This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
onto K (S). From the definition of the operation in X x G x Y we see immediately that W is a homomorphism. By Theorem 1.42 we have that L = XG and R = GY. By Theorem 1.53, K(S) = LR = XGGY = XGY = rp[X x G x Y]. Thus it suffices to produce an inverse for V.
For each t E K(S), let y(t) be the inverse of ete in eSe = G. Then ty(t) _ ty(t)e E Se = L and
ty(t)ty(t) = ty(t)etey(t) = tey(t) = ty(t), SO ty(t) E X. Similarly, y(t)t E Y. Define r : K(S)  X x G x Y by r(t) = (ty(t), et e, y(t)t). We claim that r = V  1 . So let (x, g, y) E X x G x Y. Then
r(W(x, g, y)) = (xgyy(xgy), exgye, y(xgy)xgy)
1.7 Minimal Left Ideals with Idempotents
29
Now
xgyy(xgy) = xxgyy(xgy)
(x = xx) (x = xe and y(xgy) = ey(xgy))
= xexgyey(xgy) = xe = X.
Similarly y(xgy)xgy = y. Since also exgye = ege = g, we have that r = (pt as required.
The following theorem enables us to identify the smallest ideal of many semigroups that arise in topological applications.
Theorem 1.65. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Let T be a subsemigroup of S and assume also that T has a minimal left ideal with an idempotent. If K(S) fl T 0, then K(T) = K(S) fl T.
Proof. By Theorem 1.51, K(T) exists so, since K(S) fl T is an ideal of T, K(T) C K(S) fl T. For the reverse inclusion, let x E K(S) fl T be given. Then Tx is a left ideal of T so by Corollary 1.47 and Theorem 1.56 Tx contains a minimal left ideal Te of T for some idempotent e E T. Now X E K(S) so by Theorem 1.51 pick a minimal left ideal L of S with x E L. Then L = Sx and e E Tx C_ Sx so L = Se so x E Se so by
Lemma 1.30, x = xe E Te C K(T). We know from the Structure Theorem (Theorem 1.64) that maximal groups in the smallest ideal are isomorphic. It will be convenient for us later to know an explicit isomorphism between them.
Theorem 1.66. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Let e, f E E(K(S)). If g is the inverse of of e in eSe, then the function rp : eSe * f Sf defined by cp(x) = fxgf is an isomorphism. Proof. To see that W is a homomorphism, let x, y E eSe. Then
co(x)co(Y) = fxgffygf = fxgfygf = fxgefeygf
(ge = g, ey = y)
= fxeygf = fxygf = co(xY) To see that W is onetoone, let x be in the kernel of W. Then
fxgf = f
efxgfe = efe efexgefe = efe efexe = efe
(ex = x, ge = g)
efex = efee
x=e
(left cancellation in eSe).
30
1
Semigroups and Their Ideals
To see that cp is onto f Sf , let y E f Sf and let h and k be the inverses of f gf and f e f respectively in f S .f. Then ekyhe E e Se and
rp(ekyhe) = fekyhegf
= fefkyhgf = fyhgf = fyhfgf
= fyf = Y.
(fk = k, eg = g)
(fefk = f) (h = hf)
(hfgf = f) 11
We conclude the chapter with a theorem characterizing arbitrary elements of K (S).
Theorem 1.67. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Lets E S. The following statements are equivalent.
(a) s E K(S). (b) For all t E S, S E Sts. (c) For all t E S, S E stS. (d) For all t E S, S E stS f1 Sts. Proof. (a) implies (d). Pick by Theorem 1.51 and Lemma 1.57 a minimal left ideal L of S and a minimal right ideal R of S with s E L fl R. Let t E S. Then is E L so Sts is a left ideal contained in L so Sts = L. Similarly stS = R. The facts that (d) implies (c) and (d) implies (b) are trivial. (b) implies (a). Pick t E K(S). Then S E Sts C K(S). Similarly (c) implies (a).
Exercise 1.7.1. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Prove that if K(S) 0 S and X E K(S), then x is neither left nor right cancelable in S. (Hint: If x is a member of the minimal left ideal L, then
L = Sx = Lx.) Exercise 1.7.2. Identify K(S) for those semigroups S in Exercises 1.4.2 and 1.4.3 for which the smallest ideal exists. Exercise 1.7.3. Let S and T be semigroups and let h : S s T be a surjective homomorphism. If S has a smallest ideal, show that T does as well and that K(T) = h[K(S)].
Notes Much of the material in this chapter is based on the treatment in [39, Section II. I]. The presentation of the Structure Theorem was suggested to us by J. Pym and is based on his treatment in [202]. The Structure Theorem (Theorem 1.64) is due to A. Suschkewitsch [231] in the case of finite semigroups and to D. Rees [210] in the general case.
Chapter 2
Right Topological (and Semitopological and Topological) Semigroups
In this (and subsequent) chapters, we assume that the reader has mastered an introductory
course in general topology. In particular, we expect familiarity with the notions of continuous functions, nets, and compactness.
2.1 Topological Hierarchy Definition 2.1. (a) A right topological semigroup is a triple (S, , T) where (S, .) is a semigroup, (S, T) is a topological space, and for all x E S, Px : S a S is continuous. (b) A left topological semigroup is a triple (S, , 7) where (S, .) is a semigroup, (S, T) is a topological space, and for all x E S, Ax : S + S is continuous. (c) A semitopological semigroup is a right topological semigroup which is also a left topological semigroup. (d) A topological semigroup is a triple (S, , 7) where (S, ) is a semigroup, (S, T) is a topological space, and : S x S S is continuous. (e) A topological group is a triple (S, , T) such that (S, ) is a group, (S, 7) is a topological space,  : S x S  S is continuous, and In : S + S is continuous (where In(x) is the inverse of x in S). We did not include any separation axioms in the definitions given above. However, all of our applications involve Hausdorff spaces. So we shall be assuming throughout, except in Chapter 7, that all hypothesized topological spaces are Hausdorff. In a right topological semigroup we say that the operation is "right continuous". We should note that many authors use the term "left topological" for what we call "right topological" and vice versa. One may reasonably ask why someone would refer to an operation for which multiplication on the right is continuous as "left continuous". The people who do so ask why we refer to an operation which is continuous in the left variable as "right continuous". We shall customarily not mention either the operation or the topology and say something like "let S be a right topological semigroup".
2 Right Topological Semigroups
32
Note that trivially each topological group is a topological semigroup, each topological semigroup is a semitopological semigroup and each semitopological semigroup is both a left and right topological semigroup. Of course any semigroup which is not a group provides an example of a topological semigroup which is not a topological group simply by providing it with the discrete topology. It is the content of Exercise 2.1.1 to show that there is a topological semigroup which is a group but is not a topological group. It is a celebrated theorem of R. Ellis [84], that if S is a locally compact semitopological semigroup which is a group then S is a topological group. That is, if S is locally compact and a group, then separate continuity implies joint continuity and continuity
of the inverse. We shall prove this theorem in the last section of this chapter. For an example of a semitopological semigroup which is a group but is not a topological semigroup see Exercise 9.2.7. It is the content of Exercise 2.1.2 that there is a semitopological semigroup which is not a topological semigroup. Recall that given any topological space (X, 7), the product topology on XX is the topology with subbasis { r 1 [U] : x E X and U E 7), where for f E XX and X E X, it (f) = f (x). Whenever we refer to a "basic" or "subbasic" open set in XX, we mean sets defined in terms of this subbasis. The product topology is also often referred to as the topology of pointwise convergence . The reason for this terminology is that a
net (f,)IEI converges to fin XX if and only if (f,(x)),EI converges to f (x) for every
xEX. Theorem 2.2. Let (X, T) be any topological space and let V be the product topology on XX.
(a) (XX, o, "V) is a right topological semigroup. (b) For each f E XX, k f is continuous if and only if f is continuous.
Proof. Let f E XX. Suppose that the net (gL61 converges to g in (XX, V). Then, for
any x E X, (g,(f(x))),Ef converges to g(f (x)) in X. Thus (g, o f),Ef converges to g o f in (XX, V), and so pf is continuous. This establishes (a). Now,X f is continuous if and only if (f (g,(x))),E! converges to f(g(x)) for every
net (g,),El converging to g in ft, V) and every x E X. This is obviously the case if f is continuous. Conversely, suppose that X f is continuous. Let (x,),E1 be a net converging to x in X. We define g, = x,, the function in XX which is constantly equal
to x, and g = T. Then (g,),E1 converges to g in XX and so (f 0 g,),Ej converges to f o g. This means that (f (x,)),EJ converges to f (x). Thus f is continuous, and we have established (b).
Corollary 23. Let X be a topological space. The following statements are equivalent. (a) XX is a topological semigroup. (b) XX is a semitopological semigroup.
(c) For all f E XX, f is continuous. (d) X is discrete.
2.2 Compact Right Topological Semigroups
33
Proof. Exercise 2.1.3
if X is any nondiscrete space, it follows from Theorem 2.2 and Corollary 2.3 that XX is a right topological semigroup which is not left topological. Of course, reversing the order of operation yields a left topological semigroup which is not right topological. Definition 2.4. Let S be a right topological semigroup. The topological center of S is the set A(S) = {x E S : AX is continuous}. Thus a right topological sernigroup S is a semitopological semigroup if and only if A(S) = S. Note that trivially the algebraic center of a right topological semigroup is contained in its topological center.
Exercise 2.1.1. Let T be the topology on IR with basis B = {(a, b] : a, b E R and a < b}. Prove that (R, +, T) is a topological semigroup but not a topological group. Exercise 2.1.2. Let S = R U {oo}, let S have the topology of the one point compactification of IR (with its usual topology), and define an operation * on S by X*
_ Jx+y ifx,yER 00
if x
 ooo ry oo. 
(a) Prove that (S, *) is a semitopological semigroup. (b) Show that * : S x S * S is not continuous at (oo, oo).
Exercise 2.1.3. Prove Corollary 2.3.
2.2 Compact Right Topological Semigroups We shall be concerned throughout this book with certain compact right topological semigroups. Of fundamental importance is the following theorem. Theorem 2.5. Let S be a compact right topological semigroup. Then E(S)
0.
Proof. Let A = {T C S : T ¢ 0, T is compact, and T T. T C T}. That is, A is the set of compact subsemigroups of S. We show that A has a minimal member using Zorn's Lemma. Since S E A, A 0 0. Let e be a chain in A. Then C is a collection of closed subsets of the compact space S with the finite intersection property, so n C' # 0 and n e is trivially compact and a semigroup. Thus n C E A, so we may pick a minimal member A of A. Pick X E A. We shall show that xx = x. (It will follow that A = {x}, but we do not need this.) We start by showing that Ax = A. Let B = Ax. Then B 0 and since B = pr [A], B is the continuous image of a compact space, hence compact. Also
2 Right Topological Semigroups
34
BB = AxAx C AAAx C Ax = B. Thus B E A. Since B = Ax C AA C A and A s minimal, B = A.
Let C = {y E A : yx = x}. Since X E A = Ax, we have C # 0.
Also,
C = A fl px t [{x}], so C is closed and hence compact. Given y, z E C one has yz E AA c A and yzx = yx = x so yz E C. Thus C E A. Since C C A and A is minimal, we have C = A so x E C and so xx = x as required. In Section 1.7 there were several results which had as part of their hypotheses "Let S be a semigroup and assume there is a minimal left ideal of S which has an idempotent" Because of the following corollary, we are able to incorporate all of these results.
Corollary 2.6. Let S be a compact right topological semigroup. Then every left ideal of S contains a minimal left ideal. Minimal left ideals are closed, and each minimal left ideal has an idempotent. Proof. If L is any left ideal L of S and X E L, then Sx is a compact left ideal contained in L. (It is compact because Sx = px [S].) Consequently any minimal left ideal is closed and by Theorem 2.5 any minimal left ideal contains an idempotent. Thus we need only show that any left ideal of S contains a minimal left ideal. So let L be a left ideal of S
and let A = {T : T is a closed left ideal of S and T c L}. Applying Zorn's Lemma to A, one gets a left ideal M minimal among all closed left ideals contained in L. But since every left ideal contains a closed left ideal, M is a minimal left ideal. We now deduce some consequences of Corollary 2.6. Note that these consequences apply in particular to any finite semigroup S, since S is a compact topological semigroup when provided with the discrete topology.
Theorem 2.7. Let S be a compact right topological semigroup. (a) Every right ideal of S contains a minimal right ideal which has an idempotent.
(b) Let T C S. Then T is a minimal left ideal of S if and only if there is some e E E(K(S)) such that T = Se. (c) Let T C S. Then T is a minimal right ideal of S if and only if there is some e E E(K(S)) such that T = eS. (d) Given any minimal left ideal L of S and any minimal right ideal R of S, there is an idempotent e E R fl L such that R fl L = eSe and eSe is a group.
Proof (a) Corollary 2.6, Lemma 1.57, Corollary 1.47, and Theorem 1.56. (b) and (c). Corollary 2.6 and Theorem 1.58. (d) Corollary 2.6 and Theorem 1.61.
Theorem 2.8. Let S be a compact right topological semigroup. Then S has a smallest (two sided) ideal K (S) which is the union of all minimal left ideals of S and also the union
of all minimal right ideals of S. Each of (Se : e E E(K(S))}, {eS : e E E(K(S))}, and (eSe : e E E(K(S))} are partitions of K(S). Proof. Corollary 2.6 and Theorems 1.58, 1.61, and 1.64.
2.2 Compact Right Topological Semigroups
35
Theorem 2.9. Let S be a compact right topological semigroup and let e E E(S). The f vowing statements are equivalent. (a) Se is a minimal left ideal. (b) Se is left simple. (c) eSe is a group.
(d) eSe = H(e). (e) eS is a minimal right ideal. (f) eS is right simple. (g) e is a minimal idempotent.
(h) e E K(S). (i) K(S) = SeS. Proof. Corollary 2.6 and Theorem 1.59.
Theorem 2.10. Let S be a compact right topological semigroup. Let S E S. The following statements are equivalent.
(a) s E K(S). (b) For all t E S, S E Sts. (c) For all t E S, s E stS. (d) For all t E S, s E stS f1 Sts.
Proof. Corollary 2.6 and Theorem 1.67. The last few results have had purely algebraic conclusions. We now obtain a result with both topological and algebraic conclusions. Suppose that we have two topological spaces which are also semigroups. We say that they are topologically and algebraically
isomorphic if there is a function from one of them onto the other which is both an isomorphism and a homeomorphism.
Theorem 2.11. Let S be a compact right topological semigroup. (a) All maximal subgroups of K(S) are (algebraically) isomorphic. (b) Maximal subgroups of K (S) which lie in the same minimal right ideal are topologically and algebraically isomorphic. (c) All minimal left ideals of S are homeomorphic. In fact, if L and L' are minimal left ideals of S and Z E L', then PzIL is a homeomorphism from L onto L'. Proof. (a) Corollary 2.6 and Theorem 1.66.
(b) Let R be a minimal right ideal of S and let e, f E E(R). Then eS and f S are right ideals contained in R and so R = eS = f S. Then by Lemma 1.30, of = f and f e = e. Let g be the inverse of of e in the group eSe and define cp : eSe + f Sf by cp(x) = fxgf. Then by Theorem 1.66, rp is an isomorphism from eSe onto fSf. To
2 Right Topological Semigroups
36
see that cp is continuous, we show that rp is the restriction of p8 f to eSe. To this end, let
x E eSe. Then w(x) = fxgf
= fexgf
(x = ex)
= exgf
(fe = e)
= xgf
(ex = x).
Now let h and k be the inverses in f Sf of fgf and f of respectively. We showed in the proof of Theorem 1.66 that if y E f Sf, then rp'1 (y) = ekyhe. Thus
W'(y) = ekyhe = fefkyhe
= fyhe = yhe
(fk = k and fe = e)
(fefk = f). (.fY = Y).
So (P1 is the restriction of Phe to f Sf and hence is continuous.
(c) Let L and L' be minimal left ideals of S and let z E L'. By Theorem 2.7(b), pick e E E(K(S)) such that L = Se. Then PzIL is a continuous function from Se to Sz = L' and pz[Se] = L' because Sez is a left ideal of S which is contained in L'. To see that pZ is onetoone on Se, let g be the inverse of eze in eSe. We show that for x E Se, pg(pz(x)) = x, so let x E Se be given. xzg = xezeg
(x = xe and g = eg)
= xe = X. Since PzIL is onetoone and continuous and L is compact, PzIL is continuous.
Recall that given any idempotents e, f in a semigroup S, e
fe = e. Theorem 2.12. Let S be a compact right topological semigroup and let e E E(S). There is a
Proof Let A = {x E E(S) : e <_R x}. Then A
0 because e E A. Let C be a
Given e, f E E(K(S)) and an assignment to find an isomorphism from eSe onto f Sf, most of us would try first the function r : eSe  f Sf defined by r (y) = f yf . In fact, if eS = f S, this works (Exercise 2.2.1). We see now that this natural function need not be a homomorphism if eS f S and Se Sf
2.2 Compact Right Topological Semigroups
37
Example 2.13. Let S be the semigroup consisting of the eight distinct elements e, f, e f, f e, efe, fef, efef, and fefe, with the following multiplication table. e
f
of
fe
efe fef efef fefe
e
e
of
of
efe
efe efef efef
f
fe
of
efe
fe
fe efe efef efef
efe
f fef fe fefe fef e efef of efef efe fef fef fefe fefe f
fef fefe fef
efef
e
f
efef of
fefe fefe
f
f
e
e
fefe fe
f
e
fefe
of
e
f
fe
of
of
efe
f
fef
fe
e
efe
of efef of e
fe
fe
fef fef fefe
Then S with the discrete topology is a compact topological semigroup and fef is not an idempotent so the function r : eSe s fSf defined by r (x) = fxf is not a homomorphism.
Once again, the simplest way to see that one has in fact defined a semigroup is to produce a concrete representation. In this case the 2 x 2 integer matrices e = (1
and f = 1
11
2) generate the semigroup S.
0 00)
The topological center of a compact right topological semigroup is important for many applications. The following lemma will be used later in this book.
Lemma 2.14. Let S and T be compact right topological semigroups, let D be a dense subsemigroup of S such that D C A(S), and let i7 be a continuous function from S to T such that
(1) n[D] C A(T) and (2) 771D is a homomorphism.
Then ri is a homomorphism.
Proof. For each d E D, n o Ad and Aq(d) o q are continuous functions agreeing on the dense subset D of S. Thus for all d E D and all y E S, rl(dy) = n(d)n(y). Therefore, for all y E S, n o py and pq(y) o q are continuous functions agreeing on a dense subset of S so f o r all x and y in S, t (xy) = rt(x)rt(y). Exercise 2.2.1. Let S be a compact right topological semigroup and let e, f E E (K (S)) such that eS = f S. Let g be the inverse of efe in the group eSe and define cp : eSe *
f Sf by cp(x) = fxgf. Define r : eSe + f Sf by r(x) = fxf. Prove that r = V. Exercise 2.2.2. Let S be a compact right topological semigroup, let T be a semigroup with topology, and let cp : S * T be a continuous homomorphism. Prove that V[S] is a compact right topological semigroup.
2 Right Topological Semigroups
38
Exercise 2.2.3. Prove that a compact cancellative right topological semigroup is a group
2.3 Closures and Products of Ideals We investigate briefly the closures of right ideals, left ideals, and maximal subgroups of right topological semigroups. We also consider their Cartesian product.
Theorem 2.15. Let S be a right topological semigroup and let R be a right ideal of S. Then ct R is a right ideal of S. Proof This is Exercise 2.3.1. On the other hand, the closure of a left ideal need not be a left ideal. We leave the verification of the details in the following example to the reader.
Example 2.16. Let X be any compact space with a subset D such that ct D # X and I cf D\D) > 2. Define an operation on X by X
__
y
Y ifyED I x if yoD
Then X is a compact right topological semigroup, A(X) = 0, and the set of left ideals of X is {X} U {B : 0 0 B C D}. In particular, D is a left ideal of X and cf D is not a left ideal.
Often, however, we see that the closure of a left ideal is an ideal.
Theorem 2.17. Let S be a compact right topological semigroup and assume that A(S) is dense in S. Let L be a left ideal of S. Then cf L is a left ideal of S. Proof. Letx E cf L and let y E S. To see that yx E c2 L, let U be an open neighborhood of yx. Pick a neighborhood V of y such that V x = pX (V } c U and pick z E A (S) fl V.
Then zx = A (x) E U so pick a neighborhood W of x such that zW c_ U. Pick
wEWfL.Then zwEUflL. In our applications we shall often be concerned with semigroups with dense center. (If S is discrete and commutative, then f3S has dense center  see Theorem 4.23.)
Lemma 2.18. Let S be a compact right topological semigroup. The following statements are equivalent.
(a) The (algebraic) center of S is dense in S. (b) There is a dense commutative subset A of S with A C A(S)
2.3 Closures and Products of Ideals
39
proof. The fact that (a) implies (b) is trivial. To see that (b) implies (a), let A be a dense commutative subset of S With A C A(S). For every x E A, the continuous functions px and Az are equal on A and are therefore equal on S. So A is contained in the center of S.
Theorem 2.19. Let S be a compact right topological semigroup with dense center. (a) If R is a right ideal of S, then ct R is a two sided ideal of S.
(b) If e E E(K(S)), then ci(eSe) = Se. Proof Let A be the center of S. (a) By Theorem 2.15, ct R is a right ideal of S. To see that S S. (ci R) C c2 R, let y E ct R be given. Then given any x E A one has py (x) = xy = yx E (cf R)x C ct R. Thus py [A] C ct R, so py [S] = py [cf A] C ct py [A] C c2 R. That is Sy C ct R as required. (b) Since Se is closed (being the continuous image of a compact space), ci(eSe) C
Se. On the other hand, since Pe is continuous, Se = (ci A)e = ce(Ae) = ci(eAe) C cf (eSe).
Theorem 2.20. Let S be a compact right topological semigroup with dense center. Assume that S has some minimal right ideal R which is closed. Then R = K (S) and all maximal subgroups of K(S) are closed and pairwise algebraically and topologically isomorphic.
Proof. By Theorem 2.19(a), ct R is an ideal so K(S) c c2 R = R C K(S). Given e E E(K(S)), H(e) = eSe = R f1 Se by Theorems 2.9 and 2.7 so H(e) is closed. Any two maximal subgroups of K(S) lie in the same minimal right ideal and so are algebraically and topologically isomorphic by Theorem 2.11. Theorem 2.21. Let S be a compact right topological semigroup with dense center. The following statements are equivalent.
(a) K(S) is a minimal right ideal of S. (b) All maximal subgroups of K (S) are closed. (c) Some maximal subgroup of K(S) is closed.
Proof. (a) implies (b). Let e E E(K(S)). Then by Theorem 2.9 eS is a minimal right ideal and Se is a minimal left ideal so by Theorem 2.7(d), eSe = eS fl Se. Since K(S) is a minimal right ideal eS = K(S). Since Se C K(S), eSe = eS fl Se = Se, and Se is closed. The fact that (b) implies (c) is trivial. (c) implies (a). Pick e E E (K(S)) such that eSe is closed. Let R = eS. By Theorem
2.19(b), ce(eSe) = Se. So Se = eSe C eS = R. Now any other minimal right ideal of S would be disjoint from R so would miss Se, which is impossible by Lemma 1.29(b). Thus R is the only minimal right ideal of K(S), which is the union of all minimal right
ideals, so K(S) = R.
2 Right Topological Semigroups
40
We have seen that the Cartesian product of semigroups is itself a semigroup under the coordinatewise operation. If the semigroups are also topological spaces the product is naturally a topological space.
Theorem 2.22. Let (S1)iEI be a family of right topological semigroups and let S = X iE/Si. With the product topology and coordinatewise operations, S is a right topological semigroup. If each Si is compact, then so is S. If x' E S and for each i E 1, .lxi : Si
Si is continuous, then Ag : S ) S is continuous.
Proof This is Exercise 2.3.5. Note that the following theorem is entirely algebraic. We shall only need it, however, in the case in which each Si is a compact right topological semigroup. In this case, each Si does have a smallest ideal by Theorem 2.8.
Theorem 2.23. Let (Si)iE1 be a family of semigroups and let S = X iEI Si. Suppose
that, for each i E 1, Si has a smallest ideal. Then S also has a smallest ideal and K(S) = X iE1 K(Si). Proof. We first note that X i,1 K (Si) is an ideal of S. Let u E K(S). Then
(X;EIK(Si)) u (XiE1K(Si)) = XiE,(K(Si) ui K(Si)) = X ie,K(Si) so by Lemma 1.52(b), K(S) = XiEIK(Si). While we are on the subject of products, we use Cartesian products to establish the following result which will be useful later.
Theorem 2.24. Let A be a set and let G be the free group generated by A. Then G can be embedded in a compact topological group. This means that there is a compact topological group H and a onetoone homomorphism yl : G + H. Proof Recall that 0 is the identity of G. For each g E G\{0} = G', pick by Theorem 1.23 a finite group Fg and a homomorphism Og : G + Fg such that 4'g(g) is not the identity of Fg. Let each Fg have the discrete topology. Then H = XgEGI Fg is a compact topological group. Define a homomorphism .p : G ). H by stating that V (h)g = (g (h). Then, if g E G', we know that cp(g)g is not the identity of Fg. So the kernel of cp is {0} and hence V is onetoone as required.
Exercise 2.3.1. Prove Theorem 2.15.
Exercise 2.3.2. Let S be a right topological semigroup, and let T be a subset of the topological center of S. Prove that ce T is a semigroup if T is a semigroup. Exercise 2.3.3. Verify the assertions in Example 2.16. Exercise 2.3.4. Let X and D be as in Example 2.16 except that D is open and ce D\D = (z). Prove that A(X) = {z}.
Exercise 2.3.5. Prove Theorem 2.22.
2.4 Semitopological Semigroups
41
2.4 Semitopological and Topological Semigroups We only scratch the surface of the theory of semitopological semigroups and the theory
of topological semigroups, in order to indicate the kinds of results that hold in these settings but not in the setting of right topological semigroups. We shall see many examples of right topological semigroups that have no closed minimal right ideals, including our favorite (ON, +). (See Theorems 6.9 and 2.20.) By way of contrast we have the following theorem.
Theorem 2.25. Let S be a compact semitopological semigroup and let a E S. Then aS, Sa, and aSa are closed. Proof Exercise 2.4.1. Corollary 2.26. Let S be a compact semitopological semigroup with dense center. Then S is commutative and K(S) is a group. Proof. For any x E S, Ax and px are continuous functions agreeing on a dense subspace of S and therefore on the whole of S. So S is commutative. By Corollary 2.6 and Exercise 1.6.5, K(S) is a group.
In fact a stronger conclusion holds. See Corollary 2.40.
Theorem 2.27. Let S be a semitopological semigroup and let T be a subsemigroup of S. Then cf T is a subsemigroup of S. Proof. This follows from Exercise 2.3.2.
By way of contrast we shall now see that in a right topological semigroup, the closure of a subsemigroup need not be a semigroup. (This also holds, as we shall see in Theorem 8.21, in (ON, +).) Theorem 2.28. Let X be the one point compactification of N and let T = (f E XX f is onetoone). Then ce T is not a semigroup. Proof. For each n E N, define gn : X + X by
gn(x) 
x+n ifx E `N 00
ifx = oo.
Let g : X  X be the constant function oo. Then each gn E T and
tog inXXsogEciT. Now define f : X + X by f (x)
x+1 ifxEN 1
ifx = oo.
converges
2 Right Topological Semigroups
42
Then f E T while f o g 0 ct T because {h E XX : h (1) = h (2) = 1) is a neighborhood of f o g in XX which misses T. We see, however, that many subsemigroups of XX do have closures that are semigroups.
Theorem 2.29. Let X be a topological space and let S c XX be a semigroup such that for each f E S, f is continuous. Then ct S is a semigroup. Proof This follows from Theorem 2.2 and Exercise 2.3.2. The closed semigroups of Theorem 2.29 are important in topological dynamics and we will have occasion to refer to them often.
Definition 2.30. Let X be a topological space and let S be a semigroup contained in XX such that for all f E S, f is continuous. Then cB S is the enveloping semigroup of S.
If T E XX is continuous and S = (T" : n E NJ, then cf S is the enveloping semigroup of T. By Theorem 2.29, if T E XX is continuous, then cZ{T" : n E N) is a semigroup. We see that one cannot add a single point of discontinuity and expect the same conclusion.
Example 2.31. Let X = [0, 1] with the usual topology and define f E XX by 1
f W  x/2
ifx=0 if 0 < x < 1.
Then ct{f":nENJ={f":nEN}U{0} Theorem 2.32. Let S be a compact topological semigroup. Then all maximal subgroups of S are closed and are topological groups.
Proof. Let e E E(S). By Lemma 1.19, an element x E eSe is in H(e) if and only if there is an element y E eSe for which yx = xy = e. Now eSe is closed as it is the continuous image of S under the mapping Pe 0 ke. Suppose that x is a limit point of a net (x, ),EI in H (e). For each c E I there is an element y, E H(e) for which x, y, = y,x, = e.
By joint continuity any limit point y of the net (y,),0 satisfies xy = yx = e and so x E H(e). To see that the inverse function In : H(e) + H(e) is continuous, let x E H(e) and let (x,),EI be a net in H(e) converging to x. As observed above, any limit point y of the net (x,I ),EI satisfies xy = yx = e and so x1 is the only limit point of (X,'),El' That is, (x,  ' ),E I converges to x 1 .
We see that we cannot obtain the conclusion of Theorem 2.32 in an arbitrary compact semitopological semigroup.
2.5 Ellis' Theorem
43
Example 2.33. Let S = IR U (oo} with the topology and semigroup operation given in Exercise 2.1.2. Then S is a semiropological semigroup, E(S) = {0, oo} and H(O) = R. Thus H(O) is not closed. Note that in Example 2.33, H (oo) = {oo}, which is closed. In fact, by Theorem 2.25, in any compact semitopological semigroup, all maximal groups in K(S) are closed.
Exercise 2.4.1. Prove Theorem 2.25 Exercise 2.4.2. Let X be any set. We can identify .% (X) with X{0, 1} by identifying each Y E .P(X) with its characteristic function X y, where,
Xy(x)=
1
0
ifxEY ifxEX\Y.
We can use this identification to give .' (X) a compact topology. Prove that (Y(X), U) and (,p (x), n) are topological semigroups. Prove also that (Y(X), A) is a topological group which can be identified topologically and algebraically with XX2 (where Y A Z = (Y U Z)\(Y n Z), the symmetric difference of Y and Z). Exercise 2.4.3. Given a set A C JR" and X E R" define d(x, A) = inf {d (x, y) : y E A) Let 3e(R") denote the set of nonempty compact subsets of R". The Hausdorff metric on df (R") is defined by
h(A,B)=inf{r:d(a,B)
2.5 EUis' Theorem We now set out to show, in Corollary 2.39, that if S is a locally compact semitopological group, then S is in fact a topological group. While this result is of fundamental importance to the theory of semitopological semigroups, it will not be used again in this book until Chapter 21. Consequently, this section may be viewed as "optional".
In a metric space (X, d), given x E X and e > 0 we write N(x, e) = {y E X d(x, y) < e}. Lemma 2.34. Let X be a compact metric space and let g : X x X > JR be a separately continuous function. There is a dense GS subset D of X such that g is jointly continuous at each point of D x X.
Proof. Let d be the metric of X. For every e > 0 and every S > 0, let
ES, = (x e X : whenever y, y' E X and d(y, y') < 3 one has Ig(x, y) g(x, y')j < E}.
2 Right Topological Semigroups
44
We first observe that each Es,, is closed.
We next observe that for each e > 0, X = Uno° 1 Ei/n,E Indeed, for each x the function y H g(x, y) is uniformly continuous, so for some n, x E El /n,E Now let Ube a nonempty open subset off. For each n E N, ifUflintx(Ei/n,E) = 0, then UflEI/n,E is nowhere dense. Thus, by the Baire CategoryTheorem, there exists n E N such that U flintx(El/n,E) 0. Let HE = U00 intx(EI/n,E) = U8>o intx(ES,E). Then HE is a dense open subset of X. Let D= fE>o H E = n 1 Hi/,,,. By the Baire Category Theorem, D is dense 1
in X. To see that g is continuous at each point of D x X, let (x, y) E D x X and let E > 0. Since x E D e HE/2, pick S > 0 such that x E intx(Es,E/2). Also pick a neighborhood U of x such that I g(x, y)  g(z, y)I < E/2 for all z E U. Let (z, w) E (intx(E8,E/2) fl U) x N(y, 8). Since Z E Es,,/2 and d(y, w) < 8, one has Ig(z, y)  g(z, w)I < Z. Since Z E U, one has Ig(.v, y)  g(z, y)I < Z. Thus Ig(x, Y)  g(z, w)I < E. Lemma 2.35. Let (X, ) be a compact metrizable semitopological semigroup with identity e and let x be an invertible element of X. Then is jointly continuous at every point of ({x} x X) U (X x {x}).
Proof. By symmetry it suffices to show that is jointly continuous at every point of [x) x X. To this end, let y E X and let (x)1 and (yn )n° 1 be sequences in X converging to x and y respectively. We shall show that (xnyn)n 1 converges to xy. To see this, let z be any limit point of the sequence (x yn )n° 1 and choose a subsequence (xn,yn,)°_1 of (xny, )' 1 which converges to z. Suppose that z 96 xy. Then x1z 0 y so pick a continuous function 0: X
such that ¢(x' z) = 0 and 0(y) = 1. Define g : X x X * R by g(u, v) = 4(uv) and note that g is separately continuous. Pick by Lemma 2.34 a dense subset D of X such that g is jointly continuous at each point of D x X. Choose a sequence (dn)n° 1 in D converging to e. For each m E N, lim dmxixn, = dm so, since g is jointly continuous at (dm, y),
roo
lim 0(dmx1xnrYnr) = r.oo lim g(dmx1xn,, Y",) = g(dm, Y) = 0(dmY) roo
Also, for each m, lim 0(dmx'xn, yn,) =
roo
0(dmx'z). Thus
lim 0(dmx'z) _ ¢(x'z), (y) = lim 0(dmy) = mpoorgoo lim lim 0(dmx'xn,yn,) =m+oo m>oo
a contradiction.
We shall need the following simple lemma.
Lemma 2.36. Let X be a semitopological semigroup, let C be a dense subset of X, and let 0 be a continuous function from X to a topological space Z. Let x, y E X. If 4(uxv) _ 0(uyv) for every u, V E C, then 4(uxv) _ ¢(uyv) for every u, v E X.
45
2.5 Ellis' Theorem
.Proof. This is Exercise 2.5.1. Lemma 2.37. Let X be a compact semitopological semigroup and let 0 be a continuous real valued function defined on X. We define an equivalence relation on X by stating
that x = y if 0 (uxv) _ 0 (uy v) for every u, v E X. Let Y denote the quotient space Y denote the canonical projection. Then Y can be given a X/ = and let r : X semigroup structure for which Y is also a compact semitopological semigroup and 7r is a homomorphism. Furthermore, if X is separable, Y is metrizable.
Proof First notice that Y is compact since it is the continuous image of a compact space. To see that Y is Hausdorff, let x, y E X and assume that 7r(x) 0 7r(y). Pick
u, v E X such that a = cb(uxv) qA 4,(uyv) = b and let e = la  bl. Let U = Z)]. ThenU
(4'oAuopv)1[N(a, 2)]and let V = (4,o
=7r1[7r[U]]
=7r[7r[V]] so7r[U] and7r[V] are disjoint neighborhoods of7r(x) and7r(y). and V Define an operation on Y by 7r(x)7r(y) = 7r(xy). To see that the operation is well defined, assume that x a x' and y = y' and let u, VEX. Then 4, (uxy v) = ¢ (ux'yv) = 0 (ux'y'v). The operation is clearly associative. It follows from Exercise 2.2.2 and its dual that Y is a semitopological semigroup.
X * CXCR be Now suppose that X has a countable dense subset C. Let Y CxCR by defined by i/r(x)(u, v) = cb(uxv). Then i/i is continuous. Define ,lc (7r (x)) = * (x). Then is clearly well defined. To see that is continuous, let U be an open subset of C>CR and let V = 1 [U]. Then 7r'[V] _ *1 [U] so V is open. is onetoone. To see this, suppose that >li(7r(x)) = {n(y)}. We claim that Then0(uxv) = 4,(uyv)forevery u, v E C. Itfollowsthat 4,(uxv) = 4,(uyv)forevery u, v E X (by Lemma 2.36) and hence that 7r(x) = it (y). We can now conclude that Y is homeomorphic to *[Y]. This is metrizable, as it is a subspace of C I C R, the product of a countable number of copies of R. Theorem 2.38. Let (S, .) be a compact semitopological semigroup with identity e and
let x be an invertible element of S. Then
is jointly continuous at every point of
({x} x S) U (S x {x}).
Proof Let y E S. To show that is jointly continuous at (x, y), it is sufficient to show that the mapping (w, z) H y(wz) is jointly continuous at (x, y) for every continuous real valued function y. Suppose, on the contrary, that there exists a continuous real valued function y for which this mapping is not jointly continuous at (x, y). Then there exists S > 0 such that every neighborhood of (x, y) in S x S contains a point (w, z) for which Iy(wz) Y(xy)I > S. 1 and (yn)R° 1 in S. We first choose We shall inductively choose sequences x1 and Y1 arbitrarily. We then assume that m E N and that xi and yi have been chosen f o r every i E {1, 2, ... , m}. Put
C , n = {TI;" 1 wi :foralli, wi E
{e,x,x,y}U{x1,x2,...,xm}U[Y1,Y2,...,yn}}. 1
2 Right Topological Semigroups
46
We note that Cm is finite and that for each u, v E Cm, y o Xu o p is continuous. We can therefore choose xm+1 and ym+1 satisfying I y (uxm+1 v)  y (uxv) I < M+1 and
Iy(uym+l v)y(uyv)I < +forevery u, v E Cm, while Iy(Xm+tym+l)y(Xy)I > S m1 Let C = UM' I Cm. Then C is a subsemigroup of S, because, if u E Ck and V E Cm,
then uv E Ck+m. Let X = ct C. Then X is a subsemigroup of S by Theorem 2.27. Let 0 = )IX and let Y be the quotient of X as described in Lemma 2.37. Then Y is a compact metrizable semitopological semigroup with identity 7r(e), and 7r (x) has the inverse 7r(x1) in Y. So, by Lemma 2.35, is jointly continuous at (7r(x), 7r(y)). We claim that 7r(xn) + 7r(x). Suppose instead that there is some z E X such that 7r(z) # 7r(x) and 7r(z) is an accumulation point of the sequence (7r(xn))n°1. Choose
by Lemma 2.36 some u, v E C such that 0(uzv) # 0(uxv) and pick k E N such that 1¢(uzv)  0(uxv)I > J. Pick m > 2k such that u, v E Cm. Let V = {w E X I0(uwv)0(uzv)I < L). Then V =7r1[7r[V]] so7r[V1is a neighborhood of7r(z)so choosen > msuchthatl0(uxnv)¢(uzv)I < L. SincealsoI (uxnv)0(uxv)I < n we have that Ib(uzv)  cb(uxv)I < k, a contradiction. Similarly, 7r(yn)  7r(y). So 7r(xn)7r(yn)
7r(xy) in Y. Let W = {w E X : I0(w)  4(xy)I < 8}. Then
7r[W] is a neighborhood of 7r(xy) in Y, so pick n such that 7r(xnyn) E 7r[W]. Then Iy(xnyn)  y(xy)I < 3, a contradiction. This establishes that is jointly continuous at every point of {x} x S. By symmetry, it is also jointly continuous at every point of S x {x}. We can now prove Ellis' Theorem.
Corollary 2.39 (Ellis' Theorem). Let S be a locally compact semitopological semigroup which is algebraically a group. Then S is a topological group.
If S is compact, let S = S. Otherwise, let S = S U {oo} denote the onepoint compactification of S. We extend the semigroup operation of S to S by putting s oo = oo s = oo for every s E S. It is then simple to check that S is a compact semitopological semigroup with an identity. So, by Theorem 2.38, the semigroup operation on S is jointly continuous at every point of (S x S) U (S x S). In particular, S is a topological semigroup. be a net To see that the inverse function is continuous on S, let x E S and let in S converging to x. We claim that (xa1)aEI converges to x1. Let z be any cluster point of (xa1)aEl in S and choose a subnet (xs1)6EJ which converges to z. Then by the continuity of at (x, z), (xsxs1)3EJ converges to xz. Therefore z = Proof.
x1.
Corollary 2.40. Let S be a compact semitopological semigroup with dense center. Then K(S) is a compact topological group.
Proof. By Corollary 2.26, K(S) is a semitopological group. Thus K(S) has a unique idempotent e and so K(S) = Se by Theorem 2.8. Thus K(S) is compact and so by Corollary 2.39, K(S) is a topological group. Exercise 2.5.1. Prove Lemma 2.36.
Notes
47
Notes Theorem 2.5 was proved for topological semigroups by K. Numakura [ 187] and A. Wallace [240, 241, 2421. The first proof using only one sided continuity seems to be that of R. Ellis [86, Corollary 2.10]. Theorem 2.9 is due to W. Ruppert [215]. Example 2.13 is from [38], a result of collaboration with J. Berglund. Example 2.16 is due to W. Ruppert [215], as is Theorem 2.19(a) [218].
Lemma 2.34 is a special case of a theorem of J. Christensen [63]. As we have already remarked, Corollary 2.39 is a result of R. Ellis [84]. The proof given here involves essentially the same ideas as that given by W. Ruppert in [216]. (An alternate proof of comparable length to that given here is presented by J. Auslander in [5, pp. 5763].) Theorem 2.38 was proved by J. Lawson [173]. For additional information see [216] and [174].
Chapter 3
BBD  Ultrafilters and the Stonetech Compactification of a Discrete Space
There are many different constructions of the Stonetech compactification of a topological space X. In the case in which the space is discrete, the Stonetech compactification of X may be viewed as the set of ultrafilters on S and it is this approach that we adopt.
3.1 Ultrafilters Definition 3.1. Let D be any set. A filter on D is a nonempty set U of subsets of D with the following properties:
(a) IfA,B E U, then Af1B E U.
(b) IfAE'andACBCD,then BEU. (c) 00U. A classic example of a filter is the set of neighborhoods of a point in a topological
space. (We remark that a neighborhood of a point is a set containing an open set containing that point. That is neighborhoods do not, in our view, have to be open.) Another example is the set of subsets of any infinite set whose complements are finite. We observe that, if U is any filter on D, then D E U.
Definition 3.2. Let D be a set and let U be a filter on D. A family .4 is a filter base for 2( if and only if A C U and for each B E U there is some A E A such that A'c B.
Thus, in a topological space, the open neighborhoods of a point form a filter base for the filter of neighborhoods of that point.
Definition 3.3. An ultrafilter on D is a filter on D which is not properly contained in any other filter on D. We record immediately the following very simple but also very useful fact about ultrafilters.
3.1 Ultrafilters
49
Remark 3.4. Let D be a set and let U and V be ultrafilters on D. Then U = V if and only if U c V. Those more familiar with measures may find it helpful to view an ultrafilter on D as a (0, 1)valued finitely additive measure on P(D). (The members of the ultrafilter are the "big" sets. See Exercise 3.1.2.)
Lemma 3.5. Let U be a filter on the set D and let A C D. Either
(a) there is some B E'U such that A fl B = 0 or (b) (C C D : there is some B E U with A fl B C C) is a filter on D. Proof. This is Exercise 3.1.3. Recall that a set A of sets has the finite intersection property if and only if whenever
F is a finite nonempty subset of A, n P * 0.
Theorem 3.6. Let D be a set and let U C P(D). The following statements are equivalent.
(a) 2( is an ultrafilter on D. (b) 2( has the finite intersection property and for each A E P (D)\U there is some
BE U such that AflB=0. (c) 21 is maximal with respect to the finite intersection property. (That is, U is a maximal member of (V C Y(D) : V has the finite intersection property).) (d) 2( is a filter on D and for all .T' E Pf (P (D_)), if U F E U, then F fl 1,l 96 0. (e) 2( is a filter on D and for all A c D either A E U or D\A E U.
Proof (a) implies (b). By conditions (a) and (c) of Definition 3.1, U has the finite intersection property. Let A E D (D)\'U and let V = (C c D : there is some B E 2(
with A fl B C C}. Then A E V so U Z V so V is not a filter on D. Thus by Lemma 3.5, there is some B E 2( such thatA fl B = 0.
(b) implies (c). If'
V C_ P(D), pick A E V\U and pick B E 2l such that
A fl B = 0. Then A, B E V so V does not have the finite intersection property. (c) implies (d). Assume that 2t is maximal with respect to the finite intersection property among subsets of R (D). Then one has immediately that 2l is a nonempty set of subsets of D. Since 2( U {D} has the finite intersection property and 'L c U U {D}, one has 2( = 2l U {D}. That is, D E U. Given A, B E 2(, U U (A fl B) has the finite
intersection property so A fl B E U. Given A and B with A E 2( and A C B C D, U U { B) has the finite intersection property so B E U. Thus 2C is a filter. Now let F E Pf (J" (D)) with U F E U, and suppose that for each A E .T', A 0 U.
Then given A E F, U Z U U (A) so U U {A} does not have the finite intersection property so there exists 9 A E Pf (') such that A fl n 9A = 0. Let 3e = UAEF gA. Then 3e U {U F) C 2l while (U F) f1 n 3e = 0, a contradiction. (d) implies (e). Let F = (A, D\A).
3 OD
50
(e) implies (a). Assume that ti is a filter on D and for all A C D either A E U or D\A E U. Let V be a filter with U C V and suppose that U 0 V. Pick A E V\U. Then D\A E U C V while A f1 (D\A) = 0, a contradiction.
If a E D, (A E .P(D) : a E A) is easily seen to be an ultrafilter on D. This ultrafilter is called the principal ultrafilter defined by a. Theorem 3.7. Let D be a set and let U be an ultrafilter on D. The following statements are equivalent.
(a) U is a principal ultrafilter. (b) There is some F E J" f(D) such that F E U. (c) The set (A C D : D\A is finite) is not contained in U.
(d) nt(#0. (e) There is some x E D such that n t(_ {x}. Proof. (a) implies (b). Pick X E D such that U = {A C_ D : x E A). Let F = {x}. (b) implies (c). Given F E J" f(D) f1 U one has D\F 0 U. (c) implies (d). Pick A C D such that D\A is finite and A gE U. Let F = D\A. Then F E U and F = U { {x) : x E F } so by Theorem 3.6, we may pick x E F such that {x} E U. Then for each B E U, B f1 {x} ,E 0 so x E n t(.
(d) implies (e). Assume that n t(# 0 and pick x E n U. Then D\{x} 0 U so
{x}EU so nUC{x}. (e) implies (a). Pick X E D such that n t(= {x}. Then U and (A C D : x E A) are both ultrafilters so by Remark 3.4 it suffices to note that t( C (A C D : x E A). It is a fact that principal ultrafilters are the only ones whose members can be explicitly defined. There are no others which can be defined within ZermeloFraenkel set theory.
(See the notes to this chapter.) However, the axiom of choice produces a rich set of nonprincipal ultrafilters on any infinite set.
Theorem 3.8. Let D be a set and let A be a subset of R(D) which has the finite intersection property. Then there is an ultrafilter U on D such that A C U.
Proof Let r = (2 C J"(D) : A C S and 2 has the finite intersection property). Then A E r so r, 0 0. Given a chain e in r one has immediately that A c U C. Given F E JPf( U C) there is some £ E C with F C 2 son F # 0. Thus by Zorn's Lemma we may pick a maximal member U of I'. Trivially U is not only maximal in r, but in fact U is maximal with respect to the finite intersection property. By Theorem 3.6, t( is an ultrafilter on D.
Corollary 3.9. Let D be a set, let A be a filter on D, and let A C D. Then A 0 A if and only if there is some ultrafilter t( with A U {D\A} C U. Proof. Since one cannot have a filter U with A E U and D\A E U, the sufficiency is trivial.
3.1 Ultrafilters
51
For the necessity it suffices by Theorem 3.8 to show that A U {D\A} has the finite intersection property. So suppose instead that there is some finite nonempty F C A
such that (D\A)flnF=0.ThennFCAsoAEA. The following concept will be important in the combinatorial applications of ultrafilters.
Definition 3.10. Let R be a nonempty set of sets. We say that R is partition regular if and only if whenever F is a finite set of sets and U F E J2, there exist A E F and B E J2 such that B C A. Given a property W of subsets of some set D, (i.e. a statement about these sets), we say the property is partition regular provided (A C D :'P(A)) is partition regular. Notice that we do not require that a partition regular family be closed under supersets (in some set D). (The reason is that there are combinatorial families, such as the sets of finite products from infinite sequences, that are partition regular under our requirement (by Corollary 5.15) but not under the stronger requirement.) See however Exercise 3.1.6
Theorem 3.11. Let D be a set and let J2 C '(D) be nonempty and assume that 0 V R. Let .Rf = {B E .?(D) : A C B for some A E R}. The following statements are equivalent. (a) 92 is partition regular.
(b) Whenever A C R(D) has the property that every finite nonempty subfamily of A has an intersection which is in R , there is an ultrafilter U on D such that
AC_UC_JRt. (c) Whenever A E R, there is some ultrafilter U on D such that A E U C J2
Proof. (a) implies (b). Let 2 = (A C D : for all B E R, A fl B 96 0} and note that 2 # 0 since D E S. Note also that we may assume that A 0, since {D} has the hypothesized property. Let C = A U S. We claim that e has the finite intersection property. To see this it suffices (since A and 2 are nonempty) to let F E ,P1(A) and 9. E J"f(2) and show that n F n n 9 o 0. So suppose instead that we have such F
and 9.withnFfln9,=0. PickB E,Rsuch that B CnF.Then Bfln9,=0and so B = UAeg(B\A). Pick A E 9. and C E ,R such that C C B\A. Then A fl C = 0, contradicting the fact that A E S. By Theorem 3.8 there is an ultrafilter 2( on D such that C C U. Given C E U,
D\C 0 S(since Cfl(D\C) = 0 0 U). SopicksomeB E J2suchthatBfl(D\C) = 0. That is, B C C. (b) implies (c). Let A = (A). (c) implies (a). Let F be a finite set of sets with U F E R and let U be an ultrafilter on D such that U F E U and for each C E U there is some B E R such that B C C. Pick by Theorem 3.6 some A E F fl U.
Definition 3.12. Let D be a set and let 'U be an ultrafilter on D. The norm of 2( is IIUII = min(IAI : A E 2l).
3 OD
52
Note that, by Theorem 3.7, if U is an ultrafilter, then 119111 is either 1 or infinite.
Definition 3.13. Let D be a set and let K be an infinite cardinal. A Kuniform ultrafilter
on D is an ultrafilter U on D such that 119111 ? K. The set UK(D) = {91 : 91 is a Kuniform ultrafilter on D}. A uniform ultrafilter on D is a Kuniform ultrafilter on D,
whereK = IDI. Corollary3.14. Let D be any set and let A be a family of subsets of D. If the intersection of every finite subfamily of A is infinite, then A is contained in an ultrafilter on D all of whose members are infinite. More generally, if K is an infinite cardinal and if the intersection of everyfinite subfamily of A has cardinality at least K, then A is contained in a Kuniform ultrafilter on D. Proof The property of being infinite is partition regular, and so is the property of having cardinality at least K. 0
If the intersection of every finite subfamily of A is infinite, A is said to have the infinite finite intersection property. Thus, Corollary 3.14 says that if A has the infinite finite intersection property, then there is a nonprincipal ultrafilter U on D with A c U.
Exercise 3.1.1. Let A be a set of sets. Prove that there is some filter U on D = U A such that A is a filter base for U if and only if A 54 0, 0 it A, and for every finite nonempty subset F of A, there is some A E A such that A c n ,F.
Exercise 3.1.2. Let D be any set. Show that the ultrafilters on D are in onetoone correspondence with the finitely additive measures defined on .P (D) which take values in {0, 1) and are not identically zero.
Exercise 3.1.3. Prove Lemma 3.5.
Exercise 3.1.4. Let D be any set. Show that the ultrafilters on D are in onetoone correspondence with the Boolean algebra homomorphisms mapping (.P (D), u, fl) onto the Boolean algebra ({0, 1}, v, n).
Exercise 3.1.5. Let D be a set with cardinality c. Show that there is no nonprincipal ultrafilter U on D with the property that every countable subfamily of U has an intersection which belongs to U. (Hint: D can be taken to be the interval [0, 1] in R. If an ultrafilter 91 with these properties did exist, every a E [0, 1] would have a neighborhood which did not belong to U.)
Exercise 3.1.6. Let ,9 be a set of sets and let D be a set such that U . C D. Let J2T = {B C D : there is some A E R with A C B}. Prove that the following statements are equivalent.
(a) JZ is partition regular.
(b) If F is a finite set of sets and there is some C E R with C c U T, then there
exist AE.Fand BED with BCA. (c) JZ t is partition regular.
(d) If F is a finite set of sets and U F E. t,thenFfl.R1 54 0.
3.2 The Topological Space #D
53
3.2 The Topological Space 8 D In this section we define a topology on the set of all ultrafilters on a set D, and establish some of the properties of the resulting space.
Definition 3.15. Let D be a discrete topological space. (a)13D = (p : p is an ultrafilter on D).
(b)Given AcD,A={pE,BD:AEp}. The reason for the notation 13 D will become clear in the next section. We shall hence forward use lower case letters to denote ultrafilters on D, since we shall be thinking of ultrafilters as points in a topological space.
Definition 3.16. Let D be a set and let a E D. Then e(a) = (A C D : a E A). Thus for each a E D, e(a) is the principal ultrafilter corresponding to a.
Lemma 3.17. Let D be a set and let A, B C D.
