then there exists q' E Z* such that q' lpl = Is1 . Irl, where Ipl has its obvious meaning. Thus the questions are nearly the same. However, it is conceivable that one could have p, r, s E N* and q E -N* such that 9 p = s .r but not have any q' € W* such that q' p = s . r. Accordingly. we address both questions.

.

+

+

+

+

Lemma 13.8. Let p. q, r

E Z* and let

+

p = s - r. We note thaf, by Lemma 13.6, thereisatmostonea E Zf0rwhicha.r E BZ+p. Thus,if S = (n E Z : n.r f pZ+pJ. we have S E s. Suppose that H $ q and let B = Z\H E q. Since q p E ce(B p) ands .r E cC(S.r), it follows fmmTheorem3.40. thatm + p E BZ.r forsomem E B. or else n . r E pZ p for some n E S. Tbe first possibility can be ruled out because it implies that m E B n H. The second can be ruled out by our choice of S. Sufficiency. We note that there is at most one m E H for which m + p E Z . r (by Lemma 13.7). Let H' = (m l Z : m p E iZ' - r ) . Then H' E q and hence q p E c e p ( H ' p) E cepz(Zg .r ) = Z ' .r. 0

P m f Necessity. Picks E Z* such that q

+

+

+

+

+

+

Theorem 13.9. k t p, r E Z*. let

.

G = (q E Z*:thaeexistss € Z ' s u c h t h a t q + p = s

.r)

and let

H = { m EZ:thereexisfst ~ f i Z s u c h t h a t m + p = t . r l . Then thefollowing statemnfs om equivalent.

(a) G # 0. (b) IHI = o. (c) /GI = ZC.

(a) implies (b). Pick q E G. By Lemma 13.8. H E q so,since q E Z*, IHI = o. (b)implies (c). By Theorem 3.58. l(q E Z' : H E qJI = 2' so by Lemma 13.8, IGl = 2'.

P&

That (c) implies (a) is trivial

Theorem 13.10. Lct p. r

E N*. let

G=[q~N*:thmcxistss~N*suchthatq+p=s~r) and let

H = (m ~ N : f h e m e x i s t s tE fiNsuchrhatm + p = I . r ) . Then rhefollowing srafemenlsa n equivolent. (a) G

# 0.

(b) IHI = o. (c) IGI = 2'.

.

,

Pmof: (a) implies (b). Pick q E G. By Lemma 13.8, (m EZ:thereexistsr ~ f i Z s u c h t h a t m + p = t - r le q . Since also N E q, one has (m ~ M : t h e r e e x i s t s r ~ f i Z s u c b t h a t m + p = r . r ] ~ q . Now.givenm ~ N a n d ~t p Z s u c h t h a t m + p = t . r . a n e h a s m + p E ~ * , s o o n e a l s o must have t E fiN. (Otherwise t r E -N* U 101.) Thus H E q and therefore IHI = o. (b) implies (c). By Theorem 3.58. Hq E N* : H E q)l = 2'. Further,

.

H ~ [ m ~ Z : t h e r e e x i s t ~s ft i Z s u c h t h a t m + p = t . r ) . So by Lemma 13.8, if q E N* and H E q, then there is some s E Z' such that q + p = s . r . S i n c e r a n d q + p m i n N * , s ~ N * , s o h t q E G. 0 set

Ta put T h e o m s 13.9 and 13.10 in context. notice that we do not know whether the H e v n has two members.let alone infinitely many members. Notice also thaf if H,

p. and r m a s in Theorem 13.10. then H + p the existence of such H.

G fiW.

r. Theorem 13.12 characterizes

262

13 Multiple Shucnnes in BS

Ddinition l3.11. Let 0 # H

c N. Then

7~r [A g N : there exists F E [HICusuch that N = UaEF(-n +A)). C

Tbeonm 13.12 Lct H s N with IHI > 1. Then then &st p, r E N* such fhot ~+p~~N.rifrmdonlyifthe~uo&iceofx~ENforeochAET~such (XA-'A : A E TH) hos the infinitefinite intersection propeny. '

Prwf: Sufficiency. Pick by CmUary 3.14 r E N* such that {XA-'A : A E TH) E r. Recall that C(r) m {A s N : f o r d x E N, *.-'A E r ] and that, by Themem 6.18. &W.r = ( p ~ f i N : C ( r ) c p ) .Weclaimthat[-n+B: B ~ C ( r ) a a d nH ~) h a the finite. inhrsection properly. Since C(r) is a 6k.n it suffices to Iel B E C(r) and F E Pf(H) and show that (-n B) # 0. So s u p instead that (-n + B) = 0. Let A = N\B. Then A E THso XA-'A E r so B 6 C(r), aconh'adiction. B : B E C(r) and n E H ) 2 p. Then for each Pick p E fJN such that (-n n E H, C(r) g n p so by Themem 6.18. n p E BN - r. Since, by Comllary4.33, ,8R. r 5 PI*, it follows that p € W. Necessity. Pick p and r in W* such that H p E BN. r. Given A E TH,it suffices to show that there is some XA E such that XA-IA E r. SOk t A E TH and pick F E [H]<*such that N = UmEF(-n +A). Pick n E F such that -n + A € p and picks~~Nsuchthatn+p=s.r.~hen~€n+~=s.rso{x€N:x-'~€r)Es 0 so pick x~ E W such that X;'A E r.

nmeF +

+

neF

+

+ +

By Lemma 13.8 and Th13.10, ifthere exist p, q. r. ands in W with q + p = s.r,then{g E FP : q + p = s . r f o r s o m c p , s , r E N*)hasnommptyintcriorin KT. By way of conhast, the set of points occupying any of the other positions of such an equation must be topologically small.

+

P m j To see that P is nowhere dense in No, we show that P g Z N' . N*. (We have that N* W* is nowhere dense in N* by Theorem 6.35 and so Z + N* N* is nowhere dense in N* by Corollary 3.37.) Let p E P and pick q , s. and r in W such thatq+p=s.r.ByLemma13.8.(m~N:m+p~BW.r)~qandbyLemma 13.7 at most one m E W has rn p E N r. Thus p E Z W* N* as claimed. To see that S is nowhere dense in N'. suppose instead that t h m is an infinite subset B of N such that B* c c C p ~S and choose x E B*. Choose a sequence (b.)zl in B withtheproperrythath,(b,,) = h,(x)foreveryn E Wandeverym 6 11.2. ..., n). Let A = [b. :n E W). Then forevery s E A* and nmy m E N, h^-,(s)=&(XI.

.

.

+

-

+ .

.

P i c k s ~ A * n S p a d p i c k q , p , a n d r i n W w h t h a t q + p = s . r . ByLrmms 13.6. t h a is at IUOSt one n E N for which n .r € pN p. Lf such an n exists. let c ~ A \ [ n ) . a n d i f n o t l e t C = A . L e t D = [ m E N : m + p E Cg.r]. Weclaimthat ~ ~ q s o s u p p ~ x b s t e a d t h a t M \ DTehqe.n q + p E CFJ\D)+pands.r byTheorem3.40eithern-r E (N\D)+pforsomen E C ~ m + p = s ' . r f o r s o m e nr E N\D and some s' E C*. Both possibilities are excluded so D E q azalm i ed. Chaosem,kE Dwithkcm.~owh^-,isconsranton~*.r.soh~(m+~)=h,(k+p) and sa h, (m) = h. (k), a contradiction. To see that R is nowheredense in N* we show first that R g en-'[Z+W .W]. Let r E R and let q p = s - r, where q. p. s E N*. By Lemma 13.8, we can choose epN.Puning m <m'inNsuchthatm+p=x.randm'+p=y.rforex,y a = m l - m , w e h a v e a + x . r = y . r . N o w a + x . r ~ a + N . r a n d ~ . r ~ ~So. it follows from Theorem 3.40 that we may suppose that the equation a + x r = y . r holds with x E N M y E N. By Lemma 13.7. this equation cannot hold with x and y bothinN. ( L e t p = x . r . I f a + x . r = y . r . t h e n a + p = y . r a n d O + p = x . r . ) S o thereexistsn E Nf0rwhichn.r E Z + N * . N ' . T ~ ~ ~ R ~ U . ~ ~ L ; ' [ Z + M I . M I ] as claimed. As we have observed, Z W .MI is nowhm dense in W. Now if M is a nowhere d e n s subset of N* and n E N. then !;'[MI is also nowhere denx in N*. Otherwise, if (;'[MI were dense ia A* for some infinite subset A of N. M would be dense in f,[A*I = (n. A)'. So UnEN (7' E + N * . No] is nowhere dense in W. by Corollary 3.37.

+

+

.

+

0

The following result etablishes chat one cannat find p. q, r, and s in Z* with

q

+ p = s .r in the most familiar places.

PmaJ Suppose instead that q +p = s .r. By T h m m 13.9. pick m # n in Z such that m_+ p E BZ. r e n + p = BZ:r. Pick 4 s max[lml, In11 such that aZ E r. Then h.(r) = 0 and so h,(m + p) = h.(n + p ) = 0. This implies that h.(m) = h,(n), a

conhadiction.

Pmof: Suppose one has some r E K(BN, .) n K(pN, +). Pick a miminimal left ideal L of (BN. .) such that r E L. Then by Lemma 132. L = L r so r = s . r for some s E L C N*. Similarly r = q + r for some q € N'. But nEl5 is an ideal of m (BN, .). Thus K(BN. .) G and consequently r E r),=' nN. But then by Theorem 13.14. q r # s r, acontradiction. 0

+

.

nWm, a

-

By way of contrast with Corollary 13.15. we shall see in Cmllary 16.25 that KVN, .) nce K(BN, +) 0. We conclude this section with a result in a somewhat different vein. showing that a farain linear equation cannot be solved.

+

13 Multiple ShucnueJ in BS

264

Lemma 13.16.. Let u # v in Z\(Ol. There do nof exist p 5

n,,Naand q E No such 1

thnr uq+p=vq+p.

+

P m j Suppose we have such elements p and q and pick a prime r r luJ Ivl. F a i E (1.2,. ,r - 11, let Ci = [rn(rk i) : n, k E o ) . Then Ci is the set of x E N whose righhnost nonzero digit in the base r expansion is i. Pick i such that Ci E q. For each m E N. let f (m) denote the largest integer in o for which rftm)is a facta of m. Since rf")+'N 6 p. it foUows from 'heorem 4.15 that (um s : m Ci and s E rf('")+'N) E uq p and (vn t : n E Ci and t E rf("'+'N) E vq p. Thus thereexist m, n E C;, s E rf(")+'N,and r E rf(")+'N such that um + S = un + I . We observe that f (um + s) = f (m), because rf("') is a factor of um s, but rf('")+' is not. Similarly. f (vn + t ) = f (n). So f ( m ) = f (n) = k..say. We have m I irk (mod rk+') and n = irk (mod rk+'). So um s r uirk (mod rk+') and vn + t = virk (mod rk+'). This implies that u v (mod r), a canhadiction.

..

+

+

+

+

+

-

+

+

Theorem 13.17. k t u # v in Z\[O). There do nor exist p , q E N* such that uq+p=uq+p.

+ +

Pmoj Suppose we have such elements p and q. Pick r 6 K(BN, +). Then p r E K(BN. +) so by T h e m m 1.64 there is some maximal group G in K(6N. +) with p + r E G. Let e be the identity of G and let s be the (additive) inverse of p r in G. Then adding r s on the right in the equation uq E = uq p, one obtains uq + e = vq e.>s conmdicts Lemma 13.16, because h. (e) = 0 for every n E N. and so e E nN. 0

+

+

+

+

nnen

Exercise 13.1.1. Given p E BZ, define

Rove that if p, r, s E Z* and there exists q exists q' E Z* such that q' lpl = 1st. lrl.

+

€

Z* such that q

+ p = s .r , then then

13.2 The Distributive Laws in 8% WesawinLemma13.1thatifn ~ Z a n d p . q~BZ,thenn.(p+q)=n.p+n.q.This is the only nontrivial instance of the distributive law known to hold involving members We establish in this section that any other instances. if they exist, are rare indeed. of Z*.

Theorem 13.18. Let n E Z\{O] and let k, m E Z Thefollowing statements are equip aknt.

.

.

Pmof: (a) implies (b). Assume that we have p E 2'. such that m p + n . p = k p. As in the proof of Theorem 13.5 and Lemma 13.6, we can assume that p E W and n, k E N. We show first that k = n so supposejnstead-that k # n. Choose a real numberb > 1 forwhichnbtc korkb3 6 n. SoI#b(n)-#b(k)l ) 3. By-mnma13.4. E 6b(k)+ @b(p)+(O.1). &b(m.p++n.p) E 4b(n)+@b(p)+(-1.0,1.2)md@b(k.1) It follows from Lmuna 6.28, that I#&) #b(n)l 5 2. This contradiction shows that k=n. Suppose now that rn_f 0. Let r E N-k a prime sati-g r r Iml n. For any a E W, we have h+(m)hp(p) +hp(n)h,-(p) = h+(n)h,a(p). Sincer is not afactor of Iml, it followsthat h,n(p) = 0. That is, P N E p for each a e N. Forie[l,2 r-l],let

-

+

....,

m u s Ai is the set of positive integers whose rightmast nonzero digit jn the base r expansion is i.) Pick i E (1.2. .. . r 1) for which Ai E p. For each x E N,let f (x) denote the largest integu in o for which rf(') is a factor ofx. Wehave[mx+s : I EAiands ~rf"'+'N)€m.p+n.~(b~Thcaem4.15) and nAi E n . p. Thus there exist x, y E Ai and s E rf(')+'W for which mx s = ny. We note that f (rnx s) = f (x) and f (ny) = f (y). So f (x) = f ( y ) = t . say. NOW x t irc (mod r"') and y E ir' (mod rf+l). So mx s = mire (mod rt+') and ny Inir' (mod #+I). It follows that m I n (mod r ) and hence that m = n. (b)implies(a). Assumethat k = n. Ifm = O.thenrn.p+n.p = k.pforall p E @Z.

.-

+

+

+

Km=nmdp+p=p,thenbyLemma13.1,n.p+n.p=n.(p+p)=n.p.o Corollary W.19. Let p E N ' .

Then p # (-p)

+ p.

Pmof This is immediate from Theorem 13.18. Corollary 1320. Let p E Z* and kt n, m E Z\{O]. Then (n ondp.(n+m)#p.n+p.m.

0

+ m) .p # n .p + rn .p

P m j Since, by Theorem 4.23. Z is contained in the center of (pZ,.), it suffices to s h o w t h a t ( n + m ) . p # n . p + m . p . Supposeinsteadthat(n+m)-p = n . p + r n . p . 0 Then by Theorem 13.18.11+ m r rn. so n = 0. acontradiction. Theorem 13.18 has special consequencesfor the semigroup 8. Theorem 1331. Let q, r E H a n d let p E 2'. Then p . (q

+

.

+ r ) # p .q + p . r and

. ( p q) .r # p r + q .r. In particular there are no instances of the validity of either distributive law in W.

13 Multiple Stmclmcs in pS

266

+

. +

+

r E A. so by Theacm 13.14. p (q r ) g Z' Z*. Since (Z\(O). .)is cancellative. we have by CoroUaty 4.33 &at Z*is a subaemigroup

P m f By Lemma 6.8, q

o f ( ~ Z . . ) a n d s o p p q + p per E * + Z W . S i m i l a d y ( p + q ) . r f Z * + Z * a n d p . r + q . r ~ Z * f 2'.

0

We now set out b establish the topological rarity of instances of a distributive law in W, if inany such instancesexist at all.

Delhritlon 13.22. Let A s N. A is s h u b $ thin set if ind only if A is infinite and whenever(m. n) and(&,1) andistinctelementsofh'xo,onehas I(n+mA)n(l+kA)I c W.

Lemma 13.23. Let B be rm infiite subset of N. 7'hem is o doubly thin sef A

E B.

Pmqf Enumerate N x o as ((m(k), n(k)))zl and pick sl e B. IDductively k t k E N and assume that q.Q, . ,s k have been chosen. tick st+, z st such that

..

st .m(l)+n(l) -n(i) : i , I , t ~ ( 1 , ,..., 2 k)). Let A = {sk :k E h'). Tben A is infinite. Suppose that one has some I # i such thal

+

k x ( m O , n(1)) f (m(i), n(i)), thwc is at most one x E A such that n(l) m(l)x = n(i) +m(i)x. If there issuch an x = s., let b = max(u, i. I), and otherwise let b = maxfi, I). Pick z E (n(1) m(l)A) n (n(i) m(i)A) with z r n(l) m(l)sb and z > n(i) +m(i)sb and pick t, k E R such that z r n(1) +m(l)s, = n(i) +m(i)sk and notice that t > b and k > b. Then st # s t so assume without loss of generality that t < k. But then n ( 0 mWsr n(i) m(i) o a contradiction.

Now, s

+

+

*= +

+

-

Lnmnn 1324. Let A be a doubly thin subset of N. let p E A*. and let B 5 N x o. ~nce{n+m.p:(m,n)EB)nct(n+m-p:(m,n)E(h'xo)\B]=0. Pmof Supposethatq e cC[n+m.p : (m, n) E B)ncefn+m.p : (m. n) E (Nxo)\B). Order N x w by < in order type o.m a t is. < linearly orders El x o so that each element has only finitely many predecessors.) Let

-

Notice thac C f Dl = 0. Thus witbout loss of gemdity we have N\C E q. Pick (m,n) ~ B s u c h t h a t n + m . p e N \ C . Nown+mA € n + m . p . n + m . p ~ N * , and forevery (n', m') < (n,m). I(n' m'A) fl( n mA)1 c a,since A is doubly thin.

+ (n + mA) \ U(n,,n7<(m.n, (n' + m'A) E n + m .p +

mu3

We have one final preliminary.

.

Lemma 13.25. Lct A be a doubly thin subset of A and let p E A*. Then N. p. p, and N. (N. p) hnvc painvise disjoint clomres.

N*+ W'

.

. +

Pnwf: By Lemma 13.24, it suffices to obsmre that

The first observation is immediate. For the wondjust A c e that if q , r E W ,then

. +

For the last observation, let q , r E FP, let B E q (r p), and pick m E N such that m-'B E r + p. Then (n E N : -n m-'B E p) E r so pick n such that

+

-n+m-'B~~.~henm.n+m.~=m.(n+~)~B.

Theoreml3.26. ThesetIp E N* : N * . p , N . + W . p , painvise disjoint closures)has dense interior in N*.

0

andW.(FP+p)hmtc

Pnwf Let B be an infinite subset of N. F'ick by Lemma 13.23 a doubly thin subset A of B. Then apply Lemma 13.25. 0

Pmof: Given p, q, r

E A*.

so Theorem 13.26 applies.

13 Multiple S m d u m in BS

13.3 Ultrafilters on R Near 0 We have dealt throughout this book with the S t o d c h compactification of a discrete space. especially a discrcte semigroup. One may naturally w ~ d e why r we restrict our attention to discrete spaces, especially given that the StoncCech compactification exists for a much larger class of spaces. We shall see in Theorem 21.47 that for many nondiscrete semigmups S. including (R.+), the StonbCech compactification of S cannot be made into a semigroup compactification of S. Of course, if S is a topological semigroup, one can give the set S the discrete topology and p&eed as we always have. It would seem that by doing that one would lose any information about the original topological aemigmup. We shall see hawever in Section 17.5 that we can get information about partitions of the real interval (0, m). every member of which is measurable, or every member of which is a Baire set. by first giving (0, m) the discrete topology and passing to its ~tone4?echcompactification. Because of these interesting applications, we shall investigate the algebra of (B%, +) and (B&. .).where 4 is the set W with the discrete topology. We shall be primarily concerned with (5&, +).

Definition 13.28. (a) Given a topological space X. Xd j'S the set X endowed with the discrete topology. (b)Let u :B& + 1-w, m1be the continuous extension of the identity function. (c) B(W = { P a B% :d p ) $ {-w, mt). (c) Given x E W. x+={p€B@):cr(p)=~aod(x,m)€~}and

x- = {p a B(R) :a ( p ) = x and (-m, x ) a p). (d) U = UXpR x C and D = UrERX-. The set B(W) is the set of "bounded" ulmfilters on W. That is, an ultrahlter p a 8 4 is in B(R) if and only if there is some n E N with [-n, n] E p. We collect some routine information about the notions defined above.

Lemma 1339. (a) Let x E W and let p a

Then p E xC ifand only iffor every r z 0, ( x , x 6 ) a p. Also, p E X - ifand only iffo? evev 6 > 0, (x r, x) E p. (b) Lcr p, q a B(R), l e t x = a(p), and let y = a(q). I f p a x+. then p q e (X y)+. lfp E X - , then p + q E ( x y)-. In (c) B(W)\R = U U D and U and D a m disjohf right ideals of (Em), particular; B(W) is not commutative. (dl The set 0+ is a compact subsemigroup of ( B e ) ,+). (e)Ifx € W a n d p a B i l & , t h e n r + p = p + x . (0The set 0' is a nvo sided ideal ofthe semigmup (B(0, l)d. .).

+

+

-

+

+

+).

Pmof: The pmof of (a) is a routineexercise. (e) is a consequence of Theorem 4.23, and (c) and (d) follow from (b). We establish (b) and (0. To verify (b) assume first that p E x+. To see that p + q E (x + y)+. let r > 0 be given and let A = (x + y, x y 6). To see that A E p q. we show that

+ +

+

.-. ?

.:

13.3 Ultnfltcn on 1Near 0

269

(x,x+c) G IZ E R : -z+A €9). Soletz E (x,x+E). btts =min{z-X,X+E-z). eq Sincea(q) =y.wehave(y-6,y+6) ~ q a n d ( y - 6 . y + 6 ) S -z+Aso-z+A as required. The proof that p q E (x y)- if p E x- is nearly identical. Toveri&(0,letpEO+andletq €/3(o,l)d.To~eeLbtp.q€Of andq.pEOC, let c 7 0 be given. Since (0. c) (0.1) = (0, E). we have (0, c) t (x E (0, 1) : x-'(0, E) E 4) (SOthat ( 0 . ~ 1E p - q) and (0. I) G (x E (0.1) : x-'(O, 0 E p ) (SO that (0, E) E q . p). 0

+

+

.

We shall restrict our attention to 0+, and it is natural to ask why. On the one hand, it is a subsemigroup of (BRd. +) and for any otherx E R,x+ and x- are not semigroups. On the other hand we see in the following theorem that O+ holds all of the algebraic structure of (B(W). +) not already revealed by B.

Theorem 1330. (a) (O+, +) and (0-. +) a n isomorphic. (b) Thefunction ~p : IIb x ((0) U O+ U 0-) + B(B) defined by (p(x, p) = x + p is a continuous isomo~hismonto (B(R). +). Pmofl (a) Define r : O+ -t 0- by r(p) = -p, where -p = (-A : A E p). It is mutine to verify that r takes 0+ one-to-one onto 0-. Let p, q G O+. To see that r ( p + q ) = r(p) +r(q) itsuffices, since r(p+q) and r(p) r(q) are bothulkafilters, to show that r ( p + q ) 2 r(p) r(q). So let A E r ( p q). Then -A E p q so

+

+

+

+

B=(xEIR:-x+-AE~}E~

+

+

and hence - B E r(p). Then -B c {x G B : -x A E r(q)) so A E r(p) r(q) as required. (b) To see that (p is a homomorphism. let (x, p) and (y, q) be in & x ((0)uO+UO-). Since p + Y = Y p. we have (p(x, P) V(Y,4) = (P(X Y,P 4). To see that (p is one-to-one, assume we have (p(x. p) = ~ ( yq). , By Lemma 13.29b)

+

+

+

+

wehavex=rr(x+p)=u(y+q)=y.Thenx+p=x+qsop=-x+x+q=q. To see that ~p is onto B(W), let q E B(W) and let x = ~ ( 4 )Let . p = -x + q . Then

q=x+~=W,p). To see that p is continuous, let (x, p) E 4 x ( ( 0 )U OC U 0-) and let A E x p. Then -x A E p so (x) x (cCpb(-x + A ) ) is a neighborhood of (x, p) contained in

+

+

~p-'lctBWd A]. As we have remarked, OC has an interesting and useful multiplicative structure. But much is known of this shucture because. by Lemma 13.29(0. 0+ is a two sided ideal of (B(0, l)d. .),so K(p(0, I),+..) G 0+ and hence by Theorem 1.65. K(O+, .) = KWO, l)d, .I. On the other hand, OC is far from being an ideal of (B(S). +),so no general results apply to (O+, +) beyond those that apply to any compact right topological semigroup. We first give an easy characterization of idempotents in (O+. +).whose pmof we leave as an exercise.

Theorem 1331. Then exists p = p + p h O+ with A E p ifand only ifthere is some sequence ( x n ) z lin (0, 1) such thor x. c o n v c g e s a n d F S ( ( ~ , ) ~2~A. )

xEl

13 Multiple Smmna in BS

TI0

PI&

n

Thisis Exercise 13.3.1.

As a compact rigbt topo1ogical semigroup, 0+ has a smallest two sided ideal by

Theorcm 2.8. We now twn our attention to characterizing the smallest ideal of 0+ and its closure. If (S. is a disnete semigroup we know From Thcornn 4.39 tha~a point p E BS isinthesdlesti&alofBSifandonlyKforeachA E p.(x E S : - x + A E p ) is syndctic (and also if and only if for all q e BS, p E BS g p). We obtain in ' k o ~ m 13.33 a nearly identical charactcrizahon of K(O+, +).

+)

+ +

Delinition 1332. A subset B of (0.1) is syndetic -0 if and only if forevery c > 0 thcn exist some F E J',((O, c)) and some 6 > 0 such that (0.8) G UIeF(-r 8).

+

Tbconm 1333. La p E Of. Tlvfollowing starmrents om equiwht.

+

Pmt$ (a) implies (b). Let A E p, let B = {x E (0.1) : -x A E p). and suppose that B is not syndetic near 0. Pick c z- 0 such that for sll F E P,((O, a)) and all 8 > 0, (0, 8)\ U,,F(-r + B) 0. Let 9 = ((0. 8)\UCEF(-r B) : F E Pf((O, a)) and 8 > O\. Then g has the finite intersectionproperty so pick r € B(0. l)d with $ G r. S i ((0.8) :8 > 01 G r we have r E O+. Pick a minimal left idcal L of O+ with L E 0+ r, by Corollary 2.6. Since K(O+) is the union of all of the minimal right ideals of O+ by Theorem 2.8, pick a minimal right ideal R of Of with p E R. Then L nR is a groupby Thwrem 2.7, so let q be the identity ofLilR. T h e n R = q + O + , s o p ~ q + O + s o p = q + p s o B ~ q . Alsoq e O + + r s o pickw€O+suchthatq = w + r . Then(0,c)E wand(r E ( 0 . 1 ) : - r + B ~ r ) E w so pick t E (0, a) such that -1 B E r. But (0, I)\(-[ B) E $ S r, a contradiction. (b) implies (c). Let r E Of. For each A E p, let B(A) = {x E (0,l) :-x A E p) and let C(A) = (r E (0.1) : -1 B(A) E r). Observe that for any A], Az E p. one has B(AI n Az) = B(A1) fl B(A2) and C(A1 fl At) = C(Al) fl C(A2). We claim that for every A E p and every c z 0. C(A) fl (0, c) # 0. TO see this. let A E p and c > 0 be given and pick F E Pf((O. a)) and 8 > 0 such that (0.8) C UreF(-f B(A)). Since (0.8) E r we have UleF(-I B(A)) E r and B(A) E r. Then r E C(A) fl(0.r). hencethere is some r E F with -t Thus ((0. r) n C(A) : a > 0 and A E p ) has the finiteintersection pmpeny so pick q E B(O, l)d with ((0.6) flC(A) : s 0 and A E p ) G q. Then q E OC. We claim that p = q r p for which it suffices (since both are ulhdlten) to show that

+

+

+

+

+

+

+

+

+

+

+ +

p~q+r+p.~AEpbegiven.Then(tE(O,1):-t+B(A)Er)=C(A)Eq

soB(A)Eq+r,soA € q + r + p a s r e q u i d (c) implies (a). Pick r E K(O+).

If (S. +) is any d k e t c semipup, we lmow from Cnollary 4.41 +hat,given any discrete semigroup Sand any p E BS. p is in thec~OSM of the smallest ideal of @ Sif and ifeach A E p is piecewise syndetic. We obtain a similar result in O+. Modifying the definition of "piecewise syndetic" to apply to O+ is somewhat less stmightfonuard than was the case with "syndetic".

Dcenitiw 13.34. A subset A of (0.1) is piecmise syndctic near 0 if and only if there exist sequrnws

(F.)zland (6.)z1 such that

(1) for each n E N, F. E Pf (@. l/n)) and 6. E (0. lln), and (2) f a all G E Pf((O.l)) and all p > 0 there is some x E (0, p) such that for all x S Ute,(-r A). n E PI, (G n (0.6.))

+

+

Theorem 13.35. Lcf A E (0.1). 7 7 ~ K(O+) 1 n ctpp~.~),A # 0 ifand only if A is piecewise &etic near 0. Proof. Necessity. Pickp E ~(O+)flcC~(~,l), AandletB = (x E (0. 1) : -x+A E p]. By Theorem 13.33, B issyndeticnearo. Inductively forn E N pick F. E Pf((0.1 In)) and 8. E (O..l/n) (with 6. 5 &,-I if n > 1) such that (0.6.) G Ute,(-t B): Let G E Pf ((0, 1)) be given. IfGfl(O.61) = 0, the conclusion isaiviaL so assume G n ( 0 , 6 1 ) p 0 a n d l a H =Gn(0,6,). Fareachy E H,let

+

m(y) = max{n E N :y

.

4

6").

..

+ ,....

E (1.2,. m(y)}, we haw. y E UrgFa(-r B) so pick r ( y , n ) ~ F ~ s u c h t h M y ~ - t ~ , n ) + B . ' I h c n p i v r n y ~ H a n d n ~ { 1m(y)). .2 onehas-(r(y,n)+y)+A~p. Now let P > 0 be given. Then (0, p ) E p so pick

For each y E H and each n

x E (0. P)

nn ,

n:3)c-ctlv9

+ Y)+ A).

Thengivean ~ N a n d yE Gn(O.&).onehasy E Handn m(y)sorCy,n)+y+x E A so' -NY, n) + A G U,,F.(-' A). Y+x Sufficiency. Pick and (6.)zl satisfying (I) and (2) of Definition 13.34. Given G E Pf ((0.1)) and p z 0. let

(F.)zl

C(G, p) = lx E (0. a) :for all n E M. (G n (0.6,))

+

+ x c U,,,

(-1

+ A)).

By assumption each C(G. p) # 0. Further. given G I and Gz in q((0.1)) and

C ~ I ~2 . > 0. one has C(G1 U Gz. min(p1. p21) G C(GI. P I ) n C(G2, ~ 2 so)

W G . p ) :G E Pf((O.1)) and p > 01

hasthefiniteintmectionpropmysopickpE p(O,l)dwith{C(G. s): G E dp/((0.1)) and p > 0) g p. Note that since each C(G, p) G (0. p), one has p E 0+.

13 Multiple Strucnms in BS

272

Now we claim that for each n E N. O+ n E Nand let q E O+. To show that U,,F"(-t

+ p G clp
+

UceFa(-t + A ) . So let Y E (0, an). Then W Y ) . 8,) E P and W Y ) .8.) S -Y Now pick r E (O+ p) f l K(O+) (since OC p is a left ideal of O+). Given n E W. UrEFm(-t A ) 6 r SO pick I,, E F,,such that -f,, A E r. Now each . E (0, l / n ) so lim t,, = 0 so pick q E O+ r l c l p ~ , l ) ~ { t:. n E A). Then t. E F

+

+

+

+

n-m

q+r~K(O+)and(t.:n~N)E{t~(O.l):-t+A~r]soA€q+r.

0

+)

Since (O+, is a compact right topological semigroup, the closurc of any right ideal is again a right ideal. Consequently do+K(O+) = cLp(o.l), K(O+) is a right ideal of O+. On the other hand, if S is any discme semigroup, we know from Theorem 4.44 that the closurc of K(pS) is a two sided ideal of pS. We do not know whether' ct, K(o+) is a left ideal of 0+. but would conjefNIe that it is not.

Exercise 13.3.1. Rove Theorem 13.31. (Hint: Consider Lemma 5.1 1 and either proof of Theorem 5.8.)

13.4 The Left and Right Continuous Extensions of One Operation Throughout this book,we have taken the operation. on pS to be the operation making (BS. .) a right topological semigroup with S contained in its topological center. Of course, the extension can also be accomplished in the reverse order to that used in Section 4.1. That is, given x E Sand p E BS we can define p o x = p-lim y .x (so YES

thatifpES,thenpox=p~x)andgivmp.qE~S.

p o q = q - l i m p o x =q-lim(p-1imy.x). rsS

IES

YES

Then (pS, o) is acompact left topological semigroup and for each x E S,the function r, : @S+ BS defined by r,(p) = p o x iscontinuous. Funher, Tbeosems 2.7 through 2.1 1 apply to (BS, o) with "left" and "right" interchanged. Notice that, given p, q E BS and A G S.

~ ~ ~ o ~ i f a n d o n l ~ i f ( Ex p€) ~ E q:, ~wxh e- r~e ~ x - ' = { y € S : y x ~ A ] . Definition 13.36. Let (S, .)be asemigmup. Then o is theextension of. to @Swith the property that for each p E BS, the function t,, is continuous, and for each x E S, the function r, is continuous, where for q E BS. &(q) = p o q and ~ . ~ (= q )q o x .

1

!

1'

13.4 Ldl and Right Continuous Extensions

273

We see that if S is commutative, then there is no substantive difference between (BS, .)and (BS, 0). Theorem 13.37. Lef (S. .) be a commutative semigmup. Then for all p, q E BS. p o q = q .P. h particular. K W S , .) = K (BS, o) and, if H is a subser of BS, t k n H is a subsmigmup of (BS, -)i f and only if H is a subsemigmup of (BS. 0).

PmoJ This is Exercise 13.4.1.

0

We show now that ifSis notcommutative. then bothof the%panidar"conclusions may fail.

Theorem 1338. Let S be t h e f i e semigmup on countably many generators. Then there is a subset H of BS such that (1) H is a subsemigmup of (BS, .) and (2) for every P, q E BS. P 0 q (t H. Pmof Let the generators of S be (y. : n E N). For each k E N.let

nEl

cL Mk. and let H = To see that H is a subsemigroup of (6s..), let p, q € H and let k E N. We need to show that M, E p - q. To see this, we show that Mk G [ I E S : X - I Mk E q ] , so let x E Mk and pick 1 E N and n ( l ) , n(2). ...,n(1) E N such that k 5 d l ) < n(2) < . . < n o ) and1 = Y R ( I ) Y ~ ( ~.).. ~ n ( tThen ) Mm+r E q and Mnct)+~5 x - l ~ t . To verify conclusion (2). let p, q E BS. For each k E N.let Rk=(yk)USykUykSUSykS=(x ~ S : y k o c c u r s i n x ) . Then Rk is an ideal of S so by (the left-right switch of)Corollary 4.18, %is an ideal of (BS, 0). Since Rk n MI+! = 0 for each k. one has that p o q $ H if p E So assume that p $ U& We claim that MI p o q, so suppose instead that MI E p o q and let B = ( x E S : MIX-I E p). Then B E q so B # 0 so pick x E B and pick k E N such that yk occurs in x. Then MIX-'\ UfS, R, E p so pick z E MIX-!\ Rt and pick j E N such that y, occurs in z . Then j r k so rx $ Mr. a contradiction. 0

z.

z.

For the left-right switch of Theorem 13.38, see Exercise 13.4.2. We see now that we can have K O S , o ) # K(flS, .). Recall that after Definition 4.38 we noted that there are really two notions of "piecewise syndetic", namely the notion of rightpiecewise syndetic which was defined in Definition 4.38, and the notion of lefr piecewise syndetic. A subset A of the semigroup S is lefr piecewise syndelic if and only.if there is some G E P f ( S )such that for every F E P f ( S )there is some x E S with x - F E U,EG AI-'.

13 Multiple Srmetmes m BS

274

Pm$ LetthegmaatorsofSbeaandb. Foreachn E N.letW. = [x e S : l ( x ) I n ] . where I(x) is the length of x. Let A = UE, b" W.a. Suppose that A is right piecewise syndetic and pick G E Tf (0such that for every F E SPf(S)thereissomex E SwithF-x E UleG r - ' ~ . Letm = max[l(r) :r e G)+1 and let F = (am). Pick x E S such that a m x E UIeo;-'A aod pick r E G such that ramx E A. Pick n E W such that ramx E bnW.a. Then r = b" v for some v E S U (0)

+ + +

son 5 1(t). But now, the length of r#x is at least n m 1 while the length of any element of b" W,a is at most 2n 1. Sincc m ? l(r)+ 1 2 n 1, this is aconwdictioa To see that A is left piecnvise syndetic, let G = (a) and let F E SPf (S) be given. Pickn e NsuchthatF E W.andletx =b". T h c n x . F . a g b"W,a g A s o x F C Aa-' as required. 0

+

.

Fora eonIrast to Lemma 13.39. see Exercise 13.4.3

Tb-

13.40. Lcr S be rhcfrrc smigmup on rwo genemrors. Then

P m $ We establish the h t statement, leaving the second as an exenise. Pick by Lemma 13.39 a set A E S which is left piecewise syndetic but not right piecewise syndetic. Then by Thmnm4.40, ; ifl K(B.7, .) = 0 so ; in cf_K(BS, .) = 0. On the other hand. by the left-right switched version of 'Ihmrem 4.40, A n K (BS, o) # 0. We do not know ðer K (BS. o) n K(BS, .) = 0 for some semimp S or even whether this is m e for the free semigroup on two generators. The following theorem tells us that in any case the two smallest ideals arc not too far apaR

T h e o m l3.41. Let S be any semigmup. Then K(BS, o) n cf K(BS, .) # 0 and ce ~ ( p s . 0 n ) KVS. .) 0.

+

Pmofl We establish the &st statement only, the other pmof being nearly identical. Let p E K(BS. .) and let q 6 K(BS, 0). Then p o q l K(BS. 0). We show that p o q E ce K (BS, .I. Givens E S, the continuous Functions p, and r, agree on S, hmcc cn BS. so p o s = r,(p) = p,(p) = p .s. Thus

= eP(q-ris) SES

= q-limpos IES

= q-limp .s. SES

Exercise 13A.1. R m e Theorem 13.37.

1

Ex& 13.42. Let Sbe the free semigroup on countably many generators. Rwclhat -. mere is a subset H of BS such that

&,

Exerebe 13A3. Let S be any semigroup and assume that r E N and S = Ai. Prove that there is some i E (1.2,. .. r ) such that Ai is both left pisnvise syndetic and right piecewise syndetic. (Hit: Use Theorems 4.40 and 13.41.)

.

Notes Most of the materid in Section 13.1 is based on results from [124], except for Theorem 13.17 which is b d on arcsult of B. Balcarand P. Kal&k in (131 and on a personal communication from A. Malcki. In (1241 it was also pmved that the equation q p = q . p has no solutions in N*. That proof is quitecomplicated so is not included here. Theorem 13.18, ComUary 13.20, and Theorem 13.21 arc from [I381 and Corollary 13.19 is from [117]. Theorem 13.26 and Comllary 13.27 are due to E. van Douwen in (801. Most of the material in Section 13.3 is from [143]. results obtained in collaboration with 1. leader. Theorem 13.38 is from 1871. a result of collaboration with A. El-Mabhouh and I. Pym. Theorems 13.40 and 13.41 are due to P. Anthony in [2].

+

Part III Combinatorial Applications

Chapter 14

The Central Sets Theorem

In this chapter we derive tbe powerful Central Sets Theorem and some of its con* quences. Other consequences will be obtained in later chapters. (Recall that a subset A of a semigroup S is central if and only if A is a member of some minimal idempotent in PS.) The notion of "central" originated in topological dynamics. and its connection with concepts in this field will be discussed in Chapter 19. We shall introduce the ideas of the proof gently by proving progressively stronger theorems.

14.1 Van der Waerden's Theorem Om first intmduction to the idea of the proof of the Central Sets Theorem is via van der Waerden's Theorem Recall from Theorem 4.40 that given a subset A of a dismte semigroup S, 'ji fl K(fl.7) # 0 if and only if A is piecewise syndetic. Thus. in particular, any central set is piecewise syndetic.

T ~ ~ ~ I141. W I Lrr I A be a piecewise syndetic set in the aemigmup (N. +)and let C E N. Theree.rista,d E Nsuchthat[a,a+d,a +2d o + C d ) A.

,....

PmoJ Let Y = x:,@N. Thenby Th-m2.22, (Y. +)isacornpactright topological semigmup and if ?. E X :+,K then :A is continuous. Consequently Y is a semigroup t

compactification of X ,&'. Let

and let

19=((a.a+d,a+2d

,....a + C d ) : o , d € N ) .

~ eEt t cey Ea and let I = c t y I". Observe that Ea is a subsemigroup of X : ~ N and that I' is an ideal of E4. 'IhUs by Theorem 4.17 E is a subsemigroupof Y and I is an ideal of E.

280

14 The Cenwl Sets Thenem

By Themm 2.23, K(Y) = x:~K(&N). We now claim that E n K(Y) f 0 so that by Theorem 1.65, K(E) = E n K(Y). In fact we show that

.

.

To establish (*I, let U be a neighborhood of B and for t E [O, 1,. . , P ) pick At E p such that X:,Z G U. Then AI E p so Al # 0 so pick a c A,. Theh(a,a ,.... Q ) E U ~ E . Since K(E) = E flK(Y) = E n x : ~ K ( ~ ? N )and I is an ideal of E, we have t En X,,,K(BN) E I. Now since A is piecewise syndetic, pick by 'Iheorem4.40 some p c x n K ( B N ) a n d l e t j i = ( p,p,..., p ) . T h e n b y ( * ) F E E n ~ : ~ ~ ( f l m g ~ . Let V = X Then U is a neighborhood of so U n ! '# 5 so pick a , d E N suchthat(a,o+d,af Zd .....a + l d ) c U. 0

&,

n:=,,

fax.

?he abon proof already illustrates the most startling fact about the Central Sets Theorem proof, namely how much one gets for how little. That is, one establishes the statement (*) based on the trivial fact that E4 contains the diagonal of X:,~PI Then one concludes immediately that F E I and hence all members of p contain interesting configurations. It is enough to make someone raised on the work ethic feel guilty. Note that if r E N and N = U ,: Ai, then some Ai is piecewise syndetic. This can be seen in at least two ways. One can verify combinatorially that the union of two sets which are not piecewise syndetic is not piecewise syndetic. Somewhat easier h m our point of view is to note that BN = so some n K(BN) # 0 so Theorem 4.40 applies. C o d a r y 143 (van der Wanden [2381). Let r c N and let N = U;=, Ai. For ench L ~ N t h r r e e x i s t iE (1.2 ,....r ) a n d a , d ~ N s u c h t h a t ( a . a + ...., d o+Ld)gAi.

Pmqf Some Ai is piecewise syndetic so Theorem 14.1 applies.

0

In Exercise 5.1.1 the reader was asked to show that the above version of van der Waerden's Theorem implies the apparently smnger version wherein one i is chosen which "works" for all l . In fact the current method of proof yields the stronger version directly. ComUary 14.3 (van der Waerden 12381). Let r E N and let N = lJ:=l Ai. There . .. ,r ) such thar for each l E N there exists a. d E N such that (a.a+d a+Cd)EAi.

exists i E (I, 2.

.....

Pmof: Some Ai is picecwise syndetic so Theorem 14.1 applies.

0

One may make the aesthetic objectionto the proof of Theorem 14.1 thar one requires a different space Y for each length of arithmetic progression. In fact,one may do the proof once for all values of e by starting with an infinite product. (See Exercise 14.1.2.) We shall use such an approach in our proof of the Central Sets Theorem.

As a consequence of van der Waerden's Theorem, we establish the existence of a combinatorially intensting ideal of (BN, +) andof (BN, .),namely the set of ulhatiltm every member of which contains arbitrarily long arithmetic progressions.

Delinition 14.4. A 9 = ( p E /3N : for every A E p and every f E N, there exist n , d ~ N s u c h t h a t ( a , a + d , a + Z d....,a + L d ) & A ) .

Theorem 145. A 9 is an ideal of (BN. +) and of(pN. .). Pmof: By Theorems 4.40 and 14.1 K@N, +) E A 9 so A 9 # 0. To complete the p m f , l e t p € A P a n d l e t q € pN. GivenA E q + p a n d e E N,pickx E Nsuch that-x+A ~ pP .i c k a , d ~ N s u c h t h a t ( a , a + d , a + 2 d ,....a + e d ) c -x+A. T h e n ( x + a , x + a + d , x + a + Z d , ..., x + a + e d ) E A. GivenA ~ p + q a n d L ~ N , p i c k a , d ~ N s u c h t h a t ( a . a + d , a + Z d , a+!d) (x E N : - X + A E q). Then r):J-(a td) A) E q so pick x E nfd(-(a rd) A). Then ( x + a , x + a + d , x + a + Z d ,....x + a + L d ) c A . GivenA f q p a n d C ~ N , p i c k xE Nsuchthatx-'A E p. P i c k a . d ~ N s u c h t h a t (a, a+d. a + X . . a+ed) E x-'A. Then (xa.xa+xd, xa+2xd,. . ,xa+(xd) A G i v e n A ~p q a n d t ~ N . p i c k a , d~ N s u c h t h a t ( a , a + d , a + t d ,.... a+Cd) r d ) - l ~ E q so pick x E &(a td)-'A. (X E N : x-'A E qJ. Then &(a Then (xu, xa xd, xa 2xd. . .. ,xu lxd) c A.

....

+ +

c

+

+

..

+

.

+

+

c

+

+

Exerdse 14.1.1. Using compactness (see Section 5.5) and Corollary 14.2 prove that given any k, C E N t h m exists n E N such that whenever (1.2, ... n) is r-colored, there must be a length C monochrome arithmetic progression.

.

Exerdse 14.12. Prove T h m m 14.1 as follows. Let Y = X E o p N . Let Ea = ((a.a+d,a+Zd ,...) : a E W a n d d ~ N U ( O ) ] a n d l e t I "= ( ( a , a + d , a + 2 d ....) : a , d E N). Let E = c f y E4 and let I = cfu fa. Rove that E is a subsemigroup of Y and that I is an ideal of E. Then pmve

...).then 7 E E. Pick p E K (pN) nx,let p = (p. p, p. .. .). and show that p E I. Given L E N let U = [i E Y : for all i E (0.1,... el, qi E 7i)and show that U n I' p 0. (*) if p E K(pN) a n d 7 = (p, p, p,

.

14.2 The Hales-Jewett Theorem In this section we prove the powerful generalization of van der Waerden's theorem which is due to A. Hales and R. Jewett [113]. In this case it is worth noting that even though we work with a very noncommutative semigroup, namely the free semigroup on a finite alphabet. the proof is nearly identical to the proof of van der Waerden's .Theorem. (There is a much more significant contrast between the commutative and noncommutative versions of the Central Sets Theorem.)

-

282

14 TheCenaalSm'Ibemrn

Deilnhn 14.6. Let S be thc fne semigroup on the alphabet A and let u be a variable which is not a member of A. (a) A variable word w(v) is a member of the f m semigroup on the alphabet A U (v) in which v occurs. (b)S(v) = (w(v) : w(v) is a variable word). (c) Given a variable word w(v) and a mmbet a of A, w(a) is the word in which each occurrence of v is replaced by an Occumnce of a. Recallthat thefnesemigmupon thealphabet A wasfomallydefined(Ddinition 1.3)

.

as the set of functions with domain (0.1,. .. n ] for some n E w and range contained in A. Given a member w of the fm semigroup on an alphabet A and a 'letter" a E A. when we say that a occurs in w we mean formally that a is in the range of (the function) w. Given w(v) E S(v), and a E A, one then has fomrally that.w(a) is the function with the same domain as w(v) such that for i in this domain.

InfomaUy,ifA = (1.2.3,4.5)andw(v) = 13vv3251~4,thenw(3) = 1333325134 and w(2) = 1322325124.

Theorem 147. Let A be afulite nonen~pryalphabet. let S be the free semigmup over A, and let B be a piecewise syndetic subset of S. Thm there is a variablc word w(v) such that (w(a) :a E A}

B.

..

t

Proof: Let L = IAl and write A = (a~.az.. .at). Let Y = X,,lfiS. Denote. the operation of Y (as well as that of S) by juxtaposition. Then by Theorem 2.22. Y is a compact right topologifal semigroup and if ?. E X :,,s, then A? is continuous. l b u s t Y is a semigroup compactificationof X S. Let I" = ((w(al), w(ad.. , ad) : W(V)E S(v)) andlet

..

,=,

~ eEt = r e y E* let I = CCY I-. Ob-e that Ea is a subsemigroup of X:=~S and that I" is an ideal of E*. Thus by Theorem 4.17 E is a subsemigroup of Y and I is an ideal of E. By Theorem2.23 wehave K(Y) = x:=~ KUS). Wenow claimthat EnK(Y) # 0 so that by Theorem 1.65. K(E) = E fl K(Y). In fact we show that

.

To establish (+), let U be a neighborhood of and for r E (1.2,. .,C} pick B, E p f suchthat X,=,Bt E U. Thenn:=IBt E ~ S O ~ : , ~ B~I ~ s o p i Ec ~~: =~, Bw, . Then(w.w ,...,w ) s U n E . ) I is an ideal of E, we have Since K(E) = E n K(Y) = E n x ~ ~ K ( @ sand E n x : ~K(@S) , E I. Now since B is piecewise syndetic, pick by Theo1em4.40 sow

-

14.3 The Cnnmutativc Cenaal Sea lheorrrn

283

C o d u g 14.8 (Hales-Jewett [1131). Let A be a h i r e nonempty alphabet. let S be the jkesemigmupovrrA,letr ~ N a n d l e t S = U ~ = , B Z'henthereuisti i. E (1.2. ....r ) anda variable word w(v) w h rhat (w(a) :a E A ) E Bi.

Pmof: Some Bi is piecewise syndetic so Tbeonm 14.7 applies.

Exercise 14.2.l. Using compactness (see Section 55) and Corollary 14.8 prove chat given any finite mnempty alphabet A and any r E N there is some n e N such that whenever the length n words over A an r-colored there must be a variable word w(v) such that (w(a) :a E A ) is monochmme. Exerdsc 142.2. Dcrivc van dcr Waden's Theonm (Corollary 14.2) from the HalesJewen Thwrem (Corollary 14.8). (Hint: Consider for example the alphabet {O,l, 2, 3.4) and the following length 5 arithmetic progression written in base 5: 20100010134, 20111011134,20122012134,20133013134,20144014134.)

14.3 The Commutative Central S& Theorem In this section we derive the Central Sets Theorem for commutative semigroups, as well as some of its immediate corollaries. (One is Corollary 14.13 and two more are exercises.) We fust encode some of the important ideas that underlie the proof of the Central Sets Theorem, some of which we have illustrated in the proofs of van der Waerden's Theorem and the Hales-Jnuen Theorem. The lemma is more general than needed here because we shall use it again in later chapters. (We remind the reader that we have been using the notation P for a function constmtly equal to 1.)

Lemma 14.9. Let I be a set, ler (D. 5 )be a directed set, and let (S, .) be a semigroup. y of S such that for each i E D Let (Ti)i,o be a decreasing family of ~ n e m p l subsets and each x E 'I;. there b some j E D such rhar x Ti G Ti. Let Q = ctsS Ti. Then Q is a compacr subsemigroup of BS. Let (Ei)ico and (Ii)iEDbe decreasing families of nonempry subsets of X with thefollowing pmperties:

.

nieD

,

(a) foreachi E D. Ii C_ Ei E X,,j'I;.. (b) foreachi ~ D a n d e a c h ~ ~ l i t h e r e ~ t s j ~ D s u c h r h a It ii -, Ed j E (c) for each i E D and each . i E Ei\ It there exists j E D such that I E, G Ei and 2 . I, G Ii.

.

Lcr Y = XrejBS, let E = nieDc&Ei. and let I = n i E D c L yIi. Then E is a subsemigmup of X Q and I is an ideal of E. Y: in oddition, either

,

14 'Ihc Central Set8 'Iheorem

284

(d) for each i E D, Ti = Sand (a E S :ii f Ei) is not piecewke syndetic, or (e) f o r e a c h i ~ D a n d e a c h a E T , . i iEi, ~

thmgivmanyp~K(Q),onehns7€EnK(X~~~Q)=K(E)~I. Proof: It foUows from Thwrrm 4.20 that Q is a subsrrnigroupof @S. By condition (a) and the fact that ( l i ) ; , ~is decreasing we have 0 # I G E X,,J Q. To complete the proof that E is a subsemigroup of X l s l Q and I is an ideal of E, we let p', ? E E and show that q' E E and if either E I or E I, then p' .? E I. To this end, let U be an open neighborhood of p' .? and let i E D be given. We show that U n Ei # 0 and if if.E 1 or ? E I. then U nli # 0. Pick a neighborhood Vofp'suchthat V.? G Uandpicki E E i n V w i t h i E I i i f i f E I. I f i E I; pick j E D such that 2 . Ej E Ii. If i E Ei\Ij, pick j E D such that 2 . E, g Ei and 2.1, SIi. Now?.? E Usopickaneighborhood Wofq'suchthati. W G Uand pick?€ W n E j w i t h ? ~ I j i f $ E I . T h e n 2 . y ' ~U n E i a n d i f e i t h e r p ' ~I o r E 1 , t h e n I . ~E U n I ; . To complete the proof. assume that (d) or (e) holds. It suffic ' s to establish

c.

i

(*) if p E K(Q), thenF E E.

,,

Indeed, assumewehaveestablished(*).Then 7 E E n X ,,J K(Q) and X K(Q) = K(X,,J Q) by Theorem 2.23. Then by Theorem 1.65. K(E) = E n K(X,,J Q)and, since I is an ideal of E, K(E) g I. To establish (*),let p E K(Q) be given. To see that P E E. let i E D be given and let U be a neighborhood of F. Pick F E Pf((J and for each t E F pick some At E p ir;'[clas A,] G U. such that Assume now that (d) holds. Since p E K ( @ S ) and (a E S : ii f Ei) is not piecewise syndetic, we have by Theorem 4.40 that (a E S : ii f Ei ] p and hence ( a € S : i i E E i ) ~ pP. i c k a ~ ( n , , ~ A , ) n { a ~ S :Ei; i) ~ .Thenii~UnEi. F i l y assume that (e) holds. Since p E c t 6, pick a E A,) n f i . Then 0 ii€Un&.

n,,,

(nlcF

Definition 14.10. 9 is the set of all functions f : N + N for which f (n) 5 n for all n E N.

We are now ready to prove the C e n t d Sets Theorem (for commutative semigroups). 'lleorem 14.11 (Central Sets Theorem). Let S be a commutative semigmup. let A be a central subset of S, andfor each e € N. let (yt,. be a sequence in S. There &sra sequence ( a n ) g lin S andosequence (H,)Z, in Pf(N) such that max H. c min H.+I

)Zl

for each n

E

N andsuch thatfor each f

E

9. FP((an

.n,,~,yf(,,).~)~=~) E A. m

P m f Let Y = xZIpS. By Thmm2.22. Y isacompactrighttopoIogical semigroup and

X Z ~h (S~ ) .~~ i v e n and^ a ~~f(N)let

7

?

~.

( ~ ~ t h a t l ( a X, ~~ )~~S E Y .) . Foreachi ~ N , l e t T i=S.letIi = ( x ' ( a , H ) : a € S , H € P f ( N ) , a n d m i n H ~ i ) . and let Ei = Ii U {ii : a E S). We claim that Ti, Ei. and Ii satisfy hypotheses (a), (b). (c), and (d) of Lrmma 14.9. Indeed. (a) and (d) hold trivially. To verify (b), let i ~ N a n d ?E Ii begiven. K c k a E S a n d H ~ P ~ ( O s u c h t h a t r n i n2H iand i = ;(a, H). Let j = max H 1. Then. since S is commutative, one sees easily that i E j E I;. To verify (c). let i E N and x' E Ei\Ii be given. Pick a E S such that

+

.

I=Z.Tbeni.E;gEiandi.IigIi. Since A is central in S, pick an idempotent p E K (BS) such that A E p. By Lemma 14.9E ~ I = n,,,.ey I ~ . We now construct the sequences ( a , ) z l and (H,,)z, inductively. The construction is reminiscent of the proof of Theorem 5.8: Let A' = A'(p) = ( x E A : X - I A E p) and let U = {p' E Y : qql G F). By Theorem 4.12 A* E p so U is a neighborhood of j7. Since also E I, pick i E U f l I1. Pick a1 E Sand HI P f ( O such that i = x'(al, HI). Since ;(a,, Hi) E U we have YI., E A'. that01 Inductively, let n r 1 and assume that we have chosen (am)z:ll and (H),:;~ so t h a i f m ~(1,2..;.,n--Z).thenmaxH~ <minHm+t.andsothatforany f E rD, 1. Let FP((am. ~ , S H . ~ f ( ~n-l) . ~E )A*. ~ = Let~k)= max

.n ~ ,

+

nb,,

By Lemma4.14 we have for each b E L that b-I A* E p. Let B = A' n b-'A'. Since L is finite ((although 0 isn't) we have that B E p. Let U = [p' E Y : for d l t! E (1,2... .,n), qt E @. Then U is a neighborhood of j7 so pick i E U n Ix. Pick a. E S and H, E PJ(N) with min H, 2 k such that Z = ?(an. H.). To complete the proof we need to show that for any f E rD.

n,,~, n1.&

so assume that n E F. Now a. . YJ(n),l E B since ?(a,, H,,)= i E U. If F = In), then we have n m E ~ ( a.m ~f(~)= . , )a, . yfc.,., E B E A'. Otherwise, let G = F\(n) and let b = nmec(am nreHm~f(,,,).~).Then b E L so ~, = b . a n . n t s ~ .Yf@)., E an.ntef, Yf(").t E B E b - l ~ * s o n m c ~ ( a m . n r s~/(n).l)

.

nltg

A*.

0

Of course we have the conrsponding panition result.

Corollary 14.11 Let S be a commutativesemigroupandfor each t E N. let (~t..):=~ be a sequence in S. k t r E Nand let S = U ,: Ai. There uist i E (1.2. ... r ) .a sequence (a.)zl in Sandasequence (H.)zI in Pf[(Nsuch thatmax H. c min H.+I m foreach n E Nandforeach f E rD, FP((an. yj(,,).~),,=~) E Ai.

.

nre~.

286

14 TbcCenbalSetsTknmn

PmqJ Some At is cenaal.

0

A special ~ a 3 eof the Cenhal Sets T h e m is adequate for many applications. S* Exncise 14.3.2. It is not surprising, since the pmof rws an idempoteat that one gets the Finik Roducts T h m (ComUary 5.9) as a cmUary to the Cenhal Sets Theorem (in the case S is commutative). For a still simple. but more interesting, corollary consider the followiog extension of van &r Wacrden's Theorem, in which the increment can k chosen from the finite sums of any prespecified sequence.