(a) (AFB)=AFB; (b) (AUB)=AUB; (c) (D\A) = flD\A;
(d) A=0ifand only ifA=0; (e) A = PD if and only if A = D; (f) A = B if and only if A = B. Proof. This is Exercise 3.2.1. We observe that the sets of the form A are closed under finite intersections, because
X n B = (A fl B). Consequently, {A : A C D} forms a basis for a topology on f D. We define the topology of f3D to be the topology which has these sets as a basis. The following theorem describes some of the basic topological properties of PD.
Theorem 3.18. Let D be any set. (a) fi D is a compact Hausdorff space. (b) The sets of the form A are the clopen subsets of,BD. (c) For every A C D, A = ce,6D e[A].
(d) For any A C D and any p E D, P E ctfD e[A] if and only if A E p. (e) The mapping e is injective and e[D] is a dense subset of /3D whose points are precisely the isolated points of PD. (f) If U is an open subset of /3D, c9,6 D U is also open.
Proof. (a) Suppose that p and q are distinct elements of PD. If A E p\q, then D\A E q. So A and (D\A) are disjoint open subsets of PD containing p and q respectively. Thus PD is Hausdorff. We observe that the sets of the form A are also a base for the closed sets, because
,BD\A = (D\A). Thus, to show that PD is compact, we shall consider a family A
3 flD
54
of sets of the form A with the finite intersection property. and show that A has a non
empty intersection. Let 2 = {A C_ D : A E A}. If T E Rf(2), then there is some p E nAE.F A and so n J' E p and thus n F # 0. That is, 2 has the finite intersection property, so by Theorem 3.8 pick q E PD with 2 c q. Then q E n A. (b) We pointed out in the proof of (a) that each set A was closed as well as open. Suppose that C is any clopen subset of PD. Let A = (X: A C D and A c C}. Since C is open, A is an open cover of C. Since C is closed, it is compact by (a) so pick a finite subfamily of R(0) such that C = UAE.c A. Then by Lemma 3.17(b), C = U F.
(c) Clearly, for each a E A, e(a) E A and therefore cepD e[A] c A. To prove the reverse inclusion, let p E A. If B denotes a basic neighborhood of p, then A EX and B E p and so A fl B 56 0. Choose any a E A fl B. Since e(a) E e[A] fl B, e[A] f1 B 0 0 and thus p E ct5D e[A]. (d) By (c) and the definition of A,
4 AEp. (e) If a, b E D are distinct, D\{a} E e(b)\e(a) and so e(a) # e(b). If A is a nonempty basic open subset of PD, then A : 0. Any a E A satisfies e(a) E e[D] fl A and so e[D] fl A # 0. Thus e[D] is dense in $D. For any a E D, e (a) is isolated in PD because {a} is an open subset of PD whose only member is e(a). Conversely if p is an isolated point of PD, then {p} fl e[D] # 0 and sop E e[D]. (f) If U = 0, the conclusion is trivial and so we assume that U i6 0. Put A = e1 [U].
We claim first that U C ciSD e[A]. So let p E U and let B be a basic neighborhood of p. Then U f1 B is a nonempty open set and so by (e), U fl B f1 e[D] # 0. So pick b E B with e(b) E U. Then e(b) E B fl e[A] and so B fl e[A] 0 0. Also e[A] C U andhence U C ci#D e[A] c ci,D U. Thus cifD U = ciflD e[A] _ A (by (c)), and so cifD U is open in PD. We next establish a characterization of the closed subsets of 0 D which will be useful later.
Definition 3.19. Let D be a set and let A be a filter on D. Then A = (P E PD A c p}. Theorem 3.20. Let D be a set. (a) If A is a filter on D, then A is a closed subset of PD. (b) If A C PD and A= n A, then A is a filter on D and A = ci A.
Proof. (a) Let p E PD\.A. Pick B E A\p. Then D\B is a neighborhood of p which misses A. (b) A is the intersection of a set of filters, so A is a filter. Further, for each p E A,
A C p so A C A and thus by (a), c2 A C_ A. To see that A c cP A, let p E A and let B E P. Suppose B fl A = 0. Then for each q E A, D\B E q so D\B E A C_ p, a contradiction.
3.3 Stonetech Compactification
55
It is customary to identify a principal ultrafilter e(x) with the point x, and we shall adopt that practice ourselves after we have proved that PD is the Stonetech compactification of the discrete space D in Section 3.3. Once this is done, we can write the following definition as A* = A\A. For the moment, we shall continue to maintain the distinction between x and e(x).
Definition 3.21. Let D be a set and let A C D. Then A* = A\e[A].
The following theorem is simple but useful. Once e(x) is identified with x the conclusion becomes "U fl D E P". Theorem 3.22. Let p E 0 D and let U be a subset of PD. If U is a neighborhood of p in PD, then a1[U] E p.
Proof. If U is a neighborhood of p, there is a basic open subset A of $D for which p E A C U. This implies that A E p and so a1 [U] E p, because A c e' [U]. Recall that a space is zero dimensional if and only if it has a basis of clopen sets.
Theorem 3.23. Let X be a zero dimensional space and let Y be a compact subset of X. The clopen subsets of Y are the sets of the form C fl Y where C is clopen in X. In particular, if D is an infinite set, then the nonempty clopen subsets of D* are the sets of the form A* where A is an infinite subset of D.
Proof. Trivially, if C is clopen in X, then c fl Y is clopen in Y. For the converse, let B be clopen in Y and let A = {A fl Y : A is clopen in X and A fl Y C_ B}. Since X is zero dimensional, A is an open cover of B. Since Y is compact and hence B is compact, pick a finite set Y of clopen subsets of X such that B = UAE.F (A fl Y) and
letC=UT.
The "in particular" conclusion follows from Corollary 3.14 and Theorem 3.18(b).
0 Exercise 3.2.1. Prove Lemma 3.17.
3.3 Stonetech Compactification In this section we show that 9D is the Stonetech compactification of the discrete space D. Recall that by an embedding of a topological space X into a topological space Z, one means a function W : X > Z which defines a homeomorphism from X onto rp[X]. We remind the reader that we are assuming that all hypothesized topological spaces are Hausdorff. Definition 3.24. Let X be a topological space. A compactification of X is a pair ((p, C) such that C is a compact space, cp is an embedding of X into C, and (p[X] is dense in C.
3 flD
56
Any completely regular space X has a largest compactification called its Stonetech compactification.
Definition 3.25. Let X be a completely regular topological space. A StoneCech compactifrcation of X is a pair (rp, Z) such that
(a) Z is a compact space, (b) (p is an embedding of X into Z, (c) V[X] is dense in Z, and (d) given any compact space Y and any continuous function f : X  Y there exists a continuous function g : Z a Y such that g o rp = f . (That is the diagram
z
f
X
Y
commutes.)
One customarily refers to the StoneCech compactification of a space X rather than a StoneCech compactification of X. The reason is made clear by the following remark. If one views rp as an inclusion map, then Remark 3.26 may be viewed as saying: "The StoneCech compactification of X is unique up to a homeomorphism leaving X pointwise fixed."
Remark 3.26. Let X be a completely regular space and let (rp, Z) and (r, W) be StoneCech compactifications of X. Then there is a homeomorphism y : Z + W such that
yore=r.
Theorem 3.27. Let D be a discrete space. Then (e, PD) is a StoneCech compactiftcation of D.
Proof Conditions (a), (b), and (c) of Definition 3.25 hold by Theorem 3.18. It remains for us to verify condition (d).
Let Y be a compact space and let f : D a Y. For each p E fD let AP = {cl y f [A] : A E p}. Then for each p E fiD, AP has the finite intersection property (Exercise 3.3.1) and so has a nonempty intersection. Choose g(p) E n AP. Then we have the following diagram.
OD
3.3 Stonetech Compactification
57
We need to show that the diagram commutes and that g is continuous.
For the first assertion, let x E D. Then {x} E e(x) so g(e(x)) E cty f [{x}] _ cfy[{f (x)}] = {f (x)} so g o e = f as required. To see that g is continuous, let p E f3D and let U be a neighborhood of g(p) in Y. Since Y is regular, pick a neighborhood V of g (p) with cty V c U and let A' [vi. We claim that A E p so suppose instead that D\A E p. Then g(p) E cty f [D\A]
and V is a neighborhood of g(p) so V fl f[D\A] 0 0, contradicting the fact that A = f'' [V ]. Thus X is a neighborhood of p. jWe claim that g[;?] c U, so let q E A and suppose that g (q) 0 U. Then Y\ ct y Visa neighborhood of g(q) and g (q) E cE y f [A] so (Y\ ct y V) fl f [A] # 0, again contradicting the fact that A = f ' [V ]. o
Although we have not used that fact, each of the sets n An is a singleton. (See Exercise 3.3.2.) We have shown in Theorem 3.27 that 3D is the Stonetech compactification of D. This explains the reason for using the notation fi D for this space: if X is any completely
regular space, fiX is the standard notation for its StoneCech compactification. X is usually regarded as being a subspace of OX.
Identifying D with e[D] It is common practice in dealing with 13D to identify the points of D with the principal ultrafilters generated by those points, and we shall adopt this practice from this point on. Only rarely will it be necessary to remind the reader that when we write s we sometimes mean e(s). Once we have identified s E S with e(s) E_fD, we shall suppose that D C 1D and shall write D* = flD\D, rather than D* = i4D\e[D]. Further, with this identification, A = cEOD A for every A E JP D, by Theorem 3.18. So the notations A and A become interchangeable. We illustrate the conversion process by restating Theorem 3.27.
Theorem 3.28 (Stonetech Compactification  restated). Let D be an infinite discrete space. Then (a) fD is a compact space,
(b)DcfD,
(c) D is dense in fD, and (d) given any compact space Y and any function f : D > Y there exists a continuous
function g : fiD * Y such that g1D = f.
Identifying fi T with T for T c S In a similar fashion, if T C S, we shall identify p E ? (which is an ultrafilter on S) with the ultrafilter (A fl T : A E p) (which is an ultrafilter on T) and thus we shall pretend
that f T C 0S.
3 OD
58
Exercise 3.3.1. Let D be a discrete space, let Y be a compact space, and let f : D > Y.
For each p E liD let Ap = {cfy f[Al : A E p}. Prove that for each p E fl D, Ap has the finite intersection property.
Exercise 3.3.2. Let the sets AP be as defined in the proof of Theorem 3.27. Prove that for each p E fD, n Abp is a singleton. (Hint: Consider the fact that two continuous functions agreeing on e[D] must be equal.)
Exercise 3.3.3. Let D be any discrete space and let A C D. Prove that cfOD A can be identified with ,8A.
3.4 More Topology of D If f is a continuous mapping from a completely regular space X into a compact space Y, we shall often use f to denote the continuous mapping from ,6 X to Y which extends f , although in some cases we may use the same notation for a function and its extension. (Notice that there can be only one continuous extension, since any two extensions agree on a dense subspace.)
Definition 3.29. Let D be a discrete space, let Y be a compact space, and let f : D > Y.
Then f is the continuous function from ,5D to Y such that fjD = f. If f : X  * Y is a continuous function between completely regular spaces, it has a continuous extension f 0 : lX + 13 Y. The reader with an interest in category theory might like to know that this defines a functor from the category of completely regular spaces to that of compact Hausdorff spaces, and that this is an adjoint to the inclusion. functor embedding the second category in the first.
Lemma 3.30. Let D and E be discrete spaces and let f : D k E C: j6E. For each p E ,BD, f (g) = (A C E : f '[A] E p). In particular, if A E p, then f [A] E T(p);
and if B E f (p), then f[B] E P. Proof. It is routine to verify that {A C E : f  '[A] E p} is an ultrafilter on E. For
each p E fD, let g(p) = (A C E : f1[A] E p). Now, given x E D we have g(x) = (A C E : f (x) E Al. Recalling that we are identifying x with e(x) (and f (x) with e(f (x))) we have g(x) = f (x). To see that g is continuous, let A be a basic open set in ;BE. Then g1 [A] = f1 [A]. Since g is a continuous extension of f, we have
.f =gLemma 3.31. Let D and E be discrete spaces, let f, g : D E, and let p E OD satisfy f (p) = g(p). Then for each A E p, (x E D : f (x) E g[A]} E P.
By Lemma 3.30, g[A] E g(p) = 7(p) so, again applying Lemma 3.30, .f' [g[A]] E p. Proof.
3.4 More Topology of P D
59
Lemma 3.32. Suppose that D is any discrete space and that f is a mapping from D to itself The mapping f : /D * #D has a fixed point if and only if every finite partition
of Dhas acell Cfor which Cfl f[C]#0. Proof. Suppose first that D has a finite partition in which every cell C satisfies C n f [C] = 0. Let p ,E PD and pick a cell C satisfying C E p. This implies by Lemma 3.30 that f [C] E f (p) and hence that f (p) p. Conversely, suppose that f has no fixed points. For each p E flD pick Ap E
pff(p), pick Bp E p such that f[Bp] fl Ap = 0, and let Cp = Ap fl Bp. Then {Cp : p E flD} is an open cover of PD so pick finite F C 81) such that (Cp : p E F) covers fiD. Then {Cp : p E F} covers D and can thus be refined to a finite partition F of D such that each C E F satisfies C fl f [C] = 0. Lemma 3.33. Suppose that D is a set and that f : D ± D is a function with no fixed points. Then D can be partitioned into three sets Ap, A 1, and A2 with the property that Ai fl f [Ai ] = O for each i E {0, 1, 2}.
Proof. We consider the set
9 _ {g : (i) g is a function, (ii) dom(g) C D, (iii) ran(g) C {0, 1, 2}, (iv) f [dom(g)] C dom(g), and
(v) for each a E dom(g), g(a) 0 g(f (a)) }. We observe that 9. is nonempty, because the function 0 is a member of .. Then 9. is partially ordered by set inclusion and, if e is a chain in 9, then U C E 9, so Zorn's Lemma implies that 9, has a maximal element g. We shall show that dom (g) = D. We assume the contrary and suppose that b E D\ dom(g). We then define a function h extending g. To do so, we put h(a) = g(a) for every a E doin(g). We choose h(b) to be a value in (0, 1, 2), choosing it so that h (b) 0 h (f (b)) if h (f (b)) has already been defined. Now suppose that h (fI (b)) has been defined f o r each m E {0, 1, 2, ... , n). (Where f ° (b) = b and f :+1(b) = f (f I (b)).) If h (f"+1 (b)) has not yet been defined,
we choose h (f n+1(b)) to be a value in (0, 1, 2) different from h (f" (b)) and from h(f"+2(b)), if the latter has already been defined. In this way, we can inductively define a function h E 9, with dom(h) = dom(g) U {f(b) : n E w}. Since g h, we have contradicted our choice of g as being maximal in 9. Thus dom(g) = D. We now define the sets Ai by putting Ai = g1 [{i }] for each i E (0, 1, 2}.
Theorem 3.34. Let D be a discrete space and let f : D * D. If f has no fixed points, neither does f : f3 D > 0 D. Proof. This follows from Lemmas 3.32 and 3.33.
Theorem 3.35. Let D be a set and let f : D > D. If f : 6 D ' 13 D and if p E P D, then 7(p) = p if and only if (a E D : f (a) = a) E p.
3 OD
60
Proof. Let E denote (a E D : f (a) = a). We first assume that T (P) = p. We want to show that E E p so suppose instead that D\E E p. We choose any b E D\E, and D by stating that g(a) = f (a) if a E D\E and g(a) = b if a E E. define g : D Then g has no fixed points. However, f = g on D\E and so f = g on ce(D\E). Since p E cf(D\E), g(p) = T (p) = p. This contradicts Theorem 3.34 and hence E E P. Conversely, suppose that E E p. Let t : D * D be the identity function. Since
f =ton E, f =Ton ce E. Since p E ct E, f (p) =T(p) = p. Theorem 3.36. If D is any infinite set, every nonempty Gssubset of D* has a nonempty interior in D*. Proof Suppose that, for each n E N, U is an open subset of PD and that fn= 1(Un f1 D*) # 0. Choose any p E f n_ 1(Un f1 D*). For each n E N, we canchoose a subset A of D for which p E An* c Un. We may assume that these sets are decreasing, because we can replace each An by n"_ 1 A; . Observe that each An is infinite, for otherwise we would have An* = 0. We can thus choose an infinite sequence (an)n of distinct points of D for which an E An for each n E N. Put A = (an : n E N). If q E A*, then q E An for every n E N, because A\An is finite. Thus the nonempty open subset A* of D* is contained in nn°_1(U. f1 D*). 1
Corollary 3.37. Let D be any set. Any countable union of nowhere dense subsets of D* is again nowhere dense in D*. Proof. For each n E N, let An be a nowhere dense subset of D*. To see that U0° An is nowhere dense, it suffices to show that B = U0° cf A. is nowhere dense. Suppose instead one has U = intD* cI B 96 0. By the Baire Category Theorem D*\B is dense 1
1
in D* so U\B ,{ 0. Thus U\B = f n_1(U\ cf An) is a nonempty Gs which thus, by Theorem 3.36, has nonempty interior, say V. But then v fl c2 B = 0 while V c U, a contradiction.
A point p in a topological space is said to be a Ppoint if the intersection of any countable family of neighborhoods of p is also a neighborhood of p. Thus an ultrafilter p E N* is a Ppoint of N* if and only if whenever (A,)n is a sequence of members of p, there is some B E p such that IB\An I < w for alll n E N. We see now that the Continuum Hypothesis implies that Ppoints exist in N*. It is a fact that their existence cannot be proved in ZFC. (See the notes to this chapter.) 1
Theorem 3.38. The Continuum Hypothesis implies that Ppoints exist in N*.
Proof. Assume the Continuum Hypothesis and enumerate R(N) as (Ca)a« with Co = N. Let ,o = {Co}. Inductively let 0 < a < w and assume that we have chosen Aa for all a < a such that (1) Aa has the infinite finite intersection property,
(2) either Ca E Aa or N\Ca E Aa, (3) if 8 < a, then As c Aa,
3.4 More Topology of ,5D
61
(4) IA. I < w, and (5) there exists A E Aa such that for each .F E Jlf (A.), I A\ n .T' I < co. Let (Bn)O°_1 list the elements of U., Aa (with repetition if necessary). If there is some n such that I Ca f, nm_ 1 B,n I < w, let D = hl\Ca and otherwise let D = Ca . Then for each n E N, I D fl f nn1 B I = w so we may choose a onetoone sequence (x n )°O n1
witheachxn E DflnmBn,. LetA = (xn : n E N) and let A, = (A, D) UUa
Let p = U0<1 Aa. Then by hypotheses (1) and (2), p E N*. Given a sequence (En)00 1 of members of p, pick a < wl such that {En : n E N} c (Ca : a < a). By hypothesis (5), there is some A E Aa C p such that IA\En I < w for each n E N. Definition 3.39. A topological space is said to be extremally disconnected if the closure of every open subset is open.
We showed in Theorem 3.18(f) that PD is an extremally disconnected space. Since
we often work with D*, the reader should be cautioned that D* is not extremally disconnected [104, Exercise 6W]. The following theorem will be very useful in our algebraic investigations of countable seniigroups.
Theorem 3.40. Let D be a discrete space and let A and B be acompact subsets of
PD. IfAflciB=ciAf1B=0,thenctAflciB=O. Proof. Write A = Un° An and B = Un_1 Bn where An and Bn are compact for each 1
n. Since PD is a compact (Hausdorff) space, it is normal. For each n E N, An and ce B are disjoint closed sets and ct A and Bn are disjoint closed sets so pick open sets Tn, Un,
Vn, and Wn such that Tn flUn = Vn nWn =0,An C 7n,c2B C Un,cfAC Vn,and Bn C W. For each n E N, let Gn = Tn fl nk=1 Vk and let Hn = Wn fl nk1 Uk. Then for each n, one has An C Gn and Bn C Hn. Further, given any n, m E N, Gn fl Hn, = 0.
Let C = Un_1 Gn and let D = Un_1 Hn. Then C and D are disjoint open sets
(so D fl ac= 0), A C C, and B C D. By Theorem 3.18 (f) ce C is open, so ci B fl ce C = 0. Since ce A C ce C we have ce A fl ct B = 0 as required. The following equivalent (see Exercise 3.4.2) version of Theorem 3.40 is also useful.
Corollary 3.41. Let D be a discrete space and let A and B be acompact subsets of PD. Then ct A fl ct B C ce(A fl ct B) U ce(B fl ct A). Proof. Let p E ce A fl ct B and suppose that p 0 ce(A fl ce B) U ct(B fl ce A). Pick
H E psuch that Hfl(AnctB)=OandHfl(BlctA)=0. Let A'=HflAand B' = H fl B. Then A' and B' are acompact subsets of fiD and p e ct A' fl ct B' so by Theorem 3.40 we may assume without loss of generality that A' fl ct B'
0 so pick
q E A' fl ce B'. Then q E H fl (A n ct B), a contradiction. We isolate some specific instances of Theorem 3.40 that we shall often use.
3 OD
62
Corollary 3.42. Let D be a discrete space. (a) Let A and B be acompact subsets of D*. If A n ct B = ct A n B = 0, then
ciAnciB=0.
(b) Let A and B be countable subsets of ,8 D. If A n c2 B = c8 A n B = 0, then
ciAnc2B = 0. (c) Let A and B be countable subsets of D*. If A n cf B = ct A n B = 0, then
ciAnciB=0.
Proof. Countable sets are acompact. By Theorem 3.18 (e) D* = #D\e[D] is a closed subset of OD so acompact subsets of D* are acompact in O D. The following notion will be used in the exercises and later.
Definition 3.43. Let X be a topological space and let A C X. Then A is strongly discrete if and only if there is a family (U.I)XEA of open subsets of X such that x E Ux for each x E A and Ux n Uy = 0 whenever x and y are distinct members of A.
Exercise 3.4.1. Suppose that' : D > E is a mapping from a discrete space D to a discrete space E. Prove that f : OD  13E is injective if f is injective, surjective if f is surjective and a homeomorphism if f is bijective. Exercise 3.4.2. Derive Theorem 3.40 from Corollary 3.41.
Exercise 3.4.3. Prove that any infinite regular space has an infinite strongly discrete subset.
Exercise 3.4.4. Let X be an extremally disconnected regular space. Prove that no sequence can converge in X unless it is eventually constant.
Exercise 3.4.5. Let D be any set. Prove that no proper Fesubset of D* can be dense in D*.
Exercise 3.4.6. Let D be any set. Prove that no zero set in D* can be a singleton. (A
subset Z of a topological space X is said to be a zero set if Z = f
[{0}] for some
continuous function f : X + [0, 1].) (Hint: Use Theorem 3.36.)
Exercise 3.4.7. Let D be any discrete space. Prove that every separable subspace of !4D is extremally disconnected. (Hint: Use Theorem 3.40.)
3.5 Uniform Limits via Ultrafilters We now introduce the notion of plimit. The definition of plimit is very natural to anyone familiar with nets. We would like p lim xS = y to mean that x, is "often" SED
3.5 Uniform Limits via Ultrafilters
63
"close to" Y. Closeness is of course determined by neighborhoods of y while "often" is determined by members of p. (Recall that an ultrafilter can be thought of as a {0, 1 }valued measure.) As we shall see, the notion is as versatile as the notion of nets, and has two significant
advantages: (1) in a compact space a plimit always converges and (2) it provides a "uniform" way of taking limits, as opposed to randomly choosing from among many possible limit points of a net.
Definition 3.44. Let D be a discrete space, let p E ,BD, let (xs)SED be an indexed family in a topological space X, and let y E X. Then pl o x5 = y if and only if for every neighborhood U of y, IS E D : Xs E U} E p. Recall that we have identified A with 0A for A C D. Consequently, if A E p and x5 is defined for s E A we write p lim xs without worrying about possible values of SEA x5 for s E D\A. There is a more general concept of limit in a topological space, which may well be familiar to the reader. We shall show that plimits and limits coincide for functions defined on PD.
Definition 3.45. Suppose that X and Y are topological spaces, that A C X and that f : A * Y. Let X E cfx A and Y E Y. We shall write lint f (a) = y if and only if, for every neighborhood V of y, there is a neighborhood U of x such that f [A fl U] C v. We observe that lim f (a), if it exists, is obviously unique. a.x Theorem 3.46. Let D be a discrete space, let Y be a topological space, and let p E 13 D
and Y E Y. If A E p and f : A + Y, then p lim f (a) = y if and only if aEA lim f (a) = y. a>p
Proof. Suppose that p lim f (a) = y. Then, if V is a neighborhood of y, f aEA
[V ] E p.
Let B = f [V ]. Then B is a neighborhood of p by Theorem 3.18 and f [B fl A] _ f [B] C V. Conversely, suppose that lim f (a) = y. Then, if V is a neighborhood of y, there
ap
is a neighborhood U of p in PD for which f [A fl U] C V. Now u fl A E p and so, since U f1 A C f'[V], it follows that f ' [V] E p. Thus plion f (a) = y. Remark 3.47. Let D be a discrete space and let p E PD. Viewing (s)SED as an indexed
family in fiD one has p lim s = p. sED
Theorem 3.48. Let D be a discrete space, let p E PD, and let (XS)SED be an indexed family in a topological space X. (a) If p lim x5 exists, then it is unique. sED
(b) If X is a compact space, then p lim xs exists. sED
3 OD
64
Roof. Statement (a) is obvious. (b) Suppose that p lira (xs)sED does not exist and for each y E X, pick an open neighborhood Uy of y such that Is E D : Xs E Uy} 0 p. Then {Uy : y E X} is an open
cover of X so pick finite F c X such that X = UYEF U. Then D = UyEF{s E D : Xs E Uy ) so picky e F such that {s E D : xs E Uy } E p. This contradiction completes the proof.
Theorem 3.49. Let D be a discrete space, let p E i8D, let X and Y be topological spaces, let (XS)SED be an indexed family in X, and let f : X ). Y. If f is continuous and p IEm xs exists, then p l f (xs) = f (p Em xs).
D
Proof. Let U be a neighborhood of f (p limb xs) and pick a neighborhood V of p lii o xs
such that f [V] c U. Let A= IS E D: xs E V). Then A E p and A c (s E D: .f(xs) E U). We pause to verify that Theorems 3.48 and 3.49 in fact provide a characterization of ultrafilters.
Definition 350. Let D be a discrete space. A uniform operator on functions from D to compact spaces is an operator O which assigns to each function f from D to a compact space Y some point Off) of Y such that whenever Y and Z are compact spaces, g is a function from D to Y, and f is a continuous function from Y to Z, one
has f(0(g)) =O(f o g). Theorems 3.48 and 3.49 tell us that given a discrete space D and p E ,BD, "plim" is a uniform operator on functions from D to compact spaces. The following theorem provides the converse. Theorem 3.51. Let D be a discrete space and let O be a uniform operator on functions
from D to compact spaces. There is a unique p E fl D such that for every compact space Y and everyfunction f : D .* Y, 0(f) = plim f (s). sED
Proof. Let t : D ). D c #D be the inclusion map. (Before we had identified the points of D with principal ultrafilters, we would have said t = e.) Now the choice of p E j6D is forced, since we must have 0(t) = p li Bs = p, so let p = 0(t). Now
let Y be a compact space and let f : D > Y. Let f : PD ). Y be the continuous extension of f . Then
p Siam f(s) = p sD .f(s) = f (p lim s) SED
= f(p) = f(0(t)) = O(fot) = O(f).
0
3.5 Uniform Limits via Ultrafilters
65
We now verify the assertion that the notion of plimit is as versatile as the notion of nets by proving some of the standard theorems about nets in the context of plimits. Theorem 3.52. Let X be a topological space. Then X is compact if and only if whenever (xs)SED is an indexed family in X and p is an ultrafilter on D, p llint exists.
proof The necessity is Theorem 3.48(b). Sufficiency. Let A be a collection of closed subsets of X with the finite intersection
property. Let D = {n F: ,T' is a finite nonempty subset of A}. For each A E D let BA = {B E D : B C Al. Then ( A : A E D} has the finite intersection property. (The proof of this assertion is Exercise 3.5.1.) So by Theorem 3.8, pick an ultrafilter p on D such that { BA : A E D) c p. For each A E D, pick xA E A. Let y = p l o xA. A
We claim that y E n A. To see this let A E A and suppose that y 0 A. Then X \A is a neighborhood of y so {B E D : XB E X\A} E p. Also SA E p so pick B ELA such that XB E X\A. Then (X\A) fl B 0 while B C A, a contradiction. Theorem 3.53. Let X be a topological space, let A c X and let y E X. Then Y E ce A if and only if there exists an indexed family (xs)sED in A such that p urn xs = Y. SED
Proof. Sufficiency. Let U be a neighborhood of y. Then is E D : xs E U) E p, so
(sED:x5EU) ,a<0. Necessity. Let D be the set of neighborhoods of y. For each U E D pick XU E
U fl A and let SU = {V E D : V C U). Then { Bu : U E D} has the finite intersection property so pick an ultrafilter p on D with {;BU : U E D} c p. To see that p i ii D xU = y, let U be a neighborhood of y. Then Su {V E D : xv E U) so
{VED:xVEU)EP. Theorem 3.54. Let X and Y be topological spaces and let f : X * Y. Then f is continuous if and only if whenever (XS)SED is an indexed family in X, p E BBD, and
p lim xs exists, one has f (p lim xs) = p lim f (xs). SED
SED
sED
Proof. The necessity is Theorem 3.49
For the sufficiency, let a E X and let W be a neighborhood of f (a). Suppose that no neighborhood U of a has f [U] c W. Then a E ct f 1 [Y\W] so pick by Theorem 3.53 an indexed family (xs)sED in f 1 [Y\W] such that p lim xs = a. Then sED
p sin f(x5) = f (a) so is r= D : f (xs) E W} is in p and is hence nonempty, a contradiction.
Comment 3.55. The concept of limit is closely related to that of continuity. Suppose that X, Y are topological spaces, that A C X and that x E A. If f : A > Y, then f is continuous at x if and only if for every neighborhood V of f (x), there is a neighborhood U of x for which f [U] C V. So f is continuous at x if and only if lim f (a) = f (x). a x
3 OD
66
Also, if x E cE A, then f has a continuous extension to A U {x} if and only if lim f (a) aix exists. Exercise 3.5.1. Let A be a collection of closed subsets of the topological space X with
the finite intersection property. Let D = in F : F is a finite nonempty subset of A}. For each A E D let QA = {B E D : B C A}. Prove that {SA : A E D} has the finite intersection property.
Exercise 3.5.2. Let (x,,)n° 1 be a sequence in a topological space X and assume that
Rlimo x = a. Prove that for each p E flN\N, p lim x = a. nEN
3.6 The Cardinality of $D 22iDi.
We show here that for any infinite discrete space D, I3DI = We do the proof for the case in which D is countable first. The proof uses the following surprising theorem which is of significant interest in its own right. Recall that c is the cardinality of the continuum. That is c = 2W = IRI = )I. Theorem 3.56. Let D be a countably infinite set. There is a c x c matrix ((Aa,, ) a <)s of subsets of D satisfying the following two statements.
«
(a) Given a, 3, r < c with 3 0 r, IAa,s fl A,,,I < co. (b) Given any F E .'P1(c) and any g : F + c, I ICEF Aa,g(o)I = 0) I
Proof. Let
S={(k, f):kENandf :J"({1,2,...,k}), J'({l,2,...,k})}. Then S is countable so it suffices to produce a c x c matrix of subsets of S satisfying statements (a) and (b). Enumerate ,P(N) as (Xa)o<. Given (Q, 3) E c x c , let Aa,s = 1(k, f) E S : f (Xa fl [1, 2, ..., k}) = Xa fl 11, 2, ... , k} }.
To verify statement (a), let a, 3, r < c with 3 # r. Pick M E N such that x6 fl ( 1 , 2, ..., m) # Xt n( 1 ,2 , . . . , m}. Then Xa,s fl Xa,T C {(k, f) E S : k < m}, a finite set.
To verify statement (b), let F E .Pf (c) and g : F ). c be given. Pick m E N such that given any a # r in F, Xa fl { 1 , 2, ... , m) # X7 fl { 1, 2, ..., m). Then given any k > m, there exists
f :, ({1,2,...,k}) J"({1,2,...,k}) such that (k, f) E
I
OEF
Aa,g(a). Consequently IaEF Aa,g(a) is infinite. I
3.6 The Cardinality of fiD
67
Corollary 3.57. Let D be a countably infinite discrete space. Then 1,8 DI = 2`.
Proof. Since 6D c J'(.P(D)), one has that I1DI < 2`. Let ((Aa,a)a« )a« be as in Theorem 3.56. For each f : c ). {0, 1), let A f = {Aa,f (a) : a < c}. Then by Theorem 3.56(b), for each f , A f has the property that all finite intersections are infinite, so pick by Corollary 3.14 a nonprincipal ultrafilter p f on D such that A f c pf. If f and g are distinct functions from c to {0, 11, then since p f and pg are nonprincipal, one has by
Theorem 3.56(a) that pf # pg. We now prove the general result, using a modification of the construction in Theorem 3.56.
Theorem 3.58. Let D be an infinite set with cardinality K. Then IUK(D)I = 22K.
Proof Again we note that BD C .P (Y (D)) and so I UK(D)I < I P D I < 22K.
We define S = {(F, f) : F E Pf(D) and f :.P(F) > J"(F)}. Note that ISI = K, and so it will be sufficient to prove that I UK (S) I > 22'
.
For each X C D, we put Ax,o = {(F, f) E S : f(F n x) = F n X} and Ax,i = {(F, f) E S : f(F n x) = F\X}. Then Ax,p n Ax,j = 0 for every X. Let g :.P(D) * 10, 1}. We shall show that, for every finite subset e of 5'(D), I nxEc Ax,g(x) I = K. We can choose B E J" f(D) for which the sets of the form x n B with X E C, are all distinct. (For each pair (X, Y) of distinct elements of C, pick a point b(X, Y) in the symmetric difference X A Y. Let B = {b(X, Y) : X, Y E C and X ; Y}.) Then,
whenever F E J'f(D) and B C F, the sets x n F for X E C are all distinct. For each F E Pf(D) such that B C F, we can define f : J" (F) + R(F) such that, for
each X E C, f (X n F) = X n F if g(X) = 0 and f(Xf1F)=F\X if g(X) = 1. Then (F, f) E Ax,g(x) for every X E C. Since there are K possible choices for F, I nxEe Ax,g(x) I = K.
It follows from Corollary 3.14 that, for every function g : .P(D) + (0, 11, there is an ultrafilter p E UK(D) such that Ax,g(x) E p for every X E Y (D). Since Ax,p n Ax,I = 0 for every X,22K different functions correspond to different ultrafilters in UK(D). Since there are functions g : ,P(D) . (0, 1), it follows that I` LK(D)I > 22K.
We now see that infinite closed subsets of BBD must be reasonably large.
Theorem 3.59. Let D be an infinite discrete space and let A be an infinite closed subset of BBD. Then A contains a topological copy of #N. In particular I AI ? 2`. Proof. Choose an infinite strongly discrete subset B of A. (The fact that one can do this is Exercise 3.4.3.) Let (p, be a onetoone sequence in B. Since { p : n E N) is
strongly discrete choose for each n some C E p such that c,, n C,,, = 0 whenever Ont.
3flD
68
Define f : N * fiD by f (n) = p,,. Then f : #N + A is a continuous function with compact domain, which is onetoone and so f [14N] is homeomorphic to ^ (The proof that f is onetoone is Exercise 3.6.1.) We conclude by showing in Theorem 3.62 that sets which are not too large, but have finite intersections that are large, extend to many xuniform ultrafilters.
Definition 3.60. Let A be a set of sets and let K be an infinite cardinal. Then A has the Kuniform finite intersection property if and only if whenever Y, E Rf (A), one has
IflFl>K. Thus the "infinite finite intersection property" is the same as the "wuniform finite intersection property".
Lemma 3.61. Let D be an infinite set with cardinality K and let A be a set of at most K subsets of D with the Kuniform finite intersection property. There is a set 2 of K pairwise disjoint subsets of D such that for each B E 2, A U {B} has the Kuniform finite intersection property. Proof. Since IRf(A)1 < K we may presume that A is closed under finite intersections. Since K I A I = K, we may choose a xsequence (Ao )a
1(a
(1) for ally
(3) for ally
such that IYI=lrlandY1fl(Uy
Having chosen (YO)a
each IB,,I = K. Further, if 3 < rf < K, then BS fl B,, = 0. (Since (y,,,,),,<, is an y, we enumeration of Ya we can't have yQ,s = ya and since Ya fl Yy = 0 for a can't have ya,,7 = yy,s.)
Finally, let n < K and F E Rf(A) be given. Let
C={a:J1
I{pEUK(D):AC p}I =2K. 2
Notes
69
Proof. Choose by Lemma 3.61 a family B of x pairwise disjoint subsets of D such that for each B E 2, A U {B} has the xuniform finite intersection property. Choose a onetoone function 9 : D > 2 and for each x E D, pick by Corollary 3.14 some f (x) E UK(D) such that A U {9(x)} C f (x). Notice that UK (D) is a closed, hence compact, subset of PD. Thus, by Theorem
3.28, there is a continuous extension f : PD * UK(D). Since for each x E D, A C f (x), we have for each q E PD that A C T(q) To complete the proof, it suffices to show by Theorem 3.58 that f is onetoone. For this it in turn suffices to show that for each q E PD and each A E q, UXEAB(X) E
.7(q). (For then, given q# r and A E q and C E r with A fl C= 0 one has that (UXEA9(X)) n (UXEc9(x)) = 0.) The proof of this assertion is Exercise 3.6.2.
Exercise 3.6.1. Prove that the function fin the proof of Theorem 3.59 is onetoone. Exercise 3.6.2. Let D and E be discrete spaces, let
9:DAP(E) and f:D+8E be functions such that for each x E D, 9(x) E f (x), and let f : /3D + fl E be the continuous extension of f . Prove that for each q E fiD and each A E q, UXEAO(X) E
.f (q) Exercise 3.6.3. Prove that N* contains a collection of c pairwise disjoint clopen sets. Exercise 3.6.4. Prove that N* contains exactly c clopen subsets. Exercise 3.6.5. Prove that every clopen subset of N* is homeomorphic to N*.
Notes The standard reference for detailed information about ultrafilters is [69].
We used the Axiom of Choice (or to be precise, Zorn's Lemma) in our proof of Theorem 3.8, which we used to deduce that nonprincipal ultrafilters exist. Such an appeal is unavoidable as there are models of set theory (without choice) in which no nonprincipal ultrafilters on N exist. (See the discussion of this point in [69, p.162].) The Stonetech compactification was produced independently by M. Stone [227]
and E. tech [60]. The approach used by Cech was to embed a space X in a product
of lines. A method that is close to the one that we chose in defining PD, is due to H. Wallman [244]. Suppose that X is a normal space. Let Z denote the set of closed subsets of X. The points of 1X are defined to be the ultrafilters in the lattice of closed subsets of X, and the topology of PX is defined by taking the sets of the form {p E ,3X : Z E p}, where Z E Z, as a base for the closed sets. The space X is embedded in 0X by mapping each x E X to {Z E Z : x E Z}, this being an ultrafilter
70
3 6D
in the lattice of closed sets. The approach used here for discrete spaces is based on the treatment in [104]. See the notes in [104, p. 269] for a more detailed discussion of the development of this approach. Lemma 3.33 is due to M. Katetov [167]. Ppoints were invented (and shown to exist in N* under the assumption of the continuum hypothesis) by W. Rudin [214] who used them to show that N* is not homogeneous. It is a result of S. Shelah [222, VI, §4] that it is consistent relative to ZFC that there are no Ppoints in N*. The reader may be interested to know that extremally disconnected spaces have remarkable properties. For example, any continuous function mapping a dense subspace of an extremally disconnected space into a compact space, has a continuous extension to the entire space. [104, Exercise 6M2] Extremally disconnected spaces can be characterized among completely regular spaces by the following property: suppose that X is a completely regular space and that Ca (X) denotes the set of bounded continuous realvalued functions defined on X. Then X is extremally disconnected if and only if every bounded subset of Ca(X) has a least upper bound in Cx(X). [104, Exercise 3N6} For the reader with some acquaintance with category theory, we might mention that it is the extremally disconnected compact Hausdorff spaces which are the projective objects in the category of compact Hausdorff spaces. [107] The notion of plimit is apparently due originally to Z. FrolIk [95] Theorem 3.56 is due to K. Kunen [ 170], and answered a question of F. Galvin. The proof given here is a simplification of Kunen's original proof due to P. Simon.
Closing Remarks More general constructions of compact spaces. Our construction of 6 D is a special case of more general constructions in which compact Hausdorff spaces are obtained by using sets which are maximal subject to having certain algebraic properties. We shall briefly mention two of these constructions, because they are relevant to the Stonetech compactification. (i) Any unital commutative C*algebra is associated with a compact Hausdorff space, called its spectrum, whose points could be described as maximal ideals of the algebra or, alternatively, as homomorphisms from the algebra to the complex numbers. Let X be a completely regular space and let C(X) denote the C*algebra formed by the continuous bounded complexvalued functions defined on X, with each f E C(X) having the uniform norm 11f 11 = sup{ I f (X) J : x E X). Then'6X can be identified with the spectrum of C(X). In this framework, if D denotes a discrete space, PD is the spectrum of the Banach algebra 1'O(D) C(D). See [166].
Closing Remarks
71
(ii) Any Boolean algebra B is associated with a totally disconnected compact Hausdorff space, called its Stone space, whose points could be described as the ultrafilters of B. All totally disconnected compact Hausdorff spaces arise in this way, as any such space can be identified with the Stone space associated with the Boolean algebra formed by its clopen subsets. This theory displays the category of totally disconnected compact Hausdorff spaces as being the dual of the category of Boolean algebras. The extremally disconnected compact spaces are those corresponding to complete Boolean algebras. If D is a discrete space, /3D could be described as the Stone space of the Boolean algebra ,P (D), while D* could be described as the Stone space of the quotient of ,P (D) by the ideal of finite subsets of D. See [165].
Compactifications and subalgebras of CR(X). Suppose that X is a completely regular space and that CR(X) denotes the subspace of C(X) which consists of the realvalued functions. A subalgebra A of CR(X) is said to separate points and closed sets if, for each closed subset C of X and each x E X\C, there is a function f E A for which f (x) 0 cE f [C]. The compact spaces in which X can be densely embedded correspond to the closed subalgebras of CR(X) which contain the constant functions and separate points and closed sets. For any such algebra A, there is a compactification (Y, 40) of X which has the following property: a function f E Ca (X) can be expressed as f = g o cp for some g E CR(Y) if and only if f E A. Y could be defined as the spectrum of A, and, for each x E X, tp(x) could be defined as the homomorphism from A to the complex
numbers for which rp(x)(f) = f (x). Conversely, if (Y, gyp) is a compactification of X, it can be identified with the com
pactification arising in this way from the subalgebra A = {g o (p : g E CR(Y)) of CR(X). The Stonetech compactification of X corresponds to the largest possible subalgebra, namely CR(X) itself. Any other compactification is a quotient of AX. If Y is such a compactification, corresponding to the subalgebra A of CR(X), Y can be identified with the quotient of fiX by the equivalence relation p q if and only if T (p) = T (q)
for every f E A. We shall address this issue again in Chapter 21.
FSpaces. Extremally disconnected spaces are examples of spaces called Fspaces. A completely regular space X is said to be an Fspace if any two disjoint cozero subsets A and B of X are completely separated. This means that there is a continuous function f : X * [0, 11 for which f [A] = {0} and f [B] = {1}. These spaces are significant in the theory of the Stonetech compactification, because, if X is locally compact, $X\X is an Fspace (See [104].) It is quite easy to see that a compact Hausdorff space X satisfies Theorem 3.40 if
and only if it is an Fspace. (See for example [154, Lemma 1.1].) In particular, if D is a discrete space, every compact subset of fiD is an Fspace. Thus D* is an Fspace, although  unlike 0 D  it need not be extremally disconnected.
Chapter 4
S  The StoneCech Compactification of a Discrete Semigroup
4.1 Extending the Operation to CBS We shall see in this chapter that we can extend the operation of a discrete semigroup to its StoneCech compactification. When we say that (S, ) is a semigroup we intend also that it have the discrete topology. We remind the reader that we are pretending that S c S. Without the identification of the point s of S with the principal ultrafilter e(s), conclusions (a) and (c) of Theorem 4.1 would read as follows: (a) The embedding e is a homomorphism. (c) For all s E S, Ae(s) is continuous.
Theorem 4.1. Let S be a discrete space and let be a binary operation defined on S. There is a unique binary operation * : 1S x PS + PS satisfying the following three conditions: (a) For every s, t E S, s * t = s t.
(b) For each q E fl S, the function pq : PS + PS is continuous, where pq (p) = p *q. (c) For each s E S, the function As : fiS ) fIS is continuous, where Is(q) = s * q.
Proof. We establish uniqueness and existence at the same time, defining * as we are forced to define it. We first define * on S x fl S. Given any s E S, define Ls : S a S C
fJ S
by is(t) = s t. Then by Theorem 3.28, there is a continuous function Is : ,5S + PS such that Asks = is. If S E S and q E ,BS, we defines * q = As(q). Then (c) holds and so does (a), because As extends is. Furthermore, the extension As is unique (because continuous functions agreeing on a dense subspace are equal), and so this is the only possible definition of * satisfying (a) and (c). Now we extend * to the rest of 13S x PS. Given q E PS, define rq : S + )SS by rq(s) = s * q. Then there is a continuous function pq : 1S > 14S such that pqI s = rq. For p E PS\S, we define p * q = pq (p) and note that ifs E S, pq (s) = rq (s) = s * q. So for all p E fS, pq (p) = p * q. We observe that (b) holds. Again, by the uniqueness
4.1 Extending the Operation to ,BS
73
of continuous extensions, this is the only possible definition which satisfies the required conditions. It is customary to denote the operation on fl S by the same symbol as that used for the operation on S and we shall adopt that practice. Thus if the operation on S is denoted by
we shall talk about the operation on PS. If the operation is denoted by +, we shall talk about the operation + on fS. (There will be exceptions. If we are working with the semigroup (Yf (N), U) we shall not talk about the semigroup (P JPf (N), U), since the union of two ultrafilters already means something else, because ultrafilters are sets.)
We do not yet know that (VS, ) is a semigroup if (S, ) is a semigroup, as we have not yet verified associativity. We shall do this shortly, but first we want to get a handle on the operations. The operation on PS has a characterization in terms of limits. The statements in the following remark follow immediately from the fact that X is continuous for every s E S and Pq is continuous for every q E S. (Here s and t represent members of S.)
Remark 4.2. Let be a binary operation on a discrete space S. ES E PS, t h e n  q =lim s  t. t+q (b) If p, q E PS, then p q = lim (lim s t). s*p tq (a)
The following version of Remark 4.2(b) will often be useful.
Remark 4.3. Let be a binary operation on a discrete space S, let p, q E ,S, let
P E p, and let Q Eq. Then p q = p lim(qlim s t). SEP
tEQ
We remind the reader that if f : X > Y is a continuous function and lim f (xs) S*p
and lim xs exist, then lim f (x,.) = f (lim xs), by Theorem 3.49. We shall regularly s+p s+p s P use this fact without mentioning it explicitly.
Theorem 4.4. Let (S, ) be a semigroup. Then the extended operation on /3S is associative.
Proof. Let p, q, r E /3S. We consider lim lim lim(a b) c, where a, b and c denote a P b+q c>r
elements of S. We have:
lim lim lim (a b) c = lim lim (a b) r
aWpb.q c>r
aWpbq
= lim ap(a q) r
_ (p q) r
(because ,la.b is continuous) (because pr 0 Aa is continuous)
(because pr o pq is continuous).
Also:
lima (b r) lim lim lim a (b c) = lim a.pbgcsr a+pb>q
lima (q r)
a p p
= p (q r)
(because Xa 0 lb is continuous). (because Xa 0 Pr is continuous) (because Pq r is continuous).
4 fiS
74
(q r).
The following theorem illustrates again the virtue of the uniformity of the process of passing to limits using plimit. Theorem 4.5. Let (S, ) be a semigmup, let X be a topological space, let (x,.),5Es be an indexed family in X, and let p, q E 3 S. If all limits involved exist, then (p q)lim x _
mq lr P liSES Proof.
VES
xst
Let z = (p q) lim x and, for each s E S, let ys = q lim xst. Suppose
that p m ys # z and pick disjoint open neighborhoods U and V of p lim ys and z SES respectively. Let A = {v E S : x E V) and let B = {s E S : ys E U). Then A E p q and B E P. Since A is a neighborhood of pq(p), pick C E p (so that C is a basic neighborhood of p) such that pq [C] S; A. Then B fl C E p so picks E B fl C. Since
SEB,qlimxsrEU.LetD=(tES:xstEU).Then DEq.Since tES
Then D fl E E q so pick t E D fl E. Since t E D, xst E U. Since t E E, St E A so xst E V and hence U fl V # 0, a contradiction. so A is a neighborhood of As (q). Pick E E q such that As
Observe that the hypotheses in Theorem 4.5 regarding the existence of limits can be dispensed with if X is compact by Theorem 3.48. Observe also, that if X is contained in a compact space Y (i.e. if X is a completely regular space) the proof is much shorter. For then let f : S > X be defined by f (s) = xs and let f be the continuous extension
off to PS. Then one has lim
= (P.q)lim f(v) = .7((p . q) lim v)
=.f(P.q)
vEs
= f(plimqlimst)
SES tE, = p limq lim f (st)
SES
tES
= p lim q lim xst SES
ZES
In the case in which (S, ) is a semigroup, conclusion (c) of Theorem 4.1 may seem extraneous. Indeed it would appear more natural to replace (c) by the requirement that the extended operation be associative. We see in fact that we lose the uniqueness of the extension by such a replacement.
Example 4.6. Let + be the extension of addition on N tofN satisfying (a), (b), and (c) of Theorem 4.1. Define an operation * on fN by
ifgEfN\N
P*q={qp + q ifgEN.
4.1 Extending the Operation to (3S
75
Then * is associative and satisfies statements (a) and (b) of Theorem 4.1, but if q E pN\N, then 1 * q # 1 + q. The verification of the assertions in Example 4.6 is Exercise 4.1.1. As a consequence of Theorems 4.1 and 4.4 we see that (74S, ) is a compact right topological semigroup, so that all of the results of Chapter 2 apply. In addition, we shall see that (fS, ) is maximal among semigroup compactifications of S in a sense similar to that in which 78S is maximal among topological compactifications.
Recall that, given a right topological semigroup T, A(T) = {x E T : kx is continuous}.
Definition 4.7. Let S be a semigroup which is also a topological space. A semigroup compactification of S is a pair ((p, T) where T is a compact right topological semigroup, (p : S ). T is a continuous homomorphism, (p[S] C A(T), and (p[S] is dense in T. It is customary to assume in the definition of a semigroup compactification that S is a semitopological semigroup. However, as we shall see in Chapter 21, this restriction is not essential. Note that a semigroup compactification need not be a (topological) compactification because the homomorphism (p need not be onetoone.
Theorem 4.8. Let (S, ) be a discrete semigroup, and let i : S + 73S be the inclusion map.
(a) (t, PS) is a semigroup compactification of S. T is a continuous (b) If T is a compact right topological semigroup and (p : S homomorphism with cp[S] C A(T) (in particular, if ((p, T) is a semigroup compactification of S), then there is a continuous homomorphism ri : 8S a T such that 771s = (p. (That is the diagram 1S
commutes.)
Proof. Conclusion (a) follows from Theorems 3.28, 4.1, and 4.4. (b) Denote the operation of T by *. By Theorem 3.28 choose a continuous function p : CBS + T such that P71s = (p. We need only show that r7 is a homomorphism. By Theorems 3.28 and 4.1, S is dense in jS and S C A(,8S), so Lemma 2.14 applies. Comment 4.9. We remark that the characterization of the semigroup operation in 0 S in terms of limits is valid in all semigroup compactifications. Let S be any semigroup with topology and let ((p, T) be any semigroup compactification of S. For every p, q E T, we have: pq = lim lim (p(s)(p(t) W(s)gyp w(t)+q
4 PS
76
where s and t denote elements of S. (The notation lim. xs = y means that for any rp(s)>p
neighborhood U of y, there is a neighborhood V of p such that whenever s E S and cp(s) E V, one has xs E U.) Since the points of 16S are ultrafilters, we want to know which subsets of S are members of p q.
Definition 4.10. Let (S, ) be a semigroup, let A c S, and lets E S.
(a)s'A=(tES:StEA}. (b)As'={tES:tsEA}.
Note that s' A is simply an alternative notation for XS [A], and its use does not imply that s has an inverse in A. If t E S, we may use to to denote {ta : a E A}. This does not introduce any conflict, because, ifs does have an inverse s1 in A, then JAS' [A] = {s'a : a E A}. However, even if S can be embedded in a group G (so that s1 is defined in G), s1A need not equal {s'a : a E A}. For example, in (N, ), let
A=N2+1={2n+1:n EN}.Then 21A={teN:2tEA}=0. We shall often deal with semigroups where the operation is denoted by +, and so we introduce the appropriate notation.
Definition 4.11. Let (S, +) be a semigroup, let A C S, and let s E S.
(a)s+A= It E S : s + t E A}.
(b)As={tES:t+sEA}.
Theorem 4.12. Let (S, ) be a semigroup and let A c S.
(a) For any sESand
ifsAEq.
(b) For any p, q E PS, A E p q if and only if {s E S: S_' A E q} E p. Proof (a) Necessity. Let A E s q. Then A is a neighborhood of),, (q) so pick B E q such that As[B] C A. Since B C s1 A, we are done. ,
Sufficiency. Assume sIA E q and suppose that A 0 s q. Then S\A E s q so, by the already established necessity, s
(S\A) E q. This is a contradiction since
s' A n s' (S\A) = 0. (b) This is Exercise 4.1.3.
The following notion will be of significant interest to us in applications involving idempotents in PS.
Definition 4.13. Let (S, ) be a semigroup, let A c_ S, and let p E S. Then A* (p)
{sEA:s'AEp}..
_
We shall frequently write A* rather than A*(p). As an immediate consequence of Theorem 4.12(b), one has that a point p of 8S is an idempotent if and only if for every A E p, A* (p) E p. And, of course, if A E p and s E A*(p), then s1 A E p. In fact, one has the following stronger result.
4.1 Extending the Operation to ,6S
77
Lemma 4.14. Let (S, .) be a semigroup, let p p = p E AS, and let A C S. For each s E A*(p), st (A*(P)) E p.
Proof Lets E A"(p), and let B = si A. Then B E p and, since p is an idempotent, B*(p) E p. We claim that B*(p) C s1 (A*(p)). So let t E B*(p). Then t E B so st E A. Also tt B E p. That is, (st)' A E p. (See Exercise 4.1.4.) Since st E A and (st)' A E p, one has that st E A'(p) as required.
Theorem 4.15. Let (S, ) be a semigroup, let p, q E AS, and let A C S. Then A E p q if and only if there exists B E p and an indexed family (CS), IB in q such that USEB SC5 C A.
Proof. This is Exercise 4.1.5.
Lemma 4.16. Let (S, ) be a semigroup, lets E S, let q E AS and let A C S.
(a)IfAEq,then (b) If S is left cancellative and sA E s q then A E q. Proof (a) We have A C s  t (s A) so Theorem 4.12(a) applies. (b) Since S is left cancellative, s' (s A) = A. The following results will frequently be useful to us later. Theorem 4.17. Let S be a discrete semigroup, let (cp, T) be a semigroup compactification of S, and let A C B C S. Suppose that B is a subsemigroup of S. (a) ct(cp[B]) is a subsemigroup of T. (b) If A is a left ideal of B, then cf(cp[A]) is a left ideal of ct(cp[B]). (c) If A is a right ideal of B, then ct(cp[A]) is a right ideal of ct(([B]).
Proof (a) This is a consequence of Exercise 2.3.2. (b) Suppose that A is a left ideal of B. Let X E ct(cp[B]) and y E ct(cp[A]). Then xy = lim X lim y cp (s)cp (t ), where s denotes an element of Band t denotes an element of A. Ifs E B and t E A, we have ip(s)cp(t) = cp(st) E W[A] and so xy E ct (P[A]. (c) The proof of (c) is similar to that of (b).
Corollary 4.18. Let S be a subsemigroup of the discrete semigroup T. Then ct S is a subsemigroup of AT. If S is a right or left ideal of T, then cf S is respectively a right or left ideal of f3T.
Proof. By Theorem 4.8, (t, ,BT) is a semigroup compactification of T, so Theorem 4.17 applies.
Remark 4.19. Suppose that S is a subsemigroup of a discrete semigroup T. In the light of Exercise 4.1.10 (below), we can regard AS as being a subsemigroup of AT. In particular, it will often be convenient to regard AN as embedded in #7G.
78
4 flS
Theorem 4.20. Let (S, ) be a semigroup and let A C P (S) have the finite intersection property. If for each A E A and each x E A, there exists B E A such that x B C A, then nAEA A is a subsemigroup of 8S.
Proof. Let T= nAE.A A. Since A has the finite intersection property, T # 0. Let p, q E T and let A E A. Given X E A, there is some B E A such that xB c A and hence x1A Eq. Thus A C {x E S : x'1A E q} so {x E S : xtA E q} E p. By Theorem 4.12(b), A E p q. Theorem 4.21. Let (S, ) be a semigroup and let A C .I (S) have the finite intersection property. Let (T, ) be a compact right topological semigroup and let cp : S + T satisfy rp [S] C A (T). Assume that there is some A E A such that for each x E A, there exists B E A for which rp(x y) _ (p (x) (p (y) for every y E B. Then for all p, q E nAE.A A,
W(p  q) = W(p) W(q) Proof. Let p, q E nAEA A. For each x E A, one has
lim rp(x  y)
y'q = lim yq(P (x)
(P (Y)
= p(x) lim (P(y) y+q
because W o Ax is continuous
because p(x y) = W(x) sp(y) on a member of q because O(x) E A(T)
= cp(x) W(q) Since A E p one then has
W(p  q) = W((z
lim W(x q)
x+p
because ip o pgis continuous
= 1i P(gq(x) W(q))
_ (lim xpp (x)) W(q)
by the continuity of po(q)
= W(p) W(q>
0
Corollary 4.22. Let (S, ) be a semigroup and let V : S * T be a homomorphism to a compact right topological semigroup (T, ) such that (p[S] C A(T). Then ip is a homomorphism from 6S to T.
Proof. Let A=( S). Exercise 4.1.1. (a) Prove that if q E ISN\N and r E N, then q + r E fN\N. (b) Verify that the operation * in Example 4.6 is associative. (c) Verify that the operation * in Example 4.6 satisfies conclusions (a) and (b) of Theorem 4.1. (d) Let q E ,BN\N and let A = N2 = {2n : n E N}. Prove that A E q = 1 * q if and only if A 0 1 + q.
4.1 Extending the Operation to fiS
79
Exercise 4.1.2. Show that one cannot dispense with the assumption that cp[S] c A(T) in Theorem 4.8. (Hint: consider Example 4.6.) Exercise 4.1.3. Prove Theorem 4.12(b).
Exercise 4.1.4. Let S be a semigroup, let s, t E S, and let A C S. Prove that s' (t I A) = (ts) 1 A. (Caution: This is very easy, but the reason is not that s I t(ts)1, for no such objects need exist.) Exercise 4.1.5. Prove Theorem 4.15. We can generalize Theorem 4.15 to the product of any finite number of ultrafilters. The following exercise shows that we obtain a set in pi P2 Pk by choosing all products of the form a i a2 ... ak, where each ai is chosen to lie in a member of pi which depends on a 1 , a2, ... , ai_1.
Exercise 4.1.6. Let (S, ) be a semigroup, let k E N\{1}, and let p,, P2, ... , Pk E 8S. Suppose that D c {0} U Si, With 0 E D. Suppose also that, for each v E D, there is a subset AQ of S such that the following conditions hold:
(1) A0 E PI, (2) A0 C D, (3) AQ E Pi+I if O E D n Si, and (4) if k > 2, then f o r every i E { 1, 2, ... , k2) and every a = (al, a2, ... , ai) E D,
a E Aa implies that (a1, a2, ... , ai, a) E D.
ak:aI EA0and(a1,a2,...,ai1)EDandai EA(aj,a2....,a1 for every i E {2, 3, ... , k} }. Prove that P E PI P2 by induction on k, using Theorem 4.12 or 4.15).
...
)
Pk. (Hint: This can be done
Exercise 4.1.7. Let P E N* + N* and let A E P. Prove that there exists k E N such that {a E A : a + k E A} is infinite. Deduce that A cannot be arranged as a sequence (x n )' , for which Xn+1  Xn  no as n > no. Exercise 4.1.8. Let (S, ) be a semigroup, let A C_ S, let s E S and let q E 6S. (a) Prove that A E q s if and only if As I Eq. (b) Prove that if A E q, then As E q s. (c) Prove that if As E q s and S is right cancellative, then A E q.
Exercise 4.1.9. Let (S, ) be a semigroup and let s E S. Prove that the following statements are equivalent.
(a) For every
sp pE (c) The function As : S  S is surjective.
E p}.
4 PS
80
Exercise 4.1.10. Recall that if S c T, we have identified /3S with the subset S of ,T. Show that if (S, .) is a subsemigroup of (T, ) and p, q E PS, then the product p  q is the same whether it is computed in fS or in T. That is, if r = {A C S : {x E S x1A E q} E p} (the product p  q computed in MS), then {B C T : B fl s E r} = {A c T : {x E T : x1A E q} E p} (the product p q computed in PT). (The notation xIA is ambiguous. In the first case it should be {y E S : xy E A} and in the second case it should be (y E T : xy E A).) Exercise 4.1.11. We recall that the semigroup operations v and A are defined on ` i by n v m = max{n, m} and n A m = min{n, m}. As customary, we use the same notation for the extensions of these operations to 8N. (a) Prove that for p, q E fiN,
pvq
if q E fN\N p if q E N and p E $N\N.
Jq
(b) Derive and verify a similar characterization of p A q.
4.2 Commutativity in fiS We provide some elementary results about commutativity in fiS here. We shall study this subject more deeply in Chapter 6. Theorem 4.23. If (S, ) is a commutative semigroup, then S is contained in the center
of (PS, ) Proof. Lets E S and p E ,8S. Then
s  q = lim st t.q = lim tq is = (lim t) t>q
by Remark 4.2(a)
s
since p,s is continuous
Theorem 4.24. Let S be a discrete commutative semigroup. Then the topological center of fiS coincides with its algebraic center.
Proof. Suppose that p E A(PS) and that q E fiS. Since AP is continuous, it follows that
pq= q lim(t p) tES
=q.p
(by Theorem 4.23)
4.2 Commutativity in 0S
81
Thus p is also in the algebraic center of fS. Conversely, if p is in the algebraic center of $S, ,lP is continuous because AP = PP. Thus p E A($S). We have defined the operation on PS in such a way as to make CBS right topological.
It is not clear whether this somehow forces it to be semitopological: In the case in which S is commutative, we see that this question is equivalent to asking whether /3S is commutative.
Theorem 4.25. Let (S, ) be a commutative semigroup. The following statements are equivalent.
(a) (fS, ) is commutative. (b) (,BS, ) is a left topological semigroup. (c) (CBS, ) is a semitopological semigroup. Proof. The fact that (a) implies (b) is trivial as is the fact that (b) implies (c). It follows from Theorem 4.24 that (c) implies (a).
Lemma 4.26. Let (S, ) be a semigroup, let and (yn)n° I be sequences in S. and let p, q E fiS. If {{xn : n > k} : k E N} C q and {yk : k E N} E p, then Proof. This follows from Theorem 4.15.
We are now able to characterize when $S is commutative. (A moment's thought will tell you that this characterization says "hardly ever".) Theorem 4.27. Let (S, ) be a semigroup. Then (P S, ) is not commutative if and only if there exist sequences (xn)n°_t and (yn)n_i such that
ENandk
and y 1
B p and C E E C. Inductively given xt, x2, ... , xn , y2, ... , y,,, choose xn+t E B fl lk=I yk' (S\A) and yn+t E C fl lk= xk A.
k
As a consequence of Theorem 4.27, we see that neither ($N, +) nor ($N, ) is commutative, and hence by Theorem 4.25 neither is a left topological semigroup. (As we shall see in Chapter 6, in fact the center of each is N.) Exercise 4.2.1. Prove that if S is a left zero semigroup, so is $S. In this case show that 13S is not only a semitopological semigroup, but in fact a topological semigroup.
4 6S
82
Exercise 4.2.2. Prove that if S is a right zero semigroup, so is #S. In this case show that 6 BS is not only a semitopological semigroup, but in fact a topological semigroup. (Note that because of our lack of symmetry in the definition of the operation on IBS this is not a "dual" of Exercise 4.2.1.)
4.3 S* For many reasons we are interested in the semigroup S* = ,9S\S. In the first place, it is the algebra of S* that is the "new" material to study. In the second place it turns out that the structure of S* provides most of the combinatorial applications that are a large part of our motivation for studying this subject. One of the first things we want to know about the "semigroup S*" is whether it is a semigroup. Theorem 4.28. Let S be a semigroup. Then S* is a subsemigroup of fiS if and only if for any A E J"f (S) and for any infinite subset B of S there exists F E J'f (B) such that
nXEF x' I A is finite. Proof. Necessity. Let a finite nonempty subset A and an infinite subset B of S be given.
Suppose that for each F E P1(B), nXEF xIA is infinite. Then {xIA : x E B} has the property that all of its finite intersections are infinite so by Corollary 3.14 we may
pick p E S* such that {xIA : x E B} c p. Pick q E S* such that B Eq. Then A E q p and A is finite so by Theorem 3.7, q p E S, a contradiction. (Recall once again that we have identified the principal ultrafilters with the points of S.)
Sufficiency. Let p, q E S* be given and suppose that q p = y E S, (that is, precisely, that q p is the principal ultrafilter generated by y). Let A = {y} and let
B = {x E S: xIA E p}. Then B E q while for each F E J'f(B), one has nXEF
xIA E p so that
fXEF
xI A is infinite, a contradiction.
Corollary 4.29. Let S be a semigroup. If S is either right or left cancellative then S* is a subsemigroup of ,S. Proof. This is Exercise 4.3.1. We can give simple conditions characterizing when S* is a left ideal of $S. Definition 4.30. A semigroup S is weakly left cancellative if and only if for all u, v E S, {x E S : ux = v} is finite. Similarly, S is weakly right cancellative if and only if for all
u,vES,{xES:xu=v}is finite. Of course a left cancellative semigroup is weakly left cancellative. On the other hand the semigroup (N, v) is weakly left (and right) cancellative but is far from being cancellative.
4.3 S*
83
When we say that a function f is finitetoone, we mean that for each x in the range of f, f 1 [{x}] is finite. Thus a semigroup S is weakly left cancellative if and only if for each x E S, Ax is finitetoone.
Theorem 4.31. Let S be an infinite semigroup. Then S* is a left ideal of $ S if and only if S is weakly left cancellative.
Proof. Necessity. Let x, y E S be given, let A = Ay1 [{x}] and suppose that A is infinite. Pick P E S* fl A. Then y p = x, a contradiction. Sufficiency. Since S is infinite, S* # 0. Let P E S*, let q E 8S and suppose that So picky E S such that
y1 {x}
is hence nonempty. E p. But y1 {x} = ,ly1[{x}] so Ay1[{x}] is infinite,
a contradiction.
The characterization of S* as a right ideal is considerably more complicated. Theorem 4.32. Let S be an infinite semigroup. The following statements are equivalent. (a) S* is a right ideal of 6S. (b) Given any finite subset A of S, any sequence (Zn)n 1 in S, and any onetoone
sequence (xn)n_1 in S, there exist n < m in N such that x zm V A. (c) Given any a E S, any sequence (zn)n°_1 in S, and any onetoone sequence
(xn)n°_1 in S, there exist n < m in N such that xn zm # a. Proof. (a) implies (b). Suppose {xn Zm : n, m E N and n < m) c A. Pick P E 8S such that {{Zm : m > n} : n E N} c p and pick q E S* such that (xn : n E N) E q, which one can do, since {xn : n E N) is infinite. Then by Lemma 4.26, A E q p so by Theorem 3.7, q p E S, a contradiction. The fact that (b) implies (c) is trivial. (c) implies (a). Since S is infinite, S* 0 0. Let p E ;BS, let q E S*, and suppose that q p = a E S. Then IS E S : s1 Jul E p} E q so choose a onetoone sequence (x,,)' 1 such that {xn : n E N} C IS E S : s {a) E p}. Inductively choose a sequence (Zn)no 1 in S such that for each m E N, Zm E nn 1 xn1 (a) (which one can do since nn 1 xn 1 {a } E p). Then for each n < m in N, xn Zm = a, a contradiction. We see that cancellation on the appropriate side guarantees that S* is a left or right ideal of PS. In particular, if S is cancellative, then S* is a (two sided) ideal of PS.
Corollary 4.33. Let S be an infinite semigroup. (a) If S is left cancellative, then S* is a left ideal of PS. (b) If S is right cancellative, then S* is a right ideal of 6S. Proof. This is Exercise 4.3.2.
Corollary 4.34. Let S be an infinite semigroup. If S* is a right ideal of fS, then S is weakly right cancellative.
4 OS
84
Proof Let a, y E S and suppose that pyt[{a}] is infinite. Choose a onetoone sequence (xn)n_I in pyI [{a}] and f o r each M E N let zm = Y. Corollary 4.35. Let S be an infinite semigroup. If S is weakly right cancellative and for all but finitely many y E S, Ay is finitetoone, then S* is a right ideal of jS. Proof Suppose S* is not a right ideal of f3S and pick by Theorem 4.32(c) some a E S, a onetoone sequence (xn )R° I in S. and a sequence (zn) I in S such that Xn zm = a
; for all n E for all n < m in N. Now if {zm : m E N} is infinite, then '[(a)] is infinite, a contradiction. Thus {z. : m E N} is finite so we may pick b such that
{m E N : zm = b} is infinite. Then, given n E N one may pick m > n such that zm = b and conclude that xn b = a. Consequently, pb'I [(a)] is infinite, a contradiction.
Somewhat surprisingly, the situation with respect to S* as a two sided ideal is considerably simpler than the situation with respect to S* as a right ideal. Theorem 4.36. Let S be an infinite semigroup. Then S* is an ideal of CBS if and only if S is both weakly left cancellative and weakly right cancellative.
Proof Necessity. Theorem 4.31 and Corollary 4.34. Sufficiency. Theorem 4.31 and Corollary 4.35. The following simple fact is of considerable importance, since it is frequently easier to work with S* than with PS. Theorem 4.37. Let S be an infinite semigroup. If S* is an ideal of fl S, then the minimal left ideals, the minimal right ideals, and the smallest ideal of S* and of CBS are the same. Proof. For this it suffices to establish the assertions about minimal left ideals and minimal right ideals, since the smallest ideal is the union of all minimal left ideals (and
of all minimal right ideals). Further, the proofs are completely algebraic, so it suffices to establish the result for minimal left ideals. First assume L is a minimal left ideal of PS. Since S* is an ideal of CBS we have L C S* and hence L is a left ideal of S*. To see that L is a minimal left ideal of S*, let L' be a left ideal of S* with L' C L. Then by Lemma 1.43(b), L' = L. Now assume L is a minimal left ideal of S*. Then by Lemma 1.43(c), L is a left
ideal of PS. If L' is a left ideal of ,BS with L' C L, then L' is a left ideal of S* and consequently L' = L. Exercise 4.3.1. Prove Corollary 4.29. Exercise 4.3.2. Prove Corollary 4.33. Exercise 4.3.3. Give an example of a semigroup S which is not left cancellative such that S* is a left ideal of 0 S.
4.4 K(8S) and Its Closure
85
Exercise 43.4. Give an example of a semigroup S which is not right cancellative such that S* is a right ideal of CBS.
Exercise 43.5. Prove that (N*, +) and (_N*, +) are both left ideals of (AZ, +). (Here _N* = {p : p E N*} and p is the ultrafilter on Z generated by {A : A E p}.)
Exercise 43.6. Let T = fl EN Cepz nZ. (a) Prove that T is a subsemigroup of (flZ, +) which contains all of the idempotents. (Hint: Use the canonical homomorphisms from Z onto 7L,1.)
(b) Prove that T is an ideal of (flZ, ).
4.4 K($S) and Its Closure In this section we determine precisely which ultrafilters are in the smallest ideal K (V S) of 9S and which are in its closure. We borrow some terminology from topological dynamics. The terms syndetic and
piecewise syndetic originated in the context of (N, +). In (N, +), a set A is syndetic if and only if it has bounded gaps and a set is piecewise syndetic if and only if there exist a fixed bound b and arbitrarily long intervals in which the gaps of A are bounded by b. Definition 4.38. Let S be a semigroup. (a) A set A C S is syndetic if and only if there exists some G E Rf(S) such that
S = UtEG t'A. (b) A set A C S is piecewise syndetic if and only if there is some G E Pf(S) such that {a1(UIEG t1 A) : a E S} has the finite intersection property.
Equivalently, a set A C S is piecewise syndetic if and only if there is some G E p1(S) such that for every F E .9'f (S) there is some x E S with F x c UtEG t'A. We should really call the notions defined above right syndetic and right piecewise syndetic. If we were taking 46S to be left topological we would have replaced
"S =
t1A" in the definition of syndetic by "S = U:EG At1". We also
would have replaced "a' (UIEG t1 A)" in the definition of piecewise syndetic by "(UIEG At1)a"". We shall see in Lemma 13.39 that the notions of left and right piecewise syndetic are different. The equivalence of statements (a) and (c) in the following theorem follows from Theorem 2.10. However, it requires no extra effort to provide a complete proof here, so we do.
Theorem 4.39. Let S be a semigroup and let p E equivalent.
(a) p E K(8S). (b) For all A E p, {x E S : x A E p} is syndetic.
(c) ForallgE,8S,
pEfSqp.
S. The following statements are
4 6S
86
Proof. (a) implies (b). Let A E p and let B = {x E S : x1 A E p). Let L be the minimal left ideal of 6S for which p E L. For every q E L, we have p E 16S q = ctps(S q). Since A is a neighbourhood of p in /3S, t q E A for some t E S and so q E t1A. Thus the sets of the form ttA cover the compact set L and hence L C UIEG tt A for some finite subset G of S. To see that S c UIEG t1B, let a E S. Then a p E L so pick t E G such that
a p E t'A. Then t'A E a p so that (ta)1A = a1(t'A) E p and so to E B and thus a E t1 B.
(b) implies (c). Let q E 8S and suppose that p 0 /3S q p. Pick A E p such that A fl /3S . q p = 0. Let B = {x E S : x1A E p} and pick G E ?j(S) such that S = UrEG t1 B. Pick t E G such that t1 B E q. Then B E tq. That is, {x E S : x1A E p} E tq so A E tqp, a contradiction. (c) implies (a). Pick q E K(/3S). Theorem 4.40. Let S be a semigroup and let A C S. Then An K(/BS) i4 0 if and only if A is piecewise syndetic.
Proof Necessity. Let P E K(/3S) fl A and let B = {x E S : xA E p}. Then by Theorem 4.39, B is syndetic and so S = UIEG t1 B for some G E .?f(S). For each a E S, a E t1 B for some t E G and so a1(t'A) = (ta)'A E P. It follows t1A) : a E S} has the finite that a'1(U,EG tI A) E p and hence that {a1(UfEG intersection property.
Sufficiency. Assume that A is piecewise syndetic and pick G E J'f (S) such that (a1 ( UrEG t_'A) : a E S) has the finite intersection property. Pick q E /3S such
that {a1(UrEG t' A) : a E S} c q. Then S q c UIEG t1 A. This implies that (/3S) q S UrEG t1 A. We can choosey E K(,BS) fl (,lS q). We then have y E t1 A
for some t E G and sot y E A n K(/3S). Corollary 4.41. Let S be a semigroup and let p E 8S. Then p E cB K(PS) if and only if every A E p is piecewise syndetic. Proof. This is an immediate consequence of Theorem 4.40. The members of idempotents in K (/3S) are of particular combinatorial interest. (See Chapter 14.)
Definition 4.42. Let S be a semigroup and let A C_ S. Then A is central in S if and only if there is some idempotent p E K (/3 S) such that A E P. Theorem 4.43. Let S be an infinite semigroup and let A C S. The following statements are equivalent.
(a) A is piecewise syndetic. (b) The set {x E S : x1 A is central} is syndetic. (c) There is some x E S such that x1 A is central.
4.4 K(fS) and Its Closure
87
proof. (a) implies (b). Pick by Theorem 4.40 some p E K(66S) with A E p. Now K (0S) is the union of all minimal left ideals of ,BS by Theorem 2.8. So pick a minimal
left ideal L of $S with p E L and pick an idempotent e E L. Then p = p e so pick y E S such that y'1A E e. Now by Theorem 4.39 B = (z E S : zt(y'A) E e} is syndetic, so pick finite G C S such that S = UtEG tt B. Let D = {x E S : x1A is central}. We claim that S= U,E(y.G)tt D. Indeed,letx E Sbegivenandpickt E Gsuch that E B. Then
(t . x)(y1 A) E e so (t x)'t (y't A) is central. But (t x)t (yt A) = (y t x)t A. Thus y t x E D so x E (y . t)t D as required. (b) implies (c). This is trivial. (c) implies (a). Pick X E S such that x'A is central and pick an idempotent p E K ($ S) such that x t A E p. Then A E x p and x p E K (,B S) so by Theorem 4.40, A is piecewise syndetic. Recall from Theorem 2.15 and Example 2.16 that the closure of any right ideal in a right topological semigroup is again a right ideal but the closure of a left ideal need not be a left ideal.
Theorem 4.44. Let S be a semigroup. Then ct K(18S) is an ideal of $S. Proof. This is an immediate consequence of Theorems 2.15, 2.17, and 4.1.
Exercise 4.4.1. Let S be a discrete semigroup and let A CS. Show that A is piecewise
syndetic if and only if there is a finite subset F of S for which ctos (UrEF t' A) contains a left ideal of $5.
Exercise 4.4.2. Let A _: N. Show that A is piecewise syndetic in (N, +) if and only if there exists k E N such that, for every n E N, there is a set J of n consecutive positive integers with the property that A intersects every subset of J containing k consecutive integers.
Exercise 4.4.3. Prove that {EfEF22i (N, +).
:
F E JPf((w)} is not piecewise syndetic in
Exercise 4.4.4. Let A c N. Show that K($N, +) c A if and only if, for every k E N, there exists nk E N such that every subset of N which contains nk consecutive integers must contain k consecutive integers belonging to A. Exercise 4.4.5. Let G be a group and let H be a subgroup of G. Prove that H is syndetic if and only if H is piecewise syndetic. (Hint: By Theorem 4.40, pick p E K($G) such
that H E p and consider {x E G : x' H E p}.) Exercise 4.4.6. Let G be a group and let H be a subgroup of G. Show that the index of H is finite if and only if ctPG(H) fl K($G) # 0. (By the index of H, we mean the number of left cosecs of the form a H).
4,BS
88
Exercise 4.4.7. Let G be a group. Suppose that G can be expressed as the union of a finite nuinber of subgroups, H1, H2, ... , H. Show that, for some i E {1, 2, ... , n}, Hi has finite index. Exercise 4.4.8. Let S be a discrete semigroup and let T be a subsemigroup of S. Show that T is central if (ceps T) f1 K(,8S) 0 0. (Apply Corollary 4.18 and Theorem 1.65.) Exercise 4.4.9. Show that if S is commutative, then the closure of any right ideal of,3S is a two sided ideal of PS. (Hint: See Theorems 2.19 and 4.23.)
Notes We have chosen to extend the operation in such a way as to make (PS, ) a right topological semigroup. One can equally well extend the operation so as to make (fS, ) a left topological semigroup and in fact this choice is often made in the literature. (It used to be the customary choice of the first author of this book.) It would seem then that one could find two situations in the literature. But no, one instead finds four! The reader will recall that in Section 2.1 we remarked that what we refer to as "right topological" is called by some authors "left topological" and vice versa. In the following table we include one citation from our Bibliography, where the particular combination of choices is made. Obviously, when referring to the literature one must be careful to determine what the author means. Called Right Topological
Called Left Topological
pp Continuous
[192]
[45]
A,, Continuous
[215]
[178]
Example 4.6 is due to J. Baker and R. Butcher [6]. The existence of a compactification with the properties of P S given by Theorem 4.8 is [40, Theorem 4.5.31, where it is called the £NCcompactification of S. The fact that an operation on a discrete semigroup S can be extended to t3S was first implicitly established by M. Day in [75] using methods of R. Arens 13]. The first explicit statement seems to have been made by P. Civin and B. Yood [64]. These mathematicians tended to view 13S as a subspace of the dual of the real or complex valued functions on S. The extension to i6S as a space of ultrafilters is done by R. Ellis [86, Chapter 8] in the case that S is a group. Corollary 4.22 is due to P. Milnes [ 184]. The results in Section 4.3 are special cases of results from [130] which are in turn special cases of results due to D. Davenport in [72]. The notion of "central" has its origins in topological dynamics. See Chapter 19 for the dynamical definition and a proof of the equivalence of the notions.
Notes
89
Exercise 4.4.7 was suggested by I. Protasov. This result is due to B. Neumann [ 186], who proved something stronger: if the union of the subgroups H; is irredundant, then every H; has finite index.
Chapter 5
fiS and Ramsey Theory  Some Easy Applications
We describe in this chapter some easy applications of the algebraic structure of CBS to the branch of combinatorics known as "Ramsey Theory".
5.1 Ramsey Theory We begin our discussion of the area of mathematics known as Ramsey Theory by illustrating the subject area by example. We cite here several of the classic theorems of the field. They will all be proved in this book. The oldest is the 1892 result of Hilbert. It involves the notion of finite sums, which will be of continuing interest to us in this book, so we pause to introduce some notation. Recall that in a noncommutative semigroup we have defined r1°_1 x, to be the product in increasing order of indices. Similarly, we take r]nEF Xn to be the product in increasing order of indices. So, for example, if
F=(2, 5, 6, 9), then
X,
We introduce separate notation ("FP" for "finite products" and "FS" for "finite sums") depending on whether the operation of the semigroup is denoted by or +. Definition 5.1. (a) Let (S, ) be a semigroup. Given an infinite sequence (xn)n° 1 in S,
FP((xn)n' 1) = {IInEFxn : F E . f(N)}. Given a finite sequence (xn)n1 in S, FP((xn)n 1) = {nnEF Xn : F E J" f({1, 2, ..., m})}. (b) Let (S, +) be a semigroup. Given an infinite sequence (xn)n in S, FS((xn)n_1) = {EnEFxn : F E Pf(N)}. Given a finite sequence (xn)n 1 in S, FS((xn)n 1 ) = {EnEFxn : F E 5 ( ( l , 2 . . . . . m})). 1
Theorem 5.2 (Hilbert [ 116]). Let r E N and let 1\Y = Ui=1 A,. For each m E N there exist i E 11, 2, ..., r), a sequence (xn) n in N, and an infinite set B C N such that for 1
each a E B, a+FS((xn)n1) C A,. Proof. This is a consequence of Corollary 5.10.
The next classical result is the 1916 result of Schur, which allows one to omit the translates on the finite sums when m = 2.
5.1 Ramsey Theory
91
Theorem 5.3 (Schur [221]). Let r E N and let N = U;=1 Ai. There exist i
E
{1,2,...,r}andxandyinNwith {x,y,x+y}cAi. Proof. This is a consequence of Corollary 5.10.
One of the most famous results of the field is the 1927 result of van der Waerden guaranteeing "monochrome" arithmetic progressions. (The statement "let r E N and let N = Ui=1 Ai" is often replaced by "let r E N and let N be rcolored", in which case the conclusion "... C Ai" is replaced by "... is monochrome".)
Theorem 5.4 (van der Waerden [238]). Let r E N and let N =
Ur
i=1
Ai. For each
f E Nthereexisti E (1,2,...,r) anda,d E Nsuch that {a,a+d,...,a+fd} S; Ai. Proof This is Corollary 14.2. Since the first of the classical results in this area are due to Hilbert, Schur, and van der Waerden, one may wonder why it is called "Ramsey Theory". The reason lies in the kind of result proved by Ramsey in 1930. It is a more general structural result not dependent on the arithmetic structure of N. Definition 5.5. Let A be a set and let K be a cardinal number.
(a)[A]"`=(B CA: JAI =K}. (b)[A]`K ={B CA: JAI
Notice that the case k = 1 of Ramsey's Theorem is the pigeon hole principle. We mention also that another fundamental early result of Ramsey Theory is Rado's Theorem [206]. It involves the introduction of several new notions, so we shall not state it here. We shall however present and prove Rado's Theorem in Chapter 15. Notice that all of these theorems are instances of the following general statement:
One has a set X and a collection of "good" subsets 9, of X. One then asserts that whenever r E N and X = U;=1 Ai there will exist i E {1, 2, ... , r) and G E 9, with G C Ai. (Consider the following table.) Theorem
X
GE9,
Hilbert
ICY
UaEB(a +FS((xn)n1)1
{x, y,x+y}
Schur
van der Waerden Ramsey
N
{a,a+d,...,a+fd}
[Ylk
[B ]k
It would not be entirely accurate, but certainly not far off the mark, to define Ramsey Theory as the classification of pairs (X, 9,) for which the above statement is true. We
92
5 6S and Ramsey Theory
see now that under this definition, any question in Ramsey Theory is a question about ultrafilters.
Theorem 5.7. Let X be a set and let 9 c R (X). The following statements are equivalent.
(a) Whenever r E N and X = U =1 A,, there exist i E 11, 2, ... , r} and G E 9 such that G C Ai. (b) There is an ultrafilter p on X such that for every member A of p, there exists
GE9with GcA. Proof. (a) implies (b). If 0 E 9, then any ultrafilter on X will do, so assume 0 0 9. Let A = {B C X : for every G E 9, B fl G # 0). Then A has the finite intersection property. (For suppose one has (B,, B2, ..., Br) c A with n;=1 Bi = 0. Then X = Ui=1(X\Bi) so there exist some i E (1, 2, ... , r) and some G E 9 with G fl B; = 0, a contradiction.) Pick by Theorem 3.8 some ultrafilter p on X with A c p. Then given any A E p, X\A 0 A so there is some G E 9 with G fl (X\A) = 0. (b) implies (a). For some i E (1, 2, ... , r), A; E P. Exercise 5.1.1. Prove that Theorem 5.4 implies the following superficially stronger statement.
Let r E N and let N = U =1 A,. There exists i E (1, 2, ... , r) such that for each
f E N, there exist a,d E Nsuch that (a,a+d,...,a+fd) c Ai.
5.2 Idempotents and Finite Products The first application of the algebraic structure of fS to Ramsey Theory was the GalvinGlazer proof of the Finite Sums Theorem (Corollary 5.10). It established an intimate relationship between finite sums (or products) in S and idempotents in l3 S. We present two proofs. The first of these is the original and is given for historic reasons as well as the fact that variations on this proof will be used later. The second, simpler, proof is of recent origin.
Theorem 5.8. Let S be a semigroup, let p be an idempotent in fS, and let A E p. There is a sequence (xn)°O in S such that FP((xn)n° 1) C A. 1
First Proof. Let A, = A and let B, = {x E S : x1 Al E p}. Since Al E p = p p, B1 E P. Pick x, E B, fl Al , let A2 = A, fl (x,' 1 A,), and note that A2 E p. Inductively
given A E p, let Bn = {x E S : xlA, E p}. Since A E p = p  p, B E p. Pick (xn1 An). xn E Bn fl An and let An+, = An fl To see for example why x2  X4  X5  X7 E A, note that x7 E A7 S; A6 c x51 A5 so that x5 x7 E A5 c x41A4. Thus x4 x5 x7 E A4 C A3 C X2_1 A2 so that
x2 x4 x5 x7 E A2 CA, =A.
5.2 Idempotents and Finite Products
93
More formally, we show by induction on IF I that if F E P f (N) and in = min F Xn = Xm E Am. Assume IFI > 1, let G = then fInEF Xn E Am. If IFI =1, then F\(m}, and let k = min G. Note that since k > m, Ak C Am+i. Then by the induction hypothesis, I1nEGXn E Ak C Am+t cxm1 Am SO IInEFxn =Xm  nnEGXn E Am.
Second Proof. Recall that A*(p) = (x E A : x' A E p) and write A* for A'(p). Pick xl E A*. Let n E N and assume we have chosen (xt)i_1 such that FP((xt)r=1) c A*. Let E = FP((zt)t_1). Then E is finite and for each a E E we have by Lemma 4.14 that al A* E p and hence n0EE a1 A* E p. Pick xn+l E A* fl naEE at A*. Then X,,+1 E A* and given a E E, a xn+1 E A* and hence FP((xr)n+,) 13 t=1 C A*.

Corollary 5.9. Let S be a semigroup, let r E N and let S = Ui=1 A. There exist i E {1, 2, ... , r} and a sequence (xn)n° 1 in S such that FP((xn)n° 1) C A,. P r o o f . Pick by Theorem 2.5 an idempotent p E CBS and pick i E ( 1 , 2, ... , r) such that
Ai E p. Apply Theorem 5.8.
Note that in the proof of Theorem 5.8, the fact that we have chosen to take our products in increasing order of indices is important. However, the corresponding version of Corollary 5.9 taking products in decreasing order of indices remains true. The hard way to see this is to redefine the operation on fiS so that it is a left topological semigroup. In this case one gets that A E p q if and only if {x E S : Ax 1 E p } E q, and mimicking the proof of Theorem 5.8 produces sequences with products in decreasing order of indices contained in any member of any idempotent.
The easy way to see this is to consider the semigroup (S, *) where x * y = y x. We isolate the following corollary for historical reasons. We also point out that this corollary is strong enough to derive all of the combinatorial results of this section. (See the exercises at the end of this section.)
Corollary 5.10 (Finite Sums Theorem). Let r E N and let N = U,=1 A. There exist i E {1, 2, ... , r} and a sequence (xn)n in N such that FS((xn)n°1) C A,. 1
Proof. This is a special case of Corollary 5.9.
It turns out that the relationship between idempotents and finite products is even more intimate than indicated by Theorem 5.8.
Lemma 5.11. Let S be a semigroup and let (xn)' 1 be a sequence in S.
Then
n°m°__1 FP((xn)n°_,n) is a subsemigroup of,6S. In particular, there is an idempotent p in PS such that for each m E N, FP((xn )n°_m) E p.
Proof. Let T = nm
1
FP((xn I
°_m ). To see that T is a semigroup, we use Theorem 4.20.
Trivially, {FP((xn)nm) : m E N) has the finite intersection property. Let m E N and let s E FP((xn )n° m) be given. Pick F E pf (N) with min F > m such that s = fInEF xn. Let k = max F + 1. To see that s FP((xn)n°k) 9 FP((Xn)nm), let
5 fiS and Ramsey Theory
94
t E FP((xn)n°_k) be given and pick G E Pf(N) with min G ? k such that t = l inEG Xt Then max F < min G so st = l inEFUG X. E FP((Xn)°_m) For the "in particular" conclusion note that by Theorem 2.5, E(T) 0.
Theorem 5.12. Let S be a semigroup and let A c S. There exists an idempotent p of PS with A E p if and only if there exists a sequence (xn) n
1
in S with FP((xn )n° 1) c A.
Proof. The necessity is Theorem 5.8 and the sufficiency follows from Lemma 5.11.
One should note what is not guaranteed by Theorem 5.8 and Lemma 5.11. That
is, given an idempotent p and A E p one can obtain a sequence (xn)n°1 with FP((xn)n1) c A. One can then in turn obtain an idempotent q with FP((xn)n°_1) E q, but one is not guaranteed that p = q or even that FP((xn)n°1) E P. (See Chapter 12 for more information on this point.) We now show that sets of finite products are themselves partition regular. Definition 5.13. (a) Let (S, ) be a semigroup and let (xn)n°1 be a sequence in S. The sequence (yn)n° is a product subsystem of (xn)n if and only if there is a sequence °_1 in JPf(N) such that for every n E N, max Hn < min Hn+1 and Yn = ntEH Xt (Hn)1j (b) Let (S, +) be a semigroup and let (xn)n° be a sequence in S. The sequence (yn)n° is a sum subsystem of (xn) ° 1 if and only if there is a sequence (Hn)n°1 in p1(N) such that for every n E N, max Hn < min Hn+1 and yn = xt. 1
1
Theorem 5.14. Let S be a semigroup, let (xn)n°1 be a sequence in S and let p be an idempotent in, S such that for every m E N, FP((xn)n°_m) E p. Let A E p. There is a product subsystem (y, )n°1 of (xn)n°1 such that FP((yn)°°1) C A. Proof. Let A* = A*(p) and pick y1 E A* fl FP((xn)n°1) Pick H1 E .Pf(N) such that Y1 = ftEH, X,. Inductively, let n E N and assume that we have chosen (y;)" and (H;)" such that:
(1) each y; = F LEH; x,, (2) if i < n, then max Hi < min H;+1, and (3) FP((y,)n=1) C A*.
Let E=FP((y;)"=1) and let k =maxHH+ 1. Let
B = FP((x;), k) fl A' fl naEE aIA'. By Lemma 4.14 aI A' E p for each a E E so B E P. Pick Y,,+1 E B and Hn+1 with min H,+1 > k such that yn+I = Xt As in the second proof of Theorem 5.8 we see that FP((y; )j +11) C A'
Corollary 5.15. Let S be a semigroup, let (xn)n1 be a sequence in S, let r E N, and let FP((xn)n°_1) = U;1 A;. There exist i E {1, 2, ... , r} and a product subsystem (Yn)n'=1 of (xn)n°1 such that FP((yn)n° 1) C A;.
5.2 Idempotents and Finite Products
95
Pick by Lemma 5.11 an idempotent p E fS such that for each m E N, FP((xn)n°,n) E p. Then Ui1 A; E p so pick i E {l, 2, ... , r} such that A; E p. proof.
Now Theorem 5.14 applies.
The semigroup (.J f(N), U) is of sufficient importance to introduce special notation for it, and to make special mention of its partition regularity.
Definition 5.16. Let (Fn)0°1 be a sequence in Rf(N). Fn : G E fPf(N)}. (a) FU((Fn)n°_1) = (b) The sequence (Gn)n° 1 is a union subsystem of 1 if and only if there is a sequence (Hn)n 1 in p1(N) such that for every n E N, max H,, < min Hn+1 and
Gn = U,EH F,. Corollary 5.17. Let (Fn )n° 1 be a sequence in J" f (N), let r E N and let FU((Fn) n° 1)
Ui=1 A j. There existi E { 1, 2, ..., r} and a union subsystem (Gn )n° 1 of(F)1 such that FU((Gn)('° 1) C A;. In particular, if r E N and JPf(N) = Ui=1 A;, then there exist i E {1, 2, ... , r} and a sequence (Gn)n° 1 in J" f(N)such that FU((Gn)00°1) C A; and for each n E N, max Gn < min Gn+1 Proof. The first conclusion is an immediate consequence of Corollary 5.15. To see the "in particular" conclusion, let Fn = [n) for each n E N, so that FU((Fn )n° 1) = Pf (N). 0
We conclude this section with a sequence of exercises designed to show that the Finite
Sums Theorem (Corollary 5.10) can be used to derive the combinatorial conclusions of this section. (That is, Corollaries 5.9, 5.15, and 5.17.) Since Corollary 5.9 is a consequence of Corollary 5.15, it suffices to establish the last two of these results. The intent in all of these exercises is that one should not use the algebraic structure of 1BS.
Exercise 5.2.1. Let (xn)n° 1 be a sequence in N and let t E N. Prove that for each m E N there exists H E Rf({n E N : n > m}) such that 2` X, Exercise 5.2.2. Let (xn)' 1 be a sequence in N. Prove that there is a sum subsystem (yn)n° 1 of (xn)n° 1 such that for each n and t in N, if 2' < y,,, then 21+1 Iyn+1
Exercise 5.2.3. Prove, using Corollary 5.10, that if r E N and . f(N) = U;=1 A;, then there exist i E 11, 2, ... , r} and a sequence (Gn)n°_1 in PPf(N) such that FU((Gn)n° 1) C A; and for each n E N, max Gn < min G,,+1. (Hint: Consider the function r : J" f(N) # N defined by r(F) = EnEF 2n1.)
Exercise 5.2.4. Prove Corollary 5.15.
5 fiS and Ramsey Theory
96
5.3 Sums and Products in N The results of the previous section are powerful and yet easily proved using the algebraic structure of PS. But their first proofs were combinatorial. We present in this section a simple result whose first proof was found using the algebraic structure of j8N and for which an elementary proof was only found much later. (In subsequent chapters we shall encounter many examples of combinatorial results whose only known proofs utilize the
algebraic structure of fS) One knows from the Finite Sums Theorem that whenever r E N and N = U,_1 A,, there exist i E (1, 2, ..., r) and a sequence (xn)R° 1 in N such that FS((xn)n° 1) c A;. It is an easy consequence of this fact that whenever r E N and N = Ut=1 A;, there
exist j E (1, 2, ..., r) and a sequence (yn)R°_1 in N such that FP((yn)n° 1) c Al. (Of course the second statement follows from Corollary .9. It is also an elementary consequence of the Finite Sums Theorem in the fashion outlined in the exercises at the end of the previous section. However, it follows much more quickly from the Finite Sums Theorem by a consideration of the homomorphism rp : (N, +) > (N, ) defined by co(n) = 2".) Two questions naturally arise. First can one choose i = j in the above statements? Second, if so, can one choose (x,,) ° 1 = (y,,) n_ ? We answer the first of these questions affirmatively in this section. The negative answer to the second question is a consequence of Theorem 17.16. 1
Definition 5.18. r = ct{p E ON : p = p + p} We have the following simple combinatorial characterization of r.
Lemma 5.19. Let p E fN. Then P E F if and only if for every A E p there exists a sequence (x)1 such that FS((xn )n° 1) c A. Proof Necessity. Let A E p be given. Then A is a neighborhood of p so pick q = q +q in fN with q E A. Then A E q so by Theorem 5.12 there is a sequence (xn)n 1 in N
with FS((xn)n 1) c A. Sufficiency. Let a basic neighborhood A of p be given. Then there is a sequence (xn)rs°1 in N with FS((xn)n 1) C A, so by Theorem 5.12 there is some q = q + q in fN with A E q. We shall see as a consequence of Exercise 6.1.4 that r is not a subsemigroup of (fiN, +). However, we see that r does have significant multiplicative structure.
Theorem 5.20. F is a left ideal of (fN, ). Proof. Let P E F and let q E fN. To see that q p E F, let A E q . p be given. Then ( ZEN: z' A E p) E q so pick z E N such that z' A E p. Then by Lemma 5.19 we may pick a sequence (xn )' 1 in N such that FS((xn) °O 1) C z' A. For each n E N, let yn = zxn. ThenFS((yn )n° ,) C A so by Lemma 5.19, q p E r.
5.4 Adjacent Finite Unions
97
We shall see in Corollary 17.17 that there is no p E ON with p + p = p p. However, we see that there are multiplicative idempotents close to the additive idempotents.
Corollary 5.21. There exists p = p p in F. Proof By Theorem 5.20, (1', ) is a compact right topological semigroup so Theorem 2.5 applies.
Corollary 5.22. Let r E N and let N = U;=1 A;. There exist i E {1, 2, ... , r} and 1 and (yn)n° t in N with
sequences
1) U FP((yn)n 1) C A,.
P r o o f Pick by Corollary 5.21 some p = p p in r. Pick i E {1, 2, ... , r} such that Ai E P. Since p = p p there is by Theorem 5.8 a sequence (yn)n°1 in N with FP((yn)o°1) C A,. Since p E r there is by Lemma 5.19 a sequence (xn)n_ 1 in N with FS((xn)no°_1) C Ai.
5.4 Adjacent Finite Unions In Corollary 5.17, there is no reason to expect the terms of the sequence (Gn)n° 1 to be
close to each other. In fact, if as in Corollary 5.17 one has max Gn < min G,,+1 for all n, we can absolutely guarantee that they are not. Theorem 5.23. There is a partition of J"f (N) into two cells such that, if (Gn )n_ is any sequence in J f(N) such that FU((Gn)' 1) is contained in one cell of the partition and max Gn < min Gn+t for all n, then (min Gn+1 max Gn : n E N) is unbounded. 1
Proof. Given a E N and F E .i f(N), define
rp(a,F)=I{xEF:{x+1,x+2,...,x+a)fF#0}I. For i E {0, 1}, let
A, = {F E Pf(N) : cp(min F, F)  i (mod 2)). Suppose that we have a i E 10, 1) and a sequence (Gn)n_1 such that
(1) max Gn < min Gn+t for all n E N, (2) FU((Gn)n0_1) C Ai, and (3) (min Gn+1  max Gn : n E N) is bounded.
Pick b E N such that for all n E N, min G, +I  max Gn < b, and pick k E N such that min Gk > b. Let a = min Gk and pick f such that min Ge  max Gk > a, noticing that f > k + 1. Let F = utk+l Gt. Then each of Gk, Gk U F, Gk U Ge, and
GkUFUGeisinA;and min Gk =min(GkUF)=min(GkUGe)=min(GkUFUGe)=a
5 fiS and Ramsey Theory
98 so
(p(a, Gk) = (p(a, Gk U F) = (p(a, Gk U Ge)
rp(a, Gk U F U Ge) (mod 2).
Now sp(a, Gk U F) = w(a, Gk) + w(a, F) + 1, V(a, Gk U Ge) = W(a, Gk) + cp(a, Ge), and
rp(a, Gk U F U Ge) = W(a, Gk) + p(a, F) + V(a, GI) + 2 so
(p(a, Gk)
(p(a, Gk) + V(a, F) + 1 W(a, Gk) + (p(a, Ge) cp(a, Gk) + cp(a, F) + rp(a, Ge) + 2 (mod 2),
which is impossible.
We show in the remainder of this section that, given any finite partition of 30f (N),
we can get a sequence (H,)', with any specified separation between max H and min (including min Hn+1 = max Hn + 1) and all finite unions that do not use adjacent terms in one cell of the partition. In the following definition "naFU" is intended to represent "non adjacent finite unions".
Definition 5.24. Let (Hk)k° 1 be a sequence in .% f(N).
(a) naFU((Hk)k_1) = { UfEF HI : F E Pf(N) and for all t E F, t + 10 F}. (b) If n E N, then naFU((Hk)k1) _ { U1 Ht : 0 F c {l, 2, ... , n} and for
alltEF, t+lF}.
(c) naFU((Hk)°_1) = 0.
We shall be using the semigroup (JPf(N), U) and the customary extension of the operation to fl(Pf(N)). However, we cannot follow our usual practice of denoting the extension of the operation to fl (,Pf (N)) by the same symbol used to denote the operation
in Pf(N). (If p, q E fl(Pf(N)), then p U q already means something.) So, we shall denote the extension of the operation U to P(Pf(N)) by W.
Lemma 5.25. Let f : N + N be a nondecreasing function. Define M : Pf (N) + N
andm : Pf(N) > NbyM(F) = max F + f (max F) and m (F) =minF. ThenMisa homomorphismfrom (,P f (N), U) to (N, v) andm is a homomorphism from (.Pf (N), U) to (N, A). Consequently, M is a homomorphism from (,8(Pf(N)), U) to (,8N, v) and m is a homomorphism from (f (Pf(N)), l+J) to (flN, n).
Proof. That m is a homomorphism is trivial. Given F, G E Yf (N), one has
f (max(F U G)) = f (max F) v f (max G) since f is a nondecreasing function, so that M(F U G) = M(F) v M(G). The fact that M and in are homomorphisms then follows from Corollary 4.22 and Theorem 4.24.
5.4 Adjacent Finite Unions
99
Lemma 5.26. Let f : N + N be a nondecreasing function, define M and m as in Lemma 5.25, and let Y = (a + f (a) : a E N). If X E Y\N and
T={P Efi(?f(N)):M(P)=in(P)=x}, then T is a compact subsemigroup of (0(.Pf(N)), W).
proof. Trivially T is compact. Further, by Exercise 4.1.11, x v x = x A x = x, so by Lemma 5.25, if p, q E T, then p l+1 q E T. Thus it suffices to show that T # 0. Let A = {M1 [X] fl m1 [X] : X E x}. We claim that A has the finite intersection property, for which it suffices to show that M1 [X] fl mt [X] # 0 for every X E X. So let X E x be given, pick a E X, and pick c E X fl Y such that c > a + f (a). Since C E Y, pick b E N such that c = b + f (b). Let F = {a, b}. If one had b < a, then
one would have c = b + f (b) < a + f (a), so max F = b. Then m(F) = a E X and Pick p E 6(Pf(N)) such that A c p. By Lemma 3.30 p E T. We can now prove the promised theorem yielding adjacent sequences with non adjacent unions in one cell.
Theorem 5.27. Let f : N > N be a nondecreasing function and let .Pf (N) _ u,=1 A. Then there exist some i E (1, 2, ..., r) and a sequence (H,,)n 1 in ?f (N) such that
(i) for each n, min Hn+1 = max Hn + f (max H,,) and (ii) naFU((Hn)n° 1) C A,. Proof. Let M and m be as defined in Lemma 5.25 and let Y = (a + f (a) : a E N). Pick x E Y\N and by Lemma 5.26 and Theorem 2.5 pick an idempotent p in (f(.Pf(N)), W)
such that M(p) = m(p) = x. We show that for each U E p, there is a sequence (Hn )n° in .P f (N) such that 1
(a) for each n, m(Hn+1) = M(Hn) and
(b) naFU((HH)' 1) C U. Then choosing i E { 1, 2, ... , r) such that A, E p completes the proof.
So let U E p be given. For each G E ,l"f(N) and any V C J"'f(N) let G1 V = (F E Yf(N) : G U F E V} and let U* = (G E 2l : G1 U E p}. Then by Lemma 4.14, U* E p and for each G E 2(*, G1 U* E P.
We claim now that, given any H E 5'f (N), {J E .Pf(N) : m(J) > M(H)} E p. Indeed, otherwise we have some t < M(H) such that (J E .Pf(N) : m(J) = t) E p so that 11(p) = t, a contradiction. Now, by Lemma 3.31, (H E J" f(N) : M(H) E m[U*]} E p so pick H1 E U* such that M(H1) E m[U*]. Inductively, let n E N and assume that we have chosen in 3 ' f (N) such that for each k E { 1, 2, ..., n), (1)
C 2(
5 fiS and Ramsey Theory
100
(2) if k > 1, then M(Hk) E m[`U* n n {G' U* : G.E
and
(3) if k > 1, then m(Hk) = M(Hk_1). Hypothesis (1) holds at n = 1 and hypotheses (2) and (3) are vacuous there.
If n = 1, pick H E U* such that m(H) = M(H1 ). Otherwise, by hypothesis (2), pick
HEU*nn{G'U*: GEnaFU((Hj)f_,I)} such that m(H) = M(H,,). Let V be the family of sets J E ?f(N) satisfying
(i) m(J) > M(H), (ii) H U J E U*, (iii) GUHUJ E U*forallG EnaFU((Hj)n_1),and (iv) M(J) E m[U* n n {G'U* : G E We have seen that {J E .Pf(N) : m(J) > M(H)} E p and HU* E p because H E U*. If n > 1 and G E then H E G' U* so G U H E U* so (G U H)' U* E p. Thus the family of sets satisfying (i), (ii), and (iii) is in p. then G E U* by hypothesis (1), so G' U* E P. Finally, if G E Thus U* n n {G' U* : G E E p so
{J E JPf(N) : M(J) E m[U* n n {G'U* : G E by Lemma 3.31. Thus V E p and so is nonempty. Pick J E V and let H,,+1 = H U J. Then m(H) = so hypothesis (3) holds. Further, J E V so
EP
M(J) and also
M(Hn+i) = M(J) E m[U* n n {G'U* : G E naFU((Hj)j%,)}] so hypothesis (2) holds.