*

u=,

.

Corollary 14.13. L t (x,,)zl be a sequence in N,let r E N, and let N = A,. Then then is some i E (1,2, ... r ) such that for d L E N them misf a E N and d ~ F S ( ( x , ) z ~ ) w i & ( a , a + d , a +,..., 2 d a+fd)EA;.

.

..

r ) such that Ai is cenhal in (o.+). For each k, n E N. let yt,. = (k 1) x,. Pick q u e n c t s (a,)z, and (H,,)zl as guaranteed by Theorem 14.11. Given C E N, pick any m s L and let a = n. and let d = &H, x,. Now,given k E (0.1.. .. ,C), pick any f E O such that f (m) = k I. Then

Pm$ Pick i E {I, 2..

- .

+

Ai. Exercise 14.3.1. Let ( x m ) z 1be a sequence h N, let r E N. and let N = Prove that there is some i E (1.2, ,r ) such that for all L E N thue exist a E N and d m((-G)El) and c € FS((xmz)zl)with {a.a d , a +2d. ..., n + ed) u (a + c , a + 2 c ,.... a + C c ) c A i .

...

+

.

Exercise 143.2. Let S be a commutative semigroup, let A be cenhal in S,let k E N and for each L E (1,2, ... k). let (yt,JZ, be a sequence in S. Rove (as a cmUary to Themern 14.11) that there exist a sequence (a.)zl and for each C E (1.2. ,k) a product subsystem (zl..)zl of (yt,.)z, such that faeach L E {I, 2,. ,k), FP((a.. zt..)gl) 2 A. Show in fact that the product subsystems can be chosen in a uniform . < mis H.+I for fashion. That is, there is a sequence ( H , ) z l in Pf(N) with max H each n such that for each L E (1.2. . k ) and each n E N,zc.,, = &H. ye.,.

..

...

.. .

14.4 The Noncommutative Central Sets Theorem In this section we generalize the Central Sets Theorem (Theorem 14.1 1) to arbinary semigroups. The statementof the Central SetsTh-in this generality isconsiderably more complicated because the "a" in the wnclusion must be split up into many paw. (In the event that the s e m i p u p S is commutative, the conclusion of Theorem 14.15 reduces to that of Theorem 14.11.y

.

~finilton1414 L a m E N. Then

Theorem 14.15. Let S be a smigmup, let A be a cmml subsn of S, and for each I E W,let be a sequence in S. Given t ,m E N,ii E Sm+', mrd 2 E I,, let

nyzl

(Obsem that if S is eommutativc. a = ai. H = UrmlHi. and .?(a, H) is as defined in the proof of Theorem 14.11, then ?(a. H_) = ?(ii. H).) For each-i E N, let 4 = S, let li = (?(ii, H) : there exists m E N such that . ~ ~ E S " ' + ~~ ,l H , , a n d m i n H ~2 _ i ) a n d I e t E i = I i U ( Z : a ~ S ) WeclaimthatT,. Ei, and Ii satisfy hypotheses (a), (b), (c), and (d) of Lemma 14.9. Indad. (a) and (d) hold wivially. To verify @), let i E N a*.? E 5 be given. Pick m G N, ii E Sm+', and 2 E 1. suchthatminH1 z i a n d Z = i ( i i , H ) . Letj=rnPxHm+1.Toscethati.Ej cIj. let j E E,. If y' = F for some s E S, let g = ( a ~ , ~ z , .am. . a,+!. s). Then i j = I(& 2) E li. SOassume j E Ij, pick k E N. b E St+'. and G E It such am.a,+l -bt.-bz bk+~) thatmin91 z j aady = ? & & L e t t = (a1.a~ Hm.G I , Gz, Gt). Then t E Sk+m+l,K E 1m h t and and let K = (HI, Hz.. i . j = i(t. E ti. To verify (c), let i E N and x' E Ei\li be given. Pick a E S such that x' = Z. Then as above one easily sees that i .Ei G Ei and x' li G Ii. Since A is central in S, pick an idempotent p E K (0s)such that A E p. By Lemma y 14.9p E I = n i G N c lli. inductively. Now we construct the sequences (m(n))zl, ( & ) z l , and Write A* = A*(p). Let U = [q' E Y : ql E El. By Theorem 4.12 A' E p so U is a neighborhood of F. Since also j5 E I, pic! j E U n 11._Pick m(1) E N, & E s'"(')+', and HI E such that y' = 36.HI). Since i(&, HI) E U we have that w(6. dl, 1) E A*. Inductively. let n > I and assume that we have chosen (m(j)):::. ( ~ ) ~ : ~and .

..

.

a)

.

..

,....

....

.

,...,

14 me Cmtd Sea Theorem

288

+

Let k = maxH.-~.,(.-t) 1. Let E = Ufeom(4w(4.Gi, f~i)));:). By Lemma 4.14 we have for each b E E that b - ' ~ ' E p. Let B = A' n b-!~'. Since E is fmite we have that B E p. Let U = (G E Y : for all L E (1,2,. . ,n). qc E B. Then U is a neighborhood of P so pick y' E U n Ik. Pick m(n) E N. & E Sm(n)+land & E I,(.) with min H.,t z k such that j = :(a'.. e n ) . To complete the proof we need to show that for any f E @.

nbGE .

z.,,

so assume that n E F. Now w(&, f (n)) E B since ?(&.&) = y' E U. If F = b),then we have n j Ew(4. ~ fij, f (j)) = w(&, fin, f (n)) E B E A'. Othrrwise, IetG = F\{nlandletb= ni,cw(4.Ej/, f(j)). Thenb E E~~W(~,I?,,,f(n)) E B G b-'~'so ~ ( 4zi, , f ti)) = b w(&. f (n)) E A*. 0

.

nj#

6.

Exercise 14.4.1. Derive the Hales-Jewett Theorem (Theorem 14.7) as a corollary to V' the noncommutative Central Sets Theorem (Theorem 14.15).

145 A Combinatorial Characterization of Central Sets Recall fmm Theorem 5.12 that members of idempotents h,9S are completely characterized by the fact that they containFP((x.)~,) for some sequence (x.)z,. Accordingly, it is natural to ask whether the Cenkal Sets Theorem characterizesmembm of minimal idempotents, that is, central sets. We see now that it dots not.

Lemma 14.16. Lcr n.m,k E Nand for each i E {1,2, .... n). let (yi.,)z, be a sequence in N Then there exists H E Pf (N) with min H r m such that for each E WZk. i E (1.2. .. ..n). &.If~i.r P~tmf: Chooseaninfiniteset G I c Nsuchthatforallt.s E GI. y1.1 I y t , (mod zk). Inductively, given i E {I, 2, . .. n 1) and Gi, chwse an infinite subset Gi+l of Gi

.-

.

such that for all t, s E Gi+l. yi+l., i yi+l., (mod 2'). Then for all i E (I. 2. ... n) and all t . s E G, one has yi., I yi., (mod Zk). Now pick H 2 G, with min H > m and I H I= zk. 0 Recall from Definition 6.2 that given n E N, supp(n) E P f ( o ) is defined by n = C~G,.~~ 2'.( . I

Lemma 14.17. There is a set A E W such that (a) A is notpiecewise syndetic in (PI, +).

5 "

P m o f : F o r e a c h k ~ N I c t B t = ( 2 ~ . 2 ~ + 1 , 2 ~. +. .2. ,2 k + 1 - I ) a n d l e t ~ = ( n ~ N : for each k E N. Bk\ supp(n) # 0). Tben one racognivcs that n E A by looking at the binary expansion of n and noting that then is at least one 0 between positions 2' and 2'+' for each k E M. To show that A is not pikewise syndetic we need to show that for each g E N t h m issomeb E Nsuchthatforanyx ~ N t h e r e i s s o m e yE ( x + l , x + 2 , ..., x + b ) with(y+l,y+Z. . . . . y + g ) n A =O.Tothismdletg EPdbegivenandpickk~M such that 2* > g. Let b = z2'+'. Let x E M be given and pick the least a E N such thata.2*+'-2* s x a n d l e t y =a-22t+'-221.~henx < y Sx+bandforeach ~ + Zk+I 2 - 1) g supp(y+t)so t E 11.2,.... z2' 1)onehas ( ~ ~ , 2 ~ + 1 . 2,.... y+tg!A. To verify conclusion (b), let r E A and pick k E N such that z2'-l > x. Let n = 2'. ThenO#AnNZnE-x+A. Fhlly. verify (c) letn E N and let sequences ( Y I J ) ~(n.r),,l,. ~ . m .. (yn.r)zl be given. We first observe that by Lemma 14.16 we can choose H E Pf(W) such that for each i E (1.2.. . ..).X l e yi., ~ E N2"+I. Next we observe that given any zl. 22.. .. zn in N and any k with Zk > n, then exists r E Bk such that Bx\ supp(2' 2;) # 0 for each i E (1.2, .. n). Indeed, if r E Bk and Bk G supp(Zr z) then supp(z) Tr Bk = Bk\(r). Consequently I(r E Bt : there is some i E (1.2,. .. .n) with Bk E supp(2' zi))l 5 n. For i E (1.2.. ... n). let z0.i = &H yi.,. Pick the least L such that 2' s n. Now given i we have 2"+Ilzo.i and 2'-' c n 1 so 2'-' E Be-'\ supp(z0.i). Pick ro E Be such that Bt\supp(Z'Q z0.i) # 0 for each i E (1.2. . , n ) and let 21.; = zo,i 2Q. Inductively choose r j E such that supp(ZrJ 2j.i) f 0 for each i E (1.2.. . . n) and let z,+l.i = z,,; 2'j. Continue the induction until t + j = k Z r eyi., ~ foreach i E (1.2,. n) andleta = 2" +Zrl +. .+Zrt-'. where z2'

-

.

+

+

.

+

+

+

+

.

..

+

.

..

.. +

.

Theorem 14.18. There is aset A g N such that A satisfiesthe conclusion of the Centml Sets Zkorem (Theorem 14.1 1)for (N. +) bur K(j3M. +) n cC A = 0. Pmof Let A be as in Lemma 14.17. By conclusion (a) of Lemma 14.17 and Theorem 4.40 we have that K(j3N. +) n cl A = 0. For each e E N, let ( ~ t . , , ) ~ ,be ~ a sequence in N. Choose by condition (c) of Lemma 14.17 some a1 and H I with a [ & r ~ y1.1 I E A Inductively, given a,, and H., let L = max H. and pick m > n in N such that for all i E (1.2,. ,n). a. &H. yi.~< z2-. Pick by condition @)of Lemma 14.17 (applied to the sequences m ( ~ ~ . e + t )(~2,t+r),,~. z~. ... ( ~ ~ + ~ . t + ~ )sot mthat , ~ min . H.+I z e = max H.) some i (1.2 .....n + 1 ) , a . + 1 + Z ~ ~ ~ .y+i ~ . ~E A ~ N P . a.+l and^,+^ s u c h t h a t f o r a ~ E

+

+

.

..

We n m proacd toderive a combhatorial&mc&imtion of central sets. This characterization involvr~the following generalization of the notion of a pieawise syndetic set.

DeRnitlon 1419. Let (S,-) be a semigroup and let A E P(S). Thm A is c o l k c ~ . wisepiecewise syndetic if and only if there exist functions G : Pf ( A )+ P f (S)and x : Pf ( A )x Pf ( S ) -+ S such that for all F G 9fPf.S) and all 3 and X in Pf (A)with 3 E 3one has F .X ( X ,F ) E U I G C ( F , t - l ( n3). Note that a subxt A is piecewise syndetic if and only if ( A ) is collationwise p i m i x syndetic. An alternate characterization is: A is coUcctionwise piecewise syndetic if and only if there exists a function G : P f ( A )-+ P f ( S ) such that

has the finite intRsection property.

ere

Y - ~ ( G ( P ) ) - ' (3) ~ = u , , c , , , y - l t - l ( n ~ ) , sa Exercise 14.5.1.) j/ In the event that the family and the semigroup are both countable. we have a considerably simpler characterization of collectionwise piecewise syndetic.

Lemma 14.20. Assume the semigmup S = (rr, : n

E N) and (A. : n E N ) g P(S). Thefmnily [A. :n E N ) is colleciionwisepiecewisesyndetic ifand only i f t h &a sequence ( g ( n ) ) z lin Nand a sequence ( y n ) z lin S sueh thatfor dl n , m E N with m p n. I(") -I hy., azym,. .. am~mlE Uj=1 aj Ai).

.

(n:*'

Pmof: For the necessity, pick functions G and x as guaranteed by Definition 14.19. Given n E N,let

mdy, = ~ ( ( A Az.. I,

..,A"), (a~.az.....an)). Tben if m 2 n, letting

For the sufficiency, let ( g ( n ) ) z land ( y . ) z , be as in the statemem of the lemma. Given 3 E * ( [ A n : n E N)). let rn = max(k E N : At E 3)and let G ( 3 ) =

T b e o ~ m14.21. Let (S. .) be an inhire srmigmup and kt A C P(S). Them aisu p E K (PS) with A p ifand only ifA is coUectionwiscpiecewisesyndefic.

Sufficiency. PickG :Pf(A) + Pf(S)andx : P f ( A ) x g ( S ) by the definition. For each 3 E 9f((A and each Y E S, let

-+

Sa~gmlced

Then (D(3. y) : 3 E +(A) and y E S) has the finite intasection propmy so pick u E 8.9 such that (D(3. y) : 3 E Pf (A) and y E S) G u. We claim that

To!hisend,let3

E

Pf(A)andkty

€

S. ThenD(3.y)

U,Eccr,f-'(nO so U t e o ( ~ , t - ' ( n 3 )E Y .

G

(x : y . x E

=roqni=d.

By (*) we have that for each 3 E Pf(A). 0s. u G u,,o(p,t-'(nF). Since PS - u is a left ideal of PS, pick a minimal left ideal L of PS such that L c /3S .u and pick q E L. Then for each JC € Pf(A) we have UrEG(mt-'( X) E q so pick r(X) E G(X) such that ( I ( x ) ) - ' ( ~ x ) E 4. .

n

292

14 The Central Sets Theorem

For each 3 E Pf(A). let E ( 3 ) = M X ) : X E PfW and 9 G XI.nK. ( E ( 3 ) : F E Pf (A)] has the finite intersection propmy. So pick w E pS such tha ( E ( F ) : F E P ~ ( A ) } C W . L ~ ( ~ = ~ . ~ . T ~ ~ ~ € L G K1 ( A g p, let A E A. Since E((A)) E w , it suffices to show that E({A)) {r E S : r - ' ~ E q]. so let 3f E P f ( A ) with (A] G 3f. Then (t(X))-'(nd() E q and Ourcombinatorialcharacterization of central is basedon an analysis of the proofs of Theore.mS.8. The imponant thing to notice about these proofs is that when one chooses x, one in fact has a large number of choices. That is. one can draw a tree, branching infinitely often at each node, sothat any path through that trec yields a sequence ( x , ) z t withFP((x.)zl) G A. (Recall that in FP((x,,)El), theproducts are takenin increasing order of indices.) We formalize the notion of 'Wbelow. We recall that each ordinal is the set of its prcderrssors. (So 3 = (0,1,2) and 0 = 0 and, iff is the function ((0,3), ( I , 5 ) , (2,9), ~ I(0,3). (1.5). (2.9)t.) (3,7), (4,5)t, then f i =

DeAnition 14.22. T is a tree in A if and only if T is a set of functions and for each f E T, domain(f ) E o and range(f ) G A and if domain(f) = n > 0, then E T. Tisatreeifandonlyifforsome A. Tisatreein A. The last requirement in the definition is not essential. Any set of functions with domains in w can be converted to a tree by adding in all reshictions to initial segments. We include the requirement in the definition for aesthetic reasons - it is not nice for branches at some late level to appear from nowhere.

Ddinitlon 14.23. (a) Let f be a function with domain(f) = n E w and let x be given. Then f-x = f U ((n, x)). @)GivenaaeeTandf € T , B f = B f ( T ) = ( x : f-XET]. (c) Let (S, .) be a semigroup and let A E S. Then T is a *-me in A if and only if T is a tree in A and for f E and n E B/. B/-, x-I B/. (d) Let (S, .) be a semigroup and let A 5 S. Then T is an FP-tree in A if and only if T is a tree in A and for all f E T, BJ = (17,,Fg(t) : g E T and f g and 0 # F G dom(g)\dom(f)).

c

The idea of the terminology is that an FP-tree is a tree of finite products. It is this notion which provides the most fundamental combinatorial characterization of the notion of '*central". A *-tree arises more directly from the proof outlined above.

Lemma 1424. k t (S, .) be an injnite sernigmup and let A G S. Let p be an idemporent in BS with A E p. There is an FP-tree T in A ~ u c hthat for each f E T. Bf E P. Pmof: We shall define the initial segments T" = (f E T : dom(f ) = n) inductively. Let To = (0) (of course) and let Cn = A'. Then Cn E p by Theorem 4.12. Let TI = (((0.x)) : x E Cn).

I I

b

293

14.5 Combmatorial Characterization

Inductively assume that we have n E N and have defined T. so that for each f E T. one has FP((f (t))y,-d) E A'. Given f E T,. write Pf = FP((f (t))y;d),let Cf = ~'n~,,,,x-~~*,andnorethatby~emma4.14,C~ E p. LetT.+l = I f - y : f E T, and y E C f )T h m given g E T.+I. onehas FP((g(t))Y4) E A'. T. Then T is a tree in A. Om The induction being complete. let T = sees immediately from the construction that for each f E T . Bf = C f . We need to show that for.each f E T one has Bf = ( n , , ~ g ( t ): g E T and f P g and 0 f F E dom(g)\dom(f)). Given f E T and x E B f , let g = f-x and let F = dom(g)\ dom(f ) (which is a singleton). Then x = ll,,F g(t). For the other inclusion we first observe that i f f ,h E T with f E h thm Pf E Ph so Bh E B f . Let f E T,, and let x E g(t) : g E T and f $ g and 0 # F c dom(g)\dom(f)). Pickg E T with f gandpickFwith0 # F dom(g)\dom(f) such that x = & g g ( t ) . First assume F = ( m ) . Then m 2 n. Let h = g,,. Then f E h and h-x = glm+l E T . Hence x E Bh E Bf as required. Now assume . g(t), and let IF1 1 I , let m = max F , andlet G = F\(m). Let h = gl,. let w = y = g(m). Then y E Bh. Let Pf = FP((f (1)):;) and P . = FP((h(t))ysl).Weneed toshowthatw.y ~ B f . T h a t i s , w e n e e d w . y ~ A ' a n d f o r a l l zP~f , w . y E Z - I A * . N o w w ~ ~ a n d ~ ~ B h s o y ~ w - ~ A ' s o w ~ y ~ A * . L e t z ~ P ~ . T h e n z . w and y E Bh soy E ( Z . w)-'A' and so w . y E z-~A*.

Us

(n,,~

n,,G

Theorem 1425. Let (S, .) be an injinite semigmup and let A c S. Statements (a). (b). (c),and ( d )are equivalent and are implied by statement (e). I f S is countable, then aN jive statements are equivalent. (a) A is centmL (b) There is a FP-tree T in A such that ( B f : f E TI is collectionwise piecewise

syndetic. (c) There is a *-tree T in A such that ( B f : f E TI is collectionwise piecewise syndetic. (d) There is a downwarddirecredfamily ( C F ) F of ~ Isubsets of A such that (i)foreach F E I andeachx E C F thereexists G E I with Cc E x - I c F and (ii) {CF: F E I ) is collectionwise piecewise syndetic. (e) There is a decreasing sequence (C. )Elof subsets of A such that (i)for each n E Wand each x E C. there exists m E N with C, E x-I C,, and (ii) (C. :n E N)is collectionwisepiecewise syndetic. Pmof: ( a ) implies (b). Pick an idempotent p E K ( B S ) with A E p. By Lemma 14.24 pick an FP-tree with { B f : f E TI E p. By Theorem 14.21 ( B f : f E T ) is collectionwise piecewise syndetic. (b) implies (c). Let T be an FP-me. Then given f E T and x E B f . we claim that Bf-x E x - ' B f . Tothisendlety E Bf-,andpickg E TandF Sdom(g)\dom(f-X) such that f - x g and y = g(t). Let n = dom(f ) and let G = F U In). Then g(t) and G E dom(g)\dom( f ). so x . y E Bf as required. x .y =

n,,c

nrE~

(c) implies (d). Lct T be the given *-Ira%S i (Bf : f E T)is cOUe*i~nwi~ piefnvise sjmdetic, so ia Bf : F E Pf(T)). Obis can be seen directly or by hmking Theorem 14.21.) Let I =T (O /$' and for each F E I . let CF = Bf. Then [CF : F E I ) is coUeCtio~wisepiecewise syndehc, so (u) holds. Let F E I and l e t x ~ C pL . e t G = [ f - x : f E F ] . T h e n G ~ I . N o w f o r e a ~ b ~f F w e b a v e Bf-= C x - ' ~ f so Co E x-~CF. cL CF. By Themem 4.20 M is a subsemigroup of (d) implies (a). Let M = @S. Since [CF : F E I) is collectionwise piecewise sylldetic, we haw by ThtOnm 14.21 that MnK(pS) jA 0 so wemay pickaminimal Ieftidcal L of pS with LflM # 0. Tbm L fl M is a compact subsemipup of pS which thus contains an idempotent, and this idempotent is necessarily minimal. That (e) implies (d) is trivial. Finally assume that S is countable. We show thaI (c) implies (el. So let T be the given *-tree in A. Then T is countable so enurnexate T as (f.)zl. For each n E N. Bfk. Then [C, : n E M) is collcaionwise piecewise syndetic. Let let C, = n E N a n d l e t x ~ C , , . Pickm €Nsuchthat(fk-x:kE(1.2 ....,n)] E [ f , : t E (1.2 .....m)]. Then C, = Bfi E Bfi-, E f l , l x - l ~ f i =x-'c.. 0

tnfEF

nfEF

n,,,

n;=,

n:!,

n;=,

We close this section by pointing out a Ramscy-theontic consequence of the charsacrization. Ai. Them CoroIhry 14.26. Let Sbem infinite semigroup, let r E N, and k t S = mit i E [I. 2,. I ) mufan FP-tree T in Ai such that (Bf :f E T)is collectionwise

.

..

piecewise syndctic.

Pmofi Pick an idempotent p E K(pS) and pick i E (1.2, Apply Theorem 14.25.

.

... r ) such that Ai

E p. 0

Ex& 145.1. Let Sbeasemigroup. Rovethatafamily A g Z(S)is colleztionwia piecewise syndetic if and only if there exists a function G :Pf (A) + Pf ( S ) such that [y-' (G(P))-'(n F) : y E S and F E Pf (A)) bas the finite intersection property. (Where Y - ~ ( G ( F ) ) - ' ( ~ r)= u,,,,,,Y-lt-'(n ~ 1 . )

Notes The original Central Sets Theorem is due to Furstenberg [98, Proposition 8.211 and

applied to the semigroup (N.+). It used a different but equivalent definition of c d . SeeThmrem 19.27 fora proof of the equivalenccof the notions of central. This original to take values in Z.Sina version also allowed the (finitely many)sequences (yc,,,)~=l any idempotent minimal in (PW. +) is also minimal in (BZ. +). and hence any c e n d set in (W. +) is central in (Z, +),the original version foliows from Theorem 14.1 1.

'Iheidea forthe @of theCenaal Sc& Tbmrcm(aswell as the proofs in this chapter of vsn der WaerdetfsThaonm and the Hales-JewenThearem) isdue to Furstenberg and Katmelson in [W]. where it was developtd in the context of enveloping semigrmps. The idea to convat this proof into a proof in @ Sis due to V.Bergelson, and the constmction in this context first appeared in [27]. Corollary 14.13 can in fact be derived from the Hales-Jcwen Theorem (CoroUary 14.8). See faexample [28, p. 4341. The material in Section 14.5 is from [148]. a d t of coUaboration with A. Malelti. except for Theorem 14.21 which is from [147l, a result of collaboratioa with A. Lisan.

Partition Regularity of Matrices

In this chapter we present several applications of the Cenaal Sets Theorem

waxem

14.11)andofitsproof.

15.1 Image Partition Regular Matrices Many of the classical results of Ramsey nKory arr natutally stated as instances of the following problem. Given u, v E N and a u x v matrix A with non-negative integer entries, is it hue that whenever N is finitely colored there must exist some .? E Nusuch that the entries of A.i are monochrome? Consider for example van der Waerden's Theorem (Corollary 14.2). The arithmetic progression [a, a + d , a + 2 4 a + 3d) is precisely the set of entries of

Also Schur's Theorem (Theorem 5.3) and the case m = 3 of HiIben's Theorem (Theorem 5.2) guarantee an affirmative answer in the case of the following two matrices:

This suggests the following natural definition. We remind the reader that for n E N and x in a semigroup (S. +),nx meansthesumofx withitself n times, i t . the additive version of x". We use additive notation here because it is most convenient for the matrix manipulations. Note that the requirement that Shave an identity is not substantive since

F

15.1 Image Partition Regular Matrices

297

one may be added to any semigroup. We add the requirement so tbat Ox will make sense. (See also Exercise 15.1.1, where the reader is asked to show that the central sets in S are not affected by the adjoining of a 0.) Deenition 15.1. Let (S, +) be a semigroup with identity 0, let u, v E N, andlet A be a u x u matrix with entries from o.Then A is image panition regular over S if and only if whenever r E N and S = U:=, Ei, there exist i E (1,2,. ,r ] and i E (S\(O))" such that A? E EjU.

..

It is obvious that one must =quire in Definition 15.1 that the vector x' not be constantly 0. We make the stronger requirement because in the classical applications one wants all of the entries to be non-zero. (Consider van der Waerden's Theorem with increment 0.) We have restricted our matrix A to have nonnegative enlries, since in an arbitrary semigroup -x may not mean anything. In Section 15.4 where we shall deal with image partition regularity over W we shall extend the definition of image partition regularity to allow enaies from Q. Deanition 152. Let u, v E N and let A be a u x v matrix with *tries from Q. Then A satisfies thefirsf entries condition if and only if no row of A is 0 and whenever i , j E (1.2,. . . , u ) and k = min{t E (1.2 v) : a i , ~# 0) = min{f E (1.2, .... v) : aj., # 0). then ai,k = aj.k > 0. An element b of Q is afirst entry of A if and only if # 0). thereissomemwi of A such that b = aLkwhere k = m h { t E [I, 2. . .. u ) :

,...,

.

' of all sets that Given any family R of subsets of a set S, one can define the set 2 meet every member of R. We shall investigate some of these in Chapter 16. For now we need tbc notion of central* sets. IMMtion 153. Let S be a xmigroup and let A G S. Then A is a centraP set of S if and only if A n C # 0 for every central set C of S.

Lemma 15.4. Let S be a semigmup and kt A G S. Then thefollowing statements a n equivalent. (a) A is a cennar set. (b) A is a member of every minimal idempotem in BS.

(c) A n C is a cenml set for every central set C of S.

Pmoj: (a) implies (b). Let p be a minimal idempotent in BS. If A $ p then S\A is a cenual set of S which misses A. (b) implies (c). Let C be a central set of S and pick a minimal idempotent p in flS suchthatCEp. ' I h c n a l s o A e p s o A n C e p. That (c) implies (a) is trivial. 0 In the following theorem we get a conclusion far stronger than the assertion that mahices satisfying the first entries condition are image partition regular. The stmnger conclusionisof some interest in itsown right. Moreimportantly, thestrongerconclusion is needed as an induction hypothesis in the proof.

298

I5 Partition Regularity of Maaices

TbcMcm 155. k t (S. +) be an infinite c o m smigmup ~ with identity 0, let u, v E N, and &t A be n u x v matrix with nfriesfrom co which sutigksthefirst m e condition Let C be cenfml in S. for everyfirst entry c of A, cS is a cmtml* set. then there exisfsequmces ( X I , ~ ) E(x~.d,,~. ~ , d(l .... (xvr)z1 in S such that for eve^ F E pf ((N, ?F E (S\(O))" and A i F E Cu. whem

Pmof: S\{O) is an ideal of S so by Corollary 4.18.0 is not a minimal idempotent. and thus we may prcsume that 0 p C. We proaed by indnction on u. Assume f a t u = 1. We can assume A has no repeated w s . so in this case we have A = (c) for some c E N such that cS is central*. Then C n c S is a central set so pick by Theorem 14.1 1 (with the sequence y l , = 0 for each n and the function f constantly equal to 1) some sequence (k,,)E1 with FS((k,)=,) C C rl cS. (Infact here we could get by with an appeal to Theorem 5.8.) For each n € N pick some XI, E S such that .k = cxl,,,. The sequence XI..)^^ is asrequired. (Given F € 4 ( W . &F .k f 0 so % € F X1.n f 0.) Now let u E W and assume the theorem is m e for v. Let A be a u x ( v 1) mmix with entriesfrom o which satisfies the first enhies condition, and assume that whenem c is a first entry of A, cS is a central* set By m g i n g the w s of A and adding additional w s to A if need be. we may assume that we have sow r E {I, 2, ..,u - 1) and some d E N such that

+ .

0 i f i € (1.2. ..., r ] ai.1 = d i f i E { r + l , r + 2

I

.....u).

Let B be the r x u matrix with entries bi,, = ai.j+!. Pick sequences (ZI.~):~, ( z ~ . ) ~...~, (. z ~ , ~ )inE S , as guaranteed by the inducuon hypothesis for the matrix B . F o r e a c h i e { r + l , r + Z ...., u J a n d e a c h n ~ W . l e t

and let y , . = 0 for all n E N. (For other values of i, me may let y i , take on any value in Sat all.) Now CfldSiscentral, sopickbyTheorem 14.11 sequences h S a c d (H.)zl in Pf(N) such that m a H. c min H,+I for each n and for each i E (r, r + 1. u).

.

...

(If (kL)zl and (H;)zl are as given by Theorem 14.1 1, let k,, = V,+=and H, = HL, foreveryn.andfori E ( r , r + I . ...,u l , l e t f i ( n ) = n i f n < i a n d l e t f i ( n ) = i i f n 2 i.)

15.1 Image Parrition Regollr M d a s

299

+

Note in parcicularthat each S, = k,, E,.H, y,., e C C dS, so pick X I , S such bat k. = d x ~ , " .For f .E 12.3,. ,v 11. let xj, = E ~ E Hzj-I.,. . We claim that the scquCWXS (x~..):=~ are as reqoired To see this, let F E Pf(N)be given. We need to ~ E (1.2 u). s h a v t h a t f o r e a c h j ~11.2 ..... ~ + l ) . & ~ x j fOmdforeachi

.. +

....,

For the lira &on m e that if j > 1. then = xrEczj-1.1 where G = U I E ~ H m I. f j = 1 . h d X n E ~ x = ~ ,Cns.=(km E t E ~yr,d . E C. To establish the second assenion, let i E ( 1 . 2 . . u) be given. . Case 1. i 5 r. Then

...

+

where G = UnEFH.. C a s e 2 . i > r . Thm

We now preseat the obvious partition regularity m U q . But note that with a little

effort a stmnger result (Theorem 15.10) follows. Corollary 15.6. k t (S. +) be an in@ite c o m m u ~ v semigmup e and let A be afinite matrix with entriesfmm o which sah'sfies the first rnm.es condition. Iffor each first erury c of A, cS is a central* set, then A is image pam'tion regular over S.

u,,

P m J Let S = Ei. Then some Ei is cenbal in S. I f S has a two sided identity. thenTheomn 15.5 applies directly. Otherwise,adjoin an identity 0. Then, by Exmise 15.1.1. Ei is central in SU (0). Fmrher, given any first entry cof A, c(SU (0)) iscentral* in S U ( 0 ) so again Themm 15.5 applies. Some common, well behaved. semigroups fail to satisfy the hypothesis that cS is a central* set for each c E N. For example, in the semigroup (N..), [x2 : x E W) (the multiplicative analogue of 2 s ) is not even a central set. (See Exercise 15.1.2.) Consequently B = W\{x2 : x f N) is a central* set and A = ( 2 ) is a first entries matrix. But thm does not exist a E N with aZ E B. To derive the stronger partition result, we need the following immediate corollary. For it, one needs only to recall that for any n E N, nN is central*. (In fact it is an IP* set; that is nN is a member of evcry idernpotent in (BN, +) by Lemma6.6.) Corollary 15.7. Any finite manix with eniriesfrnn o which sofisjies thefirst enrries condition is image panition regular over N. 0

15 Wltition R q u h t y of Mabins

300

Corollary 15.8. La A be a h i r e marrix with entries from. w which sorisfirs the fim entries condrtion and let r E N. There exl'st9 k E N such that whenever (1.2. ...,k ) & r-colon4 the= exiits .? E (1.2, ...,k)" such char the entries of A? a n rnonochmmc

Pmof: This is a standard compactness argument using CoroUary 15.7. (See Section 5.5 or the proof that (b) implies (a) in Theorem 15.30.) See also Exercise 15.1.3. 0

'

As we nmrnked aftex Definition 15.1, in the definition of image panition ngularity we demand that the entries of .? me all non-zero because that is the natural vmio. for the classical applications. However, one may reasonably ask what happens if one weakem the conclusion. (As we shall see. the answer is "nothing".) Likewise, one may strengthenthe definition by requiring that all entries of A? be non-zero. Again, we shall see that we get the same answer.

+)

Definition 15.9. Let (S. be a semigroup with identity 0, Ie4 u, v E N,and let A be a u x v matrix with entries from o. (a) The matrix A is weakly image pam'tion regular over S if and only-if whmever r E N and S = U;=l E,,there exist i E (1,2. .. . ,r ) and i E (Su\(0)) such that A? E EiU. (b) The matrix A is strongly imagcpammtionregular over S if and only if whemvcr r E N and S\(O) = U{=l El,there exist i E (1,2, . r ] and .? E (S\(O))" such that

..

Ax' E

EiU.

.

Theorem 15.10. Let (S, +)be a commutativesemigroup with identify 0. 7 hfollowing statements a n equivalent. (a) WheneverA is a h i r e matrix with entriesfmnr o which

satisfies thefirst entries condition. A is stmngly image partition regular over S. (b) WheneverA is afinite matrix with entriesfmm o which sotisfis thefirst enhies condition A is image partition ngular over S. (c) WheneverA is afvrite ma& with enniesfmm o which sariipes thefirst entries condition, A is weakly image partition regular over S. (d) For each n E N,nS # (0).

P m t That (a) implies (b) and (b) implies (c) is trivial. (c) implies (d). Let n E Nand assume that nS = (0).Let

Then A satisfies the first enmes condition. To see that A is not weakly image partition regular over S, let Et = (0) and let E2 = S\{O). Suppose we have some.? =

'1 .

153 Kcmcl Partition Regular Matrices

301

~ ~ \ {and 6 ] some i < (I. 2) such that A?

E (Ei)3. Now O = n q so i = 1. Thus +nxz =OsoLhatxz = X I + X Z =Oandhence.?=O,acontradietim. (d) implies (a). Let A be a finite matrix with entries from o which satisfies the first entries condition. To see that A is strongly image partition regular over S,let r E N and let S\[O)= Ei. Pick by Corollary 15.8 some k E W such that whenever (1.2,. . , k) is r-colored, there exists .? E ((1.2, ... ,k))" such that the entries of A.? are m~ochrome. There exists z E S such that [z, 22.32,. .,kzl n (0)= 0. Indeed. otherwise one would have k!S = (0).So pick such z and for i E (1.2,. . . r ) . let Bi = (t E (1.2.. .. ,k) : rz E E i } . Pick i E (1.2,. ...r ] and 9 E ((1,2, ...,k))" such that A$ E (Bi)". Let.? = jz. Then A? E (Ei)'. 0 XI = X I

UT=,

.

.

.

Exercise 15.1.1. Let (S,+) be a semigroup without a two sided identity, and let C be central in S. Adjoin an identity 0 to S and pmve that C is central in S U (0). Exerdse 15.1.2. Prove that (x2 : x E central) in (N..).

N) is not piecewise syndetic (and hence'not

Exercise 15.13. Prove Corollary 15.8. Exerdse 15.1.4. Let u, v E N and let A be a u x v matrix with entries from o which is image partition regular over N. Let T denote the set of u l M t e r s p E @N with the property that. for every E E p. there exists .? E PT for which all the entries of A.? are in E. Prove that T is a closed subsemigroup of BN.

15.2 Kernel Partition Regular Matrices If one has a group (G.+) and a matrix C with integer entries, one can define C to be over G if and only if whenevqr G\(O)is finitely colored there will exist i with monochrome entries such that C.? = 0.Thus in such a situation, one is saying that monochrome solutions to a given system of homogeneous linear equations can always be found. On the other hand, in an arbitrary semigroup we know that -x may not mean anything, and so we generalize the definition. kernelplrt'fion regular

Definition 15.11. Ln u , u E N and let C be a u x v matrix with entries from Z. Then C+ and C- are the u x v matrices with entries from o defined by czj = (IciVjI + q j ) / 2 and c; = (lci,jI - q j ) / 2 .

Thus, for example, if

c=(.

1 -3 -2

0 2)g

1 0 0 then

Note that C = C+ - C-

c+q0

2)

0 3 0 and C-=( 0 2 0

)

'

m

I5 PartitimRgolarityof~0~1

+)

IkMUm 15.U Let (S. be a scmigmup with identity 4 l a u. v E Nand let C be a u x' v matrix with enhies fmm Z.Then C is kemlpam'tion regular o w S if and d g if whenever r E N and S\{0) = U:=, Di, then exist i E (1.2,. ,r ] and x' E (Di)u

..

such that C+x' = C-x'.

The condition which guamntees kernel partition regularity owr most semigroup & known as the columnr condition. It says that the columns of the mahix can be g a t h a into groups so that the sum of each group is a linear combii$on of columns horn prcecding groups (and in particular the sum of the first group is 0).

-- .

Dcgnitlon 15.13. Let u , v E N. l a C be a u x v matrix with enhies horn Q rmd l a cl. a,. . . E; be the columns of C. Let R = Z or R = Q. The matrix C satisfies the columns condition over R if and only if there exist m E Nand 11.Iz. . I,,,such that

..

.

.....Im)isapartitionof(1.2.....v].

(1) {11,12

(2) &I, 6 = 6. (3) If m > 1 and r E ( 2 . 3 . . .. m]. let JI = R such that XI,,, 6 = Xi,,, 61.r.cj.

.

u!-'

1-1

Ij. Then there exist (61.i)i,,, in

Observe that one can effectively check whether a given maaix satisfies the cohmms condition. me problem is, however. NP complete because it implies the ability to determine whether a subset of a given set sums to 0.)

Lemma 15.14. Supposethat C isa u x m matrix with entriesfmm Z which sati$es the firnenbiescondition. L . e t k = m a x { ~ c i ~ ~j :) (Ei {. 1 , 2,...,u)x{1,2 ,...,m ) ] + l , and let E be the m x m matrir whose entry h mw t Md column j is

Thm C E is a ntolrix with enhiesfmn, o which also satisfiesthefirst entries condition Pmef Let D = CE. To see that D satisfies the first enhies condition and has entries horn o,let i E (1.2,. ,u] and lets = minlt E (1.2,. . m ] : ci., # 0 ) . Thm for j < s. d;:.j= 0 and di,, = ci,,. If j s s, then

..

..

A connection between mahices satisfying the columnscondition and those satisfying

the first enhies condition is provided by fhe following lemma

a-

Lem~15.15.Lnu.v E N a n d I r t C b e a u x vmafrixwith&sfmnrQwhich satisfiesrhrcol~condirionowrQ.Themadnm e (1,2. vJmdavxmmatrix B with enrriesfmma,thatsari.$es thefirst enfries conditiomsuch that C B = 0,where 0 is the u x m matrix with all zero mnies. IfC ssolirfus the columns condition over Z then the morriX B can be chosen so that its onlyfirst entry is 1.

...,

Pmof: tick m E N, (It):=,. (J&. md ((&,i)isr,)E2as gu~fanreedby the columns condition for C. Let B' be the v x m matrix whose enby in row i and column t is given by - 6 if i E J: 0

ifi

CUj,, 4.

We observe that 8' satisfies the first eohies condition, with the frrst non-zem enby ineachmbeingl. WealsoobsmethatCB'= 0. Ondeed,letj E (1.2. ..., u)and I E (1.2 ml. I f f = 1,then &cj.i .bi,, = &r, cj.i = Oandift w 1,then Er-1 Cj,i ' b;,, = &J, -8t.i ' cj.i Cis,, cj,i = 0.) We can choose a positive integer d for which dB' has entries in 2. ?hen dB' also satisfie8the first entries condition and the equation C(dBf) = 0. If all of the numbers 6i. are in 2,let d = 1. L c t k = m a ~ ( l d b ~ , ~ l : ( i , j ) ~...., [ 1 , 2v ) x ( 1 . 2 ,...,m)]+l.aadletEbethe m x m matrix whose enby in IUW i and column j is

.....

+

By Lemma 15.14. B = dB'E has entries in o and satisties the lirst enhies condition. Clearly. CB = 0. Now,giveni ~ ( 1 . 2 v),choosese(1.2 ,...,m)suchthati E I,. I f t < s , then b;, = 0 while if t > j, then e:,j = 0. 'Ihus if j c s, then binj= 0 while if j 2 s. then bi, 3 E/=, dbi,Ikj-f. In particulrn the first nonzero entty of row i is d.

....,

We see that, not only is the columns condition over Q sufficient for kernel partition regularity over most semipups. but also that in many cases we can guarantee that solutions to the equations C + i = C-x' can be found in any central set.

+)

Theorem 15.16. Ln (S. be an injinire commutative semigroup wirh idem.ty 0, let u, v E R and let C be a u x v nrofrir with enm'esfivm Z. (a) If C sarisfis the columns condition over Z thenfor any cenmrl subset D of S, them aists ? i D" such that C+i = C-L (b)If C sofisfies the columns condition over Q andfor each d E N dS is a cmtml*set in S. thenfor any central subset D of S. there &IS 2 E DVsuch that C+x' = C-2. (c) IfC satisfies the columnr condition over Q and for each d E N, dS # (01, then C is kernelpm'rion regular over S.

P n $ (a) Pick a u x m mamix B as guaranteed for C by Lequna 15.15. Then the only firstentry of B is 1 (and IS = S is a central* set) so by Theomn 15.5 we may pick some 2 E Sm such that B i E DU. Let 2 = Bi. Now CB = 0 so C+B = C-B (and all entries of C+B and of C-B are non-negative) so C f i = C+Bi = C-BZ = C-i. (b) This pmaf is nearly identical to the proof of (a). (c) F'ickamauix Bas g u m t e e d for C by Lemma 15.15. Let r E N andlet S\(O) = D,. By Theorem 15.10chwseavecta~E (S\(ODm and i E (1.2.. .. , r ] such that BE E (D;)", and let i = LIZ. As above, we wnclude that C + i = C+Bi = C-B? = C-i. 0

Exercise 15.21. Note that the matrix (2 -2 1) satisfies the columns condition. Show that there is a set which is cenhal in (N,.) but contains no solution to the quation x12- x3 = ~ 2 ' . (Hint: Consider Exercise 15.1.2.)

15.3 Kernel Partition Regularity Over N Rado's Theorem In this seetion we show that matrices satisfying the columns condition over Q arc precisely those wbicharekemel partition regularover (N, +)(which is Rado's Theorem) and are also precisely those which are kernel partition regular over (N, .). Given a u x v ma& C with integer entries and given 2 E NVand 9 E NUwe write iC = j: tomean that for i E (1.2,. ..,u], xjC,-; = yi.

ny=,

Theorem 15.17. La u. v E R and let C be a u x v ma;& following sratements are equivalent.

with entriesfmm 2 The

(a) The mnrrix C is kernel partition regular over (N, +). (b) The mar& C is kemelpa&ion regular over (N..). That is, whenever r E N a@ w\(l] = Ui=,D;, thereexist i E (I. 2,. . ., r ] a n d l E (Di)"such thnriC = I.

us,

P m f (a)irnplies(b). Letr E WandIetR\{I] = D;. Foreachi E (1,2, .. ., r!, letA;= (n E W : T e Di). Picki E (1.2.....r J c $ ~ dEj (Ai)"suchthatC~=O. Foreach j e (1,2,..., u}.letxj = 2 " J . T h e n i C = 1. (b)implies (a). Let ( p ; ) z , denote the sequenceof prime numbers. Each q E Q\(O)

can be expressed uniquely as q = nE1p?, where e; E Z for every i. We define 9 : Q\{O] -+ Z by putting +(q) = C z l ei. (Thus if q E R. O(q) is the length of the prime factorization of q.) We extend 9 to a function I) : (Q\(O])" + Zuby putting *(f); =#(xi) foreach i E (Q\(OI)" and each i E (1.2.. . . v]. Since 4 is a homomorphism from (Q\{O], .) to (Z, +),it follows easily that @ ( i') = C g ( l ) for evety l E (Q\(0]IV. Let {Ei)l=lbe a finite partition of N. Then (@-'fEi] OR):=, is a finite partition of N\(IJ, because 9[N\(I]] G W. So pick i E (1.2.. ,k) and

.

..

2'

15.3 Kaml Partition Regularity Over N

2 E_ (#-f[Eil n Nusuch that; *(I) = 0.

= i. Then $(?)

E

305

(EiY and C@(.Z)= $(zC) = 0

Definition 15.18. Let u, v € N, let C be a u x v matrix with entries from Z.and let J and I be disjoint nonempty subsets of (1.2,. . . v ) . Denote the columns of C as

.

- - ..,

+

Cl, C2,. C. (a) If there exist (zj)jcj in Q such that t, = EjEJzj+ then E(C, J , I ) = 0. (b) If there do not exist (zj)jtj in Q such that 3 = EjeJz,Zj, then

xjCi,, xjE1

E ( C , J. I ) = {q : q is a prime and there exist (zj)j.j in o, aE(1.2 q-L),d~o.a n d j ~ P such that EjEjzjZj aqd Cjer 3 = q d + l j ).

.....

+

Lunma 15.19. Let u . v E R. let C be a u x u mufrix wirh entriesfrom Z, and let J and I be disjoint mnempfysubsetsof (1.2, ... v). Then E(C, J , I ) isfinite.

.

Pmof: If (a) of Definition 15.18 applies, the result is trivial so we assume that (a) does not upply. Let b = Ej.r i,. Then b is not in the vector subspace of Q" spanned by (Z,),EJ SO pick some 3 E Qu such that 3 Z, = 0 for each j E J and 3 i # 0. By multiplying ky a suitahre m e m e of Z we may assume that all entries of 3 are integers and that lir . b s 0.Lets = ti! . 6. We show now that if q E E(C. J, I), then q divides

.

.

5.

Letq E E(C,J. I)andpick(zj)j.jino.a ~ ( 1 . ..... 2 q-1),d c o , a n d y ' ~ Z " such that Ej,j z j 3 aqd EjE,Zj = qd+I?. Then

+

.

Since 3 - Zj = 0 for each j E J we then have that aqds = qdt'(G j) and hence a s = q ( 3 7). Since a E (1.2, . . .,q 1). it follows that q divides s as claimed. ir

-

The equivalence of (a) and Ic) in the following theorem is Rado's Theorem. Note that Rado's Theorem pmvides an effective computation to determine whether a given system of linear homogeneous equations is kernel partition regular. Theorem 1530. Let u. v E R and Iet C be a u x v morrir with entriesfmnt Z. The following statements are equivalent. (a) The matrix C is kernel panition regular over (N. +). @) The marrir C is kenelpartition regular over (R. .). That is, whenever r E N a* N\{I] =U:=, Di,thereerisri E (1,2, .... r l a n d l ~(Di)"suchthariC = 1. (c) The matrix C satisfies rhe columnr condition over Q. Pmofi The equivalence of (a) and (b) is Theorem 15.17,

(a) implies (c). By Lrmma 15.19, whenevez J and I are disjoint wnempty s u b of 11, 2, v), E(C, J, I ) is finite. Consequently, we may pick a prime q such that q>-(lZjGr ~ , j I : i E ( 1 . 2 ,.... ~ ) a n d B #J E ( 1 , 2 v))andwhcmmJ andI;1redisjointnomrnpcywWof(1.2.....v ) , q f E(C.J.1). Given x E N, pick a(x) € 11.2.. ,q 1) and ((x) and b(x) in a, such that x = a(x) .qtCX) b(x) .qf(x)+ That is, in the base q expansion of x, L(x) is the number of rightmost zeros and a(x) is the rightmost noavm digit. Foreacha E (1.2. ....q - 1 ) l e t A. = (x E W\{lJ : a(x) = a ) . Picka E (1.2.. ..,q 1) and XI, XZ, ,xa E A. such that C i = 0. Phtion(1.2 ,....vlaccordingtoL(x,). T h a t i s , p i c k m ~ N . s e t s I 1 . 1...., ~ I, aodoumbmll
....

,...,

'.

+

.. -

...

-

.....

....

=a

(I) CjEll 4 and (2) given n E (2.3. ,m), CjErq

...

,...,

6is a limarcombination over Q of (4)j.1..

so that C satislies the columns condition over Q. To establish (I). let d = [I. Then for any j E I1 l a e j = b(xj) and note that = a . qd + ej qd+l. For j E (1.2,. ,v)\Il we have t(xj) > d SO xj = ej qd+l for some ej E W. Then

XI

.

..

.

so, since q Xa we have ql Zj.1, ci.1. conmdicting our choice of q. Now consida (2). Let n E (2.3. . ,m ) be given and let d = I,. For j E I., la el = b(xj) andnotethatxj = a .qd+ej .qd+'. For j E UY-+, 4.ifany.pickej E N such that xj = ej qd+l. Thus

..

.

Since q f E(C. J,, I.) we have that Xjel, 4 is a linear cornhination of (4)jpjn as required. 0 (c) implies (a). This fouows fmm Theorem 15.16(c).

15.3 Kcmcl Partitioa Regnlarity Over N

307

To illusmte the use of Theorem 15.20 we establish the case f = 4 of van der Waerdm's Theorem. 'Ihat is, we show that wbeamr N is finitely colored we can get somea,d E PIwitha,a+d,a+2d.a+3dmowehmme. Sooacletsxl = a . x2 =a+d.x1=a+2d,andx4=a+3d.Thenxz-x1 =dsox~=x1+2(x~-x1) and xr = X I 3(x2 -XI). That is we have the equations

+

so we arc asking to show that the matrix

is kernd pattit@ ~gular.Indeed it s&fics the columns condition bsceuse the sum of its colulnns is 0. Bur there is a problem! The assignment XI = xz = XI = x4 = 1 (or any other number) solves the equations. That is, these equations allow d = 0. So we strengthen the problem by asking that d also be the same color as the terms of the progression. Let XI =a,xz = d . x 3 = a + d , x 4 = a + 2 d , a n d x ~ = a + 3 d . Thenwegettheequations

so we need to show that the matrix

is -el @tion regular. Far this. let 11= (1.3,4,51, Iz = {2),82.1 = 0.82.1 = -1. 62.4 = -2. and 82.5 = -3.

Exercirc 153.1. M v e the finite version of the Finite SumsT h m m as aconsequmce of Rado's Theorem. That is, show that for each n. r E N. whenever N = UiSl Ai, At. For example, the there exist i E [1,2. . . r Jand (x,):=~ such that FS((X,):,~) case n = 3 requires that orrc show that the matrix

..

is kernel partition regular.

.

15 M t i o n Regularity of Matrices

308

15.4 Image Partition Regularity Over W

.

f

We concluded the last section with a verification of the length 4 version of van da Waerden's Theorem via Rado's Theorem. To do this we had to figure out a set of equations to solve and had to strengthen the pmblem. On the other hand. we have already seen that the original problem 1snanvally seen as one of determining that the matrix

is image panition regular. (And the strengthened problem clearly asks that the mahix

be image panition regular.)

Since many problems in Ramsey Theory are naturally stated as deciding whether certain matrices are image partition regular, it is natural to ask for a determination of precisely which matrices are image partition regular. We do this in this section. Since we are now dealing with the semigroup (N, +),we extend Definition 15.1 to allow entries from Q. Delinition 1521. Let u. v E N,and let A be a u x u mahix with entries from Q. Then A is imngc parrition regular over W if and only if whenever r E W and N = U ,,: Ei. thereexist i E {1,2. .. ., r ) and? E WV suchthat A? E Eiu. We need a preliminary lemma which deals with a specialized notion from linear algebra.

,....& b e i n Q Y , a n d l e t I C (1-2,...,v).

Delinition15.22. L e t u . u ~ W . l e t c i , i i The I-resm'cted span of (&, &, ... G ) is

.

lzr=,ai .

: each ai E

lR and if i E I, then ui 2 0).

Lmnma15.23. L c r u , v € N , l e t e t ~ . ..... i &beinQ".andlerI s(1.2,..., v). Let S be the I-restricted span of (G, 6.... &). (a) S is closed in R". @)If?E SnQ".rhenrkreexirt8t.82 ..... 8,inQsuchrhntji= Xy=16i.Gand 8i 2 0for each i E I.

.

Pmof (a)We proceed by induction on II I (for all u). If I = 0, this is simply the assertion that any vector subspace of RU is closed. So we assume I # 0 and assume without loss

,

F 15.4 Image Pmitioa Regularity Om N

309

..

of geoerality that 1 E I. Let T be the (I\(l))-restricted span of (G,6 , . ,?& By the induction hypothesis, T is tosed. ToreethatSisclosedInb ~ c C S ~ ~ e s h o Iwf 66 =~i ~ j , .t h m 6 ~ c t ~ = cL T = T = S, so assume that ci # 0. For each n E N, pick (ai(n)):=l such that ai(n) 2.0foreachi E I and I I -~ ELIai(n) ,611 < l/n. Assume k t that (a1(n) : n E W) is bounded. Piik a l i t point 8 of the sequence (al(n))gI-and note that 8 2 0. We claim that b 8 6 E T. To see this we show that b - 8 . 6 E CC T. So let E > 0 be given and pick n > 2/< such that lal(n) -81 < d(2ll6n). Then

- .

ttS-6.6

- EiV=zai~n).B~~ 5 116- E b l a i ( n ) . d ~ ~11al(n)6 + - 6 6 1 1 < c.

Sinczi-8.6 E ~ . w e h a v e ~ ~ 8 . ~ + ~ ~ ~ . Now assume that (al (n) :n E M)is unbounded. To see that S is c l o d it sufficesto show that -6 E T. F a then S is in fact the (I\(l))-restricted span of ( 6 . 6 . . G) which is closed by the induction hypothesis. Toseethat- (1 Ilbll)/c. For i E (2.3,. ,V) let 8; = ai(n)/al(n) and note that for i E [\(I). 6i > 0. Then

...

..

+

I1 - 6 - Xy=28i .dl1 5 llZ/a~(n)-

- Ey=z(ai(fi)h(n)) ;GI1 + ll~/a~(n)ll < l/(n . a ~ ( n ) + ) II~II/~I(~)

= ( l / ~ l ( n ) ) Ilb - xyTlsib) - zil+ Ibll/a~(n) c (1

+ ll~ll)/a1(n)

C 6..

(b)Again we p m c d by induction on (11. The case I = 0 is immediate, being mmly the assertion that a rational vector in the linear span of some other rational vectors is actually in their linear rational span (which is true because one is solving linear equations with rational coefficients). E IR" : Ey=lxi . 6 = 7). Thus X is an Bffine So assume I # 0. Let X = (i subspace of W U ,and we are given (by the fact that y' E S) that there is some i E X such that xi 2 0 for all i E I. Also (by the case I = 0) the= is some i E X such that zi E Q for all i E (1,2. . .. v). If zi 1 0 for all i E I , then we are done. so assume there is some i E I such that zi c 0. Given i E 1. (t E [O.11 : ( I t)xi tz, E 0) is [O. 11 if zi > 0 and is 10, xi/(xi zi)] if zi c 0. Lett be the largest member of [O. I] such that thevectorG=(l-!).:+I .isatisfieswi ~ O f o r a l l iEI.Thenforsomei E I we have that wi = 0. We assume without loss of generalily that 1 E I and wt = 0. Then 7 = E:=2 wi . K so 9 is in the (I\(]))-resuicted span of (6.6,. , G ) so by 61 .G and the induction hypothesis we may pick 82.63, . ..,8, in Q such that = 8i 2 0 for each i E 1\11]. Letting 81 = 0, we are done. 0

.

-

-

+

.. xy=2

Theorem 15.24. Let u. v E W and let A be a u x v mhix with entriesfrom Q. The following statements a n equivalent.

(a) A is image pam'rion regular over N.

310 (b)

15 PartitianRegularity of Mahim

,....c;,bethecolum~~ofA.T h e n a i s r s l . s 2.....s.

LutG,fi

E{X EQ:X>

0) such rhar t k matrix

/

-1

0

...

0 \

is kernel pam'tim rc$ujar over N. (c) Then exist m E N and a u x m matrix B wirh enm'esfmm Q which satisfies tkfirsz eniries condition such that given any 7 e W' rhem is some 2 e Nu with

A i 5 89. (d) Then exisr m E tiand a u x m mawh C with entries+ Z which satisfies the first entries condition such that given any 3 E Mm there is some 3 E Nu wirh Ax' = Cj7 (e) ntnrrxirtm~NMdauxmnronirDwirhmtriesfmnrowhichsan'sfiestk first mtries condition such thor given any 3 E Nmthere is some f E NYwith A? = 07. (0There exist m G N. a u x m matrix E wirh mtricsfmm o,and c E PI such that E satisfiesrhejirst enfries condition. c is the onlyjirst entry of E, and given m y

y'~N*t(lcmiSsomei€N"xti~hA?=E~.

Pm# (a) implies (b). Given any p E N\(I), we let the stan h e p coloring of N be the function

.. -

x:=o

defined as follows: given y E N, write y = a,pJ, where each at 6 (0.1. . ,p 1) and a. # 0: i f n > 0, up(y) = (a.,an-1.i) WheZT i E n (mod 2); if n = 0, up(y) = (q.0.0). (For example, given p z 8. if x = 8320100, y = 503011, z = 834. and w = 834012. all written in base p. then op(x) = a&) = (8.3.0). J+(W)= (8.3.1). and up(?)= (5.0.1). L e t 6 . i ,.... 6 bcthecolumnsof~nradletd;.d;, d;,dmotetbccocolmnnsof the u x u identity matrix. Let B be the matrix

....

..;

whefe SJ, 52. s, aze as yet unspecified positive ~~tionak. Dmotc the columns of B by bi, b2. b.+,. Then

....

.

/ ,

15.4 Image Pmiticn Regularity Over N

311

.

.

..

( ~ l ( p )&(PI). . .. ,~ ~ ( pof) (I, ) 2,. u) as f011ow.s. M i E w such that *i is monochrome with respect to the start base p coloring and let y' = A.?. Now divide up (1.2,. ..,u) according to which of the yi's stan furthest to the left in their base p representation. That is, weget Dl(p), &(p), ...,Dm(p)so that

.