To complete the proof, we show that
{1,2,...,n+1} such that foreacht E F,t+10 F. Ifn+1
W. So let 0 0 F c F, thenUfEF H, E U*
by hypothesis (1) at n, so assume that n + 1 E F. If F = In + 1), then we have H,,+i = H U J E U*. Thus, assume that F # In + 1),
letK=F\{n+1}, andletG=UfEK H,. ThenUrEF Hr=GUHUJEU*. Exercise 5.4.1. Let f : N + N be a nondecreasing function and let Z be any infinite subset of (a + f (a) : a E N). Show that the conclusion of Theorem 5.27 can be strengthened to require that min H,, E Z for each n E N. (Hint: In Lemma 5.26, choose
x E Z\N.)
5.5 Compactness
101
5.5 Compactness Most of the combinatorial results that we shall prove in this book are infinite in nature. We deal here with a general method used to derive finite analogues from the infinite versions. This version is commonly referred to as "compactness". (One will read in the literature, "by a standard compactness argument one sees...".) We illustrate two forms of this method of proof in this section, first deriving a finite version of the Finite Products Theorem (Corollary 5.9). For either of the common forms of compactness arguments it is more convenient to work with the "coloring" method of stating results in Ramsey Theory.
Definition 5.28. Let X be a set and let r E N. (a) An rcoloring of X is a function (p : X ). {1, 2, ... , r}. (b) Given an rcoloring tp of X, a subset B of X is said to be monochrome (with respect to cp) provided cp is constant on B.
Theorem 5.29. Let S be a countable semigroup and enumerate S as (sn)' . For each r and m in N, there exists n E N such that whenever { st : t E (1 , 2, ... , n}} is rcolored, there is a sequence (x,)1 in S such that FP((xt) m t) is monochrome. Proof. Let r, m E N be given and suppose the conclusion fails. For each n E N choose an rcoloring cp,t of {st : t E 11, 2, ..., n j } such that for no sequence (xt) m in S is FP((xt )'1) monochrome. Choose by the pigeon hole principle an infinite subset B1 of N and an element a(1) E 11, 2, ... , r} such that for all k E B1, tpk(s1) = a(1). Inductively, given f E N with f > 1 and an infinite subset Bt_ i of N choose an infinite subset Be of Be_ 1 and an element a(e) E (1, 2, ... , r) such that min Be ? f and for all k E Be, tpk(se) = a(f). 1
Define an rcoloring r of S by r(se) = a(f). Then S = Ui=1 r1 [{i}] so pick by Corollary 5.9 some i E 11, 2, ..., r} and an infinite sequence (x,) °O 1 in S such
that FP((xt)_1) c r1[{i}]. Pick k E N such that FP((x,)m1) c {si, s2, ... , ski and pick n E Bk. We claim that FP((x,)m,) is monochrome with respect to tp,,, a contradiction. To see this, let a E FS((xt )m,) be given and pick £ E (1, 2, ..., k) such that a = se. Then n E Bk C Be SO Qn(a) = a(2). Since a E FP((x,)°O 1) c r1 [{i }], i = r(a) = r(se) = a(t). Thus tpn is constantly equal to i on FP((x,)m1) as claimed.o The reader may wonder why the term "compactness" is applied to the proof of Theorem 5.29. One answer is that such results can be proved using the compactness theorem of logic. Another interpretation is provided by the proof of the following theorem which utilizes topological compactness.
Theorem 5.30. For each r and m in N there exists n E N such that whenever {1, 2, ... , n} is rcolored, there exist sequences (x,)m, and (yr)!', in N such that
FS((x,)m,) and FP((y,)"are contained in (1,2,...,n} and FS((xt)mi) U FP((y,)mi) is monochrome.
5 QS and Ramsey Theory
102
Proof. Let r, m E N be given and suppose the conclusion fails. For each n E RI
choose an rcoloring con of 11, 2, ..., n} such that there are no sequences (xt)m1 and 1) and FP((yt)m1) are contained in {1, 2, ..., n} and (yt)m1 in N for which FS((xt)'c"__t) U FP((yt)m 1) is monochrome.
Let Y = X n_1 { 1 , 2, ... , r}, where 11, 2, ... , r) is viewed as a topological space with the discrete topology. For each n E N define J[.n E Y by An (t)
Icon (t) I
ift
Then (µn)n' l is a sequence in the compact space Y and so we can choose a cluster point r of (An)n° in Y. Then N = U,=1 r1[{i}]. So by Corollary 5.22 there exist i E 11, 2, ... , r} and sequences (xn)R° 1 and (yn)n_1 in N such that FS((xn)rt° 1) 1) C r1[{i}]. 1)) andlet U = {g E Y :forallt E {1, 2, ..., k}, Letk = 1
g(t) = r(t)}. Then U is a neighborhood of r in Y, so pick n > k such that µn E U. But then An agrees with r on FS((xt)m 1) U FP((yt)m ). Also An agrees with con on {1, 2, ... , n} so FS((xr)m 1) U FP((yt)m 1) is monochrome with respect to con, a contradiction. Exercise 5.5.1. Prove, using compactness and Theorem 5.6, the following version of Ramsey's Theorem. Let k, r, m E N. There exists n E N such that whenever Y is a set
with JYJ = n and [y]k = 1J;=1 A;, there exist i E {1, 2, ... , r} and B E [Y]"' with [B]k C A1.

Notes The basic reference for Ramsey Theory is the book by that title [111]. Corollary 5.10 was originally proved in [ 118], with a purely elementary (but very
complicated) combinatorial proof. A simplified combinatorial proof was given by J. Baumgartner [15], and a proof using tools from Topological Dynamics has been given by H. Furstenberg and B. Weiss [100].
The first proof of the Finite Sums Theorem given here is due to F. Galvin and S. Glazer. This proof was never published by the originators, although it has appeared in several surveys, the first of which was [66]. The idea for the construction occurred
to Galvin around 1970. At that time the Finite Sums Theorem was a conjecture of R. Graham and B. Rothschild [110], as yet unproved. Galvin asked whether an "almost translation invariant ultrafilter" existed. That is, is there an ultrafilter p on N such that
whenever A E p, one has {x E S : A  x E p} E p? (In terms of the measure µ corresponding to p which was introduced in Exercise 3.1.2, this is asking that any set of measure 1 should almost always translate to a set of measure 1.) Galvin had invented the
Notes
103
construction used in the first proof of Theorem 5.8, and knew that an affirmative answer to his question would provide a proof of the conjecture of Graham and Rothschild. One of the current authors tried to answer this question and succeeded only in showing that, under the assumption of the continuum hypothesis, the validity of the conjecture of Graham and Rothschild implied an affirmative answer to Galvin's question. With the subsequent elementary proof of the Finite Sums Theorem [118], Galvin's almost translation invariant ultrafilters became a figment of the continuum hypothesis. Galvin
was interested in establishing their existence in ZFC, and one day in 1975 he asked Glazer whether such ultrafilters existed. When Glazer quickly answered "yes", Galvin tried to explain that he must be missing something because it couldn't be that easy. In fact it was that easy, because Glazer (1) knew that $N could be made into a right topological semigroup with an operation extending ordinary addition and (2) knew the characterization of that operation in terms of ultrafilters. (Most of the mathematicians who were aware of the algebraic structure of PN did not think of 8N as a space of ultrafilters.) In terms of that characterization, it was immediate that Galvin's almost translation invariant ultrafilters were simply idempotents. Theorem 5.12 is a result of F. Galvin (also not published by him). An elementary proof of Corollary 5.22 can be found in [30], a result of collaboration with V. Bergelson. Theorem 5.27 is due to A. Blass in [44] where it was proved using Martin's Axiom, followed by an absoluteness argument showing that it is a theorem of ZFC.
Part II Algebra of PS
Chapter 6
Ideals and Commutativity in fS
A very striking fact about $N is how far it is from being commutative. Although (fN, +) is a natural extension of the semigroup (N, +), which is the most familiar of all semigroups, its algebraic structure is amazingly complicated. For example, as we shall show in Corollary 7.36, it contains many copies of the free group on 21 generators. It is a simple observation that a semigroup (S, ) which contains two disjoint left ideals or two disjoint right ideals cannot be commutative. If L 1 and L2, say, are two disjoint left ideals of Sand if s1 E L 1 and s2 E L2, then sl  s2 E L2 and s2 sl E L1. S O S l $2 # S2 sl. We shall show that (f N, +) contains 2` mutually disjoint left ideals and 2` mutually disjoint right ideals, and that this is a property shared by a large class of semigroups of the form 18S.
Using this observation, it is fairly easy to see that OZ is not commutative. In fact, no element of Z* can be in the center of fiZ. We first observe that N* and (N)* are both left ideals of,6Z, as shown in Exercise 4.3.5. Thus, if p E N* and q E (N)*, q +P E N* and p + q E (N)*. So q + p ¢ p + q and neither p nor q is in the center of #Z. (Here, as elsewhere, if we mention "the semigroup 6Z" or "the semigroup fiN" without mention of the operation, we assume that the operation is addition.) We shall show in Theorem 6.10 that N is the center of (fN , +) and of (#N, ), and we shall show in Theorem 6.54 that these facts follow from a more general theorem.
6.1 The Semigroup IHI We shall use the binary expansion of positive integers to show how rich the algebraic structure of (f N, +) is. This tool will yield information, not only about f N, but about all semigroups fiS which arise from infinite, discrete, cancellative semigroups S.
Definition 6.1. H= nfEN cPfrq(2nN). Recall that each n E N can be expressed uniquely as n = EiEF 2', where F E P f (co). (Recall that w = N U {0}.)
Definition 6.2. Given n E N, supp(n) E J" f (w) is defined by n = Ei Esupp(n) 2'
108
6 Ideals and Commutativity in 6S
In our study of ,BIN it will often be the case that functions which are not homomor
phisms on N nonetheless extend to functions whose restrictions to H are homomorphisms. We remind the reader that the topological center A(T) of a right topological semigroup T is the set of elements t E T for which At is continuous. Lemma 6.3. Let (T, ) be a compact right topological semigroup and let tp : N * T with tp[IN] C A(T). Assume that there is some k E w such that whenever x, y E N and max.supp(x) + k < min supp(y) one has tp(x + y) = V(x) cp(y). Then for each p, 9 E H, W(p + 9) = i (p) . (q) Proof. This follows from Theorem 4.21 with A = {2"N : n E N}. The following theorem shows that homomorphic images of H occur very widely. In the following theorem, the possibility of a finite dense topological center can only occur if T itself is finite, since we are assuming that all hypothesized topological spaces are Hausdorff.
Theorem 6.4. Let T be a compact right topological semigroup with a countable dense topological center. Then T is the image of H under a continuous homomorphism. P r o o f . W e enumerate A(T) as {t; : i E 1}, where I is either co or 10, 1 , 2, ... , k} for some k E w. We then choose a disjoint partition {A; : i E 1) of (2" : n E w), with each
A, being infinite.
We define a mapping r : N . T by first stating that r (2") = t, if 2" E A, . We then extend r to N by putting r(n) = fIresupp(n) r(2'), with the terms in this product occurring in the order of increasing i. It follows from Lemma 6.3, that i : fiN + T is a homomorphism on H. We must show that ?[H] = T. To see this, let i E I and let x E Ai *. Then X E H and i(x) = ti. So ?[H] J A(T) and thus ?[H] c8(A(T)) = T. Corollary 6.5. Every finite (discrete) semigroup is the image of H under a continuous homomorphism.
Proof. A finite discrete semigroup is a compact right topological semigroup which is equal to its own topological center. The following simple result will frequently be useful.
Lemma 6.6. Let p be an idempotent in (fN, +). Then for every n E N, nN E p. Proof. Let y : N 3 Z, denote the canonical homomorphism. Then 7 : 8N + 7Gn is also a homomorphism, by Corollary 4.22. Thus y (p) = y (p) + T (p) and so y (p) = 0. It follows that y 1 [{0)] is a neighborhood of p, and hence that j71 [{0}] n IN = nN E P. 11
Definition 6.7. We define 0 : N + w and 0 : N * w by stating that 0(n) _ max(supp(n)) and 0(n) = min(supp(n)).
6.1 The Semigroup H
109
Lemma 6.8. The set II II is a compact subsemigroup of (ON, +), which contains all the
idempotentsof(fiN, +). Furthermore, forany p E,3Nandanyq E H, 4(p+q) = ¢(q)
andO(p+q) =9(p). Proof H is obviously compact. Now, if n E 2kN and if r > i(n), n + 2'N C 2kN. Thus it follows from Theorem 4.20 that IHI is a semigroup. By Lemma 6.6, H contains all of the idempotents of (ON, +).
Now suppose that p E ON and q E H. Given any m E N, choose r E N satisfying
r > ¢(m). Then, if n E 2'N, 0(m + n) _ fi(n) and 6(m + n) = 6(m). Hence since
2'NEq,
q(m+q) = q lim q(m+n) nE2'N = q lim 0(n) nE2'N
=
q(q).
It follows that 4(p + q) =m+p lim 4(m + q) = (q). Similarly, 6(p + q) = 6(p). Theorem 6.9. (ON, +) contains 2` minimal left ideals and 2` minimal right ideals. Each of these contains 2` idempotents.
Proof. Let A = (2n : n E N). Since 0(2n) = 0(2n) = n for each n E N, 4IA = : A > N is bijective and so 0,A = 01A : A > ON is bijective as well. Now if
BIA
q1, q2 E A* and (ON + q1) fl (ON + q2) 54 0, then by Lemma 6.8, q1 = q2. (For if By Corollary 3.57 r + qt = s + q2, then O(ql) = 4'(r + qi) = 0(s + q2) = IA*1 = 2`, so fiN has 21 pairwise disjoint left ideals, and each contains a minimal left ideal, by Corollary 2.6. Now observe that any idempotent e which is minimal (with respect to the ordering of idempotents where e < f if and only if e = e + f = f + e) in IIi[ is in fact minimal in
ON. For otherwise there is an idempotent f of ON with f < e. But every idempotent of ON is in H and so f E H, contradicting the minimality of e in H. Given any q E A*, q + H contains an idempotent e(q) which is minimal in II3[ by Theorems 2.7 and 2.9, and is hence minimal in ON. Thus by Theorem 2.9, e(q) + ON is a minimal right ideal of ON. We claim that if q1 0 Q2 in A*, then e(ql) + ON
e(q2) + ON. For otherwise, e(42) E ON and so e(q2) = e(qi) + e(q2) by Lemma 1.30. However, 6(e(Q2)) E 0[q2 + H] = {42} (by Lemma 6.8), while 9(e(gi) +e(42)) = 9(e(gi)) = q1, a contradiction. If L is any minimal left ideal and R is any minimal right ideal of (14N, +), L f1 R contains an idempotent, by Theorem 1.61. It follows that L and R both contain 21 idempotents. As we shall see in Theorem 6.54, the following theorem is a special case of a theorem that is far more general. We include it, however, because the special case is important and has a simpler proof.
Theorem 6.10. N is the center of (ON, +) and of (ON, ).
6 Ideals and Commutativity in f3S
110
Proof. By Theorem 4.23 N is contained in the centers of (ON, +) and of (ON, ).
Let A = {2" : n E N}. Suppose that p E N* and q E A*. By Lemma 6.8, 4'(p + q) = (q). On the other hand, suppose that in, n E N and that n > m. Then 0(m+n) = 0(n)or0(m+n) = 0(n)+1. Foreachm, {n E N : 0(m+n) _ 0(n)} E p or {n E N : 0 (m + n) = 0 (n) + 1 } E p. In the first case,
pEct{nEN:cb(m+n)_O(n)} andso
(m+p)_gy(p).
In the second case, ¢(m + p) _ gy(p) + 1. Now {m E N : ¢(m + p) = O(p)} E q or
{m E N : 0(m + p) = (p) + 1} E q. So (q + p) = gy(p) or O(q + p) = gy(p) + I. Since there are 2` different values of o(q), we can choose q E A* satisfying 0(q) f
((p),0(p)+1). Thenq E 1HIso0(p+q) _ (q) 00(q+p)sop+q 96 q+p, and thus p cannot be in the center of (ON, +).
Now for any m, n E N, g(m2") _ ¢(m) +n = 0(m) +0(2"). Now given p E N* and q E A*, one has
(p ' q) =
lim lim q5 (m 2") 2n* q
lim lim (0(m) + 0(2")) m*p 2"'q lim (0(m) + lim 0(2")) msp 2".q nli P(0(m) + 0(q))
0(p) + 0(q) And similarly, (q ' p) = (q) + gy(p). Now let p E N*. We show that p is not in the center of (ON, ). Since 0 is finite
toone, 0(p) E N* so pick r E N* such that 0(p) + r # r + ¢(p). Since 0 maps A biectively to ON, pick q E A* such that t(q) = r. Then 0(p q) = ¢(p) + r # 13
Remark 6.11. It follows from Theorem 4.24 that N is the topological center of (ON, +) and of (ON, ).
We shall in fact see in Theorems 6.79 and 6.80 that there is a subset A of N* such that A and N*\A are both left ideals of (14N, +) and of (P N, ). Theorem 6.12. H contains an infinite decreasing sequence of idempotents. Proof. Let (An) R° , be an infinite increasing sequence of subsets of N such that An+t \An
is infinite for every n. Let Sn = {m E N : supp(m) c A" }. We observe that, for every n, r E N, we have k+m E 2'Nf1Sn wheneverk, m E 2'Nf1 Sn and supp(k)flsupp(m) = 0. Thus it follows from Theorem 4.20 (with A = {Sn n 2'N : r E N)) that Tn = Sn f1H is a subsemigroup of ON. We shall inductively construct a sequence (en)n i of distinct idempotents in N satisfying en+1 < en and en E K(Tn) for every n.
6.1 The Semigroup H
111
We first choose el to be any minimal idempotent in T1. We then assume that e, has been chosen for each i E {1, 2, ..., m}. By Theorem 1.60, we can choose an idempotent em+i E K (Tm+i) for which em+i < em. We shall show that em+t : em by showing that em 0 K(Tm+i). To see this, we choose any x E N* fl {2" : n E Am+i\Am}. Let
M = {r + 2" + s : r, s E N, n E Am+t \Am and max supp(r) < n < min supp(s) }.
Since m fl Sm = 0, em 0 M. However, it follows from Exercise 4.1.6, that y +x + z E M for every y, z E H. So M contains the ideal Tm+i +X + Tm+i of Tm+i and therefore contains K(Tm+i). Hence em f K(Tm+i)
We have seen that it is remarkably easy to produce homomorphisms on H. This is related to the fact that H can be defined in terms of the concept of oids, and the algebraic structure of an oid is very minimal. Theorem 6.15 shows that the only algebraic information needed to define H is the knowledge of how to multiply by 1.
Definition 6.13. (a) A set A is called an id if A has a distinguished element 1 and a multiplication mapping ({ 1 } x A) U (A x { 1 }) to A with the property that la = a 1 = a for every a E A. (b) If for each i in some index set I, A, is an id, then
®iEt A,={xE X;EfA,:{iEI:xi
1} is finite}
is an oid.
(c) If S = ®iEI Ai is an oid and x E S, then supp(x) = {i E I : x; # 1}.
Notice that we have already defined the notation "supp(x)" when x E N. The correspondence between the two versions will become apparent in the proof of Theorem 6.15 below.
Definition 6.14. Let S = ®iEJ A, be an oid. If x, y E S and supp(x) fl supp(y) = 0, then x y is defined by (x y)i = xi yi for all i E I. Notice that in an oid, one does not require x y to be defined if supp(x) fl supp(y) # 0.
Theorem 6.15. Let A = {a, 1} be an id with two elements and, for each i E w, let
Ai = A. Let S = ®iE,,, Ai and let H= n,. ce,6s{x E S : min(supp(x)) > n}. Extend the operation to all of S x S arbitrarily and then extend the operation to,6 S as in Theorem 4.1. Then H is topologically and algebraically isomorphic to H. Proof. Define cp : N * S by
_ `p(x)`
1
a
if i 0 supp(x) if i E supp(x)
and notice that cp is onetoone, W[N] = S\(1), and for all x E N, supp(cp(x)) = supp(x). In particular, for all n E N, cp[2"N] = {x E S : min(supp(x)) > n} so that i7[1EII] = H.
6 Ideals and Commutativity in fiS
112
Since rp is onetoone, so is W by Exercise 3.4.1. Thus Pjn is a homeomorphism onto
H. To see that iWIH is a homomorphism, let p, q E H. Then by Remark 4.2,
D(p+q) =' (p lim q lim x + y) = pzl XEN
YEN
qyliEmrp(x+y).
Now, given any x E N, if n = max(supp(x)) + 1, then for all y E 2"N, V(x + y) _ sp(x) cp(y), so that, since q E H and A (x) and W are continuous,
q lim rp(x + y) = q lim cp(x + y) = q lim ((p(x) w(y)) = V(x) V(q) YEN yE2"N yE2"N Thus rp(p + q) = p lim(W (x) w(q)) = j (p) w(q) as required. xEN Notice that in the statement of Theorem 6.15, we did not require the arbitrary extension of the operation to be associative on S. However, the theorem says that it nonetheless induces an associative operation on H.
Exercise 6.1.1. Show that H has 2` pairwise disjoint closed right ideals. (Hint: Consider {9(p) : p E A*} where A = (2" : n E N).) Exercise 6.1.2. Show that no two closed right ideals of ON can be disjoint. (Hint: Use Theorem 2.19.)
Exercise 6.1.3. Show that no two closed right ideals of N* can be disjoint. (Hint: Consider Theorem 4.37 and Exercise 6.1.2.)
Exercise 6.1.4. Show that there are two minimal idempotents in (ON, +) whose sum is not in r, the closure of the set of idempotents. (Hint: Let S denote the semigroup described in Example 2.13. By Corollary 6.5 there is a continuous homomorphism h mapping H onto S. Show that there are minimal idempotents u and v in H for which h(u) = e and h(v) = f. Observe that idempotents minimal in 1H[ are also minimal in ON.)
6.2 Intersecting Left Ideals In this section we consider left ideals of ,BS of the form J6S p.
Definition 6.16. Let S be a semigroup and let s E S. The left ideal Ss of S is called the semiprincipal left ideal of S generated by s.
Note that the semiprincipal left ideal generated by s is equal to the principal left ideal generated by s if and only if s E Ss.
6.2 Intersecting Left Ideals
113
Definition 6.17. Let S be a semigroup and let p E f3S. Then C(p) = {A C S : for all
XES,x IAE p}. Theorem 6.18. Let S be a semigroup and let p E fiS. Then /3S p = (q E /3S C(p) C q).
Proof. Let r E /3S. Then given A E C(p) one has S = {x E S : x1A E p} so For the other inclusion, let q E /3S such that C(p) c q. For A E q, let B(A) _ (x E S : X1 A E p}. We claim that {B(A) : A E q} has the finite intersection property. To see this observe that {B(A) : A E q} is closed under finite intersections and that if
B(A) = 0, then S\A E C(p)
q. Pick r E 3S such that (B(A) : A E q) c r. Then
q=r  p. Theorem 6.19. Suppose that S is a countable discrete semigroup and that p, q E fiS. If 14S p fl fS q # 0, then sp = xq for some s E S and some x E 3S, or yp = tq for some t E S and some y E /3S.
Proof. Since 3S p = Sp and PS q = Sq we can apply Corollary 3.42 and deduce that sp E Sq for some s E S, or else tq E Sp for some t E S. In the first case, sp = xq for some x E /3S. In the second case, tq = yp for some y E /3S. Corollary 6.20. Let S be a countable group. Suppose that p and q are elements of PS
for which /3S p fl /BS q 0 0. Then P E fl S q or q E /BS p. Proof. By Theorem 6.19, sp = xq for some s E S and some x E $S, or tq = yp for some t E S and some y E /3S. In the first case, p = sI xq. In the second, q = tI yp. Corollary 6.21. Let p and q be distinct elements of ON. If the left ideals ON + p and ON + q are not disjoint, p E ON + q or q E ON + p. Proof. Suppose that (ON + p) fl (fiN + q) 0 0. It then follows by Theorem 6.19, that m + p = x + q for some m E N and some x E ON, or else n + q = y + p for some n E N and some y E ON. Assume without loss of generality that m + p = x + q for some m E N and some x E ON. If X E N*, then m + x E N*, because N* is a left ideal of 7L*, as shown in Exercise 4.3.5. Sop = m + x + q E ON + q. Suppose then
that x E N. Then x # m, for otherwise we should have p = m + x + q = q. If x > m, m + x E N and again p = m + x + q E ON + q. If x < m, x + m E N
andsoq=x+m+p EON+p.
By Corollaries 6.20 and 6.21, we have the following.
Remark 6.22. If S is a countable group or if S = N, any two semiprincipal left ideals of /3S are either disjoint or comparable for the relationship of inclusion.
6 Ideals and Commutativity in 0 S
114
Recall frgm Chapter 3 that if T C S we have identified PT with the subset ci T of X45.
Corollary 6.23. Let S be a countable semigroup which can be embedded in a group G,
and let e, f E E(S).
f 00, of = e or f e = f.
Proof. By Theorem 6.19, we may assume that se = x f for some s E S and some x E f S.
This implies that e = stxf in,6G, and hence that of = s'xf f = stxf = e. We can now show that, for many semigroups S, the study of commutativity for idempotents in fS is equivalent to the study of the order relation <. Corollary 6.24. Suppose that S is a countable discrete semigroup which can be embedded in a group, and that e, f E E(,6S). Then e and f Commute if and only if e < f
or f < e. Proof. By definition of the relation <, if say e < f , then e = of = f e. Conversely, suppose that of = f e. Then flS e n pS f 0 0, and so, by Corollary 6.23, we may suppose that of = e. Since of = f e, this implies that e < f . Corollary 6.25. Suppose that S is a countable discrete semigroup which can be embed
ded in a group, and that Cisasubsemigroupof6S.Ife, f E E(C)satisfy CenCf =0,
f =0. Proof. If ,S e n flS f # 0, we may suppose by Corollary 6.23, that of = e. But then e E Ce n C f, a contradiction.
Exercise 6.2.1. Let S be a semigroup and let p E $5. Prove that C(p) is a filter on S an infinite semigroup and let p E S*. Let C,,(p) = (A c S : {x E S : xA gE p} is finite). Prove that C,,, (p) is a filter on S, CU,(p) = n(5* p),
and
6.3 Numbers of Idempotents and Ideals Copies of IHI The semigroup III arises in $N. We shall see, however, that copies of ]EII can be found in S* for any infinite cancellative semigroup S. We introduced the notation FP( (x )n° t) in Definition 5.1. We extend it now to apply to sequences indexed by linearly ordered sets other than N.
Definition 6.26. Let S be a semigroup, let (D, <) be a linearly ordered set, and let (xs)SED be a Dsequence (i.e., a set indexed by D) in S.
6.3 Numbers of Idempotents and Ideals
115
(a) For F E ?f (D), I1sEF xs is the product in increasing order of indices. (b) FP((Xs)sED) = {nSEF Xs : F E .' f(D)}. (c) The Dsequence (XS)SED has distinct finite products if and only if whenever F, G E pf(D) and 11SEF xs = fSEG xs, one has F = G.
Thus, if F = {sl, s2, s3, s4} where sl < s2 < s3 < S4, then IZSEF Xs = xsl x$3
xs2
XS4
If additive notation is being used we define ESEFxs, FS((Xs)SED), and "distinct finite sums" analogously.
Theorem 6.27. Let S be a discrete semigroup and let (xn )°O 1 be a sequence in S with distinct finite products. Let T= nm 1 CB(FP((Xn)n_,n)). Then T is a subsemigroup of PS which is algebraically and topologically isomorphic to H. Proof. It was shown in Lemma 5.11 that T is a subsemigroup of 6S. We define a
mapping f from FP((xn)n 1) to 2N by stating that f (11nEF xn) = EnEF 2' for every F E JPf(N). Since f is a bijection (because of the distinct finite products assumption),
f cf(FP((xn)n° 1))  P(2N) is a homeomorphism. We shall show that fIT is a homomorphism from T onto H. Let Tm = FP((Xn)n°_m). Then f [T,n] = 2'N and so
f[T]=f[nm
1Tm]=nm
If[Tm]=1 mo lf[T.i=nm
To see that TIT is a homomorphism, it suffices by Theorem 4.21 to show that for
each s E FP((xn)n 1) there is some m E N such that f(st) = f (s) + f (t) for all t E FP((xn)n_n,). Solets E FP((xn)n 1) andpick F E Rf(N) suchthats = I1nEFXn Let m = max F + 1. 0 We now turn our attention to the effect of certain cancellation assumptions. Much more will be said on this subject in Chapter 8.
Lemma 6.28. Let S be a discrete semigroup and let s be a cancelable element of S.
For every tESand pEPS (i) if S is right cancellative and sp = tp, then s = t, and (ii) if S is left cancellative and ps = pt, then s = t. Proof. (i) Suppose that S is right cancellative. Let t E S, let p E 6S, and suppose that s # t. We shall define a function f : S > S by stating that f (su) = to for every u E S and that f (v) = s2 for every v E S\sS. We note that f is welldefined because every element in sS has a unique expression of the form su, and that f has no fixed points. We also note that since f o As and At agree on S, one has that f (sp) = tp. Now by Lemma 3.33 we can partition S into three sets A0, A,, A2 such that A, fl f [A;] = 0 for each i E {0, 1, 2}. Choose i such that sp E A, . Then by Lemma 3.30, f [A, ] E f (sp) = tp,
so sp 0 tp.
6 Ideals and Commutativity in 6S
116
(ii) This is proved in essentially the same way.
Recall that a semigroup S is weakly left cancellative if and only if for all u, v E S, (x E S : ux = v) is finite, and that S is weakly right cancellative if and only if for all
u,vES,{xES:xu=v}is finite. Lemma 6.29. Let S be an infinite weakly left cancellative semigroup and let I denote the set of right identities of S. Let K be an infinite cardinal with K < IS I, and let (sx)x
be a onetoone Ksequence in S. If T is a subset of S with cardinality K, then there exists a onetoone Ksequence (tx)x
(b) for everyA < µ < K and every u, v E 1 U {s, and utx # vtu, and (c) I n FP((tx)x
:
i < µ} U FP((tt)t<, ), u # utµ
Proof. For each u, v E S, let Au,v = {s E S : u = vs}. Since S is weakly left cancellative, Au,v is finite. We note that this implies that I is finite, because I C A,,,t, for every u E S. We construct (tx)x
Let G = I U is, : t < v} U FP((t,),
u, v E H}. If v < w, then H and K are finite so we can choose
tVET\(HUK). If v > w, then IHI = IvI
Then there is a subset V of T with cardinality K such that the set P of uniform
ultrafilters on V has the following properties:
(1) IPI = 22k, (2) for each pair of distinct elements p, q E P, cl, s (Rp) and ct fs (Rq) are disjoint, and (3) if S is also right cancellative, then for each p E P, ap # by whenever a i6 b in R and Rp is strongly discrete in 6S. Proof We apply Lemma 6.29, taking (SOX
6.3 Numbers of Idempotents and Ideals
117
W e have P 1 = 22` , by Theorem 3.58, so (1) holds.
Suppose that a, 0, .l, A E K satisfy a < X and fl < lt, and that satx = sptµ. We claim that k = A. To see this, suppose instead without loss of generality that A < p Then the equation satx = situ, contradicts condition (b) of Lemma 6.29.
For each le < K, let VN, = (t : li, < v < K). Then VN, E p for every p E P. Let p and q be distinct elements of P. We can choose disjoint subsets A and B of V satisfying A E p and B E q. For any a, P E K, sa (Va n A) E sap and s,s (Vp n B) E spq and we have seen that these two sets are disjoint. Thus, if X = U0
Y=Ufl
cips(Rp) c X and clps(Rq) c Y so (2) holds. Now suppose that S is right cancellative and p E P. If a < a. < K and < µ < K, the equation satx = spt,, implies that .L = A and therefore that a = 4. For any a E K, and so Rp is strongly discrete. we have Sc, Va E sap. Now sa Va n sp Vp = 0 if a ,
13
Lemma 6.31. Let S be an infinite discrete semigroup, which is right cancellative and weakly left cancellative. Let T be an infinite subset of S, and let K = IT I. There is a Ksequence (tx)x
sequence (t)I)A
To see that (t),)x
as possible. We assume without loss of generality that max F < max G = t. The assumption that max F < it contradicts condition (a) of Lemma 6.29 if G = {µ} and contradicts condition (b) of Lemma 6.29 otherwise. So we also have max F = µ. If I FI > 1 and I G I > 1, we contradict the assumption that I F U G I is small as possible by cancelling t, from our equation and deducing that fIAEF\(N.) ti. = nAEG\(k} tx. Thus
we assume without loss of generality that IFI = 1 and IGI > 1. Let x = IIAEG\(µ) 4
Then t, = xt,2 and so st/, = sxt,, for every s E S. Hence, by our right cancellative assumption, s = sx for every s E S and so x E I. This contradicts condition (c) of Lemma 6.29.
Theorem 6.32. Let S be an infinite discrete right cancellative and weakly left cancellative semigroup. Then every Gs subset of S* which contains an idempotent contains a copy of III.
Proof. Let p E E(S*) and let A be a GSsubset of S* for which p E A. We may suppose that A = fn__i An, where (A,)n° i is a decreasing sequence of subsets of S. We shall inductively construct a sequence (x n) n t of elements of S with the property that, for every m E N, FP((xn)n_,n) c Am . We first choose xi to be any element of A I*. We then suppose that k E N and that xi, x2, ... , xk have been chosen so that FP((xn )n;) g Ai* for every i E {1,2..... k}.
118
6 Ideals and Commutativity in fiS
Now Ak+t` E p and, for each i E (1,2..... k) and each y E FP((xn)n_i), we have y1 Ai * E p (by Lemma 4.14). We can therefore continue the induction by choosing xk+1 E Ak+l* n n{y1Ai* : i E {1, 2, ... , k} and y E FP((xn)n_i)}\{x1, X2, ... , Xk}.
By Lemma 6.31, we may suppose that (xn)' 1 has distinct finite products, because we can replace (xn)n° 1 by a subsequence with this property. (Any sequence with distinct finite products is necessarily onetoone. A onetoone sequence in {xn n E N) can be thinned to be a subsequence of (xn)n° 1.) Then, by Theorem 6.27, nm t ct(FP((xn)n_m)) is a copy of H. Since ct (FP(xn)n_m) C Am for every m E N, nml ct (FP(xn)n°_m) C A.
Corollary 6.33. Let S bean infinite discrete right cancellative and weakly left cancellative semigroup. Then every Gssubset of S* which contains an idempotent contains 2` nonminimal idempotents.
Proof. Let A be a G8 subset of S*. By Theorem 6.32, A contains a copy of H. So it suffices to show that H contains 2` nonminimal idempotents. Let T= noo 1 ct'#N(FS((22n)n__m)). Trivially T C H and by Theorem 6.27 T is a copy of H. By Exercise 4.4.3 and Theorem 4.40, T fl K (PN) = 0. Thus by Lemma 6.8 and Theorem 1.65, T fl K(H) = 0. By Theorem 6.32, T contains a copy of H so by Lemma 6.8 and Theorem 6.9 T has 2` idempotents. Corollary 6.34. Let S be an infinite discrete semigroup which is right cancellative and weakly left cancellative. Then every Gs subset of S* which contains an idempotent, contains an infinite decreasing chain of idempotents. Proof. This follows immediately from Theorems 6.32 and 6.12. We now see that, for many familiar semigroups S, S* S* is nowhere dense in S*.
Theorem 6.35. Let S be a countable semigroup which is weakly left cancellative and right cancellative. Then S*S* is nowhere dense in S*. Proof. We enumerate S as a sequence (sn)n 1 of distinct elements. Let T denote any infinite subset of S. We shall produce an infinite subset V of T such that V* fl S* S* = 0. To this end we shall inductively choose a sequence (tn)n° of distinct elements of T 1
with the property that the equations smx = to and sm'x = t, have no simultaneous
solution in Sif m
Let V = (tn : n E N). We claim that, for any p, q E S*, pq f V*. Suppose, on the contrary, that pq E V*. Then, if P = {s E S : sq E V), P E P. Ifs E P, it follows from Theorem 4.31 that sq E V*. Pick m < m' with sm,sm' E P. If
6.3 Numbers of Idempotents and Ideals
119
Q = {x E S :smx and sm'x are in {t, : i > m'}}, then Q Eq. So we can choose x E Q and will then have smx = to and sm,x = to for some n > m and some n' > m'. Since S is right cancellable, n n'. But then we have a contradiction to the choice of
the sequence (t)1. Notice that the hypotheses of Corollary 6.34 and Theorem 6.35 cannot be significantly weakened. The semigroup (N, v) is weakly right and weakly left cancellative, but every member of N* is a minimal idempotent in (ON, v) and N* v N* = N*. We have need of the following topological fact which it would take us too far afield to prove.
Theorem 6.36. There is a subset L of N* such that I L I = 2` and for all p ¢ q in L
and all f : N > N, 7(p) # q. Proof. This is a special case of [213].
Lemma 6.37. Let p,,g E N*. If there is a onetoone function f : N ). ON such that f [N] is discrete and f (p) = q, then there is a function g : N N such that g(q) = p. Proof. For each n E N choose An E f (n) such that An n f [N\(n)] = 0. For each n > I in N, let Bn = An\ U',' Ak and let B1 = N\ Un° 2 B. Then (Bn : n E N} is a partition of N and for each n E N, Bn E f (n). Define g : N * N by g(k) = n if and only if k E Bn. Then for each n E N one has g is constantly equal to n on a member of f (n) sog{ f (n)} = n. Therefore g o f : ON o. ON is the identity on ON. Thus
g(q) = g(f(p)) = P Recall that two points x and y in a topological space X have the same homeomorphism type if and only if there is a homeomorphism f from X onto X such that f (x) = y. See [69, Chapter 9]. Theorem 6.38. Let D be a discrete space and let X be an infinite compact subset of 6D. Then X has at least 2` distinct homeomorphism types. Proof. By Theorem 3.59, X contains a copy of ON, so we shall assume that ON C X. Let L C N* be as guaranteed by Theorem 6.36. We claim that the elements of L all have different homeomorphism types in X. Suppose instead that some two elements of L have the same homeomorphism type in X. We claim that in fact
(*) there exist some homeomorphism h from X onto X and some p # q in L such that h(p) = q and q E cf(h[N] n ce N). To verify (*), pick a homeomorphism h from X onto X and p # q in L such that
h(p) = q. Now q E ce N n ct h[N] so by Corollary 3.41, q E ce(N n ce h[N]) U ce(h[N] n ce N). If q E ce(h[N] n ce N), then we have that (*) holds directly, so assume that q E ce(N n cl h [N]). Then p = h1(q) E h' [ce(N n ce h[N])] =
120
6 Ideals and Commutativity in 6S
ce(h' [N] n ct N). Thus (*) holds with p and q interchanged and with h1 replacing h.
Thus (*) holds and we have q E ce(h[N] n MN). Let A = h[N] n N and let
B =h[N]nN*. Then q E ctAUctB. Assume first that q E cf A. Define f : N > N by
if n E h'' [A] f(n) = J h(n) N\h1 [A]. 1 if n E Then p = h' (q) E h' [c2 A] = ct h1 [A] and f and h agree on h1 [A] so f (p) _ h(p) = q, contradicting the choice of L. Thus we must have q E ct B. Since h[N] is discrete, and no countable subset of N* is dense in N* by Corollary 3.37, we may pick a onetoone function f : N > 1`Y*
such that f (n) = h(n) for all n E h1 [B] and f [N] is discrete. Then p = h' (q) E h' [c8 B] = cf h' [B] and f and h agree on h1 [B]. So T (p) = q. But then by Lemma 6.37 there is a function g : N > N such that g(q) = p, again contradicting the choice of L.
Recall from Theorem 2.11 that the minimal left ideals of fiS are homeomorphic. Thus the hypothesis of the following theorem is the same as stating that some minimal left ideal of fS is infinite.
Theorem 6.39. Let S be a discrete semigroup. If the minimal left ideals of 15S are infinite, f3S contains at least 2` minimal right ideals. Proof. Suppose that L is an infinite minimal left ideal of 1BS. Then L is compact. Now two points x and y of L which belong to the same minimal right ideal R belong to the same homeomorphism type in L. To see this, observe that R n L is a group, by Theorem 1.61. Let x1 and y1 be the inverses of x and y in R n L. The mapping pX y is a homeomorphism from L to itself by Theorem 2.11(c) and pX_i (x) = y.
By Theorem 6.38 the points of L belong to at least 2' different homeomorphism classes. So 6S contains at least 2` minimal right ideals. Lemma 6.40. Let S be an infinite discrete cancellative semigroup. Then every minimal left ideal of flS is infinite.
Proof. Let L be a left ideal of PS and let p E L. If s, t are distinct elements of S, then sp ,f tp, by Lemma 6.28. Corollary 6.41. Let S be an infinite discrete cancellative semigroup. Then OS contains at least 2` minimal right ideals. Proof. This follows from Theorem 6.39 and Lemma 6.40. Theorem 6.42. Let S be an infinite discrete semigroup which is weakly left cancellative. 22K pairwise disjoint left ideals of PS. If BSI = K, then there are
6.3 Numbers of Idempotents and Ideals
121
proof This is an immediate consequence of Theorem 6.30, with R = S. Corollary 6.43. If S is a weakly left cancellative infinite discrete semigroup with car22' dinality K, then fi S contains minimal idempotents.
Proof Each left ideal of 1S contains a minimal idempotent. Theorem 6.44. Let S be an infinite cancellative discrete semigroup. Then fS contains at least 2` minimal left ideals and at least 2` minimal right ideals. Each minimal right ideal and each minimal left ideal contains at least 2` idempotents. Proof. By Theorem 6.42 and Corollary 6.41, 0S contains at least 2° minimal left ideal and at least 21 minimal right ideals. Now each minimal right ideal and each minimal left ideal have an intersection which contains an idempotent, by Theorem 1.61. So each minimal right ideal and each minimal left ideal contains at least 2` idempotents.
Since minimal left ideals of PS are closed, it is immediate that, if S is weakly left cancellative, there are many pairwise disjoint closed left ideals of fS. We see that it is more unusual for S* to be the disjoint union of finitely many closed left ideals of ,SS.
Definition 6.45. Let S be a semigroup and let B c S. Then B is almost left invariant if and only if for each s E S, sB\B is finite. Theorem 6.46. Let S be a semigroup and let n E N. Then S* is the union of n pairwise disjoint closed left ideals of PS if and only if
(a) S is weakly left cancellative and (b) S is the union of n pairwise disjoint infinite almost left invariant subsets.
Proof Necessity. Assume that S* = U;=, Li, where each Li is a closed left ideal of ,S and Li fl Lj = 0 for i 96 j. Then S* is a left ideal of $S so by Theorem 4.3 1, S is weakly left cancellative.
Since Li fl Lj = 0 when i # j, each Li is clopen so by Theorem 3.23, there is some Ci c S such that Li = C, . Let Bi = Ci. For i E (2, 3, 4, ... , n  1) (if any) let Bi = Ci\ UU= C3, and let C = S\ U'=1 Then for each i, B7 = C! = Li and
B, flBj=Owheni # j. Finally, let i E 11, 2, ... , n}, lets E S, and suppose that sBi\Bi is infinite. Since (Bi \s' Bi ), it follows that Bi \s' Bi is infinite and so there exists p E (Bi \s' Bi)*. Then P E Li while sp Li, a contradiction. Sufficiency. Let S = U; i Bi where each Bi is infinite and almost invariant and Bi fl B j = 0 when i 0 j. For each i E (1, 2, ... , n), let Li = Bl . By Theorem 4.31, S* is a left ideal of S, so it suffices to show that for each i E 11, 2, ..., n}, each p E Li, and each s E S, Bi E sp. T o this end, let i E 11, 2, ... , n}, p E Li, and s E S be given. Then sBi\Bi is finite and Bi\s'Bi S U.XES8,\B; (Y E S : sy = x} so, since S is weakly left cancellative, Bi \s ' Bi is finite. Thus s Bi E p so Bi E sp as s Bi \ Bi = s
required.
6 Ideals and Commutativity in,6S
122
Exercise 6.3.1. Show that, for any discrete semigroup S, the number of minimal right ideals of 6S is either finite or at least 21.
Exercise 6.3.2. Suppose that S is a commutative discrete semigroup. Show that, for any minimal right ideal R and any minimal left ideal L of fS, R n L is dense in L. Deduce that the number of minimal right ideals of fS is either one or at least 21. Exercise 6.3.3. Several of the results of this section refer to a right cancellative and weakly left cancellative semigroup. Give an example of an infinite right cancellative and weakly left cancellative semigroup which is not cancellative. (Hint: Such examples can be found among the subsemigroups of (NN, o).) Exercise 6.3.4. Show that there is a semigroup S which is both weakly left cancellative and weakly right cancellative but )9S has only one minimal right ideal.
Exercise 6.3.5. Show that N* is not the union of two disjoint closed left ideals of (16N, +).
Exercise 6.3.6. Show that Z* is the union of two disjoint closed left ideals of (f Z, +) but is not the union of three such left ideals.
6.4 Weakly Left Cancellative Semigroups We have shown that the center of (,BIN +) is disjoint from N*, and so is the center of (,6N, ). In this section, we shall show that this is a property shared by a large class of semigroups. We shall use K to denote an infinite cardinal and shall regard K as being a discrete space. Lemma 6.47. Let S be an infinite weakly left cancellative semigroup with cardinality K and let T be a given subset of S with cardinality K. Then there exists a function
f : S + K such that
(1) f[T]=K, (2) for every A < K, f and
[XI is finite if k is finite and I f '[),] I < ICI if X is infinite,
(3) for everys, s' E S,iff(s)+l < f(s'),then f(ss') E {f(s')1, f(s'), f(s')+1} if f (s') is not a limit ordinal and f (ss') E [f (s'), f (s') + 1} otherwise. Proof. We assume that S has been arranged as a Ksequence (sI )a
for y
6.4 Weakly Left Cancellative Semigroups
123
(ii) T fl Ua
(iv) ify=a+ 1,sEEa,and ss'EEa,then s'EEy; (v) if y <w,then IEaI <w;and (vi) if y> co, then I Ey I< I Y I
We choose any t E T and put Eo = {so, t). We then assume that 0 < y < K and that we have defined Ea for every a < y. Let U = Ua
For every fl < K there exists t E T fl Ef\ Ua
Finally, suppose that f (s) = a, f (s') = 8 and a + 1 < P. Let y = f (ss'). Since s, s' E Efi, we have ss' E Ep+1 and soy < 0 + 1. We now claim that y + 1 > P. So suppose instead that y + 1 < P and let A = max(a, y). Then s, ss' E EN, and so s' E EN,+1. Thus P < µ + 1 < ,B, a contradiction. Definition 6.48. Let S be a discrete semigroup. We define a binary relation R on U,, (S)
by stating that pRq if (jS)p fl (i3S)q # 0. We extend this to an equivalence relation on U,K (S) by stating that p q if pRnq for some n E N. By R' we mean the composition of R with itself n times. Thus, given p, q E UK (S),
p q if and only if there exist elements xo, x 1 , X2, ... , Xn E U (S) such that p = xo, q = xn and xi Rxi+1 f o r every i E {0, 1 , 2, ... , n  11. Lemma 6.49. Let S be a discrete infinite weakly left cancellative semigroup with cardinality K and let T be any subset of S with cardinality K. Let f be the function guaranteed
by Lemma 6.47 and let f : 6S  & be its continuous extension.
(i) IfK=w, we put A={2' :m E N}. (ii) IfK > w, we put A equal to the set of limit ordinals in K. If B and C are disjoint subsets of A with cardinality K and if B E T (p) and C E T (q) for some p, q E UK (S), then p f q.
Proof. (i) We first suppose that K = w and that A = {2' : m E N}. We shall show by induction that, for every n E N, every x, y E S* and every X C N, if X E 7(x) and xRny, then U2n2n(i + X) E 7(y). Suppose first that n = 1. If xRy then ux = vy for some u, v E PS. Let U = {ss'
S, s' E S, f (s) + 1 < f (s'), and f (s') E X). Then U E ux by Theorem 4.15, and
124
6 Ideals and Commutativity in fiS
f[U] C (X 1)UXU(X+1)bycondition (3)ofLemma 6.47. So (X  1) U X U (X + 1) E T(UX). Similarly, if Y E f (y), then (Y  1) U Y U (Y + 1) E f (vy). So Y n U2i=2(' + X) 0 0. Thus U83=20 + X) E f (y) Now assume that our claim holds for n E N. Suppose that x Rn+l y Then x R' z and z Ry for some z E S*. LetZ = UZ__2n (i +X ). By our inductive assumption, Z E f (Z). By what we have just proved with n = 1, U13=2('+ Z) = 00+1) +1) (i + X) E f(y). So we have established our claim.
We now observe that, for every n E N, B fl ( U?__zn (i + C)) is finite and so U 2n (i + C) 0 f (p). Thus we cannot have pR'q. (ii) Suppose that K > co and that A is the set of limit ordinals in K.
Let B' = {A + n : A E B and n E W). We shall show that, if x, y E UK (S), B' E T (x), and xRny, then B' E T(Y) .
Suppose first that xRy. Then ux = vy for some u, v E fS. Now, if U = {ss' s, s' E S, f (s) + 1 < f (s'), and f (s') E B'), then U E ux (by Theorem 4.15). We also haveJ[U] c B', by condition (3) of Lemma 6.47. So B' E f(ux). Similarly, if K\B' E f (y), then K\B' E f (vy). This contradicts our assumption that ux = vy and so B' E f (y). It follows immediately, by induction, that xRny implies that B' E f (y). Since B' E T (p) and B' 0 f (q), it follows that p q. Lemma 6.50. Let S be a discrete infinite weakly left cancellative semigroup with cardinality K. For every p E UK(S), {q E UK(S) : q p} is a nowhere dense subset of UK (S).
Proof. Suppose instead that we have some subset T of cardinality K such that UK (T) c cZ{q E UK (S) : q p}. Let f be the function guaranteed by Lemma 6.47. If K = w, we put A = {2'n : m E N). If K > co, we put A equal to the set of limit ordinals in K. Let B and C be disjoint subsets of A with cardinality K. It follows from Lemma 6.49 that {q E UK (S) : q p} is disjoint from c2 (f 1 [B]) fl UK (S) or from cf (f 1 [C]) fl UK(S). So either cf(f 1 [B]) fl UK(T) or ct(f 1 [C]) fl UK (T) is a nonempty open subset of UK(T) disjoint from {q E S* : q p}, a contradiction. o
Corollary 6.51. Let S be a discrete infinite weakly left cancellative semigroup with cardinality K. For every p E UK(S), {q E UK(S) : qp = pq} is nowhere dense in UK (S).
Proof If qp = pq, (CBS) p fl (P S)q ¢ 0 and so qRp. Corollary 6.52. Let S be a discrete infinite weakly left cancellative semigroup with cardinality K. Then the center of UK (S) is empty.
Proof. This follows immediately from Corollary 6.51.
6.5 Semiprincipal Left Ideals
125
Theorem 6.53. Let S be a discrete infinite weakly left cancellative semigroup with cardinality K. Then there exists a decomposition I of UK (S) with the following properties:
(1) Each I E I is a left ideal of fiS. (2) Each I E I is nowhere dense in UK (S).
(3) Ill = 22". Proof. We use the equivalence relation described in Definition 6.48 and denote the equivalence class of an element p E UK(S) by [p]. We put I = {[p] : p E UK(S)). (1) By Exercise 6.4.1, UK (S) is a left ideal of JS. So, for each p E UK (S) and each
x E fS, pRxp. Thus [p] is a left ideal of fS. (2) This is Lemma 6.50.
(3) Let f denote the function guaranteed by Lemma 6.47, with T = S, and let f : ,BS > f K denote its continuous extension. We note that f is surjective because f is surjective. Furthermore, 7'[UK (K)] c UK (S). If K = co, we put A = 121 : m E N). If K > co, we put A equal to the set of limit ordinals in K. For each u E UK (A), we can choose pu E UK (S) for which 7(p.) = u. If u and v are distinct elements of UK (A), there exist disjoint subsets B and C of A for
which B E u and C E V. Since B E f (pu) and C E f (p ), it follows from Lemma 6.49 that [Pu] 0 22K 22'. Now J A I = K and so I UK (A) I =
(by Theorem 3.58), and so I I I =
Theorem 6.54. Let S be a weakly left cancellative semigroup. Then the center of,BS is equal to the center of S. The center of S* is empty.
Proof. First, if s is in the center of S, then s is also in the center of fiS. Indeed, for every p E 6S, we have sp = p lim st = p lim is = ps. zES IES To see the reverse inclusion, let p be in the center of fS and suppose p V S. (Of course, if p E S, then p is in the center of S.) Let K = II P II . We choose any B E p with I B I = K and let T denote the subsemigroup of S generated by B. Since I T I = K and since we can regard p as being in UK (T), it follows from Corollary 6.52 that p is not in the center of $T. So p is not in the center of CBS.
Exercise 6.4.1. Let S be a discrete infinite weakly left cancellative semigroup with cardinality K. Prove that UK(S) is a left ideal of $S.
6.5 Semiprincipal Left Ideals and the Center of p(PS) p We show in this section that in N* there are infinite decreasing chains of semiprincipal left ideals. We also study the center of p($S)p for nonminimal idempotents p. If S is embedded in a group G_6 S is also embedded (topologically and algebraically)
in PG by Remark 4.19. We shall assume, whenever it is convenient to do so, that
$ScPG.