Thereareonly Gnitely many ordnedpmtitionsof (1, 2.. .. u). Therefore wemay pick an infinite subset P of N\(l), m E El, and an ordmd partition (Dl. &. , Dm)of (1.2, ...,u)sothatforallp€ P,m(p)=mand

...

...

..

..

We shall utilize (Dl, 9 , ,Dm)to find q.q,. . s and a padtion of (1.2.. , u 1))85 required for the columns condition. We proceed by induction. F i t we shall find ZI G (1,2, ... v), spcify 4 E-Q+ (s E Q : s z 0) for each i E El, let 11= E LU ( v Dl), and show that &,I, bi = 0. That is, we shall show that &E, 4 .6 = &o, &. In order to do this, we show that Ci,nl d is in the (1.2,. . .,v)-restricted span of (-5.6,. ..,-5).For then by Lemma 15.23(b) one has 4 = C:=, ai 6 , where each ai E Q and each cri 2 0. Let ,V ) :ai z 0) and for i E Ei. lets; = at. El = (i E (1.2. Let S be the (I, 2,. ,v)-restricted span of (6.6, &). In orderjo show that &b, 2. is in S it sufii&s, by Lemma 15.23(a). to show that &D, di is in d S. To this end. let c z 0 be given and pick p E P with p > u/c. Pick the i E Nu and y' E Nu that we used to get (Dl (p), &(p). . ,Dm(p)). That is, A? = y'. y' is monochrome with respect to the start base p wloring, and (Dl, &, ,Dm) = (Dl (p), Dz(p), . .,D,(p)) is the ordered @tion of (1.2,. ,u ) induced by the starting positions of the yi's. Pick y so that for all i E Dl. y(p, yi) = y. Pick (a. b, C) E (1.2.. .. p 1) x (0.1.. . . p - 1) x (0.1) such that up(yi) = (a, b. C) foralliE(1,2 ~).LecC=a+b/pandobsmethatl~L~p.ForifD~. y; = a ~ P Y + b . p y - 1 + ~ i . p ~ - 2 w h m O j z c ip,andhenceyi/p~=L+zi/p2;let Ai = zi/p2 and note that 0 5 Ai c l/p. For i 6 U;L=2D,. we have y (p, yi) 5 y 2; let Ai = yi/pY and note that 0 c Ai c Ilp. Now A i = y' so

+

.

+

xiED, ... ..

.

.

...

..

.

..

.

.....

-

...

.

-

S i c e xy=l (xi/(CpY)) .6 is in S,we have that &D,

4 E ct S as required.

15 Partition Regularity of Matrices

312

.

Now let k r (2.3.. .. rn) and =same we have. chosen El, Ez, s (1.2.. .,v), si E Q+ for i E E ~ and , Ij = El U (v Dj) as quired for the columns condition. Let Lt = Ej and let Mk = Dj and mumerate Mk in order as q ( l ) ,q(2). ...,q(r). We claim that it suffices to show that CieD, is in the (1.2.. .. v)-restricted span of (?I.&. ,G,Liq(1). &2), ....&). which we will again denote by S. Indeed, assume we have done this and pick by L e n t 15.23(b) a1.q.. . .,a, in ( x E Q : x 2 0 ) and e ( l ) , 6,(z), ,Sq(,) in Q such that C i + o k 4= X ~ = I a i . G + C & , 8 , ( i ) . ~ qk( ti )E. * = (i E ( 1 . 2 ,..., v)\Lk :ai s 01 and for i E Ek, k t si = ai. Let I k = Ei U (v DI). Then

.

u:;

uz+:

uzj

.

.?

4

..

...

+

and

as required for the columns condition. In order to show that C i e ~di, is in S, it suffices. by Lemma 15.23(a) to show that 4 is in EL S. TOthis end, let w 0 be given and pick p E P with p > u/e. Pick the iE Nu and E Nu that we used lo get (Dl (p). Dz(p), ..., D,(p)). Pick y so p-11 x { O , l , ...,p t h a t f o r d l i e D t , y ( p . y i ) = y . Pick(a,b,c) E (1,Z 1) x ( 0 , l )such that a,(yi) = (a. b, c ) for all i E (1,2,. .. ut. Let C E a blp. For i E Dk, yi = a . pY + b pr-l zi pY-Z where 0 + zi < p, and hence Dj. y;/pY = L zi/pZ; let Ai = zilpZ and note that 0 5 A; < Ilp. For i E we have y (p, yi) 5 y 2: let Ai = yi/pY and note that 0 iAi -z I l p . (Of course we have no control on the size of yilpv for i E Mk.) Now A i = 9 so

,...,

+

.

-

+ .

.

+

u=k+l

I 15.4 Image Partition Regularity Over N

Thus ELl (xi/pY). G = & ~ ~ b / p Y .& ) Consequently.

313

+ Xiah e . 4 +

&,D~ Ai

.&

11 XidIk 4 - (2;-1 (xi/(lpv)) .G + Xie&f, (-yi/[lpY)) .&)I! = II xy=t Xis~~(Ai/!) 5 Xy=k XiGD,

k/tl

< U/P C

c.

.. .... +

which suffices, since xi/(tpy) > 0 for i E (1.2.. ,v). u v) = Having chosen I I , Iz, .,I,, if (1.2, Ij, we me done. So Now assume(l.2 ,..., u + u ) f L-&Ij. LetIm+1 = ( 1 . 2 ....,u+v)\l$=lIj. (1.2 U) = Uy=l Dj.so[-dl. - d ~ . c (& : i E U;(sl Ij)andhencewe can write XiEL+, as a linear combination of (6 : i E Uy=, I,). (b) implies (c). By Theorem 15.20 M satisfies the columns condition over Q. Thus by Lemma 15.15. t h m exist some m E (1.2, ...,u v) and a (u v) x m matrix F with entries from o which satisfies the first entries condition such that M F = 0, the u x m matrix whose entries are all zero. Let S denote the diagonal v x v matrix whose diagonal entries are sl,4.. . .,s,. Then M can be written in block formas ( A S - I )

..

,....

6

- - ....-4)

+

and F can be written in block form as

+

( ),where I denotes the u

x r identity matrix

and G and H denote v x m and u x rn matrices respectively. We observe that G and H are first entries matrices and that ASG = H, because M F = 0.We can choose d E N such that all the entries of dSG are in o.Let B = d H . Then B is a first entries matrix. Let y' E Nm be given and let I = dSGy'. Then Ax' = By' as required. (c) implies (d). Given B. let d E N be a common multiple of the denominators in . such that Ax' = B i = Cy'. entries of B and let C = Bd. Given y', let ? = d y and pick ? (d) implies (e). Let C be given as guaranteed by (d). By Lemma 15.14, there is an m x rn matrix E for which D = C E has entries from o and also satisfies the fint entries condition. Given y' E Nm,we define i E Nm by i = Ey'. Pick x' E Nu such that A? . = Ci. Then Ci = CEy' = D j . (e) implies (9. Let D be given as guaranteed by (e). and for each j E [I, 2, . .. m), pickw, E Nsuchthatforanyi E (1,2. .... ulif j =min(r E (1.2. ....m) : di,, $ 0 ) . then di,j = wj. (That is, w, is the first entry associated with column j. if there are any ,w,). Define first entries in column j.) Let c be a common multiple of (WI,wz, the u x m matrix E by, for (i. j ) E (1.2, ..., u) x (1.2, .. .,m), ci,j = (c/wj)di,j. Now, given 7 E Nm,w e d e h e ? E N m by, for j E (1, 2.. ..,m). zj = (c/wj)yj. Then Di = E j . ( 0 implies (a). For each c E N, cN is a member of every idempotent in pN by Lemma 6.6. and is in particular a central* sel. Thus by Corollary 15.6 E is image partition regular over W. To see that A is image partition regular over M, let r E N and let M = U:=! Fi. Pick i E (1, 2,. . .,r ) and y' E Am such that Ey' E Flu. Then pick i E MU such that A? = Ey'.

.

...

314

.

15 Partitism Regulariv of Matrim

Statement (b)of T l m 15.24 is a computable condition. We illustrate i b use by determining whether the mahica

.

an image partition regular over N. Consider the matrix

0 3s1 IS2 -1 2.~1 3 s 2 0 -1 where sl and ~2 arc positive ration!^. One quickly sees that the only possible choice for a set I1 of columns summing to 0 is 11 = ( 1 . 2 . 3 . 4 ) and then, solving the equations

3s1 2.q

+ 32

=1

+ 352 = 1

om g a sl = 2/7 and 32 = 117 and one has established that

is image partition regular. Now consider the matrix

48,

0

-I

Is1 -1q

0

6S2

0

-1

where again sl and Q arc positive rationals. By a laborious consideration of cases one sees that the only non zero choices for st and sz which make this matrix satisfy the columns condition are sl = 315 and q = -215. Consequently,

(i i) is not image partition regular.

Exercise 154.1. Rmn that the matrix 2

0

2 -2

3

is image partition regular.

Exe& 15.4.2. Let A be a u x v matrix with entries from o. Rovc that A is image partition regular over (N,+) if and only if A is image partition regularover (M..). (Hint: Consider the proof of Theorem 15.17.)

I55 Maaices with Pmr*s from Fklds .

.

315

15.5 Matrices with Entries from Fields

In the g c n d situation whae one is dealing with arbitrary commutative semigroups, we restricted our coefficients to have enhies from o.In the case of (N. +),we allowed the entries of mahices to C o w from 0.Them is another natural setting in which the maies of a efficient matrix can be allowed to come from other sets. is the case in which the s e m i p u p is a vector space over a field. We show here that the appropriate analogue of the k t entries condition is sufficientfor image partition regularity in this case, and that the appropriate analogue of the columns condition is sufficient for kernel partition regularity. We also show that in the case of a vectorspace over afinire field, the columns condition is also necessary for kernel partition regularity. We begin by generalizing the notion of the first entries condition from &finition 15.2. Definition 15.25. k t F be a field let u, v E N.and let A be a u x v matrix with entries from F. Then A satisfies thefirst entries condition over F if and only if no mw of A is 6 and whenever i, j E (1.2.. ..,u) and k = m i n { t ~ { I , 2..... v ) : a i , , # O ) = m i n ( t ~ { 1 , 2,.... v):a,,,#O). then ai.k = aj.k. An element b of F is afirn enhy of A if and only if them is some row i of A such that b = ai,k where k = min{t E 11.2,. . . v) : ai., 0). in which case b is thefirst enrry of Afkm column k.

.

+

Notice that the notion of 'Hrst enhiescondition"fromDefinition 15.2 and the special case of Definition 15.25 in which F = Q not exactly the same since there is no requirement in Delinition 15.25 that the first enhies be positive (as this has no meaning in many fields). We now consider vector spaces V over arbitrary fields. We shall be dealing with vectors (meaning ordered v-tuples) each of whose enhies is avector(meaning amember of V). We shall use bold face for the members of V and continue to represent v-tuples by an arrow above the letter. Tbe following theorem is very similar to Theorem 15.5 and so is its proof. (The main difference is that given a vector space V over a field F and d E F\(O), one has dV = V so that dV is automatically central* in (V, f).) Accordingly. we leave the proof as an exercise. Theom 1536. Lct F be afeld and let V be an infinite vecror space over F . Lct u, v E Nand let A be a u x v matrix with enm~esfiumF which smisfies thcfirst entries condition over F. Le C be central in (V. +). Then there exist sequences ( x ~ . ~ ) ~ = , , (xz..)zl. ....(xu..)z, in V such rhor for every G E Pf(N), .?G E (V\{O))" and A i G E CU, where

/ L E G XI.. \

316

15 Partition Regularity of Mattices

Pm$ This is Exercise 15.5.1. The assertion that V is an infinite vector space over F reduces, of course, to the a d o n that either F is inlinite and V is nonh*ial or V is infinite dimensional over F.

Comhry 1527. Let F be afield and Id V be an infinite vector space over F. Lcr u. v E Nand Iet A be a u x u matrix with entriesfrom F which thefirst entries conditionover F. Then A is~lron#Iyimage purrition regular over V . T h t is, whenever r E Wand V\(O) = U;-, Ei. there uisr i E (1.2.. .. r ) and? E (V\(O])Y such that A? E EiY

.

PI&$ We lint observe that ( 0 ) is not central in (V. +). To see this, note that by Corollary 4.33. V* is an ideal of (PV, +) so that all minimal idempotents are in V*. Consequently om may choose i E (1.2. ... r ) such that Ei is central in (V. +). Pick. by Theorem 15.26 some i: E (V\(O))' with A? E EiU. 0

.

Now we turn our attention to kcmel partition regularity. We extend the definition of the columns condition to apply to matrices with entries from an arbitrary field. Delinition 1528. Let F be a field let u , u E N. let C be a u x u matrix with entries from F, and let &. &, ... be the columns of C. The matrix C satides the columns condition over F if and only if there exist m E ti and 11,12.. ... I , such that

.c;

(3) If m > 1 and t E (2.3,. ..,m).let JI = F such that xi,,, 6 = &I, 61.i . 4

Ij.

Then there exist (8,,i)i,l, in

Note that Definitions 15.13 and 15.28 agree in the case that F = Q. Theorem 15.29. Lct F be afield let V be an infmilc vector space over F, Iet u , v E N. ond Ier C be a u x u matrix with entriesfmm F that satiqiies the columns condition over F. Then C is keml panition rcgular over V. That is, whenewer r -E ti rmd V\(O) = U:=, Ei, then exist i E ( 1 . 2, ... r ) and? E E,! such that C i = 0.

.

...

P m f : Pick m E M, 11.12, ..., I,, and for r E (2.3,. m ) , J, and (8,,i)i,J, as guaranteed by the definition of the columns condition. Define the u x m matrix B by, for (i. r ) E (1,2.. . . v ) x (1.2,. .. ,m].

.

We observe that B satisfiesthe fint entriescondition. We also observe that CB = 0. (Indeed, let j E (1.2.. . .,u ) and t E (1.2. ...,m ) . If t = 1, then cj.i . bi., = &I, cj.i=Oandift> 1 t h e n ~ i V , , ~ j . i ~ b i , ~-8,,i.cj.i+& =~i~j, cj.i=O.)

'

15.5 Matrices with Entries fnnn Fields

3 17

Now let r E N and let V\{O) = U;=[ Ei. Pick by CmIIary 15;27 some i E (1.2, r ) an$ j E (V\(0))m such that B j E EiY. Let f = B j . Then Cr' = CBf = O j = 0. 0

....

We see that in the case that kemd partition regularity.

F is a finite field, we in fact have a characterizationof

Theorem 1530. Let F be ajinitefirld, let u, u E N,and let C be a u x v matrix with entriesfmm F . Thefollowing storements a n equivalent. (a) For each r E N then is some n E N such that whenever V is a vector space of dimension ar least n over F and V\(O) = U:=, Ei then exist some i E (1.2 ,.... r J a n d s o m e f e ~ ~ ' s u c h t h a t ~ f = 6 . @) The matrix C satisfies the columns condition over F.

-

Proof: (a) implies @). Let r = IF1 1 and pick n E N such &at whenever V is a vector space of dimension at least n over F and V\(O) = U:=, Ei there exist some i E (1.2. ... r) and same Z E EiYsuch that C f = 6. Let V = X:=I F . Since in this case we will be working with v-tuples of n-tuples. let us establish our notation. Given f E VY.we write

.

...

...,u), xi = (xi (1). xi (2), ,xi (n)). We color V accnding to the value of its first nonzero d n a t e . For each x E V\(O). let y(x) = min{i E (I. 2. ..n) : x(i) # 0). For each u E F\{O). let Em= (Ir E V\{O) :~ ( ~ ( x=)a)) . Picksomea E F\(O) andsomef E EnYsuchrhat C f = 0. Lcl D = xi) : i E (I, 2,. .. u)) and let rn = IDI. Enumerate D in increasing order as (dl. 4....,dm). For each r E (1,2.. .. m). let I, = (i E (1.2,. . . u) : y(x;) = dl). For t E (2.3,. ..,m), let JI = Uj;; Ij and for i E I,, let = -xi(d~).a-I. To see $at these are as required for the columns condition. we first show that Xi,/, ? = 0.Tothisend let j E (1.2, . . .,u) begiven. We show that Xi.l, cj.i = 0. NOW ~f = 6 so X;='=,cj,i .xi = o so in particular, X;=t cj.; . xi(dl) = O. NOW if i E (1.2..... u ) \ I ~ . t h e n d < ~ y(xi)soxi(dl) = O . ThusO= Xiell cj,i.xi(dl)= o.xi,/, cj.isozi,1, cj.i=o. 6 = &J, 4.; .?,we Nowassumem s 1 andt E (2.3, ,mJ. Toseethat &, again let j E 11.2, .. . .ul be given and show that &I, c1.i = XiEjra,.; . cj.; Then ~ ~ = l c j . i . =O~)inparticular.C,~=~cj.i.xi(d,)=O.Ifi xi E (1.2 ....,v)\(f,UJ,). thenxi(d0 = OsoO = Xie/, ~ ~ . ~ . x ~ ( d cj.i.xi(dr) ~ ) + C ~=~ Xi,,, ~ , cj.i.a+Ci,~, cj.i. and given i E {I, 2,

..

.

.

...

.

IS F'atitioa Regularity of Matrices

(b) implies (a). We use a standard compaemess apment. Let r E Nand suppose theconclusionfails. Forcachn E Nlet V, = (x E Xi,, F :foreachi > n,x(i) = 0). Then for each n. V. is an n d i n s i o n n l veclor space over F and V,, V,,+l. Now given any n E N, there is a vector space V of dimension at least n om F for which there exist El. E2.. . E, with V\(O) = Ei swh that for each i E (1.2.. ..,r ) and each 2 E EiY. C.? # 6. The same asmliou holds for any n-dimensional subspace of V as weH. Thus we can assume that we have some cp. : V,\(O)_+ (1.2, ...,r ) such that for any i E (1.2, ...,r ) and any 2 E (9,,-1[(i)])Y, cz 0. ChooseaninfinitcsubsetA~ofNsothatforn.mE A l ~ n e h a s ( p , ~= ~ ~( ~~not , , , , \ , ~ ~ . Inductively, given A,-!, choose an infinite subset A, of At-I with min A, ? t such that forn,m E A, onehas("nlK\l~l =qmlv,\,ol.Foreachrpickn(t) E A,. Let V = V. Then V is an infinite vector space over F. Forx E V\(O) pick the first t such that x E V,. Then x E V, E V,(,) because n(r) E A, and min A, 2 r. Define(p(x) = cp.(,)(x). By Theorem 15.29picki € (1.2,. .-..r) and2 E ((p-'[(i)])" V,. such that c.? = 6. Pick t E N such that (XI.x2, ... ,x.] We claim that for each j E (1.2, ,v ) one has w&j) = i. To set this, let j E (1.2....,~)begivenandpickthelesstssuchthatx~ E &.Thenn(s),n(r) EA, so q.(,,(xj) = %(,,(xi) = = i . Thus ?. E (p,,ct)-l[(ill)Y and C2 = 5, a conhadiction. 0

..

+

UF=l

...

c

Of course any field is a vector space over itself. Thus Corollary I f .27 implies thah if F is an infinite field. any matrix with entries from F which satisfies the first entries condition over F is strongly image partition regular o m F. Exrrcige 155.1. Row Theorem 15.26 by suitably modifying the proof of Theo-

rem 15.5.

Notes The tcfminology"image @tion regular" and "kernel partition regular" was suggested by W. Deuber. Matrices satisfying . . the first entries condition are based on Deuber's (m. p. c) sets (761. The columns condition was introduced by Rado [206] and he s h e d there that a matrix is kernel partition regular over (N. +) if and only if it satisfies the columns condition over Q. Other gmeralizations of this result w m obtained in [206] and [207]. The pmof that (a) and (c) are equivalent in Theomn 15.20 is based on Rado's original arguments [206]. It is shown in [162], a result of collaboration with W. Woan, that there

solutions to the systan of equations iCr 7 in any cenhal set in (N. .) if and only if C satisfies the columns condition o v a Z. The proof given here of the sufficiency of the columns mdition using the image partition regularity of matrices satisfying the first entries condition is based on Deuber's proof that the set of subsets of W containing solutions to all kernel panition regular manices is partition regular [76]. The material from Section 15.4 is taken from [142], a result of collaboration with I. Leader, where several other characterizationsof image panition regular matrices an given, including one which seems to be easier to verify. This characterization of the image partition regular matrices is quite recent Whereas Rado's Theorem was proved in 1933 [206] and (m. p. c ) sets (on which the first entries condition was based) were introduced in 1973 [76]. the characterization of image pattition regular matrices was not obtained until I993 [1421. Theorems 15.26. 15.29. and 1530 arc from [21], written in collaboration with V. Bergelson and W.Deuber, except that t k the field F was required to be countable and the vector space V was required to beof countable dimension. (At the time [21] was mitten. the Central Sets Theorem was only known to hold for countable commutative semigroups.)

-. Chapter 16

IP, IP*,Central, and Central* Sets

We saw in Chapter 14 that in any semigroup S, central sets have rich combinatonal content. Andourinwduction tothecombinatonalapplicationsof the algebraic shuctwe of p.7 came through the Finite Products Theorem (Corollary 5.9). We shall see in this chapter that sets which intersen FP((x.)El) forevery sequence (x.)z, (that is, the IP* sets) have very rich combinatorial structure, especially in the semigroups (W. +) and (N,.). Further, by means of the old and often studied combinatorial m i o n of spectra of numbm, we exhibit a large class of examples of these special sets in (Pi. +).

16.1 IP, IP*,Central, and Central* Sets in Arbitrary Semigroups Recall that we have defined a cenfml set in a semigroup S as one which is a member of a minimal idempotent in BS (Definition 4.42). As one of our main concerns in this chapter is the presentation of examples, we begin by describing a class of sets that arc central in an arbitrary semigroup.

Theorem 16.1. k t S be a semigroup andfor each F UFEP,(S)(F .XF) is cenhul in S.

E

Pf(S),let X F

E S. l k n

Then (BF : F E Pf(S)] is a set of subsets of S with the finite intersection pmpeny so choose (by Theorem 3.8) some p E BS such that {BF : F E Pf(S)) G p. We claim that PS. p g 7i for which it suffices, since pp is continuous, to show that ~ . ~ ~ 7 i . ~ o t h i s e n d ~, lSe. tTs h e n t J ~~ , ~ s - ~ ~ s o ~s~-s ol b~~ T h e o r c m 4 . 1 2 A E sp. By Corollary 2.6, there is a minimal idempotent q E p S . p and so A E q.

In the event that S is countable, the description given by Theorem 16.1 can be simplified.

CoroUarg 16.2. Let S = (a. :n E M ) be a cowttable semigmup and let (x.)z, be a sequence in S. Thcn (a. .x. : n. m E M and n j m ) is central in S.

Pmof: This is Exercise 16.1.1.

0

We now i n d u c e some terminology which is due to Furstenberg [98] and is commonly used in Topological Dynamics circles. lk6nitim163. Let S be a semigroup. A subset A of S is an IP set if and only if there is a sequence (x.)El in S such that FP((x,)gl) E A.

Actually. as defined by Furstenberg. an IP set is a set which can be written as FP((x.)z,) for some sequence {x.)El. We have modified the definition be~auscwe already have a notation for F P ( ( x , ) ~ , )and because of the nice characterizationof IP set obtained in Thwrern 16.4 below. The terminology may be remembered because of the intimate relationship between IP sets and idempotents. However, the origin as described in 198) is as an "infinite dimensional parallelepiped". To see the idea behid that term. consider the elements of FP((x,)!=,) which we have placed at seven of the vertices of a cube (adding an identity e at the origin).

Theorem 16A. Let S be a semigmup and let A be a subset of S. Then A is an IP set if and only ifthem is some idempotent p E pS such that A E p.

P m f : This is a reformulation of Theorem 5.12.

0

In general. given aclass R of subsets ofa set S, one may define the class R* of sets that meet ever member of R. We have already done so (in Definition 15.3) for central sets. DeRnitl01116.5. Let S be a semigroup and let A ifforevery IPsetB S . A n B f 0 .

s S. Then A is an IP set if and only

Recall from Lemma 15.4 that a set is a central* set if and only if it is a member of every minimal idempotent. A similar characterization is valid for IP* sets.

Theorem 16.6. k t S be a sentigmup and let A 5 S. Tlu following statements are equivalent.

16 IP. IP.,Cenml. and Cenml* Sus

(a) A is an IF" set. (b) A is a member of every idempotenf of BS.

(c) A f l B is an IP set for every IP set B of S. Pmof: (a) implies (b). Let p be an idempotent of BS and suppose that A # p. 'Ihm S\A E p soby Theorem 5.8 thereis a sequence ( x , , ) ~ in ~Swith FP((x.)zl) G S\A. Tbat iq S\A is an IF' set which misses A, a contradiction. (b) implies (c). Let B be an IP set and pick by Lemma 5.1 1 an idempotent p of BS sucbthatB E p. Then A r l B e p s o A n BisanIPsetbyTheomn5.8. That (c) implies (a) is trivial. 0 As a trivial coosequence of Lemma 15.4. Theorems 16.4 and 16.6. and the dehition of central, m e has that

IF'*

* cmtral* * central * IF'.

One also sea immediately the following:

Remark 16.7. Lrt S be a sernigmup and let A and B be subsets of S. (a) IfA and B on I P sets, then A n B is an IP* ser. (b) IfA ond B am cennol* sets, then A n B is a centml* set. We s a now that in most reasonable semigmups, it is easy to produce a specific a n d * set which is not an IF'* set.

Tbeorrm 16.8. Let S be a semigroup and a s m e that (x.)z, is a sequence in S such thatFP((x.)~,) isnotpiecnvisesyndetic. 7'hen S\ F P ( ( x i ) z l )isa central* setwhich is nor an IF set. Pmof Trivially S\ FP((x.)zl) is not an IP* set Since FP((x&) is not piacwisc syndetic. we have by Theorem 4.40 FP((x.)E,) n K(pS) = 0 so for every minimal idcm~otmtp, S\ FP((x.)z,) E p.

Of course. since the notions of central aad 1P arc c b ~ z c byd membership in an idempotent, they arc partition regular notions. Ia hivial situations (see Exercises 16.1.2 and 16.1.3) the notions of central* and IP* may also be partition regular. Lemma 16.9. Let S be o sernigmup. (a) The norion o f I P is pnition regular in S i f and only ifpS has a unique idempotent. (b) 77renotionofcennal* ispartition ~ g u h r i nS ifandonly ifK(BS) has a unique idemporenr. Pmof We establish (a) only. the other proof being very similar. Necessity. Suppose that pS has two idempocents p and q and pick A A U (S\A) is an IP* set while neithct A nor S\A is an IP* set.

p\q.

Then

16.1 Sets in Arbitmy Semipups

323

Sufficiency. Let p be the unique idempotent of PS. Then a subset A of S is an IP* set if and only if A E p. 0 As a m n s e q of~ Lemma ~ ~ ~ 16.9 we have the following.

R m u L 16.10. Lrr S be a semigroup.

If the notion of lP. is partWon

rrgular in S.

then so is the notion of cmtml*.

Exercise 16.1.3 shows that one may have the notion of central* pattition regular when the notion of IP* is not. We see now that in more civilized semigroup. the notions of IPO and central* arc not partition regular. ('heomn 16.1 1 isin fact acomllary to Corollary 6.43 and Lemma 16.9, but it has a simple selfcontained proof, so we present it.) Theorem 16.11. Lrr S be an i n i i t e weakly lcp canceIlative semigmup. Then exin disjoint centml subsets of S. Consequently, neither the notions of c e n m P nor I P am panition regular in S. Pmofl Let K = IS1 and mumerate Pf(S)as (Fa).,,. Obsem that wheBES andIBI < x a n d F E q(S)thereissomex ESsuchthat(F.x)1lB=0. (Foreach u E F and u E B, (x E S :ux = v ) is finite by thedefinition of weakly leficaacellative so I U y e ~ U y . ~E lS~: ux = u)I -= K.) Choose xo E S and choose yo E S such that (Fo . yo) n (Fo xo) = 0. Inductively, let a < K and assume we have chosen (x,),, and (yo).,. in S so that

.

LetB = (Uo,,F,.xo)U(U,,,Fo.y,)andchoosex. E Ssuchthat (F..x,)nB =0 and choosey, E S such that (F, y.) n ( B U F, x.) = 0. By 'Ihcorem 16.1 we have that (U,,,F, . xu) and ((U,,F, .y,) sre disjoint central subsets of S.

.

.

In any semigmup, we see that IP9 sets satisfy a significantly stronger combinaknial conclusion than that given by their definition.

T h e o m 16.12. Let S be a semigroup. let A be an IP* set in S. and let (x):,. be a sequence in S. There is a pmducr subsyam (y,)g, of (x.)E1 such that m u ~ . ) z , ) c A.

Proof: Pick by Lmuna 5.11 an idrmpotent p

E

flS such that FP((x,)r-)

every m E Pd. Then A € p by Theorem 16.6, so Theorem 5.14 applies.

E p for 0

Exercbe 16.1.1. Rove Corollary 16.2.

Exercise 16.13. Let S beasetandleta E S. Definexy = a foranx and y in S. Show that a subset A of S is an IP set if and only if a e A and c o n ~ u e n t l ythat the notions of IP* and central* arc partition regular for this semigroup.

16 IP. IP*, Cenhll. and Central* Sets

Ex&

16.13. Recall the semigroup (N, A) where n Am = min(n, m). (See Exercise

4.1.11.) Verify eacb of the following assertions. (a) In (BN, A) every element is idempotent and K@M, A) = (1). (b) The P s e t s in (N. A) are the nonempty subsets of N and the central sets in (N, A) are the subsets A of N with 1 E A. (c) The only IP* set in (W, A) is N, while the central* sets in (W, A) are the same as the central sets. Consequently, the notion of IP* is not partition regular in (N, A) while the notion of central* is partition regular in (N, A). Exercise 16.1.4. Recallthesemigroup (N, v ) wherenvm = max[n, m]. (SeeExmise 4.1.1 I.) Verify each of the following assertions. (a) In (BN,v ) every element is idempotent and K(BM. v) = W. (b)The IP sets in (N, v ) are the nonempty subsets of N and the central sets in (N,v ) are the infinite subsets of N. (c) The only IP* set in (N, v ) is N. while the central* sets in (N. V) are the cofinite subsets of N. Consequently, neither the notions of IP* nor the notions of central* arc partition regular in (N, v).

16.2 IP* and Central Sets in W In this section we compare the additive and multiplicative structures of N. We begin with a trivial o b ~ e ~ a t i o n .

Lemma 16.13. k t n E N. Then Nn is an IP* set in W.+).

P m f This is an immediate consquence.of Theorem 16.6 and Lemma 6.6.

0

Now we establish that central sets in (N. +) have a richer shucturr than that guaranteed to an arbitrary commutative semigroup (and in fact richer than that passessed by central sets in (N, .) ).

Theonm 16.14. Let B be centml in (N. +). let A be a u x v matrix with enm'es

Imm Q.

(a) IfA is image panition regular, then there exists j E NUsuch that all entries of Ay' are in B. (b) IfA is kernelparfition regular, then there exists j E BV such that Ay' = 6.

P m f (a) Pick by Theorem 15.24 some rn E Nand a u x m matrix D satisfying the first entries condition with entries from o such that for any i E Nm there is some 3 E NU with Ay' = Di. By Lemma 16.13. for eachn E N. Nn is an IP* set and hence acmtral* set. Consequently, by Theorem 15.5, pick Z E Nm such that all entries of D t are in B and pick y' E Nu such that A j = Di. (b) Let d E N be acommon multiple of all of the denominatorsof enaies of A. Then dA is a matrix with entries from Z which is kernel partition regular. (If N = U ,,; Ci.

4

.

16.2 IP*and Cenbal Sets in N

325

..

pick i E (1.2,. .rl and i E CiYsuch that A.? = 6. Then (dA)i = 6.) 'Ibus by Rado's Theorem Clheorem 15.20) dA satisfies the columns condition over Q so, by Lemma 15.15. pick m E N and a v x m matrix C with entries from w which satisfies such th_atall the first entries condition such that dAC = 0. As in pan (a), pick; E

ym

enmesofCiareinBandlet~=Ci.'IhendA~=dACi=Oi=OsoA~=O. In fact, c e n d sets in (a, +) not only contain images of all image partition regular matrices, but all finite sums choosing at most one term from each such image as well.

(Yn)zl

+)

Definition 16.15. Let (S. be a semigroup and let be a sequence of subsets ofS. ThenFS((Y.)zl)= [&,,,=a. : F E Pf(M)andforalln E F. a. E Y,,).

Theorem 16.16. Let B be centml in (N, +). Let (A(n))zl entuneme the imngc partition mgular matrices with entriesfmrn Q andfor each n. let m(n) be the number of columns of A(n). There cristsfor each n E W a choice of i ( n ) E Wm(") such that, if Y,, is the set of entries of A(n)?(n), then FS((Y,)zl) G B. Pmof: F'ick aminimal idempotent in (BW, +) such that B E p. F'ickby Theorem 16.14 somei(1) E Wm(" such that all entriesof A(])?(]) arein B* = (a E B : -a+B E p). Let YI be the set of entries of A(l)i(l). Inductively. let n E PI and assume that we have chosen ?(k) E for each k E [I, 2,. . . ,n) so that, with Yk as the set of entries of A(k)?(k). one has FS((Yk);_,) c B'.ByLemma4.14,foreacha E B ' , - a + B ' ~ p s o

so pick ;(n

+ 1) E Nm"+') such that all entries of A(n + I)i(n + 1) are in ~ *nc-~ n + B*

Letting Y.+I be the set of entries of A(n

: a E FS((Y~);=~)I.

+ I)i(n + 1) one bas FS((yk);:f)

g B'.

o

A similar result applies to kernel partition regular matrices. Theorem 16.17. Let B be central in (W, +). Let ( A ( n ) ) z l enumerate the kernel pariition regular matrices with entriesfmm Q and for each n, let m(n) be the number of columns $A(n). %re existsfor each n E W a choice of i ( n ) E Nm(")such that A(n)i(n) = 0 and, if Y, is the set of entries of i(n), then FS((Y.)Zl) E B. Pmf. This can be copied nearly verbatim from the proof of T h m m 16.16.

0

The contrast with Theorems 16.14, 16.16, and 16.17 in the case of sets c e p l . in fact central* in (W, .) is striking. Recall from Section 15.3 th_e notation iA = 1 which represents the multiplicative analogue of the equation A? = 0.

Thrn C is c e n M in (N, .). A is m e mrion n g u h in (N. .), ond B is kemd portirion ngulor in (N. .). Howevec for no x E N is x2 E C &for no x' E C ' is is = 1.

The reader was asked to show in Exmise 15.1.2 that (x2 :x E N) is ncd cabal in (N. .). C q t l y , its complementmust be anailla.Trivially A is image panition regular in (N..) and B satisfies the colutuns condition over Q. so is kernel partition regular in (N. .) by Theorem 15.20. 0

P&

'Ibe following lemma establishes that dtiplication p n s e m s IP* sets in (N.

Lcanaa 16.19. k t A E N ond let n

E N.

+).

Thefillowin8 srcunnmrs o n equivdm

(a) A is M I P set in (N, +). (b) n-'A ison I P set in (N, +). (c) nA is an IP* set in (N. +).

A.

(a)implics(b). ~ s s u m i t h aAt is an IP*set and let Bbean IP Picka~cqucnw fl A # 0 so n-'A n B # 0. (XI)PO,~ such that FS((x,)E,) G 8. T h m FS((~X,)PO,~) (b) implies (a). Assume that n-' A is an IP* set, and let B be an IP set. Pick a sequence (xt)El such that FS((x,)E1) G B. By Lemma 16.13, Nn is an IP* set in (N. +) so pick by Theorem 16.12 a sum subsystem (Y,)PO_~ of (X,)PO,~such that FS((Y,)PO,~) E Nn. In particular n divides y, for each t. Consequently

P&

We see now that IF" sets in (N, +) are guaranteed to have substantial multiplicative shuchue.

Tbwrmn 16.20. k t d bc the set offinite sequencesin N (inctuding the empty sequence) ond let f : d + N. Let (y.)zl be o sequence in M ond let A be on IP* set in (N. +). Then then is o sum subsystem ( x n ) z l of (y.)zl such that whenever F P,'/O. mmF-I ),ondt E (1.2. ..., L), onehart. Z n e ~ xEn A. C = f((xdk_,

Pmof: Pick by Lunma 5.11 some p = p + p in /3Nsuch that FS((ym)zO=,) E p for each m E N. Then by Lemma 16.19, we have for each t E W that t-'A is an IP* set in

nfiy'

t-lA and note ihat BI E p. (N. +) and hence is in p. Let 81= FS((y.)zl) r l Pickxl E B1' and pick HI E P,(W such that XI = &H, y,. Inductively, letn E Nandassumethat wehavechosen (xi)bl. (Hi);=', and (Bi);=' k h t h a t f o r e a c h i ~ ( 1 , 2 , n):

...,

'

Only hypotheses (I), (3). and (4) apply at n = 1 and they bold hivially. Let k = max H. + 1. By assumption FS((y,)rd) E p. Again we have by Lemma 16.19 that for each t E N, t-'A is an IF*set in (N, +) and is thus in p. Foreachm E (1.2.

...,nllet

.

Em = ~ ~ j e r :x0j# F C {1,2.. .. n) andm = min F).

.

By hypothesis (4) we have for each m E (1.2,. .. nl and each a E Emthat a E Bm' and hence. by Lemma 4.14, -a B*. E p. Let

+

B"+I= FS((YI)%) n

f((x1)y-J

rxI

1-'A

n n",=,

n.~,(-a

+ B,').

Since X.+I E FS((y,)z). choose Then B,+I E p. Choose X.+I E B.+I'. %+I E PfC9 such that min H.+I 2 k andx,+~ = &H.+, Y,. Then hypotheses (1). (2). (3). and (5) arc satisfied directly. To verify hypothesis (4). let 0 # F E (1.2.. ..,n + 1) and let m = min F. If n + l $f F,the conclusion holds by hypothesis, so assumethat n + l E F. If F = (n+ I). = xn+l E B,,+I'. so assume F # (n + 1) andlet G = F\{n + 1). Let then XjCjp~xj a = Ejecxj. Then a E Em sox,+l E -a + Bm'so X j e ~ x = j a +x.+l E B*. as

m.

We thushave that (x.)Z, is asurnsubsystem of &)El.Tocompletethe proof, let F E Pf(M). l e t t = f ( ( ~ ~ ) ~ ~ ~ - ' ) , a n Ed l{1,2, e t t ...,C]. Letm = minF. Then by hypotheses (4) and (5). X j e xj ~ E Bm E t-'A so t X.,FX. E A.

.

Corollary 16.21. Let A be an IP* set in (N. +) and let (y,)zl be a sequence in N. Then is a swn subyam ( x B ) z lof (y.)zl such char FS((xn):,)

' J F W ( ~ n ) ~Cl )A-

Pmof: Let d be the set of finite sequences in N. Define f (@ = 1 and given ( ~ j ) & ~ define f ((xi)&l) = xj. Choose (x,)El as guaranteed by Theorem 16.20. Letting t = 1, one sees that FS({xn)zl) C A. To see that FP((x,)zl) C A. let F E Pf(N)andletn = max F. IfIFI = 1,thennjIjcFxj=x,, E A.soassumeIFI > 1 ~ j . ~ E~ A.o j andletG = F\{n). Lett = nj,cxj. Thmr 5 f ( ( x j ) ~ ~ ~ ) s o =t.x,,

nJmI

Theorem 16.20 and Corollary 16.21 establish that an IP* set in @. +) must have substantial multiplicative structure. One may naturally ask whether similar results apply to central* or central sets. On the one hand, we shall see in Corollary 16.26 that sets

which a n central* in (N, +)have significant mult@cative structure, in fact are cenhal in (N. .). On the 0th- hand, central sets in (N. +) need have no multiplicative strucave at all (7korem 16.27) while central sets in (N. .) must have a significant amount of additive strucntre (Theorem 16.28).

Deflnltion 1622. M = ( p E pN : for each A E p, A is central in (N, +)). As an immediate consequence of the definition of central we have the fouowing.

~cmark1 6 2 . M = ce E (K (BN, +)).

Theorem 1624. M u a leji ideal of (BN, .).

P m J Let p E M and let q

E BN. To see that q p E M,let B E q p and pick a E N such that a-'B E p. Then a-'B is central in (N. +) so pick, by Theorem 14.25 a decreasing sequence (C.)z, such that

(i) for each n E N and each x E C. there is some m E N such that C, & -x and (ii) (C. :n E N] is collectionwisepiecewise syndetic.

+ C.

Then (aC,,)z, is a decreasing sequence of subsets of B and immediately one has thatforeachn E Nandeachx E aC..thereissomem E NsuchthataC, g -x+aC,,. It thus suffices by Theorem 14.25 to show that lac,, : n E N) is collectionwise piecewise syndetic. That is, using Lemma 14.20, that t h m exist sequences (g(n));" and (x.)z, in N such that for all n, m E N with m 2 n,

If a = 1, the conclusion is immediate from the fact that {Cn : n E N) is collectionwise piecewise syndetic so assume that a s 1. Since (C. : n E N) is collectionwisepiecewisesyndetic,choosesequences (h(n))zl and (y.)zl in N such that for all n.m E N with m 2 n.

For each n E N, let x. = ay. and let g(n) = ah(n) + a . Given n, m E N with m 2 n and given k E (I, 2, ....in), pick r E (0, I, ....a - I] such that k r = a1 for some I E (1.2 ,....m). Thenforsome j E (1.2 ,....h(n))wehavey,,,+l E -j+n:=ICi a1 E -aj aCi so x, k E -(aj r) aCi. Since so that ay, a j + r < g(n) we are done.

+

+

+ ny=,

+

+ + n;=,

We saw in Corollary 13.15 that K(pN, .) n K(pN. +) = 0. We pause to observe now that these objects are nonetheless close.

Comllary 16.25. K@N. .) flcC K@N. +) # 0.

1

16.2 P*and Central Sets in N

329

P m 4 By Theorem 16.24. M is a left ideal of (BN, .) and consequently has nonempty intersection with K(BN, .) while by Remark 16.23, M E cC K ( p N , +). Corollary 16.26. Let A be central* in (N. +). l l u n A is central in (M..). Since M is a left ideal of P m $ 'BY Lemma 15.4. E(K (BN, +)) g 7i so M c (PA, .). M r l E ( K ( p N , .)) # 0 by Corollary 2.6 so A is cenwl in (N. .). 0

Theorem 16.27. T h e n is a set A E N which is central in (W, +) such that for no y a n d z i n N i s ( y , z , y z ] S A . Inpar~icularfornoyE N ~ S { ~ , ~ ~ ) ~ A .

+

Proof: Let X I = 2 and inductively for n E N choose x.+l (x,, n ) Z . Notice in particular that for each n , x. s n. Let A = [x. k : n , k E Nand k 5 n]. By Corollary 16.2 A is central in (N. +). Suppose now we have y 5 z in N with [ y , z , y z ) G Aandpickn E Nsuchthatz G [x.+l,x,+2 x.+n]. Then y s 2 so yz s 22 s 2x. s x. n so yz 2 x.+l 1 w (x. n)' 2 z2 2 yz, a conwadiction. 0

+

+

,....

+

+

We see.now thatsets central in (N;.)must contain large finite additive structure.

Theorem 16.28. Lcr A E N be central in (N,.). For each m sequence (x,)y=' such t h a t F S ( ( ~ , ) y =)~ )A.

E

N there exists ajinite

Proof: Let

By Theorem 5.8. all idcmpotenls in (pN. +) are in T so T # 0. We claim that T is a two sided ideal of (BN, .)so that K(pN. -) G T. Let p E T and let q E pN. To see that q .p E T, let B E q . p and let m E N. Pick a E N such that a-'B E p and pick (x,):!~ such that FS((X,):=~) E a-'B. Then F s ( ( a ~ t ) ; "C _ ~B. ) Toseethatp-q E T,letB ~ p - q a n d l e t mE N . L ~ ~ C = ( ~ E N : ~ - ' B E ~ ) . Then C E p so pick (x,):=~ such that F S ( ( X , ) ~ =E~C. ) Let E = FS((x,):=,). Thrn E is finite, so n O e E a - ' B E q so pick b E a e E a - ' B . T l e n F S ( ( b ~ , )E~ B. ~~) Since A is central in (N,.), pick a minimal idempotent p in (pN. -)such that A E p. Then p E K ( p N , .) G T so for each m E N there exists a finite sequence XI)^=^ such thatFS((x~):=~)C A. It is natural, in view of the above theorem, to ask whether one can extend the conclusion to infinite sequences. We see now that one cannot.

Theorem 16.29. There is a set A N which is central in (A. .) such that for no sequence (y,,)zl does one have F S ( ( y , , ) z l ) C A.

Pm$ Let XI = 1and fmn E N pick some X.+I

z nx.,.

I

La

By Corollary 16.2 A is central in (N. :). Suppose we have a sequence ( Y . ) ~with ~ FS((y.)zl) G A. We may presmoe that the sequence (y,):. is increasing since in any m n t an incrrasing sum subsystem of (y,)g, can bechosen. Pick m, n, and r in N such that n < r, y,,, E Ix,, 2s". 3x.. ,nx.1, and ym+l E lx,, 2r,. 3xr.. rxrl. Pick k E (1.2, . r ) such that ym+l = kx,. 'Ihcn

...

.

..

...

Given any central set A in (N, +), om has by definition that there is a minirml idempotent p in (BN. +) such that A E p. Since p = p p, one has that

+

+

A is centml) is central. That is, evey central set and so. in particular, [x E N : -x often translates down to a central set. Recall that for any p E BN, -p = (-1) . p E BZ. S i by Exercise 4.3.5. -W i a left ideal of (PZ, +).one cannot have p = p (-p) for any p E N*. If ooc had p = (-p) p for some minimal idempotent p in (BN, +)one would have as above that foranyA E p.[x e N : x + A ~ p Epandhence(x ] EN:.x+Aiscentml]iscentral. Howem, according to Corollary 13.19 such an equation cannot hold. Nonetheless. we are able to establish that for any central set A in (PI, +), [x E El : x A is central] is central.

+

+

+

the om^^^ 1630. Let p E K(BN. +) Md Ier L be a aminid I@ ideal Then is an idempotenz q E L such :hot p = (-q) p.

+

of (BN. +).

P m $ LetT = {q E L : (-q)+p = p). Givenq. r E PN, -(q+r) = (-q)+(-r) by Lenuna13.1,soifq.r E T , t h a q + r E T. Dehael: p Z + BZbyL(p) = -p. (That is, in thesemigmup (gZ, .). l = A-I.) Then.! iscontinuous and T = L n ( p , ~ t ) - ' [ [ ~ ] ] so T is compact Therefore, it suffices to show that T # 0. F a then. T is a compact subsemigroup of PN. hence contains an idempotent by Theorem 2.5. Since p E K@N, +), pick a minimal left ideal L' of BW such that p E L'. Now W is a left ideal of BZ by Exercise 4.35 ro by Lemma 1.43(c), L and L' are Iefi ideals of BZ. In panicular -L p L'. We claim that -L p is a left ideal of BN so that -L+p = L'. Toseethis,letq E -Landletr E BN. Weshowthatr+q+p E -L+p. -q = -(r q ) by Lemma 13.1 we have Now -q E L so -r + -q E L. Since -r r + q E - L s o r + q + p ~ - L + p . S i n c e - L + p = L 1 a n d p € L1wehaveT#O asquirrd. 0

+

+

+ +

proof: Kck a a idempotent p in (BN.+) such that A G p and pick, by Theonm 16.30. a minimal idempotent q in (BN. +) such that p = (-q) + p. Then [ x € N : x + A E p ) ~ q a n d [ x ~ N : x + A ~ p ) ~ ( x E N : x + A i s c e n t r aol ) . We do not know wh& every centrpl* s a often translates eitha up or down to anothn central* set. However we do have following strong contrast with Corollary 16.31 for P scts.

Theorem 1632 nirm is an IP set A of (N. +) such thatfor atl n 6 N. neithur n + A nor -n +A is an IP* set. Proof: Let (D,,).,z be a sequence of pairwise disjoint infinite sets of positive even integers such that for each n, min D. > 12nl. For each n E Z enmerate' D, in increasing order as (a(n. k))& and for each k E N,let y..t = 2"(n.U)+ P(".=-l). Foreachn E Z\(O], let B. = FS((y..k)&). Let C = [n +z :n € Z\(OJ and z E 8.1. Notice that if n E Z\(O) and z E B. then In1 < .i so C g N. Let A = W\C.-Then givenany n E Z\(O).(A+n)rlB-. =O,soA+nisnotanIP*seL We now claim that A is an IP"L set, so suppose instead we have a sequence (x),: with FS((x,,)g,) n A = 0. That is. FS((x,)=,) E C. By passing to a suitable sum subsystem we may presume that the sequence ,x(:). is increasing. Wefirstobsewethat ifn E Z\(O), u E n+B,,.andu = &F 2'.thenmaxF E D,. Indeed u = n + z where z E B, and z = &.q 2' where H 5 D. and H has at least two members (becauseeach y..t has two b i i digits). Thus, if n c 0, borrowing will not reach man H. Since then min H > 12nl one has max F = max H. We now show that one cannot have n E Z\(O} and i < j with [xi. xjJ G n B., so 2' ~ where man F E Dm.Let k = max F. suppose instead that we do. Now XI = z t a Since xi < xj we have that xj +xi = C t e 2: ~ where either max H = k or max H = k + 1. But as we have just sen. given any member of C, the largest element of its biiw suppott is even, so the latter case is impossible. The former case tells us that B..soifL=minD.we xj+xi E n + & . ButnowIxi-n,xj-n,xi+xj-n]c have that 2' divides each of xi - n, xj n, and xi + x j - n, and hence 2' divides n, a conmdiction. Consequently, we may choose for each i E FI some n(i) E Z\[O] such that xi E n(i) + B,,(i) and n(i) p n(j) for i j4 j. Choose i such that In(i)l r xi. Then xi = Z t E F 2' and xi + XI = z,ac2' where max F = maxG and consequently, xi + X I E n(i) + B,,(i). But now11 = (xi + X I )-xi is a difference of two members of B,,(i) and is hencedivisible by 2'. where[ = min Dng). This is aconhadiction because L > (2n(i)l > X I .

+

-

Exerdse 16.2.1. In the proof of Lemma 16.19 we used the fact that in the semigmup (N. +), one must have n-'(nA) = A. Show that in (N. +) one need not have n(nW'A)= A. Show infactthatn(n-'A) = A if andonly if A E Nn.

16 IP. IP*. Central, and Central* Seln

332

Exercise 1622. Rove that if r € A and A = Ci, and for only one i E (1.2,. .. r ) is there a sequence ( y m ) z lin N with F S ( ( y , ) g l ) g Ci, then in fact for this i , Cj is an IP*set (so that there is a sequence ( x , ) g l in A such that FS((x,)Z,) u ~ ( ( x x , ) c c,).

.

: -

163 IP*Sets in Weak Rings In this section we extend Corollary 16.21 to apply to a much wider class, called "weak rings". obtaining a sequence and its finite sums and, depending on the precise hypotheses. either all products or almost all products in a given IP* set

+,

Definition 1633. (a) A left weak ring is a triple ( S , .) such that (S. +) and (S. .) are semigroups and the left distributive Law holds. That is, for all x. y, z € S one has x.(y+z)=x.y+x.z. (b) A right weak ring is a triple ( S , .) such that (S. +) and ( S , .) are semigroups and the right distributive law holds. That is. for all x , y, z € S one has ( x y ) . z = x.z+y.z. (c) A weak ring is a triple ( S , .) which is both a left weak ring and a right weak ring.

+,

+

+,

In the above definition, we have followed the usual custom regardig order of operations. That is x y x z = ( x y ) ( x .z). Notice that neitherof the semigroups (S. 9 )nor ( S , .)isassumed to becommutative. Of course all rings are weak rings. Other examples of weak rings include all subsets of C that are closed under both addition and multiplication.

. + .

. +

Lemma 16.34. Let ( S , +) be any semigroup and let . be the operation making ( S , .) a

+, +,

+)

has a? least one elemenf righr zem semigroup. Then ( S , .)is lefr weak ring. I f ( & which is not idempotent. then (S. .) is not a right weak ring.

P m J This is Exerci~e16.3.2. Analogously to Lemma 16.19, we have the following. Notice that it does not matter whether we define a-'Ab-' to be a - ' ( ~ b - ' ) or (a-'A)b-I. In either case, y € a-' Ab-' if and only if ayb E A.

Lermnn1635. LetSbeaset.IetA

S,andleta,b~S. (a) I f ( S . .) is a left weak ring and A isan IPS set in ( S , +), then a - ] A isan IF+ set in (S. +). (b)v ( S , .) is a right weak ring and A is an IP* set in (S. +), then ~ b - 'is an IP* set in (S. +). ( c ) I f ( S , + , . ) i s a w e a k r i n g a n d A isanIP*setin(S,+), t h e n a - ' ~ b - ' i s a n I P set in (S. +).

+, +,

.

16.3

J P Sets in Weak Ringa

333

P m f It suffices to establish (a) s i n e then (b) follows from a left-right switch and (c) follows from (a) and (b). So, let ( x , , ) z t be a sequence in S. Then

so pick F E P j ( W such that C , , ~aF .x,, E A. Then, using the left distributive law we have that &F xn E K I A . 0 Recall that in FP((X,)F=~).the products me taken in inmasing order of indices.

Definition 1636. Let (S, .) be a semigroup. let ( x . ) z l be a sequence in S, and let k E N . Then AP((x&) is the set of all products of terms of (x.):=~ in any order with norepetitions. Similarly A P ( ( x , ) E l ) is the set of all pmducts of terms of (x,,)El in any order with no repetitions. For example, with k = 3, we obtain the following:

Theorem 16.37. Let (S, +, .) be a lep weak ring. let A be an IP* ser in (S, +), and let m be any sequence in S. Then there exists a sum subsystem ( x n ) E I of ( y . ) E l (Y.),=~ such that f f m ? 2 F E P j ( N with minF 2 m, and b E A P ( ( x . ) ~ ~ ~then ), b . & F X , E A. Inpmricujan

Proof:Pick by Lemma 5.11 some idempotent p of (BS, +) with

Then by Lemma 16.35. we have for each a E S that a - l ~is an IP* set in (S, +) and hence is in p. Let B l = F S ( ( y , , ) E l )and note that BI E p. Pick X I E BI' and pick HI E P j ( N ) such that xl = C~EH, y,. Inductively. let n E Nand assume that we have chosen (xi):=, (Hi);=, and (Bi ):=, such that for each i E (1,2,. . .,n ) :

.

= & e ~ ,Yr. ( 2 ) if i > 1, then min Hi > max Hi-,

(1) xi

(3) Bi E P. (4) if 0 f F & (1.2..

(5) if i > 1. then Bi

.

.

...i) and m = min F, then X j G Fxj

E n ( a - ' ~: a E A P ( ( x , ) ~ I ~ ) ) .

E Bm.. and

16 1P.IP.Central. awl Central* Sar

334

Only hypotheses (1). (3). and (4) apply at n = 1 and they hold trivially. Let k = max H. 1. By assumption FS((y,)z) € p. Again we have by Lemma 16.35 thatfmeacha E S . ~ - ' ~ i s a n ~ P * s e t i n ( ~ , + ) a n d i s r h u s i n ~ . Foreachm~{1.2. njlet

+

...,

.

By hypothesis (4) we have for each m E 11.2,. .. m ) and each a B,' E p. Let and hence. by Lemma 4.14, -a

+

E

Em that a E Em*

m e n &+I E p. &msc &+I G B~+I'. S k *.+I E FS((y,)%), choose H.+I E Pf(N) such that min &+I 2 k and X,+I = &H.+, yf. Then hypotheses (1). (2). (3). and (5) are satisfied directly. To verify hypothesis (4). let 0 # F S {I. 2,. ,n 1) and let m = min F. If n+l $ F. theconclusion holds by hypothesis, so assume that n+ I E F. I f F = (n+l). then zj;jEF~ =jX"+I E B~+I.. SO assume F # {n 1) and let G = F\{n + 1). a = z j r ~ x j .Thena E Em Wx.+l E - a + EmgW E j G i x j =a+x,,+l E Bm*as required. We thus have that (xn)Elis a sum subsystem of (y,,)zl. To complete the proof, let m 1 2, let F E Pf(N)with min F 2 rn and let a E AF('x.);):'. Then by hypotheses (4) and (5). xj,rxj E 8. G a-' A so that a &F xt E A. o

.. + +

.

Observe that {b . x, : in E 2 and b E AF'((xn);i1)) is the set of all products (without *petition) horn (x& that have their largest index occurring on the right. Notice that the proof of Theorem 16.37 is nearly identical to that of Theorem 16.20. Essentially the same proof establishes a much shunga conclusion in the event that one has a weak ring rather than just a left weak ring.

+.

Theorem 1638. Let (3, .) be a weak ring. let A be an ZP* set in (S. +), and la (y.)zl be any sequence in X. Then then exists a sum subsysrem (x.)zl of (y,,)zl in Ssuch thntFS((x,)zl) UAP((x.)zl) E A.

Pmof Modify the proof of Theorem 16.37 by replacing induction hypothesis (5) with: (5) if i > 1, then

Thm replace the definition of

with

J

16.3 IPI Sda m Weat Rings

335

Theomus 16.37 and 16.38 raise the natural qucstion of whethex the stronger amelusion in fact holds in any left mak ring. The example of Lemma 16.34 is not a countmxampk to this question be~ause,in this left weak ring, given any sequence (x.)Z"=, one has AP((x,)zl) = (x, :n E N).

Theorrm 1639. Let S be Ihefree semigmup on the two distinct lettersa and b and kr hom(S. S)be the ser of homomorphisrns/mrn S to S. Let o be the usual composirion of fwtctions anddfie an operation 8on horn(& S)asfollows. Given f. g E h o d s , S) andu1,u2, ut E (a,bl.

....

( f 88)(ulu2.. .ut) = f (ul)"8(ul)"f

(u2)-8(~2)-.

.,-f ( ~ , ) - g ( ~ t ) .

Then@om(S. S),8,o) i s a ~ w e a k r i n g M d t h e n u i s t m I P * s e r A in @om(S. S). e) and a sequence (fm)Kl in hom(S. S) such rkor no swn subsystem (g,)gl of ( f n ) z l has AP(k")K,) G A. Proof: The wification that (hom(S. S). 8 , o) is a left weak ring is Exercise 16.3.3. Notice that in order to define a member of hom(S, S)it is enough to define its values at a and b. Definef~6 hom(S, S) by fl(a) = a b and fl(b) = b and define inductively forn E N, &+I = f ,8fi. Notice that foreachn E N. fn(a) = (ab)" and f.(b) = b". Let A = hom(S. S)\[ f, o f, :r, s E N and r > a]. Notice that, given r, s E N. I

, I

fr(f,(a)) = fr((aW)) = (f,(a)f,(b))* = ((06) 6 )

.

(Wehaveused the Fact that f, is a homomorphism.) In panicular, notice that if f, o f, = fm 0 fn, then (r, s) = (m. n). Weclaimthatifm,n,r,s.C,r~N,h=f o f n . k = f r o f r . a n d h @ k = ftof,, thene=rn=randr=n+s.Indeed

((ab)'bf)' = (h 8k)(a) = h(a)-k(a) = ((ab)"bm)"((ab)'b')'.

Suppose now mat (h),:

is a sequence in hom(S. S) with

W ( h . ) z ~ ) G [f, 0 f, :r,s

E

Nand r > sl.

Then. using the factjust established,there is some r and for each n some s(n) such that h, = f. 0 f,(.). Then h18h28--.8h,=f,of, where t = x i = I s(n) 2 r, a coneadiction. Thus A is an IPS set in (hom(S, S). +). Rnally. suppose that (g& isasum subsystemof ( withAP((g.)z,) A. Then gl = Cn,n f. for some H E .P,-(N) so g1 = fk where k = C H. Then gk+l = ft for some t w k and thus gk+l o 81= ft o fi $ A;i a conhadiction.

f,)zl

c

+.

Exmlse 163.1. Show that if (S. .) is a weak ring in which (S, +) is commutative. n E Wand the operations on the n x n mahices with entries from S an defined as usual (SO,for example, the cnhy in row i and column j of A . B is E;=I ai.k. bk j ) , then these mahices form a weak ring. Give an example of a left weak ring (S. .) for which (S, +) is commutative, but the 2 x 2 mahices over S do not form a left weak ring.

+.

16 IP, P. Central,and Central* Sar

336

Exercise 16.33. Prove Lemma 16.34.

.

Exerdse 16.35. Rove that (hom(S. S). @. o) as dcsaibed in Theorem 16.39 is a left weak ring. Errrdse 16.3.4. We know that the left weak ring (hom(S, S).@. o) ofTheorem 16.39 is not a right weak ~g because it does not satisfy the conclusion of 'Ihcorem 16.38. Establish this fact directly by producing f, g. h E hom(S, S) such that (f fB g) o h # f 0heJgoh.

16.4 Spectra and Iterated Spectra

+

Spectra of numbers are sets of the form (LnaJ : n E N1 or (Lna y J : n E N] where a and y are positive reals. Sets of this form or of the form {LnaJ : n E A ) a (Lna yJ : n E A] for some specified sets A have been extensively studied in number theory. (See the notes to this chapter for some references.) We are interested in these sets because they provide us with a valuable collection of rather explicit examples of IP* sets and central* sets in (N.+). By the very nature of their definition, it is easy to give examples of IP sets. And anytime one explicitly describes a finite partition of N at least one cell must be a central set and it is often easy to identify which cells are central. The situation with respect to IP* sets and cenhal* sets is considerably different however. We know from Themem 16.8 that whenever (x.)zI is a sequence in N such that FS((x,)=,) is not piecewise syndetic, one has N\FS((x.)E,) is a central* set which is not IPa. So, for example. (CtCF 2' : F E P,(N) and some t E F is even) is central* but not IP* because it is N\Fs((~"-')~,). We also know from Lemma 16.13 that f a cacb n E M. Nn is an IP* set. But at this point. we would be nearly at a loss to come up with an IP* set which doesn't almost contain Nn for some n. (The set of Theorem 16.32 is one such example.) Inthis section we shallbeutilizingsomeinfomationobtained in Section 10.1. Recall that we defined there for any suhsemigroup S of (R. +) and any positive real number a. functions g, : S + Z.f, : S + and h. : S -+ T by g d x ) = LXL ;J. f&) = xu g&). and h. (x) = n (f,(x)) where the circle group T = R/Zand n is the projection of R onto T.

+

+

[-4. i),

-

Mnitfon 16.40. Let a > 0 and let 0 < y < 1. The function g,.~ :N + a,is defined by g,,,(n) = Lna yJ.

+

We denote by. ;g

the continuous extension of g,,y h m pN to pw. Notice that

8e = g,,:.

Recall that for a > 0 we have defined Z = [ p E pS : ?,(p) = 0).

Lemma 16.41. Let a > 0. let 0 < y < 1. Md let p

E

Z,. Then g;(p)

= &(p).

16.4 Spsmand ltcratcd S p m

337

pmof:Lete=min(y.l-y)andletA=(n~N:-c< fdn) < ~ ] . T h e n A e p s o it suffices to show that g., and g, a p e on A. So let n E A and let rn = g,(n). Then m-s
As regards spectra, we see that "as you sew, so shall you reapP.

Theorem16.42. k t a r O , I e I O < y < 1 , a n d l e t A t N . (a) If A is an IP* set, then g+ ,. [A] is an IP* set. (b) IfA is a cenrml* set, then [A] is a central* set. (c) IfA is a centml set, then g,,,. [A] is a centml set. (d) IfA is an I P set, then g,. [A] is an IP set.

,,

is a member of every Pmof (a) By Theorem 16.6 we need to show that g..,[A] idempotent in (BN. +). So let p be an idempotent in (5N. +). Since ZI/, is the kernel of a homomorphismbyLenyna 10.3 we have that p E ZI/,. Since. by Theorem 10.12. gT, is an isomorphismfrom ZI/, to Z., gT,(p) is an idempotent and so A E gG(p). By Lemma 16.41 gTy(g;T.(p)) = &(g;J,(p)) which is p by Theorem 10.12. Since A E gG(p) and gT,(gTa(p)) = P we have that g , , CAI E P as required. (b) Let p be a minimal idempotent of (pW, +). We need to show that g,,JA] E p. Now p E K(5W) 0 Zllp and K(pW) nZ1/, = K(Zl/,) by Theorem 1.65. Thus p is a minimal idempotent in ZI/. so by Theorem 10.12 gG(p) is a minimal idempotent in Z. That is, gTa(p) E K(Z,) = K(pW) n Z so g G ( p ) is a minimal idempotent in (pN, +). Consequently, A E gTw(p)so as in part (a), g..,,[A] E p. (c) Since A is central, pick a minimal idempotenc p with A E p. Then p E K@N) n Z, = K(Z,,) so by Theorem 10.12, &(p) E K(ZI/,) = K(BW n ZI/, so &(p) is a minimal idempotent and g,,,[Al E gTy(p) = &(p). (d) Since A is an IP set, pick an idempotent p with A E p. 'Ihen p E Z. so by Theorem 10.12. &(p) is an idempotent and g,,,[A] E gyy(p) = &(p). 0 In the event that a is one to one.

> 1we get an even stronger correspondence, because then g,,,

Corollary 16.43. k t a 2 1, let 0 < y < 1, and let A N. (a) A is an IP* set ifandonly ifg,,,,[A] is an IP* set. (b) A is a centml* set ifand only ifg,,,[A] is a centml* set. [A] is a central set. (c) A is a cennal set ifand only (d) A i s m IP set ifand only ifg,,, [A] is M IP set. Pmof We establish (a) and (c). The proof of (b) is similar to that of (a) and the proof of (d) is similar to that of (c). (a) The necessity is Theorem 16.42(a). Assume that g,.,[A] is an IP* set and suppose that A is not an IP* set. Pick an IP kt B such that A r l B = 0. Since go., is

16

IP,IP,Cmtnl aad Cenhal* Set.?

one to one. g..,[Al nga.,[Bl = 0 while by Ib16.4Xdh 8a.y [Bl is an IP Jet, a conhadiction. (c) The necessity is nKonm 16.42(c). Assume that gJA] is a cenhal set and suppose that A is not a central set m e n N\A is a central* set so by Ib16.42(b), t., [N\A] is acenaal* set But this is aconaadiction becaw ~qyIAlnga.ytN\Al= 0.

0

Because fora s I the output of g., is the same kind of set as the input, one may iterate the functim at will. Thus, for example, if A is a set which is central* but not IP* (such as one given by Theorem 16.8). then

is a cenml* set which is not I F . Notice also that the description of a set ga.,[N is effective. That is,one only needs a sufliciently precise decimal approximation to a and y in order to dearmine whether a specified number is a member of g,., W.

The notions of IP. IP*. central, and central* are due to H.Furstenberg in [981 w h m central sets were defined in terms of notions from topological dynamics. See Chapter 19 for a prwf of the equivalence of the notions of central. Theorem 16.20 is from [IZO] where it had a purely combinatorialpmof. Most of the rcmainingresultsof Section 16.2 arc from [31], obtained incoUaborationwith V. Bergelson, except for Theorem 16.16 (which is from 1781 and was obtained in coUaboration with W. Dcuber) and Theorem 16.30 which, while previously unpublished, is from an early draft of [33], obtained in collaboration with V. Bergelson and B. Kra. The results of Section 16.3 are due to E. Terry in [234]. The left weak ring (horn($, S), fB,o) is due to J. Clay in [65] where it is given as an example of a left neaning which is not a right nearring. (The notion of nearring is a stronger notion than that of a weak ring.) The results of Section 16.4 arc from [33]. results of collaboration with V. Bergelson and B .Kra. H. Furstenbq 198. Proposition 9.41 gives an explicit description of c d n IP* seU in terms of topological dynamics. and other examples can be deduced from his paper [I001 with B. Weiss. Spectra of the form {[nu yJ : n 6 N) were i n d u c e d by T. Skolem [224] and given the name the y -nonho~genous spcfrmofa by R Graham. S.Lin, and C. Lia in [109]. See the introduction to 1331 for a brief history of the spectra (LnaJ : n E N) and further references.

+

-.

Chapter 17

Sums and Products

We saw in Chapter 16 that any IP* set in (N,+) has extensive fnultiplicative mucture in addition to the additive shucture that one would expect. In particular. by Corollary 16.21, if A is an IP* set in ( N , +),then there is some sequence ( x n ) z lin N such that FS((x,)z,) UFF'((x,Jzl) A. On the other hand. IF" sets in (N. +) arenotpartition regular by Thcorem 16.11 so this fact does not yield any results about finite partitions of W. In fact, we shall show in Theorem 17.16 that there is a finite partition of N such that no cell contains all painvise sums and products fmm the same sequence. We saw in Corollary 5.22 that given any finite partition of W, thm must be one cell A and sequences (x,)Z,and ( y d E l withFS((xdz,) U F W Y J ~G~A.) We am concerned in this chapter with extensions of this result in various differentdirections.

17.1 Ultrafilters with Rich Additive and Multiplicative Structure Recall that we have defined M = ( p e PW :for all A E p. A is central in (W, +)) and A = [p E PW :for all A E p.z(A) > 0).

Definition 17.1. A combirutorially rich ultmjilter is any p G ha r l A n K (BN,.) such thatp.p=p. We shall see the reason for the name "combinatorially rich" in Theorem 17.3. F i t we observe that thcy exist.

Le-

17.2 Then exists a combinatoriolly rich ultrafilter.

+)

P m f By Theorem 6.79 we have that A is a left ideal of (BPI. and so, by ComUary 2.6 contains an additive idempotent r which is minimal in (PA. +). By Remark 16.23 we have r E ha and consequently M n A # 0. Thus by Theorems6.79 and 16.24. Mn A is a left ideal of (PN. .) andhence contains a multiplicative idempotent which is minimal in (BPI. .).

17 Sums and Roducta

Recall the notionof FP-meintroduced in Definition 14.23. We cmll the corresponding additive notion an FS-tree. Since any member of a c o r n b ' i y rich ultrafilter is additivelycentral, it mustmntainby Theorem 14.25 anFS-tree T such that [B, : f E T ] is collectionwise piecewise syndetic, and in particular each Bf is piecearise syndetic. (Recall that Bf is the set of successas to the node f of T.) Notice, however, that in an arbitrary central set none of the Bf's need have positive upper density. (In fact, mall from Theorem 6.80 that W\A is a left ideal of (BN, +) and hence there are central sets in (W, +) with zero density.)

The0~llll7.3. Let p be a combinororially rich ulhofijter and let C E p. (a) C ir cenml in (W, +). (b) C is central in (N, .). (c) Let ( A ( n ) ) g lenumemfe the imagepanition regularmatrices with entrirsfrom Q andfor each n, let m(n) be the number of columns of A@). %re existsfor each n E W a choice of i ( n ) E Nm") such that, if Y. is the set of enfries ofA(n)i(n), then FS((Y")Z,) c c. (d) Let (A(n))zl enumerate the kernelpartition regularmamh?~with entriesfmm Q andfor each n, let m(n) be the number of columns g A ( n ) . There &ts for each n E N a choice of i ( n ) E Wm(") such thaf A(n).?(n) = 0 and, if Y,, is the set ofentries ofi(n), then FS((Y.)El) 5 C. (e) There is an FS-tree T in C such t hfor each f E T, a(Bf) z 0. (0There is an FP-tree T in C such t h for each f E T, Z(B,) s 0.

.

Prof. Conclusion (a) holds because p E M while conclusion (b)holds baause p p = P E K(BN. .). Conclusions (c) and (d) foUow from Thmms 16.16 and 16.17 respectively. To verify conclusion (e), notice that by Lemma 14.24 them is anFS-tree T in A such that for each f E T. Bf E p. S i c e each member of p has positive upper density the conclusion follows. 0 Conclusion (0 follows in the same way. As a consequence of conclusions (a) and @) of Theorem 17.3 we have in particular that any member of a combinatorially rich ultrafilrcr satisfies the conclusions of the Cenhal Sets Theorem (Theorem 14.11). phrased both additively and multiplicatively. There is naturally acomponding partitionresult. Notice that Corollary 17.4applies in particular when C = W. Corollary 17.4. Let C be a central* set in (W,+), let r E N, and let C = Uj=l Ci. There is some i f [I. 2, . ..,r ) such t h each of the conclusions of Theorem 17.3 hold with Ci replacing C.

Prof. By Lemma 15.4 we have ( p E K@N, +) : p = p + p) a combinatonally rich ultrafilter p. Then C E p so some Ci E p.

F so M F. Pick

-

17.2 Pairwix Sums nod Roducts

17.2 Pairwise Sums and Prpducts We know from Corollary 16.21 that any IP* set in (N, +) conlains FS((x.)E!P_!! u F P ( ( X . ) ~ ~for ) same sequence in N. It is natural to ask whether t h m is some pambon analogue of this result We give a strong negative answer to this question in this smtion. That is we produce a finite partition of N such that no cell contains the pairwise sums and products of any injective sequence. As a consequence, we see that the equation p p = p . p has no solutions in N*.

+

Detinition 17.5. Let {xn)Elbe a sequence in N. (a)PS({xn)zl)= {x. +x, : n, m E Nandn # m]. (b) PP((x.)zI) = (x. x, :n. rn E N and n # rn).

.

The partition we use is based on the binary repsentation of an integer. Recall that for any x e N, x = E,,,w(x) 2'. Definition 17.6. Let x E N. Then (a) a(x) = max supp(x). (b) E x $! (2' : t E w), then b(x) = max(supp(x)\(a(x))).

..

,a(x)l\ supp(x1). (c) C(X)= max(1-1.0.1,. (d) d(x) = min supp(x). (el If x $! (2' : t G w], then e(x) = min(supp(x)\(d(x)]). When x is written (without leading 0's) in binary, a(x). b(x), c(x), d(x), and c(x) are respectively the positions of the leftmost 1. the next to leftmost 1, the leftmost 0. the rightmost 1. and the next to rightmost 1. Remark 17.7. Let x E N\(2' : t E N] and let k E o. (a) b(x) 2 k ifandonly ifx 2 2"") 2k. (b) b(x) 5 k ifand only ifx < 2"(') + 2'"'. (c) C(X)2 k ifand only ifx c 2°(X)+' - 2'. (d) c(x) 5 k ifand only ifx 2 2"")+' 21r+'. (e) e(x) = k ifand only if k > d(x) and there is some m E o such thnt x = 2'+'m 2' 2d(r).

+

-

+ +

We now introduce some sets that will be used to define the partition that we are seeking. Definition 17.8. (a) Ao

= (x E N : a(*) is even and 20") c x c rb)+f 1.

A = (X E N : a (x) is even and 2a(x)+: c x c 24(X)+'1. A2 = (x E N : a(x) is odd and 2a(X)c x c 2a(x1+4I. A3 = (x E N : a(x) is odd and 2'(")+f c x < 2a(X)+'1. . Aq

= (2' : f E w).

(c) (CO.CI) is any partition of Nsuchthatforall k E W\(l). k + 1 E Coif and only i f 2 E CI. Notice that a partition as specified in Definition 17.8(c) is easy to come by. Odd numbers may be assigned at will, and if the numbers less than 2k have been assigned. assign 2k to the cell which docs not contain k 1.

+

Remark 17.9. (a) Ifx, y E A0 U Az, then a(xy) = a(x) + a(y). (b) Ifx, Y E AI U As. hen a(xy) = a(x) +a(y) 1.

+

We are now nady to define a partition of N. When we write x ss y (mod R)we mean. of course, that x and y are elements of the same member of 3.

+

Dellnitinn 17.11. Define a partition 92 of N by specifying that A4 and 2N 1 are cells of R and for any x, y E N\((2W -C 1) U A4). x a y (mod 3)if and only if each of the following statements holds. (1) Fori E ( O , 1 , 2 ) . x ~~ ~ i f a n d o n lE~Bi. if~ (2) For i E (0. 11, d(x) E Ci if and only if d(y) E Ci. (3) a(x) b(x) 5 d(x) if and only if a(y) b(y) 5 d(y). (4) a(x) c(x) 5 d(x) if and only if a(y) cCy) 5 dCy). (5) a(x) b(x) ia@) b(y) (mod 3). (6) a(x) 0 a ( ) (mod 2). (7) e(x) r e(y) (mod 2). (8) x = y (mod 1)6.

-

-

-

-

Lerrmu 17.12. Let ( x n ) z l be a one-to-one sequence in N. If PSI(X,,)~~) U PP((x.)zl) is contained in o m cell of thejmrtition d,then (d(x.) : n E N) is unbounded. ProOK Suppose instead that (d(xm).:n E N) is bounded and pick k € W such that for infinitely many n, d(x.) = k. If k = 0. then we would have PP((X,,)~=~)G 2N 1 so PS((x,,)zl) E 2W 1, which is impossible. Thus we may assume that k 2 1. If k > 1. then pick n < r such that d(x.) = d(x,) = k and either

+

+

Then d(x. d(xn

+ x,)

=

k

+ 1 and d(x,,x,)

+x,), d(xnxr) E Ci.

= 2k so one can't have i E (0.1) with

Thus we must have that k = 1. Suppose first that for hlinitely many n one bas

d(x,,) = 1 and e(x,,) = 2. Pick n < r and u. u E w such that u = v (mod 2) and x,, =2+4+8uandx, =2+4+8v.Thenxm+x, = 1 2 + 1 6 . ( Y ) i r 12 (mod 1)6 while x.x, = 36 48(u v) 64uv 4 (mod 116. Consequently one has infinitely many n with d(x.) = 1and e(xJ w 2. Rckn ir and u, v E o such that d(x,) = d(x,) = 1.3 5 dx,) Ie(x,), u = v (mod 2). and

+

+ +

Consequently we have some e > 2 such that e(x.) = e(x,) = C. Then x.

and x.x, = 4 so that e(xn

+ x,)

+x, = 4 +2'+' + 2"'

u +v (T)

u +v + (u + ~ ) 2 ~ - -+' ~ + 2'+' + 2'+3(2t-3 + 2

~

= L + 1 while e(x.x,) = e + 2, again a contradiction.

2

~

4 0

As aconsequenceofLemma 17.12. we know that if (x.)zl is a one-to-one sequence with P S ( ( x . ) ~ , )U PP((x,,)zl) contained in one cell of the panition 3.then we can assume that for each n, a(x.) < d(x.+l) and consequently. the= is no mixing of the bits of x. and x,,, when they are added in binary.

)

344

17 S u m and Products

Lemma 17.13. k t (x& be a on#-to-one sequence in N. If PS((xn)zl) u PP((xn)El) is cON&lined in one cell of the partition R,then (n E A : x. E Ao] infinite o r (n E N : x. E Aj] is infhite

Prm$ One cannot have PS((x,,)z1) E A4 and one c a n n O t h a ~ e P S ( { x ~ ) g ~ =2N+ ~) 1

+

so one has a(x. x,) ia(x.x,) (mod 2) whenevcr n and r are distinct m e m h of A. By the pigeon hole principle we may presume that we have some i E (0.1,2,3.4) such that (x. : n E N] E Ai. If o m had i = 4, then one wouldhave PP((xn)zl) c 4 and hence that PS((x,)zl) g 4,which we have already noted is impossible. By Lemma 17.12 we have that [d(x.) : n E Nl is unbounded so pick n E A such thatd(x.) > a(x1). Thena(x. + X I ) = a(x.). Suppose i = 1, that is {x, : n E N) g AI. Then by Remark 17.9, a(x.xl) t a(x.) a(x1) 1 so ~(x,,xI)is odd while a(x. + X I ) is wen. 0 Similarly, if i = 2, then a(x,xl) is wen while a(%, + X I ) is odd.

+

+

Lemma 17.14. Lct Cr.)El be a one-to-one sequence in N. If d(x.+l) z a(x.) for each n. PS((X.)F=~) U PP((x,)zl) is contained in one cell of the partition W and (x,, :n E N] E Ao, then (a(x.) b(x.) : n E N] is bounded

-

-

Pmof: Suppose instead that (a(x.) b(x.) : n E N) is unbounded. Pickn E N such that a(x,,) b(xJ > d(x1). Then, using the fact that d(x,,) > o(x1) we have that

-

a(x. + X I ) - b ( x . + x ~ ) =a(x.)-b(x.)

> d ( x ~ )=d(x.+xl)

while by Lemma 17.10

Lemma 17.15. Lct (x.)gl be a one-to-one sequence in H Ifd(x.+~) > a(x.) for each n, PS((x,,)zl) U PP((x.)z,) is contained in one cell of the pam'tion R,and (x. : n E N) C A,. then {a(xn) c(xJ : n E N) is bounded

-

Proof This is nearly identical to the proof of Lemma 17.14.

0

Theorem 17.16. There i s m one-to-one sequence (x.)z, in Nsuch that PS((xn)E,)U PP((x,)zl) is contained in one cellof theparfition R. Pmofl Supposeinstead we have such a sequence. By Lemmas 17.12,17.13,17.14, and 17.15 we may presume that for each n E N. d(x.+l) r a(xJ I and either

+

- b(x,) :n E N) is bounded, a - c(x.) : n E N) is bounded. Assume lint that {x. : n E Nt G Ao and [a(x,) - b(x.) : n E A) is bounded. Pick some k G N and n < r in N such that a(x.) - b(x,,) = a(x,) - b(x,) = k. Then a(x, + x.) - b(x, + x,) = a(+) - b(x,) = k. (i) (x. :n E N] E Ao and {a(x.) (ii) [x. : n E N] E A3 and {a(x.)

.;

17.2 Pairwise Sums and Rcdwts

- k +2. Thus a(xrxn)- b(xrx.) E (k- 1,k - 2 ) so a(x,x,,) + - b(x, +x.) (mod 3). a wnhadiction. Finally assume that k : n € lal) S An and [o(xd - c(x.) : n E N) is bounded.

so b(x,x.) 5 a(x,xn) b(x,x.) $ a(x, x.)

We may presume that we have some k E W such that a(x,J - c(x.) = k for all n By the pigeon hole principle, pick n c r in N such that e i k

- +

E

W.

k 1. and consequently c(x,x.) 5 a(x,x.) Assume that x,, < 2°(x.)+l(1 2-k): and i, < y @ - ) + l ( l 2 4 ) t . we have a(xr x.) - c(xr x,,) = a(x,) - c(x,) = k. Since x,x, c 2a(X,x0)+1(1- 2 - 9 = T"x")+l 2a(xrx")+1-kand hence c(x,x.) > a(x,x.) - k 1, one has c(x.xn) = a(x,x,) - k 1. But then a(x,x.) - c(x,x.) f a(x, +x,) c(x, +x.) (mod 2), a conhadiction. Thuswemusthavethatx, 2 ~ ~ ( ' ) + 1 ( 1 - 2 - ~ )andx, i ) T(Xr)+l(l-2-k)f.

+

+

+

-

-

+ -

-2-')4 anda(x,+x,) - c ( ~ , + ~ , , ) = a(xr+x.) = a h ) sox, +x. w 2"(xr+X")+1(1 a(x,) c(x,) = k. so picking i E (0. 1.2) such that i 1 I k (mod 3 ) we have that x, +x, E Bi and consequently x,x. E Bi. ~ ( k " ~ l + l-( l2-k) = 2a(x.&)+l 2&sJ+l-k SO ~ ( x r x d5 x,xm a(x,x.) k. By Lemma 17.10 a(x,x.) - c(x,x.) 5 k so a(x,x,) - c(x,x.) = k. Notice that (1 2-'-' - 2-2k-2) < (1 - 2-')f, a fact that may be verified by

-

-

+ -

-

17 Sums and Rodueta

346

Thus since x,x.

E B,

we have that k = a(x,x.)

- c(x,x,)

t

i za k

- I (mod 3). a

Recall from Theorem 13.14 that if p E BN and Nn E p for infinitely many n. then thm do not exist q, r. and s in N* such that q .p = r +S. The following corollary bas a much weaker conclusion. but applis to any p E No.

+

.

PmoJ Suppose instead that p p = p .p and pick some A E R such that A E p p. L a B = ( x ~ N : - x + A € ~ ) n ( ~ ~ N : x - ~ A ~ ~ ) a nEdBp. Inductively, ickx~ let n E N and assume we have chosen (xt);,, Pick

.

Then PS((x.)z,) U P P ( ( x , ) ~ , ) E A, contradicting Theorem 17.16. Theorem 17.16establishesthat one c?.nnotexpect any mt of combined additive and multiplicative results from an infinite sequence in an arbitrary finite partition of M. On the other b&d. the fouowing question remains wide open. Question 17.18. Ut r, n E M. IfN = U:=, Ai, must them exist i E (1.2,. a one-to-one sequence (x~):,~ such rhar FS((xt):_,) U FP((X,):,~) c Ai?

.

.. r ) and

We would conjhcturr sfmngly that the answa to Quation 17.18 is 'yes". However. the only nontrivial case for which it is known to be true is n = r = 2. (See the notes to this chapter).

17.3 Sums of Products Shortly after the original (combinatorial) proof of the F ~ t Sums e T h m m and its comllary, the Finite Products Theorem. Erd6s asked [891 whether. given any finite partition of N. there must exist one cell containing a sequence and all of its "multilinear combinations". We know, of course. that whatever the precise meaning of '4nultilinear combinations", the answer is "no". (Theorem 17.16.) We see in this section. however. that a cemin regularity can be impoxd on sums of products of a sequence. Recall that, if n E N and p E BW, then n . p is the product of n and p in the semigroup (BW. .) which need not be the same as the sum of p with itself n times. (We already know hom Theorem 13.18 that if p E W'. then p p # 2 .p.) Consequently we introduce some notation for the sum of p with itself n times.

+

Definition 17.19. Let p E pN. Then ot(p) = p and. given n E N, o,+l(p) = %(P) P.

+

-:

.

Of come, if p € N, tben for all n, n p = u,,(p). The question naturally arises ar to whether it is possible to haven .p = o.(p) for some n E N\(1) and p E N'. We shall see in CornUary 17.22 that it is not possible.

Lemma 1720. k t n E N, let p E W. let A E p and Irr B E a,,(p). Them is a om-to-one sequence ( x l ) z l in A such that, for each F E [Nln, &P xr E B. prwf: This is Exercise 17.3.1.

Tbeorem 17.21. Lcr n E N\(1). There is afrnite panition R of W such that there do not exist A E 3! anda one-to-one sequeke ( x l ) z l such that

.

(1) for each t E N, n x, E A, and (2) whenever F E [Nu. X t Ext~ E A.

and let R = (Ao, Al. Az, A,]. Suppose that one has a one-t0;One sequence (x~)PO_~ and i E 10, 1.2.3) such that

.

(1) for each t E El. n x, E Ai, and (2) whmeva F E [N", &F X, E Ai.

BY the d m n hole vrinciple. we may presume that we have some j

so that i f j

E

(0.1,2,3) such

+ 2 (mod 4). a contradiction.

0

P m J Supposethatn.p=un(p)andpickA

E RsuchthatA € 8 8 . ~ . '~henn-'A € p so. by Lemma 17.20. choose a sequence (x,)El inn-' A such that for each F E [Nln, ErcFx, E A. This contradicts Theorem 17.21. 0

Recall that given F. G E P,((N) we write F < G if and only if max F < min G. Delinition 17.23. Let ( x t ) z I be a sequence in N and let m E N. Then

17 Sums md Produns

348

P m f I f m = 1,this is just the hitepmducts theorem (Themem5.8). In theproof, we a!~dealing with two semigmups. so the notation B* is ambiguous. We shall use it to refer to the semigroup (ON..), so that B* = (x E B :x-lB E p). We do the m = 2 case separately because it lacks some of the complexity of the general thmrem so assume A E p p. Let BI = (x E N : -x A E p). m e n BI E p so BI* E p. Pick XI E BI'. Inductively, let n E Nand assume that we have chosen (~t):=~in N and BI 2 B2 2 . 2 B,, in p such that. if 0 # F 2 (1.2,. ,n). k=minF,andl=maxF.then

+

+

..

..

+

...,

n) and a E Ek. a E Bt. E BI SO -a A E p and Now. given k E (1.2. E p so B.+I E p. Choosex,,+l E B.+I'. n 1). k = min F , and To verify the induction hypotheses, let 0 # F G (1.2,. C = max F. If I < n, then both hypotheses hold by assumption. Assume that C = n. 'Ihen (1) holds by assumption and n,,,vxt E Ek so (2) holds. Finally assume that L = n + l . Then(2)isvacuous..Ifk = n+I,then &FX, = x,+l E Ifk 5 n. xt E Ek sox.+l E &+I G (&G xr)-lBke and thus then let G = F\(k). Then (1)holds. The wnshuction being wmplete, let FI.F2 E Pf(N with FI < Fz. Let k = min F2 and let L = max FI.Then

... +

n,,~

This completes the proof in the case m = 2. Now assume that m s_ 3. Let BI = (x E N : -x + A E a m - ~ ( p ) )and choose X I E El'. Inductively, let n E Nand assume that we have chosen (xk);=, in Nand (&);=I in p so that for each r E (1.2, ,n) each of the following statements holds.

...

,...,m - 2). F1,Fz ....,Ft E g ( ( l . 2 , ...,r)) with Fl < F2 < ... < Fl. and r < n, then B.+I c (XE N : -x + (- ZIn,,, xr + A) E U~-L-I(P)).

(V) If L E (1,2

At n = 1, hypothesis 0 says that XI E BI*. Hypotheses (11). 0, and (V) are vacuous. and hypothesis (m) says that -XI A E am-1(p). P o r e ~ ( 1 . ,..., 2 m-i).let

+

55 = ((FI,Fz.

...,Fe) : FI, F2,. ...FLE Pf((l.2. ....n)) and FI < F2 < ... < Fe)

Given a E Ek, we have that a E Bk* by hypothesis (I and ) so a-'BkR E p by then by (m)we have

Lemma 4.14. If (FI, Fz. ...,Fmml)E

- Xyi' nf,Rx, + A If e E (I, 2. . . . ,m

E p.

- 2) and (FI, Fz, . . .,Ff) E 35 we have by hypothesis (m)that

So we let

and notice that B.+I E p. For simplicity of notation we are taking 0 = N in the above. Thus, for example. if n = m - 3. then Fm-z = F m - l = 0 and so

n

Choose *,+I E B.+I'. To verify hypothesis 0,let 0 # F g (1, 2,. . .,n + 1) and let k = min F. If n 1 @ F, then (I) holds by assumption, so assume that n + I E F. If k = n 1. we have x.+l E B.+l' d i i t l y , so assume that k < n + 1 and let G = F\{n 11. Then n , , ~ x ,E Ek soxn+l E (nrP~xr)-lBk'. Hypothesis @) holds trivially and hypotheses (IV) and (V) hold d i i y . To verify hypothesis (In). let t E 11.2, ...,m 1) and let FI,F2, Fc E Pf((1.2.. . n + 1)) with FI < Fz < ... < Fc. If e = I, then by hypothesis

+

+

+

.

.

-

....

350

17 S u m a d Products

-

O and (n).%F, XI E BI* G BI so &F, XI + A E um-t(p). So assume that C s 1. Let k = min Fe and let j = max FM. '&a by hypotheses O and 0, nteF, XI E Bk' E B,+t and by hypothesis (V)at r = j .

..

The induction being complete. we have thu wbeneva FI, F2,. ,F,,, E Pf(N) with FI < Fz < ... -z F, if k = min F, and r = max F,-I. then by (I O. ), and

(N). n r e ~ , E ~ t E*&+I

E - Cyi;;'nlEfi XI + A

.

Cornnary 17.25. Letr, m E NandletN = U;=I Aj. Thenther~exiaj E (1.2.. .. r ) anda sequence (X~)PO,~ such that SP,,,((x,)z,) Aj.

PIV@ Pick any idernpotent p in (BN, .) and pick j E (1.2.. MP).

...r ] such that At

e 0

There is a partial converse to Theorem 17.24. In this convme, the meaning of SP.((xI)E2) should be obvious. Notice that one does not require that p .p = p.

Theorem 17.26. Lct ( x l ) z l be a sequence in N. l f p E n and k in N SP,,((x,)zk) E u,,(p).

nzIF P ( ( x I ) s ) ,thenforall

PmaJ We p e e d by induction on n, the case n = 1 holding by assumption. So let n E Nand assume that facach k E N. SP.((X,)PO,~)E a.(p). Let k E N. We claim that

so that -a

+ SP.+1((xf)%)

E

p.

Weseenowthatonecannot necessarily expect t o f i n d S P , ( ( ~ ~ )a~n~d)S P ~ ( ( x , ) z ~ ) (= FP((xI)zI))in one cell of apartition, indeed not even SP,,,((xI)z,) USPI ( ( y , ) z l )

with possibly different sequences (x,)zl and ( y l ) z l .In fact muchsmngerconclusions are known -see the notes to this chapter. Theorem 17.27. Let m E N\(1). Them is afmite pam'tion R of N such that. given any A E R. Ihem do not &st one-to-one sequences (x,)El and ( y , ) z l with SPm((xr)El)E A andSP~((yr)PO,~) C A.

35 1

17.3 Sums ofRadu*s

4 ~ 1 n-aOll)+d(yl) )

> a ( y ~ y z ) - a ( y r ) r a ( y ~ y z ) - d ( y d > a(yryz)-d(ylyz).

Also, yl t rna(Y1)+rnd(yl) and yz > ma(n)so

. .

a canhadiction. Next we claim that 4 does not contain any S P , , , ( ( X ~ ) ~so~suppose ). that we have a one-to-one sequence (X,)PO,~with S P , ( { X , ) ~ ~C) A4. If for infinitely many 1's we have d(xJ = 0. we may choose F E [NJmsuch that for any r, s c F one has xf

le x,

q4 m (mod mZ).

Then ~ ( Z ~ F X=I1 )SO & F X ~ € S P ~ ( ( X ~ ) P O , ~ ) \ A ~ . Thus we may assume that each d(xf) > 0 and hence, by passing to a product subsystem as we did whcn discussing A3, we may presume that for each t, a(x,) < d(x,+l), so that there is no canying whcn x, and x,+l are added in base rn arithmetic. (Notice that by Exercise 17.3.2,if ( y , ) z , is a product subsystem of ( x , ) E l , then S P m ( ( ~ t ) gE I ) SPm((xt)zl).)

+

+

We may further assume that f a each t 2 m. a(x,) ? a(xm-1) d(x1) 2. We claim that there is some t 2 m such that XI < ma(x"-d(xl'-l. Suppose instead

+

17 Sums and Pmductn

k

t

+

and c ~ l l ~ e q u c nacn,, tl~ XI) = Z r = I . a ( ~ , ) I. Since. for each I, d(x,+l) w XI). we have that for each s E (m, m a h + X~+...+X,-I +x,)=a(x,)an&since

+ 1,. ...k),

k k k a(x,)iseven. Alsoa(x~+x2+. ..+%,-I +n,=,x,) = U(~,=~X,),SOU(I'I,,X~) is even, a contradiction. Thus we have some t 2 m such that x, < + m"(x~)-d(xl)-'. k t = XI + X 2 + . . . + X m - ~ + X r Then.a(y) =a(xt)andd(y) =d(xl). Andhecausethen is no carrying when these numbers an added in base m, we have

contradicting the fact that y E Aq. Now we claim that As d w s not contain any FP((X,)PO_~).So suppose instead that we have a one-to-one sequence ( x , ) z l such that F P ( ( x , ) ~ , ) A s As before, we rnd(") as can may presume that for each 1, d(x,+l) > a(x,). Now XI 5 ma(")+' be seen by considering the base m expansion of X I . Also, x2 < m'(Xz)+l so that XIX2 < ,,,~(xI)+Nzz)+~ - , , , ~ ( ~ Z ) + ~ ( since X I ) +also ~ . xlxZ > , , , ~ T I ) + ~and (XZ) a h ). a(xz), and a(x1x2)arc all odd. we have a(x~xz)= a(x1) a(x2) 1. Also

-

+

+

co~nradicting the fact thatxlx2 e As.

Finally we show that A6 docs nm contain any S P m ( ( x 1 ) ~ , )so , suppose instead that we have a one-to-one sequence ( x l ) E l such that SPm((xr)PO,I)C A6. ASin the considemtion of A4, we see that we cannot have d(xJ = 0 for infi~telymany t's and hence we can assume that for all t, d(x,+l) w a(xl). We claim that there is some t 2 m such that xr s m"(xt)tL ma(x*)-d(xl). For suppose instead that for each t 2 m. x, m"(X1)+l(l m-d(xl)-'). Then for some e, (1 - m-d(X!)-l)'+l < 11," so

-

-

Pick the iirst k such that

Then

,,

nk-1

- I xr > m ~ kr-n'(*)+k-l-"'

and xk

,,,a(xd,

and hmn

.... k ) , a ( x ~+ x ~ + . . . + x . - l +x,) =a(x,)soeach c:=~a(x,) + k - 1 - m arc odd, which is impossible.

Alsoforeacht E l m , m + l , a(xd as well as

-

Thus we have some t 2 rn such that x1 > ma(*)+' m ' ( X ~ ) - d ( x ~ ) . k t Y = +xr. Thena(y) = a(xl) andd(y) = d(x1) and y w ma'rr)+'

xt +xz +...+x,-I mds,)-d(x~)

=m a ( ~ ) t l ,,,n(y)-d(y),

acon~c~on,

-

0

We see now that we can in fact assume that the partition of Theorem 17.27 has only two cells. Corollary 17.28. Let m E N\(l). There ir a set B

g N such tho1

( I ) whenever p . p = p E pN B E p, (2) whenever p . p = p E pN. N\B E um(p), (3) them is M one-ta-one sequence ( x r ) Z I in N with SPm((xr)PO,l)g B. and (4) there is M one-to-one sequence (x,)PO_,in N with SPl((x,)ZI) g W\B.

Pmof Let R be the partition guaranteed by Theorem 17.27 and let B = U(A E R : there exists ( x r ) Z l with FP((x,)EI)2 A).

To verify conclusion ( I ) , let p .p = p E pN and pick A E R such that A E p. Then by Theorem 5.8 there is a sequence ( x , ) z , such that F P ( ( x r ) z I )G A, so A G B

SO BE^.

354

17 S m aad Rodncb

To verify conclusion (2). Imp. p = p E BN and pick A E R such that A E d p ) . Thenby Theorem 17.24pickasequence (y,)zl with SPm((yt)Zl)C A. By Theorem 17.27 there does not exist a sequence (xl)El with FF'((xf)zl) E A, so A n B = 8. To verify conclusion (3). suppose that o m had a one-to-om sequence (x,)PO,, with S P m ( ( x , ) ~ ,c ) B. By Lemma 5.11, pick p p = p E ngl FP((xr)z). Then by Theorem 17.26 SP,((xr)z,) E um(p) so that B E um(p), contradicting conclusion (2). To verify conclusion (4). suppose that one had a one-to-one sequence (x,)EI with FP((x,)zI) C N\B. Again using Lemma 5.11, pick p p = p E FP((x,)zk). Then FP((xf)Po,,)E p so N\B E p. contradicting ccmclusion (I). 0

.

.

nzl

Exerdre 173.1. Pmve Lemma 17.20. (Hint: See the proofs of Corollary 17.17 and Theorem 17.24.)

EXedse 1733. Let (y,)El. Modify the proof of Theonm 17.24 to show that if p p =p E FP((yl)L) and A E am(p), then them is a product subsystem m of (x~),,, such that SP,((X~)PO,~) E A. (Hinc: Seetheproof ofTheomn 17.31)

.

ngl

bf)zk

.

Exe* 173A. Let n, r E N and let N = U:=I Ai. Prove that t h m is a function f : (I.?, ..n] -t (1.2,. r ] and a sequence ( x l ) z l such that for each m E (1,2.. ..,n]. SPm((x,)zl) C A,@). (Hint: Use the results of Exercises 17.3.2 and

..

..

17.3.3.)

-

17.4 Linear Combinations of Sums Infinite Partition Regular Matrices

In this section we show that, given a finite sequence of coefficients, one can always find one cell of a partition containing the linear combinations of sums of a sequence which have the specified coefficients. We show fimher that the cell can depend on the choice of coefficients. In many respects, the results of this section arc similar to those of Section 173. However. the motivation is significantly different. In Section 15.4, we characterizedthe (finite) image panition rrgular matrices with entries from Q One may naturally extend the notion of image partition regulariry to infinite dimensional matrices. Definition 17.29. Let A be an o x o matrix with enhies from Q such that each mw of A has only finitely many nonzero enhies. Then A is i m g e partition q u l a r over N if Ci, thereexisti E (1.2,. ,r ) and? E Nm andonly if wheneverr e Nand P1= such that all entries of A? are in Ci.

..

"

17.4 Linear Combinatim of Sum

A simple example of an in6nite padtion regular rnahix is

Hcn A is a finite sums matrix. That is. the assertion that all entries of A: are in Ci is the same as the assertion that FS((x,,)gl)C Ci. We establish here the image partition regularity of certain infinite matrices and provide a smng contrast to the situation with respect to finite image partition regular matrices. For example, any central set C in (W,+) has the pmpmty that. givmany finiteimage partition regular matrix A. there must exist x' with all entries of Ax' in C . (Theorems 15.5.15.24 and Lemma 16.13.) Consequently, given any finite partitionof N,some o m cell must conah the entries of & for every finite image partition rrgular matrix A. By way of contrast. we shall establish here that a certain class of infinite matrices an image partition regular, and then show that there arc two of these matrices, A and B, and a two cell partition of W, neither cell of whichcontains all entries of A i and B3 foranyiandj. We call the systcms we are studying "Milliken-Taylor" systems because of their nlation to the MillibTaylor Theorem, which we shall pmve in Chapter 18.

De6~itfoa1730.( a ) A = [ ( a i ) ~ = l ~ N: m m ~ N a n d f o r a l l i ~ { l..... , 2 m-I), ai # a i + ~ ) . '3)Girm o' E A with length m and a sequence ( x t ) E ,in N,

'Ihc reason for requiring that oi # ai+l in the definition of A is of course that a E,,F x, + a Etstx, = a E,,NO X, when F < G. As a consequence of Corollary 17.33 below, whenmr o' E A and N is partitioned ( x r ) z I for ) some sequence into finitely many cells, one cell must contain m(o'. ( X , ) ~ O , ~ .This is the same as the assertion that a particular infinite matrix is image partition reguiar.

356

17 S m and Rcducls

For a m k . if

and.? = (xt)zl.thenMT((I. 2). (x,)PO,,) isthesetofentriesof A.?. (Thermvsof A are all mws with entries from (0, 1.2) such that (I) only finitely many entries arc non-zero, (2) at least one entry is 1, (3) at least one entry is 2, and (4) all occurrences of l come before any occurrences of 2.)

Pmof Assume first that m = 1. Then 41-'A

E p so by Theorem 5.14 there is a sum subsystem (x1)pO,, of (yl)EI such that FS((X,)PO,~) a 1 - I ~ .This says precisely that MTG. (xr)Fl) G A. Assume now that m ? 2 and notice that

c

so that {x E N : - a 1 x + A € a ~ p + a 3 p + . . . + a , p ) Ep.

Pick X I E BI* and pick H I E Pf (N)such that xl = &.H, y,, (Here we are using the BI E p).) additive version of BI'. That is, 81'= {x E BI : -x Inductively. let n E N and assume that we have chosen ( ~ r ) ; , ~in N, (B& in p. and (HI)",=, in Pf(W) so that for each r E (1.2,. ..,n ) :

+

(D I f O # F S ( 1 , 2 .....r ) a n d k = m i n F , t h e n E l E ~ x t ~ B i . . S B, and H, < H,+I. {1,2...., m - I ) , FI.&..... FEE PfP1({1.2

(II)I f r < n. then B,+I

(Ill) I f e E

... r~ Fc.then

...., r ) ) . a n d F ~ < FZ

4

17.4 Line= Combhtiom of Sums

(IV) f f F 1 . F..... ~ Fm-I

E

357

P ~ ( ( 1 .,..., 2 11). FI < FZ < ::. < Fm-l.Bndr < n.

thm Br+t E am-'(-

~ 2ai CtGW ' X I + A).

At n = 1, hypotheses 0 and (Vl)hold dictly, hypotheses 0. (IV), and (V) are A E a l p a3p ...+ a m p which is

vacuous, and hypothesis (111) says that -alxl true because xl E BI

.

For!

E

.

{I, 2, ... m

and fork E ( 1 . 2 .

+

+

+

- 11. let

....n], let +

and so -b Bt' E p. If Givm b E Ek. we have b E B I . by hypothesis 0, ( F I .F2,. .. & - I ) E F + - I then , by (m)we have (- EL;' ai XrsFi xr A) E a m p M that a m - l ( - ~ ? = y 1 a i + A ) E p. If ! E (1.2, .... m - 2 ) and ( F I .F2. . .; F d E Ft we have by (111) that

.

+

.

Lets = max H.

+ 1. Then we have that B.+I

Hen again we use the convenrion that

E p, where

n0 = N so.for example, if m = 2. then

Cbooae *.+I

S (s,s

and pick H.+I

E B.+I*

+ 1,s + 2.. ..) such that x.+l

=

& I . + YI. ,

Hypothes'i 0can be verified as in the sswd pmof of lborcm 5.8 and hypothcs'i 0holds trivially. Hypotheses 0, (V) and (VI) hold dinctly. To verify hypothesis (W, let C E (1.2.. .m 1) and l a Fl < F2 < . < Ft in Pf((I.2 n+1)).ff~=l,wehavebyhypocheses(9and@)that~~~,x, EBI so that -a1 &IF, xt + A E a2p a3p amp as required. So assume that L > 1 andletk=RlinFcandj=manFr-1.l'hen

,....

+

X t t ~X,I E

.. + ...

4* E &+I E ( X E N

..

+

+

: -atx (- X f z : ai X I c h X I A) E at+lp ae+2p +amp)

+ ...

+

@y hypothesis (V)at r = j ) so

+

ai XISF,X I ) A E at+lp+ar+2p

-&I

asrequired

The induction being complete, let FI < F2 < min F, and j = max F,-I. Then %F.

so

XL1ai

XI

E Bk*

c Bj+l

+... +amp

-. . < F,,, in P f ( N ) and let k =

C am-'(- X:slai E r C X~I i+ A )

o

xI E A.

Just as Theorem 17.26 was a partial cnmne to Theonm 17.24, we now obtain a partial converse m Theorem 17.31. Notice that p is not rcquind to be an idempotent.

= (a!,02,. ...am) E A let (X:)PO,~ beasequmcc in N,d l n F S ( ( x , ) s ) . Then MT(& (x&) E a l p + a z p +. . .+amp.

lheomrn 1732. Let;

PE

Pmol: We show, by downward induction on C E (1.2,

.

... m ) that for each k E Q

{ ~ : = t a i ~ r e f i x t : F t . F t +..., ~ , F m ~ P f ( ( k . k +,... l DandFc < Fe+l <-.-
For e = m. we have F S ( ( x , ) z ) E p so that FS((amxI)z,) E amp. So let f ~ ( 1 , 2..., , m-1)andassumethattheasPerrionis~efwC+l.Letk~Nandlet A = IErdai

We show that

.

C E I ;:~ Fe,, F ~ + I ... .

Fm E P,((k, k and Fe < F ~ + I< . . < Ir,,,).

.

+ 1,. ..))

-

17.4 Linear C m r b h h n of Sums

359

andFt+,
+ A E ac+tp + at+zp + -:.+ a m p as required.

We see now that Milliken-Taylor systems are themselves panition regular.

+

nEl

.

pick by LC5.11 = G FS((X,)%). BY n ~ o r n n 17.32. MT(o'. (X,)PO,~) E a l p a2p ...+ a mp. where o' = (01.a2.. . am). Pick iE(1.2, ...,r)suchthatBi Eatp+azp+.~.+amp,andapplyTheorern17.31.o PWJ

+

+

.

~ e o b s e r vthat c certaindiffer&t membasof ~dwayshavc~illikcn-~a~1ors~stems in the same cell of a partition.

Assume rhat r = 7 where m. n E N and let Z = m6. Pick by CoroUary 17.33 some i E (1.2,. ,r ] and a sequence (zt)PO,, such that M ( Z . (z,)Z,) g A,. For each t E N. let x, = nz, and yt = mzt. *en MT(2. (x,)Et) = m ( b , (yt)ZI) = MT(Z. (zr)Zl). o

P&

..

~-

We show now, as we pmmised earlier, that we can separate certain Milliken-Taylor systems.

For x E N. let supp(x) be the binary support of x so that x = &,pp(x) 2'. For each x E N,let a(x) = max supp(x) and let d(x) = min supp(x). We also define a function z so that z(x) mums the number of blocks of zeros of odd length interior to the binary expansion of x, that is between a(x) and d(x). 'Ihus, for example. if in binary, x = 1011001010001100MX), then z(x) = 3. ('he blocks of length 1 and 3 w m counted. but the block of length 5 is not counted because it is not interior to the expansion of x.)

PI&

17 Sums aad Raducs

360

Let A =; (x E N :z(x) H 0 (mod 4) w z(x) E 3 (mod A)). Suppose that we have a sequence (XI)P"_~ such that either

Pick a sum subsystem (yl)zl of (xt)PO,,such that for each t, d(yt+~)> a(yt). (See Exercise 174.1.) For i E (0.1,2,3) let

B, = (x E N :z(x) = i (mod 4)). By Comlliay 17.33 we may presume that if

then m ( ( l , 2). (yl)El) 30 or MT((1.2). (yt)pO=,)E B3 and, if

~ ~ ( ( 1 . 4(xI)Po=,) ). E WA.

By the pigeon hole principle we may presume that for t, k

E

N.

Assume first that fort E W, a(y,) = d(y,) (mod 2). Now if MT((1.2). (yr)PO,,)g n 2y3 and n 2y3 are in the same Bi so

A, we have y~

+ +

+

+ +2 ~ 3 )

~ ( Y In

I

zCYZ

+ 2 n ) (mod 4).

Since 0 1 ) = d ( n ) (mod 2) there is an odd block of zeros between yl and n in Y 1 + ~ + 2 n , s o z l v 1 + n + 2 n )= z(y1)+l+z(n+2~)andconsequentlyz(y~) 1 3 (mod 4). Buld(2n) = d(y3) 1 f a ( n ) (mod 2) s o z ( n +2yd = z ( n ) + z ( n ) E 3 3 = 2 (mod 4). acontradiction to the fact that MT((1,Z). (yt)E,) c A. I f m ( ( 1 . 4 ) . ( Y I ) ~ , )c N\A, we have as above that

+

+

17.5 Sums and M c t s in (0, 1)

361

As a consequenceof Theom 17.35, we havethe foUowing conaast between infinite and finite image partition regularmahices. Corollary 17.36. Then exist inJinite image partition regular mafrices B and C and a two cell partition ofN, neither cell of which contains all of the entries of B i and of C i for any 2 , y' E Nw.

Prwf: The matrices for the M i l l i T a y I o r systems MT((1,2), (x1)PO_,) and MT((1.4). (y,)z,) are image partition regular.

In &tion 15.4, we found several characterizationsof finite image partition regular matrices. Question 17.37. Can onefind a reasonable combinatoria1 chamcterization ofinfiite

image partition regular mat&ces? Exercise 17.4.1. In Exercise 5.2.2, the reader was asked to show combinatoridly that given any sequence (xl)PO,, there is a sum subsystem ( y , ) z , of (xl)El such that Use the fact that, by Lemma 6.8, given any whenever Zk < y,, one has 2k+L~Y,+l. p p =p E F S ( ( x , ) z ) . one has p G W.to show that there is a sum subsystem m (YO,,, of (xt)PO_,such that for each t E N, a(y,) 2 5 ~ ( Y I + I ) .

+

n&

+

Exerdse 17.43. Rove that if H is a finite subset of A, r E N,and N = U;=, Ai, then there exist a function f : H -r (1.2. . . . , r ) and a sequence (x,)Z, such that for each 2 E H,MT(Z, (x1)PO_,)E Af(i,, (Hint: UseCorolhy 17.33.)

-

17.5 Sums and Products in ( 0 , l ) Measurable and Baire Partitions

We show in this section that if one has a finite partition of the real interval (0.1) each member of which is Lebesgue measurable, or each member of which is a Baire s e ~ then one cell will contain FS((x,)E,) U F P ( ( x . ) ~ , ) for some sequence (x.)Z, in 0 I). (The terminology "Baire set" has different meanings in the literature. We take it to mean membm of the smallest a-algebra containing the open sets and the nowhere dense sets.) In fact, we show that certain sums of pmducts and products of sums can also be found in the specified cell. We use the algebraic structure of the subset OC of b(0, l)d which we discussed in Section 13.3. The results that we obtain are topological in nature. so it is natural to

wonder how m e can obtain them stmting with the discnte.topology on (0.1). '& answer is that we use certain topologically charactaized subsets of O+. Recall that a subset A of R is mager (orfirst category) if and only if A is the countable union of

I

nowhnc denx sets.

Delinltion 1738. (a) A set A c (0. 1) is Baiw large if and only if for every c > 0, A f l (0.6) is not meager. @)8={pEO+:fordAEp,AisBpirClargc). Lemma 17.39. 8 u a Icp i&al

of(@,

.).

P m J Since the union of finitely many meager sets is meager, one sees easily that whenever (0, 1) is partitioned into finitely many sets, one of them is Baire large. Consequently, by ibeorem 5.7, it follows that { p E B(0. l)d :for all A E p, A is Baire large) # 0. On the other hand if p E B(O, l)d\O+ one has some r s 0 such that (c. 1) E p and (r. 1) is not Baire large. Thus S f 0 . To see that P( is a left ideal of O+. let p E 21 and q E 0+ and let A E q . p. Pick x E (0.1) such that X - ~ A E p. Given r s 0. x-' A n ( 0 . 0 is not meager so x(x-'A n (0.0) is not meager and X(X-!A n (0,0) ~n ( 0 . ~ ) . o

=

The fouowing result. and its measurable analogue Th17.46, are the ones that allow us to obtain the combined additive and multiplicative results. Rmall that a subset of R is a Baire set if and only if it can be written as the symmetric diEerenee of an open set and a meager set. (Or do Exercise 17.5.1.)

Theorem 17.40. W p be a multiplieorive idmrporent in 8 and let A be a Baire set whichisamemberofp. T h r n { x ~ A : x - ~ A € ~ a n d - x + A € ~ ) E ~ .

.

PnmJ Let B = {x E A : x-'A E p). Then since p = p p. B p. Also A is a Baire set so pick an open set U and a meager set M such that A = UAM. Now M\U is meager so M\U (t p so U\M E p. We claim that So let x E (U\M) n B and pick c s 0 such thal (x,x + r ) g U. To see that -x A E p, we obsmc that (0, I)\(+ + A) is not Bake large. Indeed one has ((0, I)\(-x +A)) n (0.6) E -x M, a meager set. (Owen y E (0, c). x + y E U so. ifx+y$A,thenx+y~M.)

+

+

Now we tom our a m t i o n to deriving the measurable analogue of Theorem 17.40. We denote by p(A) the Lebesgue measure of tbe measurable set A, and by pW(B)the outer Lebesgue measure of an arbitmy set 8. We assume familiarity with the basic facts of Lebesgue measure as encountered in any standard introductory analysis cowx. The notion corresponding to "Baire large" is that of having positive upper density near 0. We use the same notation for upper density near 0 as we used for upper density of sets of integers.

.

17.5 Suna and Roducb in (0.1)

Ddinilion 17.41. Let A (0.1). (a) The upper density near 0 @A, Z(A) is defined by

(b)A point x E (0.1) is a h i t y point of A if and only if

The following result, while quite weU known, is not a pm ofstandard introducmry analysis courses so we include a proof.

T h w ) ~ m17.42 (Lebesgue Density Theorem). Let A be a measurable subsrtof(0,l). Then p((x E A : x is not a dmirypoint of A)) = 0. Pmof: If x is not a density point of A, then there is some 6 s 0 such lhat

Since the union of countably many sets of measure 0 is again of measure 0, it suffices to let E s 0 be given. let

and show that p(B) = 0. Suppose instead that p8(B) > 0. Pick an open set U such that^ ~ U a n d p ( [ l )< We first claim that

e.

(t) if (I.),,,F is a (finite a countably infinite) indexed family of pinvise disjoint intewals contained in U and for each n E F , p(A t l I.) c (I - c)p(L), then p*(B\ UnEF In) 0. To see this, since in general p*(C U D ) 5 p'(C) p * (n ~ U,, I.) < p * ( ~ ) ~. n indeed d

+ p*(D), il suffices to show that

364

17 Sums pad Rodvcrr

...

Inductively, let n E N and assume we have chogea XI,xz, ,x, in B and Wtive hl, h2.. .. hn such that each [xi hi.xi +hi] g U,each p(A fl (xi -hi. xi +hi)) < (1 6)2hi.and[xi - h i , & +hi] Il[xi - hj,xj + hl] = Ofor i # j. Let d. = sup(h : there cxistx E B such that [x - h, x hl E U,

-

.

-

+

[ ~ - h , x + h l n U ~ ~ ~ [ ~ i - h i , x i + h i l = 0 . a n d p ( A ~ ( x - h . x< + h(1-~)2h}. )) Notice that d,, > 0. Indeed, by (t)one may pick x E B\Uf,i[xi hi, xi hi] 8) flU:,,[xi hi,xi hi] = 0 and and then pick 8 > 0 such that (x - 8 , x (x 8, x 8) E U. Then, since x E B, one may pick positive h < 8 such that @ ( A n(X h, x h)) < (1 c)2h. Pickx.+l E B and h.+l > such that [x,+l h.+~. *,,+I h.+ll g U ,

-

+ -

+

+

-

4

-

+

-

+

+

-

[ X ~ +-hm+~,xn+~ I + h , + ~ l n U : ~ ~-hi,xi t ~ i +hi1 =0.