126
6 Ideals and Commutativity in ,5S
Theorem 6.55. Let S be a countably infinite discrete semigroup embedded in a count
able discrete group G. Let q E K($S) and let A Eq. Let n E N and let pl, p2, ... , pn E S*\K(i6S). Then there is an infinite subset B of A such that, for every r E B*
and every i, j E (1,2,...,n), pi 0 (, G)rpj. Proof. We claim that, for every i, j E 11, 2, ... , n}, pi 0 (18G)gpj. To see this, we note that qpj E K(f S) and hence that qpj = gpje for some idempotent e E K(MS) (by Theorem 2.8). So if pi E (,BG)gpj, we have pi = xqpj for some x E PG and so pie = xgpje = xqpj = pi and thus pi = pie E K(PS), a contradiction. Thus there is a subset D of G such that
{Pi, P2..... pn} C D and D n (U"_l(fG)gpi) = 0. For each a E G, let Ea = {x E A* : there exists i E {1, 2, ... , n} such that axpi E D}. Notice that Ea = A* n U"_ i ()La o pp;)' [D] and is therefore closed. Now q E S*, by Theorem 4.36. Since q E A*\ UaEG Ea, it follows that naEG (A*\Ea) is a nonempty Gs set in S*. So by Theorem 3.36 there is an infinite subset B of A such that
B* C naEG(A*\Ea) Let r E B* and let i, j E {1, 2, ... , n}. If pi E (fG)rpj = c$fG(Grpj), then arpj E D for some a E G and hence r E B* n Ea, contradicting our choice of B*. Theorem 6.56. Let S be a countably infinite discrete semigroup embedded in a count.
able discrete group G. Let p E S*\K(fS), let q E K(,BS) and let A E q. There is an infinite subset B of A with the property that, for every r E B*, p 0 (flG)rp, and, whenever rl and r2 are distinct members of B*, we have (,6G)ri p n ($G)r2 p = 0. Furthermore, for every r E B*, rp is right cancelable in /3G. Proof. By Theorem 6.55, there is an infinite subset C of A with the property that r E C*
implies that p 0 (flG)rp. We choose a sequential ordering for G and write a < b if a precedes b in this ordering. We can choose a sequence (bn )n , in C with the property that ab n 0 bn whenever
m < n and a E G satisfies a < b,n. We then let B = {bn : n E NJ. We shall now show that whenever ri, r2 E B*, V E PG, and rl p = vr2 p, we have rl = r2 and v = 1, the identity of G. We choose subsets B, and B2 of B which satisfy B, E ri and B2 E r2, choosing them to be disjoint if ri # r2. Now rip E ct(B1 p) and vr2P E cf(Gr2p). Furthermore, vr2p E ce((G\{1})r2p) in the case in which v 0 1. By applying Corollary 3.42, we can deduce that by = v'r2 p
for some b E B, and some v' E PG, or else r'p = ar2p for some r' E Bi and some a E G. We can strengthen the second statement in the case in which v # 1, by asserting that a E G\{ 1 }. The first possibility contradicts the fact that p 0 (/3G)r2 p, because it implies that p = b 1 v'r2 p, and so we assume the second.
Put B,=B1n{sES:s>a'}and B'=B2n{sES:s>a}. ThenBj Er'and B2' E r2. Since r'p E cf(Bi p) and ar2p E ct(aB' p), another application of Corollary 3.42 shows that cp = awp for some c E Bi and some w E cE B', or else zp = adp
6.5 Semiprincipal Left Ideals
127
for some z E ct Bi and some d E B. Now, in the first case, w 0 (B')*, because p 0 (flG)wp if w E (B2)*. Similarly, in the second case, z 0 (Bi)*. Thus we can deduce that bm p = abn p for some bm E B, and some bn E B2. This equation implies that bm = abn by Lemma 6.28. By our choice of the sequence (bn) i, atbm # bn if n > m and abn 0 bm if m > n. Thus the equation bm = abn can only hold if m = n, and this implies that a = 1. We can thus deduce that rl = r2 and that v = 1, as claimed.
Now if (PG)rip fl (flG)r2p # 0, then rip E (#G)r2 p or r2 p E (fG)ri p, by Corollary 6.20. We have seen that this implies that ri = r2. Finally, let r E B*. If rp is not right cancelable in PG, there are distinct elements
w, z E PG for which wrp = zrp. We can choose disjoint subsets W and Z of G satisfying W E w and Z E z. Since wrp E ct(Wrp) and zrp E cf(Zrp), another application of Corollary 3.42 allows us to deduce that drp = z'rp for some d E W and
some z' E cf Z, or else w'rp = d'rp for some d' E Z and some w' E c2 W. In either case, it follows that rp E (fG\{1})rp, because diz' E fiG\{1} in the first case, and (d')iw' E fG\{1} in the second case. (If Z' E G, then diz' 1 because z 0 W. If z' E G*, then diz' E G* by Theorem 4.36.) We have seen that this is not possible. Theorem 6.57. Suppose that S is a countable discrete semigroup embedded in a count
able discrete group G. Let n E N and let pi, P2, ... , pn E S*\K(fS). Suppose that, in addition, pi 0 Gpj whenever i and j are distinct members o f 11 , 2, ... , n}. Let B be an infinite subset of S with the property that pi V (fG)rpj whenever r E B* and
i, j E { 1, 2, ... n}. Then (fG)rpi fl (fG)rpj = 0 for every r E B* and every pair o f distinct members i , j o f { 1 , 2, ... , n}.
Proof. Suppose, on the contrary, that there is an element r E B* for which (fG)rpi fl (f G)rpj # 0 for some pair i , j of distinct members of { 1, 2, ... , n). We may then suppose that rpi = xrpj for some x E PG, by Corollary 6.20. Now rpi E c2(Bpi) and xrpf E ct(Grpj). Thus, by Corollary 3.42, r' pi = arpj for some r' E ct(B) and some a E G, or bpi = x'rp1 for some b E B and some x' E PG. The second possibility can be ruled out because pi gE (PG) rpj, and so the first must hold. We now observe that r'pi E cl(Bpi) and arpj E cf(aBpj). So another application of Corollary 3.42 shows that cpi = aspj for some c E B and some s E ce(B), or else tpi = ad pj for some t E B* and some d E B. The first possibility cannot hold if s E B *, because then pi 0 (fG)spi. Neither can it hold if s E B, because pi V Gpj. So the first possibility can be ruled out. The second possibility cannot hold since pj 0 (fG)tpi.
Theorem 6.58. Let G be a countably infinite discrete group and let p r= G*\K(fG). Suppose that B C G is an infinite set with the property that, for every r E B*, p ¢ (,BG)rp, and that (fG)rp fl (fG)r'p = O for every pair of distinct elements r and r' in B*. (The existence of such a set follows from Theorem 6.56 with S = G.) Then, for every r E B*, (f G)rp is maximal subject to being a principal left ideal of PG strictly contained in (f G) p. Proof. We note that (f G)rp is strictly contained in (f G) p, because p E (f G) p and
p 0 (fG)rp.
6 Ideals and Commutativity in 6S
128
Suppose that we have (/3G)rp Z ($G)x % (f G) p for some x E ,PG.
Then rp = yx for some y E f3G and thus Bp fl Gx # 0. So by Corollary 3.42, either by = zx for some b E B and some z E fiG or else r'p = ax for some r' E B* and some a E G. The first possibility can be ruled out since p = btzx E (fG)x implies that (fG)p C (fG)x.
Therefore assume that r'p = ax for some r' E B* and some a E G. Then x = aJr'p E (fG)r'p and so (fG)rp C (fG)x C (fG)r'p. In particular (fG)rp fl (f G)r'p 54 0 and so r = r'. But then (f G)x = (f G)rp, a contradiction. We now prove a theorem about semiprincipal left ideals in N*. An analogous theorem is true if N is replaced by any countably infinite commutative cancellative discrete semigroup. However, we shall confine ourselves to the special case of N , because it is the most important and because the more general theorem is somewhat more complicated to formulate.
Theorem 6.59. Let p E N*\K(,8N). Suppose that B C N is an infinite set with the property that, for every x E B*, p 0 OZ + x + p, and that for every distinct pair of elements x, x' in B *, (P Z + x + p) fl (fl Z + x' + p) = 0. (The existence of such a set follows from Theorem 6.56). Then, for every x E B*, N* + x + p is maximal subject to being a semiprincipal left ideal of N* strictly contained in N* + p.
Proof. We first note that N* + x + p is strictly contained in N* + p. To see this,
pick x' E B*\{x}. Then x' + p E N* + p and x' + p 0 N* + x + p because
(N*+x'+p)fl(N*+x+p)=0.
Suppose that N* + x + p S N* + y S N* + p, where y E N* and x E B*. We shall show that N* + y is equal to N* + x + p or N* + p. By Corollary 6.21, x + p E fiN + y or y E ON + x + p. The second possibility implies that N* + y = N* + x + p, and so we assume the first.
Thenx+p =z+y forsomez E fN. Nowx+p E ct(B+p)andz+y E ct(N+y). We can therefore deduce from Corollary 3.42, that n + p = z' + y for some n E B and some z' E fN or else x' + p = in + y for some x' E B* and some m E N. In the first case p = n + z' + y and so N* + p = N* + y. Thus we assume that the second case holds.
The left ideals P Z + x + p and fiZ + x' + p intersect, because, for any u E N*,
u+x+p E N*+y = N*+(m)+x'+p. Itfollowsthatx' = x. Theny = m+x+p and this implies that N* + y = N* + x + p. Corollary 6.60. If p E N*\K(f N), the semiprincipal left ideal N* + p of N* contains 2` disjoint semiprincipal left ideals, each maximal subject to being a semiprincipal left ideal strictly contained in N* + p.
Proof We choose a set B satisfying the hypotheses of Theorem 6.59. The corollary then follows from the fact that IB*I = 2`. (See Theorem 3.59.)
6.5 Semiprincipal Left Ideals
129
Corollary 6.60 does not hold as it stands if N* is replaced by fiN. For example, if p is a right cancelable element of f N, there is precisely one semiprincipal left ideal of fiN maximal subject to being strictly contained in fiN + p, namely ,BN + 1 + p.
Corollary 6.61. Let p E N*\K(fN). Then N* + p belongs to an infinite decreasing sequence of semiprincipal left ideals of N*, each maximal subject to being strictly contained in its predecessor.
Proof By Theorem 6.56 choose an infinite set B c N such that for every x E B*, p 0 1476 + x + p and x + p is right cancelable in f7L and such that whenever x and x' are distinct members of B*, ,B7L + x + p fl ,876 + x' + p = 0. Pick any x E B*. Then by Theorem 6.59 N* + x + p is maximal among all semiprincipal left ideals that are properly contained in N* + p. Since x + p is right cancelable, x + p 0 K(fN) by Exercise 1.7.1, so one may repeat the process with x + p in place of p. It is possible to prove [180] although we shall not do so here that the statement of Corollary 6.61 can be strengthened by replacing the word "sequence" by "cvi sequence". Thus N* certainly contains many reverse wellordered chains of semiprincipal left ideals. It is not known whether every totally ordered chain of semiprincipal left ideals of N* is reverse wellordered. Theorem 6.62. Let S be a countably infinite semigroup, embedded in a countable group
G. If p is a nonminimal idempotent in S*, the center of the semigroup p(,BS)p is contained in Gp.
Proof. Suppose that x is an element of the center of p(fS) p. Then, for every r E fS,
we have x(prp) = (prp)x and hence xrp = prx, because xp = px = x. We first show that x 0 K(fS). Suppose instead that x E K(,6 S) and pick an idempotent e E K(fS) such that x = xe. By Theorem 6.56 there is an infinite subset B of S such that (fG)rlp fl (i3G)r2 p = 0 whenever ri and r2 are distinct elements of B* and rp is right cancelable in ,BG for every r E B*. We can choose r E B* such that x 0 (fG)rp, because this must hold for every r E B* with at most one exception. Now rp 0 (fG)x because otherwise rp = rpe E K(fS), while rp is right cancelable. (By Theorem 4.36K(,8S) c S*, so by Exercise 1.7.1 no member ofK(,BS)
is right cancelable.) Since x 0 (f G)rp and rp f (f G)x we have by Corollary 6.20 that (,G)rp fl (fG)x = 0 and hence that xrp 96 prx, a contradiction. Now suppose that x 0 Gp. Then also p 0 Gx and so by Theorems 6.55 and 6.57 with n = 2, pi = p, and P2 = x, there is an infinite subset C of S such that
(fG)rp fl (,BG)rx = 0 for every r E C*. This implies that xrp # prx, again contradicting our assumption that x is in the center of p(PS)p. Recall that if S is an infinite, commutative, and cancellative semigroup, then S has a "group of quotients", G. This means that G is a group in which S can be embedded in such a way that each element of G has the form s It for some s, t E S. The group G is also commutative. If S is countable, then so is G.
130
6 Ideals and Commutativity in 0 S
Theorem 6.63. Let S be a countable commutative cancellative semigroup and let G be its group of quotients. If p is a nonminimal idempotent in 8S the center of p(f S) p is
equal to Gp n p(PS)p. Proof By Theorem 6.62 the center of p(,3S)p is contained in Gp. Conversely, let a E G such that ap E p(48S) p. Now a is in the center of ,BG by Theorem 4.23. Thus,
if y E p(8S) p, then apy = ay = ya = ypa = yap. Corollary 6.64. If p is a nonminimal idempotent in N*, the center of p + fiN + p is equal to Z + p.
Proof. This follows from Theorem 6.63 and the fact that Z + N* C N* (by Exercise 4.3.5).
Notice that the requirement that p be nonminimal in Theorem 6.63 is important. It is not even known whether Z + p is the center of p +,6N + p for an idempotent which
is minimal in (fN, +). Lemma 6.65. Let S be a countably infinite, commutative, and cancellative semigroup, and let G be its group of quotients. Then K (f S) = K (0G) n f S.
Proof. By Theorem 1.65, it is sufficient to show that K(fG) n 'Os # 0. Let q be a minimal idempotent in fS. We claim that Gq c 8S. To see this, choose
any r E G and put r = stt for some s, t E S. Now sS is an ideal of S and so ct(sS) = sfiS is an ideal of ,BS, by Corollary 4.18. It follows that K($S) c sj6S.
Hence q E si4S so slq E fiS and so rq = s'tq = istq E t,BS c fS. Now there is a minimal idempotent p in K(fG) for which pq = p, by Theorem 1.60. Thus P E (fG)q = cl(Gq) c fiS. Exercise 6.5.1. Suppose that p E N*\(N* + N*). Prove that N* + p is a maximal semiprincipal left ideal of N*. (Hint: Use Corollary 6.21.)
Exercise 6.5.2. Let F denote the free semigroup on two generators, a and b, and let G denote the free group on these generators. Prove that K($G) n,BF = 0. (Hint:
Suppose that p E PF and that q E ct,G{b"a : n E N} n G*. Each X E G can be expressed uniquely in the form a"b" 'a n1b"z ... an'k b", , where all the exponents are in Z and all, except possibly n 1 and nk, are in Z\{0}. Define f (x) = Ek_i In[ I.
Let E = (xb"ay : x E G, Y E F, and n > f (x)). Show that E n F = 0, that (fG)gp c CEOG E, and hence that p g (fG)gp.) Exercise 6.5.3. Let S be a countably infinite, commutative, and cancellative semigroup, and let G be its group of quotients. Show that if p is a minimal idempotent in fS, then
Gp c p(PS)p. (Hint: Consider the proof of Lemma 6.65.) Exercise 6.5.4. Let S = (N, ), so that the group of quotients G of S is (Q1, ). Show that there are idempotents p E S* for which Gp ¢ S*.
6.6 Principal Ideals in PZ
131
6.6 Principal Ideals in $Z It is a consequence of Theorems 6.56 and 6.58 that, given any p E Z*\K(,8Z) there is an infinite sequence (pn )n° t with pi = p such that fl Z + pn+t Z OZ + Pn for each n. Whether there is an infinite strictly increasing chain of principal left ideals of ,8Z is an old and notoriously difficult problem. (See the notes to this chapter for a discussion of the history of this problem.) We do not address this problem here, but instead solve the corresponding problem for principal right ideals and principal closed ideals (defining the latter term below). Recall from Exercise 4.4.9 that if S is a commutative semigroup, then the closure of any right ideal of fiS is a two sided ideal of PS.
Definition 6.66. Let G be a commutative group and let p E fiG. Then ct(p + PG) is the principal closed ideal generated by p. Notice that the terms "principal closed ideal" and "closed principal ideal" mean two different things. Indeed the later can only rarely be found in fiS where S is a semigroup. The reasons for defining the term "principal closed ideal" only for groups are first, that this guarantees that p is a member, and second, because we are only going to use the notion in Z.
We now introduce some special sets needed for our construction. Recall that ®°_i to = {a E X 00 : {i : ai # 0} is finite}. 1
Definition 6.67. (a) For m, n E N, {a' E ®°O_t w :
Zi°D_
ai
1
21 = 2"
and min{i E N : ai
0) > m +n}.
(b) Fix a sequence (zk)kt_1 such that for each k, zk+1 > Ei=1 2'4z. i
(c)Form,n EN, An ={E,1aizi :a EBn }. Lemma 6.68. For each m, n E N, AR +1 C An and IAn I = w. Further, for each
a E6' and each i EN,ai <2`. Proof That An +1 C Am is trivial. For the second assertion notice that {2k"zk : k >
m + n} C An . For the third assertion notice that if ai > 2i, then 2' >
".
C]
Lemma 6.69. Let m, n E N. Then A +1 + A +1 C A"+1 Proof. Let x, y E A+1 and pick a, b E 0,,m+ such that x = E°_1 aizi and y = E°O1 bizi. Let cZ = a + b. Then x + y = E°_1 cizi so it suffices to show that
c' E On +'. First E°_° 1
2i = E 1 21 + E 1 21 = 2"+1 + 2n+1 =
2n
min{i EN:ci:0}=min({i EN :ai00}U(i EN:bi00})>n+l+m.
Also,
6 Ideals and Commutativity in PS
132
Lemma 6.70. Ifaa,b' E ®°_1 Z, E,°°1aizi = E,°=1 b:zi,andforeachi E N, JaiI <2i and I bi I < 2i, then a = b.
Proof Suppose a b. Since a and b each have only finitely many nonzero coordinates, we may pick the largest n such that an # b,, and assume without loss of generality that an > bn. Then E" 1 aizi = E°=1 bizi so
n 2i+ 11 Zi < Zn, nI (bi  aizi < Ei=1 Zn <(an  bn)Zn = E;=1 a contradiction.
Definition 6.71. For each n E N, Bn = I lm 1 cE An . Lemma 6.72. For each n E N, Bn is a nonempty GS subset of N* such that
(a) Bn+1 + Bn+1 C Bn and
(b) for each p E Bn, ct(p + oz) fl Bn+1 = 0. Proof. By Lemma 6.68, we have that Bn # 0. Since for each m, min An = 2m Zn+m, we have Bn C N*. To verify (a), let p, q E Bn+1. To see that p + q E Bn, let m E N be given. Then by Lemma 6.69, for each x E A +1, An+1 C x + An +1 so n+
C{xE7Z:x+An+' Eq}
and consequently An +1 E p +q. Since An +1 C A',, we have An E p + q as required. To verify (b), let p E Bn and suppose that we have some r E Bn+1 fl cl (p + #7G). 
Then An+1 E r so An+1 n (p + PZ) # 0. Pick q E PZ such that An+i E p + q and let C = {x E 7G : x + An+1 E q}. Then C E p so pick x E C fl An and pick a ' E 9 , such that x = E 1 aizi. Let m = max{i E N : ai # 0}. Since An E p, picky E An fl C and pick b E 0,m such that y = E,°__1 bizi . Since x and y are in C,
pick w E (x + An+1) fl (y + An+1). Then x + w E An+1 and y + w E An +1 so pick c' and d in 69n+1 such that x + w = E°_1 ci zi and y + w = E °° 1 di zi Then 1(di  bi )zi . By Lemma 6.68, for each i, I ci  ai I < 2i w = E°_ 1(ci  ai )zi and Idi  bi I < 2i so by Lemma 6.70 c"  ii = d  b. Now b E On so bi = 0 for
i < n + m and thus E°__n+m 2' = 2n i > n + m, di = bi + ci. But then 1
2n+1
 E[=1di2' > EOO[=n+m 2i
di
Since a1 = 0 for i > m we have that for E°°
[=n+m
bi + Ci > E00
2'
bi
[n+m 2i
=
1
2n ,
a contradiction.
The following lemma is quite general. We state it for semigroups written additively because we intend to use it with (flZ, +).
6.7 Ideals and Density
133
Lemma 6.73. Let (S, +) be a compact right topological semigroup and for each n r= N, let Dn be a nonempty closed subset of S such that for each n, Dn+l + Dn+l S D. Given any sequence (gn)n_1 in S with each qn E Dn, there is a sequence (pn)n° 1 with each Pn E Dn, such that, for each n, pn+l + qn+1 = Pn
proof. Pick some r E N*. For n and m in N with n > m, let tn,m = qn. By downward induction on n, for n < m, let tn.m = to+1,m +qn+1 and notice that each tn,m E Dn. For each n E N, let pn = r lim tn,m and notice that, since Dn is closed, pn E Dn. Since MES
form > n, tn,m = to+l,m + qn+1 one has that
pn = r lim (tn+1,m + qn+1) = (r lim to+i,m) + qn+1 = Pn+1 + qn+1 m>n
m>n
Theorem 6.74. There is a sequence (pn )n° 1 in N* such that (pn +
OZ)', 1
is a strictly
increasing chain of principal right ideals of fZ and (cf(pn + flZ))n° 1 is a strictly increasing chain of principal closed ideals of #7L.
Proof. Pick some qn E Bn for each n E N. By Lemma 6.72, we have the sequences (Bn)n°_1 and (qn)n° l satisfy the hypotheses of Lemma 6.73 with S = f7L so pick a sequence (pn)ono_1 with each pn E Bn and each Pn = pn+l + q,,+1 Then one has immediately that for each n, pn + 167E C pn+1 + flZ and so, of course, cf (pn +,87L) C ct(pn+1 + 18Z). Further, by Lemma 6.72(b), for each n,
Pn+1 = Pn+1 + 0 E (Pn+1 + PZ)\ ct(pn + P7L) so both chains are strictly increasing.
6.7 Ideals and Density The concept of density for a set of positive integers has interesting algebraic implications in fiN. We shall show that the set of ultrafilters in fiN all of whose members have positive upper density is a left ideal of (fiN, +) as well as of (fiN, ), and that the same statement holds for the complement in N* of this set.
Lemma 6.75. Let S be an arbitrary semigroup and let J2 be a partition regular set of subsets of S. Let AR denote the set of ultrafilters in fS all of whose members are supersets of sets in R. If .2 has the property that s A E IR for every A E .R and every s E S, then A is a closed left ideal of f S. Proof. It follows from Theorem 3.11 that 0j? is nonempty and it is immediate that 0 JZ is closed.
To see that AR is a left ideal, let p E 0.R. It suffices to show that Sp c p,R, for then (f S)p = cefis(Sp) C Ax. So lets E S and let B E sp. Then s1 B E p so pick
AEJRsuchthat ACs1B.Then
sACB.
134
6 Ideals and Commutativity in 3S
Definition 6.76. Let A C_ N. We define the upper density 3(A) of A by
d(A)=lim sup IAfI{1,2,...,n}l n
n+Oo
Remark 6.77. Let A C N and let m E N. Then
d(m + A) = d(A) and d(mA) = l d(A). m
Definition 6.78. We define A C ON by
A={pEfN:d(A)>0for all AEp}. Theorem 6.79. A is a closed left idealof (ON, +) and of (ON, ).
Proof. We apply Lemma 6.75 with .R = {A c N : d(A) > 0). It is clear that .R is partition regular. By Remark 6.77 we have m + A E .R and mA E .R for every A E .R and every m E N. Thus our conclusion follows.
Theorem 6.80. N*\A is a left ideal of (ON, +) and of (ON, ).
Proof Let P E N*\A and let q E ON. There is a set B E p such that d(B) = 0. IBf1{1,2,...,n}j So f (n) ). 0 as n > oo. For n = For each n E N we put f () n each m E N, we choose nn E N so that f (n) < whenever n > nm. We put B,n = in E B : n > nm} and observe that Bm E P. It follows from Theorem 4.15 that S = UmEPI(m + Bm) E q + p and P = UmEFI mBm E qp. We shall show that these sets both have zero upper density. It will follow that p + q and pq are both in N* \ A. Suppose then that m E N and that n E Bm and that m + n < r for some r E N. This implies that nn < r because n > nm, and hence that f (r) <  . So, for a given value of r, if f (r) i6 0, the number of possible choices of m is at most 1/f (r). Since we also haven E B fl { l , 2, ... , r}, the number of possible choices of n is at most f(r)r. Thus the number of possible choices of in + n is at most r f (r). It follows that
ISfl{1,2,...,r}I
f(r)
r The same argument shows that
1P fl {1, 2.... r}l
r
<
f(r).
This establishes the claim that d(S) = d(P) = 0 and it follows that fN*\A is a left ideal of (ON, +) and of (ON, ). Notice that it did not suffice to show in the proof of Theorem 6.80 that n + p and
np are in N*\A for every p E N*\A because N*\A is not closed. Notice also that Theorems 6.79 and 6.80 provide an alternative proof that the centers of (ON, +) and of (ON, ) are both equal to N.
Notes
135
Notes The fact in Theorem 6.9 that fiN has 2' minimal right ideals is due to J. Baker and p. Milnes [10], while the fact that $N has 2` minimal left ideals is due to C. Chou [61].
The concept of an oid (" `® id' being unpronounceable") is due to J. Pym [204], who showed that the semigroup structure H depended only on the oid structure of FS((2")rt_1). Theorem 6.27 is due to A. Lisan [178]. Theorem 6.32 is due to T. Budak (nee Papazyan) [192]. Exercise 6.1.4 is from [38], a result of collaboration with J. Berglund. Theorem 6.36, one of three nonelementary results used in this book that we do not prove, is due to M. Rudin and S. Shelah. Theorem 6.36 and Lemma 6.37 deal (without ever explicitly defining them) with the RudinKeisler and Rudin Frohlc orderings of ultrafilters. See Chapter 11 for more information about these orderings as well as a discussion of their origins. For a simpler proof that there exist two points in N* that are not RudinKeisler comparable, see [182, Section 3.4]. The proof of Theorem 6.38 is from [66]. Theorem 6.46 is due to D. Parsons in [193]. Theorem 6.53 is, except for its weaker hypotheses, a special case of the following theorem which is due to E. van Douwen (in a letter to the first author). A proof, obtained in collaboration with D. Davenport, can be found in [74]. Theorem Let S be an infinite cancellative semigroup with cardinality K. There exists a decomposition i of U,K (S) with the following properties: (1) Ill = 22`.
(2) Each I E .f is a left ideal of PS.
(3) For each I E i and each p E I, ce(pS) C I. (4) Each I E i is nowhere dense in UK (S). Theorem 6.63 is from [180], a result of collaboration with A. Maleki. The results of Section 6.6 are from [149], a result of collaboration with J. van Mill and P. Simon. Sometime in the 1970's or 1980's M. Rudin was asked by some, now anonymous, analysts whether every point of Z* is a member of a maximal orbit closure
under the continuous extension a of the shift function a, where a (n) = n + 1. This question was not initially recognized as a question about the algebra of ,8Z. However,
if p E ,87Z, then a (p) = 1 + p and so, for all n E 9L, an (p) = n + p. Thus the orbit closure of p is PZ + p and so the question can be rephrased as asking whether every point of ,BZ lies in some maximal (proper) principal left ideal of PZ. One could answer this question in the affirmative by determining that there is no strictly increasing sequence of principal left ideals of P7L. The results of Section 6.7 are due to E. van Douwen in [80].
Chapter 7
Groups in fiS
If S is a discrete semigroup, it is often quite easy to find large groups contained in fS. For example, the maximal groups in the smallest ideal of ON contain 2' elements. More generally, as we shall show in this chapter, if S is infinite and cancellative with cardinality K, fS contains algebraic copies of the free group on 22` generators. This provides a remarkable illustration of how far fiS is from being commutative. However, it can be tantalizingly difficult to find nontrivial small groups in fS. For many years, one of the difficult open questions about the algebra of fN was whether or not fiN contained any nontrivial finite groups. This question has now been answered by E. Zelenuk, and we give the proof of his theorem in Section 1 below. Whether or not ON contains any elements of finite order which are not idempotent still remains a challenging open question.
Of course, there are many copies of Z in N*, since Z + p provides a copy of Z if p denotes any idempotent in N*. It is consistent with ZFC that there are maximal groups in N* isomorphic to Z, since Martin's Axiom can be used to show that there are idempotents pin N* for which H (p), the largest group with p as identity, is just Z + p. (See Theorem 12.42). It is not known whether the existence of such an idempotent can be proved within ZFC. Because we shall be constructing several topological spaces in this chapter, some of which are not necessarily Hausdorf we depart for this chapter only from our standing assumption that all hypothesized spaces are Hausdorff.
7.1 Zelenuk's Theorem The proof of Zelenuk's Theorem uses the notion of a left invariant topology on a group.
Definition 7.1. Let G be a group. A topology 7 on G is left invariant if and only if
for every UE7andevery a E G, a U E 7. Notice that a topology on G is left invariant if and only if for every a E G, Xa is a homeomorphism. Notice also that to say that (G, ) is a group with a left invariant
7.1 Zelenuk's Theorem
137
topology T is the same as saying that (G, , 7) is a left topological group, i.e., a group which is a left topological semigroup. The next result, Lemma 7.4, is unfortunately rather lengthy and involves a good deal of notation. We shall see after the proof of this lemma how a topology on G and a set X satisfying the hypotheses of this lemma arise naturally from the assumption that fG has a nontrivial finite subgroup while G does not. Definition 7.2. (a) F will denote the free semigroup on the letters 0 and 1 with identi
ty. (b) If M E w and i E (0, 1, 2, ... , m), sm will denote the element of F consisting of i 0's followed by m  i 1's. We also write u,,, = sm (so that uo = 0 ). (c) If s E F, I(s) will denote the length of s and supp s = {i E ( 1 , 2, ... ,1(s)} si = 1) where s, is the it' letter of s. (d) If s, t E F, we shall writes < < t if max supp s + 1 < min supp t.
(e) If s, t E F, we defines + t to be the element of F for which I (s + t) _ max(I(s), l(t)} and (s + t)i = 1 if and only if si = 1 or ti = 1. Given any t E F, t has a unique representation in the form t = s o° +sm I +
+smk
where O < io < mo < i < m i < < ik < Mk (except that, if k = 0, the requirement is 0 < io < mo). We shall call this the canonical representation of t. When we write t = s o° + sm' + + smk we shall assume that this is the canonical representation. Definition 7.3. (a) Given t E F, if t = s n for some i, m E co, then t' = 0 and t* = t.
Otherwise, if t = s o° + sm' +
+ s, then t' = s o° + sm1 +
+ SMk and
t* = SMk+I 'k+1
(b) The mapping c : F > co is defined by stating that c(t)
supp ti . (c) For each p E N, define gp : F + Z,, by gp(t) = c(t) (mod p). While the proof of Lemma 7.4 is long, because there area large number of statements to be checked, the reader will see that checking any one of them is not difficult. Recall that a topology is zero dimensional if it has a basis of clopen sets.
Lemma 7.4. Suppose that G is a group with identity e, and that X is a countable subset of G containing e. Suppose also that G has a left invariant Hausdorf zero dimensional topology and that X has no isolated points in the relative topology. We also suppose that, for every a E X, aX fl X is a neighborhood of a in X. We suppose in addition
that p E N and that there is a mapping h : X  Zp such that, for each a E X, there is a neighborhood V (a) of e in X satisfying aV (a) C X and h(ab) = h(a) + h(b) for every b E V (a). Furthermore, we suppose that h[Y] = Zp for every nonempty open subset Y of X and that V (e) = X.
Then we can define x(t) E X and X(t) c X for every t E F so that x(0) = e, X(0) = X, and the following conditions are all satisfied: (1) X (t) is clopen in X.
(2) x(t) E X (t).
7 Groups in flS
138
(3) X (t"v) U X (t" 1) = X (t) and X(t"0) n X (t" 1) = 0.
(4) x(t0) = x(t). (5) x(t) = x(t')x(t*). (6) X(t) =x(t')X(t*). (7) X(t*) c V(x(t')). (8) h(x(t)) = gp(t). (9) For any s, t E F, x(s) = x(t) if and only ifs is equal tot followed by 0's or viceversa.
(10) If S, t E F ands << t, then x(s + t) =x(s)x(t). (11) Ifs, t E F ands < < t, then h(x(s + t)) = h(x(s)) + h(x(t)). (12) For every n E w, x[unF] = X(un), so that in particular x[FI = X. (13) If (Wn)n° 1 is any preassigned sequence of neighborhoods of e in X, we can choose X (un) to Satisfy X (Un+1) a Wn for every n E N.
Proof. We assume that we have chosen a sequential ordering for X. We define x(t) and X (t) by induction on I (t).
We start by stating that x(0) = e and X(0) = X. We then make the inductive assumption that, for some n E w, x(t) and X(t) have been defined for every t E F with l(t) < n so that conditions (1)(9) and (13) are satisfied and further if n >_ 1, sequences
(ak)k in X and (tk)k=o in F have been chosen satisfying the following additional hypotheses for each k:
(i) ak = min X\{x(t) : t E F and 1(t) < k}, (ii) ak E X(tk1), and (iii) if h(ak) = gp(tk1), then ak = x(tk 1). Of these hypotheses, only (8) requires any effort to verify at n = 0. For this, note
that h(e) = h(ee) = h(e) + h(e) so h(e) = 0 = gp(0). Let an be the first element of X which is not in {x (t) : t E F and l (t) < n). Now the sets X (t) with l (t) = n form a disjoint partition of X by condition (3) and the choice of X (0) and so an E X (tn) for a unique to E F with I (tn) = n. Since X (tn) = x (t,,) X (tn ), it follows that an = x(tn)cn for some cn E X (tn). F o r each s E {s" : i E {0, 1, ... , n}} choose b, E X (s) such that bs 0 x(s) and
h(bs) = gp(s"1). (To see that we can do this, note that, by conditions (1) and (2), X (s) is a nonempty open subset of X so, since X has no isolated points, X (s)\{x(s)} is a nonempty open subset of X and hence h[X (s)\{x(s)}] = 7Gp.) In the case in which s = tn, we choose b,. = cn if h (cn) = gp (tn  1). To see that this can be done, notice that if cn = x(s), then by condition (5), an = x(tn)cn = x(t,,)x(tn) = x(tn). For each
s E {s, i E {0, 1, ..., n}} define x(s"0) = x(s) and x(s"1) = bs. Notice that x(s"1) =x(s')x(s*"1). Now, given any other t E F with 1(t) = n, notice that we have already defined x(t*"' 1) E X (t*). We define x(t"0) = x(t) and x(t" 1) = x(t')x(t*"' 1). Notice that
x(t) = x(t')x(t*) : x(t')x(t*" 1) = x(t"' 1)
7.1 Zelenuk's Theorem
139
andx(t '1) E x(t')X(t*) = X(t). We can choose a clopen neighborhood U of e in G such that Un fl x c V (x W)
and x(v)Un fl x c x(v)X for every v E F with 1(v) S n. (We can get the latter inclusion because x(v)X fl x is a neighborhood of x(v) in X.) We can also require that an f x (tn) U and that Un satisfies both of the following conditions for every t E F with 1(t) = n: x(t)U,,fl X C X (t) and
x(t1)x(t)U.
Furthermore, in the case in which there is an assigned sequence (Wn)' 1 of neighborhoods of e in X, we choose Un to satisfy u, fl x C W.
Let us note now that for any v E F with 1(v) < n, we have x(v)U, fl x = x(v)(U,, fl X). To see this, notice that x(v)Un fl x C x(v)X so that
X(v)Un fl x C x(v)Un fl x(v)X = x(v)(Un fl x), Also Un fl X c V (x(v)) and so x(v)(U fl X) C x(v)V (x(v)) c X. Hence x(v)(Un fl X) C x(v)U, fl X.
We put X(t"0) = x(t)Un fl x and X(t"1) = X(t)\X(t"0). We need to check that conditions (1)  (9) and (13) and hypotheses (i), (ii), and (iii) are satisfied for elements of F of length n + 1. Suppose then that t E F and that
1(t) = n. We put v = t1 and w = t"0. Notice that v' = t' and v* = t*" 1. Notice also that t is equal to w' followed by a certain number (possibly zero) of 0's so that x(t) = x(w') and that w* = un+1. Conditions (1), (2), (3) and (4) are immediate. In particular, notice that X(Un+1)
x(un) = e. We observe that (5) holds for v because x(v) = x(t')x(t*"1) = x(v')x(v*). It holds for w because x(w) = x(t) = x(w') = x(w')e = x(w')x(w*). We now check condition (6). By the definition of X (w), we have
X(w) = x(t)Un fl
x
= x(t)(UU fl x) = x(w')X(un+i) = x(w')X(w*). We also have
x (v) = X (t)\(x(t)Un n x) = X (t)\(x(t)(UU fl x)) = x(t')X (t*)\(x(t')x(t*)(Un fl x)) = X(t')(X(t*)\x(t*)(UU n X) x(t') X (t*)\(x(t*)Un fl X)
= x(t')X(t*"1) x(v')X(v*).
To verifycondition(7)forv,wenotethatX(v*) = X(t*"1) C X(t*) C V(x(t')) = V x(v')). It also holds for w, because X(w*) = X(un+i) = Un fl x C V(x(t)) = Vx(w')).
7 Groups in fiS
140
Condition (8) for w is immediate because h(x(w)) r h(x(t)) = gp(t) = gp(w). To verify condition (8) for v, we have h (x (v))= h (x (v')x (v*)) = h (x (v')) + h (x (v*))
because x(v*) E X(v*) c V(x(v')). Also x(v*) = x(t*"1) = b,* and so
h(x(v)) = gp(v') + h(br*)
= gp(v')+gp(t*1) = gp(v')+gp(v*) = gp(v' + v*) = gp(v) because the supports of v' and v* are disjoint. To see that (9) holds, we first note that s and t can be assumed to have the same length, because we can add 0's to s or t to achieve this. By condition (4), this will not change the value of x(s) or x(t). Then x(t) and x(s) belong to disjoint clopen sets if one of the elements s or t has a 1 in a position where the other has a 0. Also, if there is 1, then X C W,, so (13) holds. an assigned sequence Now an was chosen to satisfy (i). Further, since an ¢ x (tn) Un , one has an 0 X (tn "0)
so, since an E X(tn), one has an E X(tn"1) and thus (ii) holds. To verify (iii), assume that h(an) = gp(tn[1). Observe that tn"1 = to + to 1 so gp(tn" 1) =
gp(tn) + gp(tnl). Also, an = x(tn)c and cn E X(tn) C V(x(tn)) so h(an) = h(x(t;,)) + h(cn). Thus h(cn) = gp(tn 'l) so x(tn"1) = 1,, = cn. Therefore an = x(tn)c = x(tn)x(tn"l) = x(tn"1). Thus we can extend our definition of x and X to sequences in F of length n + 1 so that conditions (1)(9) and (13), as well as hypotheses (i), (ii), and (iii) remain true. This shows that these functions can be defined on the whole of F with these conditions remaining valid. It is now easy to show by induction on k using condition (5) that if the canonical representation of t is +smk
then x(t) = x(s o°)x(s"'') ... x(smk), so that condition (10) holds. To verify condition (11) notice that for any t E F we have
h(x(t)) = h(x(t')x(t*)) = h(x(t')) +h(x(t*)) because x(t*) E X(t*) c V(x(t')). Thus one can show by induction on k that if the canonical representation of t is
t =SO°+sn'' + h (x (smk then h (x (t )) = h (x (sm° )) + h (x (sr' )) + To check condition (12), we show first that each a E X eventually occurs as a value of x. Suppose instead that X\x[F] 0 0 and let a = mm n X \x [F]. 'Men a has only finitely
many predecessors soa = an forsomen. Picket E Zp such that h(an) = gp(t,[ 1)+m. If one had m = 0, then one would have by (iii) that an = x(tn"1) so m > 1. By (i) and the assumption that an 0 x[F], one has an+1 = an. Also by (ii) an E X(t1) and
7.1 Zelenuk's Theorem an+1 E X(tn+l'1) C X(tn+l) SO g p (t 1) + m = g p h (a,,+ 1) = h
141
Then gp(tn+t'1) = gp(tn1) + 1 so
 1). Repeating this argument m times, one has h (a,,+.) = gp (tn+m' l) SO by (iii), an = an+m = x (tn+m ' l ), a (tn+11) + (m
contradiction.
To complete the verification of condition (12), we note that, for every n E co and
E X(unt) S; X(un) and so x[unF] c every t E F, we have On the other hand, suppose that a E X\x[un F]. We have already established that a = x(v) for some v E F\un F. Since e E x[un F], a # e and hence v 0 {un, : m E w} so in particular supp v 96 0. If k = minsupp v, then k < n. We have x(v) E X(sk_1) X (sk_1) fl X(uk) = 0 and so a = x(v) 0 X(un). Thus and X(un) c x[unF]. From this point until the statement of Zelenuk's Theorem, G will denote a discrete group with identity e and C will denote a finite subsemigroup of G*.
Definition 7.5. (a) C" = {x E fG : xC C C}. (b) rp is the filter of subsets U of G for which C c U. (c) (p"' is the filter of subsets U of G for which C c U.
Observe that C is a semigroup and e E C. Note also that (p = n C and V` =
n c. Lemma 7.6. We can define a left invariant topology on G for which of is the filter of neighborhoods of e. If we also have xC = C for every x E C`, then this topology has a basis of clopen sets.
Proof. Given U E (P , we put U = {a (= G : aC c U) and observe that e E U. We begin by showing that
(i) for each UEgyp,UEgyp (ii) {U" : U E cp} is a base for the filter rp, and
(iii) for all U E cp and all a E U', a1 U" E rp`.
To verify (i), suppose that x E C\U'. Since U" V x,
G\U'=(aEG:ay0Uforsome yEC)Ex. Since C is finite, there exists y E C such that {a E G : ay 0 U) E x and hence xy 0 U. This contradicts the assumption that x E C. Since (i) holds, to verify (ii) it will be sufficient to show that nUE, U = C. (For then if one had some V E gyp such that for all U E'p, U\V # 0 one would have that {U\V : U E q') is a collection of closed subsets of fiG with the finite intersection property, and hence there would be some x E nUEw U"\V.) Suppose, on the contrary, that there is an element x E nUEcy U" \C`. Then x y 0 C for some y E C. We can choose U E 'p such that xy 0 U. This implies that x 0 U", a contradiction.
7 Groups in fS
142
To verify (iii), let U E V and a E U" be giver, and suppose that a1 U" V V. Then C"\a1 U. Then there is an element y E
G\a1U"={bEG:abz Uforsome zEC}Ey. Since C is finite, we can choose z E C such that {b E G : abz U} E y. This implies that ayz 0 U contradicting the assumptions that yz E C and a E U. Now that (i), (ii), and (iii) have been verified, let .B = {a U" : U E V). We claim that B is a basis for a left invariant topology on G with rp as the set of neighborhoods of e. To see that 8 is a basis for a topology, let U, V E cp, let a, b, c E G, and assume
that c E au fl bV". Now alc E U so by (iii), c1aU" E gyp" so by (ii) we may pick W1 E rp such that W(c c 1aU". Similarly, we may pick W2 E V such that cW2 c bV" and hence c(Wj fl W2) c aU fl bV". That the topology generated by 8 is left invariant is trivial. To see that rp is the set of neighborhoods of e, notice that by (ii), each member of W  is a neighborhood of e. So let W be a neighborhood of e and pick a E G and U E such that e E aU c W. Then a1 E U so by (iii), aU" E cp and hence W E gyp`. Finally we suppose that xC = C for every x E C. We shall show that in this case, for each U E W, U is closed. G\a1 U° E (P' To this end let U E rp and let a E G\U". We show that V = so that aV is a neighborhood of a missing U"'. So let y E C. We show that y E V.
Now ayC = aC
U since a ¢ U" so pick z E C such that ayz 0 U. Then
{bEG:abz0U}Eyand(bEG:abz0U) c{bEG:abC5U}=V.
Since each Aa is a homeomorphism we have shown that `a$ consists of clopen sets.
0 Lenuna 7.7. Assume that xC = C for every x E C. The following statements are equivalent.
(a) The topology defined in Lemma 7.6 is Hausdorff
(b) {aEG:aCcC}={e}. (c) n gyp" = {e}. Proof.
(a) implies (b). Let a E G\{e} and pick U E rp such that a 0 U. Since
C _c U, one has a C. (b) implies (c). Let a E G\{e}. Then by assumption a ¢ C. Also, C  _ I
XEC
Px1 [C] so C is closed in PG. Pick U C G such that C c U and a 0 U.
Then U E gyp' so a o n gyp`.
(c) implies (a). Let a E G\{e}. By Lemma 7.6 there is some clopen U E (P such that a ¢ U. Then U and G\U are disjoint neighborhoods of e and a. We shall now assume, until the statement of Zelenuk's Theorem, that C is a finite subgroup of G*. We shall denote the identity of C by u. We note that, since C is a group, xC = C for every x E C`.
7.1 Zelenuk's Theorem
143
Lemma 7.8. if G has no nontrivial finite subgroups, then j (p" = {e}.
proof. Observe that n w = C fl G = {a E G : aC = C}. (For if a E C fl G, then
for each U E rp`, a E U. And if a E G\, then for each p E C, G\{a} E p so G\{a} E (p".) Therefore nip` is a subgroup of G so is either {e} or infinite. So suppose that n (p"' is infinite. By the pigeonhole principle, there exists a # b in G such that au = bu, contradicting Lemma 6.28.
Lemma 7.9. There is a left invariant topology on G with a basis of clopen sets such that cp" is the filter of neighborhoods of e. If G has no nontrivial finite subgroups, then the topology is Hausdorff.
Proof. This is immediate from Lemmas 7.6, 7.7, and 7.8.
Definition 7.10. (a) Fix a family (Uy)yEC of pairwise disjoint subsets of G such that Uy E y for every y E C. (b) For each y E C, we put Ay = (a E G : az E Uyz for every z E C).
Observe that Ay = nZEC {a E G : aI UyZ E z) and that e E A. Lemma 7.11. For each y E C, Ay E y, and if y and w are distinct members of C, then
Ay flA,,,=0. Proof. Let y E C. For each z E C, UyZ E yz so {a E G : a1Uyz E Z) E y. Thus nzEC {a E G: a1Uyz E Z) E y. Now let y and w be distinct members of C and suppose that a E Ay fl A, Then Uy = Up, E au and U. = U,,, E au so Uy fl U. 0, a contradiction. Definition 7.12. (a) X = UyEC Ay. C by stating that f (a) = y if a E Ay. (b) We define f : X
(c) For each zECandaEX,Vz(a)={bEA,:abEAf(a)z). (d) For each a E X, V(a) = UZEC Vz(a).
Notice that for each a E X, V (a) c X and aV (a) c X. Notice also that f (e) = u so that for each z E C, VZ (e) = AZ and consequently V (e) = X.
Lemma 7.13. For each a E X and each b E V (a), f (a) f (b) = f(ab). Proof. Let a E X and let b E V (a). Pick Z E C such that b E Vz (a). Since Vz (a) C Az, we have that f (b) = z. Then ab E Af (a)z = Af (a)f (b) SO f (ab) = f (a) f (b).
Lemma 7.14. For each a E X, V (a) E p Proof. We begin by showing that
(i) for each y E C and each a E G, a E Ay if and only if au E Ay and (ii) for all y, z E C and each a E A Y, az E ;,,Z.
7 Groups in l4S
144
To verify (i), suppose first that a E Ay. If au 0 Ay, then {b E G : ab 0 Ay} E U. Now ab 0 Ay implies that abz 0 Uyz for some z E C. Since C is finite, we may suppose that there exists z E C such that {b E G : abz gE Uyz} E u. This implies that
auz = az 0 Uy, a contradiction. Now suppose that au E Ay. Then {b E G : ab E Ay} E u so for each z E C, {b E G : abz E U,,z} E u. Thus, for each z E C, az = auz E Uvz so that a E A. a1Ayz 0 z so that G\a1 Ayz E z. Now To verify (ii), suppose instead that
G\a'Ayz =
E G : abw 0 Uyz}
so pick w E C such that (b E G : abw 0 Uyzw) E z. Then azw 0 Uyzw so a 0 Ay, a contradiction.
Now, having established (i) and (ii), let a E X. We show that V(a) E V. So suppose instead that V (a) f of and pick x E CV (a). Let y = f (a) and let z = xu. Since x E C, Z E C, so x 0 Vz(a) and hence
By(i),{bE.G:b0Azorab0Ayz}= {b G:bu 0Azorabu 0 A,z) so either {bEG:bu0Az}Exor {bEG:abu0Ayz}Ex.That is,xuf Azoraxu0 yz. Since xu = z this says z
Az, which is impossible, or az
Ay,z which contradicts
(ii).
Corollary 7.15. X is open in the left invariant topology defined on G by taking gyp` as the base of neighborhoods of e.
Proof. For every a E X, we have aV (a) c X by the definition of V (a).
Lemma 7.16. For every nonempty Y C X which is open in the topology defined by cp", we have f [Y] = C.
Proof. If U E rp", then, for every y E C, u fl Ay 0 0 because U E y and Ay E Y. So f [U] = C. If a E Y, then aU C Y for some U E (p satisfying U C V (a). By Lemma 7.13 f [aU] = f (a) f [U] so C = f (a)C = f (a) f [U] = f [aU] C f[Y]. Theorem 7.17 (Zelenuk's Theorem). If G is a countable discrete group with no nontrivial finite subgroups, then G* contains no nontrivial finite groups.
Proof. We assume that C c G* is a finite group satisfying C = Zp for some integer p > 1 and derive a contradiction. Let y be an isomorphism from C onto Z,, and for each i E Zp, let y, = y (i ). Let
h = y o f. Then h : X + Zp and h(a) = i if and only if f (a) = yj. We assume that G has the topology produced in Lemma 7.9. Then by Lemmas 7.9, 7.13, 7.14, and 7.16 and Corollary 7.15 (and some other observations made after the definitions) the hypotheses of Lemma 7.4 are satisfied. So we presume we have chosen x(t) and X(t) for each t E F as guaranteed by Lemma 7.4.
7.1 Zelenuk's Theorem
145
Now Ay, E y1 and if a E Ay,, then f (a) = Y1 so h(a) = 1. If a = x(t), then h (a) = gp (t) by condition (8) of Lemma 7.4 and so Ay, c (X(01 t E F and gp (t) = 1) and hence
{x(t):tEFand c(t)1 (mod p))=(x(t):tEFandgp(t)=1}Ey1. For r E 1(0,1..... p  1}, let Br = {x(t) : t E F and c(t) = rp + 1 (mod p2)}. Now Ay, C u : Br and by condition (9) of Lemma 7.4 Br fl Bk = 0 when r # k so we may pick the unique r E {0, 1, ... , p  1) such that Br E y1.
Now for each n E N, X(un) is open and e = x(un) E X(un) so X(un) is a neighborhood of a and hence X (un) E V= (l C C n C c y1. Given t E F and n E N one has minsupp t > n if and only if t E unF\{un, : m E N). Also, for each n E N, x[unF] = X(un) by condition (12) of Lemma 7.4 so {x(t) : t E F and min supp t > n} = X(un)\{e} E yi. For each n E N, let
An ={x(s1+s2+ +sn): for each i E (1,2,...,n), Si E F, c(s;)  rp + 1 (mod p2), and if i < n, s, < < s;+1}. We show by induction on n that An E y1n. Since Al = Br E Y1, the assertion is true for n = 1. So let n E N and assume that An E yin. We claim that
An c{aEG:a1An+iEY1} so that An+t E ylny1 = Y1n}1. So let a E An and pick s1 << s2 << ... << sn in F such that each c(s,)  rp + 1 (mod p2) and a = x(s1 +S2 +  + sn). Let D = {x (t) : t E F and min supp t > max supp sn + 1). Then Br fl D E yi so it suffices to show that Br fl D c a1 An+i . To this end, let t E F
such that min supp t > max supp s + 1 and c(t)  rp + 1 (mod p2). By condition (10) of Lemma 7.4,
+ Sn)X (t) E An+1 as required. Now y1 P+1 = yi so Ap+1 E yi so pick some a E Ap+i fl Br. Then
SO x(Si + S2 +
a =x(si +s2++sp+1) where each c(s,)  rp + I (mod p2) and s1 < < s2 < <
(r +
< < sp+1. Then
1 (mod p2)
and hence a E Br+i SO Br fl Brt1 0 0, a contradiction.
146
7 Groups in fl S
Corollary 7.18. N* contains no nontrivial finite subgroups.. Proof Since Z* contains no nontrivial finite subgroups, neither does N*. Recall that a partial multiplication on a set X is a function mapping some subset Z of X x X to X. Given a partial multiplication on X and points a and b of X, we say that ab is defined if and only if (a, b) E Z. We shall sometimes express the same fact by saying "ab E X". (The latter terminology is convenient when the partial multiplication on X is induced by a multiplication on a larger structure.)
Definition 7.19. Let X be a topological space with a distinguished element e and a partial multiplication. We shall say that X is a local left group if there is a left topological
group G in which X can be topologically embedded so that the following conditions hold:
(i) e is the identity of G, (ii) the partial multiplication defined on X is that induced by the multiplication of G, (iii) for every a E X, there is a neighborhood V (a) of e in X for which aV (a) c X, and (iv) for every a E X, aX fl x is a neighborhood of a in X. We shall say that X is a regular local left group if X can be embedded in a Hausdorff zero dimensional left topological group G so that these four conditions hold. Notice that if G is a left topological group with identity e, then every open neighborhood of e in G is a local left group. Notice also that in any local left group, one may presume that V (e) = X.