+

and P(A n &+I -hn+l. xn+l hn+l)) < (1 -€)%+I. The inductive construction being complete, let C = B\ h.]. [x. - h., x. h,l)T=l is a painvise disjoint Then by (t) p8(C) > 0. Also. since ([x" - h.. x. = , so pick m E W-such that collection contained in (0, 1). ~ ~h. converges,

+

+

zZ4+l hn

-

< P*(C)/~.

+

men P( U&,+l tx. 36,. x. 3hnl) < P*(C) so pick x E C\( U~-+l[xn 3hn,xn+3h& Thenx E B \ ~ = l [ x . - h . . x n + h . l s o p i c k h >Osuchthat[x-h.x+hlE U, [x-h,x+hlnU~=,[x,-h..x,+h,l=8.andp(An(x-h,x+h)) <(I-c)2h. Now t h m is some k-such that [x - h. x h] f7 U:=l[xn - h., x. h.] # 0 (since h. diverges) so pick the first k such that otherwise each dk 2 h and consequently [I-h,x+hIn[xk-hk,xk+hk]#0andnoticethatk>m. Thenhkz + ? so that {x x t l s h hk 5 3ht and hence x E [xt 3hr.xk 3ht1, conhdictingthe - 3h., x,, 3hJ. fact thatx t U;?+,[x.

+ CEl

-

+

+

+

-

4

+

We also need another basic result about measurable sch.

Lemma 17.43. Let A be a meosumble subset of (0.1) such thuta(A) > 0. There exists B

E A such that B U (0) is compact a n d Z ( ~ \ B = ) 0.

Pmqf Foreachn E N,letA. = Afl(l/Zn, l / ~ ~ - ' ) d l eTt = (n E N : @(A") > 0). As is well known given any bounded measurable set C and any c > 0 there is acornpact subset D of C with p ( D ) > p(C) 6 . Thus for each n E T,pick compact B, 2 An with w(Bd > @(An)For n E N\T. if any. let B. = 0. Let B = B,,. Then B U (0) is compact. < a. Pick Suppose now that &A\B) = a > 0. Pick m E N such that x < 112" s u c h t h a t p ( ( ~ \ ~ ) n ( ~ . x ) ) />x F'ickn E Wwith 1/2" 5 x i1/2"-' and note that n > m. Then

A.

-

Uzl

A.

&

2'

-

Definition 17.44. Lolnma 17-45.

P = { p E 0+ : for all A E p, a(A) > 0).

L is a I@

ideal of (O+.

-).

Pmof: It is an easy exercise to show that if a(A U B) > 0 then either &A) z 0 or Z(B) s 0. C-uendy, by lheorem 3.11. it fol~owsthat [ p e ~ ( 0 , : for all A E p, &A) > 0) # 0. On the other hand, if p E B(O. l)d\O+ one has some E > 0 such that (6.1) E p and ;i((c. 1)) = 0. Thus 6 # 0. Let p G P and let q E O+. TO see that q .p E L. let A E q .p and pick x such that x-' A E p. Another easy exercise 0 establishes that &A) = a(x-'A) > 0. Theorem 17.46. Let p be a rnultiplicarive idempotmr in 6 and let A be a mearurable member of p. Then [x E A : x-' A G p and -x +A E p) E p.

Pr00f:LecB=(x~~:x-~A~p).ThmB~psinccp=p~p.LetC=(y~A:y is not a density point of A]. By Theorem 17.42, p(C) = 0. Consequently since p E L, C $ p s o B \ C ~ p .W e c l a i m t h a t ~ \ ~ ~ [ x € ~ : x - ' ~ ~ ~ a n d - x + ~ € ~ ) . Indeed, given x 6 C one has 0 is a density point of -x A so by an easy computation, d((o. I)\(-x A)) = o so (0. I)\(-x + A ) e p SO -X A G p.

+

+

+

Now let us define the kind of combined additive and multiplicative stnrctures we obtain. Definition 17.47. Let ( x " ) z l be a sequence in (0. 1). We define FSP((xn)zl) and a : FSP((x.)zl) + 9(Pf((W) inductively to consist of only those objectsobtainable by iteration of the following:

+

+

For example, if z = x3 xs .x7 . (xu xlo - xll). then z E FSP((x.)Zl) and [3,5,7,8, 10, 11) E a(z). (Of course, it is also possiblethat z = x4+x1z.x13, in which case also (4, 12. 13) E ~ ( z ) . )Note also that (x3 + x d .x7 is noL on its face, a member of FSP((x.)El). Notice that trivially FS((xfl)zl) U FP((X.)F=~) 5 FSP((X.)T=~). The proof of the following theorem is reminiscent of the first proof of Theorem 5.8. Theorem 17.48. Let p be a ntuhiplicative idempotent in B and ler A be a Baire set which is a member of p. Then there is a sequence (x.)zl in (0.1) such that FSP((xn)zl) G A

'Iltcn Az E p and, since multiplication by XI-'

and addition of -XI are howomorphisms. A2 is a Baire set. Inductively, let n E m ( 1 )and assume chat we have chosen (x,)::: and (A,);,, so that A. is a Baire set which is a member of p. Again invoking Themem 17.40, one has that Lx E Am : x-IA. E p and -x A. E p ) E p so pick x. E A, such that x.-'A. E pand-x.+A. E p. LetA.+l =A.~x.-'~.n(-x,,+A.). lbenA.+~ is a Baire set which is a member of p. The induction W i g complete, we show that FSP((x,,)El) E A by establishing the following stronger assertion: If

+

thq z

E

A,.

.

Suppose instead that this conclusion fails, and choose

wch that z (t A, and IF(is as smaU as possible among all such counterexamples. Now if F = [m), then z = x, E A,, so we must have IF1 z 1. Let G = F\(m) and picky E FSP((x,,)z,) such that G E u(y) and either z = xm y or z = x, y. Let r = min G. Since IGI < IF],we have that

+

and hence x,

+ y E A,

and x,

.y E A,,

contradicting the choice of z.

.

0

u=,

ComUary 17A9. k t r E Nand let ( 0 , l ) = A;. Ifeach Ai is a Baire set, then thereexisti E (1.2. r ) a n d a s e q u c n c e ( ~i n~ () O ~ ~, 1 ) ~ h t h a t F S P ( ( ~GJ ~ ~ )

....

A<.

Pmoj By Lemma 17.39, S is a left ideal of (O+. .),which is a wmpact right toplogical semigroupby Lemma 13.29. Thus, by Comllary 2.6 we may pick an idempotent p E $3. Pick i E (1.2.. . . r ) such that Ai E p and apply lbeorem 17.48. 0

.

We have similar. but shvngcr. rcsulb for measurable see.

Theoma 17.50. Let p be a multipIicative idempotent in C and Irr A be a measurabIe set which is a member of p. Then then is a sequence ( x , , ) z l in (0.1) such that c e F S P ( ( x n ) z lG ) A U (0)

P+

Since A E p, &A) > 0. By Lemma 17.43, pick B compact and Z(A\B) = 0. Thm A\B $ p so B E p.

A such that B u (0) is

Now, proceeding exactly as in i h pmof of lbborrm 17.48, invoking Thcorrm 17.46 instead of lbborrm 17.40. one obtains a sequence (x.)zl such that FSP((x,,)zl) 5 B N&etbatifctFSP((x,,)L1) 5 AU{O).theninparticUlmctFS((~~)~~) 5 AU(0) d hence, if F is any subset of N. 6nite or infinite, one has &F xn E A. Corollary 1751. Icr r E N and lct(O.1) = UA1 Ai. If each A, is a mcanrmble r ) and a sequence ( x n ) z l in (0. I) such that set, then t h e &.sf i E (1.2, c ~ F S P ( ( X ~G) Ai ~ ~U)(0).

.

...

P m f By Lemma 17.45. +is a l e f t i d d of (w,.),which is acanpract right topological semigmup by Lemma 13.29. Thus, by Corollary 2.6 we may pick an idempotent p Pick i E (1.2,. ,r ] such char A, E p and apply Theorem 17.50.

..

G

L. 0

Exercise 175.1. Let X be a topological space and let 8 = (UA M : U is open in X and M is meager in X). Show that 8 is the x t of Bake sets in X.

Notes Thematerialinsection 17.1 is tiurn [25], [27L and[31] (nsultsobtained in collaboration with V. Bergelson). The material in Section 17.2 is from [127l. (In [1271. some pains were taken to reduce the number of cells of the partition so that, in lieu of the 4610 cells of the partition we used, only 7 were needed.) , It is aresult of R. Graham(prescntedin [120. Theorem4.31) that wbenevc~(1.2, 252) =A~_lUA~.thmmustbesomex # y a n d s ~ n c i€ (1.2)with(x,y,x+y,x.y) c At, and consequently the answer to Question 17.18 is "yes" for n t r = 2. Thconm 17.21 isfrom[ll9]inthecasen = 2. Itisshownin[119]thatthepartitim R can in fact have 3 cells. It is a still unsolved problem of J. Owings [I881 as to whether there is a two cell partition of W such that neither cell conlains (x, +x, : n, m E N) for any one-to-one sequence (x,)z, (where, of course, one specifically does allow n = m). Theorem 17.24 is due to G. Smith [226] and Theorem 17.27 is a special case of a much more general result from [226], namely that given any rn # n in N,there is a nvo cell papartirion of N neither cell ofwhich contains both SP.((xl)zI) and SP.((Y~)PO,~) forany sequences ( x f ) z 1mtd (y,)zl. The results of Section 17.4 are from 1791. which is a result of collaboration with W. Deuber. H. Lefmann. and I. Leader. Infact a result much stronger than Theorem 17.35 is proved in [79]. That is, $2 and b are elements of A, neither of which is a m t i o ~multiple l of the other, then there is a subse~ A of N such that for no sequence (%,)El in W,is either MT(ii. ( x f ) z I )c A or MT(b, (x,)zl) S N\A. Most of the material in Section 17.5 is from [35], results obtained in collaboration with V. Bergelson and I. Leader. The pmof of the Lebcsgue Density T h e m (Theorem

...

17.42) is horn [l891. llle idea of considuing Baire or mC8SUCBbIe partitions arises horn research of Plewk Riimel and Voigl: Given a sequence ( t n ) z l in (0.11, such that 1, converges, define the set of all s m of the sequence by AS((r.)z,) = (&p h : 0 # F E N). In 11971, Riirnel and Voigt considered the question: If (0,ll = U!=, Ai, must there exist i E (1.2.. ..,r) and a sequence (t,)zl in (0, I] with AS((t,,)=,) E Ai? As they pointed out, one easily sees (using the Axiom of Choice) that the answer is "no" by a standard diagonalization argument. They showed, however that if one adds the requirement that each Ai has the property of Baire, then the answer becomes "yes". In [I951 Plewik and Voigl reached,tbe same conclusion in the event that each Ai is assumed to be Lebesgue measurable. A unified and simplified proof of both results was prexnted in [361, a result of collaboration with V. Bergelson and B. Weiss.

.*j$ .-,~

Chapter 18

Multidimensional Ramsey Theory

Several resub in Ramsey Theory, including Ramsey's Theorem itself, nanvally apply to more than one "dimension", suitably interpreted. That is, while van der Waerden's TheoremandtheFinite Sums Theorem, for example, deal with coloringsof the elements of W. Ramsey's Theorem deals with colorings of finite subsets of W, which can be identified with points in Cartesian products.

18.1 Ramsey's Theorem and Generaliiations In this section we present a proof of Ramsey's Theorem which utilizes an arbitrary nonp~cipalultrafilter. Then we adapt that proof to obtain some generalizations, including the Milliken-Taylor Iheorem. The main feanue of the adaptation is that we utilize ultrafilters with special properties. (For example, to obtain the Milliken-Taylor Theorem, we use an idempotent in place of the arbiaacy nonprincipal ultrafilter.) While many of the applications are algebraic, the basic tools are purely set theoretic.

Lemma 18.1. Let S be a set, let p E S*, let k, r E N,and let [flk = Ai. For each i E {I, 2.. . r), each t E (1.2.. ..,k ) , &each E E [Vf-I.de/ne B , ( E , i) by downward induction on t:

..

(1) Far E E 1s)'-'. Bk(E, i) = (y E S\E : E U (y) E Ai). (2) FortE{1,2 ,..., k - l ) a n d E ~ [ S J ' - ' ,

Thenforeach t E (1.2,.

..,k ) and each E E [s]'-'.

S\E =

B d E . i)

P d We proceed by downward induction on t. If t = A, then for each y

E S\E. E U (y] E Ai fm some i. So let t E (1.2.. . . k 1) and let E E [s]'-'.Then given y E S\E, one has by the induction hypothesis that S\(E U (y)) = U:,] B,+,(E U (yt, i) so for some i. B , + I ( E U {Y). 0 E P.

.-

Theorem 18.2 (Ramscy'sThmm). Lcr Sbcan in&teMMd[nk.

r E

N. YISlk =

Ubl Ai. rhmthcncxicti E (1.2. ...,r ) M d a n i n f u r i t e s ~ s e t ~ o f ~ w iEf hA,.[ ~ ] ~

Pmof: If k = 1, this isjust the p i g m hole principle. so arsumc that k z_ 2. . Let p be any nonpiincipal ulMter on S. Deiine M E , i) as in the statement of Lcmms18.1. ThenS = U:,l Bt(0.i) sopicki E (1.2, ..., r)suchthatB1(0, i) E p. Pickxt E Br(0.i) (sothat Bz((xI), i ) E p. Inductive1y. let n E W and assume that (x~):=~ has been chosen so that whencvu t E ( 1 . 2 ...., k-1)andmt < m z < . . . < m , 5 n , O M h a s

.

.

... -

t E (1,2. k I] and To see that B,+l((xm,,,.x,,. . ..xm,), i) E p whenml < m z < . . . < m t ~ n + l , l e t s u c h r a n d m ~ . r n z,..., m , b e g i v e n . I f m , ~ n . then the induction hypothesis applies, so assume that m, = n 1. If r = 1, the conctusion holds because x.+t E Bt(0. i). If t r 1, the conclusion holds because X.+I E Br((xml.xm2. xm,-, 1. i). The sequence (x.)zl having ban chosen. let m~ s mz < ... c mr. Then 0 x,. E Bk((x,, ,xm2,. ,xm,-, 1, i) so (xml,.x,,. ... xm,1 E Ai.

+

.... ..

.

Inorda toestablish some #ations special notation.

of Ramsey's 'Ihcorem.we introduce some

DeBaitim183. LetSbeaset,letkeN.andlet(p:S+ N. (a) If (D&

is a squence of subsets of S, then

[ ( D . ) ~ ~ . - ?= I ~{ (x1.x~....,x d : x i #xi fori # j in (1.2, .. ..k) andthereexistmt e m 2 c . . . <mkinNsuchthat foreach j E (1.2. .... k), x j E Dm, andforeach j E 12.3. k). mj > -?(xi-I)].

....

(b) If t E N and (D,):=I

is a sequence of subsets of S. then

..

..

[(D&=,. PI\ = ( (XI.XZ.. ,XI} :xi # x j f m i # j in (1.2.. . k) andthereexistm~< m z c - . .c m k i n ( 1 . 2 . ...,t) such that for each j E (1.2, k), x j E D, and for each j E (2.3, . ,k), m, s p(xj-~)].

..

.

...

Forexample, if S = N, ~ ( n=) n , and Da = [m E N : rn 2 n) for each n E N, then I ( D ~ ) ~ ~ .=( ~ { (Ix\~ , x r ,.. . .xtl

G N :XI

c xz <

... < xk] = [wk.

Pa. a n d a s m r h a t [sI' =U;.,l~i. L e t p : S + N r m d ( ~ ~ ~ t h a t t h e r e i s s o n~vpCo n w h i c h p i s finite-to-one. Then thcre exist i E (1.2,. . ,r ) ando sequence (D,,)zl of members of p such that Dm+lG D, foreach n r m d [ ( ~ . ) z ~$4,: C Ai. lI1eo~m18.4.Let Sbeaninfiiteset. k t p E S, 1etk.r E

.

.

Pmqf Rrst assume that k = 1; Pick i E {l. 2, ... r ) such that E = (x E S : (x) E At] E p andlet D, = E forcachn. Then[(~.)z~.(pl:= Ai. Now assume thai k > 1 and let &(E, i) be defined for each i E (1.2,. . ,r), cach t E (1.2.. . k). and each E E [S]~-I as in Lemma 18.1. Then by Lemma 18.1, Bl(0. i ) . so pick i E (1.2, ... r ) such that B1(0, i) E p. S= For each t E N. let Ht = [x E S : p(x) s 0.l k n by assumption, for each t . Hc n C is finite. Let Dl = C n El (0,i). Inductively let n E W and assume that we have chosen (Dm):,1 in p such that for e%ch m E 11.2.. ,n):

.

..

.

..

(1) If m < n, then D.+I E Dm. (2) Ifm s 1,thenforeacht E (1,2, ....k-lland&whE E[(D~~H,,,-I)~_;~.~]:, Dm G B~+I(E, 0.

.

Notice that bMh hypotheses an vacuous when n = 1. We now claim that if r E (1.2,. k - 1) and E E [(Dj n H.)!=l. pl:. then Bt+l ( E , i) E p. SOlet such r and E be given. Then t h m exist XI,x2,. ,xt in S and ml <mz <... < m , i n ( 1 , 2 .....n)suchthatE=(x~.xz,....xtJ.xj e D m i n H n foreach j E (1.2,. .., t), and for each j E (2.3,. , t). mj s ~(xj-1). If r = 1, we havexl E Dm, E Dl E &(0. i). so &((XI).i) E p. SOnow asstunt than z 1andlet F = ( ~ 1 . ~ 2.. .,xt-l). Weclaimthat F E [(~jnH,,,_I)y';', PI:-.' Indeed. one only needs to vrrify that for cach j E (1.2.. .. ,t - 1). xi E Hm,-1. To see this.notice that p(xj) < mj+l 5 m,. Since F E [(Dj n Hm,-l)y';'. p]:-l, we have by hypothesis (2) that

..

..

..

.

xt E Dm. C BdF. i) so that Bt+l(E, i) E p as requhd. Now for each t E (1.2. ...,k - 1). [(Dj n H.);=l. (pl: is finite (because C n H. is finite), so we may c b Dn+l E p such that Dn+l C 4 a d for each t E (1.2 ....,k - 1)andforeach E E [(Dj n Hn)~=l.(pl:, D.+I E B,+I(E.~). The constnrction being complete, we now only need to show that [(D.)zl, (pl5 C Ai. SoIet E E [(D.)z I.~~~andpickx~.x~.....x~,andm~ < m2 < <mtsuch k),mj r p(xj-I). thatforeach j E (1.2. .. ..k).xj E D.,,andforeach j E (2.3. and E = [XI,XZ.. xk). Let F = (xl,xz. ....xk-11. Then foreach j E (1.2.. .. k Il.(p(xj) < y + l 5 ntx so F E [(Djn~a,-~)~:;',(p]~-l s o x ~ E DmkC Bk(F,i),so E E Ai asquired.

...

.

.... -

...

0

As the first application of Theorem 18.4. consider tbe following mUmm which is a common generalization of Ramsey's Theorem and van dcr Waerden's Theorem (Corollary 14.2). To prove this generalization, we use an ultrafilter which has the property that every member contains an arbitrarily long arithmetic progression.

Tbmrem 185. Let k, r E N and let [MIk = U t l AI. l l e n then exist sequences CO and (d,)El in such that whenever nl < n l c . < nk and for each

.. ....,n , ) . o n e h {a,, + td., .am,+ Q&,, ...,an, f tkdnr1 E AI.

j E (1,2.....k),tj E (0.1

G AB (which is nonempq by Theorem 14.5). Pick i E (1.2.. .. ,r ) and a sequence (D.)zl,of members of p as guaranteed by Theorem 18.4. Pickal anddl suchthat(a1, al+dl) E DI andlaC(1) = 1. Inductively assumethat (am):,,, (dm):=,, and (l(m));=l have beenchosen sothat foreachm E (1,2. ...,n), am+mdmlg Dt(,,. Lett(n+l) =max{l(n)+l,a,+ (a,,am+dm.am+2dm. nd,, 1) and picka.+l and d.+l such that

P m f Let (p :N + N be the identity map and pick p

+

....

Pn+ltan+l +d.+i.a.+~

+%+I..

Assume now that n~ < n2 < (0.1, ...,nj). Then

... -z

...%+I + (n + I)d,+11 E DI(.+I). nk and for each j E 11.2,.

..,k) rj

E

E[ ( D n ) ~ jE , ~Ai. l~ (anl +tldnl,an2+ t ~ d n ~ . . . . , a +trdnk} n~

The case k = 2 of Theorem 18.5 is illushated in Figure 18.1, where pairs of integers are identified with points below the diagonal. In this figure arithmetic progressions of length 2.3, and 4 are indicated and the points which are guaranteed to be monochrome are circled. It is the content of Exercise 18.1.1 to show that Theorem 18.5 cannot be extended in the case k = 2 to require that pairs from the same arithmetic progression be included. A significant gencralizstion of Ramsey's Theorem, which has often been applied to obtain otherresults. is the Millie-Taylor Theorem. To state it, we introduce some noIation similar to that introduced in Definition 18.3. We state it in three forms. depending ".",or "U". on whether the operation of the semigroup is written as

"+".

Ddinition 18.6. (a) Let (S, +) be a semigroup. let ( x , ) z , be a sequence in S, and let k E W. Then

.....

F S ( ( X ~ ) ~ ~= ) I(${CsH,X,, %fix, &H~XI): foreach j E (1.2 ,...,k), HI E $(N) andif j -z k, then maxHj c minH,+I].

(b)Let (S. be a semigroup. let (x,):=~ [FP((X~)Z~)I:= {

be a sequence in S, and let k E N.Then

cn,,~, n r s ~ l nrsnI

XI, x,, . .., xr) : foreach j E (1.2.. ..,k), H, d Pf(F1) a n d i f j e k , thenmaxHj <minHj+l}.

Figure 18.1: Monochrome Products of Arithmetic Rogrr~sions (c) Let (Fn)E1be a sequence in the semigroup (3'f(W), U) and let k E N.Then

The Milliken-Taylor Theorem is proved by replacing the arbitrary nonprincipal ultralilter used in the proof of Theorem 18.2 by an idempotent. Theorem 18.7 (Milliken-Taylor Theorem -Version 1). Let R, r E N,let ( x , ) z l be a sequence in N,and assume thnt [Nlk = U:=l Ai. Then there exist i E (1.2. ... r ) nnda sum subsystem ( y . ) z l of ( x . ) z l such thar

.

Proof By passing toa suitable sum subsystem if necessary. wemay assume that (x,,)zl satisfies uniquenessoffinitesums. (Tharis.if F, G E Pf (0and C.,F x, = x., then F = G.) Define 9 : N -+ N so that for each F E Pf(N), ~(C,,,FX.)= max F. defining cp arbitrarily on W\ FS((x,,)z, ). Notice that 9 is finite-to-one on FS((X.)~=~). Pick by Lemma 5.1 1 an idempotent p E pW such that FS((x,,,)r=,,) E p for every n E W. Pick by Theorem 18.4 some i E (I, 2. ..,.r) and a sequence (D.)r=l of members of p such that D,+I E D, for each n and [ ( D , ) z l , 91: E Ai.

Let BI = DI n F S ( ( x , ) ~ , ) d pick YI E BI'. Inductively, letn E Nand assume that m have chosen (yr);=, in Nand (B,):=! in p. Let

and pick Y.+I E %+I*. S i ,for each n E N, y,+~ E FS((X.)&,(,,+~), we have immediately that (y.)z, is a sum subsystem of (%,)El. One cm verify as in the first proof of Theorem 5.8 that whemver F E P r o and m = min F,onehas E.,F y. E B,. To complete the pmof, it suffices to show that

.

So, let o E [PS((y.)~,)]$ and pick FI.F2,.. ,Fk E PI (Nsuch that rnax Fj < min F/+I

for each j e (l,2..

..,k - 1)

and

a = (Cr.Ft Yt. E f & ~ l ,.... C!€F~ y,). For each j E (1.2,. ,k), let tj = min Fj. let L j = maxfj, and let mj = (P(Y,,-I)+ 1. Then E,,F, yf E B,, ED,,. Further.if j E (2.3. ...,k),thm

..

-

ComII~ry18.8 (Milliken-Taylor Theorrm Version 2). Let k. i E N and k-i [ p f( N ) ] ~= U:=l Aj. Then there &st i E (1.2,. . r] and o sequence ( F , ) z l in P,(W such rhnrmax F,,< min F . + ~ f ~ r a l l n [PU((FJz,)l$ E Ai.

..

.

...

Then FI]" = Ur Bi so pick by Theorem 18.7, i € {I, 2. r ) anda sum subsystem (y,,)z, of (2n-i& suchthat[FS((y.)~,)l$ E Bi. Since (y.)zl isasurnsubsystem of (2n-1)z0,,, pick for each n, some F. e *(N such that y. = CrEFa and rnax F, - G i n &+I. To ~ e that c [FU((F.)~,)I$ C A,. let HI. Hz.... Hi E Pf 0 with max Hi < min Hj+l for each j E [1,2,.. ,k - I]. Then

.

.

Now. given j E (1.2. . .,k), let Gj = UneHjF". Thea &H, if j < k. then maxGj < minGj+l. Thus

..

(&K,

..

2I-I. x t € ~ ~ Z I - l ,., X,&, 2I-I) = ( ~ Ezt-I. G ~ &c2 2'-l,

..

yI = &cj 2'-'

...,&G,

and

2'-')

. Gx] = (Kt. Kz, . .Kk) (althaugh, we don't lolow for example that GI = Kr) SO that (GI. GI.. GI) E Ai as q u i d . 0

so (GI, Gz..

..,

It is interesting to note that we obtain the following m i o n of the Millien-Taylor nKorcm (which trivially implies the first two versions) as a corollary to Corollary 18.8. which was in nun a corollary to Theorem 18.7, but cannot pmve it using the shaight fmward modification of the proof of Theorem 18.7 which simply replaces sums by products. The reason is that the function (p may not be definable if the sequence (x.)zl in the semigroup (S. .) does not have a product subsystem which satisfies uniqueness of finite products.

-

ComUary 18.9 (Milli-Taylor Theorem Version 3). Lct k, r E N. let (S, .) be o semigmup, let ( x a ) z , be o sequence in S, ond arsumr that [S]' = Ai. Thm there &t i E (1.2. ,r] and o product subsystem (yn)Kl of (x.)zl such that [ W ( Y ~ ) ~ ~G )Ai. I$

...

P m f For each i E (1.2.

.

... r), let

.... ~ , E F , x E~ ]A ~ ] . Bi so pick by CoroUary 18.8 i E (I. 2.. ...r ) and a sequence Then [Pf(N)Jk = F & (, in Pf 0such that ma%F. < min F.+l for all n and PW((F.)~=l)I: G Bi. For each n E N. let y, = n,,~, Thm l F P ( ( y . ) ~ l ) lE~ Ai. (The verification si= {(FI.FZ. ....F ~ E) [

P ~ ( :~cnlsfixI, I ~ ntsf2xI,

XI.

of this assertion is Exercise 18.1.2.)

0

Exercise 181.1, Let ~ ~ = { ( x , y ] E [ ~ ~ : x ~ y ~ ~ x ] and let AZ = (IX. YI E

I N I ~: Y ? 2x1.

m e n [M2 = A, U A2. Rove that there do not exist i E (1.2) and sequences (o.)zl and (d,,)z, in N such that

.

(I) whenever nt .c nt and for each j E (1.2) and E ( I , 2, ... nj) one has (on1 +tldn,ron, +Qdnnl € 4and Ai. (2) whenevern E Nand 1 5 t .c s 5 n one has (0. td., on +sd.]

+

Exercise 18.1.2 Complete the proof of Corollary 18.9 by verifying that [~p((y~)z~E ) l '4. ,

.

.

.Errrdse 18.1.3. Apply Theorem 18.4using a minimal idempotent in acountable semigroup S. See what kind of statements you can come up with. Do the same thing using a combiitorially rich ulbafilter in BN. (See Section 17.1.) (This is an open ended exercise.)

183 IP* Sets in Pmduct Spaces We show in this section that IF'" sets in any finite pmduct of commutative semigroups mntain products of finite subsystems of sequences in Ihe coordinates. We first extend the notion of product subsystems to finite &uences.

De6nition 18.10. Let (y.)zl be a sequence in a semigroup S and let k. rn G N.Then (x.);=:=, is a product subsystem of FP((y,)Ed) if and only if there exists a sequence (H.);=,in Pf(E"d) such that (a) min HI 2 k, (b) max H, c min H.+I for each n E (1.2, .,m - I), and (c) xn = y, for each n E [I, 2,. . .,m).

..

nrE~.

The proof of the following theorem is analogous to the pmof of Theorem 18.4.

.

Theorem 18.11. Let L E N\(l J and for each i E (1.2, . . ,L) let S; be a semigmup f and le: (yi,.)zl be a sequence in S;. Let m, r E PI and let Xi=IS; = UJpl Cj. There cn'sts j E (1,2, ,r) andfor each i E 11.2.. ..,f - 1). t h m exists a pmducr

...

subsystem (x;,.);=~ of FP((yi,.)EI) and them exists apmduct subsysfem (xt,.)zl of FP((yt..JEI) such t h t

Proof: Given i E (1.2.

.

... L), pick by Lemma 5.1 1an idempotent p; with

.-

Now given t E 12.3,. .. e 1). assume that B,+l(x~.12,. ..,x,, j ) has been defined foreach (XI.x2,. .. ,x,) E and each j E (1.2,. .. r ] . Given

xI=~s~

.

18.2 iP Sets in Ftuduct Spsces

Finally, given that B ~ xj .) has been defined for m h x E SI and eath j 6 (1,2, ...,r ) , let B t ( j ) = ( x E SI : &(x, j ) E m). We show by downward induction on t that for each t t [2. 3.. .. ,() and each ( x I , x z , . . .,xt-l) E X ; ; : S ~ . S ~= U;-l B,(XI.XZ,.. .;xt-I, j ) . Tbisisaiviallytme for t = L. Assume t E (2.3, ...,l 1) and the statement is true for t 1. Let 1-1 ( X Ixz, , . . xt-l) E X i=ISi. Given y E S,, one has that

.

-

.

+

nKn Dt+l E pt+l so pick by Theorem 5.14 a product subsystem

c

F P ( ( Y ~ + I . ~such ) E ~that ) F P ( ( x l + ~ , , ) z l ) DI+I. Then FF'((X~."):=~)) x FP((Xt.")Zl) G

(xZ

(~~+l,)r=~ of

c,

Incontrast withTheorem 18.11. where the partition conclusion applied to products of arbitrary scmigroups, we now restrict ourselvesto products of commutative semigroups. We shall see in Theorem 18.16 that this restriction is necessary.

Definition 18.12 Let S be a semigroup. let ( y . ) z l be a sequence in S, and k t m E M. The sequence (x,,):=~ is a weak product subsystem of (y,,)El if and only if there exists ,m ) and a sequence (H,,):=l in P,(N such that H, n Hx = 0 when n # k in (1.2, x. = &H. yt foreachn E (1.2 m).

,....

...

Recall by way of contrast, that in a product subsystem one requires that max H. < min H.+I.

Then L b a compacf subsmigroup of B( x

SO.

P m f Since any pmduct subsystem is also a weak product subsystem we have by Thuncms 18.1 l and 5.7 that P # 0. Since C is defined as Ihe set of ultrafilters all of whose members satisfy a given property. P is closed. hence compact. To see that P is a semigroup, let p, q E P, let A E p .q and let rn, k E N.Then (2E X :ii -LA E q ) E p so choose for each i E (1.2,.. . f) a weak product subsystem (q,,);=l of fl((yi.n)ZA) such that

.

. n,.~,,

...,

rn), pick Hi, E $',@ with minHi,. I) 2k Given i E (1,2....L) and n E (1.2. such that xi,. = yi,, and if 1 5 n < s 5 rn, chen Hi,. n H i , = 0. Let r = m a ~ ( U : = ~E=, Hi,.) 1 and let

+

.

..

Then B E q so choose for each i E (1,2.. L ) a weak product subsystem (~i.~);=~ o f ~ ~ ( ( ~ ~ , . ) ~ ) s u c h~tft=t ~ a t~ ~ ( ( z i , . ) E ; = B. ~ ) mi E (1,2 ,....eland n E (1.2...., m ) , pick K i , E +(PO with min Ki,. 2 r such that ti,. = yi,, and if 1 5 n < s 5 m, then Ki.. n Kiss = 0. For i E (1.2.. ,L ) and n E (1.2, ml,let Li,. = Hi,, U Ki,,. Then

n,,~#,

.

..

...

....,

n < s 5 m. then Li.,, i l Lic = 0. Thus for each i E (1.2 1). (xi,, .~i.~):'~ is a weak product subsystem ofFP((yi,JZA). t Finally we claim that Xi,, FP((xi, a,,,);=:,,) C A. To this end let Z E e Xi=, FP((xi,. -zi,.);=',,) be given. For each i E [I,2,...,L), pick Fi G (1,2,...,m) such that ci = nlEfi(xi,, . zi.,) and hi= xi,, md b; = l l , . ~ti.,. ~ Then t m((xi,)L,) one has that 6 E X f = l F P ( ( Z ~ , & ~so~ E B . Since ii E xi,, i E 2-IA so that r3 b E A. Since each Si is commutative. we have for e3ch i E (1.2. 9.) that nrtfi(xi.r - ti,,) = ( n f s ~xi.,) , .(n,,fi zi,,) sothat: = r i . b as IequiRd 0

and if 1 5

.

nIEC

....

.

..

Tbeorrm 18.14. Let f E N and for each i E (1,2,. . 8). Ict Si be o commutative s~ .remifmup and Irt (yi,,)E, br o sequence in St. Ln C be on IF" set in x ~ , and lerm E N. Thenforeach i E (1.2.. . e) then isa weokpfuductsubsystmt (xi,.);=, of FP((yi.tt)El)

&

..

Xi,, FP((~i,m)r=,) C C.

xI-~s~

so Pmof: Ld be as in Lnnma 18.13. Then L is a compact subsemigroup of by Theorem 25 there is an idempotent p E L.By Theorem 16.6, C E p. Thus by the definition of C. for each i E (1.2,. . 8 ) c h m is a weak pmduct subsyskm (xi,),"=,

.

.

ofW(~i.n):~)

C C. such that x;=~ FP((X~.~)?=~)

Time natural questions are raised by Th-

18.14.

(1) Can one obtain infinite weak product subsystems (definedin the obvious fashion) smh that XI-1 W((*i.n)zl) C C? (2) Can one replace ''weak pmduct subsystems" with "product subsystem"? (3) Can one omit the requirement that the semigroups Si be commutative?

We answer all three of these questions in the negative. The hrst two questions are answered in Theorem 18.15 using the semigroup (N x N,+). Sincetheopetationis addition we referto"sumsubsystems"rather(han ''product subsystems".

Thmrem 18.15. Then is an IP* set C in N x N such that: (a) There do not &a z E N ond a sequmce (x.)zl in N such tkaf either ( z ] x FS((x,)zl) 5 C or FS((x,,)zl) x (z} E C. (In particulnr then do not exist infinite weok sum n*bsystcm XI..)^, a& (xz..)zl of FS((2')Zl) with FS(G~I..):,) x FS((xz.n)Zl) E C.) (b) 77wn do not exist sum s u b s y a m (x& . and (Y& ofFS((2')El) with lx1.xd x ly1.n) c c. Pmaf: Let

Tosee that C is anIPe set in NxN. suppose insteadthatwe have asequenee ((x.. y,,))Zl inN xNwith

PickFl andGI i n P ( o ) suchthatxl = &F, 2'andyl = &c, 2'. Letk = max(F1U GI). Choose.byLemma14.16.H E .F',(N)suchthatminH z 1 , 2 k + ' 1 ~ n E ~ ~ n , d Zk+'l Z n Ey.. ~ Pick F', G' E 9 ( m ) such that &H xn = X,,p 2" and C n eyn~ =

380

18 Multidimensional Ramsey Theory

Cn6,yP. Then k + 1 i min(Ft U G'), so XI Cnm~n = LO~M 2n. Also

+

+ E~EHX. .=

2' and y1

+

+

&all)u~(xnrYPZ) (xl &r;H *n. Y1 %wH ~ n ) E [ ( C ~ E F ~En& " . 2") : F. d E 9(@) and maxF < minGor manG < minF). T h u s k + l i m a x ( F I U F t ) c m i n ( G ~ ~ G ' ) k' io r k + l ~ m a x ( G ~ u G < ') min(F1 U F') i k, a contradiction. To establish (a), suppose that one has z E N and a sequence ( x R ) z lin N such that either ( 2 ) x FS((x.)?=,) 5 C or FS((xn)zl) x (2) G C and assume without loss of generality that (z] x FS((x,)zl) E C. Pick F E P ( o ) such that z = & ~ 2 ' and let k = max F. Pick H E P f ( W such that 2'+'I &H xn. Pick G E B ( o ) such that &HX. s C I E2'.~ Tbcn max F < min G so (z, Z n exn) ~ f C. To establish @) suppose that one has sum subsystems (x,,):,~ and ( ~ . ) ; f =of ~ FS((Zn)Zl) with [XI,xz) x [yl. n ] E C. Pick FI.GI, Fz, Gz E B(w)such that X I = C n e ~2", , xz = Ench2". yl = C n E 2". ~ , y2 = C n e ~2".2 man FI < min Fz. and max GI < min Gz. Without loss of generality. max FI 5 max GI. But then we o have max GI i max Fl < min Fz SO (XZ.YI) 6 C. a contradiction. A shiking contrast is provided by Theorem 18.15@) and the case C = 2 of Theorem 18.11. That is IP*se.ts are not in general patition ttgular. It is easy to divide most semigroups into two classes, neither of which is an IF* set. Consequently it is not too swprking when one finds a propem that must be satisfied by an IP* set a semigroup S (such as containing a seipence with all of its sums and products when S = W which need not be satisfied by any ceU of a partition of S. In this case we have a property, namely containing FS((XI..)~=~) x FS((xz..)zl) for some sum subsystems of any given sequences, which must be satisfied by some cell of a panition of N x N,but need not be satisfied by IP*sets in N x N. The third questionraised by Theorem 18.14isanswered with anexample verysimilar to that used in the p m f of Theorem 18.15.

Theorem 18.16. Lcr S be rhefree semigmup on the alphabet ( y ~y2, , n,...). Then is an IP* set C in S x S such that them & not exisr weak product subsystems (x.)f a n d ( ~ . ) i = o~ f ~ ( ( ~ ~ ) ~ ~h ) t sl xu~c, xhz x l @ I , wzl C C.

Pmof: Let

To see that C is an IP* set, suppose one has a sequence ((x., w n ) ) E l with

183 Spces of Variable Words

381

Givenany i c jinN.pickFi, .&, GI, Gj, Fi.j, Gi.) e Pf( 0 suchthatxi = llneFiy,,, X j = nnEq Yn. Wi = nneGi Yn. W j = Y m xixj = &&,, Yn. Wiwj = ll,,,Gi,y,,. Sincexixi = nnEF.,jy..wehavethatmaxFi minFj and Fi,, = F i u F j and similarly max Gi < min GI and Gi,j = Gi U Gj. lhus we may pick j E N such that ~ ( FU IG I ) 4 min(Fj U G,). Then

nns~,

a contradiction. Now suppose we have weak pmduct subsystems (x,):=, and (w.):,~ of FP((yd?= such that (xl.xz) x ( W I , W ~G C. Pick FI, F2. GI. G2 E g)l(W) such yn. WI = LG, yn, and w = yn. since that xc = h n a F , Y. x2 = ~ 1 x E2 FP((yn)zl).wemust havemax FI < min F2 and similarly maxG1 c minG2. Without loss of generality, max FI2 max GI. But then we have max G I _c mar Fl .E 0 min F2 so (xz. w I ) $4 C. a contradiction.

nns~2

nnfol

18.3 Spaces of Variable Words In this section we obtain some infinite dimensional (or infinite parameter) extensions of the Hales-Jewett Theorem (Corollruy 14.8) including the "main lemma" to the proof of Carlson's Theorem. (We shall prove Carlson's Theorem itself in Section 18.4.) These extensions involve sequences of variable words. Recall that we have defined a variable word over an alphabet I to be a word over I U (u) in which v actually occurs (where u is a "variable" not in I ) . We i n d u c e new notation (diering from that previously used) far variable words and nonvariable words. since we shall be dealing simultaneously with both.

Doflnition 18.17. Let Q be a (possibly empty) set (alphabet). (a) W(*; u) is the set of variable words over P. @) W ( I ) is the set of (nonvariable) words over 4 ' .' Notice that W(Q; u) u W(8) = W(I

u (uJ).

Definition 18.18. Let O be a set and la (s.(u))zl be a sequence in W(P; v). (a) The sequence (&(u))zl is a variable reduction of (s. ( u ) ) z l if and only if there exist an increasing sequence of integers ( n i ) E l with nl = 1 and a sequence (ai):, in I U (u) such that for each i.

ot (s.(u))z, if end only if is a rariable reduction of some subsequence of (s,,(v))~,.

(b)The sequence ( t . ( ~ ) ) is ~ ~a variobk amcfion (t.(v))zl

Thus. a Mliable reduction of (s,,(v))~,is any infinite sequence of variable words obtained from (sn(u))zl by replacing each si(v) by one of its instances, dividing the resulting sequence (of words over W U (u)) into (infinitely many) finite blocks of consecutive words, and concatenating the members of each block, with the additional requirement that each of the resulting words has an occurrence of v. F a example, if { ~ , , ( v )were ) ~ ~the sequence (avbu, u, bau, abub, vavu. . .) then (t.(u))zl might be (aabaubab, abubbabb. . ..).

.

DeRnition 18.19. L* W be a sct and kt (sn(v))z, be a sequence in W(9; v). (a) The word w is a reduced woni of (s. ( u ) ) z l if and only if them exist k E N and at,@.. . . a k in W U [u) such that w =sl(al)-s2(a2)-. ..-*(ad. (b) A variable reduced word of (sm(u))z,is a variable word which is a reduced word of (s"(v))zl. (c) A connant reduced word of ( S . ( U ) ) ~is~a reduccd word of (s.(u))z, which is not a variable reduced word.

.

Reduced words, variable redufed words and constant reduced words of a 6nite sequence (S.(U))~,~am defined analogously. Notice that a word is a variable reduced word of ( s , , ( v ) ) ~ ,if and only if it ocam as the first tmn in some variable reduction of (s.(u))z, . De5nition 18.20. Let w be a set and let (sn(u))El be a sequence in W(W; v). (a) The word w is an emaeted word of (s,(u))z, if and only if w is a reducedword of some subsequenceof (s. ( u ) ) z l . (b) A variable extracted word of {s,,( @ ) E l h a variabk word which is an extracted word of (sn(v))EI. (c) A constanr Lnmctedword of (s. (u):, is an extractedword of (s. ( u ) ) z l which is not a variable extracted word. Notice that w is an extracted word of (s.(u))~=, if and only if there exist k nl < n z < . . . < n t i n N . a n d a ~ , a l . a ~WU(vJsuchthat

...,

w

6

N,

= s., ( ~ I ) ~ S . , ( O ~ )...^snl(ak). -

Extracted w a d s am defined in the analogous way f a a finite sequence (s,(v));_!. We now introduce some special notation which will be used in the proof of the mau~ theorem of this section. The notions defined depend on the choice of the alphabet W, but the notation does not reflect this dependence. Definition 18.21. Let 9 be a seh let k e W U (w) and let (s.(v))~=, be a sequence in W(W; v). Let S bc the semigmup (W(W U (u)).-). (a) For each m E N. ~((s.(v))!,) is the set of extracted words of (s,(u))i=,. (b) For each rn E N. v((s.(u)):,) is the set of variable extracted words of k (s.,(u))"-.

183 Spaces of Variable Wadr

383

(c) For each m E A, ~((s.(v)):,) is the set of constant extracted wads of k .. (s"(u))"=,,,. etas A((sn(u))%). (d) A(bn(v))ZD,o = (e) V((sn(v))z1)= hmSl clps V ( ( S . ( V ) ) ~ ) . (0e((&(v))L,) = ceas c((s.(v))F-).

nrsl

n:=,

Lemma 1832. Let 8 bc a set, let (s.(v))~, be a scqumcc in W(W: v), and let S be rfir semigroup (W(8 U (v)). 7.l7k-n A ( ( S ~ ( U ) is ) ~a ~compact ) subscmigmup of BS. and V ( ( S . ( ~ ) ) ~ism ~ ) Meal of A ( ( s d v ) ) z l ) . If* # 0. a compact subsemigmup of BS.

e ( ( s n ( v ) ) z l )is

Pmof: WeuscLwnms14.9withl =(l),D=N.andT, =Sforeachm EN. To secthat A((s.(u))zl) isasubsemigroupof /3S and V((s.(v))zl) is an ideal of A((a(v))?=,), foreach m E N let Em = A((s,(v))&) andlet I, = V((s.(v))&). la order to invoke Lemma 14.9 we need to show that (I) foreachm e N, .I g Em E..T (2) for each m E N and each w E I,, there is some k E N such that w-Ek G Im. and (3) foreachm E Nandeachw E E,\l,,thereisscmek~ Nsuchthatw-Ek S Em and W-lk C 1,. Statement (1) is trivial. To verify statement (2). let m E N and w e I,,, be given. P i c k t s N . n l c n z c . . . c n , i n N w i t h n l ~ m , a n d a ~...., . a ~ a, ~ O U { v J ( w i t h at least one ai = v) such that w = sn,(a~)-snr(az)-.

Let k = n,

+

..%,(at).

1. Then w-Ei C I, as required. The verification of statement (3) is identical except that a l , az. . ,a, E V. Now assume that W # 8. To show hat C((s.(v))Z1) is a subsemigroup it sufiices by Theorem 4.20 to show that for each m E N and each w E C((h(v))Fq) there is some k E N such that w-C((s,,(v))zk) E C((s.(v))~,). The verification of this assertion is nearly identical to the verificationof statement (2) above. 0

..

By now we are accustomed tousing idrmpotcntsto obtain Ramsey-theoretic results. In the following theorem, we utilize two related idempotents in the case V # 0.

Theorem 18.23. Lcr 8 bcafbritesctandlct (s.(v))z, beaseqw?nce in W(W v). kt 7 and B. befinite panitions of W(V; v) and W(W) rerpcctively. There exist V E F and C E $ and a variabk czfmction (w.(v))zl of (s.(v))zl such thn, all variable e*tracfcdwordsof (w.(v))zl an in V andall constanIu~mcfcdwordrof( W . ( U ) ) ~ ~ are in C.

384

18 MultidimensionalRamsey Theory

Prwj Assume first that O = 0. Then I(*; v) = (v* : k E N) and W(0) = (0). Given n E N. pick k(n) E N such that s,,(v) = vk("). For each V E F let B(V) = [k E M : vk E V). Pick by Corollary 5.15 V E F and a sum subsystem (L(n))zl of (k(n))gl such that PS((L(n))zI) c B(V). For each n e N let w.(v) = van). Then (w,(v))El is a variable extraction of (s++(v))El and the variable extracted words of (w.(v));p,, correspond to the elements of FS((t(n))zl) so that all variable exmcted words of ( W , ( U ) ) ~ =arc ~ in V. The conclusion about constant exfracted words of ( W . ( U ) ) holds ~ ~ vacuously. Now assume that O # 0. By Lemma 18.22. C((s.(v))gI) is a compact subsemigroup of pS. where S is the semigroup (W(W U [v)),-). Pick by Theorem 1.60 an idempotent p which is minimal in C((S.,(U))~~).Then p E A((s.(v))zl) so pick again by Theorem 1.60. an idempotent q which is minimal in A ( ( S , ( U ) ) ~such ~ ) that q j p. Since. by Lemma 18.22. V ( ( S . ( U ) ) ~ ~is)an ideal of A((s,,(u))zl), we have that q E V((s&))z1). Now, for each a E W, define the evaluation map E. : W(O U {v)) + W(W) as follows. Given w E W(W U [ v ) ) , d w ) is w with all occurrences of v (if any)replaced by a. Thus, if w = w(v) E W(P; v), then ea(w(v)) = w(a), while if w E W(P), then t,(w) = W . Notice that c. is a homomorphism from W (W U (v)) to W(O) and hence, by Cotullary 4.22, & is a homomorphism from p(W(Q U {v))) to B(W(O)). We claim that for each a E P. c(q)= p. Since G is a homomorphism, we have that t",(q) 5 G(p). Since c. is equal to the identity on W(*), we have G ( p ) = p and consequently G(q) + p. Sincep isminimalin C((s&))zl), toverify that <(q) = p we need only show that G(q) E C((s.(v))zl). To see that &(q) E C((S,,(U))F=~),let m E N be given. Now V((s.(v))T') E q so ~ , [ V ( ( s , ( v ) ) ~ ) E] G(q) by Lemma 3.30.It is mutine to verify that G[v((sn(u))Zm)I E c((sn(v))EP,,)

-

E G(q) =required. We shall. according to our usual custom, use for the binary operation on pS which extends the operation on S. L e t V ~ F n q a n d l e t C ~ $ nWeputvfl=[y ~. E V : y - p ~V a n d y - q ~ v ) a n d ~ U = ( x ~ C : x - ~ ~ T a n d x S- i~n c~e qV- )q =. p - q = q n p = q a n d p-p = pi it foltows that V h q and Cu e p. We claim that for each y E V h d each x E C < yy1(Vu) E q, y-l(V" E p, x-'(Vg) E q, and x-'(en) E p. To see this,let y E Vu and x E Cs be given. To see that y-'(vD) E q, we observe that A;'p;' (71 is a neighborlld of q in pS, and so is l;'p;'[V]. Also y-'V E q. Since y-I(VD) = y-IV n h;lp;l[~] n ~;lp;~[V], it follows that y - l ( ~ " E q. Likewise, since p E l;lppl[V] n I;I~;~[V], and y-IV E p,y-l(Vu) E p. Similarlyx-'(VU) = ~ - ~ v n ~ ; ~ ~ ; ~ [ V ] n l E ; q~ ~ , - ~ p. ; \ , - I ~ ; ~ [ V ] and X-I(CU) = X - ~ C ~ A ; ~ ~ ~ ~ [ T ] ~E A If x, y E A((S.(U))~=~),we shall write x << y if there exists m E N for which x E A ( ( S . ( U ) ) ~ ~and ~ ) y E A ( ( s . ( u ) ) ~ ~ + , ) .We observe that, for any given x E A ( ( s n ( ~ ) ) z l )IY, E A ( ( s d ~ ) ) $ ~ :)x -=
so C((s.(v))&)

-

-

18.3 S p x c s of Variable W d

385

We npv inductively wnsuuct a variable e d o n ( W . ( V ) ) of ~ ~(s.(v))zl for which V((wdv))E1) C Vn C((w.(v))E,) C Cn. e;' [Cn]. We note that this is possible because We first choose w!(v) E vPr l €o-'[CPI E q forevery. E 8. We then let n E Pa and mume that we have chosen wi (v) for each i E (1.2.. ..,n) so that wi(v) ce wi+,(v) forevery i E (1.2. ..n - 1). V((wi(~))y=~) C VU and C((wi(v))ySl)C CP. We can then choose w,,+l(v) satisfying the following conditions.

new

..

This choice is possible, because each of these conditions for w.+l (v) is satisfied by all the elements of some member of q. To see that V((wi(v))y_:') 5 V', let u E ~ ( ( w i ( v ) )and ~ ~pick ~ ) k and m; c r n 2 < ... cmtinMandal.a2, .... a k ~ 8 U ( v ) s u c h t h a t s o m e a ~ = v a n d

+

+

If mr c n 1, then u E vn by the induction hypothesis so assume that mk = n I. ~f k = 1, then u E V h Ycondition (i). so assume that k > 1. If at E 8 , then u E VP by condition (iv). If ak = v and la^. a2.. . ,a,+!) S 8,then u E Vg by condition (v). If ak = v and v E (at, az, ...a&-,), then u E V' by condition (iii). That ~ ( ( w i ( v ) ) E ~~ ~followsfromtheinduction ) CP hypothms and conditions (i) and (vi). Fmally, notice that (w.(v))Kl is a variable extraction of ( s , ( v ) ) ~ , by conditions 0 (i) and (ii).

.

.

The following coroliary includes the "main lemma" to the proof of Carlson's Theorem.

ComUary 18.24. Let 8 be afnite set andlet (s,,(v))zt bea sequence in W(8: v). Let F and $ befinitepam'tions of W(8; v) and W(W respectively. Them exist V E F and C E $ and a variable reduction (r,,(v))zl of (S,,(V)):=~ such tluu all variable reduced words of ( r . ( ~ ) )are ~ ~in V andall constant reduced words of([. ( v ) ) z l are in C.

P m f Define 9 : W ( 8 U (v)) + W(8 U (v)) by whereal.a?, . . ., a t arein8U[v). Notethat(p[W(8; v)] 5 W(8; v)and(p[W(W)] 5 W(@). Let X = (9-'[B] : B E F) and X = ((p-'[B] : B E $1. Then X and X

18 Multidimeasiod Ramaey Theory

386

are finite patlitions of W(W: v) and W(W) r;espeCtively. Pick V' E X.C' E X,and a variable exbaction (w.(v))L, of thesequence (v. v, v.. ..) as guaranteed by Theom 18.23. (That is, (w,,(v))El is simply asexpenceof VariabIe words. all of whose variable exwcted words are in V' and all of whose constant exwords are in C'.) Pick V E F and C e $ such that V' = (p-l[V] and C' =(p-'[Cl. For each n E N pick .k E W and a . ~ .an,2. ...,amst,E W U {v)such that

and notice that (r.(v))zl is a variable reduction of (s.(v))Z,. Notice funher that if bl, h,. ,b, E W u { v ] . then

..

Theorem 18.23pmducesa sequence (w,,(v))Zl with a stronger homogeneity prop erly than that in Corollary 18.24, but to obtain such a (w.(v))zl we must take an extraction of (s,,(v))zl; a reduction won't do. (To see why it won't do, consider (s,,(u))El = (av, bv. bv, bv, .. .) andpartitionthcvariablewordsaccording towhether the first letter is a or not.) Exercise 183.1. Derive the Finite Sums R ' I(Corollary 5.10) as a- c of Corollary 18.24. (Hint: Consider the function f : W(W; v) + W defined by the condition that f (w) is the number of occunences of v in w.)

18.4 Carlson's Theorem In 1988 T. CarIson published [58] a theorem which has as corollaries many urlier Ramsey-theoretic results. We prove this theorem in this section. DefiniIfon 1825. Let W be a (possibly empty) finite set and let d be the set of all infinite by 3. For 3 E 8. let sequencs in W(W: v). Denote the sequence

(s.)z,

~ ~ ( =3 {i ) E d : i i s a variable reduction of 31, and, for each n

E N.

let

Ihe following lemma is an easy cons#luencc of Remark 18.26.

Luoma 18.27. A topology can be deJfnedon 6 by choosing [B.(a :? 3 d a d n E 01 as a basisfor the open sets. Pmo]T Let 3.; E d , lctm. n E o , and let ii E B,@n Em(?).Let k = max(m, n). Then by Remark 18.26 B&) B,@ n ~ " ( 3 . Deealtion 1838 The Ellen& dandn~o).

topology on d is the topology with basis (B& :;E

For the remainder of this d o n , we shall assume that d has the EUentucktopology.

Dcfinltiaa 18.29. A s* X C d is said to be completely RMUcy if and only if for every n E o and every 3 E d , there is some i E B.(3 such that either B.(i) X or Bn(i)nX=0. Recall that we take the Baire sets in a topological space to be the u-algebragenerated by the open sets and the nowhere dense sets. (Other meanings for the tenn Baire sets also exist in the literatwe.) Recall also (see Exercise 17.5.1) that a set is a Baire set if and only if it can be written as the symmetric difference of an open set and a meager set, where the mca8er (orfirst category)sets are those sets that are the countable union of nowhere dense sets. Carlson's Theorem Crawrern 18.44) is the asserrion that a subset of d is wmpletely Ramsey if and only if it is a Bairc set (with rrspect to the Ellentuck topology). 'Ile proof in one direction is simple.

Thcorrm 18.30. IfX is a compleuly Rtvnsey s u b m o f d , then X is a Baire set. Infacr X is the union of an open set ond a nowhere dense set.

P m j L* Xo denole the interior of X. We shall show that X is a Bake set by showing that X\XO is nowhere dense so that X is the union of an open set and a nowhere dense set. If we assume the contnuy, there exists 3 E d and n E w fm which B.0 C ct$(X\Xo). SinceX iscompletelyRamSey,theree~stsi E B,,(a suchthat &(i) cx or B.(?) E d\X. However. if B d i ) C X, then B.(i) G X0 and hence B,,(i) n clr(X\Xo) = 0. This is a wnhadiction, because B.(?) G B.B). If B.(i) G d\X. then B.(i) n& ( X ) = 0. This is again a contradiction. because B,V) G B A c~c e d ( x ) . Lemma 18.31. The completely Ramsey subsets of d form a subalgebm of P ( d ) .

388

.

18 Multidimensional Ramscy Theory

PmoJ It is immediate fromthe definition of a completely Ramsey set that a subset of d is completely Ramsey if and only if its complement is. So it is su5icient to show that the completely Ramsey subsets of 6 are closed under finite unions. To this end let X and Y be completely Ramsey subsets of 8. To see that X U Y is completely Ramsey,let 3 d and n w be given. Pick? E En(?) such that B.(i) X or E,,(i) G d\X. If B,(j) c X, then En(;) G X U Y and we an done, so assume that B.(T) 5 8\X. Since Y is completely Ramsey,picku' E En(?) such that B.(u') C Y a Em(;) G d\Y. Now B.(u') c En(?) and so if En(;) E Y. then B.(C) X U Y and if g d\Y,then B.(ii) g 8\(X U Y). if.(;)

Remark 1832. Let (?')Elbe a sequence in 8 with the pmperfy that ?"+I 6 B,,,+I(?') for every m E N,nnd let i = ( i l l ,t22, t33,. ..). Then ? e B,(ik) for every m e w. We now need to inmduce some mom notation.

Delinition 1833. Let n

w.

.

(a) If 3 E d and n > 0, then ?. = (sl ,sz. . .. s.) and Zlo = 0. (b) If3 8 , then tail.(?) = ( S ~ + I , S . +.. Z.). . (c) If 3 e W(Q; u)", then lil = n . (d) I t 3 is a finite sequence and i i s a finite or infinite sequence. then (3, i ) will denote the sequence in which ? is followed by i. (Thus, if 3 = ( $ 1 . $2, . . . s,,,) and i = ( t ~12,. . ..),then (s,t ) = ( s l , s2,. .. s, rl, t ~..,.).) (e) A subset X of d is said to be dmosrdense in En(?) if X r l En(?) # 0 whenever i E En(?).

--

.

.

Notice that the notation indiffers slightly from our formal viewpoint since formally n = (0,1, ,n - 1) and (0.1. ... n - 1) is not a subset of the domain of the tixlction 3.

.

...

DeRnftion 1834. We define a condition (*)on a sequence (Tn).,, of subsets of d as follows: (*) For every n

w and every 3 E 8

(a) B.(3)f7Tnf Oand T".then Em(?)E 7".

@) if 3 E

We make some simple observations which will be useful in the proof of Carlson's Theorem. R-k

1835. Let 3 E 8. ( a ) I f X isalmostdensein En(?),then X isalmusrdense in En(;) forevery? En(& @) I f X n B . ( a = 0,then X n B.(i) = 0for every i E En(;). (c) F o r ~ iy E a. i E Em(?)fondonly ifii. = 3,. and tail.(i) E Bo(lail.(3)).

Lemma 1836. Let (Tn).,, be a sequence of subsets of 8 which satiq5es (r) and let 3 E 8. Thenfor each k E o,then exists i E B& such that,for every n r k in N and e v e v variable reducria i we have (7, rail.(?)) E.!'IT P m f : We first show that

(t) forevery 7 E 8 and evcry n E N.there exists G E Be(?) such that (7. tail.(G)) . Tlrl forevery variable nduaion ? of $.

E

..

We enumerate the finite set of variable reductions of iW as (7 I . iZ,.,im). We put ? O = tail. (?) and then inductively choose i I . ? . .., f E 8 so that the foUowing properties hold for each i E [I. 2. .. . ,m):

',

(i) iiE &(? i-l) and (ii) (;i,ii) T~i'l,

..

Suppose that i E (0.1.2, . ,m - 1) and that we have chosen ?'. Let 1 = [?'+'I. By condition (a) of (*). we can choose ri E 6,((ii+'.i i ) ) n T r . Let ?'+I = tailr(;).

men?'+' E Bo(?')and ( i i + ' , i ' + l ) = I E T ~ i i + ' ~

(6.. i "). Xiice i " E Bo(tail.(i)). we have Now let i be a variable reduction of 6.. Then i = i' for some i E ( 1 . 2 . . .. .m). Having chosen im. we p u t d =

a E 6"(i).

E Bo(ii),and hencethat ( i i , i m ) E ~ ~ ~ i i~i ) )(. So ( i (ii,?") ~ , = Wenotethat i m

( i i , taiI.(l)) E TF'I, by condition (b) of (*). Thus (t)is established. Now let k E o be given. By (7) we can inductively define elements To, i ' , of 6 with the following properties:

i2....

-

(i) ... = t ' = s . (ii) inE B.(in-') forevery n > k, and (iii) forevery n > k and every variable reduction 7 of

(7, tail.(;")) E Tlil.

Let i = (1'1, t22. 1'3,. ..). Then ? E B&. Let n z k be given and let ? be a variahle reduction of 4., Now = in-' 1, so i is a variable reduction of in-',1 and thus (i, tail,(?")) E rlil. Also (i, tail.(?)) E Blil((i.tail(in))) and so by condition @)of (*). (i, tailn(i)) E Tli1. o

4,

Corollary 1837. Let ? E 8. k E o. and X G 6. Then there exists i E B&) such that,for eveiy n s k and every variable reduction i of ii., either X is olmost dense in Blil((i,tail.(?))) orehe X n B1;1((7,tail.(?))) = 0. Pmof: For each n E o w e define Tn G 8 by stating that i E T nif and only if either X is almost dense in B.(i) or X n B,(i) = 0. We shall show that (Tn)., satisfies (*). and our claim will then follow from Lemma 18.36. Letn E oand i E 6. If X isahostdensein B.(?).theni E B,,(i)nTn. Otherwise there exists 2 E 6. ( i ) for which B, (I) n X = O and so I E B.(i) r l T". Thus satisfies condition (a) of (*).

To see that (Tn)., satisfies condition (b)of (*), let n E o.let ? E 1"a d let i E 4 ( i ) . Then B d i ) E B.(i). Consequently:if X is almost dense in B&), 0 X is almost dense in En(;) and if X n Em(?)= 0. then X f l B&) = 0.

Ddiattioa 18.38. (a) Let ;E d. then ~ ( =3{w E W(B: v ) : w is a variable redaced

a. ...,

word of ( b ) L e t ; ~ 6andletw E R(3. 'ibe.N(w.i)isihatI E Nforwhichthnruist al.az. a! E O U [v) suchthat w = Il(a~)-rz(az)-. ..-R(al).

Lemma 18.40. Lrr 3 E d. k E o and X E d. Thrn there exists 7 E B& such that, for every n E N sarirfying n > k and every variable reduction i of either

a,

Pmof:Wedetke a seq(T").,, of subsets of d by stating that i e T" ifandonly if either X n Bn+~((il..w , tailN(w,&(o)+n(i))) = 0 for every w E R(tail.(i)), or else X is almost dense in B.+l ((il,,. w , t a i l ~ ( , , ~ ] ~ ( a )(5)) + . forew E ~(tai~., (5)). We shall show that (Tn)., satisfies (*I. The claim then follows from Lemma 18.36. Let n E o and r' E d. We shall show that T" and i satisfy conditions (a) and (b) of (*). By Corollary 18.37, there exists 3 E Em(?)such that, for every rn r n in W and every variable reduction i of Zl,. X n Bl;l((i, fail,(;))) = 0 or else X is almost dense in Blil((i.tail,(;))).

Let

vo = (w E R(um(u')) : X n Bn+l((u'lm W, tail~(rn.a(ri))+n(Z))) = 0). VI = R(tail.(i))\Vo. and V2 = W(W v)\~(tail,(i)). We note that, if w E R(tail.(i)), then w E VI if and only if X is almost dense in B ~ + I ( @ w, I ~taih(w.&~<,i))+n(z))). . By Corollary 18.24, there exists i E (0.1.2) and E Bo(tail,,(Z)) such that R(5) G K. Since R(B) E R(tail.(;)). i # 2. Let G = (31.,1). Then F E Em(?)n Tn. So T" and i satisfy condition (a) of (*). To verify that T' and i satisfy condition @) of (*). assume that i Tn and l a e En(?). Then tail,(;) E Bo(tail.(?)) and R(tail"(3)) g R(taib(i)). If w E R(taiL(6)). then

so that It foUows easily that 3 E T".

a

Luuun 18.41. Every c & s d subset X ofd is cmpfctety Rams9. Pmqf L e t i E d a n d k E o . W e n c e d a s h o w t h a t ~ e x i s hE~Bk(aforwfiich Bk(2) E X or B d b ) n X = 0. We suppose, on the contrary, that no such elemmt b exists. So X is almost dense in B&. Let i E B t ( 9 be the element guaranteed by Lemau 18.40. We shun show that.for every u' E &(?) and every n 2 k in o, X is almost d e w in B&). This is m e if n = k. We shall assume that it is m e for n and deduce that it is also rme for n + 1. Suppose then that u' E Mi)and let m E W be the integer f a which El, is a variable By our inductive assumption. there exists ?. E X n B,, (ii). Let w = reduction of x.+l E ~(tail,(i)). Since I E B.+~((u'l., ~ , t a i l ~ ( ~ . ~ l ~ ( i n ) )X, ~ (itifollows ))) from our choice of ithat X is almost dense in B.+l((ril., d,tailN(,.M,il.c;,,+,(i))) forewry W' E R(taiL(i)). In particular, this holds if we put w' = u.+l. We then have ii E B.+I((~'~., w', tailN(d,dm(;,,+m(i))). and so X is almost dense in Bn+l(i). ) n > kin o, B,(u')nX # 0. We have thus shown that. forevery ii E ~ k ( iandevery Since X is closed, this implies that ii E X . Thus & ( i ) c X, a conhadiction. 0

5..

Lemma 18-42. Let ( Fn)Fa be an incremingsequence of c l o s c d ~ ~ h cdnue r e subsets of 4 and let N = F. Then N is nowhere dense in 4.

Uz4

E o, let T" = ( i E d : En(?) r l F . = 0). Sime each F. is completely Ramsey (by Lmuna 18.41) and nowheredense (T").E, satisfies condition (a) of (*). It clearly satisfies condition (b) of (4. To sct that N is nowhere dense in 4, suppose instead that we have some s' in the interior of cl N. Pick k E o such that Bk(3 g cL N. By Lemma 18.36, there exists i E & ( a such that, for every rn > k in o and every variable reduction iof.,6 we hive ,;( tailm(<)) E ~ ' ~ 1 . Then t E cL N so pick u' E & ( a r l N and pick n w k such that u' € F. Pick m > k such that u',. is a variable reduction of Thm ii E B.((iiI., tail,&)) and (u',,,,tail,(i)) E T" so ii F., a contradiction. 0

P m J Pot each n

e

4,.

CoroUmy 18.43. I f N is a meager subset of4. then N is complctrlyR m u 9 .

P m J We have that N is the union of an increasing sequeace (X.)r4 of nowhere dense subsets of 8 . Let X = c t Then X is nowhere dense, by Lemma 18.42. and X is wmpletely Ramsey, by Lemma 18.41. It follows that. for each E d and n E o, there exists if B. (3for which B. (i) n X = 0. Since N X, this implies that B.(i) n N = 0. o

(Uz4q.

Theorem 18.44 (Carlsw's Theorem). A subset of d is completely Rams9 ifand only if if is Bairr.

P m J By 'lheorem 18.30, every completely Ramsey subset of d is Bairc. Now assume that X is aBaire set and pick an open set U and a meager set M such that X = U A M . By Lemmas 18.41 and 18.31. U is completely Ramsey and by Corollary 18.43. M

392

E

18 Multidimensional RPmMy Thcoq

is completely Ramsey. Applying Lemma 18.31 again, one has that X is completely m y . 0 As an amusing c m m of Theorems 18.30 and 18.44 one oees that in 4 , every Bake set is the union of an open set and anowhcre dense set Notice that one has an immediate partition corollary to Csrlson's Theorem.

u=,

Corollnr). 18-45. k t r E Nand a r s m that 4 = Xi. Ifeach X iis a Baim set, then for every n E OJ mtd every E 4. there exist some i E (I. 2. . r ] and some i E B. ( 3 such that ~ . ( i )g xi.

a

. ..

...

Prwf: Let n G o and 3 E d and suppose that for each i E (1.2.. r ) and each i E Bn(3 one does not have &(i) 2 Xi. Pick i1E B. ( 3 as guaranteed by the fact thatX1 iscompletely Ramsey. ( S o B , ( i l ) n X ~= 0.) Fori E (1,2,. . ..r-l),assume that iihas been chosen and pick Fi+' E B.(ii) as guaranteed by the fact that Xi+l is completely Ramsey. (So B,,(ii+') r l Xi+, = 0.) Then B.(T') s 0, aconfradiction. 0 An extensive m y of theorems in Ramsey Theory an a consequence of Carlson's Theorem. For example. Ellentuck's Theorem [82] is the special case of Cadson's Theorem which has the alphabet rlr = 0. We have already seen in Exercise 18.3.1 that the Finite Sums Theorem is derivable horn the "main lemma" to Carlson's Theorem. To illusmte a m m typical application, we shall show how the Milliken-Taylor Theorem (Theorem 18.7) is dexivable from Carlson's Theorem. As is common in such applications. one only uses Ule fact that open sets are completely Ramsey.

Corollary 18.46 (Miliken-Taylor Theonm). k t k, r E N, lct(xn)z1 be a sequence in N,and assume that [N]' = U ,! Ai. Then them exist i E (1.2, ,r ) and a sum

...

subsysfm {y,)Z, of Cx.)zl

such that

[ ~ s ( ( y . ) ~ ~G)Ai. l~

P m f Let '# = (a) and define the function f : W ( W u) + Nby letting f ( w ) be the number of occumnces of u in w . Foreachi E (1,2...., r]let Xi = (jE d : (f(s1). f h ) ,

and let

..., f h ) )

E

Ail

Xr+1 = Q E d : I(f ( a ) , f (sz), .. ., f (sk))I < kl.

+

1) and i E Xi, one has that Then 8 = u~L;xi. Now, given i E (1.2,. .. ,r B&) 2 Xi so Xi is open. For each n E N. let s, = us". so that f (s.) = x.. Pick by Corollary 18.45 some i E (1.2,. . . r 1) and some i E B& such that Bo(i) E Xi. Since Bo(i) C Xi. i # r 1. For each n E N, let yn = f (fa). Then (y.)Z, is a sum subsystem of m (Xn).,l.

+

.

+

.

To see.that F S ( ( y n ) Z l ) l :C Ai, let H I .Hz.. .. Hi E PfPf be given such that man Hi < min H,+l for each j E (1.2,. ..,k - 1). For each j E [I. 2, . k). let C j = man Hi and let lo = 0. Foreachn E (1.2,. . ,l k ) , let

.. .

.

a ifn f lJf=l b m = [ v if n E ~

Hi

f H;= ~

For j > k, let wj = tr,+j-r Then G E ~ o ( i )so If ( w d , f (w2). Since

...,f (wk)) E Ai.

Notes The ultralilter proof of Ramsey's Theorem (Theorem 18.2) is by now classical. See [67, p. 391 for a discussion of its origins. Version 1 of the Milliken-Taylor Theorem is essentially UK version proved by K. Milliken (1831 while Version 2 is that pmved by A. Taylor [233]. The rest of the results of Section 18.1 are from 1261and were obtained in collaboration with V. Bergelson. The results of Section 18.2 are from [32], a result of collaboration with V. Bergelson. Theorem 18.23 and Comllary 18.24 are from [201, a result of collaboration with V. Bergelson and A. Blass. The pan of Comllary 18.24 that corresponds to W ( 8 ) is 128. Corollary 3.71, a result of collaboration with V. Bergelson. It is a modification of a result of T. Carlson and S. Simpson (59. Theorem 6.31. The pan of Corollary 18.24 that refers to W ( 8 ;v ) is due to T. Carlson. Carlson's Theorem is of course due to T. Carlson and is from [58]. The Ellentuck topology on the set of sequences of variable words was intmduced by T. Carlson and S. Simpsonin 159. Section 61. It is analagous to a topology on the set of infinite subsets of o which was introduced by E. Ellentuck in 1821. Given a finite alphabet O , call L an infinite d i m e n s i o ~subspace l of W ( 8 )if and ~ ~( 8 ;v ) such that L is the set of all constant only if there is a sequence ( S , , ( I J )in) W = ~W ( 8 ;v ) . reduced wordsof ( s . ( v ) ) z l .Similarly. given a finitesequence ( ~ . ( v ) ) i in call L a d-dimensional subspoce of W ( 8 )if and only if

L = Iw1(a1)-~2(a2)--..-wd(ad):a1.a~.

....ad E W).

It is a result of H. Furstenberg and Y. Katznelson (99. Theorem 3.11 that whenever the collection of all d-dimensional subspaces of W ( 9 ) are partitioned into finitely

many pieces, there exism aa infinite dimensional subspace of W(W) PU of whose ddimensional subspaces lie in ths samecell of the partition. This result can be established by methods similar to (hose of Section 18.3. (See [201.)

Part IV

Connections With Other Structures

Chapter 19

Relations With Topological Dynamics

We have alnady seen that the notions of syndetic andpiecewise syndetic, which have their origins in topological dynamics, are important in the theory of 6.5 for a discrete semigroup S. We have also remarked that the notion of central, which is very important in the theory of 5S and has a very simple algebraic definition, originated in topological dynamics. In this chapter we investigate additional relations between these theories. In particular, we establish the equivalence of the algebraic and dynamical definitions of "central".

19.1 Minimal Dynamical Systems The most fundamental notion in the study of topological dynamics is that of dynamical system. We remind the readex that we take all hypothesized topological spaces m be Hausdorff.

Definition 19.1. A dynonrical system is a pair (X. (T,),,s) such that (1) X is a compact topological space (called the phase space of the system); (2) S is a semigroup; (3) for each s E S, 1,is a continuous fhction from X to X; and (4) for all s . t E S. T, o T, = TI,.

If ( X , (T,),,r) is a dynamical system. one says that the semigroup S acts on X via (TS)SES. We observe that we an indeed familiar with certain dynamical systems

Remark 192. Let S be a disc& semigmup. Then (6s. (A$)SE~) is a dynamic01 system Definition 193. Let (X.(T,)& be a dynamical system. A subset Y of X is invariant if and only if for evey s E S,T,[Y] & Y .

398

19 Rehtioas W~tbTopological Dynamics

Lemma 19.4. Lct S be a semigmup. l%eclosed i& subsets of the dynamical arepncisely the cldsed lep idcab of BS. system (BS. (l.),,~)

P m J Given a leti ideal L of pS, p E L, and s E S. one has A,(p) = s p ideals an hvarianL Given a closed invariant sabset Y of BS and p cCY = Y .

E

E

L sa leti

Y m e has that ( B S ) p = c l ( S p ) c 0

De6nition 195. The dynamical system (X.(T,)*Es)is minimal if and only if there are no nonempty closed proper invariant subsets of X. A simple application of Zom's Lemma shows that given any dynamical system (X.(T,),,s), X contains a minimal nonempty closed invariant subset Y, and conse-

quently (Y, (T&,s) is a minimal dynamical system, where Ti is the restriction of T, to Y .

Lunms 19.6. Let S be a semigmup. The minimal closed invariant subsets of the d y ~ m i c asystem l (BS, ( A s ) , E ~a)n pncisely the minimal lqY ideals of BS.

Pm$ This is an immediate consequence of Lemma 19.4 and the fact that minimal left ideals of BS are closed (Comllary 2.6). Of course. a given aemigroup can act on many differcllt topological spaces. For example, if X is any topological space, f is any continuous fuaction from X to X. and for each n E N. T. = f", then (X,(T.).,N) is a dynamical system. And since my dynamical system contains a minimal dynamical system. a given semigroup may act minimally on many different spaces.

De6nilion 19.7. Let S be a semigroup. l%en (X.(T,),,s) is a wrivemal minimal dynamical systemfor S if and only if (X,(T,),.s) is a minimal dynamical system and. whenevcr (Y, (R.).,s) is a minimal dynamical system. there is a continuous function cp from X onto Y such that for each s E S one has that R, o (p = (p o T,. (That is, given any s E S. the diagram

X

-x

Theorem 19.8 Let S be a semigmup. let L be a minimal 1qY ideal of BS, and for each s E S, let A: be the nsm'ction of A, to L. Then (L. (I;),,s) u a universal minimal dyMmica1 system for S.

I I

PnmF Let (Y,(R,),.s) beaminimaldynarnicalsystemandfix y E Y. D e f i m g :S -c Y by g(s) = R,@) and let (p = &. Then immediately we have that p is a coatinuous functionfimmLtoY. Tocompktetheproofits~cerrtoshowthatforallsE Sand d l p E L . (p(A:(p)) = R&P)). (For then R , [ ~ L IG] v [ L I so B[LI is invariant and thus by minimality q [ L ] = Y .) To see that (p(li:(p)) = R,((p(p)) for all s E S and all p E L , it suffices to show that for each s E S. 5 o A, = R, o on pS for which it in turn suffices to show that ~ o A , = R , o ~ o n S . S o l e t t €Sbcgiven.Then

We now wish to show that (L, (A:JSPs) is the unique univeml minimal dynsmical system for S. For this we need the following lemma which is interesting in its own right.

Ltmm 199. Let S k a s m i g m u p and let L ond L' be minimal I@ &ah ofpS. I f ( p is a continuousfunctionfmm L to L' such that As o (p = (p o A: for all s E S (when A: is the restriction of As to LJ, then then is some p E L' such that (p is the restriction of pp to L.

P m f Pick by Corollary 2.6 some idempotent q E L and let p = p(q). Then pp and to L'. Further, given any s E S,

(p o 4 are continuous functions from pS

and thus pp and (p o pq are continuous functions agrreing on S and are thmfore 6qual. In pamcular. p p l ~= ((p o pq),& Now. by Lemma 1.30 and Theorem 159.q is a right o L required. identity for L, so ((p o p q ) l ~= (p and thus (p = P ~ I as The following theoremis inmany respects similarto Rcmark3.26which asserted the uniqueness of the ~tone-&chcom~actification.However, unlike that remark,the proof of Theorem 19.10 is not trivial. That is, given another universal minimal dynamical and givens E S, one obtains the following diagram system (X,

19 Relations W~thTopological Dynamia

400

and would like to use it to show that (p is one to me. However, if for erample S is a left so thal f3S is also a left zero semigroup. then A,(p) = A,(q) for all p and q in pS, and the above diagram is of no use in showing that cp is one to one.

zm, semigroup.

Theorem 19.10. Let S be a sernigmup und kt L be a minimal lej? ideal of 6s. 'Ihm, up to a homeomorphism respecting the action ofS. (L. (A:),,s) is the unique universal minimal dynarnicalsystemfor S. Thnt is, givenmy w i v e r y ~ l m i n i m a l d y n a m i c a l ~ (X. (TS)JE~) for Sthere is a homeomorphismv : L -+ X such rhnt (p o A: = T, o (pfor everys E S.

PmoJ Pick (p : L + X as guaranteed by the fact that (L, (Ai),,s) is a universal minimal dynamical system for S and pick r : X + L as guaranteed by the fact that (X, (T,),GS) is a universal minimal dynamical system for S. Then o m has immediately that (p o A; = T, o (p for every s E S and that (p is a continuous function on a compact space to a Hausdorff space and is thus closed. Therefore, it suffices to show that (p is one to one. Now T o (p : L + L and for each s E S, A,o(ro(p)=A~o(ro(p)=(ro(p)oA~

so by Lemma 19.9 there is some p E L such that r o (p is the restriction of pp to L. Since, by Themem 2.1 I@), the restriction of pp to L is one to one, we are done.

19.2 Enveloping Semigroups We saw in T h a m m 2.29 that if X is a topological space and 7is a semigroup contained in X ' (with the product topology) and each member of 7 is continuous, then the closure ' is a semigroupunder o, called the enveloping semigmup of 7.In panicular of 7 in X (and this is the origin of the notion of enveloping semigroup). if (X. (T,JXss) is a dynamical system, then the closure of (T, : s c S ) in ' X is a semigroup which is referred to as the enveloping semigroup of tbc dynamical system.

Theorem 19.11. Let (X, (T,),,s) be a dynamical system and defrne q : S -+ X ' by ~ ( s = ) T,. Then is a continuous homomorphismfrom f3S onto the enveloping semigmup of (X. ('),,s).

PmoJ Immediately one has that ?is continuous and that WBS] = d(T,: s E S). Also by Theorem 2.2. each T, is in the topological center of 'X so by Corollary 4.22. p is a 0 homomorphism. The following notation will be convenient in the next section.

Definition 19.12. Let (X, (T,),,r) be a dynamical system and define (p : S + X ' by (p(s) = T,. Foreach p E BS. let Tp = ?(p).

'

19.2 Enveloping Semigroup

401

As an inunediate consequence of Thwrem 19.11 we have the following. Be cautioned however that Tp is usually not continuous, so ( X . ( T p ) p E Bis~ usually ) not a dynamical system.

Remark 19.13. Let ( X . (T,).,s) k a dynamical sysfem and let p, q E pS. Then TPo Tq = TM and for each x E X. T p ( x )= p - l i T,(x). rss

How close the enveloping semigroup comes to being a copy of pS can be viewed s S on X. W~thcertain weak as a measure of the complexity of the action ( T ~ ) , Eof cancellation requirements on S, we see that t h m is one dynamical system for which pS is guaranteed to be the enveloping semigroup.

Lemma 19.14. Lst Q hasemigroup. lerSkasubsemigroupof Q, andlet0 = Q(0,I) with the pmduct topology. For each s E S, d&e T, : R + R by T,(f ) = f o p,. Then ( R , (T,),,s) is a d y ~ m i c asysrem l Funhermom, for each p € @S,f E R. and s c S, ( T p ( f ) ) ( s ) = 1 0 It E S : f ( t ) = 1) s s p . P m f : Let s E S. To see that T, is continuous it is enough to show that n, o T, is continuous for each t E Q. So, let t E Q and let f E R . Then

That is to say that R, o TS = n,, and is therefore continuous. Now lets. t E S. To see that T, o T, = T,,, let f E R. Then

Rnally, let p E pS, f E R and s E S. Then

Notice that by Lemma 8.1, the hypotheses of the following theorem hold whemver S has any left cancelable element. In particular. they hold in the imponant cases in which S is W or Z.

Theorem 19.15. Let S be a semigroup, let R = S(O, 1)- and for each s E S, drfrne T, : R -+ R by T,(f ) = f o p,. If for every pair of distinct elements p and q of BS t h e n is some s c S such that sp # sq, then f3S is topologically and algebraically isomorphic to the enveloping semigmup of ( R . ( T , ) d .

402

19 Relations WI&Topological Dynamics

Pmof: By Lemma 19.14. (C1. ( T s ) I ~is~a dynamical ) system Define p : S + *C1 by (PO)= T,. ?hen by Th19.1 1, is a continuous homomorphism from BS onto the enwloping semigroup of (Sa, ( Z ) r ~Thus ~ ) .we mtd only show that is one lo one. So let p and q be distinct membus of BS and picks e S such that s p # sq. Pick A E sp\sq and let X A be the charaaeristic h o d o n of A. 'Iben by Lemma 19.14. ( T P ( X d ) ( s )= 1 and (T,(xA))(s) = 0. k f n e = Tp # Tq = Uq).

a

a@)

We have seen that if L is a closed left ideal of BS and A: is the restriction of A, to L, thm ( L , (A:),,s) is a dynamical system. We know fuaher (by 'Ihe0mn.s 19.8 and 19.10) that if L is a minimal left ideal of BS, then ( L , (A:),E~) is theuniversal minimal dynamical system for S. It is a nanual question as to whether BS can be the enveloping semigroup of ( L . (A:),Es). Lemma 19.16. L ~SYbe a semigmup andlet L be a closed lcfr ideal of BS andforeach s E S, k t A: be the restriaion of A, to L. D&e (p : S + L~ by q ( s ) = A:. Thenfor ) A~IL. d l P E BS. 6 % ~ =

-

Pmof: Let p E BS. let q E L. and let Tp F(p). nKTL b y Remark 19.13. a(p)(q) = 0 Tp(q) = p- rcs lim sq = pq. Notice that if L contains any element which is right cancelable in BS then the hypotheses of the following theorem are satisfied. (SeeChapter 8 for characterizations of &ht cancclability.) If S is a countable semigroup which can be embedded in a group and if L g K(BS), thm L does contain an element which is right cancelable in js (by Theorem 6.56). So, in this case, the enveloping semigroup of (L.(A,)& is topologically and algebraicallyisomoq-hic to BS.

Theorem 19.17. Let S be a semigmup. let L be a closed I@ ideal of BS, andfor each s E S, let At be the restriction of A, to L. Thm BS is topologically and algcbmicully isomarphic to the enveloping semigmup of ( L , (A,),.s) via a map which takes s to A: ifand only i f whenever q and r a n distinct elements of BS. there is some p E L such that q p Z rp.

a

Proof: Define p :S + L~ by q ( s ) = A:. By 'ihamm 19.11 is a continuous homomorphism onto the enveloping semigroup of ( L , (AS),es) (and is the only continuous 0 function extending q ) . The conclusion now follows from Lmuna 19.16.

I f L is a minimal left ideal of BS, we have the following superficially weaker condition. T b m m 19.19. Let S be a semigmup, let L be a minimal I@ idealof BS, andfor each s E S. let A: be the restricrion of A, to L. Thm BS is topologicalIy and dgebmically isomorphic to the enveloping semigmup of (L. (A,),d) via a map which takes s to A: ifand only ifwhenever q and r are distinct elements cfBS. there is some p E K ( B S ) such that q p # rp.

PmoJ Since L E K(BS). the necessity follows immediately from l'h19.17. For the sufficiency. I n q and r be disfinct members of #S and pick some p E K (PS) such that q p # rp. Pick by Theorem 2.8 some minimal left ideal L' of PS such tbat p E L' and pick same z E L. By Theorem 2.11(c). the restriction of p, to L' is one to one and 0 q p and r p are in L' so qpz # rpz. S i e pz E L. Theorem 19.17 applies. We conclude this section by showingmat PA is theenveloping semigroup of certain natural actions of N on the circle group T. which we take here to be the quotient W/Z

under addition.

T h e o m 19.19. Let a E [2,3,4,. ..) andfor each n E N d q h e T, : T -+ T by T.(Z + x) = Z + amx.Then (T,(Tn).,n) is a dynamical sptm 4 9 :W + T ' is defined by(p(n) = T,. then F i s a continuousisomorphismfmm PW onto rhe enveloping Semkmu~of (T.(Tn)n~Nh

P m J Since a E N. the functions T. are well defined. Let n E N, let l :B + B be multiplication by a'. and let n : R + +be the projection map. Then T. o n = n o t so T,, is continuous. Trivially, if n.m E N, them T . o T, = T,+,. Thus fl.(Tn).,n) is a dyaamical system. 19.11, to show that 5 is one to one. To complete the proof it soffices,by TbTo this end, let p and q be distinct members of PN. Pick B E ( W .2N 1) such that B E p and pick A E p\q. Define x E (0.1) by

-

x = Z F l aja-j

where

1 0

ifj-l€AnB ifj-14AnB.

Let

+

Notice that (since a2 2 c a'). C flD = 0. We claim that F(p)(x) E C and ?(q)(x) E D. To see this it suffices to show. since C and D are closed, that for each n~N,ifn~AnB.thenT.(x)~C.andifngAnB,thenT.(x)eD. So let n E W be given. Then

since a,an-1 E 2 if j 5 n. Now assume that n E A n B. Then am+'= 1and (because Elm=,+,ajan-1 so n 1 $ B) a.+z = 0. Thus Zg,+, ajan-j = a-'

+

and hence Tn(x) E C as claimed.

+

404

I9 Relations W~thTopological Dynamics

Next sJsume that n $ A I 7 B. Ihen u,,+I = 0 and either a.+z = 0 or an+, = 0. Conxquently m 1 2 0 5 Xja+l ajan-I 5 - a2 a4 and hence T.(x) E D as claimed. a

+

Jf one views the circle group as (z E C : lzl = 1). thm Theorem 19.19 says is isomorphic to BBW, where that the emeloping s e m i p u p of the system (T. (f "),,N) f(z) =za.

19.3 Dynamically Central Sets I h e term "cenhal set" was originally defined by Furstenberg [98. Definition 8.31 for subsets of W. Sets defined algebraicallyas elementsof minimal idempotentswerecalled "cennal" because many of the same theorems could be proved about them. The original definition (restricted to the semigroup Wand applied only to metric dynmical systems) is as follows. We add the modifier "dynamically" because it is by no means obvious that the concepts are the same, even for subsets of N.

Ddinition 1930. Let S be asemigroup. A set C 5 S is dy~micallycentml if and only if there exist adynamical system (X. (T,),,s), pointsx and y in X. and a neighborhood U of y such that (I) y is a u n i f d y m n t point of X, (2) x and y are proximal, and (3) c = { s ~ S : T , ( x ) E U]. W e terms ' b i f o d y recurrent" and 'pmnimal" are notions from topological dynamics. defined as follows.

Definition 19.21. Let (X, ( T , ) , E ~be) a dynamical system. (a) A point y E X is unifbrmly recurrent if and only if for every neighborhood U of y. [s E S : TJy) E U )issyndetic. (b)Points x and y of X are proximal if and only if there is a net (s,),,! in S such that the nus (T,(x)),,r and (T,,(Y)),,I converge to the same point of X. The definition of "proximal" as standardly given for a dynamical system with a menic phase space (X.d) says that there is some sequence (s.)gl such that lim d(T,(x). T,(y)) = 0. It is easy to see that this definition is equivalent to the n-m one given above. The characterizationof proximality provided by the following lemma will be very convenient for us.

19.3 Dynamically Central Sets

405

Lunma 1 9 3 2 Let (X, (T,),,s) be a dynamical system and let x, y E S. Then x and y arepmximal ifand only ifthere exists a point p E BS such that Tp(x) = Tp(y). Pmof: Notice that if p E BS, (s,),,~is a net in S which amverges to p, and r E X, then Tp(z) = L i r n , ~Z,(z). ~ Thus if p E BS andTp(x) = Tp(y). then picking any net (s~)(EI in S which converges to p, one has Iim,,~T,,(x) = l h , r T,, (y). Conversely, assume that one has a net (&I in S such that lim,,~T,,(x) = lim,,! Z,(y). Let p be any limit point of (s,),,~ in BS. By passing to a subnet we may prsume that (s,),,~ converges to p. Then T,(x) = lim,,l T,<(x) = lim,,! T,,(y) = TP(Y). o

In order to establish the equivalence of the notions of "eenhal" we investigate relationships between some algebraic and dynamical notions. Theorem 19.23. Lct (X. (T,),,s) be a dynamical system. let L be a minimal leji ideal of BS, and let x E X. Thefollowing statements a n equivalent. (a) Thepointx is a wtifomly recumntpoint of (X. (T$),,s). L such rluu T&) = x. (c) There exists an idempotenr e E L such thar TJx) = x. (d) There mists y E X and an idempotent e E L such that T.(y) = x.

(b) There ex&s u E

Pmof: (a) implies (b). Choose any v E L. Let # be the set of neighborhoodsof x in X. For each U E X let BU = (s E S : T,(x) E U].Sineex is uniformly recwrent, each Bu is syndetic and so there exists a finite set Fu S such that S = Ur,F,,(t-l Bu). Let u be a limitpoint of the net Pick to E Fu such that ru-'Bu E v so that tuv E ( r u v ) u , ~in BS and notice that u E L. We claim that T,(x) = x. By Remark 19.13, it suffices to show that Bv E u for each V E X . So let V E X and suppose that Bv P u. Pick U E V such that tuv E S\BV. Then ruv E iT; g 5, a contradiction. (b) implies (c). Since u belongs to a group contained in L (by Theorem 2.8). there is an idempotent e E L for which eu = u. So (using Remark 19.13) T,(x) = T,(T. (x)) = Te.(x) = Tu(x) = x . That (c) implies (d) is trivial. (d) implies (a). We note that T,(x) = T,(T,(y)) = T,(y) = T,(y) = x. Let U be a neighborhood of x. We need to show that [s E S : T,(x) E U )is syndctic. Pick a neighborhood V of x such that ct V E U and let A = [s E S : TS(x)z V). Since x = e-limT,(x) by Remark 19.13. A E e. Let B = [s E S : s e E A). Then by

K.

reS

Theorem 4.39, B is syndetic. We claim that B g [s E S : T,(x) E U].Indeed. if S E B.thenTs(x)=T,(Te(x))=T,.(x)~(T,(x):t~A)E~EU. 0 Theorem 1934. Let (X, ( T , ) I ~ ~ be) a dynamical system and let x E X. ntcn there is a uniformly recumntpoint y E ce(TS(x): s E S) such that x and y are pmximal. Pmof: Let L be any minimal left ideal of BS and pick an idempotent u E L. Let y = T,(x). Then trivially y E cC[Ts(x) : s E S). By Theorem 19.23 y is a uniformly

406

19 Relations With Topological Dyanmig

recumnt point of (X. ( T s ) r ~ By ~ ) Remark . 19.13 we hpvetbat T.(y) = T.(T.(x)) T..(x) = T.(x) so by Lemma 19.22 xand y are proximal. Theorem 19.25 k-t (X. ( T h s )be a &-cd

= 0

system and letx, y E X. I f x and y

Pmof: By Lemma 19.22, ( p E @S : T p ( x )= TpQ)l # 0. It is a left ideal in BS, bsause, for every p. q E @S, Tp(x) = Tp(y)implies that T,,(x) = T 4 ( ~ & ) ) = T&(Y))

=WY).

Theorem 19.26. k-r (X. (T,),,s) be a dymmical symm and lef x. y E X. Then is a minimal idempotmt u in 8s such that T&) = y ifand only i f x and y are proximal and y is uniformly r e c u m .

Proof. Necessity. Since u is minimal. there is a minimal left ideal L of p S such that u E L. Thus by Theorem 19.23 y is uniformly recumnt. By Remark 19.13 T.Q) = T.(T.(x)) = T..(x) = T.(x) so by Lemma 19.22 x and y arc proximal. Sufficiency. Pick by Theorem 19.25 aminimal left ideal L of @Ssuch that T.(x) = T.(y) forall u E L. Pickby Theorem 19.23 an idempotentu E L such that Tv(y) = y.0 We are now prepared to establish the equivalence of the notions of central. Theorem 19.27. Let S be a semigroup and let B g S. Then B is central ifand only if B is dynamically cenmI.

P m j Necessity. Let Q = S U (e) w b e is a mw identity adjoined to S (even if S already has an identity). Let R = Q(0.1) and for s E S define T, : Q -P Q by T,( f ) = f o p,. Then by Lemma 19.14, (Q, (T,),,s) is a dynamical system. Let x = X g, the chancteristic function of B E Q. Pick a minimal idempoteat u in BS such that B E u and let y = T&). Then by Themem 19.26 y is uniformly recurrent and x and y an proximal. Now let U = ( z E R :z(e) = y(e)). Then U is a neighborhood of y in Q. We note that y(e) = 1. Indeed, y = T d x ) so ( s E S : T,(x) E U )E u so choose some s E B such that T,(x) E U.Then y(e) = T,(x)(e) = x(es) = 1. Thus given any s E S, s e B e x(s)=l 9 Ts(x)(e)= l 9 T,(x)EU. Suf6ciency. Choose a dynamical system (X. (T,),,s), points x, y E X. and a neighborhood U of y such that x and y are pmximal, y is uniformly recurrent, and B = (s E S : T&) E U).Choose by Theorem 19.26 a minimal idempotent u in p S 0 such that T&) = y. Then B E u.

It is not obvious from the definition that the notion of "dynamically central" is c1oscd undet passage to supersets. (If B = Is G S : TJx) E U )and B g C. one can let V = U U (T,(x) : s E C J .But there is no reason to believe that ( s E S :T,(x) E V ) g C.)

407

19.4 Dynamically Gcnmad iP.Sm

CoroUary 1928. Let S be a semigroup and let B G C G S. If B b dynamically cemal, then C is d y ~ ~ ' c a lcentml. ly Pmof. This follows h m Theorem 19.27 and the fact thaI supersets of central sets are central. 0 Exe193.1. Let (X. (T,),,s) be a dyMmical system and let x E X. The point x E X is periodic if and only if there is some s E S which is not an identity such that T,(x) = x. The point x E X is recurrent if and only if every neighborhood of x contains a point of the form T&) for some s E S other than an identity. Assume that x is not a periodic point of X and that S* is a subsemigmup of BS. (See Theorem 4.28.) Rove that the following statements are equivalent: (i) (ii) (iii) (iv)

x is recurrent. T&) = x for some p E S. T&) = x for some idempotent e E S. Given amighborbood U of x, there exists a sequrnce,:)s,( of S such that T,(x) E U forevery s E F P ( ( h ) z l ) .

of distinct elements

19.4 Dynamically Generated IP* Sets In this section we consider certain dynamically defined sets that are always IP9Ws. In conhast with the results of Section 19.3, we see that the notion of aUdynamicalIP* set" is not equivalent to that of "IP*set", but is rather significantly stronger. The dynamical notion with which we shall be concerned in this secIion is that of a measure preserving system. De6nition 1929. (a) A measure space is a triple (X. 91. p). w h m X is a set, 51 is a a-algebra of subsets of X, and p is a countably additive measwe on 33 with p(X) finite. (b) G i v e n a m u r e space (X, S,p). afunction T :X + X isamcamrepreserving tmnsformorion if and only if for all B E 91, T-'[B] € 33 a n d p ( ~ - ' [ ~ ]= ) p(B). (c) Given a semigroup S and a measure space (X, 33, p). a measurz preserving action of S on X is an indexed family (T,),.s such that each T, is a measure preserving transformation of X and T, o T, = T,, for all s , t E S. It is also required that if S has an identity e , then T. is the identity function on X. (d) A measurz preserving system is a quadruple (X,91,p, (T,lSGs) such that (X, d.p) is a measure space and (Ts),E~is a measure p m e d n g action of S on X.

We shall med a preliminary result about the semigroup (?,(N),

U).

Definition 1930. Let (X, 51, p) be a measure space. A monotone action of Pf (N)on X is an indexed family ( T ~ ) ~ ~ p such , ( r qthat

19 Relations Wlth Topologicd Dynamics

408

(a) each TF is a measure preservingtransformation of X and (b) if F, K E P f O and max F c min K, then TF o TK = TFux. Notice that, because of the "max F F min K" requirement, a monotone action of Pf 0 on X need not be a measure preserving action of Pf(N) on X.

Lmuna 1931. Lct ( T F ) F E P ,bC ~ a monolone M i o n of .!Pflf on the measure space (X,91. @) ~d let A E 8 with p(A) > 0. Thenforeach k E Nthere exists H E Pf(N) with min H s k such that p(A n T ~ - ' [ A ] )s 0.

.

E N such that p(A) > p(X)/m. F a n E [ I ,2.. . ,m), let F. = [k n , k n 1.. ,k m]. Then each ~ ( T F , - ' [ A ] )= p(A) s p ( X ) / m so pickn.! E [1,2, ....m ) such that n < e a n d p ( ~ ~ m - ' [ A l n T ~ t - ' [ r A I0.) Let H = {k+n.k+n+l, ....k+t-1). ThenTnoTFt = T F ~ s o T F ~ - I [ =AT] ~ ~ - ' [ T H - ' [ A ] ]

P m J Pick m

+

+ + .. +

so ~ ( T H - ' [ An ] A) = ~ ( T F ~ - ' [ T H - ' [ An] A ] )

= IL(TF~-'[TH-'[AI]n T ~ - ' [ A I ) = ~ ( T F ~ - ' [n AT ] ~ - ' [ A ]> ) 0.

0

Lemma 1932. Let (X,8 ,p, (T,),,s) be a measure presrmMn8 system ~d let A E 91 satisfip(A) > 0. Forevery sequence (s,,)gl andeveryk E El. thereexists F E .Pf (N) such that Ir < min F and p(A n T;'[Al) > 0,where s = n m € Fs,,. P m f For F E Pf(IV), let RF = ThEFs,,.Then (RF),~&,'I(Pois a monotone action of Pf(N) on X. So we can choose a set F with the required properties by Lemma 19.31. 0

c

Theorem 1933. Let(X, 8,p, (Tr)rt~)bCame~represe~~ingsyslemandIet C S. Ifthere is some A E S with @(A) s 0 such that 1s E S : p ( A n T ; I [ A ] ) > 0 ) C C. then C is an I P set. Pmof: This follows immediately from Lemma 19.32.

0

Deanition 1934. Let S be a semigroup. A subset C of S is a d y m i c a l I P set if and only if there exist a measure preserving system (X. 21, p, (T,),,s) and an A E 21 with p ( ~ >) 0 such that [s E s : p ( n~ T;'[A]) > 0)E C.

+)

Recall that by Theorem 16.32. there is an IP* set B in (N, such that for each n E N, neither n B nor -n B is an IP* set. Consequently. the following simple result shows that not every P set is a dynamical IP* set.

+

+

Theorem 1935. k t B be a dynamicat I P set in (N, +). T h e n is a dynamical I P set C E B such thatfor each n E C. -n + C is a dynamic01 IP* set (and hence -n B is a dynamical IP* set).

+

19.4 Dynamically GwratedI P . Sets

409

Proof Pick a measure space (X, S , r ) , a measure preserving action (Tn)nEN of N on X,andasetA E Ssuchthatp(A) =-Oandln €N:p(AnTn-'[AI) s o ] s B . I . e t C = (n € N : p(A r l T,-'[A]) > 0). To see that C is as rrquircd, let n E C and let D =A~T.-'[A]. Weclaimthat [m E N : p(DilT,-'[DI) > 01 E -n+C. Tothis end, let m E N such that p(D i l T,-'[D]) r 0. Then DnT;'[D]

=A~T~'[A]~T~'[AI~T~~[A]]

r A r l T;'[T;'[A]]

= A n (T, o T,)-'

[A]

= A n T,&,[A] son+m€C.

.

0

R e d from Theorem 18.15 that an IP* set in M x N need not contain [w] x FS((x,)E,) for any sequenw (x.)zl. In the final result of this section we show that a dynamical IP* rat in the product of two semigroups with identities must contain sets of the form FF'((x,JEl) x FF'((y,,)zl). Indeed, it has a much stronger property: given any sequences (w,):=~ and (z.)El one can choose infinite product subsystems m (xn)n=l ~f(w.):=~ and ( y n ) E l of (z.)El with FP((x.)=,) x lWy.)El.) contained in the given dynamical IP* set. More than this, they can be chosen in a parallel fashion. That is if x,, = w,, then y. = zr. We only discuss the product of two semigroups for simplicity and because the generalization to arbitrary finite products is straightforward. In the proof of Theorem 19.36 we utilize product measure spaces. In doing so we shall use the customary notation. If (X. 8 ,p ) is a measure space and m E N, then (Xm,P, p m )denotes the measure space defined as follows: is the usual Cartesian denotes the a-algebra generated by the sets Xi='Ai where Ai E S for product, Sm each i E (1.2.. . . m], and pmdenotes the countably additive measure on 41'" such that p m ( ~ ~ = = l ~n:=.=, i ) p(Ai) for every sequence (Ai)rCl in d8. Any transformation T : X -+ X defines a transformation T : Xm + Xm for which T(xl,x2, .. ..x,) = ( T ( x l ) , T(xl), . . . , T(x,)). If T is measure preserving, so is T, because T)-'[X~=~AA =~X] ~ = ~ ( T - ' [ A ~for ] ) every Al.Az A, G X .

n,,~,

n,,~,

.

n,

.....

n,

n.

(n,

Theorem 1936. Lct SI and S2 be semigmups with identities and lei C be a dynamicd IF" set in S1 x S2. Lcr (w.):=~ be a sequence in SI and let (z.)?=~ be a sequence in S2. There exists a sequence ( H , ) z l in P,(W) such that (a) for each n, max H. < min H.+I and @) iffor eachn, x. = nreH, w, and y. = &H,

FP((x"):,)

zr, then

r c.

x FF'((Y")~')

Proof: Let S = SI x S2. Pick a measure space (X. 41.p), a measure prcsming action (T(I.l))(I.~)~~,~~Z of S m X, and B E a) with p(B) > 0 such that,

D = ((s. t) E s : P(B

n ~,,,)-I[BI) > 0) 5 C.

410

19 Relations Wlth Toplogid Dynamics

I

!

411

Notu

Notes

.

-

The n d o n of 'dyllamical system'' is often defined only for canpact metric spsas. 'Ihe greater generality that we have chosen (which is essential if one is going to lake flS as the phase space of a dynamical system) is also wmmon in the literature of dynamical systems. General references for topological dynamics include the books by R. Ellis [86] and 1. Auslander [S]. Theorem 19.8 and Lemma 19.9 are due to B. Balcar and E Franek in [I21 as is the proof that we give of Theorem 19.10. Theorem 19.10 is proved by R Enis [86] in the case that S is a p u p . Theorem 19.15 is due to S. Glasner in [IDS], w h m it is stated in the case that S is a countable abelian group. Theorems 19.17 and 19.18 are from [141], a result of collaboration with I. Lawson and A. Lisan. Theorem 19.19 is due to W. Ruppert in [219]. S. Glasner has recently published [I061 a p m f thal the envelophg semigroup of a minimal left ideal in pZ is not topologically and algebraically isomorphic to flZ via a map taking n to A:. As a consequence, the condition of Theorem 19.18 docs not hold in BZ. The equivalence of the notions of "ccnhal" and "'dynamically central" was established in the case in which S is countable and the phase space X is metric in [Dl. aresult of collaboration with V. Bergelson with the assistance of B. Weiss. The equivalence of these notions in the general case is a result of S. Hong-ting and Y. Hong-wei in [164]. Theorem 19.24 is due to 1. Auslander [4] and R. Ellis [851. Most of the results of Section 19.4 an from an early draft of [321, results obtained in collaboration with V.Bergelson. Lemma 19.31,is a modification of standard results about Poinc& recurrence. Theorem 19.33 for the case S = Z is due to H. Furstenberg

WI.

Chapter 20

Density

-Connections with Ergodic Theory

As we have seen, many multr in Ramsey Theory assat that, given a finite partition of some set, one cell of the panitin, must contain a specified kind of shucture. One may ask instead that such a smcture be found in any "large" set. For example. van der Waenlen's Theorem (Corollary 14.3) says that whenever N is divided into Gnitely many classes, one of these contains arbitrarily long arilhmetic progressions. S z e m d i ' s Theorem says that whenever A is a subset of N with positive upper density. A contains arbitrarily long progressions. Szemen%i's original proof of this theorem 12321 was elementary, but very long and complicated. Subsequently. using ergodic theory, H. Furstenberg [97] provided a shorter proof of this result This proof used his "correspondence principle" which can be viewed as a device for translating some problems involving sets of positive density in W into problems involving measure preserving systems, the primary object of study in ergodic theory. We present in Section 20.2 a proof of Furstenberg's CorrespondenaPrinciple using the notion of p-limit and in Section 20.3 a strong density version of the Finite Sums Theorem obtained using the algebraic structure of (BW,+).

20.1 Upper Density and Banach Density We have already dealt with the notion of ordinary upper density of a subset A of N, which was defined by Z(A) = lim sup n-m

IA n(1.2. .... n)l n

Another notion of density that is more useful in the context of ergodic themy is that of B a ~ c densiry h which we define here in a quite general context. Ddinilion 20.1. Let (S,.) be a countable semigroup which h s been ~ enumerafcd as ( s , , ) ~and , let A E S. The right Banoch densify of A is d:(A) = sup(@E B : for all n E

N there exist m 2 n and x € S

20.1 Uppa Density and Banach Dtnsity

If S is commutative, one has d: = d; and we simply write d9(A) for the Banach density of A. Notice that the Banach density of a subset A of S depends on the fixed enumeration of S. For example. in the semigroup (N,+),if N is enumerated as

i.

then d*(2N) = 1, while with respect to the usual enumeration of M, d*(2N) = When working with the semigroup (N, +) we shall always.assume that it has its usual enumeration so that

such that

IAn[x+l,x+2, m

.... x+mH

r a).

Lemma 20.2. Let (S, .) be a countable semigmup which has been enumerated as

+

m (sn),,=, and let A, B G S. Then d:(A U B ) 5 d:(A) d:(B) and d:(A U B) 5 d:(A) d;(B). B S is right cancellative, then d:(S) = 1 and if S is lefr cancellative. then d;(S) = I

+

Pmof: This is Exexcise 20.1.1.

Recall that we have defined A = ( p E fJN : for all A E p. Z(A) r 0). We introduce in a more general context the similar sets defimd in terms of d: and d;. Definition 203. Let (S, .) be a countable semigmup which has been enumerated as (s,,)z,. Then A:(S, -)= ( p E fJS: for all A € p. d:(A) s 0) and A;@, .) = ( p E fJS : for all A E p, d;(A) s 0).

Lemma ZOA. Let (S, .) be a countable semigmup which has been enumerated as (s.)z, andletA g S. 0, then n A;(S. .) # 0. (a) Ifd;(A) (b) .'fd;(A) r 0, then Ti r l A;(S, .) # 0.

x

Proof: We establish (a) only. By Theorem 3.1 1 it suffices to show that i f 3 is a finite nonempty subset of P ( S ) and d : ( U 9)s 0. then for some B E 3.e ( B ) r 0. This follows fmm Lemma 20.2.

414

20 Density

-Conoectims with Ergdc Theory

.

Recall fromlleorems 6.79 and 6.80 that both A andN.\A an leftidealsof (BN. +) and of (BN, .). By way of contrast, we me in thefollowing theorems that S*\A: is far from being a left ideal of 0.9 and S\A; is far fmm being a right ideal of /IS.

Theorem 20.5. Lct (S. .) be a countable right crnueIlativc semigmp which kas been enumerated as ( s n ) g l . Thcn A:(S. .) is a right ideal of BS.

P M One has that A:(S. .) # 0 by Lemma 20.4 and the fact from Lnnma 20.2 that d:(S) s 0. La p E .I. let 9 € PS, and let A E p . q. We need to show that d:(A)~O.LaB=(y€S:y-'AEq).ThenB~pso~(B)~O.Picka~Osuch thatforalln ~ N t h e r e c x i s t mz n a n d x ~ S s u c h t h a It B n

@I.* ....,s

~ .IX I

m To see that d:(A) 2 a , let n E N and pick m > n and x E S such that I B ~ I S I , S ~S .~ I . X I 2 a. LetC = B f l ( s 1 . q s,}.xandpickz E ,r. Y-lA. Then, since S is right cancellative. p,l~ is a one-to-one h c t i o n from C to A f l (st,ST. ... sm) xz so ~ A n ' ' ' ~ s"r ~ - ~ ' ~ ' ' x 2z' a. 0

n,

......

.....

.

.

T h e o m 20.6. Let (S. .) be a countable kfl cancellorive semigmup which has been enumemted as ( s , ) z l . Then A;(& .) i s a I& ideal of BS.

Pmof L* p E A;(S, .), let q E BS, and let A E q . p. We need to show that d;(A) z 0. Now ( y E S : y - l ~E p ) E q so picky E S such that y-IA E p. Then d;@-lA) > 0 so pick a > 0 such that for all n

€

N then exist m s_ n and x E S such

nx " S ~ ' s 2 ' " ' r s m l l z a. Now givcnm andx. sinceSisleftcancellative. m A, is aone-tmne function from Y - I A fix. ($1. sz. . . .,s, ] to A fl yx .(sl,s2, .. s,). and consequently, d; ( A ) 5 a. 0

that ''-IA

..

As a consequence of Thwrcms 20.5 and 20.6, if S is commutative and cancellative. then A*@) is an ideal of BS.

Theorem 20.7. Let (S. .) be a countable right cancellorive semigmup which has been enumerated as ( s n ) g l .For any A S S,i f A is piecewise syndetic, then then is some G E P J ( S ) such thmd:(UlpG 1-'A) = 1. I f ( S . .) = (N. +), then the converse holds. Consequently.

c

Pmof Let A S , assume that A is piecewise syndetic, and pick G E P f ( S ) such that for each F E $',(S) there exists x e S with F x E U,,G t-' A. To see that d:(U,,G ' - ' A ) = 1,letn E Nbegivenandpickx E Ssuch that

.

I

20.1 Uppx Lknsity and B a n d Density

Now assume that (S, .) = (N. +) and we have G E 3f' (N) such that dg(U,,, -t A) = 1. Let F E Pf(M) be given and pick k E Pi such thar F (1.2 ....,k). Weclaimthatthereissomex E Nsuchthat{x+I.x+2 x+k) c (Ute, -1 +A). To see this. pick some m > 2k2 and some y E N such that

+

....,

+

and pick v E N such that vk < m 5 (u 1% noting that v ? 2k. If for some i E (0, I ,...,y-l)onehas(y+ik+l,y+ik+2 .....y+(i+l)kI -t+A. then we arc done, so suppose instead that far each i E (0. I, . ,u 1).

.. -

-

so, sincem > uk,omhasthat (1 &)vk < v(k- l)+k sothat 21.1 v. acontradiction. 0 The final conclusion now follows f m Corollary 4.41.

The following result reflects the interaction of addition and multiplication in ON which was m e d in Chapter 13.

+)

E A*(N, and let q E /3N. To see that q . p E A*(N. +), let A E q p. Then [x E N : x-'A E p ) E q so pick x such that x-'A E p. Then d * ( x - ' ~ ) > 0 (where here, of course. we arc referriog to the additive version of d*) so pick u > 0 such that for all n € N there exist m ? n and y € R such that

Prooj Let p

.

'x-1An(y+1'y+2"..'y+m)' m IAnIxy

?a.Theng*ensuchn,m,andy.ooeharthat

+ I,xy+2. ....xy+xm]I m

so that dW(A)?

?U

-.a X

Recall that, given a subset A of a s e m i p u p (S. .) that we have defined a me T in A whose nodes are functions. Recall further that given a node f of T. the set of successors to f is denoted by 9, and that T is an FP-tree provided that for each node

416

20 Density

-Connectionswith Ergodic 'lheny

f of T , Bf consists of all finite products of'entries on paths extending f which occur after f. (By a path in T we mean a function g : o + A such that for each n E o,the restriction of g to (0.1, ,n - I) is in T.)

...

Theorem 20.9. Let ( S , .) be a countabli c m l l a t i v e $mrigmup which has been enumerated as (s.)zl. Let r E A and ler S = Ai. 77ure &st i E {I, 2,. . ,r ] and a tree T in Ai such thatforeachpath g in T , EF'((g(n))>) S Ai, andforeach node f ofT.d:(Bf) > Oandd,'(Bf) > 0.

.

Pmof: By Theorems 205 and 20.6, -) is a right ideal of 5 S and A;(S, -) is a left ideal of 5s. By Corollary 2.6 and Theorem 2.7 pick an idempotent p E A:(& .)n A;(S. .)and pick i E {I. 2. . ..,r ) such that Ai E p. Then by Lemma 14.24 thereisan P - k c T in Ai such that for each f € T. Bf E p. Thus, in particular, for each f E T , dF(B,) > 0 and d i ( B f ) > 0. 0 The following theorem is not a corollary to Theorem 20.9 because d * ( A ) > 0 does not imply that Z(A) > 0. The proof is essentially the same, however, so we leave that proof as an exercise. Theorem 20.10 is an interesting example of the strength of combinatorid results obtainable usingidempotentsin~N.Althoughmany of theRamscy-theoretic results which we have presented were first obtained by elementiuy methods or have subsequentlyk e n given elemenmy proofs. we doubt that an elementary proof of Theorem 20.10 will he found in the near future.

Theonm 20.10. Let r E A and let A = Ai. The= exist i E (1.2,. ..,r ) and a tree T in Ai such thatfor each path g in T.F S ( ( g ( n ) ) z o )S A;;and for each node f of T , Z ( B / ) > 0. Pmof This is Exemse 20.1.2. The following lemma will he needed in the next section.

Lemma 20.11. Let A G W such that k ( A ) = w r 0. Then there exists some sequence ( I k ) K 1 of intervals in W such that lim I l k 1 = m and k+ca

-

lim

IA n I k I = a.

141

Pmof This is Exercise 20.1.3. Exercise 20.1.1. Rove Lemma 20.2. Exercise 20.13. Using the fact that A is aleft ideal of (BPI. +) (Theorem 6.79),prove Theorem 20.10. Exercise 20.13. Rove Lemma 20. It.

20.2 'Ibe Comspondence Principle

~.

20.2 The Correspondence Principle

Recall that a meanrre preserving system is a quadruple (X. 8,p, (T,),,s) w h m (X, 21, p ) is a measure space and (T,),,s is a measure preserving action of a semig r w p S o n X. If the semigroup is (N. +). the action (T.).en is generated by a single function TI and we put T = TI and refer to the measurepnserving system (X. S ,p. T).