Definition 7.20. Let X and Y be local left groups. We shall say that a mapping k : X > Y is a local homomorphism if, for each a E X, there is a neighborhood V(a) of e in X such that b E V (a) implies that ab E X, k(a)k(b) E Y and k(ab) = k(a)k(b). We shall say that k is a local isomorphism if it is a bijective local homomorphism and k1 is a local homomorphism.
Lemma 7.21. Let X and Y be local left groups with distinguished elements e and f respectively and let k : X * Y be a local homomorphism. Then k(e) = f. If k is continuous at e, then k is continuous on all of X.
Proof. Pick groups G and H containing X and Y respectively as guaranteed by the definition of local left group. For each a E X pick V (a) as guaranteed by the definition of local homomorphism. Since e E X and e E V (e), one has k(e) = k(ee) = k(e)k(e)
so k(e) is an idempotent in H and thus k(e) = f. Now assume that k is continuous at e and let a E X. Let W be a neighborhood of k(a) in Y and pick open U C H such that k(a) E u fl Y c_ W. Then Y fl k(a)1 U is a neighborhood of f in Y so pick a neighborhood T of e in X such that k[T] C Y fl k(a)1 U and pick open R C G such that e E R n X C V (a) fl T. Then aR fl x
7.1 Zelenuk's Theorem
147
is a neighborhood of a in X and, by the definition of local left group, ax n x is a neighborhood of a in X. We claim that k[aR n ax n X] C W. To see this let c E aR n ax n X and pick b E R n X such that c = ab. Then b E V(a) n T so k(a)t U. Thus k(c) E U n Y c W. k(c) = k(ab) = k(a)k(b) and k(b) E Theorem 7.22. Let X and Y be countable regular local left groups without isolated points. Then there is a local isomorphism k : X > Y. If Y is first countable, then k can be chosen to be continuous. If X and Y are both first countable, then k can be chosen to be a homeomorphism. proof. We shall apply Lemma 7.4 with p = 1, so that the functions h and gp are trivial.
Let e and f denote the distinguished elements of X and Y respectively. We can define x(t) E X and X (t) C X for every t E F, so that the conditions in the statement of Lemma 7.4 are satisfied. We can also define y(t) E Y and Y(t) C Y so that these conditions are satisfied with x replaced by y and X replaced by Y.
Define k : X . Y by k(x(t)) = y(t) for each t E F. By condition (9) of Lemma 7.4, x(t) = x(s) if and only if y(t) = y(s) so k is well defined and onetoone. By conclusion (12) of Lemma 7.4, k is defined on all of X and k[X] = Y.
Now let a E X, pick t E F such that a = x(t), and let n = 1(t) + 1. Pick (since X is a local left group) a neighborhood V1 (a) such that ab E X for all b E VI (a) and let V(a) = VI (a) n X(un). By conditions (1) and (2) of Lemma 7.4 and the fact that x(un) = e, V (a) is a neighborhood of e in X. Let b E V(a). Since b E Vt (a), ab E X. By condition (12) of Lemma 7.4 pick v E u F such that b = x(v). Then by condition
(10) of Lemma 7.4, x(t + v) = x(t)x(v) = ab and y(t + v) = y(t)y(v) = k(a)k(b) and consequently k(ab) = k(a)k(b) as required. Thus k is a bijective local homomorphism. Since kt : Y ? X is characterized by k(y(t)) = x(t) for each t E F, an identical argument establishes that k1 is a local homomorphism.
Now assume that Y is first countable, and let t W : n E N) be a neighborhood base at f. Then we may assume that for each n E N, Y(un+t) C W by condition (13) of Lemma 7.4. Given any n E N, by condition (12) of Lemma 7.4, Y(un+t) _ y[un+l F] = k[x[un+t F]] = k[X (un+t )] so X (un+t) is a neighborhood of e contained in k1 [Wn]. Thus k is continuous at e so, by Lemma 7.21, k is continuous on X. Similarly, if X is first countable, we deduce that k1 is continuous. Exercise 7.1.1. Let G be a countable group with no nontrivial finite groups. Show that G* contains no nontrivial compact groups. (Hint: An infinite compact subset of,6G cannot be homogeneous by Theorem 6.38.) Exercise 7.1.2. Let G be an infinite commutative group which does contain a nontrivial finite subgroup. Show that G* also contains a nontrivial finite subgroup.
Exercise 7.1.3. Let p E N*. Let pt = p and for n E N, let p,+1 = pn + p. (We cannot use np for the sum of p with itself n times because np is the product of n with p in (/3N, ). As we shall see in Corollary 17.22, np is never equal to the sum of p with
148
7 Groups in PS
itself n times, if n > 1 and p E N*.) Show that, if pn = p for some n > 1 in N, p must be idempotent.
7.2 Semigroups Isomorphic to IHI We remind the reader that IH[ denotes nni ceflN(2"N). Recall from Chapter 6 that a good deal is known of the structure of H. (More information will be found in Section 7.3.)
We show in this section that copies of H arise in many contexts. In particular, G* contains copies of H whenever G is a countable abelian group (Corollary 7.30) or a countable free group (Corollary 7.31).
Definition 7.23. Let X be a subset of a semigroup. A function t/r : to + X will be called an Hmap if it is bijective and if i/r(m + n) _ lr(m)*(n) whenever m, n E N satisfy max supp(m) + 1 < min supp(n). (We remind the reader that, if n E w, supp(n) E Pf(w) is defined by the equation n = E{2` : i E supp(n)}.) In the following theorem, and again in Theorem 7.28, we shall be dealing with two topologies at the same time. The spaces ,9G and fiX are constructed by taking G and X to be discrete, while the "neighborhoods" refer to the left invariant topology on G and the topology it induces on X.
Theorem 7.24. Let G be a group with a left invariant zero dimensional Hausdorff topology, and let X be a countable subspace of G which contains the identity e of G and has no isolated points. Suppose also that, for each a E X, a x fl x is a neighborhood of a in X and that, for each a E X, there is a neighborhood V (a) of e in X, with V (e) = X,
for which aV (a) C X. Then there is a countable set {Vn : n E N} of neighborhoods of e in X for which
Y = n,_, cefX V"\(e) is a subsemigroup of fG. Furthermore, there is an Hmap w + X such that defines an isomorphism from 1111 onto Y. In the case in which the filter of neighborhoods of e in X has a countable base, Y can be taken to be n{cEox W : W is a neighborhood of e in X )\[e).
Proof. We apply Lemma 7.4 with p = 1 so that IZpI = 1. We define h : X + Zi to be the constant map. We then observe that the hypotheses of Lemma 7.4 are satisfied, and hence x(t) and X (t) can be defined for every t E F so that the conditions stated in this lemma will hold. We put V" = X(un)andY = n,°O_, cP1X X(un)\{e}. We show thatYisasemigroup
by applying Theorem 4.20 with A = (X(un)\{e} : n E N). Now for each n E N, X(un)\{e} = x[unF]\{e} = {x(t) : t E F and min supp (t) > n} by condition (12) of Lemma 7.4. Given any m E N and any v E un, F\{e}, let n = max supp (v) + 2. Then
if w E u,,F\{e} we have u + w E uF\{e} and x(v + w) = x(v)x(w) by condition (10) of Lemma 7.4 so x(v)(x[u F]\{e}) 9 x[u,n F]\{e) as required by Theorem 4.20.
7.2 Semigroups Isomorphic to IIli
149
We define 8 : F  co by stating that 0(t) = E{21 : i E suppt}. By condition (9) ,f Lemma 7.4, for every t1, t2 E F, we have 8(t1) = 0(t2) if and only if x(ti) = x(t2). Thus we can define a bijective mapping t/r : w + X for which t(r o 8 = x. The napping t/r : fl w ). fl X is then also bijective (by Exercise 3.4.1). By condition (10) of Lemma 7.4, t/r(m + n) = t/r(m)t/r(n) whenever m, n E w satisfy max supp(m) + 1 < min supp(n), for we then have m = 8(s) and n =0(t)for some s, t E F with s < < t.
So tr is an Hmap. By Lemma 6.3 i (p + q) = *(p)*(q) for every p, q E H. Thus *I i is a homomorphism. Furthermore,
nn=1 ctsx *[2"N] = nn=1 cefXx{unF]\{e} = Y. This establishes that t%r defines a continuous isomorphism from El onto Y. Finally, if there is a countable base { W : n E N} for the neighborhoods of e in X, the sets X (un) can be chosen so that X (un+1) c Wn for every n E N (by condition (13) of Lemma 7.4). So Y = n{ce,sX W : W is a neighborhood of e in X}\{e}.
Lemma 7.25. Let T be a left invariant zero dimensional Hausdotftopology on (Z, +) with a countable base and assume that w has no isolated points in this topology. Then the hypotheses of Theorem 7.24 hold for G = Z and X = w. Proof. Given any a E X, a + X is a cofinite subset of the Hausdorff space X and hence is open in X. For each a E X let V (a) = X.
Theorem 7.24 has several applications to subsemigroups of fN . The following is one example and others are given in the exercises. Corollary 7.26. The set nn,= 1 ctS1y (nN) is algebraically and topologically isomorphic to H.
Proof. Let S = (a + n7L :a E Z and n E N}. Given a, b, c E Z and n, m E N, if c E (a + nz) fl (b + m7L), then c E c + nmZ C (a + n7L) fl (b + mZ) so .B is a basis for a left invariant topology 7 on Z. To see that 7 is zero dimensional, notice that if c E Z\(a + nZ), then c + nZ is a neighborhood of c missing (a + n7L). To see that 7 is Hausdorff, let a and b be distinct members of Z and pick n E N with n > ja  bi. Then (a + n7L) fl (b + n7L) = 0. Thus by Lemma 7.25, Theorem 7.24 applies. So n(cfp,, (W\{0}) : W is a neighborhood of 0 in co) = nfEN(cepN nN) is algebraically and topologically isomorphic to H.
Lemma 7.27. Let G be a countably infinite subgroup of a compact metric topological group C. Then with the relative topology, G is a Hausdorff zero dimensional first countable topological group without isolated points.
Proof. Let d be the metric of C. That G is a Hausdorff first countable topological group is immediate. To see that G is zero dimensional, let x E G and let U be a
150
7 Groups in $S
neighborhood of x in G. Since G is countable, for only countably many r E R is
{yEC:d(x,y)=r}f1G#0.Pick rEIFsuch that {yEC:d(x,y)=r}f1G=0 and {y E G : d(x, y) < r} c U. Then (y E G : d(x, y) < r) is aclopen neighborhood of x in G which is contained in U. To see that G has no isolated points, it suffices to show that the identity e of G is not isolated. Let U be an open neighborhood of e in C. Then {xU : x E G} is an open
cover of ct G. (Given Y E ct G, yU' fl G # 0. If X E yU' fl G, then y E xU.) Pick finite F C G such that G c UXEF xU. Pick X E F such that xU fl G is infinite. Then U fl G = U fl x'G is infinite. In the following theorem, the condition that C be metrizable is not really necessary, because any countable topological group which can be mapped injectively into a compact topological group can also be mapped injectively into a compact metrizable topological group. However, the proof of this fact would take us rather far from our subject; and it is not needed in any of our applications.
Theorem 7.28. Let G be a countably infinite discrete group which can be mapped into a compact metrizable topological group C by an injective homomorphism h. Let V = G* fl h'[(111, where 1 denotes the identity of C and h : PG ). C denotes the continuous extension of h. Then V is a compact G8 subsemigroup of G* which contains
all the idempotents of G*. Furthermore, there is an Hmap ik : w ± G such that i/r : fiw * PG defines an isomorphism from H onto V. In addition, for every pair of distinct elements a, b E G, a V fl V b= 0. Proof. Let e denote the identity of G. By Lemma 7.27, with the relative topology, h [G] is a Hausdorff zero dimensional first countable left topological group without isolated points. Thus, giving G the topology (h' [U] : U is open in C), h becomes a topological embedding and G enjoys all of these same properties.
For each a E G, let V (a) = G. Then G satisfies the hypotheses of Theorem 7.24
with X = G. Let V= n{ctfG W : W is a neighborhood of e in G}\{e}. (Notice that one is again considering two topologies on G. The space fiG is constructed taking
G to be discrete, while the "neighborhoods" of e are with respect to the topology just introduced.) By Theorem 7.24, V is a semigroup algebraically and topologically isomorphic to H. Further, since G is first countable, V is a G8 in PG.
To see that V = G* fl h'[{l}], note that, if p E G*, then p E V if and only if h' [U] E p for ever neighborhood U of 1 in C, and h' [U] E p if and only if h (p) E U. Then, since h is a homomorphism by Corollary 4.22, V contains all of the idempotents of G*.
Now let a, b E G and assume that aV fl Vb # 0. Pick x and y in V such that ax = yb. Since h : fiG > C is a homomorphism and h(x) = h(y) = 1, we have h(a) = h(b). So a = b, because h is injective. Lemma 7.29. Let G be an abelian group with identity e. For every a 0 e in G, there is a homomorphism h from G to the circle group T = {z E C : I z I = 1) for which h(a) 1.
7.2 Semigroups Isomorphic to H
151
proof. Let C = {a" : n E 7L} be the cyclic group generated by a. We define f on C by 2n7ri stating that f (a") = exp(in) if C is infinite, and f (a") = exp ( ) if C has order k k. We shall show that f can be extended to G. Let A = {(h, H) : H is a subgroup of G, C S H, h is a homomorphism from H to T and f h}. Ordering A by inclusion on both coordinates, we obtain a maximal member (h, H) of A by Zorn's Lemma.
We claim that H = G. Suppose instead that H
G and pick x E G\H. Let
H' = {x" y : n E 7G and y E H}. Then H' is a subgroup of G properly containing H. We show that h can be extended to a homomorphism h' from H' to G, contradicting the maximality of (h, H). Assume first that there is no n E Z\{0} such that x" E H. Since members of H' have a unique expression of the form x" y with n E Z and y E H, one may define a homomorphism h' on H' by h'(x"y) = h(y).
Otherwise, we choose m to be the first positive integer for which x' E H and choose 0 E R satisfying exp(i6) = h(xm). We observe that, for any n E Z, we have x" E H if and only if n is a multiple of m. We can now extend h to H' by niB stating that h'(x"y) = exp ( i )h(y). To see that this is well defined, suppose that
x"y=xkzwhere n,kEZandy,zEH. Thenk  n=qm for some geZand so y = (x"`)qz. Thus h(y) = (h(xm))9h(z) = exp(gi9)h(z) = exp (kiB)h(z). (niB)h(y) Thus exp
m
= exp
(
(k  00 m
)h(z).
m
Corollary 7.30. Let G be a commutative countable discrete group. Then there is a compact GS subsemigroup V of G* which contains all the idempotents of G*. Furthermore, there is an Hmap i,r : w G such that i/r : ow ). 6G defines an isomorphism from 1H[ onto V. In addition, V has the property that a V fl V = O for every a E G other than the identity.
Proof. Let e denote the identity of G. By Lemma 7.29, for every a # e in G, there is a homomorphism ha : G  T for which ha (a) 0 1. We put C = GT, and observe that C is a compact topological group and, being the countable product of metric spaces,
is metrizable. We define h : G * C by (h(x)),, = ha(x). Then the hypotheses of Theorem 7.28 are satisfied and so the conclusion follows.
Corollary 7.31. Let G denote the free group on a countable set of generators. There is a compact G8 subsemigroup V of G* which contains all the idempotents of G*. G such that i,1r : flco + PG defines an Furthermore, there is an Hmap 4r : to isomorphism from H onto V. In addition, aV fl Vb = 0 for every pair of distinct elements a, b E G. Proof. We shall use 0 to denote the identity of G. By Theorem 1.23, for every a E G \ (0}
there is a finite group Fa and a homomorphism ha : G + Fa for which ha(a) # 0. We put C = X aEG Fa, where each Fa has the discrete topology. Then C is a compact topological group and, being the countable product of metric spaces, is metrizable. We
7 Groups in $S
152
define h : G * C by (h(x)),, = ha(x). Then the hypotheses of Theorem 7.28 are satisfied and so the conclusion follows.
Theorem 7.32. Let G be an infinite countable discrete group which can be mapped by an injective homomorphism into a compact metrizable topological group. Then the minimal left ideals of $G are homeomorphic to those of flw. Furthermore, if L is any minimal left ideal of fiw, there is a homeomorphism from L onto a minimal left ideal M of fiG which maps L fl H isomorphically onto a subsemigroup of M containing all the idempotents of M.
Proof. By Theorem 7.28, there exist a compact subsemigroup V of G* which contains all the idempotents of G* and an Hmap * : w ). G for which fiw + #G defines a continuous isomorphism from H onto V. By Theorem 2.11, all the minimal left ideals of fiw are homeomorphic to each other, and so are all those of PG. Thus it suffices to establish the final statement of the theorem. Let L be a minimal left ideal of fiw and pick an idempotent p E L. By Theorem 1.38, p is a minimal idempotent in fw. Since all idempotents of iew are in H bb Lemma
6.8, all idempotents of fiG are in V, an isomorphism onto V, ,lr(p) is a minimal idempotent in SG. Thus M = (l4G)*(p) is a minimal left ideal of fiG.
Weclaimthatforeachm E w, *(m+p) = Indeed, *oAm and k*(m)oY' are continuous functions agreeing on {n E N : min supp(n) > max supp(m) + 1 }, which is a member of p. So *[w + p] = Gi/r(p). Thus
[L] = r[fiw+P] = cf9G ar[w+P] = cf,SG(G1/r(p)) = (fG)i(P) = M. Since *r is a bijection (by Exercise 3.4.1), L is homeomorphic to M. Since %rIH is an isomorphism and V contains all of the idempotents of fiG, one has *[L fl H] is a subsemigroup of M containing all of the idempotents of M. 0 Exercise 7.2.1. Show that n,,ez cfpz(2"7G) is algebraically and topologically isomorphic to H.
Exercise 7.2.2. Show that n,,e cdfN(3"N) is algebraically and topologically isomorphic to H. Exercise 7.2.3. Show that nfEN cf fz(3"7G) is algebraically and topologically isomorphic to H.
Exercise 7.2.4. Show that N* is not algebraically and topologically isomorphic to H. (Hint: Consider Exercises 6.1.1 and 6.1.2.)
Exercise 7.2.5. Suppose that G = ®iEN Gi, where each Gi is a nontrivial countable discrete group. (The direct sum of the groups Gi, is the subgroup of X i IN Gi containing the elements equal to the identity on all but finitely many coordinates.) Let ei denote the identity of Gi and let 7ri : G > Gi denote the natural projection map. If Ui =
{a E G : rrj (a) = ej for all j < i }, show that niEN cfsG Ui \{e} is algebraically and topologically isomorphic to H.
7.3 Free Semigroups and Free Groups in PS
153
7.3 Free Semigroups and Free Groups in $S We shall show that there is a high degree of algebraic freedom in 6S. If S is any cancellative discrete semigroup with cardinality K, then S* contains algebraic copies of 22K generators. the free group on Recall that a sequence (xn )nt in a semigroup S has distinct finite products provided
that, whenever F, G E ?f(N) and IIneF xn = nneG Xn, one has F = G. Theorem 7.33. Let S be a discrete semigroup and let (xn )n_ t be a sequence in S which has distinct finite products. If A = {xn : n E N), then I A* I = 2` and the elements of A* generate a free subsemigroup in S*.
Proof By Corollary 3.57 IA*j = 21. We define a mapping c : FP((xn)n_1)
N and mappings fi : FP((xn)n° 1) s A for each i E N as follows. Given Y E FP((xn)n° 1), there is a unique F E Pf(N) such that y= flmEF xn,. Let c(y) = IFI and write F = {MI, M2,..., mC(y)}, where ml < m2 < ... < mc(y). If i E {1, 2, ..., c(y)}, let fi(y) = xmi and otherwise let fr(y) = X1 
Nowletk E Nandletpl,P2,...,Pk E A*. Ifi E {1,2,...,k},then lim
f (Pl P2 ... Pk) = fi ( lim
Xn1P1 Xn2*P2
= =
lim
urn
Xn1 " PI Xn2P2
lim
lim
xn2 +P2
...
lim Xn,Xn2... xnk)
Xnk_Pk
lim fi(Xnlxn2... xnk)
Xnk"'*Pk
lim xni
XnkPk
= A. Similarly,
c(PIP2... pk) = F( lim
lim
XnlPI X2P2
...
lim XnIXn2...Xnk) Xnk'Pk
= k.
Now assume that k, m E N, P1, P2, ... , Pk, q i , q2,
, qm E A*, and further
P1P2...Pk = glg2...qm _Then k = c(pip2..Pk) = c(glg2...q,,) = m and given i E(1,2,...,k),pi Thus A* generates a free subsemigroup of OS.
As a consequence of Theorem 7.33 (with S = N) we have the following example.
Example 7.34. If A = 12n : n E N), the elements of A* generate a free subsemigroup of (ON, +).
Theorem 7.35. Let K be an infinite cardinal and let (ax)x
T=
na
cI(FP((au))L
Then T is a compact subsemigroup of , S and every maximal group in the smallest ideal 22' generators. of T contains an algebraic copy of the free group on
7 Groups in flS
154
Prof. To see that T is a subsemigroup of ,9S, we apply Theorem 4.20. So let I < K and let x E FP((a,,,)a
A
We choose any set in oneone correspondence with UK (A), denoting by aq the element in this set corresponding to the element q in UK (A). (Of course we can choose UK (A) itself, but it may be helpful to think of the set as distinct from UK (A).) Let F be the free group generated by {aq : q E UK (A)). By the universal property of free groups
(Lemma 1.22), there is a homomorphism g : F * G for which g(aq) = pqp. We shall show that g is an isomorphism. It will be sufficient to prove that g is onetoone on every subgroup of F which is generated by finitely many elements of {aq : q E UK (A)), because any two elements of F belong to such a subgroup of F.
Let ql , 42, ... , q be distinct elements of UK (A) and let D be the subgroup of F generated by {aq, , a q . , .... aq } . For each i E { 1 , 2, ... , n) choose B, E qi such that
Bi n B; = 0 if i 0 j. We may assume U°=1 Bi = A because we can replace B, by A\ U"=2 B;. We define a mapping h : FP((a,,)N,
We shall show that hIT is a homomorphism. By Theorem 4.21 it suffices to show that for each x E FP((aµ)u,
h(x y) = h(Il, ELUM au) = f , ELUM h(aµ) 11 h(ai,,) . fl EM h(au.)
= h(x) h(y). _ Thus h og is a homomorphism. Further, given i E (1, 2, ..., n), one hash (g (aq, )) = h_(p qj p) = h (p) h (q,) h (Q) = h (qi) = aq; . (Since p is an idempotent in T, h(p) is the identity of F.) Since h o g agrees with the identity on the generators of D, it agrees with the identity on D and consequently g is injective on D as required.
7.3 Free Semigroups and Free Groups in /3S
155
Corollary 7.36. Every maximal group in the smallest ideal of H contains a free group on 21 generators. Proof. The sequence
has distinct finite sums and
I = nn
Corollary 7.37. Every maximal group in the smallest ideal of (fiN, +) contains a free group on 2' generators. Proof. By Lemma 6.8, H contains all of the idempotents of ($N, +) and consequently, K (f N) fl H 0 0. Thus by Theorem 1.65, K(H) = K ($N) fl H.
Corollary 7.38. Let S be an infinite right cancellative and weakly left cancellative semigroup. Every neighborhood of every idempotent in S* contains an algebraic copy of the free group on 2' generators. Proof. This follows from Theorem 6.32 and Corollary 7.36. Note that we do not claim that every idempotent in S* is the identity of a free group
on 2' generators. In fact, according to Theorem 12.42 it is consistent with ZFC that there are idempotents p E N* for which the maximal group H(p) is just a copy of Z. Corollary 7.39. Let S be an infinite discrete semigroup with cardinality K which is right cancellative and weakly left cancellative. Then $S contains an algebraic copy of the 22' free group on generators.
Proof. By Lemma 6.31 there is a sequence (ax)x
Proof By Theorem 7.28, there is a subsemigroup G of 3G which is isomorphic to H and contains all of the idempotents of G*. Thus by Theorem 1.65, K (G) = _G fl K (,6G).
Thus, if p is any idempotent in K(PG), p is minimal in a copy of H so pGp contains a free group on 21 generators by Corollary 7.36.
We remind the reader that a subspace Y of a topological space X is said to be C*embedded in X if every continuous function from Y into [0, 1] can be extended to a continuous function defined on X.
Lemma 7.41. Let S be a discrete semigroup and let p be an idempotent in PS. Then Sp is extremally disconnected and is C* embedded in (8S) p. In fact, any continuous
156
7 Groups in PS
function from Sp to a compact Hausdorff space extends to a continuous function defined
on (PS)p.
Proof. Let U and V be disjoint open subsets of Sp. Define f : S  {1, 0, 1) by stating that f (s) = 1 if sp E U, f (s) = 1 if sp E V, and f (s) = 0 otherwise.
Let x E U. Then xp = x E U so IS E S : sp E U} E x so f (x) = 1, where f : 8S ) {1, 0, 1} denotes the continuous extension of .7. Similarly, 1(y) = 1 for every y E V. Hence U fl V = 0, and so Sp is extremally disconnected. To show that Sp is C*embedded, let V : Sp > Y be a continuous mapping into a compact Hausdorff space. The map r : S  Y defined by r(s) = cp(sp) extends to a continuous map T : /3S  Y. Let t E S. Then lim tsp = tpp = tp so s* p
T(tp) = SlimP r(ts) = lim W(tsp) S P
= V(lim tsp) s>p
= W(tp) so the restriction of T to (,BS)p is an extension of (p.
Let S be an infinite discrete semigroup. We know that, for every idempotent p in 1BS, the subsemigroup p(5S)p of 8S is a group if and only if p is in the smallest ideal of, S by Theorem 1.59. In this case, it is the maximal group containing p. We have also seen that the maximal groups in the smallest ideal of ,BS are disjoint and that they are all algebraically isomorphic (by Theorem 1.64). We shall show that, in spite of being algebraically isomorphic, they lie in at least 2` different homeomorphism classes, provided that they are infinite. If S is weakly left cancellative and has cardinality K, then there are 22` maximal groups in the smallest ideal of fS (by Corollary 6.43). Theorem 7.42. Let S be a discrete commutative semigroup and let L be a minimal left ideal in PS. If L is infinite, the maximal groups in L lie in at least 2` homeomorphism classes.
Proof Notice that if e is an idempotent in L, then the group e(fS)e = eL. Suppose that e and f are idempotents in L and that there is a homeomorphism p : eL a f L. We shall show that, given any x E eL and any y E f L, there is a homeomorphism of L to itself which maps x to y. Notice that by Theorem 4.23 Se = eS and Sf = f S. In particular, Se = See = eSe C eL. We know, by Lemma 7.41, that Vise has a continuous extension iW defined on L.
Now eS is dense in eL, because cfps(eS) = ctps(Se) = (,S)e = L and eL C L. So iW coincides with V on eL.
Now r = 01 I S f also has a continuous extension i defined on L. Since i o iW and iW o T are the identity maps on eL and f L respectively and since these sets are dense in L, it follows that they are also the identity maps on L. So iW is a homeomorphism.
Notes
157
There is an element z of f L for which W(x)z = y because f L is a group. Now the map pz defines a homeomorphism on L (by Theorem 2.11), and so pZ o iW is a homeomorphism of L to itself which maps x to y. The conclusion now follows from the fact that the points of L lie in at least 21 different homeomorphism classes by Theorem 6.38. Exercise 7.3.1. Let G be any given finite group. Show that there is an infinite discrete semigroup S for which the maximal groups in the smallest ideal of f3S are isomorphic to G and are equal to the minimal left ideals of OS. (Hint: Choose T to be an infinite right zero semigroup and put S = T x G.)
Exercise 7.3.2. Let S denote an infinite discrete commutative group. Prove that none of the maximal groups in the smallest ideal of 6S can be closed. (Hint: Use Theorem 6.38 and the fact that each maximal group in the smallest ideal of ,S is infinite.)
Notes The material in Section 7.1 as well as Theorem 7.24 are due to E. Zelenuk [246]. 1. Protasov has generalised Zelenuk's Theorem by characterizing the finite subgroups of fiG, where G denotes a countable group. Every finite subgroup of fiG has the form Hp for some finite subgroup H of G and some idempotent p in fiG which commutes with all the elements of H. [201] Our proof of Lemma 7.29 is based onthe proof in [114, p.441] Corollary 7.36 is from [152], a result of collaboration with J. Pym. Corollary 7.38 extends a result of A. Lisan in [178].
Chapter 8
Cancellation
It may be the case that $S does not have any interesting algebraic properties. For example, if S is a right zero or a left zero semigroup, then $S is as well and there is nothing very interesting to be said about the algebra of PS. However, the presence of cancellation properties in S produces a rich algebraic
structure in $S. To cite one example: we saw in Chapter 6 that, if S is an infinite 22.
cancellative semigroup with cardinality K, then fS contains disjoint left ideals. We 22* also saw in Chapter 7 that $ S contains copies of the free group on generators. In the first three sections of this chapter, we shall describe elements of $S which are right or left cancelable. If S is a cancellative semigroup with cardinality K, we 22K shall show that $S contains right cancelable elements. If we also suppose that S is countable, the set of right cancelable elements of $S contains a dense open subset of S*. In the important case in which S is N or a countable group, we can strengthen this statement and assert that the set of elements of $S that are cancelable on both sides contains a dense open subset of S*.
In the final section, we shall suppose that S is N or a countable group and shall discuss the smallest compact subsemigroup of S* containing a given right cancelable element. We shall show that these semigroups have a rich algebraic structure. Their properties lead to new insights about $S, allowing us to prove that K($S) contains infinite chains of idempotents, as well as
We could choose an idempotent p in L1. Then for any x E $S, the equation xp = xpp would imply that x = xp and hence that CBS = L1. This would contradict our assumption that 0 0 L2 C C8S\L1.
8.1 Cancellation Involving Elements of S Lemma 8.1. Suppose that S is a discrete semigroup. If s is a left cancelable element of S, it is also a left cancelable element of $S. The analogous statement holds for right cancelable elements as well.
8.1 Cancellation Involving Elements of S
159
Proof. Let S E S be left cancelable. Since the map As is injective on S, its extension to pS is injective as well (by Exercise 3.4.1). In the same way, if s E S is right cancelable, ps is injective on PS. Recall that we showed in Lemma 6.28 that if s is a cancelable element of the semigroup S, t E S, and p E flS one has:
(i) if S is right cancellative and sp = tp, then s = t, and (ii) if S is left cancellative and ps = pt, then s = t. As a consequence the following holds.
Corollary 8.2. Suppose that S is a discrete cancellative semigroup. If s and t are distinct elements of S, then, for every p E PS, sp # tp and ps # pt. Proof. This is an immediate consequence of Lemma 6.28.
We now consider some of the properties of left cancellative sernigroups. In the following example, we show that the assumption that S is left cancellative is not enough
to imply that, for every pair of distinct elements s, t E S and every p E 8S, ps # pt. Example 8.3. There is a countable left cancellative semigroup S containing distinct elements g and h such that pg = ph for some p E 0S. Proof. We take S to be the set of all injective functions f : N + N for which there exist
m, r E N such that f (n) = 2'n whenever n > m. We observe that S is a semigroup under composition and that S is countable and left cancellative.
We define g, h E S as follows: if n > 2, g(n) = h(n) = 2n, while g(1) = 1 and
h(l) = 2. We shall show that, for every finite partition F of S, there exists A E F such that
AgflAh#0. Let E = (f E S : f (1) and f (2) are odd, f (1) < f (2), and f (n) = 2n for all n > 3}. For each A E F, let A = {(f (1), f (2)) : f E E fl A). Given odd integers a < b, there is a unique member f of E with (f (1), f (2)) = (a, b). So
{(a,b):a,bE2Nlanda
c with a < b < c and {(a, b), (b, c)} C A. There exist f, k E E n A such that (a, b) =
(f (1), f (2)) and (b, c) = ((k(1), k(2)). Thenk(g(1)) = k(1) = b = f (2) = f (h(1)) and, for every n > 2, k(g(n)) = k(2n) = 4n = f (2n) = f (h (n)). So kg = fh and hence Ag fl Ah # 0. By Theorem 5.7, there is an ultrafilter p E PS such that Bg fl Bh # 0 for every
B E p. Since {Bg : B E p} is a base for pg and (Bh : B E p) is a base for ph, it follows that pg = ph.
160
8 Cancellation
We saw in Theorem 3.35 that if f : S S, p E. PS, and T (P) = p, then {x E S : f (x) = x} EJ. It might seem reasonable to conjecture that if f, k : S * s, p E ,6S, and f (p) = k(p), then {x E S : f (x) = k(x)} E p. However, the functions PS, Ph : S * S and the point p E fS of Example 8.3 provide a counterexample. In the next lemma we show that an example of this kind cannot occur with two commuting elements of S. Lemma 8.4. Let S be a left cancellative discrete semigroup and let s and t be distinct elements of S such that st = ts. Then, for every p E fl S, ps # pt.
Proof. We can define an equivalence relation  on S by stating that u  v if and only if us" = vs" for some n E N. (To verify transitivity, note that if us" = us" and vsm = wsm, then us"+m = ws"+'".) Let 6 : S * S/  denote the canonical projection. We shall define a mapping f : 6[S] > 6[S] which has no fixed points. For each u E S, we put f (6 (us)) = 6 (u t). To show that this is well defined, suppose
that B(us) = 6(vs). Then us" = vs" for some n E N, and hence us"t = vs"t so that uts" = vts". So 6(ut) = B(ut). Thus f is well defined on B[Ss]. To complete the definition of f, we put f (6(w)) = 6(ss) for every w E S\01 [6[Ss]]. We observe
that f (8(ps)) = 8(pt) for every p E flS, where 8 : PS a f (S/ ). (One has that f o 8 o ps and 6 o pt are continuous functions agreeing on S, hence on 0S.)
We claim that f has no fixed points.
Certainly if w E S\01[6[Ss]], then
f(9(w)) zA6(w). Now suppose that 0 (us) = 8(ut) for some u E S. Then uss" = uts" for some n so that us"s = us"t and hence s = t, contradicting our assumption that s and t are distinct. It now follows from Lemma 3.33, that 8[S] can be partitioned into three sets A0, A1, A2, such that f [Ai] fl A, = 0 for every i E {0, 1, 21. Suppose that p E PS
satisfies ps = pt. We can choose i such that 01[Ai] E ps. Then Ai E 6(ps) and f [Ail E f (8(ps)) = 8(pt) = 8(ps) so that Ai f1 f [Ail # 0, a contradiction. Lemma 8.5. Let S be a discrete left cancellative semigroup and let s and t be distinct elements of S. If p E PS, then sp = tp if and only if {u E S : su = tu} E p.
Proof. Let E = {u E S : su = t u}. We suppose that sp = tp and that E 0 p. We shall define a mapping f : S + S which has no fixed points.
For every u E S\E, we put f (su) = tu. We observe that this is well defined, because every element in sS has a unique expression of the form su. Furthermore, f (su) , su. To complete the definition of f, for each v E S\s (S\E), we choose f (v) to be an arbitrary element of s(S\E). This is possible, because our assumption that E 0 p implies that S\E # 0.
By Theorem 3.34, f : 8S > ,BS has no fixed point. However, f (su) = to if u E S\E and hence f (sp) = tp = sp, a contradiction. For the converse implication observe that ) and A, are continuous functions agreeing on a member of p.
8.2 Right Cancelable Elements in PS
161
Comment 8.6. Let be a semigroup and define an operation * on S by x * y = y x. Then for all x E S and all p E PS one has x * p = plim(x * y) = plim(y x) = p x yES
yES
and, similarly, p * x = x p. Consequently, the leftright switches of Example 8.3 and Lemmas 8.4 and 8.5 remain valid. Exercise 8.1.1. Let S be any semigroup. Show that the set of right cancelable elements of S is either empty or a subsemigroup of S, and that the same is true of the set of left cancelable elements. Show that the complement of the first is either empty or a right ideal in S, and that the complement of the second is either empty or a left ideal of S.
8.2 Right Cancelable Elements in PS In this section, we shall show that cancellation assumptions about S imply the existence of a rich set of right cancelable elements in S*. We shall also show that right cancelability is closely related to topological properties and to the RudinKeisler order. The following theorem gives an intrinsic characterization of ultrafilters which are right cancelable in,9 S for any infinite semigroup S. It is intriguing that this is an intrinsic property, characterizing a single ultrafilter without reference to any others, since right
cancelability is defined in terms of the set of all ultrafilters in fS. (Another intrinsic characterization will be given in Theorem 8.19 in the case that S is either (N, +) or a countable group.)
Theorem 8.7. Let S be an infinite semigroup, let p E fS, and let T be an infinite subset of S. Then sp # rp whenever s and r are distinct members of T if and only if for every
A C T there is some B C S such that A = {x E T: x1 B E p}. In particular, p is right cancelable in 18S if and only if for every A C S, there is some B C S such that
A = {XES:X1BE p}. Proof. The sufficiency is easy. Given r
Af1T={xET:x1BEp}.Then
s in f, pick A E r\s. Pick B C S such that
To prove the necessity, assume that pp is injective on f and let A C T. Since pP,T is a homeomorphism,A p is a clopen subset of T p. So by Theorem 3.23 there exists a set B C T for which Ap = B fl T p. If x E T, we have
xEA. XP EB qX iBEp soA={xET:xiBEp}. Now right cancellation in PS can be made to fail in trivial ways. For example, if S has two distinct left identities a and b, then for all p E 13 S, ap = bp. Consequently, one may also wish to know for which points p E S* is p right cancelable in S* (meaning of
162
8 Cancellation
course that whenever q and r are distinct elements of S*, qp # rp). The proof of the following theorem is a routine modification of the proof of Theorem 8.7, and we leave it as an exercise.
Theorem 8.8. Let S be an infinite semigroup and let p E S*. Then p is right cancelable
in S* if and only if for every A C S, there is some B C S such that I A D {x E S x1B E p}I < w. Proof This is Exercise 8.2.1. We remind the reader that a subset D of a topological space X is strongly discrete if and only if there is a family (Ua)aED of pairwise disjoint open subsets of X such that
a E Ua for every a E D. It is easy to see that if D is countable and X is regular, then D is strongly discrete if and only if it is discrete.
Lemma 8.9. Let S be a discrete semigroup, let T be an infinite subset of S, and let p E CBS have the property that ap # by whenever a and b are distinct elements of T. (i) If qp rp whenever q and r are distinct members of T, then Tp is discrete in,8S. (ii) If Tp is strongly discrete in 8S, then qp 0 rp whenever q and r are distinct members of T. Proof. (i) If Tp is not discrete in 8S, there is an element a E T such that ap E
cf((T\{a})p). Since pp is continuous,ct((T\{a})p) = (ci(T\{a}))pand soap =xp for some x E (T\{a}) = T\{a}. (ii) Suppose that Tp is strongly discrete in ,9S and pick a family (Ba )aET of subsets
of S such that Ba n Bb = 0 whenever a and b are distinct members of T and ap E Ba
for each a E T. Let A C T and let C = UaEA Ba. Then A = (s E T : s1 C E p).
(Ifa E A, then a1Ba C a1C andifa E T\A, then a1Ban a1C=0.) Thus by Theorem 8.7, qp 54 rp whenever q and r are distinct members of T.
Theorem 8.10. Let S be an infinite discrete right cancellative and weakly left cancellative semigroup with cardinality K. Then the set of right cancelable elements of PS 22K contains a dense open subset of UK (S). In particular, S* contains elements which are right cancelable in 6 S.
Proof. We apply Theorem 6.30 with R = S. By this theorem, every subset T of S with I T I = K contains a subset V with I V I = K such that for every uniform ultrafilter p on V, by whenever a and b are distinct members of S and Sp is strongly discrete in 8S. By Lemma 8.9, this implies that every uniform ultrafilter on V is right cancelable in flS.
ap
Since the sets of the form f n UK (S) provide a base for the open subsets of UK (S), it follows that every nonempty open subset of UK (S) contains a nonempty open subset of UK(S) whose elements are all right cancelable in $S. Thus the set of right cancelable elements of,6S contains a dense open subset of UK(S). 22' Finally, the fact that there are right cancelable elements of CBS follows from the 22' fact that I UK (V) I = if IV I = K (by Theorem 3.58.)
8.2 Right Cancelable Elements in,6S W..uv now show u1 i at
i bility in o blie , iif, CanGeiaiuiy 111 PO o.5right is countable,
163
is equivalent to several
other properties. (If S = T in Theorem 8.11, then statement (3) says that p is right cancelable in PS.)
Theorem 8.11. Let S be a semigroup, let p E ,6S, and let T be an infinite subset of S. Consider the following statements. Statements (1) and (2) are equivalent and imply each of statements (3), (4), (5), and (6) which are equivalent. Statements (7), (8), and (9) are equivalent and are implied by each of statements (3), (4), (5), and (6). If T is countable, all nine statements are equivalent.
(1) ap
by whenever a and b are distinct members of T and T p is strongly discrete.
(2) There is a function f : S + S such that for every q E T, T (q . p) = q. (3) qp # rp whenever q and r are distinct members of T. (4) For every A C T there exists B C S such that A= Is E T s1 B E p}. (5) For every pair of disjoint subsets A and B of T, Ap fl Bp = 0. (6) pplT : T  T p is a homeomorphism. (7) PpIT : T  Tp is a homeomorphism. (8) ap # by whenever a and b are distinct members of T and T p is discrete. (9) For every a E T and every q E T \{a}, ap # qp. Proof. To see that (1) implies (2) assume that (1) holds. Then there is a family (Ba)aET
of pairwise disjoint subsets of T such that ap E Ba for every a E T. We define f : S + S by stating that f (s) = a if s E Ba, defining f arbitrarily on S\ Uaes Ba. Since, given a E T, f is identically equal to a on Ba and since ap E W,,, it follows that f (ap) = a for every a E T. Hence, for every q E T,we have To Pp agrees with the identity on a member of q and thus f (qp) = q. To see that (2) implies (1) pick a function f as guaranteed by (2). For each a E T
one has that 70 Xa(P) = a and {a} is open in S so pick a member Ba of p such that f o .la [Ba] = (a). If a # b, then f [aBa] = {a} 0 {b} = f [bBb] so aBa fl bBb = 0. That (1) implies (3) is Lemma 8.9 (ii). That (3) and (4) are equivalent is Theorem 8.7. To see that (4) implies (5), let A and B be disjoint subsets of T and pick C C S such
thatA={sET:s1CEp}.ThenApCCandBpCS\C.
To see that (5) implies (3), let q and r be distinct members of T and pick A E q\r. Then q p E (A fl T) p and rp E (T \A) p so q p 0 rp. Statements (3) and (6) are equivalent because a continuous function with compact domain (onto a Hausdorff space, which all of our hypothesized spaces are) is a homeomorphism if and only if it is one to one. That statement (6) implies statement (7) is trivial as is the equivalence of statements (7) and (8). To see that statement (8) implies statement (9), let a E T and let q E T \{a}. Then
qp E (T\{a})p = (T\{a})p = (T\{a})p = Tp\{ap} andap 0 Tp\(ap). To see that statement (9) implies statement (8), notice that trivially ap 0 by whenever a b in T. If Tp is not discrete then for some a E T one has ap E Tp\{ap} _ (T \ (a }) p, a contradiction.
164
8 Cancellation
Finally, if S is countable one has that statement (8) implies statement (1).
We do not know whether the requirement that S is countable is needed for the equivalence of all nine statements in Theorem 8.11.
Question 8.12. Do there exist a semigroup S (necessarily uncountable) and a point p of fiS such that p is right cancelable in fiS but Sp is not strongly discrete? Question 8.13. Do there exist a semigroup S (necessarily uncountable) and a point p of $S such that ap # by whenever a and b are distinct members of S and Sp is discrete, but p is not right cancelable in ,S? Recall that a point x of a topological space X is a weak Ppoint of X provided that whenever D is a countable subset of X\{x}, x 0 ct D. Corollary 8.14. Assume that S is a countable cancellative semigroup. Then every weak Ppoint in S* is right cancelable in PS.
Proof. Suppose that p is a weak Ppoint in S*. We shall show that for every a E S and every q E j S\{a} one has ap # qp. The required result will then follow from Theorem 8.11.
Suppose instead that we have a E S and q E flS\(a) such that ap = qp. By Corollary 8.2. q E S*. By Lemma 8.1, ka is one to one, and hence;,alS' is a homeomorphism from S* onto aS* so that ap is a weak Ppoint of aS*.
Let B = {s E S\{a} : s'aS E p}. Since aS E ap = qp, one has B E q so that ap = qp E c2(Bp). But now, given s E B one has sp E aS = a9S so Bp g apS. By Corollary 4.33 Bp C S* so Bp c aS*. Since ap 0 Bp we have a contradiction to the fact that ap is a weak Ppoint of aS*. In certain cases we shall see that the following necessary condition for right cancelability is also sufficient.
Lenuna 8.15. Let S be a weakly left cancellative semigroup and let p E PS. If p is right cancelable in 13S, then p 0 S* p.
Proof. Suppose that p = xp for some x E S* and pick a E S. Then ax E S* by Theorem 4.31, so ax
a while ap = axp, a contradiction.
We pause to introduce a notion which will be used in our next characterization.
Definition 8.16. Let S be a discrete space. The RudinKeisler order
8.2 Right Cancelable Elements in flS
165
Theorem 8.17. Let S be a discrete space. The following statements are equivalent.
(a) P
q.
(b) p 
(c) There exist f : S > S and Q E q such that 7(q) = p and f IQ is injective. (d) There is a bijection g : S . S such that g(q) = p.
Proof. (a) implies (b). Let f : S  S such that T (q) = p. Since q <_RK p, pick g : S * S such that g(p) = q. Then go f (q) = g(f(q)) = q so by Theorem 3.35,
Q={aES:g(f(a))=a} Eq.
That (b) implies (c) is trivial. (c) implies (d). By passing to a subset if necessary we may presume that I S\ Q I _ IS\f [Q]I. Thus we may choose a bijection g : S > S such that g1Q =
(d) implies (a). Since g(q) = p, p
Theorem 8.18. Suppose that S is either (N, +) or a countable group. For every p E S* the following statements are equivalent.
(1) p is right cancelable in ,6S.
(2) p 0 S* p. (3) There is no idempotent e E S* for which p = ep. (4) For every x E S*, X
Theorem 8.11 to show that for each a E S and each q E $S\{a}, ap ,£ qp so let a E S and q E CBS\{a} and suppose that ap = qp. In case S is a countable group, we have p = aI qp and by Corollary 4.33 atq E S*. In case S is (N, +), the equation
becomes a + p = q + p so, in $7G, p = a + q + p. By Exercise 4.3.5, N* is a left
ideal of$7Lsoa+q E N*. The equivalence of (2) and (3) follows from the fact that {x E S* : p = xp} is a compact subsemigroup of S* if and only if it is nonempty. Thus, by Theorem 2.5, this set is nonempty if and only if it contains an idempotent. To show that (1) implies (4), suppose that p is right cancelable in ,BS. By Theorem 8.11, there is a family (Ua )aES of pairwise disjoint open subsets of CBS such that ap E Ua
for every a E S. We put Pa = {b E S : ab E Ua) and observe that Pa E p for every a E S. Every elements E can be expressed uniquely in the form s = ab with a E S and b E Pa. We define functions f, g : S  S by stating that f (s) = a
166
8 Cancellation
and g (s) = b ifs is expressed in this way, defining f and g arbitrarily on S\ UaES a ra Let X E S*. Since f (ab) = a and g(ab) = b if a E S and b E Pa, we have
f (xp) = f (xlim plim ab) = xlim plim f (ab) = xlim plim a = x aES
bEPa
aES
bEPa
aES
bEPa
and
g(xp) = g(xlim p lim ab) = xIim p lim g(ab) = xlim p lim b = p. aES
bEPa
aES
bEPa
aES
bEPa
SOX
Let X = {a E S : a1 B E p}. Then X E x so is nonempty. Pick a E X and, since p E S*, pick distinct elements b and c of a1B fl Pa. Then ab and ac are distinct
elements of B so f (ab) # f (ac). But ab, ac E a Pa so f (ab) = a = f (ac), a contradiction.
Similarly, if p " RK xp, then there is a set C E xp such that f c is injective. Let
Y = {a E S : a1C E p}. Then Y E x and x E S*, so pick distinct a and b in Y. Pick C E a1 C fl b1C fl Pa fl Pb. Then ac and be are distinct members of C so g(ac) i4 g(bc). But ac E aPa and be E bPb, so g(ac) = c = g(bc), a contradiction. Trivially (4) implies (5) and (5) implies (2).
The equivalence of statements (1), (4), and (5) of Theorem 8.18 will be established under weaker assumptions in Theorem 11.8.
We saw in Theorem 8.7 that for any semigroup S and any p E ,BS, p is right cancelable in ,BS if and only if for every A _: S there is some B C S such that A = {x E S : x1 B E p}. We see now that, in the case S is a countable group, it is sufficient to know this for A = {e}.
Theorem 8.19. Suppose that S is either (N, +) or a countable group. An element p E fJS is right cancelable in 6S if and only if there exists B E p such that, for every a E S distinct from the identity, a1 B f p.
Proof. Necessity. In case S is a group with identity e, let A = {e} and pick B as guaranteed by Theorem 8.7. In case S is (N, +), apply 8.7 with A = { I), and if C is such that (1) = {x E N : X + C E p}, let B = 1 + C. Sufficiency. Pick such B and suppose that p is not right cancelable. Then by
Theorem 8.18 pick x E S* such that p = xp. Since B E p, one has {a E S : a1 BEp} E x so there is some a distinct from the identity such that aB E p, a contradiction.
In the next theorem, we show that every ultrafilter in f8N is right cancelable on a large open subset of ON.
Theorem 8.20. Suppose that S is either (N, +) or a countable group. Let p E PS. There is a dense open subset U of PS such that u fl S* is dense in S* and xp # yp. whenever x and y are distinct elements of U.
8.2 Right Cancelable Elements in SS
167
proof For each a E S, let Ca = {x E S* : ap = xp}. Then Ca is a closed subset of S*. We claim that it is nowhere dense in S*. To see this, suppose we have some infinite subset A of S such that X n S* c Ca. Then by Corollaries 8.2 and 4.33, Ap is an infinite subset of S* so we can choose an infinite subset T of A\ (a) for which Tp is discrete in
/3S. Now by Theorem 8.11, qp # rp whenever q and r are distinct members of T, so for at most one q E T is qp = ap and thus there exists r E T n S* such that rp 0 ap so that r 0 Ca, a contradiction. Now UaES Ca is nowhere dense in S* by Corollary 3.37. Let U = /BS\UaES Ca. Then U is a dense open subset of /iS for which U n S* is dense in S*. It is immediate that, for every a E S and every x E U, ap # xp. Suppose that x and y are distinct elements of U. We can choose disjoint subsets Xand Y of S satisfying
X E X, Y E y, X C U, and Y _c U. We observe that xp E Y p and yp E F p. So the equation xp = yp would imply, by Theorem 3.40, that ap = y' p for some a E X and some y' E Y or that x'p = by for some x' E X and some b E Y. Since this is not possible, xp yp. We now show that the semigroup generated by a right cancelable element of S* provides, under certain conditions on S, an example of a semi group whose closure fails dramatically to be a semigroup.
Theorem 8.21. Let S be a discrete countable cancellative semigroup and let p E S* be a right cancelable element of,6S. Let T = f p' : n E N} and X = Cf pG T \T. Then the elements of X generate a free subsemigroup of /3S.
Proof. We shall use the fact that S* is an ideal of /3S (by Theorem 4.36). We first note that, if a E S, n E N, and y E X, then ap" 0 (l4 S) y. This follows from
the observation that y E ce{p' : r > n} c S*p" and that, since p" is right cancelable
in/3S,ap"0S*p". We now show that the equation ax = uy, where x, y E X, a E S and U E /3S, implies that a = u and x = y. We may assume that a # u, because a = u implies that x = y by Lemma 8.1. We note that ax E ce(aT) and uy E ce((S\{a})y). It follows from Theorem 3.40 that ap" = vy for some n E N and some v E /3S, or else az = by for some z E X and some b E S\{a}. The first possibility cannot hold because ap" 0 (,8S)y, and so we may assume the second.
Now az E ce(aT) and by E ce(bT). So another application of Theorem 3.40 shows that az' = by', where z', y' E ce T and either z' or y' is in T. We have seen that z' E T implies that y' E T and viceversa. So we have apm = bp" for some m, n E N. However, this cannot hold if m = n (by Lemma 6.28). It cannot hold if m < n, because this would imply that a = by"` E S*. Similarly, it cannot hold if m > n. So ax = u y does imply that a = u and x = y. We can therefore assert that the elements of X are right cancelable in ,B S, by condition
(9) of Theorem 8.11. Furthermore, for any x, y E X, x ($S)y = 0, by Theorem 6.19.
y implies that ($S)x n
8 Cancellation
168
To see that X generates a free subsemigroup of 48S, suppose that m, n E N and that x 1x2 ... xm = y 1 y2 ... yn where each xi and each y j is in X and the sequences (xl, x2, ... , xm) and (yl, Y2..... y,) are different. We also suppose that we have chosen an equation ofthis kind for which m +n is as small as possible. Since(u3S)xmn(flS)y _ 0 if xm # y, , we have that xm = y, . We claim that m, n > 1. To see this, we note that m = 1 implies that n > 1 and that axl = aY1Y2 ... Yn for any a E S. Since xl is right cancelable in ,S, this results in the contradiction that a E S*. Using the fact that x, is right cancelable once again, we have xlx2 ... xm_i = YlY2 . . . Y, I This contradicts our assumption that m + n is as small as possible. We shall show in Corollary 8.26 that right cancelable elements can occur in the closure of the smallest ideal of PS. Theorem 8.22. Let S be a discrete countably infinite cancellative semigroup and let T be an infinite subsemigroup of S. Then there are elements p in E(K0 T)) for which
SpnS*S*=0. Proof We recall that S* is an ideal of PS (by Theorem 4.36). Let f : S > w denote the function guaranteed by Lemma 6.47. Since f [T] = w, f [fT] = fw. We claim that
(*) f o r all p E S* and all x E fl S, f (xp) E {1 + T (P), T (p), 1 + T (p)).