Theorem 20.12 (Furstenberg's Compondence Rinciple). Let A g N with d'(A) > 0. There ex& a measure preserving system (X, B.p , T)(in which X is a compact metric space and T is a homeomorphismfmm X onto X) anda set A' E 8 such rhat (1) p(A1) = d*(A) and (2) forevery F E d*(n ~

nnSF(-n

+A))

z p ( ~ ' nnnEF. T-~[A'I).

Proof: Let R = '(0, I1 with the product topology and let T : S2 -, S2 be the shift 1). (By Exercise 20.2.1 one has in fact that S2 is a defined by T(x)(n) = x(n metric space.) L e t t be the charafteristic function of A (viewed as a subset of Z). Let X = cC(Tn(f) :n E Z), the orbit closure of 6. Then X is a compact mehic space and (the restriction 00T is a homeomorphism from X onto X. For each n E Z, let D. = X n 1r.-~[(1]] = ( 8 E X : b(n) = 1) and notice that Dmis clopen in X. Let d be the Boolean algebra of sets generated by {D,, : n E Z] and let B be the a-algebra generated by {D. : n € Z). Notice that for each n E Z. TCD.1 = D.-1 and T-'[D.] = D.+I. Consequently, if B E 8,then T - ~ [ B ]E 8 (and T[Bl E 8 ) . Define rp : P ( X ) -, P ( N ) by q(B) = {n E N : Tn(S) E B). Let a = d*(A) and pick by Lemma 20.1 1 a sequence ( I t ) & of intervals in N such that lim Ilk I = m and

+

k-co

Pick any p E N* and define v : 9 ( X ) -+ [O, 11 by

Notice that v(X) = 1. Further, given any B, C S X, one has cp(Bfl C) = p(B) nrp(C) (sothat,inparticular,if BnC = O.thenrp(B)nrp(C) = O)andrp(BUC) = rp(B)Urp(C). Thus, if B and C are disjoint subsets of X. then

= p- lim I M B )n iki+ IMC)n keN lid

20 Density

418

(-I

-Connections with Ergodic Thmry

Next we claim that for any B V ( B ) )n so

+

w

G X , v(T-'[El) = v ( B ) . Indeed, ~ ( T - I [ B ]=)

+

because for any k E N,I(-1 q ( B ) ) n It1 and Ip(B) n It1 differ by at most I. Thus v is finitely additive and T-invariant on P ( X ) and hence in particular on A. Funher all members of A arc clopen in X,so if one has a sequence in A such thar each &+I E B. and B. = 0, then since X is compact there is sow k such that for all n 2 k. B. = 0 and hence lim v(B.) = 0 . n-m Therefore, we are in a position to apply Hopf's Extension Theorem. (This is a standard result in first year analysis cowses. See for example [115, Exercise 10.371 or [14. Satz 3.21.) The finitely additive measure v on A can be extended to a countably additive measure p on 8 where for each B E 8.

nLl

A, v ( T - ' M ) = v(C).ooeseca immediately that for all B E S. p ( ~ - ' [ B l = ) p ( B ) . Thus we have that (X.8.p . T ) is a measure p m e ~ n system. g Let A' = Do = {6 E X : 8(0) = 1 ) and observe that p(A1) = A and for each n E N.T-"[A'] = Dn and (p(D.) = (-n A) r l N . Thus

- Usingthisdescriptimofp(B)andthefactthatfod E

+

and given any

F

E

Pf(N).

= p-lim

k€N

IcWO)n rXEF d D n ) n hl lltl

When proving Szemerkdi's Theorem. one wants to show that. given a set A with P ( A ) s 0 and e E N,there is some d E N such that

( a ~ N : ( a , a + d , a + 2 ..... d a+td)SA)#O. As is typical of ergcdic theoretic proofs of combinatorial facts, one establishes this fact by showing that the set is in fact large. The proof of the following theorem require. extensive background development, so we da not present it here.

C o m h q 20.14 ( S z e W s Theorem). Ln A

E N m'rh d g ( A ) r

0. Then A con-

rnim arbitrarily long anthmeric prugnssimrr.

Pmof: PieL by Theorem 20.12 a measure pnsaving system ( X . 93, p , T) and a set A' E

8 such that

(1) p(A1)= dg(A)and (2)for emy F E Pf(FI). d * (n ~

Let C E

N be

n,

c-n

+ A)) r M A ' n a,,T-"~A'I).

given and for each i E (1.2,.

...CJ,let Ti

= T i . Notice that

f i . Tz.. ..,Tc are measure presming transformationsof ( X , 8,p ) that commute with each other. Pick by Theorem 20.13 some d E N such that p(A' f i ~ ; . - ~ [ Ar' ]0) and notice that for each i. Ti-d = T - ~ L.~ et .F = {d. 2d, 3 4 ...,Cdt. Then

&

Exercise 20Zl.Let Q = '(0, 1 ) and for 6 # 9 in Q. d e k e p(#, q ) = what k = minllrl : 1 E Z and C(r) # ~ ( 0 (and 1 of p(6.F) = 0). Rave that p a a mehic on R and the metric and product topologies on Q agree.

203 A Density Version of the Finite Sums Theorem The straightforward density version of the Finite Sums Theom (which would assen or perhaps d * ( A ) r 0 - would contain that any set A E N such that &A) s 0 F S ( ( x , , ) ~ , for ) some sequence { x , , ) ~is~ obviously ) false. Consider A = 2N + 1. However. a consideration of the proof of S z m d d i ' s Theorem via ergodic theory reminds us that the pmof was obtained by showing that a set which was only requiredto be nonempty was in fact large. Viewed in this way, the Finite Sums Theorem says that whenever r E N and N = I& A,. there exist i E (1.2.. .. r) and a sequence ( x J ~ such , that for each n, x.+l E Ai n C,,FX, Ai : 0 # F E (1.2, n ] ) . In this section we show that not only this set, but indeed many of its subsets. can be made to have pasitive upper density and specify how big these sem can be made to be.

-

nf-

+

.

...,

B

20 Density -Connections with Egodie Thmry

420

Lunma 20.15. Let A E N such t h n r z ( ~> ) 0 and let B be an i y k i t c subset of N. For mery 6 w 0 then uisr x c y in B such that $(A n (-(y x ) A)) > 6.

- +

Notice that for every t E N,;Imm

l ( - t + ~ ) n ( l , 2 , ....x.11

-

=a.

Xn

Enumerate B in increasing order as ( y i ) E l . Let c > O be given. If e 2 a2. the conclusion is hivial, so assume that c c aZ and let b = J-. Pick k E N such 40 thatk w -andpickC~Wsuchthatforalli E (1.2, kJandalln >C,

....

It sfices to show that far some pair ( i , j ) with 1 5 i c j 5 k, one has

-

+

since d((-yi + A ) r l (-yj + A ) ) = $(A n (-(yj - yi) A)). Suppose instead that this conclusion fails. Then in particular for each pair (i,j ) with 1 5 i j 5 k, there is some vi,j E N such that for all n 2 vi,j

Let n = max((L) U (v1.j : 1 5 i c j 5 k ] ) and let m = x.. For each i E (1.2, ...,k).

. -. -.

let Ai = (-yi + A ) n (1.2.. .. .m)and forany C E (1.2.. .. ,m).let p(C) = ILI m Thenforeachi~(1.2. ..., k ) , b c p ( A i ) c h , a n d f o r l s i < j i k , p ( A i n A j ) c a2 - c. Notice also that for C E {I. 2. .. . , k].

We claim that it suffices to show that

Indeed, once we have done so. we have k2b2 c 2ka

a conhadiction.

+k(k - 1)(a2 - c) so that

To establish the desired inequality, observe first that

'

so that

Next one establishes by a routine induction on m that for any function f : (1.2 ,..., rn]+R,

and thus by (I) we have

Now, for any t,

so that

Now, using (2) the mquLed inequality is established.

0

Notice that in Lemma 20.15, one cannot do better than Indeed,choose a set A 5 W by randomly assigning (with probability f ) eachn E W to A or its complement. Then for any r E W,Z(A fl (-r + A)) = f = Z ( A ) ~ .

Lemma 20.16. Let

p E A such that p

+ p = p, let A € p, and let e > 0.

+

-

P~w$ La B = (x E N : ; i (n~(-x A)) 2 Z(A)* 6/21. Since by Thenem 4.12, A. = (x E A : (-x A) E p) E p, it Suffices to show that B E p. Suppose instead &at N\B E p and pick by 'Iheorcm 5.8 a sequence (x.);, in N such that FS((x.)Z,) E N\B. Let C = {C:,, xf : n E W) and pick by Lemma 20.15 some n irn such that C;",, x, C:, xf E B. Since

+

-

this is a muadidon.

0

The proof of the following theorem is reminiscent of the proof of Lemma 14.24.

Notice that, following our usual convention that concerned with at the time. one has D(0, g) = A. Let To = (0) and inductively for n E w, let

n0 is whatever semigroup we sre

La T = TT,.Then trivially T is a aee in A. (Given x E B(g). x E C(0. g) D(0, g) = A.) Notice that forn E W andg E Tn.Bs = (x E B(g) : x > g(n - 1)).

Weshowfintbyindudonoanthatforallg E T.andall F C (0.1. .... n - 1). C(F, g) E P and Z(D(F, g)) > 62'F1. O ~ S C Wthat ~ (*) ifk = max Fandhisthe~snidonofgto(0,I, ....k).thenD(F,g) = D(F, h) and C(F, g) = C(F, h). To ground the induction assume that n = 0. Then g = F = 0 and D(F. g) = A so that a(D(F. g)) s 8 = @.Let c = Z(A)' - 6'. Thm

20.3 A Density Version of tbe Fbite Slnnr l?muun

423

so that by Lemma 20.16. C(F. g ) E p. N w l e t n ~ o a n d a s s u m t h a t f o t a l l g ~ T ~ a n d a 10.1 I l F,..., ~ n-1). C(F, g) E panda(^(^, g)) r 6''". Let g E Tm+land F E (0. I.. ...n) be given. By the observation (*). we may assume that n E F. Let h be the rrJtriction of g to (0. I.. ,n 1). let H = F\[n), and let x = g(n). Then h E T. and x E Bfh). We now claim that

.. -

+ D(H. h)). To see that D(H, h) r l (-x + D(H, h)) E D(F, g), let z E D(H, h) n (-x + D(H, h)). Then z E D(H, h) S A. Let Y E m((g(t))tsF) and pick G E F such that y = &c g(0. We show that z E -y + A. If n G. then y E FS((h(t))tEH)and h e m . since z E D(H, h). z E -y + A. We thus assume that n E G. If G = (n), then z E -x + D(H. h) E -x + A = -y + A as required We therefon assume (**I

D(F, g) = D(H, h) n (-x

that K = G\(n) # 0. Then K E H. Let v = & ~ g ( r ) = X,," h(t). Then ~EFS((h(t))~c~)~~x+z~D(H,h)~-v+Aadthusu+x+z=y+z~Aas required. To see that D(F, g) E D(H. h) r l (-x D(H, h)). let z E D(F. g). Since m ( ( h ( t ) ) t s ~-E ) m ( ( g ( f ) ) t ~ ~z ) E , D(H. h). Since x = g(n) E FS((g(t)),e~), x z E A. To see that x z 'E D(H, h). let y E FS((~(~)),,H).Then y x E FS((g(f))re~) SO Z E -(y x) + A SO x z E -y A. Thus ('*)is established. D(H, h) E p. Thus Since x E B(h) E C(H, h), we have -x

+

+

+

+

+

D(H, h) n (-x

+

+

+

+ D(H, h)) E p. +

That is D(F, g) E p. Now since x E C(H. h) and D(F, g) = D(H, h) n (-x D(H. h)), ~ ( D ( F 8, ) ) r 6Z'H'f'= 8"". Let y = a(D(P. g)) s~"' and let c = ~ ~ 8 2 " ' y2. Let

-

+

E={zED(F,g):-z+D(F.g)~pmd -~ ( D ( F .g) n (-2 D(F,~))) > I(D(F.~))' - CI.

+

-

+

.

BY emm ma 20.16. E E p. since Z(D(F. g)12 = (82IF' y)2 - c = 62'm+1 we have that E = C(F, g) so that C(F, g) E p. The induction is complete. Now let f be any path of T. We show by induction on IFJ,using essentially the first proof of Theorem 5.8. that if F E Pf(cu), n = min F. and h is the restriction of f to 10. I , . n - 1). then X r E ~ f ( f )E D([O, 1. .n 1). h). first assume that F = ln1,letgbetherestrictionof f to(O.1, ....n),andlethbetherestrictionof f to(0, 1. .... n - I). Theng= hU((n, f(n)))with

... -

...

f (n)

E

.

B(h) G CUO, I,. .. n - I), h)

as required.

E DUO, I , . ..,n - 1). h)

.

Now assumethat IF1 > I. let G = F\(n),andlet m = min G. Let h, g, and k bethe restrictions off to (0. I , . .. n I), (0.1. .. . n). and (0. I , . ... m 1) respectively. men & z f ( f ) E DUO, 1, ..., m - 11.k) G D(I0, 1 n1.g)

.-

-

.....

424

20 Iknsity

...,

-Connectioar with Ergodic lbcory

andD(IO,l, n I , g ) = D ( l O , l , ... , n - l I , h ) n ( - f ( n ) t ~ ( ( o , l , . . . . n-11.h)) by (**). Thus &G f (1) f (n) E D((0. I. . ,n I), h) as required. In particular. conclusion (1Xa) holds. To establish conclusion (I)@), let F s Pf((o), pick n w max F. and let h be the restriction off to ( 0 . 1 , ... n 1). The conclusion of ( I ) @ )b precisely the statemen\ proved above, that ~ ( D ( Fh)) , s 621F'. As we observed above, for n € N and g E T,,.Bg = [x E B(g) : x r g(n 1 ) ) and thus B, E p so. since p E A. J ( B ~s) 0. 0

+

.. -

.-

-

Notes The notion which we have called "Banach density" should perhaps be called "F'olya density" since it appears (with refmnce to N) explicitly in [1961. To avoid proliferation of terminology, we have gone along with Furstenberg [98]who says the notion is of the kind appearing in early works of Banach. The definition of Banach density in the generality that we use in Section 20.1 is based on the definition of "maximal density" by H. Umoh in [237]. It is easy to consttuct subsets A of N such that for all n € N,a(U:=, -1 A) = d ( A ) > 0 . On the other hand, it was shown in 11261 that if d * ( A ) > 0, then for each E > 0 there is some n E N such that dl ( U,: -t A ) s 1 - e. Consequently, in view of Theorem 20.7, it would seem that A*(N, is not much larger than c! K @ N , +). On the other hand it was shown in [73],a result of collaboration with D. Davenport, that ct K(@N,+) is the intersection of closed ideals of the form GUN' p ) lying strictly between ce K(@W,+) and A*(N, +). Theorem 20.13, one of three nonelementaq results results used in this book that we do not prove, is due to H. Furstenberg. Lemma 20.15 is due to V. Bergelson in [18]. Theorem 20.17 is from 1241. a result of collaboration with V. Bergelson.

+

-

+)

+

+

Chapter 21

Other Semigroup Compactifications

Throughout this book we have investigated the s t r u m of pS for a discrete semigroup S. According to Theorem 4.8, OS is a maximal right topological semigroup containing S within its topological center. In this chapter we consider semigroup compactifications that am maximal right topological, semitopological, or topological semigroups, or topological groups. defined not only for discrete semigmups,but in fact for any semigroupwhich is also a topological space.

21.1 The ZAnie, WAS', As), and a A 9 Compactifications Let S be a semigmup which is also a topological space. We shall describe a method of associatingS with acompact right topological semigroupdefinedby aunivmalproperty. The names for the compactifications that we intmduce in this section are taken fmm [39]and [40]. These names come from those of the complex valued functions on S that extend continuously to the specified compactification. The four classes of spaces in which we are interested are defined by the following statements. DeRnitton 21.1.

(a) 8 1 ( C ) is the statement "C is a compact right topological semi-

groupP*. (b) rlr*(C) is the statement "C is a compact semitopological semigmup". (c) rlr3(C) is the statement "C is a compact topological semigroup". (d) 84(C) is the statement "C is a compact topological group".

Lemma 21.2. Let S be a set with IS1 = K r o. let X be a compacr space, and let f : S -t X. Iff [Sl is dense in X, then 1x1 j 22'. Pmofl Give S the discrete topology. Then the continuous extension j o f f to PS takes PS onto X M that 1x1 IpSI. By Theorem 3.58. IpSl = 2.' In the following lemma we nced to be concerned with the technicalities of set theory more than is our custom. One would like to define Fi = [(f, C) : Oi(C). f is a

continuous homomorphism from S to C . and f [S] E A(C)). However,there is no such s& (Ie existence would lead quickly to Rusell's Pafadox.) Notice that the requiremenu in ( 1 ) and (2) concerning the topological center are dundant if i # 1.

Lcmm;.21.3. k t S be an infbtite semi~mupwhich is also a topological space and k t i E ( 1 , 2 , 3 . 4 ] . There is a set Fi ofordcnd pairs ( f . C ) such that (1) jC(f, C) E Fi, thcn WC),f is a cominuour honwmorphisrn~5~1 S to C, and f ~ s rl ~ ( c )rmd . (2) given any D such that rVi(D) and given any contimow homomorphism g :S + D with gtS1 S A(D), thcn u i s r ( f . C ) E Fi and a continuour one-to-one honmorphim (p : C -r D such thm (p o f = g.

$ = [ ( f , ( C , 7. .)) : C c X , 7 is a topology on C . .is an associative binary operation on C . and f : S -r C ) .

so 9 is a set (Moreformally, the axiom schema of separation applied to the set

.

and the statement T C X,7 is a topology on C. is an associative b i m q operation on C , and f : S + C" guarantees the existence of a wt 9 as we have defined it. In defining 9 we have, out of necessity, &paned from the custom of not specifically mentioning the topology or the operation when talking about a semigroup with a topology. In the definition of the set F, we renxn to that custom writing C instead of (C. 3". 9. Let

Fi = { ( f ,C ) E $ : \Yi(C). f is a continuous homomorphism from StoC,andf[Sl c A(C)). (Notioe again that the requirement that f [S] c A(C) is redundant if i # 1.) Trivially F; satisfies conclusion (I). Now let g and D be given such that Uri(D)and g is a continuous homomorphism from S to D. Let B = ce g[S]. By Exercise 2.3.2, since g[Sl G A(D). B is a subsemigmup of D. Further, if D is a topological p u p (as it is if i = 4). then by Exercise 2.2.3 B is a topological group. Since the continuity requirements of Wi are hereditary, we have Y i ( B ) . Now by Lemma 21.2, IBI 5 '2 = 1x1 so pick a one-twne function r : B -+ X and let C = r [ B ] . Give C the topology and operation making r an isomorphism and a

21.1 The LAC?,WAS', AS', and 8APComprctibcmions

homeomorphism fromB onto C. Let f = r o g and let p = r-I. Then ( f , C ) E is a continuous one-to-one homomorphism from C to D, and p o f = g.

427

E , (p

Theo~m 21.4. Lrr S be an i n w t e semigmp which is also a topological space and Ieti E (1.2.3.4). 7 l r c m c n c n s t a p i r ( q i . y i S ) ~ h t h o r (1) Q i ( y i 9 .

(2) qi is a continuou~homomorphismfrom S to yi S. (3) qi [Sl is dense in yi S. (4) qi[Sl G N y i S ) , and (5) given ony D such that \Iri(D)and any continuous homomorphism g : S + D with g[Sl G A(D), them ainr a continuous homomorphism p :yiS + D such tlurtpoqi=g. So thefollowing diagmm commutes:

S

-

D.

Proof: Pick a set Fi as guaranteed by Lemma 21.3 and let T = X ( / , o + F ; C. Define qi : S + Tby qi(s)(f. C) = f ( s ) . LetyiS = ctrqi[Sl. Byfheonm2.22T(withthe product topologj. and coordinatewise operations) is a compact right topological semigroupand w[Sl A(T). If i € {2,3,4)oneeasily verifies theremainingrtquimnents needed to establish that *,(TI. By Exercise 2.3.2 yiS is a subsemigroup of T and, if i = 4 , by Exercise 2.2.3 yiS is a topological p u p . Consequently Oi(yi S). For each ( f , C) E F,, n(~,c) o qi = f so qi is continuous. Given s , t E S and ( f ,C ) 6 Fi.

s

so qi is a homomorphism.

Trivially qi [Sl is dense in yi S and we have already seen that qi[Sl G A ( T ) so that 4i[S1 G A(yiS). Finally, let D be given such that Wi(D) and let g : S + D be a continuous homomorphism. Pick by Lemma 21.3 some ( f , C ) E Fi and a continuous one-to-one homomorphism (p : C -r D such that (p a f = g. Let p = p a q1.c).Then. given s E S,

We now o b s m e that the compactificationswhose existence is guaranteed by Theorem 21.4 an essentially unique.

.

Theonm 215. Ler S be m injinite semigroup which is also a ropological space and let i E [I. 2.3.4). Let (qi. yiS) be as guaranteed by Theorem 21.4. Assume that also the pair (p, T) satisfies (1) (2) (3) (4) (5)

*i(T), q is a continuous homomorphismfmm S to T ,

q[Sl is dense in T,

dS1 E NT). and given any D such that rlri(D) and any continuous homomorphism g : S + D wirh g[S] g A(D), then exins a continuous homomorphism p : T -+ D such rharpoq=g.

Then then is afunction 8 : yiS + . Y which is both an isomorphism and a homeomorphism such that 6 takes qi[S] onto q[S] and 6 o q = q.

Pmof: Let 8 : yiS -+ T be as guaranteed by conclusion (5) of 'Ihaorem 21.4 for T and q and let p : T -+ yiS be as guaranteed by conclusion (5) above for yiS and qi. Then 6 and p ate wntinuous homomorphisms, 8 o q; = q , and p o q = qi. Now G[yiS] is a compact set containing q[S] which is dense in T so 8[yiS] = T: Also @ o 6 o qi = p o q = qi so it o 6 agrees with the identity on the dmse set qi[S] and thus p o 6 is the identity on yiS. Consequently @ = 6-' so 6 is a homeomorphism and an isomorphism. 0 As a consequence of Theorem 21.5 it is reasonable to speak of "the" C M @ compactification, and so forth.

LMlnition 21.6. Let S be an infinite semigroup which is also a topological space and for each i E (I. 2.3.4) let (q;. yiS) be as guaranteed by Theorem 21.4. (a) (I,. ytS) is the GMC-compactification of S and LMC(S) = n S. (Q. n S ) is the WAP-compactification of S and W A P ( S ) = n S . (c) (m. n S ) is the AP-compactification of S and A P ( S ) = nS. (d) (q4, y4S) is the 8AP-compactification of S and I A Q ( S ) = y4S.

(b)

Notice that by Theorem 4.8, if S is a discrete semigroup, then (1.6s) is another candidate to be called "the" PMC-compactification of S. We remind the rrader that semigroup compactifications need not be topological compactifications because the functions (in this case qi) are not required to be embeddings. In Exercise 21.1.1. the reader is asked to show that if it is possible for qi to be an embedding. or just one-to-one, then it is. Exerripe 21.1.1. Let S be an infinite semigroup which is also a topological space and let i E (1.2,3,4). Let Y he a semigroup with topology such that $i(Y) Let r : S -+ Y be a continuous homomorphism with r[S] 2 A(Y). (a) Rove that if r is one-to-one, then so is qi. (b) Prove that if r is an embedding. then so is qi.

21.2 Right Topological Compaetiflcations

429

Exercise 21.13. If Sis a group (whichis alsoa topalogicalspaceXshow that AB(S) b A B ( S ) . (Hint: Rovc that A P ( S ) is a group and apply Ellis' 'Iheorem (Comllary 2.391.) Exercisetl.l.3. Let C denote any of the semigmups W A P ( N ) . A P O , or b A P ( N ) . Let x E C and let m and n be distinct positive integers. Show that m+x#n+x. Exerdse 21.1.4. Show that each of the compac!ifications W A P ( W ) , A P ( N ) and b A P ( N ) has ZC elements. (Hint: By Kronecker's Theorem. FJ can be densely and homomorphically embedded in a product of c copies of the unit circle.)

21.2 Right Topological Compactifications The requirements inconclusions (4) and (5)of Theorem 21.4 referring to the topological center seem awkward. They arc redundant exceptwhen i = 1, when the property being considered is that of being a right topological semigroup. We show in this section that without the requirement that g[S] 5 A ( D ) , there is no maximal right topological compactification of any infinite discrete weakly rightcancellative semigroup in the sense of Theorem 21.4.

Theorem 21.7. k t S bean infinitediscme weakly rightcancelkztivesemigmup and let K bean infinite cardinal. There exist a compact right topological semigroup T and an injective homomorphismT : S + T such that any closedsubsemigroupof T containing r [ S ]has cardinalify at least K .

Pmof Let 00 be apoint not in K x S and let T = ( K x S)U {m).(Recall that the cardinal is an ordinal, so that K = {a : a is an ordinal and a < K).) Definean openlion . on T as follows. (0, S ) . (ti.5') = (1'. 5s'). (1, s ) .(t', s') = (I' r, s ) if t # 0, and

K

+

(t,s).w=w.(t,s)=a,-00=00,

+

where r' t is ordinal addition. We leave it as an exercise (Exercise 21.2.1) to verify that the operation is associative. Now fix an element a E S and define a topology on T as follows.

.

(1) Each point of ( K x (S\(a))) U {(O, a ) )is isolated. (2) Basic open neighborhoods of a,arc of the form

where I' c K and F is a finite subset of S with a E F.

(3) If t = r'

+ 1. then basic opn neighborhoods of (r, a ) an of the f o m W. all U @'I

X;S\F))

w h F is a finite subset of S with a E F. (4) If r is a nolimit ordid. then basic open neighbarhoods of (1, a ) at of the form {(t. a)) u ({r" :r' c t" < r) x S) U ([r'] x (S\F))

wh~r'ctandFisafinitewbsetofSwithaEF. That is to say. a basis for the topology on T is 9)

= [((r. s)I :t < K ands E S\{al} U (((0. d l ) ~ ( ( mU) ((1'' :I' < t" < K) x S) U ({f'] x (S\F)) : ~ ' c K , F ~ 9 f ( ( S ) , a n d aF ~]

+ I. a ) ] U ((1')

~(((1'

x (S\F))

:I' < K. F E 9,(S), and a E F)

u{{(t,a)] U ({t" :t' < rn < I] x S) U ((1') x (s\F)) : t is a limit, t' c r < K. F E 3',(S), and a E F).

The verification that 9) is a basis for a Hausdorff topology on T is Exercise 21.2.2. To sce thal with this topology, T is compact, let P1 be an open cover of T. Pick

U

E P 1 s u c h t h a t m ~ U a n d p i c k f 1< ~ s u c h t h a t { r " : t ~ < t " c w ] x S ~ U .If a finite subfamily of P1 covers (t : r 5 11) x S, then we arc done. So assume that no

Iinite subfamily of e( covers {f : r 5 111 x S and pick the last to such that no finite subfamily of W.covers (r : f j to] x S. Pick V E e( such that (to 1,a) E V and pick F E Pf(S) such that {to] x (S\F) C V. Pickfinite $ G V such that {to] x F S U $. Assume first that to = t' + 1 for some r' and pick finite P G U such that {r : r 5

+

r')xS~UF.ThenFU~U(VJeovm{r:rs$]xS,acmnadjctMn. Thus ro is a limit ordinal. If ro = 0, then $ U (V] coven (0) x S. Thus # 0. Pick W E P1 such that (10, a ) E W and pick r' < ro such that (1'' :r' c r" c to] x S G W. Pick finite 3 E e( such that {r : r 5 t'] x S U 3. Thcn P U $ U (V, W] coven x S, a contradiction. Thus T is compact as claimed. Wemify now that T is aright topo1ogical semigroup. Since p, is constant it is continuous. Let (r', s') E K x S. Trivially p(,t,q is continuous at each isolated point of T, so we only need to show that p(,r,,) is continuous at m and at each point of (K\[O]) x {a). To see that p(,,..r) is continuous at m , l b U be a neighborhood of w and pick u < K such that (r" : u c r" < K J x S E U. Let ( t : t j 10)

W = {m) U ({r" : u

+ l c r" < K ) x S ) u ({u + I] x (S\{a])). +

Then p(,,,,r)[W] c U since r" 5 r' r" for dl r" c K. Next let 0 c r < K and let U be a neighborhood of (r. a ) . (rl,s') = (r' + r, a). A s s u m c h t that1 = to+] andpick F E Pf(S) witha E F suchthatfr'+to)x (S\F) E

21.2 Right Topologial Compactif~cati~~

431

U. t c t G = { s E S : s s ' € F)wdaotiathat,sinrrSis~eaLlyright-rrIlui~~ is finite. Let W = Kt. a)lU(Ifol x (S\(F u G))). Then&?,7[WI E U.(Ifto = Oands E S\G,tbenssf f F.) with a E F Now assume that t is a Limit ordinal and pick to < t and F E Pf(S) smb that {(tf+t,a))U({t":t'+to

< t " < t l + t ) x S ) ~ ( [ t ' + t o } x (S\F)) E U

and let

W = ((t.a)} U ((t" :to

4

1' < t) x S)U ((to) x

(S\F)).

Then ~(z.+)[WlE (1.

Thus T is a compact right topological semigroup. Define T : S -t T by r(s) = (0. s). Trivially r is an injective homonmphisrn. Now let H be iclosed subsemigroup of T containing r[S]. Weclnimthatr x {a) E H. Suppose not. and pick the least r c r such that (t. a ) f H. Notice that t > 0. Pick a neighborhood U of ( I , a ) such that U n H = 0. Then there exist some t' < t and some F E Pf(3such thal (t'l x (S\F) S U.Then (r'. a ) € H so

.

((0)x S) (ti, a ) = (1') x Sa S H.

Since S is weakly right canceIIative. So is infinite, so ((['I x S a ) n U # 0, a contradiction.

0

The foUowing corollary says in tbe strongest tnms that without the requirement that r[S] E A(T) (or some o k requirement), t k c could not be a maximal right topological compactification of S. Notice that the corollary docs not wen danand any relationship between the topology on X and the semigroup operation, nor are. any requirements, either algebraic or topological, placed on the function (p. (Although, presumably. the kind of result one would be looking for would have X as a compact right topological semigroup and would have cp as a homomorptusm with (p[S] dense in X.) Corollsv 21.8. Let S be an infiite discrete weakly right wuellative semigmup, let X be a sernigmup with a compact topology. and let (p : S -r X. There exist a compact right topological semigmup T and an injective homomorphism r : S -r T such thnt there is no continuous homomorphism q : X + T with q o (p = r. Pmof: Pick a cardinal K > 1x1and let T and r be as gusnurtced by Theorem 21.7 for S and K. Suppose one has acontinwus homomorphism q : X + T such that q o (p = r. Then q[X] isaclosedsubJnnigroupof T containing r[S] andthusr > 1x1 2 q[X1 2 x . a contradiction.

0

F%erdse 21.2.1. Rove that the operation on T defined in the proof of lkomn 21.7 is associative.

Exerdse 21.2.2. Let 91 and T be as in the prclof of T h e m m 21.7. Rove that 33 is a basis for a Hausdorfftopology on T.

213 Periodic Compacti6cations as Quotients Given a semigroup S with topology, the WAF-. 047-.' and dA9-compactifications are all compact right topological semigroups that are equal to their topological centers andthus, by Theorem4.8 each is a quotient of BSd, where Sd denotes S with the discrete topology. In this section we shall identify equivalence relations on /3Sd that yield the W A P and AP-compactifications. We shall use the explicit descriptions of these compactifications as quotients to characterize the continuous functions from a semitopological semigroup S to a compact semitopological or toplogical semigroup that extend to the W A P - or AP-compactifications of S. We shall also show that, if S is commutative, BAP(S) is embedded in WAP(S) as the smallest ideal of WAP(S). We first give some simple 'well-known properties of quotient spaces defined by families of functions. De6nition 21.9. Let Xbe a compact space and let g (f, Y) be a statementwhich implies that f is a continuous function mapping X to a space Y. We define an equivalence relation on X by stating that x I y if and only i f f ( x ) = f ( y ) for every f for which @( f, Y) holds for some Y. Let X / i denote the quotient of X defined by this relation. and let n : X + X / E denote the canonical mapping. We give X/I the quotient topology. in which a set U is open in XP if and only if n - ' [ U ] is open in X. Ofcourse, the space X / idepends on the statement& although wehavenot indicated this in the notation.

Lemma 21.10. k t X be a compact space and let #( f, Y)be a statement which implies that f is a continuourfwtction mapping X to a space Y. Then X/c has the following properties: (1) Ifg : X -* Z is a continuousfunction and if8 : X/I + Z is afunction for which g = p o n, then p is continuous. (2) X/3 is a compact Hausdo#space. (3) A net ( R ( X < ) ) converges ~E~ to n(x) in X/= ifand only if(f (x,)),,, conveges to f(x) in Y forevery (f, Y) forwhichg(f, Y)hol&. Pmof. (1) If V is an open subset of Z. then n-I[p -'[V]]= g-'[VJ, which is open in X and s o p -'[V]is open in X/I. (2) Since X/I is the continuous image of a compact space, it is compact. To see that it is Hausdorff, suppose that x , y E X and that x(x) # n(y). Then there is a p i r

21.3 Periodic Compneti6cationsas Quotients

433

( f , Y ) such that $ ( f , Y ) holds and f ( x ) jC f (y): For every u , n E X , a @ ) = n ( v ) implies f ( u ) = f (v). So there is afunction f : X/i+ Y for which f = f o a. By (I). f is continuous. I f G and H are disjoint subsets of Y which are neighborhoods o f f ( x ) and f ( y ) rrspcctively, then -'[GI and f -'[HI are disjoint subsets of X / I which arc neighborhoods of n ( x ) and n ( y ) respectively. So XI= is Hausdofl. (3) As in (2). for each ( f , Y ) for which $(f, Y ) holds, let f : X/= Y denote the continuous function for which f = f o a . The set ( f -'[[I] : $(f, Y ) and U is open in Y )is a subbase for a topology r on X /= which is cthan the quotient topology and is therefore compact. However, we saw in ( 2 )that, for every x , y E X , n ( x ) # ~ ( y ) implied that n ( x ) and n ( y ) had disjoint r-neighborhoods. S o T is Haudorff and is therefore equal to the quotient topology on XI=. Let ( x J I E tbe a net in X and let x E X. I f ( f (&)),GIconverges to f ( x ) in Y whenever $ ( f , Y) holds, then (z(x,)),,r converges to n ( x ) in the topology r. So (x(x,)),,l converges to n ( x ) in the quotient topology. Now assume that (n(x,)),,r converges to n ( x ) and!et ( f , Y ) be given such that @( f. Y ) holds. I f U is a neighborhood of f ( x ) . then f -'[Ul is a neighborhood of n ( x ) . Thus ( X ( X ~ ) ) , is ~ Ieventually in j - ' [ ~ ]and so ( f ( x , ) ) , , ~ is eventually in U.

-

We mall that we defined in Section 13.4 a binary operation o on /IS,+by puning x o y = lim lim st, where s and t denote elements of S. Note that, in defining x o y, t-+"*-.Z

we have reversed the order of the limits used in wr definition of the semigroup (pd, .). The semigroup (BSd,o) is a left topological semigroup while (BSd. .).as we defined it, is a right topological semigroup. The following lemma summarizes the basic information that will be used in our conshuction of the W A P - and APcompactificationsof a semitopologicalsemigroup.

Lemma 21.11. k t S be a semitopological semigmup and let B ( f , C ) be a starentent which implies that f is a continuousfunction f mapping S too compact semitopological semigmup-C. Let $(g. C ) be the statement that there exists f for which B ( f , C ) holds and g = f : BSd + C. Suppose that, for every ( f , C )for which B( f , C ) is m e , we have: (1) B(f o I., C ) andB(f op., C ) holdforeverya E S(where A..p. (2) f o r d l x . y E B S ~.T(x. , Y) = .T(x o y).

:S

+ S)and

Let B S d / r and n be defined by the statement @ as dkcribed in Definirion 21.9. We dejine . on pSd/= by purring n ( x ) - n ( y ) = n ( x - y).

(a) The operation. on BSd/= is well dcfined. (b) With this operation. @Sd/=is a compact semitopological semigmup and n is a continuous homomorphismfmm (BSd, .) andfmm (BSd. 0). (c) The restriction xp is continuous. P m f : Notice that for x , y ~ i 3 S . dT. e have x r y if and only if for every pair ( f , C ) satisfying B( f , C). one has f ( x ) = f ( y ) .

-

-

(a)Givena € S a n d x , x ' € p S d w i t h x gnx'.weclaimthatx.a = x l . a a n d a x I a x'. To see this, let (f. C) be given satisfying B(f, C). Then B(f o pa. C) %dB(f o b , C ) h o l d . Sjf op%= T o & and f o L = f o G wehavethat f (x .a) = f (x' .a) and f ( a . x) = f (a. x') as requid. Now assume that x = xi and y E BSd. We claim that x y I x' y. Recall rhat we denote by 1, the continuous function from BSd to pSd d e w by &(z) = x o z. Now foreacha E S and each (f, C) for which B(f. C) holds.

.

.

-

-

70- 70 .

.

and so L, and L,, are continuous functions agreeing on S. hence on pSd. Thus f (X.y) = f (x' y). Similarly (using p,(z) = z x), y - x Iy x'.

.

.

Nowletx~x'andy=y'.Thenx.y~x'.y~x'.y'. @) By Lmuna 21.10, )%?dl' is compact and Hausdorff. By (a), x is a homomorphism from @Sd, .), and by (a) and (Z), n is a homomorphism from (BSd. 0). By Exercise 2.2.2. since p S d l r is the continuous homomorphic image of @Sd, .),it is a right topological semigroup, and since it is the continuous homomorphic image of (BSd, 0). it is a left topological semigroup. (c) To see that n l j : S + &I= is continuous, k t s E S and let (s,),,, be a net in S converging to s in S. To see that ( ~ ( s , ) ) ,converges ~r to n(s) it suffices by Lemma 21.10 to let (g. C) be given such that &(g. C) holds and show that (g(s,)),,l converges to g(s). Pick f such that f?( f. C) holds and g = f . Then f :S + C is continuous so ~l to g(s) as required. (f (s,)),~Iconverges to f (s). That is. ( g ( ~ , ) ) ,converges The numbering 62 and & in the foUowingdefinitionis intended to cornpond to the function 02 : S + WAP(S). The statem& 82 and & depend on the semigroup S. but the notation docs not reflect this depeodence. Delinition 21.12. Let S be a semitopologid scrnigrwp. &(f, C) is the statement "C is-a compact s_emitopologicalsemigroup, f is a continuous function from S to C, and f (x o y) = f (X .y) for every x. y E &". &(g. C) is the statement "them exists (f, C) for which &(f, C) holds and g = f :BSd += C". Lemma 21.13. Let S be a semiropologicalsemigroup. If&(f, C) holds and a E S, then 6M f o I., C) and &(f o pa. C) holdas well ( w k n A., p. :S + S).

-

Proof Since A,, is continuous(bsause S is a semimpological semigroup). we have that f o A. is continuous. Notice also that f o Aa = G. Now. let x, y E PSd. W. sineeaES.aox=a.xandso

-

70

foAa(xoy) = f " ( a o ( ~ ~ ~ ) ) = ~ ( ( a ~ ~ ) ~ ~ ) = ~ ( ( a . x ) o ~ = ?((a. x) Y) = j ( a (x y)) = f=(x . y).

.

Similarly &(f o pa. C) holds.

. .

The m d w may wo& why we now rrqaire S to be a semitopological semigroup. after pointing out in Section 21.1 thst we w e n not demanding this. -The reason is that we need this fact forthe validity of Lrmma 21.13.

Theorem 21.14. Ler S be a semitopological semigroup. kt q5 =

andlet flSd/= and n : flSd + f l S d / ~be defied by Definition 21.9. Then (ny,flSd/=) b a WAQcompactifiation ofS. '?'hat is: (a) (b) (c) (d)

*2'2(Bsd/'), nls is a continuous homomorphismfmm S to !S~/S. n [ q is dense in BSdlzs and given m y D such that Wz(D)and any continuo^^ homomorphism g : S + D. then aLns a continuoushomomorphismfi :f l S d / ~+ D such that r onp = g.

P m J Conclusions (a).

(b), and (c) are an immediate conssqumce of Lemmas 21.11

and 21.13. Letg and D be given such that V+(D) and g is a continuous homomorphism from Sto D. We claim that &Or. D) holds. So let x. y E BSd. Notice that g'is a homomorphism r ~ Thus from (flSd. .) to D b i ~ o r n l l a 4.22.

L*fi : f l S d / ~-f D denote the function for which won = % Then fi iscontinuous, by Lemma 21.10. and is easily seen to be a homomorphism. 0 We now turn o w attention to c h w t i o n s of functions that extend to WAQ(S).

Theorem 21.15. Ler S be a semitopological semigroup, let C be a compocr semitopological semigroup, and lct f be a continuousfunctionfrom S to C. +re isa co@mous function g : WAQ(S) + C for which f = g o qz ifandonly i f f (x y) = f (x o y) for every x. y E flSd.

.

P m $ Necessity. Pick amtimousfunctiong : WAB(S) + C such that f =gem. Then

Sufficimcy. Define zs as in 'Iheaem 21.14. Then by Th-s 21.5 and 21.14. it suffices to show that t h exists ~ a continuous function g : f l S d / i -r C for which = g o nls. Since &(f, C ) holds, t h m is a function g : f l S d / ~+ C for which f = g o n . ByLemma21.10,giscontinuous. 0

t

If S is commutative, we obtain a characterization in terms of (BSd. .) alone.

436 ,

21 Ocher Semigroup Compactifiatims

Corollary 21.16. Let S be a commutative semitopalogical sernigmup, let C be a compact semitopological semigmup, and let f be a continuousfmctionfmm S to C. %re i~ a c o n t i n y s function g : W J I P ( S ) -P C for which f = g o qz if and only if f ( x . Y) = f ( .x)forrvery ~ x. y E BSd. Pmof This is an immediate consequence of Theorems 13.37 and 21.15.

0

RecaU that C ( S )denotes the algebra of bwnded wntinuous complex-valued functions defined on S.

Definition 21.17. Let S be a semitopological semigroup and let f E C(S). (a) The function f is weakly almostperiodic if and only if the& is a continuous g : W A P ( S )+ C such that g o w = f . (b) The function f is almost periodic if and only if there is a continuous g : JI.?'(S) + C such that g o q3 = f .

Theorem 21.18. Let S be an infinite semitopological semigmup and let f E C(S). The following statements are equivalent: ( 1 ) Thrfunction f is weakly almostperiodic. (2) Wheneverx. y E BSd, f"(x. y ) = f"(x o y). (3) Whenever (a& and (b,,):, are sequences in S and all indicated limits exist, lirn lirn f (a, .b,) = lirn lim f (a, .bn). n-rmn-m tt-mm-m

PmoJ By Theorem 21.15, (1) and (2) are equivalent. We shall show that ( 2 ) and (3) are equivalent. (2)implies (3). Let ( a , ) z l and (b,):, be sequences in S for which all the limits indicated in the expressionsm ; mn!m f (a,. b.) and ;irnmrnnw f (a, .b.) exist. Let x and y be Limit points in BS,j of the sequences ( a , ) Z 1 and (b.)$, respectively. Then lim lim f (a, - b.) = f"(x 0 y ) and lim lirn f (a. .b,) = f ( x .y). So (2)implies n-rrnm-m n-mn-m that these double limit$are equal. (3) implies (2). Let x.y E BSd. let k = y ( x . y), and k t C = a x Q y). For each nENletA.=[a~S:Jf(a~y)-k]c$)andlet~.={b~~:1f(x~b)-t1c$). Since (BSd. .) is a right topological semigroup. (BSd. o)is a left topological semigroup, andx.b=xobforbES,A.~xandB. ~ y . Choose a , E A1 and bl E B I . Inductively, let r E N and assume that we have chosen ( a l ,az.. .. a,) and { b ~bz, . .. b,] so that forall m , n E (1.2.. . rl

.

(i) a, E A, and b,, E B. (ii) if m > n. then ( f(a, - b.)

. .

- Ll c i, (ii) if n > m , then If (a, . b,,) - kl < A. (iv) if m > n , then If(am . b,) - f"(x. bJ c 4, and (v) i f n > m, then If (am. b.) - f"(a,,, . y)l c ;I.

..

A}

Nowforeachn E (1.2. .... r1.b. E B.so[a E S: (f(a.b.)-tl < E X . ALso. foreafhn E (1.2. r ) . ( a s S : lf(a.b,)-?(x.b.)l < $1 ~ x s o p i c k a , + l s u c h thatforeachn E (1,2. rl, I f (a,+~.b.)-el < :andl f(a,+l.b.)-f"(x.b,,)l -= Similarly. choose

..., ....

&.

f (am . b.) = ?(x .b.) and given m E N o m has lim f (a, b.) = ?(am .y). Also Aimm "5%f (a, - b.) = k and lim tim f (a,, . mm -. "-mm-m b.) = I . SO e = k. o Now. given n E

R,w e has , I ,

.

Theorem 21.19. Let S be a commutative semitopological semigroup. Then W A P ( S ) is also cornmutorive. F u n h e m r e , ($2. WAP(S)) is the mnrinral commutative semigmup compactificarion of S in rhefollowing sense: If(v. T) is any commutative semigmupcompactificationof S. then there isa conrinuoushomomorphismp : W A P ( S ) + T for which v = p o $2. Pmof By Theorems 21.5 and 21.14, we may assume that W A P ( S ) = bSd/l and $2 = q s . Now given (fLC) such that @(f, C), one has by Theorem 13.37 that for all X, Y E bSd. f (X - y) = f (y . x) and thus x(x). s ( y ) = x ( y ) . ~ ( x ) . Now, if (v, T) is a commutative semigroup compactification of S. then T is a semitopological semigroupso there exists acontinuous homomorpsm p : W A P ( S ) + T for which v = p o m. 0

Theorem 2130. LQI S be an infinite commutative semiropologicalsemigmup. Then K ( W A P ( S ) ) can be identified with dAP(S).

Proof Let K = K (WAP(S)). By Theorem 21.19 and Corollary 2.40. K is acornpact topological group. Let e denote the unique minimal idempotent of WAP(S). We define a continuous homomorphism r : W A P ( S ) + K by r(x) = xe and define a continuous homomorphism p : S + K by p = r o w. Let T be a compact topological group and let v : S + T be acontinuous homomorphism. By the defining property of WAP(S). there is a continuous homomorphism 6 : W A P ( S ) + T for which 6 o $2 = v . We note that &(el is the identity of T and so $Ix o p = v. This shows that we can identify BAP(S) with K (by Theorem 21.5). We now discuss some simple properties of WAP(PI). We use Theorems 21.5 and 21.14 to identify WAP(H) with @N/= and $2 with nlm If S = NU (oo), the one point compactificationof N. and oo+x = x +m = oo for all x E S, then S is a semitopological semigroup. Thus, by Exmise 21.1.1, $2 : N + WAP(N) is an embedding. Consequently. we shall regard N as being a subspace of WAP(N). Lemma 21.21. Lcr x be in rhe interior in p N of BN\(N*

+ N*). Then la(x)l = 1.

P m f . Let y E BW\(x). A n (N*+N*) = 0.

We can choose a subs* A of Nforwhich x E Si. y !+ .

.

X,and

Ln f = X A : N + (0.1)sothat j = x S i . ~ e c l a i m t h o c f " ( ~ + q ) = j ( ~ + ~ ) f o t 4 l p . q e p N . I n d 9 i f p ~ N o r qsN.thenp+q =q+p.whileifp,q ~ W , t h e n f ( p q) = 0 = f (q + p). Thus by Corollary 21.16, thae is a continuous function g : W A P + {0,1) such that f = g o q~: Then f and g o R-a continuousfunctions from BN to {O,l) agreeing on N and so f = g o R. Since f ( x ) = 1 and f ( y ) = 0, ~ ( x#) ~ ( y )l l. n t is, y $ n ( x ) .

+

Theom 2122. Them isadenseopensubser U ofN.with tkpmpenythat In(x)l= 1 for every x E U.

-

P m f . By Ex&i 4.1.7, if (x~):=~is a squence in N fm wbichx.+l x. -r ca and if A = {x. :n E R),then (ctgn A ) r l (N* +PI*) = 0. Now any infinite sequence in W contains a subsequence with this property, and so the= is a dense open subset U of N ' for which U n (N* W ) = 0. 0

+

Theorem 21.23. I f x is a P-poi# in PI*, then n ( x ) is P-point in WAS(W)\N.

+

N*). We shall show that x E (I. Our P m J Let U denote the interim of BN\(N* claim will then follow from Lemma 21.21 and the observation that n [ U ]is open in W A P ( M because R - ' [ n [ U ] ] = (I. For each k E W. we can choose a set A* E x with the property that la a'( 2 k whenever a and a' are distinct integers in At. F i s can be seen from the fact that there exists i E (0. 1,2, .. k 1 ) for which Nk i E x.) Since x is a P-point in N*, there is a set A E x for which A\Ak is finite for every k. It follows from Exercise 4.1.7 that --

-

.

.-

+

Arl(W*+N*) =0.

0

We now tum our attention to representing A$'($)

ss a quotient of BSd.

Definition 21.24. Let S be a semitopological semigroup. & ( f , C) is the statement "C is a compact semitopologifal semigroup, f is a continuous h t i o n from S to C. and there is a continuous function f' : pSd x BSd + C such that for all s. r E S. C) is thestatement "there exists ( f , C ) for which & ( f , C) f(s . t ) = f*(s,t)". holdsandg= f : p S d + C " .

+&!.

We show next that h(f. C ) implies &(f. C).

Lemma 21.21 Let S be a semiropological smigmup. I f @ ( f . C) bI& and a E S, then&( f ope. C ) a d h (f oA.,, C )ho[daswell(whem &, Aa :S S' S). Funkmrore, fyisasguamnteed by this srmemenr, thenforall x , y E BSd, 7 ( x - y ) = f ' ( x , y) = f(x or)Pmof: Suppose that &(f, C ) holds aad that a € S. Let g : BSd x BSd + C be defined by g ( x , y) = f ' ( a x , y). Then g is continuous and g(s, t ) = ( f o A.)(s. r ) for every s, t E S. So &(f o A,, C ) holds. Similarly, &(f o pa. C ) holds.

Let x , y E pSd. Then

For the remainder of this section, we reuse the notations = and x for a different quotient. Theorem 2136. Ln S be a smiropological smigmup. Let and n be defined byDefinition21.9 with4 = &. Foreach x. y E pSd. l r t n ( x ) .nLy) = n ( x . y). Then (nls.flSd/') is an AP-compactificaliOn of S. That is. (a)

'h(@sd/r).

(b)

nls is a continuous homomorphismfrom S to &%ill.

(c) n [ S ]is dense in pSd/=. and

(d) given any D such that W3(D)and m y continuous hmomorphimr g : S + D. + D such that p onls = g. there exists a continuous homomorphism p :f&/i Pmof: It follows immediately from Lemmas 21.11 and 21.25 that x is wclldefined. Conclusions (b) and (c) are also immediate from these kmmas, as is the fact that ( B S d / r , .) is a compact semitopological semigroup. To complete the verification of (a). we need to show that multiplication in f i S d / t is jointly continuous. I ( y , ) , , ~are nets in @Sdfor which Suppose that x , y E BSd and that ( x , ) , ~and ( R ( X , ) ) , ~ I and (X(Y&I converge t o n ( x ) a n d n ( y )respectively in pSd/=. Weclaim I to x ( x ) . n ( y ) ,thatis that ( n ( x , y,)),,~converges to that ( n ( x , ). H ( ~ , ) ) , ~converges X ( X . y). By passing to a subnet, we may presume that (x,),,; a d (y,),,; converge in pSd to limits u and w respectively. Thenn(u) = n ( x ) a n d x ( v ) = n ( y ) . To see that ( n ( x , ;y,)),,l converges to x ( x .yJ = n ( u u), it suffices by Lcmma 21.10 to show that ( f (x, . y,)),.~converges to f (u . u) for evay ( f . C ) for which & ( f , C ) holds. So assume that & ( f , C) holds and let f'be the function guaranteed k t h i s statement. Then ( f ' Q , , y,)),,; converges to f ' ( u . v ) and so by Lemma 212.5. ( f (x, .YAEI COnVergCS 10 f ( u . U) as required. To v d y ( d ) , let g and D be given such that Y 3 ( D ) and g is a continuous homomorphism from S to D. We claim that 03(g. D) holds. To see this, d e k g* : pSd x OSd + D by g*(x,y) = z ( x ) . F(y). Then g* is continuousand. givens. t E S. one has ga(s,1 ) = g(s) .g(r) = g(s t). Let p : pSd/= + D be the function for which p o n = Then p is continuous 0 (by Lemma 21.10) and is easily seen to be a homomorphism

.

.

.

z.

Theorem 21.27. Lei S be a semitopologicalsemigmup, let C be a compact topological semigmup, and let f be a continuousfwction fmm S to C. There is a continuous fincrion g : b P ( S ) + C for which f = g a 73 ifand only ifthere is a continuous fwction f.:BSd x pSd + C such that f . ( s , 1 ) = f ( s . I ) foralls. t E S.

Piuof Necessity. Pick a conlinuous function g : A P ( S ) -+ C such that f = g o q,. Let : /3Sd + A P ( S ) denote the continuous cxtmsion of ~ 3 Definc . f . ( x . y) = g G ( x ).GCY)). Then givens, r E S one has g(ijj(s) . G ( t ) ) = g ( m ( s ) &)) = 8 q 3 b . 0 )= f ( s . 0 . To see that f is continuous, let x . y E pSd and let W be a neighbotbood of g ( G ( x ) - rji(y)). Pick neighborhoods U and V of % ( x ) and G ( y ) in A P ( S ) such that U .V c g-I[Wl. Then f . [ i j j - ' [ ~x] rl?-l[~l]c_ W . SuRiciency. Since & ( f , C) holds. there is a function g : oSd/= -+ C such that f = g o n . B y Lemma 21.10, g is continuous.

.

I

Theorpm 2128. Ln S be a semitopological semigmup and let f

E

C ( S ) . 771efollow-

ing statements are equivalent (1) Ikfunction f is almostperiadic.

( 2 ) There is a conrinwurfuncrion f': /3Sd x pSd + @ such thmfor all s , t E S, f'(3.t) = f ( s . t ) . (3) For every 6 > 0, there is an equivalence relation * on S with finitely many equivalence classes such that If ( s . I ) f (s' .f')l c r whenever s * sf and t 4 1'.

-

Pm$ The equivalence of (1) and (2)is a special case of Theorem 21.27. (2)implies (3). Let r r 0 be given. Each element of /3Sd x pSd has aneighborhood oftheformLrxV.whcreU andVareclopensubseuofflSdandIfYx, y)- f ' ( u , v)l < < whenever (x, y). ( u , u ) E U x V. We can choose a finite family 9 of sets of this form which covers BSd x pSd. Let W = ( U : U x V ~ F f o r s o m e V ) U ( V : U xV ~ 3 f a s o m e U ) . Define an equivalence relation 4 on S by s * t if and only if for all W E W, either (s,f ) E W or {s. f 1 t l W = 0. Then is an equivaknce relation with Wtely many equivalence classes. Now assume that s * s f and t * r'. Pick U and V such that U x V E 3and (s. r ) E U x V. Then sinces a s'andt * t', (s', r') E U x V andso If*(s. t ) - f ( s 1 , t')l c c. That is. If (s t ) f (s' .r')l < r . (3) implies (2). Define f ( x , y) = f"(x .y ) for all x , y E BSd. m e n hivially f'(s.1) = f ( s .t)fordls,r E S. To see that f' is continuous, let x , y E BSd and let E > 0 be given. Pick an equivalence relation * on S with finitely many equivalence classes suchthat If ( s .t ) f (s' . r')l c 5 whenever s 4 s' and t % t'. Let $2 be the set of *-equivalence classes and pick A. B E X such that x E ; iand y E We claim that for all ( u , v ) E ; ix 3. one has I f ( x , Y ) - f % , v)l c r. Since f (x y) = x-limy-lim f ( s . t ) , we can

. -

2.

.

-

rc s . . .

IFZ

choose (9, t ) E A x B such that f ( x .y) f ( s . r)l c 5. Similarly. we can choose ( s f ,t') E A x B such that 1 7 ( u.v ) f (s' . r')l c 5. Since s 4 S' and I 4 t', one has I f ( s . I ) - f(sr.t')l < 5 andso$(x .y)-?(u. v)l < r .

-

Exercise 2131. Show that WAP(N)\N contains a dense open set V disjoint from (WAP(PI)\W) + (WAP(N)\M). Exercise 2132. Show that WAP(N)\N contains weak P-points. Exercise 2133. Show that WAP(N) is not an F-space. (Hit: An F-space cannot contain an infinite compact right topological group. See [182. Comllary 3.4.21.)

Exercisc213.4. We have seen that K ( W A P O ) can be identified with 6 A P ( N ) (by Theorem 21.20). Show that the same statement holds for K(AS)(N)). Exercise 21.35. Let S be a semitopological semigroup and suppose that f E C(S) is almost periodic. Show that every sequence (s.) in S contains a subsequence (s.,) for which the sequence of functions t I+ f (s.,t) converges unifmly on S. (Hint: Use Theorem 21.28.)

21.4 Semigroup Compactifications ai Spaces of Filters We have seen that if S is a discrete semigroup, then fJS is the 1Me-compactification of S. We have also found it useful throughout this book to have a specific representation for the points of fJS.namely the ultrafilters on S. As we noted in Section 21.3, the W A F - , AP-. and dAP-compactifications are all quotients of ,SS. The points of these compactificationscorrespond t n closed subsets of fJS.Since the closed subseuof p S correspond to filters on S by Theorem 3.20, each of these compactifications -indeed any semigroup compactification can be viewed as a set of filters. In this section we characterize those sets of filters that are,in a natural way, semigroup compactifications of S.

-

Remark 21.29. Lct S be a discrete space, Ict X be a compact space. and let g be a continuousfwtctionfmm pS onto X. Then given any x E X andany neighborhood U n ~s E ~ E - ' [ I X J I . O~X g ,- l [ ~ The first step in the characterization is to define an appropriate topology on a set of filters.

Definition 2130. Let S be a discrete space, let W be a set of filters on S, and let A g S. (a)An=[(A€W:A€AJ. (b) The quotient topology on W is the topology with basis (AD: A S SJ. Observe that if A. B E S, then A ¶ nBG(A nB)# so that [An : A S] does form a basis for a topology on W. That the term "quotient topology" isappropriate is part of the content of the following theorem. Recall that, given a set W of sets, a choice function for W is a function f : W -+ W such that for all A E %, f (A) E A.

U

Tbcomn 21.31. Let S be a ducme space, Iet X be a compact space, and let g be a continuousfun&n/mn BS onto X. Let R = ( n g-'[[x]] :x e X). Then d is o set offilters on S sarisfving (a) given any choice function f for R fhere uisfs 5 E Pf(R) such thar S = U A ~fT(A) @) given distinct A and S in B, there exist A E A and B E S such that whenever C E LR, either S\A E C or S\B E e.

Further: with the quotien! topology on LR, thefunction h : X + 94defined by h(x) = ng-l[(x)] is a homeomorphism.

Pmof: Given x E X, g-'[[x)] is a set of ulbafiltm whose intersection is thenfore a filter. l&l f be a choice function for- and suppose lhat for each 3 € Pf(W),

s\ Ulh3f (A) #

'.

Then-[S\ f (A) : A E 8 ) has the finite intersection propcay so pick p E pS such that (S\ f ( A ) : A E W) S p. Let x = g(p) and let A = n8-'[[x)]. Then f (A) E A G p and S\ f (A) E p, a contradiction. Now let A and P) be distinct m e m h of R and pick x and y in X such that A = ng-'[[x)] and P) = ng-l[(y)]. Pick neighborhoods U of x and V of y such t h a t c e u n c c v = 0 . LetA = g - ' [ U ~ n s a n d l e t B =g-'[V]nS.ThmbyRemark 21.29, A E A and B E P). Now let C E Rand pick r E X such that C = ng-'[(z)]. Then without loss of generality r $ CL U so if C = g-'[X\cl U] n S we have by Remark 21.29 that C E C. Then C n A = 0 so. since C is a filter, S\A E C. Finally define h : X + % by h(x) = ng-'[[x)]. Then trivially h is one-to-one and onto !X. Since X is compact, in order to show that h is a homeomorphism it suffices to show that % is Hausdorff and h is continuous. To see that %is Hausdorff, let A and 33 be distinct members of 31 and pick A e A and B E 8 as guaranteed by (b). Then A m and B' are disjoint neighborhoods of A and 8 respectively. To see that h is continuous, let x € L a n d let A S S such that A"S a basic open neighborhood of h(x). Let U = X\g[S\A]. Now g[S\A] is the continuous image of a compact set so is compact. Thus U is an open subset of X. We claim that XLU and h[Ul G A#. For the first assertion suppose instead that we have some p E S\A such that g(p) = x. Then h(x) G p so A E p, a contradiction. Now let y E U. Then by Rernark21.29. wehave g-'[U]nS E h(y). Since g-'[U]n S E A wehave A E h(y) o so h(y) E Ad as required. The following thmrern tells us that the description of quotients pmvided by Theo-

rem 21.31 is enough to characterize them. Theorem 2132. Lrr S be a discrete space and let R be a set offilters on S satilfyng

(a) given any choice fwrcrion f for R there exists 5 E Pf(W) such thaf S = UAESf (A) and (b) given distinct A and 8 in R ! , *re exist A E A and B E 8 such thar whenever C E R either S\A E C or S\B E C'.

-.

Then, w.th the quotient topology, R is a compact Hadoflspace. Further:for each p E flS there is a unique A E R such thar A E p and IheJuncriDng : fiS + W defied by g(p) C p is a continwuc sujection.

P w J That 'R is Hausdorff is an immediate consequence of (b). That W is compact will follow from the fact that W is a continuous image of flS. Now let p E flS and suppose first that for no A E R is A E p. Then for each A E Wpick f (A) E A\p. By (a) pick some 5 E P,(W) such that S = UASS f (A). Then S E p so pick some A E 5 such that f (A) E p. a contradiction. Next suppose that we have distinct A and 8 in W with A E p and 8 E p. Pick A E A and B E 8 as guaranteed by (b). Then S \ A E S E . p and A E A E p. a contradiction. The function g is trivially onto R ! To see that g is continuous. let p E flS and let A E Swithg(p) E An. Foreach23 E W\(g(p))pickby (b)some f(23) E Sandsome D ( 8 ) E g(p) so that for all e E W, S\ f ( S ) E C or S\D(S) G C. Let f ( g ( p ) ) = A. Pick by (a) some 3 E PI(%) such that S = USES f ( 8 ) and let 8 = w(g(p)). Then S\A E Uspaf ( 8 ) . Let D = D(21). Then D E g(p) s p. We claim that g[Dl E Au. So let q E and suppose that A $ g(q). Pick by Comllary 3.9 some r E fiS such that g(q) U (S\A) E r and pick 8 E 8 such that f ( B ) E r. Then one cannot have S\ f (21) E g(r) = g(q) so S\D(S) E g(q) E 9, amtradiction.

nS,,

. .

Now we bring the algebra of S into play. The operation + on filtas is a natural generabization of the operation. on ultrafilters.

Definition 2133. Let S be a semigroup and let A and 8 be f i l m on S. Then A*S=(A~S:(XES:X-'AE~)EAJ.

Remark 2134. Lcr S be a semigmup and let A and 8 befihcrs on S. Then A * 8 is a filter on S.

Mnition 21.35. LM S be a semigroup and let Wbe a set of filters on S. If for all A and 21 in 8,there is a unique C E R such that C A * 8.then define an operation .

Theorem 21.36. Ln S be a discrete semigmup and let ((p, Y) be a semigmup com5-'-' [ ( x ) ] : x E Y}. Then W is a set offilters on S pactijcation of S. Let W = satisfying

{n

(a) given any choicefunction f for R there exists 5 E PI(%) such that S = UI~S f (4).

.(b)

given distinct A and 8 in R ! then &?.A

E

A and B .E 8 such that whenever

C~ReitkrS\A~CorS\B~e,and

(c) givmany A M d S i n W t k r e i s a w r i q w C E WsuchthatC cA*8. Funhes with the quotient topology and the operation .on !R thefunction h : Y + W F-'[[x]]is an isomorphism anda homeomorphism.

defined by h ( x ) =

n

Pmof: Conclusions(a)and (b) follow fromTheorem21.31. To establish(c),let A, 8 E 14. We note that it is enough to show that there exists C E W such that C g A * 8. (For then, i f 9)E W aod D # C pick by (b) some C E C such that S\C E D . Then C E A * 8 so S\C E D\(A * 8).)Pick x. y E Y such that A = n F - ' [ { x ) ] and 8 = F - L y ) ] and let C? = P L [ ( x Y ) ]To . see that C G A * 33. let A E C . Let U = Y\aS\Al. Then U isopenin Y andxy E U . (Forifxy $ U picksomep 6 S\A such that F(p) = xy. Then A $ p so A 4 F-' [ [ x y ) ]a, contradiction.) Since xy E U and Y is right topological, pick a neighborhood V of x such that V y G U . Then by Remark 21.29, (p-'[V] = ? ' [ V ] n S E A. W e claim that (p-'[v] c 1s E S : S-'A E 8 )SO that A E A * 8 as required. To this end, let s E (p-'[Vl. Then (p(s)yE U.Since b y , is continuous pick a neighborhood W of y such that (p(s)W G U. Then by Remark 21.29. (p-'[Wl E 8 so it sufficesto show that (p-'[~E l K I A . Lett E (p-'[Wl. Then st) = (p(s)(p(t)E U so st 6 S\A, i.e., st E A as required. B y Theorem 21.31, the function h : Y + W defined by h ( x ) = n p ' [ ( x ) ]is a homeomorphism from Y onto W. To see that it is a homomorphism. let x , y E Y . W e havejustshownthath(xy) = nF-l-'[(xyll E h(x)*h(y)sothath(x).h(y)= h(xy).o

n

n

n

The following theorem tells us that the description o f semigroup compactifications provided by (a), @). and a weakening o f (c) o f Theorem 21.36 in fact characterizes semigroup compactifications. Theorem 21.37. Let S beadiscretesemigroupMdIrfW be aset -of

(a) given any choice function f for W there exists 5 E PI(%) such that S = UA~I f ( 4 (b) given distinct A and 8 in IS there exist A E A and B E 8 such that whenever C E R either S\A E C or S\B E C. and (c) given any A and 8 in W there exists C E Wsuch that C E A * 8. Then infoct for any A and 8 in W there exists a unique e E W such that C E A * 8. Also, with the quotient topology and the operation 3 ' is a compact right topological semigroup and thefunetion g : fJS + W &fined by g ( p ) p is a continuous homomorphismfrom fJS onto W with g[Sl A(W) and g[S]dense in %. In particular: ( g p . W ) is a semigroup compactifcation of S.

.,

Proof: As in the pmof o f Theorem 21.36 we see that statements (b) and (c) imply that for any A and 8 in W there exists a unique C 6 W such that C G A * 8. so the operation . is well defined. B y Theorem 21.32, with the quotient topology W is a

.

21.5 Uniform Compactifications

445

mmpact Hausdorff space and g is a (well defined) continuous function from BS onto

W. We show mxt that g is a homomorphism. To this e n d let p, q E 0.7. To see that E ( P . q ) = g ( p ) g(q). we show that g ( p ) g ( q ) G P .q. Let A g ( p ) .d q ) . A E g ( p ) g ( q ) so B = (s E S : S - ' A E g ( q ) ) E g ( p ) G p and if s E B , then S - I A E g(q) E q SO ( S E S :s-IA E q ) E p. That is, A E p .q as required. Since g is a homomorphism onto W. we have immediately that the operation. on W is associative. By Exercise 2.2.2,'Jt is aright topological semigmup. Lets E S. To see that A,(,, is continuous, note that since g is a homomorphism g o As = A,(,, o g. Let U be open in W and let V = A (,,-'[U]. Then g-'[V] = (g o A,)-'[u] is open in pS. Then V = 'Jt\g[f3~\~-' is open in W. Since S is dense in pS and g is continuous, g [ S ] is dense i n s .

.

.

[vP]

21.5 Uniform Compactifications Evny uniform space X has a compactification (@, y,X) defined by its uniform structure. For many uniform spaces X which are also semigroups. (@. y,X) is a semigroup compaclilicationof X. It has the property that the map ( a ,x ) r @ ( a ) xis acontinuous map from X x y.X to y.X, and y,X is the maximal semigroup compactification of X for which this holds. This is m e for all discrete semigroups and for all topological groups. If X is a discrete semigroup, then y.X % pX and so the material in this section generalizes our definition of pX as a semigroup. We first remind the reader of the definition of a uniform structure. X x X, we define U-I and U V by DeBnltion 21.38. Let X be any set. If U. V U-' = [ ( y . x ) : ( x . y ) E U) and U V = ( ( x ,y ) : ( x , z) E V and (z, y ) E U forsome z E XI. We may use U' to denote UU. A will denote the diagonal ( ( x ,x ) :x E XI. A uniform srmcfure (or uniformity) PI on X is a filter of subsets of X x X with the following properties: (1) A U forevery U E P1: (2) for every U E e(, U-' E P1; and (3) for every U E U ,there exists V E PI for which v2G U . Let PI be a uniform StrucNre on X and let U E PI. For each x E X, we put U ( x ) = ( y E X : ( x , y) E U),and,foreach Y g X,weput U[Yl = U y s V U ( y ) . P1 generates a topology on X in which a base for the neighborhoods of the point x E X are the sets of the form U ( x ) ,where U E PI. If X has this topology, (X. P1) is called a uniform space and X is called a uniformizable space. If (X. P I ) and (Y,V ) are uniform spaces, a function f : X + Y is said to he uniformly continuousif, foreach V E V .thereexists U E P1 suchthat (f ( X I ) . f (x2)) E V whenever ( x ~ . x zE) U .

Mehic spaces and topological groups pmvide impntant examples of uniformizable spaces. If ( X , 4is a mMic space, the filter which has m base the sets of the form ( ( x , y) E X x X :d ( x , y ) < r ) ,wherer r 0,isauniformsaumveonX. Thisexampleincludes all discrete spaces. If X is disaete, it has the trivial uniform structure P1 = (U E XxX:AGU). If G is a topological group. its topology is defined by the right uniform shuchtrc which has as base the sets ( ( x ,y) E G x G :xy-' E V J .where V denotes ancighborhood of the identity. In this section. we shall assume that we have assigned this uniform srmclure to any topological group to which we refer. It is precisely tfK completely regular topological qaces which are uniformizable. Suppose that X is a space and that C a ( X ) denotes the subalgebra of C ( X ) consisting of the real-valued functions in C ( X ) . X is said to be completely regular if. for every closed subset E of X and every x E X\E. there is a fuoction f E C a ( X )for which f ( x ) = 0 and f [El = (I). For each f E C a ( X ) and each E z 0,we put UI.. = ((1,y) E X x X : If ( x ) - f (y)I -z E ) . 'Ihe finite intersections of the sets of the fam UJ,, then provide a base for a uniform s t ~ c t u r eon X. In paaicular, every compact space X is uniformizable. In f a X has a unique uniform structure given by the filter of neighborhoods of the diagonal in X x X (see Exercise 21.5.1). In the next lemma we establish a relation between compactifications of X and subalgebras of Ca(X).

Lmma 21.39. Let X be any topological space and let A be a norm closed subalgebm of C d X ) which contains the constantfunctions There is a compact space Y and a mnrinuoutfunnion 9 : X + Y with the pmperty thor @[XI is dense in Y and A = ( f E C=(X) : f = g o @ for some g E CR(Y)}. Thc mopping @ is an embedding 8 for every closed subset E of X and every x E X\E. there exlrrr f E A such that f ( x ) = 0 and f [El = (1).

Pmqf For each f E A, let IJ = (t E P : It1 _I Ilf 11). Let C = X f e A . l / and let 9 : X -r C be the evaluation map defined by # ( X I ( f ) = f ( x ) . Let Y = cCc 41x1. For each f E A, let n, : C -+ I f be the projection map. If f E A, we have f = r r f l y o 9. Conversely. let G = [g E C R ( Y ) : g o 9 E A). By Exercise 21.5.2, G is a closed subalgebra of C d Y ) which contains the constant functions. We claim that G also separates the points of Y . To see this, let j and Z be distinct points in Y and choose f E A such that y ~ # 21. Since n ~ l oy @ = f, we have nfi, E G and n J l y ( j ) # xfly(Z). So G separates the points of Y and it follows from the StoncWeietshass Theorem that G = CR(Y). Thus A = ( f E C a ( X ) : f = g o 9 for some g E CR(Y)]. Suppose now that, for every closed subset E of X and every x e X\E, there exists f E A such that f ( x ) = 0 and f [El = ( 1 ) . Then $I is clearly injective. To see that 9 is an embedding, let E be closed in X and k t y E @[X]\@[E].Pick x E X such that y = # ( X I and pick f E A such that f ( x ) = 0 and f [ E l = ( 1 ) . Pick g E C ( Y ) such that f = 8 09. Then g ( y ) = g ( 9 ( x ) )= Oand g [ 9 [ E l ]= (1) soy 4 c!@[El.

The following r e d t estabfishes the existence of the ~ t o d s compactification h f a any completely regular space. (Of course, since any subset of a compact space is completely regular. this is the greatest possible generality for such a result.)

Tbto~lll21.40. Let X be a completely regular space. Then X can be embedded in a compacr space B X which has thefollowing universal property: whenever g : X + C is a continuous fwtction fmm X ro a compact space, there is a conrinuousfwrction : BX + C which is an e x t e ~ ~ i 4 o n8 .

Pmof: Let A = Cn(X). By Lemma 21.39. X can be densely embedded in a compact space BX with the popclty that every function in A extends to a continuous hrnction in C&X). We shaU regard X as being a subspace of BX. Let C be a compact space and let F denote the set of a l l continuous functions £mm C to [0,1]. We define anatural embedding i : C -r [O, 11 by putting i ( u ) ( f )= f ( u ) foreveryu ~ C a n d e v q ' Ef F. Suppose that g : X + C is continuous. Thm. for every f E F, f o g has a continuous extension ( f o g r : pX + [O. I]. We define h : BX -+ I] by h(y)(f) = ( f o g r ( y ) for every y E p X and every f 6 F. We observe that h ( x ) = i(g(x))whenevirx E X. Since Xis dense in B X . hfX1 isdense in h[pX].So h[BX]c i [ C ] ,because h [ X ]E i[C]and i[Cl is compact. We can therefore dcfim g by i-'oh.

'

a=

(p

Recall that a topological compactification ((p. Y ) of a space X has the p p m y that is an embedding of X into Y.

Theorem 21Al. Let ( X , P1) be a uniform space. There is a topological compacnJcation (#. y.X) ofX such thar it is precisely the uniformly com'nuousfunctionr in CR(X) which have continuous extensions to y.X. (That is, [f E C R ( X ): f = g o q5 for some g E Cx(y,,X)) = [f E C R ( X ): f is uniformly continuous).)

Pmof: It is a routine m a w to prove that the unifody continuous functions in CR(X) are a norm closed subalgebra of CR(X). The conclusion thm follows from Exercise 215.3 and Lemma 21.39. 0 S i c e q5 is an embedding. we shall regad X as beiig a subspace of y.X. Ihe compactification yX . will be called the uniform compactiyicationof X . It can be shomthatall possible topologicalcompactificationsof acompletely regular space X arise in this way as uniform compactifications. We now see that y,X may be very large.

Lemma 21.42. Let ( X , U)be a uniform space. Suppose thar there exist a sequence (x.)~,inXnndaserU~Usuchtharx,~U(x.)whencverm n.f Letf :N-+X BPd + yX . is an embedding. be defined by f ( n ) = x. Then j:

P m f : We claim that, for any two disjoint subsets A and B of N . cCyMxf [ A ] n cL,x f [ B ] = 0. To see this, we observe that U[(x. : n E A)] and (x. : n E 8 )

are disjoint, and hence that thcrc is a uniformly wntinuous function g : X + [O, 11 E A ) ] = (0) and g[(x. : n E B ) ] = [I) (by Exercise 21.5.3). By 'lbeorem 21.41. g can be extended to a continuous fuoftion : yX . + B. Since g[ccfix f [ A ] ] = (01 andZ[~&x f [ B I ]= (1). we have C&.X f [A1n C&X f [ B l = 0. Now suppose that u and v are distinct elements of PN.-We can choose disjoint subA and B of N such that A E-u and B E v. Since f ( u ) E c&x( f [ A ] )and F(u) E d,x( f [ B ] ) it , follows that f ( u ) f (v). So f is injective and is therefore an embedding.

for which g[{x. : n

a

+

We now assume that X is a semigroup and that ( X , V )is a uniform space. We shall give sufficient conditions for (6.y.X) to be a semigroup compactification of X. We observe that these conditions are satisfied in each of of the following cases: (1) X i s discrete (with the trivial uniformity); (2) ( X ,d ) is a metric space with a metric d satisfying d(xz, yz) 5 d ( x , y ) and d ( z x , zy) I:d ( x , y ) for every x , y. z E X: ( 3 ) X is a topological p u p . Notice that requirement (ii) below is stronger than the assertion that p, is uniformly continuous fw each a E X.

Theorem 21.43. Suppose that X is a semigroup a d that ( X , P1) is a uniform space. Suppose rhar tk nvofollowing conditions are satisficd: (i) For each a E X, An : X + X is uniformly continuous and (ii) for cach U E U, them aists V E V such that for every (5, t ) E V and every a E X , one has (sa, t a ) E U .

Thrn we can define a sernigmup operation on yX . for which (6. y.X) is a semigroup compactificarion of X.

P m f For each a E X, thcrc is a continuous extension 1 . : y.X -P y.X of A. (by Exercise 21.5.4). We put ay = M y ) for each y E y.X. Given y E yuX, we define ry : X -r y,X by r y ( s ) r sy. We shall show that ry is uniformly continuous. We have observed that the (unique) uniformity on y,,X is generated by [Up., : g E Cdy,,X) and 6 r 01, where Ud.< = [(u,U) E y,,X x yUX: Ig(u) g(u)l c 6). So let g E C R ( ~ . X )and 6 w 0 be given. Now glx is uniformly continuous by Theorem 21.41 so pick W E P1 such that Ig(s) g(r)l < 5 whenever (s, t ) E W . Pick by conditia (ii) some V E P1 such that for all a E X and all (s, t ) E V , one has (so, t a ) E W . Then. given (s, 1 ) e V , one has 6 lg(syl- g(ty)l= lim (&a) g(ta)( 5 -.when a dewtes anelement of X. andso a-Y 2 (ry(s),~ ~ E ( Ug.r. 0 ) from Exercise 21.5.4 that ry can be extended to a continuous function -r y :Ity.Xfollows -c y.X.

-

-

We pow define a binary operation on yX . by putting x y = f y ( x ) for every X , y E yuX. We observe that for every a E X, the mapping y H ay from yX . to itself is ~ ~ ~ t i n u oand, u s ; for every y E yuX. the mapping x I+ x y from y.X to itself is continuous, because these are the mappings and Ty respectively. It follows that the operation defined is associative because. for every x , y, z E y.X,

where s. t , and u denote elements of X. So x(yz) = (xy)z. Thus y.X is a semigroup compactification of X. Wenow show that the semigroup operation on y.X

0

isjointly continuous on X x y.X.

Theorem 21.44. Let ( X , U ) satisfy the hypotheses of Theorem 21.43. Then the map (s,x ) H sx is a continuous mapfmm X x y.X to y.X.

Pmof:Lets E X and x E yuX. Choose any f E &(y.X) and any e > 0. Since f i x is uniformly continuous, there exists U E e( such that If ( s ) - f @')I c whenever (s. s') E U. By condition (ii) of Theorem 21.43. there exists V E 91 such that (st. s'r) E U for every (s. s') G V and every t E X. I f x' E y,X. then If (sx')-f (s'x') 1 = lim, If (st)- f (s't)l 5 c whenever (s. s') E V . Thus, ifs' E V ( s ) I-,

-

andx' E l ; ' [ ( y E yX . : If(=) f(y)I c ell. we have I f ( s x ) If (sx') - f (s1x')J5 E . SO If ($1) - f ( S ' X ' ) ~ < 2e.

- f(sx')I

c sand 0

Theorem 21.45. Suppose that ( X . 91) satis~5esthe condiIiom of Theorem 21.43. Suppose that Y isa compact right topological semigmup and that h : X -t Y isa uniformly continuous homomorphism with h [ X ]g A ( Y ) . Then h can be wtendedtoacontinuous homomorphism ii : y.X + Y.

Proof:By Exercise 21.5.4. h can be extended to a continuous function Si : yX . For every u , v E y.X, we have

+ Y.

where s and r denote elements of X. So ii is a homomorphism. We now show that, if X is a topological group, yX . is maximal among the semigmup compactifications of X which have a joint continuity propem. Theorem 21.46. Let X be a topological gmup and let (0. Y ) be a semigmup compactificafion ofX. Supposerhnrthcmapping ( x . y) r B(x)yfrmn X x Y to Y iscontinuous. Then there is u continuous homomorphism : y,X -t Y such that 0 = &.

Pmof: We shall show that 9 is uniformly continuous. Ln e denote the identity of X and let 4 be the set of neighbob& of e. Suppose that 9 is not uniformly continuous. Then there exists an open neighborhood U of the diagonal in Y x Y such that. for every V E &. tlmc are points sv. tv E X satisfying svt;' E V and (B(sv).O(tv))$ U. The net ((O(sv),B(rv)))vc.ue has a limit point ( y ,z ) E (Y x Y)\U. Now sv = evtv. where. ev E V. and so B(sv) = B(ev)B(tv). Since e v + e, 0(ev) + O(e). and B(e) is easily seen to be an identity f a Y . So our continuity'assumption implies that y = z, a contradiction. Thus 0 is uniformly continuous and therefore can be extended to a continuous homomorphism g : y.X -t Y (by Theorem 21.45). Thcorem 21.45 shows that y. X can be identifiedwith pX if X is a dimete semigroup with the hivial uniformity. However, there are many familiar examples in which y .X and pX IT different Suppose. for example. that X = (R. +). We shall see in Theorem 21.47 that pIR cannot be made into a semigroup compactification of (R, +). It is natural. given a nondiscrete topological semigroup such as (R. to anempt to proceed as we did with a discrete semigroup in Theorem 4.1. And, indeed, one can d o the first part of the extension the same way. That is, given any s E S,one can define t , : S + S E .pS by &(r) = st. Then by Theorem 21.40, there is a continuous function L, : pS + pS such that L,IS = L, so one can define for s E S and q G fiS. sq = L&). As before one still has the function r, : S + pS defined by rq(s) = sq. However. in order to invoke Theorem 21.40 to extend rq. OM would need rq to be continuous. We now see that this condition does not hold in many familiar semigroups, including (8,+). Note that in the following theorem we do not assume thatthe operation on pS is associative.

+).

Theorem 21.47. Let S be a semigrnup which is also a metric space with a metric 8 for which 8(sr,st') = 6 ( t . t') and 8(ts, 0 s ) = b(t.1') for every s , t , r' E S. Suppose that S contains a sequence ( y , ) z l such that 8(y,,,, y.) > 1 whenever m # n. Let. be a binary opemtion on BS which Mends the smigmup operation of S and has the pmperty that the mapping € n x .{ from pS to itself is continuousfor every x E S. Then then is apoim p E BS such that the mapping x n x .p is discontinuous at every non-isohfed point x of S. Pmof Lclpbeanylimit pointof ( y . ) z , andletx € Sbe thelimitofasequence (x,)z, of distinct elements of S. We may suppose Ihatd(x,. x.) e for every m , n E N. For every m , m', n , n' E N with n # n', we have

4

~(x,Y..x,+Y.~ 2 S(X,Y..X~Y,~)

-s(x,Y.,.x,,Y,,)

=

w~..Y.,) - ~ ( x ~ . x>~T.I , )

Let A = (x. y. : m < n and m is even) and B = (xmyn: m c n and m is odd). We claim that A and B have no points of accumulation in S. This can be seen from the fact thah for any s E S,[t E S : 8(s, t ) < cannot contain two points of the form xmy. and xm,y.r with n # n', and can therefore contain only a finite number of points

a]

inAUB.ThuscCsA=A,clsB=BandclsAncLsB=0.

21.5 Uniform ~

45 1

f i c 1 1 t i ~

lt f o l l m from Urysahn's Lnnmathat h ? acolltinums fimction f : S + 10.11 for which f [A] = (01 and f [Bl = [I]. Let f :B S [O,11 denote t k continuous exteasion o f f . lhen 7(x, p) = 0 if m is even and f (x, . p) = 1 if m is odd. So the sequence (x, p):, cannot converge to x . p. 0

.

-

-

We shall aow look at snm of the proprtiuof y.P when Rdmotes the topological grwpformedby therePlnumbersunderaddition.Wefirstnotethat,9Zcanbennbedded topologically andalgebraicallyinyYW.By 'lbeorcm21.45.theinclusimmapi : Z B can be extended to acontinums homomorphism 7 :BZ -+ y.B. We can show that 7 is injective and therefore an embedding, by essentially the same argument as the one used in the proof of Lemma 21.42. We shall therefore assume that BZ G y.B by identifying BZ with &9Z]. This identifies BZ with ce,~(Q, because = ce,~(i[4). The following theorem gives an expmsion for an element of y,R analogous to the expression of a real number as the sum of its fractional and integral parts. llnorrm21.48. Every x E y.B can be expressed uniquely ar x = t~[O.l)andz~BZ

r

+ z, where

-+ T = R/Z denote the canonical map. Since n is uniformly continuous, n can be extended to a continuous homomorphism if : y.B + T. We claim that F(x) E 0 if and only if x E BZ. On the one hand, T(x) = 0 if x E BZ, because n[Z] = (0). On theother hand. suppose that x $! BZ. Thm there is a continuous function f : y.R -+ 10, I] for which f (x) = 1and f [Zl = (0). Since fiz is uniformly continuous, there exists 8 s 0 such that f [UnGz[n - 8, n 611 g [O, f]. Sox f C&R (UnEZ[n 8, n 81 and therefore x E ct,~(U,,~(n a, n 1 6)) and E(x) # 0, because 0 $? CLT n UeEZ(n 6, n 1 a)]. x. Let r denote the unique number in [O. 1) for which n(t) = F(x). If z = -t d m F(z) = 0 and so z E BZ. Thus we have an expression for x of the typc required. To pmve uniqueness. suppose that x = t' z', where t' E [O. 1) and z' E BZ. Then A(*) = n(rl) and so t' = t and therefore z' = -t x = z.

Pnwf Let n : R

-

-

+

I

+

+

+

+ -

+ + -

+

+

Theorem 21.48 allows us to analyze the algebraic structwe of y.R m terms of the algebraic shoeturr of BZ.

C o m w 21.49. The 1 4 ideals of yJi have theform R + L, where L is a lefi ideal of PZ; and the right ideals of y.R lurve thrform R

+ R. where R is o right ideal of BZ.

Pmof: llis follows easily from Theorem 21.48 and the observation that R is contained in the center of y.R. 0

We omit the proofs of the following comUaries, as they arc easy consequences of Theorem 21.48.

Corollary 2151. Every idempotem of y.R is in p Z C m l l a r y 2152. y.R has 2' minimof Ieji ideals and 2' minim01 right ideals, each conmining 2' idempotmrs.

P m f This follows from Corollary 21.49 and Theorem 6.9. Exerdse 21.5.1. Let ( X , U ) be a uniform space. Show that every U U is a neighborhood of the diagonal A in X x X. If X is compact, show that 'U is the filter of all neighborhoods of A in X x X. . Exercise 21.52. Let G and Y be as in the proof of Lemma 21.39. Rove that G is a closed subalgebra of Q ( Y ) which contains the constant functions.

Exercise 21.53. Let ( X . U ) be a uniform space. Suppose that E and F are subsets of X and that U [ E ln F = 0 for some U E U.Show that there is a uniformly continuous function f : X + [O. 11 for which f [El = ( 0 )and f [ F ]= (I). in U by putting Uo = U i l U-' and choosing (Hint: Choose a sequence (U.).,, U,, to satisfy U. = LS;' and U: E U.-1 for every n > 0. Then define subsets E, of X for each dyadic rational r c [O, 11 with the following properties:

Eo= E , El = X\F. and for each n E Wand each k

E

(0, 1,2....,2"

- I), U,[E r1 i

Er

I 4-.

This can be done inductively by putting Eo = E and El = X\F, and then assuming that E r has been defined for every k E {0,1,2,. . ..2"). If k = 2m 1 form E F (0.1, . ,2" - I], E l can be defined as U.+I [ E $ ] . Once the Sets E, have been

..

+

*"+I

constructed. define f : X -+ [O, I] by putting f ( x ) = 1 if x E F and f ( x ) = inf{r : x E E,) otherwise.)

Exerdse 21.5.4. Suppose that ( X , U )and (Y,V )are uniform spaces and that 8 : X + Y is uniformly continuous. Show that 8 has a continuous extension : yX . + y.Y. (Hint: Let A = C&.Y). The mapping b : y.Y + *W defined by 9 ( y ) (f ) = f ( y ) is then an embedding. Iff E A, f i r is uniformly continuous and so f o 8 is uniformly continuous and has a co?tinuous extension f : y.X -r W. Let $ : y.X + @[y.Y] be defined by $ ( x ) ( f ) = f ( x ) . Forevery x E X. @ ( x ) = +(8(x)).)

Exerdse 215.5. A uniform space ( X , U ) is said to be totally bounded if, for every U E U , there exists a finite subset F of X such that X = U ,, U ( x ) . It ( X . U) is not totally hounded. show that y.X Lemma 21.42.)

contains a topological copy of BN. mint: Apply

Exercise 215.6. Let X be a topological p u p . Show that f E C R ( X )is uniformly continuous if and only if the map x H f o A, fmm X to C R ( X )is continuous. where the topology of C R ( X )is that defined by its norm. (For this reason, y,X may be called the left uniformly continuous compactification of X and denoted by P'UC(X).)

Notes

Notes It is customary to define the LAC. W A P . A P , and 4 A P compactifications only for semitopological semigroups S. Of course. if S is not semitopological. then the map 41 cannot be an embedding (because qtCSl is semitopological). It was shown in [I501 (a result of collabwation with P.Milnes) that there exists a completely regular semitopological semigroup S such that 41 is neither one-to-one nor open and that there exist completely regular xmigroups which are neither left nor right topological for which 4, is one-to-one and other such semigroups for which 41 is open as a map to VIISI. In keeping with our standard practice, we have assumed that d l hypothesized topological spaces are Hausdorff. However. the results of Section 21.1 remain valid if S is any semigmup with topology, without any separation axioms assumed. We defined weakly almost periodic and almost periodic functions in tmns of extendibility to the W A P - and AP-compactifications. Itis common to define them in terms of topologies on C(S). One then has that a function f € C(S) is weakly almost periodic if c.!{ f o p, :s E S) is compact in the weak topology on C(S) and is almost periodic provided ct[ f o p, : s E S) is compact in the norm topology on C(S). For the equivaknce of these c h ~ ~ ~ ~ t e r i z a with t i o nour s definitions see [MI. Theorem 21.7 and Corollary 21.8 are due to J. Berglund, H. Junghenn, and P.Mines in [39], where they &it the "main idea'' to J. Baker. Thequivalence of statements (I) and (3) inTheorem 21.18 is due to A. Gmthendieck [I 121. Theorem 21.22 is due to W. Ruppcrt [217]. Thealgebraic propertiesof W A P ( N ) are muchhardermanalyzethanthoseoffi(N). It is difficult to prove that W A P ( N ) contains more than o m idempotent. T.West was the first to prove that it contains at least two [245]. It has since been shown, in the work of G. Brown and W. Moran (541. W. Ruppert [220] and B. Bordbar [Sl], that WAP(N) has 2' idempotents. Whether the set of idempotents in WAPfW) is closed, was an open question for some time. It has recently been answered in the negative by B. Bordbar and I. S. Pym [531. The results of Section 21.4 arc from [37] and were obtained in collaboration with I. Berglund. In 1371 a characterization of the WAP-compactification as a space of filters was also obtained. Theorem 21.40 is due (independently) to E. &ch [601 and M. Stom [227]. See the notes to Chapter 3 for more information about the origins of the ~tone&ch compactification. Theorem 21.48 is due to M. Filali [90], who proved it for the more general case of

YX. Suppose that S is a semitopological semigroup and that f E C(S). It is fairly easy to prove that f is a W A P function if and only if {( f a A,)- : s E S] is relatively compact in {F:g E C(S))for the topology of pointwise canvqence on C(pSd). It follows from [81. Theorem IV.6.141 that this is equivalent to (f o I, : s E S) being weakly relatively compact in C(S).

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L i s t of Symbols

A-s76 Ax-' 76 A* 55 An 441 A* 76.93 A'(p) 76.77.93 [A]" 91 [A]<' 91 z53 A*S443 AP 281,372 an ideal, 281 compactificatioa. 428 AP(S)428 as quotient. 439 K 54 C(p) 113.262 a film, 114 C(S) 436 C ( X ) 70 C*-embcdded 155 C+ 301 C- 301 Cp 179 chains of idempotmts. 181 contains frce group on 2' generators. 181 doesnot meet K(BG). 181 in K(BG), .. . 182 minimal left and right ideals, 180 Cp(BG) COPY of Cq(BN.183 Cp(BN)185 C d X ) 71,446 cC K(W) 172 c 66

472

Li~rof Symbols

h, 207 W-map 148 W 107,150,151,361 asanoid. 111 closure of K O , 171 containsall idcmpotentsof (#IN,+), 109 contains infinite chains of i d e m p tents. 110 copiesof, 115,117,148.150,151. 152.183 homomorphic images of W, 108 bmomorphisms. 108 IP set 321 if and only if, 321 IF" set 321.339, 376 and measure preserving systems, 408 dynamical. 408 examples, 337 if and only if. 321,326 may never translate to P. 331 not a dynamical IF'* set, 408 sums and products in. 327 K ($1 a group if commuIative. 23 a p u p . 41 elements not cancelable. 30 elements of, 30 equal to SeS, 24 homomorphic image. 30 union of minimal left ideals, 21 K(T) equal to K ( S ) n T.29 K (W) closure not a left ideal of It.n. 171 K(BS) 85.291 and piecewise syndetic sets, 86 and syndetic sets, 85 closure as ideal. 87 closurr. 86 equal to K W G ) n BS. 130 not closed, 169

K(BS.:l. and K @ S , o), 274 K ( B N +) 87 closure. and Banach density. 414 K ( B N .) KG?& .) fl K V N , +) = 0.263 hits c l K(#IN.+I. 328 misses K(pW. +). 263

KO

contains right cancelableelements,

- 169

K(BW) and sums of elements outside (K(BW). 172 US)7 L, 258.272 lim 63 O'i PAC compgctification, 428 L M C ( S ) 428 PUC?453 (m,p. c) sets 318.319 M 328,339 W2 N' 69 contains copy of o.218 copies of N* in N*, 217 not a copy of W. 152 topological center empty, 170 (W, +) and (-W*. +) as left ideals of (BZ, +), 85 ( N +) center. 109.134 topological center, 110 (W. .) center. 109,134 topological center. 110 (W, V) 3.80.324 weakly left cancellative. 82 (W, A ) 3.80.324 (w*, .) -), 218 copies of (N*.+) in (W. -PI* 330 naFU 98

List of Symbols p-limit 63.74 ~+9#~.4275 P ~ R 9K164 PP((xdZ,) 341 W ( x a ) z l ) 341 P ~ R 164 K ~ p-lim 63.64.65 existence, 63 -p 85 *(A) 2 U) 95 (Pf (N, r-tdoring 101 R(S) 7 r, 258 r, 272 R+ 2 & 268 (R. A) 3 spm((x,)po=l)347

S* as ideal of BS, 84 as left ideal of BS. 83 as right ideal of BS. 84 as right ideal of BS. 83 as subsemigroup of BS, 82 minimal left ideals. 84 minimal right ideals, 84 smallest ideal, 84 s-'A 76 Sd 432 supp 107.11 1 -s+A76 6 242 6A S compactification. 428 I A P ( S ) 428 df 242 T 150,206,403 T 85 U, 207 U, 52 cardinality, 68 size of, 67 W(V) 381

W(V; v ) 381 WAP mmpaftification as quotient, 435 compactification, 428 WAP(S) 428 smallest ideal, 437 WAP(N) 437 x+ 268 x- 268 xn 11 Xd 268 X. 207 'X 4.8.32.42 a topological semigroup, 32 right topological semigroup, 32 Y, 207 2,207 subsemigroupof (BN..), 209 (Z, +) center, 107 left ideals, 85

Z*

copies of Z ' in Z*,218

BDS3.56.57 size of, 66.67 lopological properties, 53 DS as commutative semigroup, 81 as noncommutative semigroup, 81 as semigroup. 73 as universal semigroup compactification, 75 binary operation, 72 gs.p 113 BSd 432 BT copies of BT inASa. 217 identified with T. 57 additive. 107 topological copy, 67

BZ additive. I07

474

Lia of Symbols

r %, 240 and finite sums. % leti ideal of (BN,.).%

not semigmup under not semigroup, 242

+. 112

Y. 447 8 338 A 134.413 finite sum and upper density, 422 A* 413 ideal, 414 A'(K +) left ideal of (BN,.), 415 A;(& .) as I& ideal, 414 A:(& as right ideal 414 ( ? i s wS) 427 r-uniform finite intnsection propa(y 68 K-uniformulhafilter 52 A(S) 33.38.78 Ax 7 1.44) 362 p'(A) 362 Px 7 a-eompact sets 61 u~(P)346 0 284 $b 259 Wi 425 w2

.,

nglac-%~(nN) copy of IA, 149 nten cbn(3"N

a copy of IH. 152

n,. c e g z ( 3 w a copy of IN,152 n,.~c ~ s z ( ~ ~ z ) a coov of W.152

5 15 5~ 15.193

-
.

Index

aftion

wasme pnscrving, 407 monotone. 408 almost disjoint family 238 almost left invariant 121 almost periodic function 436.439 alphabet 281 Anthony. P. 275 anti-homomorphism4 anti-isomorphic4 anti-isomorphism 4 Arens. R. 88 arithmetic progression 280 associative 3 associativity of (BS, .)73 Auslander. J. 47.411 Axiom of Choice 69 Bake stt 361.387 Baku. J. viii, 88,135,204,221,453 Balcar. B. 275.411 Banach density 412 base filter, 48.52 Baumgartner, 1. 102 Bergelson. V. viii, 103,221,295.319. 338,367,368,393.41 1,424 Berglund. J. viii. 47.135.453 bicyclic 22 binary expansion 107 Blass, A. viii, 103. 185,234,257,393 Boolean algebra 52.71 homomorphism, 52 Budak. T.135 Butcher. R. 88

cancelabiity and K(S), 30 and K(BS), 169 in pZ. 176 cancelable elements in /3S 175 cancelable elements in p Z 175 cancelableelements of S cancelable in PS, 158 cancelable elements 115 cancelable right, 161 cancellative 7 pS noz 158 weakly left, 122,323 weakly righr 122 c d i a l i t y of U, 67 cardinality of p D 67 cardinality of closed sets in /3D 67 cardinal 2 Carlson's T h e m 381.386 Carlson. T. 386.393 tech, E. 69 center 7.38. 80 (Z, +). 107 of P O P , 129,130 o f p + p N + p . 130 of U.(S) empty, 124 of /3S equal to that of S. 125 central Sets Theorem 279.294 commutative, 284 noncommutative, 287 satisfied by noncentral set, 289 cenhal* sets 297 examples, 337 centralizer nowhere dense in U,(S). 124 cenaal86.279.284.320

combinatorial characterization. 293 dynamically, 404 equivalent to dynamically cenwl, 406 examples, 320 not changed by adding 0,301 often translates to central. 331 chain of ideals increasing, 133 characteristic function 43 Chou, C. 135 Christensen, 1.47 circle group 150.206.403 Civin. P. 88 Clay, J. 338 clopen 55 closed 3 closure and p-lim, 65 of subsemigroup. 40,167 collectionwise piecewise syndetic 290, 291,293,294,328 coloring 101 columns condition 302,303,305,325 over Q. 303 over Z. 303 over field, 316 comhinatorially rich ultrafilter 339 commutative semigroup S contained in center of BS. 80 commutative 7 compactification 55 A P . 428 PMC.428 d A P , 428 WA9.428 and algebra of functions, 446 semigroup, 75.77.431 Stonezech. existence, 447 Stondech, 56,57,69 topological, 447 compactness 101,281,283,300,318

compact ifiplim exists. 65 compatible 237 completely regular 446 concatenation 4 continuity 66 and p-Lim, 65 convergent ultrafilter 197 correspondence principle 417 Davcnpolt. D. viii. 88. 135.424 Day, M. 88 dense 236 density 134 Banach, 413 Deuber, W. viii. 318.319.338.367 direct product 5 direct sum 5. 152 of countable groups, 152 distinct finite products 115 semigroup isomaphic to H.115 of x sequence, 117 distinct finite sums 115 distributive laws in N*,267 in BZ.258 in BZ,264,265 in W,265 doubly thin 266 Douwen. E. van 135.275 dynamical IP* set 408 contains fw(xdE1)x F P ( ( Y " ) ~ 409 ~), dynamical system 397 minimal. 398 dynamically e e n w 404 equivalent to central. 406 El-Mabhouh, viii. A. 185.275 Ellentuck topology 387 Ellentuck. E. 392.393 Ellis' Theorem 46 Ellis, R. 32.46.47.88.41 1 embedding 55

enveloping semigroup 42,400 as image of BS, 400 ps as, 401 of actions on the circle group. 403 of closed I& ideal of BS. 402 Erdbs. P.346 extracted word 382 extraction variable. 382 examally disconnected61,62,71.155.

194.198 D* not. 61 P .61 F-space 71.441 filter base 48.52 filter 48,49,54,236 in partially ordered set, 236 finite intersection property 49.50.66 K-uniform,68 infinite, 52,262 finite products theorem 92.93.286 finite version. 101 finite products 90.11 5 and idempotents. 93 finite semigroups images of W, 108 finite sums and products in same cell, finite version, 101 in same cell. 97 F i t e Sums Theorem 93.95.96.369,

386.419 Calvin-Glazer proof. 92 finite sums 90,115 finite unions 95.97 finite-to-one83 first category 362 first entries condition 297,298,299,

free group 10,151 mapped io finitep u p , 10 on generators. 153.155 on 2' generators in C p . 181 on 2 ' generators in BN. 155 on 2 ' generators in P S . 155 subgroup of compact group, 40 universal property. 10 free semigroup 4,7,130,281,282 defined by a right cancelable element, 167 in W, 153 on 2' generatas. 153 with identity, 4 Frolik. Z. 70,235 functions equal at p not equal at any x . 160 Furstenberg,H.viii, 102,294,338,393.

z2'

4lI.412.424 Galvin. E 70.102 Garcia-Fmeira, S. viii. 234 Glazer, S. 102 Graham. R. 102,338,367 Grainger. A.viii Gmthendieck, A. 453 p u p 6,17,I8 characterizations, 17 closure, 39 equal to eSe. 20.24 in smallest ideal, 34.35 locally compact, 32 maximal groups in minimal left ideal. 19 maximal, closed, 39.42 maximal, 9.35.39 of quotients. 129 Gundersoo, D. viii

300.302.303.315 over F. 315 fixed point 59 non-existence. 59 FF-tree 292 Franek, E 41 1

Hales, A. 281 Hales-Jewett Theorem 281,282,283.

295.381 consequence of Central Sets Theorem. 288

Hawdorffmetric 43 HausdorE

allhypothuizedspaccs.~ii,31 Hilbert's Theorem 90,296 Hilbert, D. 90 Hofmanu K.vii homeomorphism classes of maximal groups. 156 homeomorphism types in pD at least 2'. 119 homeomorphism type 119 homomorphism 4.75.205 from T*to S*,216.217 from pS to G', 210 from pT to S.211 from pT to pS, 217 from ~MtoM*.Z13 image of K(S). 30 of topological center, 37 on W.108 on intersection, 78 to compact right topological semigroup. 78 Hang-ting, S. 41 1 Hong-wei. Y. 411

I-reshictcd span 308 ideal 12

equal to SxS. 14 smallest, 21,29 idempotents 8,321 2' right maximal, 187 2' strongly right maximal. 193 =- minimal idempotents. 24 SR maximal, in K(BG),182 Sn-maximal idempotena in K(BW). 184 ol chains, 201 and finite products, 93 at least 2'. 118 chains in Cp. 181 closure, % commuting, 114

fonditimsequivalentto being minimal. 35 dchning maximal topologies, 1% existence. 33 immediately below a & e n idem-

in A? iff. 94 in finite semigroup, 12 inw. 109 infinite chains in W. 110 infinite chains, 118 left maximal, 186 maximal. 186 minimal for partial orders, 15 minimal, at least 121 minimal. in intersection of left and right ideals. 34 minimal, in right id&, 34 multiplicative idempotents in r.97 nonminimal. 118 not closed. 169.242 number in minimal left and right ideals, 121 right maximal, existence. 36 right maximal, q p not in smallest ideal, 191 right maximal, fmite fibres. 189 right maximal, in Ga sek 189 right maximal. 186.188 strongly right maximal, existence. 192 sbongly right maximal, 191 sum not idempotent. 112 topologiesdefinedby stronglyright maximal idempotmu. 196 topologies defined by. 194 identity 5 adjoined. 8 in pS, 203 image partition regular matrices sums of terms. 325.340 image partition regular 297, 298, 299, 300.308.309.315.318.324

z2'.

W t e . 354 over O"I. +) and (M. .), 314 strongly. 300 veaor spaces over a Be14 316 weakly. 300 incompatible 237 infinite finite intersection property 52. 262 invariant 397 inverse 5 irrational with nspect to S, 207 isomorphic 4.35.39 isomorphism 4 of principal left ideals. 219 of principal right ideals, 220

KalZkk. P.275 Kateth, M. 70 Kaeaclson, Y. viii, 295,393 Keisler, H. 235 kernel partition regular mahices sums of terms. 325,340 kernel partition regular 301,302,303. 305.318.324 over (W, +) and (N..). 304 over a field, 316 over a finite field. 317 Km, B. 221,338 Kunen, K 70,256 Lau. A 204

Lawson. I. viii, 47.41 1 Leader. I. viii, 275,319.367 Lebesgue measurable 361 Lebcsgue density theorem. 368 Lefmann, H. viii. 367 left cancelability in pP4 176 does not imply right canoelability. 176

I& cancelable elements in pN left cancelable in pZ, 173 left caneelable elements of pS contain dense open subset of S, 174 left cancelable 7, 174 Ieft cancellative 7,82,83,224 weakly, 323 left ideal and right ideal intersection, 13 Idtideal 12.19.133 z2'. 120 S* a finite union of disjoint closed left ideals, 121 U,(9a left ideal of pS, 125 X . and Y, as left ideals, 208 closed left ideals defined by density, 134 closure. 38.77 contains minimal left ideal. 20 decomposition of U,(9 into. 125 equal to Sx. 14 infinite chains, 129 intersecting. 112 intersection, 13 isomorphicprincipalIAideals, 219 maximal principal left ideals, 127 maximal semiprincipal left ideals inN*, I28 minimal, all homeomorphic, 35 minimal, contains idempotenb 34 minimal, homeomorphic to those in po, 152 minimal, 19.3435 principal. 218 semiprincipal, 112.113 left identities 5 2'". 203 for SS.not in S*, 203 left invariant topology 136, 141 Hausdo& 142 zero dimensional. 143 Ieft inverse 5 Ieft simple 13.20.24.35

lnda

.

left topological groups partitions, 197.199 left topological scmigmup 31 left zero semigmup 26.8 1 left zem 4,5 length 4 limit 63.66 Lin,C. 338 Li,S. 338 Lisan. A. viii, 135,157,295,411 local homomorphism 146 local isomorphism 146 local left p u p 146 locally isomorphic. 147 Maleki, A. viii. 135,221,275.295 Martin's Axiom 237 MA(c) false, 239 and P-points. 239 maximal almost disjoint families have cardinality c, 239 ' have non-empty G , subsets of N non-empty interior. 239 proper F . subsets of N*not dense in W. 239 Malet, P. 257 matrices image partition regular. 309 maximal p u p s 9 in K ( S ) isomorphic, 35 in BG, 155 in pN, 155 in smallest ideal isomorphic. 29 isomorphic to Z,25 1 not closed. 157 number of homeomorphism classes, 156 maximal topologies 196 number, 197 meager 362 measurable 361 measure preserving action 407 measure preserving system 407 measure preserving transformation 407

measw space 407 measure 49 Miliken, K. 355,393 Milliken-Taylor system 355 partition regular, 359

Milliken-TaylorTheorem369,373,374. 375,392 Milnes. P. viii. 88. 135,221,453 minimal (wxt. partial order) 15 minimal dynamical system 398 universal, 398 minimal idempotent 16.20.22.24 and minimal left ideals, 16 at least 2'. 109 closure has elementsoutside S* S*. 169 existence, 23 minimal left and right ideals in Cp 180 minimal left and right ideal defined by idempoteoh, 24 intersection, 25 minimal left ideal 12.14.19.20.24 a group if commutative. 23 all homeomorphic. 35 and minimal idempotents, 16 as minimalclosedinvariantset398 at least 2'. 109, 121 infinite. 120 isomorphic, 25 same in S ' and BS. 84 without idempotenb. 23 minimal right ideal 12.20.24 at least ZC. 109.120, 121 contains idempotent. 23 number. 122 same in S' and BS, 84 monochrome 91.101 monotone action 408 multilinear combinations 346 neighborhood 48 Neuman. B. 89 nonprincipal ulmfilter 50

Index

oid 111 operator uniform, 64 order 11 ordinal 2 Owings. J. 367 P-paint 65.70.175.239 cancelable in @Z,175 cancelable, 177 defined by union ulaafilter, 247 existence. 65 independence, 248 pairwise sums and products 344 Papazyan. T.135,204 Parsons, D. 135 partial order 236 partially ordered set 236 parlition regularity and ultrafilters 51 partition regular51,52,2%. 322 image, 297. 298. 299. 300, 308, 315,318 kernel, 301.302.303.305.318 strongly image, 300 weakly image, 300 path 416 periodic 408 piecewise syndetic85.86.87.279.282. 290,301.322 and central sets theorem. 289 collectionwise,290,291.293.294 in PI. 87 left, 274 near 0.27 1 partition has left and right. 275 partition regular. 280 right pimewise syndeticvenusleft piecewise syndetic. 274 right, 274 if and only if the closure meets K(BS). 86

pigeon hole principle 91 Plewik. S. 368 poincwise convergence 32 Polya. G. 424 positive upper density 340 principal closed ideal 131 increasing chain. 133 principal ideal 14 principal ultrafilter 50.55 product in @S as ultrafilter, 76.77 product of a finitenumber of ultrafilters 79 producl subsystem 94,323,375,376 weak. 377 product topology 32 products nowhere dense in S*, 118 product ofright topological semigroups,40 proper ideal 12 ProraSov, I. vii, 89. 157, 185,192,204, 257 pmximal405 and minimal left ideals of pS, 406 Prt-me],H. 368 Pym, J.viii, 30,135,157,185,204,275 quotient spaces 432 quotient topology 441 quotients 8 group of. 129 Rado's Theorem 91.304.305 Rado. R. 91,318 Ramrey Theory 90.2% Ramsey ultrafilter 235 Ramsey's Theorem 91,369,370 finite version, 102 Ramsey, E 9 1.370 rapidly increasing sequences 79 recurrent 407 U N ~ O ~ Y 404 .

reduced word 382

482 nduction variable, 381 Rces. 0 . 3 0 right cancelability in BS and the Rudin-Keisler order. 165 characterizations. 165, 166 on dense open subs* of BS. 166 relations with discreteness, 162 right cancelability in BFI characterization, 170 does not imply left cancelability, 173 right cancelability in BZ charactnitation, 170 shvnger than in PN.170 right cancelable elements of BS contain a dense open subset of U,(S). 162 number, 162 right cancelable elements inhinsic characterization. 161 inclosureof minimal idempotents, I69 right cancelable 7. 126. 161 right cancellation characterizations, 163.234 right cancellative 7.82.83 right continuous 31 right ideals in flN closed, 1 12 right ideal 12. 14 2' disjoint in . ' N 210 closed disjoint. 112 closure as ideal, 88 closure, 38.39.77 in flS U* and D, as right ideals. 208 isomorphic principal right ideals. 220 minimal, 34,35 minimal closed, 39 of right ideal, 19 principal. 218 right identities 5

Index

22'. 202 for S. at least 2'. 204 for S,203 unique, 203 right invme 5 right maximal strongly, 248 right simple 13.20,24.35 right topological semigroup 31.32 compact, 33 right zero semigroup 26.82 finite. 187 right zero 4.5 Rothschild. B. 102 Rudin. M. vii. 135.235 Rudin. W. 70 Rudin-FrolDr order 232 K p E q implis p ~ R 4.233 and strongly right maximal idempotents. 233 transitive. 232 Rudin-Keisler order 135, 164.222 22' successors, 225 . P
maximal. 75 universal. existence. 427 universal. uniqueness. 428 semigroup 3 closure not semigroup, 167 free. 281,282 simple. 26 semiprincipal left ideal 112 at least 2' maximal, 128 characterization of members, 114 infinite chains. 129 intersecting, 113 reverse well ordered. 129 semitopological group not topological semigrwp. 199 semitopological semigroup 31 compact 41 dense center. 41 separable 62 Shelah, S. vii, 70,135,257 Simon, P. viii. 70. 135 simple 13.22.26 Simpson, S. 393 Skolem, T.338 smallest ideal of BN closure contains 2' nonminimal idempotents, 184 closure contains chains of idempotcnts. I84 smallest ideal 21,182 equal to LR. 22 in product. 40 of (pW,.) and (BN. 263 of subsemigroup. 29 smcture theorem, 28 Smith, G. 367

+).

SF'="

I-restricted, 308 special strongly summableulnaftltcr242 p=q+rimpliesthatp=q=r. 254 p=q+rimpliesthatq.r~Z+ - p. 254 existence. 244

Stone space 71 Stone, M. 69

~tollbCechcompactification56,~7.69 existence. 447 not a semigroupcompactification. 450 of completely regular space. 447 strongly discrete 6 2 162 SUM 62 strongly image padtion regular 300 stronglyright maximal idempotents 191. 248 2'. 193 defined by 'strongly summable ultrafilters, 248 defining a topology. 196 existence, 192 maximal topologies, 1% strongly summable ultrafilter 240 and union ultrafilter, 247 as idempotent, 241 as strongly right maximalidcmpotent, 248 independence. 248 maximal group isomorphic to 2. 25 1 not in c! K(BN. +).241 special. p = q r implies that p=q=r,254 special, p = q r implies that q , r E Z+p,Z54 special, existence, 244 special, 242 subgroup 8 submnigroup 8 as intmation. 78 closure in BS,77 closure not a semigroup, 41 closure. 41-77 subsystem product. 376 sum, 326 sum subsystem 94.326.373

+ +

~ummable strongly. 240 weakly, 240 sums equal to products 258,263 nowhere dense sets, 262 Suschkewitsch, A. 30 symmetric difference 43 syndetic subgroup 87 syndetic 85.86 near 0.270 piecewise, 279 Szemeddi's Thwrem vii. 419 Szerne&di, E. 412 Tang, D. viii Taylor, A. 256,355.393 tensor pmduct 223 as double limit, 223 Terry, E. viii, 338 topological center 33.40.80 equal to algebraic center. 80 of N* empty, 170 topological compactification447 topological p u p 31 topological semigroup 31 topologically and algebraically isomorphic 35 topologies defined by idempotents 194 topology left invariant, 136 transformation measure preserving, 407 m e 292.415 *, 292 FP. 292.415 in A, 292 ulaafilter48.49.50 K-uniform,52 as measure. 49.52 combinatorially rich, 339 nonprincipal. 50 principal. 50.55 uniform, 52.67

Umoh. H. 185,424 uniform compactification of R and pZ, 451 ideals. 45 1 idempotents. 452 uniform compactification universal property, 449 as semigroup, 448 contains pN. 447 existence, 447 joint continuity, 449 maximal, 449 uniform finite intersection property 68 uniform operator 64 U N ~ O I 'SI t~N C h l R 445 uniform ultrafilter 52.67 uniformity 445 uniformly continuous 445 uniformly recurrent 404 union subsystem 95 union ultrafilter 245 and strongly summable ultrafilter, 247 defines P-point, 247 universal minimal dynamical systems uniqueness. 403 universal miniddynamical system398 and minimal left ideals in pS,398 upper density positive. 340 van der Waerden's Theorem 91, 279. 280.295.296.369 consequence of Hales-Jewett. 283 increment from finite sums. 286 van der Waerden, B. 91,280 van Douwen, E. 135,204,221,256 van Mill. J. viii.135 variable extracted word 382 variable extraction 382 variable reduced word 382 variable reduction 381 variable word 282.381 Voigt, B. 368

Wallace, A. 47 Wallman, H. 69 weak P-point right cancelable. 164 weak product subsystem 377 weakly almost periodic function 436 weakly image partition regular 300 weakly left cancellative semigroup 116 wcakly left cancellative 82.122. 323 weakly right cancellative 82, 122 weakly summable ultrafilter 240 Weiss. B. viii, 102.338. 368.411

Hfm, W.viii. 318 word 4 extracted, 382 reduced. 382 variable, 282,381

Zelenuk's Thtorern 144 Zelenuk. E. 157.191 zero dimensional 137 zem set 62