To establish (*), it suffices to establish the same statement with x E S. So let x E S
and suppose that f (xp) 0 (1 + f (p), f (p), 1 + f (p)}. Pick B E f (xp) such that f (P) 0 Ui__1(i + B). Pick C E xp such that f [C] S; B and pick D E p such that f[D]nUi=_1(i+B) = 0. Picks E Dnx1Cwith f(s) > f (x) + 1, which one can do by Lemma 6.47(2). Then by Lemma 6.47(3), one has some i E {1, 0, 11 such that
f (xs) = i + f (s). Now xs E C so f (xs) E B and thus f (s) E i + B, contradicting the fact that s E D. We observe that, for any minimal left ideal L of PN, we have Z + L c L. This follows from the fact that any x E L has a left identity e E L (by Theorem 1.64). So, for every n E 7G, n + x = (n + e) + x E 6N + L C L. Thus, for any minimal left ideal L of N*, f 1 [L] is a left ideal of S*. We can choose a sequence (Ln)n° 1 of disjoint minimal left ideals of N* (by Theo
rem 6.9). For each n E N, we can choose an idempotent en E f 1[Ln] n K(fT) (by Corollary 2.6), because .f1[Ln] n f3T is a left ideal of P T. (We know that f1 [Ln]nfT 0because f[fT] = nw.) We may suppose that the set(f(en) : n E N) is discrete, because this could be ensured by replacing (en)'1 by a subsequence. Let E = (en : n E N). We claim that f is onetoone on ci E. To see this, let p and q be distinct members of c2 E and suppose that f (p) = f (y)_Pick disjoint A E p and B E q. Then 7(p) E T (A n cP E] = f [cf(A n E)] = ct f [A n E] and f (q) E cP f[B n E] so by Theorem 3.40 we may assume without loss of generality that for some em E A, f (em) E cZ f [B n E], contradicting the fact that {f(e) : n E N} is discrete. Let p be any point of accumulation of E. To show that Sp n S* S* = 0, suppose on the contrary that ap = xy for some a E S and some x, y E S*.
8.2 Right Cancelable Elements in fiS
169
Let C = {z E cl E : az E (18S)y}. Then by (*) z E C implies that i +.T(Z) = j + f (y) for some i, j E 0, 1). So C is finite, because f is injective on cf E. We can choose a clopen neighborhood U of p such that U fl (C\{p}) = 0. If M = {n E N: en E U and en # p}, then p E c2{en : n E M}. There is at most one element b E S for which ap = by (by Lemma 6.28). Let S' = S\{b} if such an element exists and let S' = S otherwise. Now ap E ct{aen : n E M} and xy E ct(S'y). It follows from Theorem 3.40 that
aen E (PS)y for some n E M, or az = cy for some z E ce{en : n E M} and some c E S. The first possibility is ruled out since if n E M, then en 0 C. So we assume the second. Now the equation az = cy implies that z E C. Since Z E ce{en : n E M} c U, one has z = p. However, the equation ap = cy is ruled out by our choice of S'.
Corollary 8.23. Let S be a discrete countably infinite cancellative semigroup and let T be an infinite subsemigroup of S. There are elements p in E(K(,6T)) which are not in S*S*.
Proof. Pick a E S. If p E S* S*, then ap E Sp fl S*S* by Theorem 4.31. Corollary 8.24. If S is a discrete infinite right cancellative and weakly left cancellative semigroup, then the set of idempotents in S* is not closed. Proof. By Corollary 8.23, the set of idempotents in flN is not closed and therefore the set of idempotents in H is not closed. Now S* is a compact semigroup (by Theorem 4.31) and therefore contains an idempotent. It follows from Theorem 6.32 that S* contains a copy of H.
Corollary 8.25. Let S be a discrete countably infinite cancellative semigroup. Then K(,6 S) is not closed in f6 S.
Proof. This follows from Corollary 8.23 (with S = T) and the observation that K ($ S) c S* S*, because S*S* is an ideal of 8S (by Theorem 4.36).
Corollary 8.26. Let S be a discrete countably infinite cancellative semigroup and let T be an infinite subsemigroup of S. There are right cancelable elements of CBS in
E(K(fiT)). Proof. This is immediate from Theorem 8.22, Corollary 8.2, and condition (9) of Theorem 8.11.
Exercise 8.2.1. Prove Theorem 8.8 by modifying the proof of Theorem 8.7.
Exercise 8.2.2. Let S be any discrete semigroup and let x, p E fS. Show that: (i) If p is right cancelable, xp E K(fS) if and only if x E K(14S). (ii) If p is left cancelable, px E K (AS) if and only if x E K (,e S). (Recall that every element in K($S) belongs to a group contained in K(48S) by Theorem 1.64).
170
8 Cancellation
Exercise 8.2.3. Prove that the topological center of N* is empty. (Hint. Suppose that x is in the topological center of N*. Show that x is in the center of ON by choosing
an element y E N* which is right cancelable in ON. For every z E ON we have
x+z+y=). (lim(n+y)) lim(x+n+y)andz+x+y= lim(n nsz nz nz+x + y), where n denotes a positive integer.)
8.3 Right Cancellation in $N and flZ In the case of the semigroups (N, +) and (Z, +), there are additional characterizations of right cancelability. Theorem 8.27. Let P E ON. Then p is right cancelable in ON if and only if there is an increasing sequence (xn)n' 1 in N such that for every k E N, {xn : xn + k < Xn+I } E P.
Proof. Necessity. Choose by Theorem 8.19 some B E p such that k + B 0 p for each k E N and let (xn 1 enumerate B in increasing order. Then given k E N, B\ uk_1(i + B) E p and B\ LJki=1 (i + B) C {xn : xn + k < xn+1 }. Sufficiency. Let B = {xn : n E N}. Then given k E N,
(k + B) fl {xn : xn + k < xn+l } = 0
sok+B0p. The proof of the following theorem is nearly identical, and we leave it as an exercise.
Theorem 8.28. Let P E ON. Then p is right cancelable in OZ if and only if there is an increasing sequence (xn) n° I in N such that for every k E N, {xn : xn_ 1 + k < xn < xn+l  k} E p. Proof. This is Exercise 8.3.1
In the next example, we show that the condition of Theorem 8.28 is really stronger than the condition of Theorem 8.27.
Example 8.29. There is an element p E ON which is right cancelable in ON but not in OZ.
Proof. We choose an idempotent q E fN for which FS((3t)°Ok) E q for each k E N. (This is possible by Lemma 5.11.) We put p = q + q and shall show that p is right cancelable in ON but p is not right cancelable in OZ. (We recall that q E fi7G is the ultrafilter generated by (A : A E q)). By Exercise 4.3.5, N* is a left ideal of OZ so p E N*.
8.3 Right Cancellation in ON and OZ
171
It is easy to check that q E E (#Z) and thus q + p = p so that p E Z* + p and hence by Theorem 8.18, p is not right cancelable in f Z. Suppose that p is not right cancelable in ON. Then p = z + p for some z E N*, by
Theorem 8.18. That is q + q = z + q + q. Let
A={ EfEF3`E(EH3`+k: kEN, F,HEJf(N), minF>maxH+1,andk<3m'"HI} and let
min F>maxH+1}. Using Theorem 4.15, one easily shows that A E z + q + q and B E q + q. Now, elements of B are easy to recognize by their ternary representations. That
is, if F, H E Y f (N), j = min F > max H + 1, i = min H, and max F = f, then EIEF 3`  EIEH 3` = El _, a,3`, where ai = ai_1 = 2, at E {1, 2} for all t E {i, i + 1, ... , j  1}, and at E {0, 11 if t > j. That is each such an element has a highest order 2 and a lowest order 2, which is its lowest nonzero ternary digit, and no 0's between these positions. No element of A can fit that description, so A fl B = 0, a contradiction. Corollary 8.26 yields some surprising information about H. We know from Example 2.16 that in a compact right topological semigroup S, ci K(S) need not be a left ideal. On the other hand, if S is a discrete semigroup, we know from Theorems 2.15 and 2.17 that ci K(flS) is always a two sided ideal of flS.
Theorem 8.30. The closure of K (H) is not a left ideal of H. Proof. By Corollary 8.26 (with T = N and S = 7G), there is an element x E E(K($N)) which is right cancelable in OZ. By Theorem 8.27, there is an increasing sequence I in N such that for each k E N, {b : ba + k} E X.
Now K(IHI) = H fl K(,M) by Theorem 1.65. So, by Lemma 6.8, E(K(IHI)) _ E(K(,BN)) and thus x E ci K(H). We shall show that there is some w E H such that w + x 0 ci K (H).
Recall that for a E N, supp(a) denotes the binary support of a. For a, b E N, write a << b if and only if max supp(a) < min supp(b). For a E N, let f (a) _ max supp(a). Let
X = (a + b,, : a E N, ba+l > b + 2f (a)+I, and a < < b,,). We claim that f N + x C X for which it suffices to show that N + x C 7. So let a E N. b,, + 2f (a)+') E x and, since x E H, {w E N : a < < w} E X. Since
Then {ba :
: b,,+i > b + 2f (a)+I } fl {w E N : a < < w} C a + X, a + x E X as required. Now each y E X has a unique representation in the form a + b with + 2f W+' so we may define g : X N by g(a + ba) = a and extend g arbitrarily
{b,,
b
to N. Let B = {bn : n E N}. Then given any y E ON, one has
g(y + x) = g(ylim xlim(a + b)) = ylim xlim g(a + b) = ylima = y. aEN
bEB
aEN
bEB
aN
8 Cancellation
172
We now show that
(*) if z E H n K(,BN) n X, then g(Z) E K(fiN).
Toseethis,letz E IIIInK(,BN)nXandpick an idempotente E K(fN)suchthatz = e+z.
Now if u, a E N and u << a, then f (u + a) = f (a) so if bn+1 > b + 2f (a)+1 and a<< bn,then u+a+bn E X anda+bn E X sog(u+a+bn) = u+a = u+g(a+bn). We now show that g(z) = e + g(z) so that g(z) E K(ON) as required. Suppose instead that g(z) 36 e + g(z) and pick disjoint open neighborhoods U of W(z) and V of
e + g(z). Pick an open neighborhood R of z such that g[R] C U. Then e + z E R so pick C1 E e such that C1 + z S R and pick C2 E e such that C2 + g(z) c V. Pick
U E CI n C2 and let m = f (u) + 1. Pick Al E z such that u + Al c R and pick a neighborhood W of g(z) such that u + W c V. Pick A2 E z such that g[A2] c W. Then Al n A2 n N2m n X E z so picky E Al n A2 n N2' n X. Since y E X, pick a and n such that y = a + b, bn+1 > b, + 2f(a)+', and a < < bn. Since a << bn, m < min supp(y) = min supp(a + bn) = min supp(a) so u << a. Now y E A'1 so u + a + bn E R so g(u + a + bn) E U. Also, y E A2 so g(a + bn) E W SO u + g(a + bn) E V. This is a contradiction since g(a + bn) = u + g(a + bn) so (*) is established.
Now pick any w E 11II\ ci K(ON) (such as w E {2" : n E N}*). We claim that w + x 0 ci K (]Hl) so suppose instead that w + x E ci K (H) = ci (J II n K (flN)). Then since ON + x c X we have w + x E ci(H n K(fN)) n X. Since X is clopen, ci(IIII n K(f N)) n X = ci(H n K(ON) n X). Thus
w = g(w + x) E g[ci(H n K(pN) n X)] C ci K(flN), by (*). This contradiction completes the proof.
Corollary 8.31. ci K (IHI) Z H n ci K (ON). Proof. We know that 1111 n ci K(fN) is an ideal of IIII by Theorems 2.15 and 2.17.
We do not know whether it is possible for the sum of two elements in f N\(K(flN)) to be in K (,BIN; nor do we know the corresponding possibility for the closure of K (ON). However, something can be said in answer to these questions if one of the elements is right cancelable. The answer for the case of K (ON) was in fact given in Exercise 8.2.2.
Theorem 8.32. Letp be a right cancelable element in N\K (N). For everyq EON, q + p E K (fN) if and only if q E K (fN). Proof. It follows from Theorem 2.15 that q + p E K(fN) if q E K(flN). So we shall suppose that q + p E K(,BN). By Theorem 8.27, there is a set P E p which can be arranged as an increasing sequence (xn )' with the property that, for every k E N, {Xn : xn + k < xn+1 } E P. It is convenient to suppose that x1 = 1. Since p 0 K(,BN), we may suppose that 1
PnK(flN)=0.
8.3 Right Cancellation in fiN and OZ
173
For each a E N, let Ca = {xn : 2a + xn < xn+1 } and let B = U0EN(a + Ca)._ By Theorem 4.15, B E q + p. We can define functions f, g : N + w by stating that
f (r) = x, if and only if xn < r < xn+l and g(r) = r  f (r). We observe that, if b E B is expressed as b = a + x, where a E N and 2a + xn < xn+1, then f (b) = xn and g(b) = a. This implies that
g(q + p) = qlim p lim g(a +xn) = qlima = q. aEN
x,EC.
aEN
We shall show that, for any z E B fl K(fN), W(Z) E K(fN). We first show that g(z) E N*. To see this, suppose that g(z) = m E N. For every b E B fl g1[{m}], we have b = g(b) + L (b) = m + f (b). Since B fl g1 [{m}] E z, it follows that z = m+.7(z). So m + f (z) E K(f N) and hence T W = m + (m + .7(Z)) E K(0N). Now T (Z) E P, and so this contradicts our assumption that K(AN) n P = 0.
Since Z E K(ON), there is an idempotent e E K(ON) for which z = e + z. We claim that for each a E N, Da = {b E N : g(a + b) = a + g(b)} E Z. To see this, let a E N. Since g(z) E N* and B E z one has (b E B : g(b) > a} E Z. We claim that {b E B : g(b) > a} c Da. To this end, let b E B with g(b) > a and let f (b) = xn. Then b = g(b) + xn and since b E B, 2g(b) +xn < xn+l and thus xn < a + b < xn+i
sog(a +b) =a+b xn =a +g(b). Thus W(z) = S(e+z) elim z lim g (a + b) aEN
bEDa
elim z lim (a + g (b)) aEN
bED,
elim z lim (Aa o g)(b) aEN
bEDa
e lim(a + g(z)) aEM
e + g(z). This shows that g(z) E K(ON), as claimed.
Since q + p E K(fN) fl B, it follows that
q = S(q +P) E 8[K(fN) fl B] c cPg[K(ON) n B] C K(fN). Exercise 8.3.1. Prove Theorem 8.28. Exercise 8.3.2. Show that there is an element of ON which is right cancelable, but not left cancelable, in PM. (Hint: Show that the element p produced in Example 8.29 is not left cancelable.) Exercise 8.3.3. Show that every left cancelable element of ON is also left cancelable in $Z. (Hint: Use the fact that there are elements of ON that are right cancelable in $Z.)
8 Cancellation
174
5.4 i4e[L 'Cancelable 'Jullements in PS Left cancelable elements in P S are harder to characterize than right cancelable elements, because the maps Ap are discontinuous and are therefore harder to work with than the maps pp. However, there are many semigroups S for which 6S does contain a rich set of left cancelable elements. These include the important case in which S = N. We shall derive several results for countable semigroups which can be algebraically
embedded in the topological center of a compact cancellative right topological semigroup, in particular for those which can be algebraically embedded in compact topological groups. We recall that this is a property of all discrete countable semigroups which are commutative and cancellative, and of many noncommutative semigroups as well, including the free group on any countable set of generators. (See the proofs of Corollaries 7.30 and 7.31.) Lemma 8.33. Suppose that S is a discrete semigroup, that C is a compact cancellative
right topological semigroup and that there is an injective homomorphism h : S A(C). If D is a countable subset of S for which h[D] is discrete, then every element in ceps D is left cancelable in 8S.
Proof. We note that our assumptions imply that S is cancellative. We also note that C is a homomorphism (by Corollary 4.22). h : PS Suppose that p E cP'§s D and that px = By, where x and y are distinct elements of
$S. Then h(p)h(x) = h(p)h(y) so h(x) = h(y). Now px E Dx and py E Dy. By Theorem 3.40, we may suppose without loss of generality that ax = qy for some a E D and some q E D. Now q a, by Lemma 8. 1, and so D\{a} E q. Now
h(q) E h[D\{a}] = cf(h[D\{a}]) = cf(h[D]\{h(a)}) and consequently h(a)h(x) = h(q)h(y) E ct(h[D]\{h(a)})h(y) and so by right cancellation h(a) E h[D]\{h(a)} contradicting our assumption that h[D] is discrete.
Theorem 8.34. Let S be a countable discrete semigroup which can be algebraically embedded in the topological center of a compact cancellative right topological semigroup. Then the set of cancelable elements of PS contains a dense open subset of S*.
Proof. Suppose that C is a compact cancellative right topological semigroup and that h : S  * A (C) is an injective homomorphism. Every infinite subset T of S contains an infinite subset D for which h[D] is discrete. By Lemma 8.33, every element of ct s D is left cancelable in ,S. So the set of left cancelable elements of 8S contains a dense open subset of S*. The same statement holds for the set of right cancelable elements of ,B S, by Theorem 8.10.
Theorem 8.35. Suppose that S is a countable discrete semigroup, that C is a compact cancellative right topological semigroup and that there is an injective homomorphism
8.4 Left Cancelable Elements in $S
175
h : S ). A(C). Let D be a countable subset of S with the property that h is constant on D fl S*. Then every element of D is cancelable in $S. proof. We note that h : 14S > C is a homomorphism (by Corollary 4.22) and that S is cancellative. Every element of D is cancelable in 6S, by Lemma 8.1. So it is sufficient to prove that every element of D n S* is cancelable in BS.
Suppose that p E D fl S* and that px = py, where x and y are distinct elements of fS. This implies that h(x) = h(y). There is at most one elements E D for which h(s) = h(p). Let B denote D with this element deleted, if it exists. Now px E Bx and py E By. By Theorem 3.40, we may suppose that ax = qy for some a E B and some q t B. Now a # q, by Lemma 8.1. Furthermore, h(ax) = h(qy) and so h(a)h(x) = h(9)h(y) so by right cancellation h(a) = h(q). This implies that q gE D and hence that h(q) = h(p), contradicting our assumption that h(a) 0 h(p). So p is left cancelable in ,BS.
We now show that, if a E S, P E D fl S*, and Z E $S\{a}, then ap 0 zp. It will then follow from Theorem 8.11 that p is also right cancelable in $S. By Corollary 8.2,
ap # zp for z E S\{a). Suppose then that ap = zp for some z E S*. Since ap E aD and zp E (S\{a}) p, it follows from Theorem 3.40 that ac = rp for some c E D and some r E S\{a} or aq = by for some q E D fl S* and some b E S\{a}. But the first possibility cannot occur because S* is a left ideal of $S by Corollary 4.33 so the second possibility must
hold. Then h(q) = h(p). So aq = by implies that h(a) = h(b) and thus that a = b, a contradiction.
Given n E N, let h : Z * Z, denote the canonical projection so that hn : $Z + 7Gn denotes its continuous extension.
Corollary 8.36. Suppose that p E_ Z* and that there is a set A E p and an infinite subset M of N such that hn (x) = hn (p) for every x E A fl Z* and every n E M. Then p is cancelable in ($Z, +). Proof. Let C denote the compact topological group X nEMZn. We can define an
embedding h : Z + C by stating that nn(h(r)) = hn(r) for every n E M. It then follows from Theorem 8.35 that p is cancelable in $Z.
Recall that a point x of a topological space X is a Ppoint if and only if whenever (U,,)', is a sequence of neighborhoods of x one has that f n_f Un is a neighborhood of X.
Corollary 8.37. Let p be a Ppoint of 7L*. Then there is a set A E p such that every
q E A fl r is cancelable in $Z. Proof. For each n E N let Bn = {m r= N : hn(m) = hn(p)}. Then each Bn E p so nn_i Bn is a neighborhood of p so pick A E p such that A fl Z* c no,, i Bn. Then Corollary 8.36 applies to A and any q E A fl V.
8 Cancellation
176
Corollary 8.38. Let (xn)n° 1 be an infinite sequence in N with the property that xn+l is a multiple of xn for every n E N. Then every ultrafilter in N* fl {x : n E N) is cancelable in (f7G, +). Proof. This is Exercise 8.4.1.
By Ramsey's Theorem (Theorem 5.6), every infinite sequence in N contains an infinite subsequence which satisfies the hypothesis of Corollary 8.38 or the hypothesis of the following theorem. So these two results taken together provide a rich set of left cancelable elements of flN. Theorem 8.39. Let (x,1)n_l be an infinite sequence in N with the property that, for every m 0 n in N, x" is not a multiple of xm. Then everyaltrafrlter in (xn : n E N) is left cancelable in ($N, +). Proof. Let P = {xn : n E N), let p E P, and suppose that u and v are distinct elements
offNforwhich p+u = p+v. We now observe that we may suppose that u, v E fl,, nN. To see this, note that there is a cancelable element r of fiN for which u + r E nnEN nN by Exercise 8.4.3. The equation p + u = p + v implies that hn (u) = hn (v) for every n E N, and hence we also have v + r E WHEN nN. Since u + r 96 v + r, we can replace u and v by u + r and v + r respectively. We define functions f : N k N and g : N + co as follows: If some xk divides n, then f (n) = xm where m is the first index such that xm divides n. If no xk divides n,
then f (n) = 1. Let g(n) = n  f (n). If s E nm=i xmN, then f (xn + s) = xn and g(xn + s) = s. For each n E N, let B n = nm=l x, N. Then each Bn E U SO
g(p + u) = p Jim u lim g(xn + s) = p lim u lim s = U. X,EP
sEB,,
SEB
Similarly, g(p + v) = v. This contradicts our assumption that p + u = p + v. Example 8.40. There is an element of ON which is left cancelable, but not right cancelable, in fN.
As in Example 8.29, we choose q to be an idempotent in fiN for which FS((3")n_k) E q for every k E N. We choose y E N* with (3" : n E N) E y,
Proof.
and we put x = q + q + y. We shall show that x is left cancelable in fN. Suppose that u and v are distinct elements of fiN for which x + u = x + v. We may suppose that u, v E nfEN nN. To see this, we observe that by Exercise 8.4.3, there is a cancelable element r of fiN such that u + r E WHEN nN. Since r is cancelable,
u + r # v + r. Since x + u = x + v implies that hn(u)hn (v) for every n E N, we also have v + r E n"ll nN. So we can replace u and u by u + r and v + r respectively.
8.4 Left Cancelable Elements in SS
177
Pick U and V disjoint members of u and v respectively. Let
A = {n+3eE,EH 3`+EIEF31.£EN, n EN3e+' flU
H,FE,lf(N), f>maxH,andminH>maxF} and
E N3t+1 n v, E B = {n + 3e 3` + H, F E J'f(N), f > max H, and min H > max F}. Using Theorem 4.15, one easily shows that A E q + q + y + u = x + u and B E q + q + y + v = x + v. Now consider the ternary expansion of any numbers of the form n + 3e  EIEH Y+
E,EF 31 where e E N, n E N3e+1, H, F E D f(N), t > max H, and min H = j > max F. Pick a number k E N such that n < Then s = Eke at 3' where 3k+1.
a,E{0,1}iftE{0,1,...,j1},aj=2,a,E{1,2}iftE{j+l,j+2,...,f1}, at = 0, and at E {0, 1 , 2} if t E it, e + 1, ... , k}. That is, knowing that s can be written in this form, one can read f just by looking at the ternary expansion, because t is the position of the lowest order 0 occurring above the lowest order 2. Since a is determined from the form of the ternary expansion, so is n. That is, n = Eke+1 a,3'. Since u fl v = 0, one concludes that A fl B = 0, a contradiction. So x is left cancelable in fiN. It is not right cancelable, because q + x = x.
Theorem 8.41. Let S be a discrete countable cancellative semigroup and let p be a Ppoint in S*. Then p is cancelable in #S. Proof. Suppose that px = py, where x and y are distinct elements of CBS.
Let D = (a E S : ax E Sy). If a E D, there is a unique element b E S for which ax = by (by Corollary 8.2). Thus we can define a function f : D > S by stating that ax = f (a) y for every a E D. We observe that f has no fixed points, by Lemma 8.1. We can extend f to a function g : S + S which also has no fixed points. By Lemma 3.33, there is a set A E p such that A fl g[A] = 0. Similarly, if E = {b E S : by E Sx}, we can choose a function h : S + S such that by = h(b)x for every b E E and B fl h[B] = 0 for some B E P. We note that the equation sx = py has at most one solution with s E S, and that the same is true of the equation px = sy (by Corollary 8.2). Hence we can choose
P E p such that P C A fl B and, for every a E P, ax # py and px # ay. For each a E P, let Ca = {u E S* : ax = uy or ux = ay}. Since each Ca is closed (Ca = py'[{ax}] U pX'[{ay}]), p 0 Ca, and p is a Ppoint of S*, it follows that P 0 UaEP Ca. So we can choose Q E p with Q C P, such that Q fl Ca = 0 for every a E P. Now px E Qx and py E Qy. It follows without loss of generality from Theorem 3.40, that ax = uy for some a E Q and some u E Q. This equation implies that u lE S*, since otherwise we should have u E Q fl Ca. So U E S and u = g(a), and hence u E A fl g[A]. This contradicts the fact that A fl g[A] = 0. We have thus shown that p is left cancelable in 8S. By Corollary 8.14, p is also right cancelable in $S. 
178
8 Cancellation
The reader may have noticed that all the criteria for left cancelability give u. ;is section, apart from that of being a Ppoint in an arbitrary countable cancellative semi. group, describe a clopen subset of S* all of whose elements are left cancelable. Thus these criteria cannot be used to find left cancelable elements in K($S). There are several questions about left cancelable elements of ON which remain open, although the corresponding questions for right cancelable elements are easily answered. Question 8.42. Is every element in N*\(N* + N*) left cancelable? Question 8.43. Are weak P points in N* left cancelable?
Question 8.44. Are there left cancelable elements in K(PS)? Exercise 8.4.1. Prove Corollary 8.38. Exercise 8.4.2. Let H° denote the interior of H in N*. Prove that H' is dense in H and that every element of ]Hl° is cancelable in OZ.
Exercise 8.4.3. Let p ,E ON. Show that there is a cancelable element r E ON for p) = hn(p + r) = 0 for every n E N. (Hint: For every k E N, which
{m E Z : hk(m) = hk(p)} E p. Thus for each n E N, one can choose m E 0k=1 {m E Z : hk(m) = hk ( p) }. Show that one can assume that this m E ICY and that
hk(mn + p) = 0 for every k E {1, 2, ..., n}. The set A = [Mn : n E N} then satisfies the hypotheses of Corollary 8.36.)
8.5 Compact Semigroups Determined by Right Cancelable Elements in Countable Groups In this section we assume that G is a countably infinite discrete group and we discuss the smallest compact subsemigroup of fiG containing a given right cancelable element of PG. We show that it has a rich structure. Furthermore, all the subsemigroups of PG arising in this way are algebraically and topologically isomorphic to ones which arise
in fN. We shall use some special notation in this section. Throughout this section, we shall suppose that G is a countable group and shall denote its identity by e. We shall also suppose that we have chosen an element p E G* which is right cancelable in fiG. We assume that we have arranged the elements of G as a sequence, (s,,) n_ 1, with e = s1. We shall write a < b if a, b E G and a precedes bin this sequence.
Given F, H _C G, let F'1 H = UIEF t1 H. We can choose P E p with the property that e 0 P and F1 P o p whenever F E J" f(G\{e}) (by Theorem 8.19). If A C G, we use FP(A) to mean the set of finite products of distinct terms of A written in increasing order. That is, if A = {st : t E B}, then FP(A) = FP((St)IEB)We shall assume that P has been arranged as an increasing sequence (b )n° 1. (That Sk, then t < k.) is, if b = st and
8.5 Compact Semigroups
179
Definition 8.45. Let n E N.
(a)Fn ={u1v:U,VEFP({aEG:a
.
T. = {bn1 bn2... bnk : for each i E {2, 3, ... , k}, bn, E Pni_, and bn, E Pn }
(c)Too=II 1T. We shall call a product of the form bn, bn2... bnk as given in the definition of T a Pproduct. Notice that P C T. (If i = 0, we define the empty product bn, bn2... bn, to be e.)
Definition 8.47. Let S be a discrete semigroup and let q E OS.
Cq = n{D C :,OS : D is a compact subsemigroup of PS and q E D}. Notice that Cq is the smallest compact subsemigroup of PS which contains q. We may denote this by Cq($S) in a context in which more than one semigroup S is being considered.
Lemma 8.48. Suppose that k, I E N, that
bn,2 ... b,nk and bn, bn2 ... bn, are P products and that abn,, bn,2 ... bn,k = bn, bn2 ... bn, for some a E G satisfying a < bn,,
anda1
if j = k  I > 0. The first possibility is what we wish to prove, and so it will be sufficient to rule out the second. This is done by noting that, if j > 1, since e, the equation ab,n, b,n2 ... b,j = e implies that bn,j E Fmj_,, which is a contradiction. If j = 1, it implies that a1 = b,n,, contradicting our assumption that a1 < bn,,. Lemma 8.49. The expression for an element of T as a Pproduct is unique.
Proof. We apply Lemma 8.48 with a = e. We observe that we cannot express e as a Pproduct bn, bn2 ... b, with i > 0. To see this, note that otherwise, if i = 1, we
8 Cancellation
180
should have bn, = e, contradicting our assumption that e f P. if i > i, we should have bn E Fn;_1, contradicting our definition of a Pproduct. Thus, if bn,, bn2 ... bn,k = bn, bn2 ... bn, , where these are Pproducts, Lemma 8.48 p implies that k =1 and that mi = n; f o r every i E { 1 , 2, ... , k}. Lemma 8.49 justifies the following definition.
Definition 8.50. Define i/r : T + N by stating that * (x) = k if x = bn, bn2 ... bnk, where this is a Pproduct. As usual i : T :+ ON is the continuous extension of 1/r.
Theorem 8.51. Let G be a countably infinite discrete group and let p E G* be right cancelable in ,B G. Then T,,. is a compact subsemigroup of G* which contains Cp. Furthermore, 1/r is a homomorphism on T,,. satisfying 1fr(p) = 1 and 1/t[Cp] = ON.
Proof. Let X E Tm and express x as a Pproduct: x = bmk. If n > Mk and if y E T,,, then xy E T. and 1/r(xy) = *(x) + 1Jr(y). It follows from Theorems 4.20 and 4.21 that T,, is a subsemigroup of G* and that 1%r is a homomorphism on Tom.
Foreveryn E NandeveryA E p,wecanchooseb E P,,nA. SoAfTnn,/<1[{1}] 54 0. It follows that
o o nAEP n,EN cl(A n Tn n; 1[{l}]) C {p} n T. n;G1[{1}].
This shows that p E T... and hence that Cp c T. It also shows that 1'(p) = I and hence that *r[Cp] = ON, because 1r[Cp] is a compact subsemigroup of ON which contains 1. Theorem 8.51 allows us to see that Cp has a rich algebraic structure.
Corollary 8.52. Let G be a countably infinite discrete group and let p E G* be right cancelable in /3G. The semigroup Cp has 2` minimal left ideals and 2` minimal right ideals. Each of these contains 2` idempotents. Proof. We apply Theorem 6.44. The result then follows because, if L is a left ideal of ON, yr1 [L] n Cp is a left ideal of Cp; and the corresponding remark holds for right ideals. We recall that every left ideal of Cp contains a minimal left ideal and every right ideal of Cp contains a minimal right ideal, and the intersection of any left ideal and any right ideal contains an idempotent (by Corollary 2.6 and Theorem 2.7).
We remind the reader that, if e and f are idempotents in a semigroup, we write
f <eifef = fe= f.
Lemma 8.53. Let el and e2 be idempotents in ON with e2 < el, and let fi be an idempotent in Cp for which 1/r (A) = e i. There is an idempotent f2 in Cp for which
f2 < ft and 1G(f2) = e2
8.5 Compact Semigroups
181
proof. First notice that l' '[(e2)]nCpf1 00, because (t 1[{e2}]nCp)fi is contained in this set. So *' [{e2}] nCp f t is a compact semigroup and thus contains an idempotent
f (by Theorem 2.5). We put f2 = fl f. It is easy to check that f2 is an idempotent satisfying *(f2) = e2 and f2 < fl. Corollary 8.54. Let G be a countably infinite discrete group and let p E G* be right cancelable in 18G.
The semigroup Cp contains infinite decreasing chains of idempotents. proof. There is a decreasing sequence of idempotents (e n ),, t in f N by Theorem 6.12. By Lemma 8.53 we can inductively choose a decreasing sequence (fn),°r° 1 of idempotents in Cp for which i%i (fn) = e for every n E N.
Corollary 8.55. Let G be a countably infinite discrete group and let p E G* be right cancelable in 6G. Every maximal group in the smallest ideal of Cp contains a copy of the free group on 2` generators. Proof. Let q be a minimal idempotent in Cp. Then 4i (q) is a minimal idempotent in fl N by Lemma 8.53 and so (*(q))fN(*(q)) contains a group F which is a copy of the free group on 2` generators (by Corollary 7.36). For each generator x in F, we can choose
y E gCpq such that /ii(y) = x. These elements generate a group in gCpq which can be mapped homomorphically onto F and which is therefore free. Comment 8.56. If we do not specify that p is right cancelable, the preceding results may no longer hold. For example, if p is an idempotent, Cp is just a singleton. Less trivially, if p = 1 + e, where e is a minimal idempotent in $Z, Cp is contained in the minimal left ideal f76 + e of fZ and contains no chains of idempotents of length greater than 1.
Theorem 8.57. Let G be a countably infinite discrete group and let p E G* be right cancelable in PG. The semigroup Cp does not meet K(0G).
Proof. We can choose x ET n G* with x # p. We can then choose Q c P such that 1 enumerate P' in increasing order. Q E x and Q p. We put P' = P\Q. Let Then define F,,, P,,, T', T,,, and T, in terms of P' analogously to Definitions 8.45 and 8.46. Notice that for each n, F,, C F,,, and hence P;, C P,,, T,' c T,,, and T, C T. We claim that T, n K(, 6G) = 0. Suppose instead that we have some y E T, n
K(fG). Then by Theorem 1.67, y E (fG)xy. So pick some u E fiG such that y = uxy. For each a E G, the set Xa = {b,,
b,, E Q, b,, > a, and bn > a1 } E X.
For each bn E Xa, we have Tn E Y. So {abn v : a E G, bn E Xa, and V E Tn} E uxy, by Theorem 4.15. Since T' E y, there must be elements a E G, bn E Xa, and V E Tn for which abn v E T'. We note that b,, v is a Pproduct and so, by Lemma 8.48, it follows that bn E P. This contradicts our assumption that bn E Q and establishes that
T nK(fG)=0.
182
8 Cancellation
No v., CP r To'o (b y Theorem 8.51 with P'in place of P) and hence Cp n K K (,6G) =.O,
Corollary 8.58. If G is a countably infinite discrete group, there is a right cancelable element p in f G for which Cp C ct (K($G))\K(f G). Proof. We can choose a right cancelable element p of$ Gin K (fl G) (by Corollary 8.26). Since K(f G) is a compact subsemigroup of 0G (by Theorem 4.44), Cp C K(f G). By Theorem 8.57, Cp fl K(,BG) = 0. 13
Lemma 8.59. Let G be a countably infinite discrete group, let p E G* be right cancelable in PG, and let q bean idempotent in Cp. Then there is some idempotent r E (fG) p which is 
every a E G, we have E 0 ap and hence Ba = aI (G\E) E p. We consider the set A of all Pproducts bn, bn2 ... bnk with the property that, for each i E { 1 , 2, ... , k}, bn, 0 E and, if i > 1, bn; E B, for every u E F, We observe that, for a Pproduct of this form, we have fly=I bnk E A for every i E {2, 3, ... , k}.
We also note that AflE=0. For each n E N, we define An to be the set of all Pproducts bn, bn2 ... bnk in A fl Tn for which bn, E Bn whenever u E F. We shall show that WHEN An is a compact semigroup which contains p. That it is a compact semigroup follows from Theorem 4.20
and the observation that, if u E An, is expressed as a Pproduct bn1 bn2 ... bnk, then
uAn C A,n whenever n > nk. Now, for each n E N, we have An E p, because {br : br E Pn and br E Ba} E P. SO p E nfEN A. Since nnEN An is a compact semigroup which contains p, Cp g WHEN An and so q E WHEN K. For each a E G, let Qa denote the set of Pproducts bn, bn2 ... bnk with
bn, > a and bn, > a'. Then Qa E q and so {au : a E E and U E Qa} E rq (by Theorem 4.15). Since A E q = rq, there exists a E E and U E Qa such that au E A. By Lemma 8.48, it follows that a E A. This contradicts the assumption that a E E. The following theorem may be surprising because all idempotents in K($N) are
Proof. By Theorem 6.56 we can choose an element x E G* for which xq is right cancelable in PG. By Corollary 8.26, there is an element p E K(fG) which is right cancelable in PG. Since pxq is also right cancelable, it follows from Lemma 8.59 that there is a
8.5 Compact Semigroups
183
Theorem 8.61. Let G be a countably infinite discrete group and let p E G* be right cancelable in 14 G. There is an injective mapping 0 : T + N with the following
properties: (1) 4)[T] C lEi[. defines an isomorphism from T,. onto 4)[T,,.]. (2) (3) gy(p) E (2" : n E N). Proof. We define 0 by stating that 0 (bn 1 bn2 ... bn,) = E; `=I 2"t whenever b" 1 b"2 ... b"k
is a Pproduct. Then 0 is well defined by Lemma 8.49, and is trivially injective. We note that, if bn, bn2 ... bnk E T, then bn, E Pn and so bn, 0 {a E G : a < bn } and hence n I> n. It follows that ¢[Tn ] C 21N and that 4)[T.] C H. Suppose that x E T is expressed as a Pproduct bn, bn2 . . . bnk . Then, if n > nk and y E Tn, we have 0(xy) = 0(x) + ¢(y). It follows from Theorem 4.21 that 4) is a homomorphism on T. Since q is injective (by Exercise 3.4.1), it follows that 4) defines an isomorphism from T,,,, onto cb[T I.
Now ¢[P] C (2" : n E N). Since P E p, {2" : n E N} E ¢(p). Corollary 8.62. Let G be a countably infinite discrete group and let p E 'G* be right cancelable in PG. There is an element q E N*, for which {2" : n E N) E q and Cp (6G) is algebraically and topologically isomorphic to Cq ($N).
Proof. We put q = ¢(p), where 0 is the mapping defined in Theorem 8.61. Then
O[Cp(f G)] is a compact subsemigroup of flN which contains q, and hence ([Cp(,8G)] 2 Cq(ON). Similarly, 41 [Cq(,BN)] fl T,, is a compact subsemigroup
of PN containing p and so Cp(,9G) C o1[Cq(^]. So [Cp(fG)] = Cq(,6N) and 4) defines an isomorphism from Cp($G) onto Cq ($N).
Theorem 8.63. Let G be a countably infinite discrete group and let p E G* be right cancelable in flG. Then Too is algebraically and topologically isomorphic to H. Proof. By Theorem 7.24 (taking X = G and V (a) = G for all a E G), it is sufficient to show that we can define a left invariant zero dimensional topology on G which has the sets {e} U Tn as a basis of neighborhoods of e. As we observed in the proof of Theorem 8.51, if x E Tmis expressed as a Pproduct b,n, b,n2 ... b,nk, then xTn C Tn, if n > MkLet 2 = { (x) U x Tn : x E G and n E N). We claim that S is a basis for a topology on G such that for each x E G, {{x} U xTn : n E N} is a neighborhood basis at x. To see this, let x, y E G, let n, k E N, and let a E ({x} U xTn) fl ({y} U yTk).
(i) If a = x = y, let r = max{n, k}. (ii) If a = x = ybn b,n, ... b,,,; where b,n, b,,2 ... b,n; is a Pproduct and bn,, E Pk, let r = max{n, m; + 1}. (iii) If a = xbj, b12 ... b4 = y where b1, b12 ... b11 is a Pproducts and b1, E Pn let
r=max{Ij+1,k}.
184
8 Cancellation
(iv) If a = xbt, b12 ... big = ybm, bm2 ... bm; where bl, b12 . . . big and b,,,, b,,,, ... b,,,. are Pproducts, bl, E P,,, and bm, E Pk, let r = max{/1 + 1, m, + 1).
Then (a) U aT, c ({x} U xTn) fl ({y} U yTk) as required. The topology generated by 2 is clearly left invariant. Suppose that a E G\Tm. If we choose n such that a < bn and a < b,, ', it follows from Lemma 8.48 that aTn fl Tm = 0. Thus the sets {e} U Tn are clopen in our topology.
We now claim that nnEtN Tn = 0, because a < bn implies that a V T. To see this, suppose that a = bn,bn2 ... b,,, where this is a Pproduct and bn, E P, . We note that bn, > b,,, since otherwise we should have bn, E Fn. So bn, > a and therefore I > 1. We then have bn, E F,,,_,, contradicting our definition of Pproduct. This shows that our topology is Hausdorff, and completes the proof. Lemma 8.64. Let G be a countably infinite discrete group and lei p be a right cance
lable element of G*. Suppose that x E PG and Y E T. Then xy E T implies that
XET. Proof. Let X E x and let Z denote the set of all products of the form abn, bn2 ... bnk, where bn, bn2 . . . bnt is a Pproduct, a E X, a < bn, and a' < bn, . By Theorem
4.15, Z E xy. Hence, if xy E T, T fl Z # 0. It then follows from Lemma 8.48, that T fl X # 0 and hence that T E X. So far, in this section, we have restricted our attention to PG, where G denotes a countably infinite discrete group. However, some of the results obtained have applications to (fN, +) and (,BIN ), the following theorem being an example. Theorem 8.65. Let S be a countably infinite discrete semigroup which can be embedded
in a countable discrete group G. Then K(PS) contains 2` nonminimal idempotents, infinite chains of idempotents, and idempotents which are
Proof. By Corollary 8.26 (with T and S replaced by S and G respectively), there is
an element p E K(fS) which is right cancelable in PG. Taking this p to be the element held fixed at the start of this section, we can choose the set P so that P C S. Consequently T C S so that T _C ,B S.
Now K(flS) is a compact subsemigroup of fS, by Theorem 4.44, and is therefore
a compact subsemigroup of PG. So Cp(fG) c K(fS). By Corollary 8.54, Cp(fG) contains infinite chains of idempotents, and so the same statement is true of K(fS).
Let el be any nonminimal idempotent in fN. Since ei is not minimal, we can choose an idempotent e2 in ,BIY for which e2 < el. By Lemma 8.53, we can then choose
idempotents fi and f2 in Cp(,BG) satisfying f2 < e,, and *(f2) = e2. Since fl, f2 E K (PS), fi is a nonminimal idempotent in K(fl S). There are 2` possible choices of ei (by Corollary 6.33) and therefore 2' possible choices of fl. So K(PS) contains 2` nonminimal idempotents. By Lemma 8.59, if q is an idempotent in Cp (f G), there is an idempotent r E (PG) p which is $Rmaximal in G* and satisfies q
from Lemma 8.64 that r E T C fS.
Notes
185
We know that r = xp for some x E PG. By Lemma 8.64, x E T and so r E (PS) p. Since K(,BS) is an ideal of fl S, it follows that r E K(fS). Most of the results in this section also hold for the semigroups Cp(flN), where p denotes a right cancelable element of N*. The proofs have to be modified, since N is not a group, but they are essentially similar to the proofs given above. We leave the details to the reader in the following exercise.
Exercise 8.5.1. Let p be a right cancelable element in ON. By Theorem 8.27, there
is a set P E p which can be arranged as an increasing sequence (b,,)', with the property that, for each k E N, Pk = {b,, : bn + k < bn+1 } E p. We define T to be the + bnk, where, f o r each i E (2, 3, ... , k), set of all sums of the f o r m bn, + bn2 + bnt+1  bn; > 1 + 2 + 3 + + bn;_, . We shall refer to a sum of this kind as a Psum. We define Tn to be the set of sums of all Psums for which bn, +1 bn, > 1 +2+ +n and we put T,,. = nnEN T. Prove the following statements:
(1) The expression of an integer in T as a Psum is unique. (2) TT is a compact subsemigroup of fiN which contains Cp. (3) There is a homomorphism mapping Cp onto ON. (4) Cp contains 21 minimal left ideals and 2` minimal right ideals, and each of these contains 2` idempotents. (5) Cp contains infinite chains of idempotents. (6) There is an element q E {2n : n E N) for which C. is algebraically and topologically isomorphic to Cp. (7) T, is algebraically and topologically isomorphic to H.
Notes Most of the results of Sections 8.1, 8.2, 8.3, and 8.4 are from [46] (aresult of collaboration
with A. Blass ), [125], [154], and [228]. Theorem 8.22 extends a result of H. Umoh in [235]. Theorem 8.30 is from [129], where it had a much longer proof, which did however provide an explicit description of the elements of cf K (H). Theorem 8.63 is due to I. Protasov. Most of the other theorems in Section 8.5 were proved for $N in [88], a result of collaboration with A. ElMabhouh and J. Pym, [157], and [228], and were proved for countable groups by I. Protasov.
Chapter 9
Idempotents
We remind the reader that, if p and q are idempotents in a semigroup (S, ), we write:
P
if if if
p.q=P; q.p=p;
p.q=q.p=p.
These relations are reflexive and transitive, and the third is antisymmetric as well.
We may write p < q if p < q and p 0 q. We may also write p
and q¢LP,and p
9.1 Right Maximal Idempotents Recall from Theorem 2.12 that right maximal idempotents exist in every compact right topological semigroup. We shall see now that there are 2` right maximal idempotents
9.1 Right Maximal Idempotents
187
in N'. This contrasts with the fact that we do not know of any ZFC proof that there are any left maximal idempotents in ON.
Theorem 9.1. There are 2` right maximal idempotents in N*.
Proof. We remind the reader that, for any n E N, supp(n) E 'f (w) is defined by the equation n = E(ESUpp(n) 2'. We define y : N * w by stating that y(n) = 2'n where m = min supp(n). If k E N and r > min supp(k), then for all n E 2'N one has y(k + n) = y(k). Thus, given p EON and q E IHI,
y(p+q) = plimqlimy(k+n) kEN nEN = P!im Y (k) kEN
= Y(P)We note that y is the identity map on {2n : n E w} and hence that y is the identity map on cesu{2n : n E N}. Let X = cZSn{2n : n E co) fl N*. Then JXi = 2` by Theorem 3.59. For each x E X, let Dx = y1 [{x}] f1 H. Then Dx is a compact subsemigroup of H and the sets Dx are pairwise disjoint. By Theorem 2.12, we can choose an idempotent qx in Dx which is a right maximal idempotent in the semigroup D. We shall show that qx is a right maximal idempotent in N*. To see this, let p be an idempotent in N* for which p + qx = qx . Then j 7(p) = y (p + qx) = y (4x) = x and so p E D. This implies that qx + p = p and thus that qx is a right maximal idempotent in N*
We now show that right maximal idempotents in Z* have remarkable properties.
Lemma 9.2. Let G be a countable discrete group and let q be a right maximal idempotent in G*. Then, if p is any element of fiG which is not right cancelable in fiG, pq = q implies that qp = p.
Proof. By Theorem 8.18, there is an idempotent e E G* for which ep = p. Assume
that pq = q. Then eq = epq = pq = q so, since q is right maximal, qe = e so
qP=qeP=eP=P Lemma 9.3. Let S be a countable right cancellative discrete semigroup. Then every compact right zero subsemigroup of fiS is finite.
Proof. We first note that, if t E S has a left identity s E S, then s has to be a right identity for S. To see this, let u E S. Then ust = ut so by right cancellation us = u. Let D denote the set of right identities of S. If q E D, then pq = p for every p E PS, because pq = p l Em g li D u v= p lim u = p. UES
Suppose that C is an infinite compact right zero subsemigroup of fS. Then q E C implies that D 0 q, because otherwise we should have xq = x for every x E C. This implies that x = q, because xq = q, and hence that ICI = 1.
188
9 Idempotents
Since C is infinite, pick an infinite sequence (pn)'t of distinct elements of C and let p be any accumulation point of the sequence. There is at most one element pn in the sequence for which pn = p, and so we can delete this element (if it exists) and assume
that pn # p for every n E N. Since p E C, p = pp E ctfs ((S\D) p) n ce,ss{ pn : n e N}. It follows from Theorem 3.40, that sp = q for some q E C fN{pn : n E N} and some s e S\D, or else pn = xp for some n E N and some x E S. The first possibility implies that sq = spq = qq = q. It follows from Theorem 3.35
that It E S : St = t} E q and hence that It E S : St = t} # 0 and that s E D, a contradiction.
The second possibility implies that p = p, ,p = xpp = xp = pn, again a contradiction.
Theorem 9.4. Suppose that G is a countable discrete group. Let q be a right maximal idempotent in G* and let C = {p E G* : pq = q}. Then C is a finite right zero subsemigroup of fIG and every member of C is a right maximal idempotent in G*.
Proof C is a compact right topological semigroup and so K(C) 0. We note that q E K(C) because {q} = K(C)q c K(C). Since {q} = Cq, {q} is a minimal left ideal of C. Thus all the minimal left ideals of C are singletons, by Theorem 1.64, and so the elements of K(C) are all idempotent. Now every minimal right ideal of C intersects every minimal left ideal of C. It follows by Theorem 1.61 that C has a unique minimal
right ideal R and therefore that K(C) = R = E(R) (by Theorem 1.64). By Lemma 1.30(b), E(R) is a right zero semigroup and so K (C) is a right zero semigroup.
We now claim that E(C) = K(C). To see this, let p E E(C). Since pq = q and q is right maximal, we have qp = p and sop E K(C). It follows that C contains no <chains of idempotents with more than one element, because all elements of K(C) are <minimal in C. Now any compact subsemigroup of G* which contains a right cancelable element, must contain infinite <chains of idempotents (by Corollary 8.54). So C cannot contain any right cancelable elements of fIG. We shall show that C = K (C). To see this, let p E C. Since p is not right cancelable in f3G and pq = q, it follows from Lemma 9.2, that qp = p and hence that p E K(C). Thus, by Lemma 9.3, C is finite.
Finally, let p E C and assume that r E G* and p
idempotent in G* and let C = {p E fG : pq = q}. Then, for every x, y E fG, the equation xq = yq implies that x E yC or y E xC.
Proof. Suppose that x 0 yC and y 0 xC. Then, for each p E C, x # yp and so there are disjoint clopen subsets Up and Vp of fIG for which x E Up and yp E Vp. If
Xp=GnupandYp=Gnpp'[Vp],then X,, Ex,Yp E yandXpn(Ypp)=0, because X p 9 Up and Yp p c Vp. Similarly, there exist X, E x and YP E y for which
9.1 Right Maximal Idempotents
189
(X'p) fl Yp = 0. Let X= npec(Xp fl Xp) and Y= fpEc(Yp fl Yp). By Lemma 6.28, C = {e} U {p E G* : pq = q), so by Theorem 9.4, C is finite and so X E x and YEy Assume that xq = yq. Since xq E Xq and yq E Yq, it follows from Theorem 3.40 that bq = uq for some b E Y and some u E X, or else vq = aq for some v E Y and some a E X. Assume without loss of generality that bq = uq for some b E Y and
_
_
some u E X. Then b'tuq = q and hence blu E C. If p = b'u, then by = u, contradicting our assumption that b E Yp and u E Y p, while Xp fl (Yp p) = 0.
Theorem 9.6. Suppose that G is a countable discrete group. Let q be a right maximal
idempotent in G*, let C = (p E fG : pq = q), and let n = JCS. Suppose that x is a given element of fG and that Y = {y E fG : yq = xq}. Then Y has either n or n  1 elements.
Proof. Let C' = C\(e), where e denotes the identity of G. We note that by Lemma 6.28, C' = (p E G* : pq = q) so C' is a finite right zero semigroup, by Theorem 9.4, and in particular all elements of C' are idempotents. For every p E C', xp is the unique element in Y fl (fG) p. To see this, we note that xp is in this set because xpq = xq.
On the other hand, if y is in this set, then y = yp = yqp = xqp = xp.
We shall show that the sets (fG)p, where p E C', are pairwise disjoint.
If
p, p' E C', then (fiG)p fl (fG)p' # 0 implies that p E (,6G)p' or p' E (fiG)p by Corollary 6.20. Now p E (fG)p' implies that pp' = p and hence that p = p', because pp' = p'. Thus IY In (,8G)C'I = IC'I = n  1. We claim that there is at most one element of Y in fiG\((f G) C'). Suppose that y and z are distinct elements of Y which are in this set. Since yq = zq, y E zC = zC' U (Z)
or z E yC = yC' U {y} (by Lemma 9.5). However, neither of these possibilities can hold, because y 0 zC', z 0 yC' and y 0 z. Thus I Y I is either n or n  1, as claimed. We now show that right maximal idempotents occur in every neighborhood of every idempotent in S*, if S is a countable cancellative semigroup.
Theorem 9.7. Let S be a countable cancellative semigroup and let U be a G, subset of S*. If U contains an idempotent, then U contains a right maximal idempotent of S*. Proof. We may suppose that S has an identity e, since an identity can be adjoined to any semigroup. We assume that S has been given a sequential ordering (i.e., a well ordering of order type co), with e as the first element. If a, b E S, we shall write a < b if a precedes b in this ordering. For each a E S, we put L(a) = {b E S : b < a}. For each F E J"f (S), II F will denote the product of the elements of F with these arranged
in increasing order. We define 110 to be e. If B c S, FP(B) will, as usual, denote
(II F:FEJ" f(B)}. Let p E U be an idempotent. For each A E p, recall that A* = {s E A : A E sp}. We remind the reader that A* E p and that, for every s E A*, s 1 A* E p (by Lemma 4.14).
9 Idempotents
190
decrcaamg sequence \,\'A,,),'=, We can choose a uwa °°=1 of infinite subsets of S satisfying An E p and An f1 S* c U for every n. We shall inductively define a sequence (an )n° in S. We choose any al E A i . We then suppose that al, a2, ..., ak have been chosen so that the following conditions are satisfied f o r every i, j E { 1, 2, ..., k}: 1
(i) If i < j,then ai
_1 I
I{b1Ai : b E FP((an)n=i)},
with the property that ak < ak+I and b ,A cak+I whenever b, c E FP(L(ak)). The induction hypotheses are then satisfied. The sequence (an )n° having been chosen, we note that, for every m E N, we have 1
FP((an)00°m) C Am.
We now show that if b E S\{e}, F, G E P ({an : n E N}) with min F = at,
FCGandb=fI(G\F).
We first establish that with b, F, and G as above, one has max F = max G. Let ai =
max F and let of = max G. If i < j, then b fI F E FP(L(aj_1)) and fI(G\{af}) E FP(L(aj_1)) (even ifG\{aj} = 0) so fI G E FP(L(aj_t))a1, contradicting hypothesis
(iii). If j < i, then fI G E FP(L(ai1)) and b fI(F\{ai}) E FP(L(a;_1)). (If F = Jai), this is because b < at_I = ai_1.) Thus, if j < i, then b fI F E FP(L(ai_1))ai, again contradicting hypothesis (iii).
Now we establish that F Z G and b = I1(G\F) by induction on IFl. If F = (at),
then bat = fI(G\{a,})a, so, since S is right cancellative, b = fI(G\{a,}). Since
b# e,G\{at}00 soFZG.
Now assume that I F I > 1 and the statement is true for smaller sets. We observe that
IGI > 1. (Indeed,ifG = (af), then wehave bf1(F\{af}) = e,soifai = max(F\{aj}), we have b 1I(F\{aj}) E FP(L(ai_I))ai f1FP(L(ai_1)), contradicting hypothesis (iii).)
Let F' = F\(af } and G' = G\{aj} where of = max F = max G. Then by right cancellation, b n F' = fI G' so F' Z G' and b = I1(G'\F') = 11(G\F). Now there is an idempotent q E nn° 1 FP((an)n_,,,) by Lemma 5.11. By Theorem 2.12, there is a right maximal idempotent r E S* for which rq = q. We shall show that
rEU.
LetR E randletm E N. Weshowthat RflFP((an)°_m) ,f 0. Foreachb E R\{e}, pick k(b) E N such that b < ak(b)_I. Then by Theorem 4.15, UbER\{e{ b FP((an)nn__k(b)) E rq = q.
Since also FP((an)n°_m) E q, pick b E R\{e), F E Pf ({an : n > k(b)}), and G E ,% f ({an : n > m}) such that b fI F = FIG. Then F Z G and b = 11(G\F) E FP((an)nm)
9.1 Right Maximal Idempotents
191
Since for each R E r and m E N, R f l FY((an))n_m) # 0, we have that
r E S* fl nm oc) 1 FP((an)n°_m) C S* (1 nm==1 Am C U. We do not know whether the sum of two elements in N* \ K (ON) can be in K (PN),
or whether the sum of two elements in N*\K(fN) can be in K(fN). We do know that, if p is a right cancelable element of ON, then, for any q E fl N, q + p E K(ON) implies that q E K(fN) by Exercise 8.2.2, and that, if p is a right cancelable element of fiN\ K (f N), then q + p E K (f N) implies that q E K (f N) by Theorem 8.32. There are idempotents p in fN which also have this property. As a consequence of the next theorem, one sees that if p is a right maximal idempotent in N* and if q E fiN\ K (f N), then q + p 0 K(fN). The corresponding statement for K(ON) is also true: if p is a
right maximal idempotent infN\K(fN) and if q E fN\K(fN), then q + p 0 K (fN). We shall not prove this here. However, a proof can be found in [159].
Theorem 9.8. Let G be a countably infinite discrete group and let p be a right maximal
idempotent in G*. Then, if q E fiG\K(fG), qp 0 K(fG).
Proof. Suppose that qp E K(fG). Then qp = eqp for some minimal idempotent e E fG by Theorem 1.64. Let C = {x E G* : xp = p}. By Theorem 9.4, C is a finite right zero semigroup. Now q 0 (,8G)C. To see this we observe that, if q = ur for some u E PG and some
r E C, then q = qr = q(pr) = (qp)r E K(fG), contradicting our assumption that q 0 K (PG). Since (fG)C is compact, wecanchoose Q E q satisfying Qfl (fG)C = 0. Since q 0 K(fG), q 0 eq and so we may also suppose that Q 0 eq.
We also claim that eq 0 (/9G)C. If we assume the contrary, then eq = ur for some u E ,BG and some r E C. By Corollary 6.20, this implies that q E (fG)r or r E (fG)q. We have ruled out the first possibility, and the second implies that p = rp E (fG)gp c K( PG). However, by Exercise 9.1.4, p 0 K(fG). So we can choose Y E eq satisfying Y fl (fG)C = 0 and Y fl Q = 0. Now qp E cI(Qp) and eqp E c4(Yp). It follows from Theorem 3.40 that ap = yp for some a E Q and some y E Y, or else xp = by for some x E Q and some b E Y. If we assume the first possibility, the fact that a 0 y implies that y E G*, by Lemma 6.28 and thus aly E G* by Theorem 4.31. The equation p = al yp then implies that a1 y E C and hence that y E (f G) C, contradicting our choice of Y. The second possibility results in a contradiction in a similar way.
Definition 9.9. Let S be a semigroup and let p be an idempotent in S. Then p is a strongly right maximal idempotent of S if and only if the equation xp = p has the unique solution x = p in S. Observe that trivially a strongly right maximal idempotent is right maximal. We now show that strongly right maximal idempotents exist in N*, by giving a proof which illustrates the power of the methods introduced by E. Zelenuk. As we shall see in Theorem 12.39, strongly summable ultrafilters are strongly right maximal. However,
192
9 Idempotents
by Corollary 12.38, the existence of strongly summable ultrafilters cannot be deduced in ZFC. It was an open question for several years whether the existence of strongly right maximal idempotents in N* could be deduced in ZFC. This question has now been answered by I. Protasov. We do not know whether it can be shown in ZFC that there are any right maximal idempotents in N* which are not strongly right maximal. However, E. Zelenuk has shown that Martin's Axiom does imply the existence of idempotents of this kind [2471.
Theorem 9.10. There is a strongly right maximal idempotent in N*. Furthermore, for every right maximal idempotent e in Z*, there is an Hmap Vr from co onto a subset of Z for which i(rt (e) is defined and is a strongly right maximal idempotent in N*. Proof. We know that there are right maximal idempotents in Z* by Theorem 2.12. Let
e be an idempotent of this kind, let C = {x E Z* : x + e'= e}, and let C" = {x E PZ : x + C C C}. By Theorem 9.4 C is a finite right zero subsemigroup of fl Z so in particular, x + C = C f o r every x E C. Then C = {0} U C. (Certainly {0} U C C C
Assume X E C"\{0}. Then by Lemma 6.28, x 0 Z. Let y = x + e. Then Y E C so
x+e=x+e+e=y+e=e.)Letcp=lC"={UCZ:CCU}. By Lemmas 7.6 and 7.7, there is a left invariant zero dimensional topology r on 7G for which cp' is the filter of neighborhoods of 0. We choose a set U C Z such that 0 and e are in U but f U if f E C\{0, e}. We
note that for every f E C"\{0, e}, In E Z : n + f E Z\U} E e because a+ f= f E 7L\U. We put
V =UnffEC\{0,e) in EZ:n+f E7G\U} and observe that 0 E V and V E e. Let X = V* = in E V : n + V E e} and note that 0 E X and X E e. Let n E X. We shall show that (i) there exists W (n) E gyp" for which n + (W (n) n x) C X, (ii) (n + X) n X is a neighborhood of n in X for the relative topology induced by r, and (iii) n is not isolated in the relative topology on X induced by r.
Before verifying (i), (ii), and (iii), note that if xw E W for each W E rp", then the net (xW)WEW (directed by reverse inclusion) has a cluster point f in C. To see this, suppose instead that for each f E C one has some U(f) E f and some W (f) E gyp"
such that xW, 0 U(f) for all W' E V with W' C W (f ). Let W' = (U fEC U (f)) n
(f fEC W (f)). Then W' E cp"'. Pick f E C such that xW' E U(f). Then since W' C W (f ), we have a contradiction.
To verify (i), suppose that for every W E (p", there exists rw E w n x such that n + rw 0 X. Pick a cluster point f E C of the net (rw)wEv. Since each rw E X
one has f E X, and thus f = 0 or f = e. Since each rw E 7G\(n + X) one has f E Z\(n + X). But 0 E n + X and by Lemma 4.14, n + X = n + V* E e so e E n + X, a contradiction. To verify(ii)weshowthatthereissome W E (psuch that (n+W)nX C_ (n+X)nX. Suppose instead that for each W E rp` there is some SW E W such that n + SW E X
9.1 Right Maximal Idempotents
193
but sW 4E X. Pick a cluster point f E C = of the net (sw)WEc. Since each SW E Z\X, f 0 X and since each sW E n + X, n + f E X. But since f V X, one has f 0 (0, e) so, since n E V, n + f U D X, a contradiction.
To verify (iii), let W E gyp'. We show that there exists m E _W\{0} such that n +m E X (so that (n + W) fl X {n}). Since W E tp"', e E C C W so W\{0} E e. Also, n E V* so by Lemma 4.14, n + V* E e. Pick m E (W\(O)) fl (n + V*). Having established (i), (ii), and (iii), for each n E X pick W (n) E cp as guaranteed by (i), choosing W (O) = X. Now we have shown that the hypotheses of Theorem 7.24
are satisfied with G = Z and V(n) = W(n) fl X. Thus there is a countable set {Vn : n E N) of neighborhoods of 0 in X for which Y = fl 1 cE,6x Vn\{0} is a subsemigroup of fl7G and there is an IHimap * : w  X such that i/rIH is an isomorphism
from H onto Y. For each n, there is some Wn E rp" such that Wn fl X C Vn and thus each Vn E e so e E Y.
Now the equation x + e = e has the unique solution x = e in X\{0}. It follows that the equation x + i/r1(e) _ f1(e) has the unique solution x = '' (e) in 1111 and hence that i/r1 (e) is a strongly right maximal idempotent in H. Since by Lemma 6.8 all idempotents of N* are in IIII, ilr1(e) is a strongly right maximal idempotent in N*. 0
Theorem 9.11. There are 2` strongly right maximal idempotents in H. Consequentl there are 2` strongly right maximal idempotents in N*. Proof. We know that there are 2` right maximal idempotents in Z* by Theorem 9.1. For each idempotent e of this kind, there is an H map i/re from w onto a subset of Z for which (e) is defined and is a strongly right maximal idempotent in H (by Theorem 9.10). For any given e, there are at most c right maximal idempotents f in Z* for which
,'(e)
because this equation implies that *rf (Vfe 1(e)) = f , so that i%rf is a tEl map taking ' e1(e) to f , and there are at most c Hmaps from w to subsets of Z.
Thus there are 2` distinct elements of the form te' (e). Corollary 9.12. Let G be a countably infinite discrete group which can be algebraically embedded in a metrizable compact topological group. Then there are 21 strongly right maximal idempotents in G*. Proof. This follows immediately from Theorem 9.11 and Theorem 7.28.
Exercise 9.1.1. Which idempotent in ($7G, +) is the unique
Exercise 9.1.2. Let S be a semigroup and let p, q E E(S). Show that p
that there is a right maximal idempotent p E fN for which p + q = q. (Hint: Use Theorems 8.18 and 2.12.)

9 Idempotents
194
The following exercise contrasts with the fact that we do not know whether idempotents in K(fZ) can be left maximal. Exercise 9.1.4. Let G be a countably infinite discrete group. Show that no idempotent in K(fG) can be right maximal in G*. (Hint: Use Theorems 6.44 and 9.4.) Exercise 9.1.5. Let p and q be any two given elements of AZ. Show that the equation x + p = q either has 2` solutions in fZ or else has only a finite number. (Hint: You may wish to use Theorem 3.59.)
9.2 Topologies Defined by Idempotents In this section we investigate left invariant topologies induced on a group G in each of two natural ways by idempotents in G*. The first of these is the topology induced on G by the map a H ap from G to Gp.
Theorem 9.13. Let G be an infinite discrete group and let p be an idempotent in G*. There is a left invariant zero dimensional Hausdorff topology on G such that the filter V` of neighborhoods of the identity e consists of the subsets U of G for which
{x E fG : xp = p) c ctfG U. This topology is extremally disconnected and is the same as the topology on G induced by the mapping (Pp)IG : G  G*. Furthermore, if (pp)IG and (pq)IG induce the same topology on G, then pjG = qfiG. Consequently there are at least 2` distinct topologies which arise in this way.
Proof Let O. = (pp)IG Notice that by Lemma 6.28, 9p is injective. In particular (a E G : ap = p} = {e} so by Lemmas 7.6 and 7.7, with C = (p) (so that C = {x E fIG : xp = p}), we can define a zero dimensional Hausdorff left invariant topology on G, for which (p is the filter of neighborhoods of e. We show first that the topology we have defined on G is the one induced by 9p. Let T be the topology defined by V and let V be the topology induced by O. In the proof of Lemma 7.6 we showed in statement (ii) that the sets of the form U = {a E G : ap E ce f3G U} where U E p form a basis for the neighborhoods of e in T and in statement
(iii) that each U E T. Hence (bU : b E G and U E p) forms a basis for T. Given b E G and U E p, bU" = BpI [kbI 1 [cePG U]] so T C_ V. To see that'V c T, let U E V and let b E U. Pick V C_ G such that b E 9p I [ceOG V] c U. Then b  ' V E p
andbEb(b IV)c U. Since Gp is extremally disconnected as a subspace of 6G (by Lemma 7.41), it follows that G is extremally disconnected. Now suppose that p and q are idempotents in G* which define the same topology on G. Then the map Vr = 9q o OpI is a homeomorphism from Gp to Gq for which
ilr(ap) = aq for all a E G. Now q = ilr(p) _ *(pp) = lim Tr(ap) = lim aq = a+p
ap
pq so q belongs to the principal right ideal of fiG defined by p. Similarly, since
9.2 Topologies Defined by Idempotents
195
9p o 9q1 is a homeomorphism, p belongs to the principal right ideal of PG defined by q. Thus pfiG = qfG. Now iG contains at least 21 disjoint minimal right ideals (by Corollary 6.41). Choosing an idempotent in each will produce at least 2` different topologies on G. The second method of inducing a topology on G by an idempotent p is more direct, taking (A U (e) : a E p) as a neighborhood base fore.
Theorem 9.14. Let G be an infinite discrete group with identity e and let p E G*. Let
.N'=(AU{e}:AE p} andletT ={UCG:forallaEU,a'UE p). ThenT is the finest left invariant topology on G such that each neighborhood of e is a member of jr. The filter of neighborhoods of e is equal to X if and only if p is idempotent. In this case, the topology 7 is Hausdotff.
Prof. It is immediate that 7 is a left invariant topology on G. (For this fact, one only needs p to be a filter on G.) Let V be a neighborhood of e with respect to T. Pick
U E T such that e E U C V. Then U E p and thus V E p. Also V = V U (e) so V E S. Now let V be a left invariant topology on G such that every neighborhood of e with
respect to V is a member of X. To see that V C T, let U E V and let a E U. Then aI U is a neighborhood of a with respect to V soa1 U = A U (e) for some A E p and thus a' U E P. Let At be the set of neighborhoods of e with respect to T. Assume first that .M = W.
To see that pp = p, let A E p. Then A U {e} is a neighborhood of a so pick U E T such that e E U c A U {e}. Then U E P. We claim that U c (a E G : a' A E p}. To
thisend,leta E U. Thena'U E panda'U C a'A U(at}soa1AU(a) E p soaA E p. Now assume that pp = p. To see that .M = J/, let V E X. Then V E p. Let B = V* = (x E V : x1 V E p). Then B E P. We show that B U {e} E T. (Then e E B U {e} C V so V E M.) To see this, let x E B U {e}. Now a1 B = B E p and, if x 96 e, then by Lemma 4.14, x I B E P. Finally assume that pp = p and let a i4 e. Then ap 96 ep by Lemma 6.28, so pick B E p\ap. Let A = B\(a B U {a, a1 }). Then aA U (a) and A U (e) are disjoint neighborhoods of a and e respectively. The following theorem tells us, among other things, that the topologies determined by pin Theorems 9.13 and 9.14 agree if and only if p is strongly right maximal. We saw in Corollary 9.12 that, if G is a countable group which can be embedded in a compact metrizable topological group, there are 21 strongly right maximal idempotents in G*.
Theorem 9.15. Let G be a group with identity e and let p be an idempotent in G*. Let 7 be the left invariant topology on G such that dV = (A U (e) : A E p} is the filter of neighborhoods of e. Then 7 is Hausdorff and the following statements are equivalent.
(a) The topology 7 is regular.
196
9 Idempotents
strongly right; in G* (c) The map (Pp)IG is a homeomorphism from (G, 7) onto Gp. (d) The topology 7 is regular and extremally disconnected. (e) The topology 7 is zero dimensional. (b) Tke idempotent p
.
Proof The topology 7 is Hausdorff by Theorem 9.14. (a) implies (b). Let q E G * such that q p = p and suppose that q # p. Pick B E q \p with e 0 B. Now G\B = (G\B) U {e} is a neighborhood of e so pick a neighborhood U of e which is closed with respect to 7 such that U C G\B. Then U E p = qp so pick b E B such that b1 U E P. Since U is closed, G\U is a neighborhood of b so b 1(G\U) E p, a contradiction. (b) implies (c). By Theorem 9.13, the topology induced on G by the map (pAG has
Proof. By Theorem 9.14, T = (U c_ G : for all a E U, a1 U E p). Suppose that T Z V and pick V E V\T. Since V 0 T pick a E V such that a1V ¢ p. Then G\a1 V E p so (G\a1 V) U {e} is a neighborhood of e with respect to the topology T a((G\a1 V) U {e}) = (G\V) U {a} is a neighborhood of a with respect to T, and thus also with respect to V. But then ((G\V) U {a}) n v = {a} is a neighborhood of a with respect to V. a contradiction.
and thus
Corollary 9.17. Let G be an infinite group. If p is a strongly right maximal idempotent in G*, then the left invariant topology T on G defined by taking {A U {e} : A E p} as the filter of neighborhoods of the identity e is homogeneous, zero dimensional, Hausdorff, extremally disconnected, and maximal among all topologies without isolated points. Distinct strongly right maximal idempotents give rise to distinct topologies. Proof. Since the topology is left invariant, it is trivially homogeneous. By Theorem 9.15,
T is Hausdorff, zero dimensional, and extremally disconnected. By Theorem 9.16, T is maximal among all topologies without isolated points. By Theorem 9.15, T is the topology induced on G by the function (Pp) I G. Given distinct right maximal idempotents
p and q, one has p # qp so p induce distinct topologies on G.
q,G. Thus, by Theorem 9.13, (Pp)IG and (pq)IG
9.2 Topologies Defined by Idempotents
197
Combined with Corollary 9.12, the following Corollary shows that there are 21 distinct homeomorphism classes of topologies on Z that are homogeneous, zero dimensional, Hausdorff, extremally disconnected, and maximal among all topologies without isolated points.
Corollary 9.18. Let G be an infinite group. Let K = I{p E G* : p is a strongly right maximal idempotent in G*}1. If2JG1 < K, then there are at least K distinct homeomorphism classes of topologies on G that are homogeneous, zero dimensional, Hausdorf, , extremally disconnected, and maximal among all topologies without isolated points. Proof.
By Corollary 9.17 distinct strongly right maximal idempotents give rise to
distinct topologies on G, each of which is homogeneous, zero dimensional, Hausdorff, extremally disconnected, and maximal among all topologies without isolated points. Since any homeomorphism class has at most JGI'GI = 2IGI members, the conclusion follows.
The use of ultrafilters leads to partition theorems for left topological groups. The following theorem and Exercises 9.2.5 and 9.2.6 illustrate results of this kind. Let X be a topological space. Recall that an ultrafilter p on X is said to converge to a point x of X if and only if p contains the filter of neighborhoods of x.
Theorem 9.19. Let G be a countably infinite discrete group with identity e. Suppose that there is a left invariant topology r on G for which there is a right cancelable ultrafilter p c G* converging to e. Then G can be partitioned into (0 disjoint subsets which are all r dense in G. Proof. Let ¢ : G x G > N be a bijection. For each b E G let Xb = {ap0(°"b) : a E G}. Notice that if a, b, c, d E G and ap0(a.b) = cp0(`Md), then 0(a, b) = 0(c, d) so that a = c and b = d. (If ap" = cp' where, say, n > m, one has by the right cancelability of p that ap"" = c E G, while G* is a right ideal of $G by Theorem 4.31.)
We claim that for each b E G, ctfG Xb fl cPfiG (UdEG\(b) Xd) = 0. Suppose instead that for some b E G, cesG Xb n cP,6G (UdEG\(b) Xd) 0 0. Then by Theorem 3.40, either Xb n cI$G (UdEG\(b} Xd) 0 0 or c4OG Xb n (UdEG\{b} Xd) # 0. Since the latter implies a version of the former, we may assume that we have some x E Xb n cl $G ( UdEG\{b} Xd). Then for some a E G, x = apO(°,b) and x E c$9G cpc(c.d) if d # b, so c E G and d E G\{b}}. As we have already observed, x E cB3G{cpm : c E G and m > 0(a, b)}. Let n = 0(a, b). Then ap" E ce$G{cpm : C E G and m > n} C G* p" so p" E ai G*p" C G*p". Thus, by Theorem 8.18 p" is not right cancelable, a contradiction. Since cf flG Xb nCI fG (UdEG\{b} Xd) = 0 for every b E G, we can choose a family (Ab)bEG of pairwise disjoint subsets of G such that Xb C ce,G Ab for each b. We may replace Ae by G\ UbEG\(e) Ab so that {Ab : b E G} is a partition of G. ap",b)
9 Idempotents
198
Now let b E G. We claim that Ab is rdense in G. Notice that for each n E N, pn converges to i because (by Exercise 9.2.3) the ultrafilters in G* that converge to e form a subsemigroup of $G. Let V be a nonempty ropen subset of G and pick a E V. Then at V is a rneighborhood of e so a' V E po(°,b). Since also Ab E apo(a,b), we have
VflAb360.
D
Corollary 9.20. Let G be a group with a left invariant topology r with respect to which G has no isolated points. If there is a countable basis for r then G can be partitioned into co disjoint subsets which are all rdense in G.
Proof Let {V,, : n E N} be a basis for the neighborhoods of the identity e of G. By Theorem 3.36 the interior in G* of nt1 is nonempty and thus by Theorem 8.10 there is some p E nn_t which is right cancelable in (3G (where PG is the Stonetech compactification of the discrete space G). Thus Theorem 9.19 applies. E3
Exercise 9.2.1. Let S be any discrete semigroup and let p be any idempotent in S*. Prove that the left ideal (fS)p is extremally disconnected. (Hint: Consider Lemma 7.41.)
Exercise 9.2.2. Let G be an infinite discrete group and let p and q be idempotents in G*. Show that the following statements are equivalent.
(a) q :5R P. (b) The function Gp ). Gq defined by /r(ap) = aq is continuous. (c) The topology induced on G by (Pp)IG is finer than or equal to the one induced by (Pq)IG.
(Hint: Consider the proof of Theorem 9.13.) Exercise 9.2.3. Let G be a group with identity e and let r be a nondiscrete left invariant
topology on G. Let Xr = n{cI$G(U\{e}) : U is a rneighborhood of e in G}. (So Xr is the set of nonprincipal ultrafilters on G which converge to e.) Show that XT is a compact subsemigroup of G*. (Hint: Use Theorem 4.20.) Show that if r' is also a nondiscrete left invariant topology on G, then r = r' if and only if XL = X1'. The following exercise shows that topologies defined by idempotents can be characterized by a simple topological property. Exercise 9.2.4. Let G be a group with identity e and let p be an idempotent in G*. Let 7 be the left invariant topology defined on G by choosing the sets of the form U U {e},
where U E p, as the neighborhoods of e. Show that for any A c G and any x E G, x E cfT A if and only if either x E A or x 1 A E p. Then show that, for any two disjoint subsets A and B of G, we have cIT A fl ceT B = (An ceT B) U (ceT A fl B). Conversely, suppose that 7 is a left invariant Hausdorff topology on G without isolated points such that for any two disjoint subsets A and B of G, one has ceT A fl ceT B = (A f1 cIT B) U (c1T A fl B). Let p = {A c G : e E ceT (A\{e})}. Show that
9.3 Chains of Idempotents
199
p is an idempotent in G* such that the Tneighborhoods of e are the sets of the form U U {e}, where U E p.
Exercise 9.2.5. Let G be a countably infinite discrete group with identity e and let p be an idempotent in G*. Let r be the left invariant topology defined on G by choosing the sets of the form U U {e}, where U E p, as the neighborhoods of e. Show that G cannot be partitioned into two disjoint sets which are both rdense in G. Exercise 9.2.6. Let G be a countably infinite discrete group with identity e and let p be a right maximal idempotent in G*. Let D = {q E G* : qp = p}. Then D is a finite right zero subsemigroup of G* by Theorem 9.4. Let v be the left invariant topology defined on G by choosing the subsets U of G for which D U {e} C c.fG U as the neighborhoods of the identity. Suppose that I D I = n. Show that G can be partitioned into n disjoint rdense subsets, but cannot be partitioned into n + 1 disjoint zdense subsets. (Hint: Use Corollary 6.20 to show that (,BG)q fl (,6G)q' = 0 if q and q' are distinct elements of C.) The following exercise provides a contrast to Ellis' Theorem (Corollary 2.39).
Exercise 9.2.7. Let p be an idempotent in N* and let 7 = {U C 7G : for all a E U,
a + U E p}. Then by Theorem 9.14, (Z, +, T) is a semitopological semigroup which is algebraically a group. Prove that it is not a topological semigroup. (Hint: Either (22,(2k + 1) : n, k E w} E p or {22i+1(2k + 1) : n, k E co) E p.)
9.3 Chains of Idempotents We saw in Corollary 6.34 that nonminimal idempotents exist in S* whenever S is right cancellative and weakly left cancellative. In this section, we shall show that every nonminimal idempotent p in Z' lies immediately above 2` nonminimal idempotents, in
the sense that there are 21 nonminimal idempotents q E Z* satisfying q < p, which are maximal subject to this condition. This will allow us to construct w1sequences of idempotents in Z*, with the property that each idempotent in the sequence corresponding
to a nonlimit ordinal is maximal subject to being less than its predecessor. Whether any infinite increasing chains of idempotents exist in V is a difficult open question. Theorem 9.21. Let G be a countably infinite discrete group, let p be any nonminimal
idempotent in G*, and let A C G with q fl K(OG) # 0. Then there is a set Q S; (E(G*) fl A)\K(5G) such that (1) IQI = 2`. (2) each q E Q satisfies q < p, and (3) each q E Q is maximal subject to the condition that q < p.
200
9 Idempotents
Proof. By Theorems 6.56 and 6.58, there is an infinite subset B of A such that. the following statements hold for every x E B*:
(i) p 0 (48G)xp, (ii) xp is right cancelable in t4 G, (iii) (13G)xp is maximal subject to being a principal left ideal of /3G strictly contained
in (flG)p, and (iv) for all distinct x and x' in B*, (,9G)xp and (f G)x' p are disjoint. Let x be a given element of B* and let C denote the smallest compact subsemigroup
of fiG which contains xp. Pick an idempotent u E C. By Theorem 8.57, u is not minimal in G*. We can choose an idempotent q E (,6G)xp which is
and p 0 (fiG)xp, and so v < p. Let D = {w E E(G*) : v < w < p}. Then D c {w E PG : wq = pq} since, if v < w < p, one has pq = v = wv = wpq = wq. Thus, by Theorem 9.6, D is finite. If D 0, choose w maximal in D. If D = 0, let w = v. We then have an idempotent w E G* which satisfies w < p and is maximal subject to this condition.
We shall show that w E (fG)xp. Since V E (fG)w fl (fG)xp, it follows from Corollary 6.20, that w E (fG)xp or xp E (fG)w. The first possibility is what we wish to prove, and so we may assume the second. We also have w E (fG)p and so (P G)xp c (f G) w c (f G) p. Now (f G)xp is maximal subject to being a principal left ideal of PG strictly contained in (P G) p. Thus (fG)w = (fG)xp or (,BG)w = (fl G) p. The first possibility implies that w E ($G)xp, as claimed. The second can be ruled
out because it implies that p E (PG)w and hence that pw = p, contradicting our assumption that pw = w. We now claim that w 0 K(,BG). To see this, since v < w it suffices to show that v 0 K(fG). We note that xvu = xpqu = xpu E C and that C fl K(fG) = 0 (by Theorem 8.57). Soxvu 0 K(fG) and therefore v ¢ K(fG). Thus we have found a nonminimal idempotent w E (46G)xp which satisfies w < p, and is maximal subject to this condition. Our theorem now follows from the fact that there are 2` possible choices of x, because 1B*I = 2` (by Theorem 3.59), and that (fG)xp and (fG)x'p are disjoint if x and x' are distinct elements in B*.
Lemma 9.22. Let G be a countably infinite discrete group. Let (q,)', be a sequence of idempotents in G* such that, for every n E N, qn+i
Proof For every n E N, we have qr E (fG)gn whenever r > n. Since (PG)q, is closed in PG, it follows that q E (PG)q,. Suppose that q is not right cancelable in PG. Then q = uq for some idempotent
u E G* (by Theorem 8.18). So q E Cf((G\{e})q) and q E c {qn : n E N}, where e denotes the identity of G. It follows from Theorem 3.40 that aq = q' for some
9.3 Chains of Idempotents
201
a E G\{e} and some q' E ci {qn : n E N}, or else qn = xq for some n E N and some
xEfiG. Assume first that qn = xq for some n E N and some x E fiG. Then q E fiGgn+1 so qn E fGgn+1 and thus, qn
allows us to deduce that aqn = q" or else aq" = qn for some n E N and some q" E ce{qn : n E N). Since the equation aq" = qn implies that a1 qn = q", we need only refute the first of these equations. Assume first that q" = qm for some m E N. Then qm = aqn = aqnqn = gmgn so qm qn. Also agngm = gmgm = qm = aqn so by Lemma 8.1, gngm = qn so that qn
Theorem 9.23. Let G be a countably infinite discrete group and let p be any nonminimal idempotent in G*. There is an cot sequence (pa)a« of distinct nonminimal idempotents in G* with the following properties:
(1) po=P, (2) for every a, 0 E co1, a < # implies that pp
satisfies pp < pp1 and is maximal subject to this condition by Theorem 9.21. Then (Pa)a<0 B has the required properties. Suppose then that fi is a limit ordinal. We can choose a cofinal sequence (an)n° 1 in
B. Let q be a limit point of the sequence (pa,) By Lemma 9.22, q E (fG)pan for every n E N and so q E (PG)p,, for every a < 0. Furthermore, q is right cancelable 1
in ,B G. So we can choose an idempotent pp E (PG)q which is <_Rmaximal in G* (by
Lemma 8.59). The fact that pp E (l4G)pa for every a < $, implies that pp
Remark 9.24. Theorems 9.21 and 9.23 are valid with (N, +) in place of G. This is an immediate consequence of the fact that N* is a left ideal of (,8Z, +) by Exercise 4.3.5. We conclude this section by listing some open questions which are tantalizingly easy to formulate, but very hard to answer.
202
9 Idempotents
We shall see, as a consequence of Theorems 12.29 and 12.45, that it is consistent
that there are idempotents p in flN such that, if p = q + r, then both q and r are in Z + p. In particular, such idempotents are both
If (p,)00 I is a sequence of idempotents in flZ such that p
Question 9.27. Is there an infinite increasing
9.4 Identities in $S If S is any discrete semigroup, then an identity e of S is also an identity of PS. This follows from the fact that the equations se = s and es = s, which are valid for every s E S, extend to 6S because the maps pe and Ae are continuous on PS. In this section, we shall see that this is the only way in which an identity can arise in PS. We shall show, in fact, that 18S cannot have a unique right identity which is a member of S*. If S is a weakly left cancellative semigroup, S* is a left ideal in /4S (by Theorem 4.31) and PS can have no right identities in S*. However, a simple example of a semigroup S in which S* has right identities, although S has none, is provided by putting S = (N, A). In this case, every element of S* is a right identity for fi S. (See Exercise 4.1.11.) In contrast, we shall see that, if S is commutative, then no element of S* can be a left identity for PS.
Theorem 9.28. Let S be a discrete semigroup. If 6S has a right identity q E S*, then 22' right identities in S*, where K = I1q JI ,8S has at least
Proof. For each s E S, let Is = {t E S : St = s}. Since sq = s, we have q E XS I [{s }].
This is an open subset of $S and so Is = S fl ASI[{s)] E q. Choose A E q with JAI = K. Then {A fl Is : s E A} is a collection of K sets with the Kuniform finite
9.4 Identities in fiS
203
intersection property. It follows from Theorem 3.62 that nseA A fl IS contains 2z` uniform ultrafilters. We shall show that every ultrafilter p in this set is a right identity for fl S.
To see this, lets E S. We can choose t E A fl I,s, because A and I. are both in q.
Then It E P. Now It c 1, because u E I, implies that su = (st)u = s(tu) = st = s. So I,s E p. Hence sp E cZ{su : u E I,s} = {s} and therefore sp = s. It follows from the continuity of pp that xp = x for every x E ,3S. Corollary 9.29. Let S be any discrete semigroup. If q E ,9S is a unique right identity for fS, then q E S.
Corollary 9.30. Let S be any discrete semigroup. If q E ,S is an identity for fS, then q E S. Theorem 9.31. Let S be a left cancellative discrete semigroup. If p is a right identity for 18S, then p E S and p is an identityfor S and hence for 8S.
Proof Let S E S. Since sp = s, we must have p E S, because S* is a left ideal of fS (by Theorem 4.31). This implies that p is a left identity for S because, for every t E S, we have st = spt and so t = pt. So p is an identity for S, and we have noted that this implies that p is an identity for flS.
Theorem 9.32. Let S be a commutative discrete semigroup and let q E S*. Then q cannot be a left identityfor PS.
Proof. If q is a left identity for flS, then qs = s for every s E S. Since qs = lim is = r.q lim st = sq, we have sq = s for every s E S. By the continuity of pq, this implies that
rq
xq = x for every x E
S. So q is an identity for CBS, contradicting Corollary 9.30.
We obtain a partial analogue to Theorem 9.28.
Theorem 9.33. Let S be a discrete semigroup. If 6S has a left identity q E S* with 22K IIq II = K, then S* has at least elements p with the property that ps = s for every S E S.
Proof. The proof of Theorem 9.28 may be copied, with leftright switches, except for the last sentence. We now see that, if S is countable, any right identity on S* is close to being a right identity on PS.
Theorem 9.34. Let S be any discrete countable semigroup. If S* has a right identity q, then S has a left ideal V such that q is a right identity on V and S\V is finite.
Proof. Let U = S*\ ct(Sq f1 S*). For each s E S, {sq} fl S* is either a singleton or empty. Thus, by Corollary 3.37, Sq fl S* is nowhere dense so that U is a dense open subset of S*.
204
9 Idempotents
Let T = It E S : tq E S}. We claim that S\T is finite. To see this, suppose the contrary. Then there is an element x E u n (S\ T) *, and we can choose X E x satisfying
X* (z U and X C S\T. We have x E X and xq E Xq. Since x = xq, it follows from Theorem 3.40 that s = yq for some s E X and some y E X, or y = sq for some s E X and some y E X*. The second possibility cannot hold because X* _C U. So we may assume the first. Then S E Xq and so, since s is isolated in 8S, we must have s = tq for some t E X. This contradicts the assumption that X c S\T. So S\T is finite and therefore T* = S*. Since q is idempotent, Tq c T. So pq maps T to T. It follows from Theorem 3.35
that, if V = It E T : tq = t}, 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.
Corollary 9.35. Let S be any discrete countable semigroup. If S* has a right identity, then S* has at least 2` right identities.
Proof 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 V. If p is one of these elements and if x E S*, then the fact that vp = v for every v E V implies that xp = x, because x E V.
Exercise 9.4.1. Let S be a right cancellative discrete semigroup and let p be a left identity for $S. Prove that p E S and that p is an identity for S and hence for $S. (This is Theorem 9.31 with the words "left" and "right" interchanged.)
Notes Theorem 9.10 was proved by I. Protasov (in a personal communication). Topologies of the kind discussed in Theorem 9.13 have been studied by T. Papazyan [191].
Theorem 9.15, Theorem 9.16, Corollary 9.17, and Theorem 9.19 are due to I. Protasov (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 introduced by E. Zelenuk in [246].
Theorem 9.28 is due to J. Baker, A. Lau, and J. Pym, in [9].
Chapter 10
Homomorphisms
In this chapter, we shall consider continuous homomorphisms defined on subsemigroups
of ,BS, where S denotes a countable discrete semigroup which is commutative and cancellative. Given two semigroups 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 H occur everywhere, because of the remarkable facility with which homomorphisms can be defined on H. (See Theorem 6.32, for example). In contrast, if S and T are countably infinite discrete semigroups which are commutative and cancellative, there are surprisingly few continuous injective homomorphisms mapping, T into fiS or T* into S*, and none at all mapping PT into S*. In particular,
there are no topological and algebraic copies of fN in N*. Of course, any injective homomorphism from T into S does determine a continuous injective homomorphism from ,B T into 0S, 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,6T into $S, and almost true that every continuous injective homomorphism from T* into S*, must arise in this way as the extension of an injective homomorphism from T into S. If S and T are groups and if L and M are principal left ideals in S* and 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 fN defined by nonminimal idempotents 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 semigroup, then S is contained in the center of fiS (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 occasion where we present two results about (N, ), because (N, +) is the most important semigroup to which our discussion applies and because this notation in convenient for some of the proofs.
206
10 Homomorphisms
10.1 Homomorphisms to the Circle Group In this section we develop some algebraic information related to natural homomorphisms
from a subsemigroup of (IR, +), the primary example being N, to the circle group T. This information is of interest for its own sake and will be useful in Section 10.2 and again in Section 16.4 where we produce a large class of examples of sets with special combinatorial properties. We take the circle group T to be R/Z and shall use ® for the group operation in T. W e define r to be the n a t u r a l projection of R onto T. If t i , 12 E T we shall write t, < t2 if there exists ui, u2 E R such that z(u1) = tI, 7r (u2) = t2 and uI < u2 < uI + 2. Lemma 10.1. Let S be an infinite discrete semigroup and let h : S  T be an injective homomorphism, with h denoting its continuous extension. Let
U = (xES*: IS ES:h(x)
Proof Let x E S*, let T1 = Is E S : h(x) < h(s)) and T2 = Is E S : h(s) < h(x)}. Then T1 fl T2 = 0. Since IS\(Ti U T2)l < 1, T1 E x or T2 E x. Thus x o u fl D and
xEUUD. We now show that u n Z and D fl Z are nonempty. Let p be an idempotent in S*. Since h is a homomorphism (by Corollary 4.22), p E Z. Suppose that p E U. Then we rr (0). can choose a sequence (tn) n _ I in h [S] for which ,r (0) < t for every n and t Now h[fS] is a compact subsemigroup of T and is therefore a group (by Exercise 2.2.3). Hence, for each n E N, we canchoose yn E PS for which h(yn) = tn. Let y E S* bea limit point of the sequence (yn)n° ,. Then y E Z and h(yn) = tn < 7r(0) foreveryn. Foreachn E N, yn E cB({s E S : h(s) < 7r (0))) and soy E ce({s E S : h(s) < n(0)}).
Thus IS E S : h(s) < n(0)} E y and y E D. This shows that p E U implies that D fl Z# 0. Similarly, p E D implies that U fl Z 0 0. So U fl Z and D fl Z are nonempty.
We now show that U is a right ideal of PS. Suppose that x E U and that y E fiS.
Choose u and v in R such that 7r(u) = h(x) and ir(v) = h(y). Let X = IS E S : h(x) < h(s)}. Then X E x. Given a E X, let a E (u, u + 1) satisfy 7r (a) = h(a). Put
Ya = IS E S : h(s) E rr[(v+ua, v+u+1a)]}. Nowir[(v+ua, v+u+1a)]is 2 2 a neighborhood of r (v) =h (y) in T, and so Ya E y. By Theorem 4.15, {a + b : a E X
andb E Ya} E x+y. We claim that{a+b : a E Xandb E Ya} C Is E S: h(x + y) < h(s)) so that x + y E U as required. To this end let a E X and b E Ya.
Pick w E (v + u  a, v + u + 1  a) such that 7r(w) = h(b). Since h and 7r are homomorphisms we have h(x + y) = 7r(u + v) and h(a + b) = n( + w). Since
u+u
10.1 Homomorphisms to the Circle Group
207
This establishes that U is a right ideal of PS. The proof that D is a right ideal of P S is essentially the same.
The following notation does not reflect its dependence on the choice of the semigroup S. Definition 10.2. Let S be a subsemigroup of (R, +) and let a be a positive real number.
We define functions ga : S ). Z, fa : S
and ha : S ). T as follows for
XES: ga(x) = Lxa + 2' J;
fa(x) = xa  ga(x); ha(x) = 7r(fa(X)). We shall use ga, fa, and ha to denote the continuous extensions of these functions which map 18S to f6Z, [1, z], and T respectively, where S has the discrete topology. These exist because the spaces i8Z, [i, ], and T are compact. Note that ga (x) is the nearest integer to xa if xa g Z + 1. We shall use the fact that, for any p E /3S and any neighborhood U of fa (p), the fact that fa is continuous implies that fat [U] is a neighborhood of p and hence that fat [U] E p (by Theorem 3.22).
Lemma 10.3. Let S be a subsemigroup of (IR, +) and let a > 0. The map ha is a homomorphism from S to T. Consequently ha : ,BS + T is also a homomorphism.
Proof. We note that ha(x) = 7r(ax) and so ha is a homomorphism. The second conclusion then follows by Corollary 4.22.
Definition 10.4. Let S be a subsemigroup of (R, +) and let a > 0.
(a) U. = {p E PS: {XES : fa(p) < fa(x)} E p}.
(b) Da={pEfS:{xES: fa(p)> fa(x)}Ep}. (c) Z. = (p E PS: fa(p) = 0). (d) Xa = Ua n Za. (e) Ya = Da n Za. Observe that Ua and Da are the set of points of 16S for which .7,, approaches its value from above and below respectively. Definition 10.5. Let S be a subsemigroup of (R, +) and let a E R. Then a is irrational with respect to S if and only if
a0{ a n b :nE7L,a,bES,anda#bIUIn:nEZandaES\{0}}. a
208
10 Homomorphisms
Saying that a is irrational with respect to S means that ha is onetoone on S and that0 0 ha[S\{0}}. Notice that "irrational with respect to N" is simply "irrational". Notice also that if ISI < c, then the set of numbers irrational with respect to S is dense in Remark 10.6. Suppose that S is a subsemigroup of (R, +) and that at > 0 is irrational with respect to S. Let U, D and Z be the sets defined in Lemma 10.1 with h = ha. Then
U=Ua,D=Daand Z=Za\{0}.
Lemma 10.7. Let S be a subsemigroup of (R, +) and let a > 0 be irrational with respect to S. Then Ua U Da = S*. Proof. This is immediate from Lemma 10.1 and Remark 10.6.
The following result is of interest because, given the continuity of pp, it is usually easier to describe left ideals of ,8S. (See for example Theorem 5.20.)
Theorem 10.8. Let S be a subsemigroup of (118, +) and let a > 0 be irrational with respect to S. Then Ua and Da are right ideals of ,9S. Proof. This is immediate from Lemma 10.1 and Remark 10.6.
In the following lemma, all we care about is that n Ua # 0 and that c2,6s(Na) n Da 0. We get the stronger conclusion for free, however.
Lemma 10.9. Let S be a subsemigroup of (R, +) and let a > 0 be irrational with respect to S. Then for each a E S\{0}, c2ps(Na) n Xa 0 and cefs(Na) n Ya 0 0. Proof. We apply Lemma 10.1 with h replaced by ha and S replaced by Na. Then
cfss(Na)nXa=UnZ#0and cess(Na)nYa=DnZ#0.
For the remainder of this section we restrict our attention to N. Notice that, while ha is a homomorphism on ($N, +), it is not a homomorphism on (fiN, ). However, close to 0 it is better behaved.
Theorem 10.10. Let S = N and let a be a positive irrational number. Then Xa and Ya are left ideals of (,ON, ).
Proof. We establish the statement for Xa, the proof for Ya being nearly identical. Let P E Xa and let q E ON. We first observe that, for any m, n E N, if l fa (n) I <
zm ,
then fa(mn) = mfa(n). To see that qp E Xa, let e > 0 be given (with e < ). For each m E N, let B. _ 2 Theorem 4.15, {mn : m E N {n E N : 0 < fa (n) < } and note that E p. Thus by
n andnEBm}eqp. since {mn: meNandneB,)9{kEN:0< fa(k)
any a > 0. Consequently, by Theorem 10.8, if a is irrational, then Xa and Ya are subsemigroups of (ON, +).
10.1 Homomorphisms to the Circle Group
209
Corollary 10.11. Let S = N and let a be a positive irrational number. Then Za is a subsemigroup of (ON, ). proof. Since a is irrational, {n E N : fa (n) = 0} = 0 so by Lemma 10.7, Za = Xa U Ya so the result follows from Theorem 10.10. The following result is our main tool to be used in Section 16.4 in deriving nontrivial explicit examples of IP* sets and central* sets.
Theorem 10.12. Let S = N and let a > 0. Then ga is an isomorphism and a homeomorphism from Za onto ZiI,, with inverse If a is irrational, then ga takes Xa onto Y1 /,, and takes Ya onto X
Proof. We show that
(1) if p, q E Za, then j,,, (p + q) = 8a(p) + Sa(q), (2) if p E Za, then ga(p) E ZI 1a, (3) if p E Za, then p,
(4) if p E Xa, then ga(p) E Yiand (5) if p E Ya, then ga(p) E Then using the fact that the same assertions are valid with 1/a replacing a proves the theorem. Notice that the hypotheses of statements (4) and (5) are nonvacuous only when a is irrational.
To verify (1), let p, q E Za and let 0 < E < 1. Let A = {m E N : Iga(m) amI < E}. If n and m are in A, then Iga(m) + ga(n)  a(m + n) I < 2E < 1 so ga (m) + ga (n) = ga (m + n). Also, A E p and A E q so
ga(p+q)=pmimqlimga(m+n)=pli
mq nm
(ga(m)+ga(n))=8a(p)+ga(q)
To verify (2) and (3), let p E Za and let q = ga(p). Let e E (0, z) and let B =
{nEN:Ifa(n)I <E}. For anynEN,wehave lfa(n)I <6ifandonly iflmnaI <E where m = ga (n). Now I m  na I < E if and only if Im  n I < . The second inequality establishes that n is the integer closest to m 1 and hence that n = g i (m). Thus we have seen that n E B implies both I f (ga(n))I < L and n = g (ga(n)). So E
I f. (q) I = p lim I.f!a (SE(n))I nEB This establishes (2) and (3).
a
 and p = plimn = p1im 8!a (g,,(n)) = 8! (q) nEB nEB a
To verify (4), let p E X, let E > 0 be given with E < 1/2, and let B = in E N E < fl/E(n) < 0). We need to show B E ga(p). Pick S > 0 with 3 < E a. Let C = (n E N : 0 < fa (n) < S). We show ga [C] c B. Let n E C and let m = ga(n) so
that m
10 Homomorphisms
210
Exercise 10.1.1. It is a fact that if F and G are disjoint finite nonempty subsets of R and F U G is linearly independent over Q, then naEF Ua fl ftEG Da 96 0. Use this fact, together with Theorem 10.8 to show that (fiN, +) has a collection of 2` pairwise disjoint right ideals. (This fact also follows from Theorem 6.44.)
Exercise 10.1.2. Let a > 0 and let p, q E fN. Prove that ga(p + q) E (ga(p) + ga (q), ga (P) + ga (q)  1, ga (P) + $a (q) + 1).
Exercise 10.1.3. Let a > 0 be irrational, let k E N with k > a, and let 8 = k  a. Prove that
(i) for all p E fN, ha (P) _ ha (P),
(ii) Z. = Za, (iii) X. = Ys, and (iv) Y. = X8.
10.2 Homomorphisms from #T into S* We know that topological copies of fiN are abundant in N*. Indeed, every infinite compact subset of N* contains a copy of this kind, by Theorem 3.59. However, as we shall show in this section, there are no copies of the right topological semigroup fiN in N*. In fact, if 0 : fiN N* is a continuous homomorphism, then 0[#N] is finite and IOfN*ll = 1. We shall show that, if S and T are countable, commutative, cancellative semigroups
and if S can be embedded in the unit circle, then we can often assert that 4[T*] is a finite group whenever ¢ : fiT > S* is a continuous homomorphism.
Lemma 10.13. Let S be an infinite discrete semigroup which can be algebraically embedded in T and let V be an infinite subsemigroup of S. Then, for every x and p in
S*, there exists y E V* for which x+ y+ p 0 y+ p+ x. Proof. Let h : S ). T be an injective homomorphism, with h : fl S * T denoting its continuous extension. Let U and D be defined as in Lemma 10.1. By this lemma,
applied to V instead of S, if U' = {y E V* : IS E V : h(y) < h(s)} E y} and D' = {y E V* : IS E V : h(s) < h(y)} E y}, then U' and D' are nonempty. Let x, p E S*. If X E U, we can choose y E D. Since D' C D and U and D are
right ideals offS,x+y+pEUand y+p+xEDand so x+y+p#y+p+x. Similarly, if x E D, we can choose y E U' and deduce that x + y + p 0 y + p + x. Lemma 10.14. Suppose that S is a countably infinite discrete semigroup which can be algebraically embedded in T and that T is a countably infinite commutative weakly left cancellative semigroup. Let 0 : 0T a S* be a continuous homomorphism. Then
K(O[T*]) is a finite group. Furthermore, for every c E T, {b E T : O(b + c) E K(O[T *])) is infinite.
10.2 Homomorphisms from fT into S*
211
proof Observe that by Theorem 4.23, if a E S, then a commutes with every member
of ,BS. Also, if t E T, then t commutes with every member of fT and hence ¢(t) commutes with every member of 0[p T].
We shall show that if p is any minimal idempotent in O[T*] and q E T* with 4,(q) = p,then (1) there exists t E T such that 0(t) E ¢[T *] + p and
(2) given any c E T and any B E q, there exists t E T such that 0(b) + 95(c) E K(O[T*]) 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 + O (q) E ¢[f T]}. We claim that V is finite. Suppose instead that V is infinite, and notice that V is a subsemigroup of S. (Given a, b E V, a + b + 0 (q) = a + b + 0 (q) +
¢(q)=a+0(q)+b+0(q).) For every Y E V*, wehave y+0(q) E O[PT], by the continuity Ofpoigi. Then foreveryy E V*, wehave ¢(c)+y+O(q) = y+O(q)+O(c) which contradicts Lemma 10.13.
Choose A E 0 (c) for which A fl v = 0. Now 0 (c) + 0 (q) _ (q) + 0 (c), 0(c) + 0(q) E C1 (A + 0(q)), and 0(q) + 0(c) E ct(O[B] + 0(b)). It follows from Theorem 3.40 that one of the two following possibilities must hold:
(i) a +0 (q) = 0 (v) +0(c) for some a E A and some v E fl T; (ii) u + 0(q) = 0(b) + 0(c) for some b E B and some u E fl S.
The first possibility is ruled out by the assumption that A fl v = 0. So we may suppose that (ii) holds. Let t = b + c. Then 0(t) E P S + p and so ¢(t) + p = 0(t). Since 0(t) = 0(t+q)+4'(q) and t+q E T* (by Theorem 4.31), 0(t) E 0[T*]+ p c K(0[T*]). Thus (1) and (2) hold.
Now let p be a minimal idempotent in 0[T*] and pick by (1) t E T such that 0(t) E 0[T*]+ p. Let F = 0[T*] +0(t). Then, since t E T, F = 4(t) +0[T*] so F is both a minimal left ideal and a minimal right ideal of 0[T*], and is therefore a group (by Theorem 1.61). Since F is compact, it must be finite, because an infinite compact subset of fS cannot be homogeneous by 6.38. If p' is any other minimal idempotent in 0[T*], there exists t' E T such that 0(t') E
0[T*] + p'. Now 0(t + t') = 0(t'+ t), and so the minimal left ideals 0[T*] + p' = 0[T*] + 0(t') and 0[T*] + p = 0[T*] + 0(t) of 0[f T] intersect. They are therefore equal. This shows that F is the unique minimal left ideal of ¢[T *] and so F = K (0[T *]) (by Theorem 1.64).
The fact that {b E T : 0(b + c) E K(¢[T*])} is infinite for each c E T follows immediately from (2).
Theorem 10.15. Suppose that S is a countably infinite discrete semigroup which can be algebraically embedded in the unit circle. Let T be a countably infinite commutative weakly left cancellative semigroup with the property that any ideal of T has finite complement and let 0 : PT + S* be a continuous homomorphism. Then 0[ f T] is finite and 0[T*] is a finite group.
10 Homomorphisms
212
Proof. By Lemma 10.14, if F = K(" [T*]), then F is a finite group. Let I =1[F]. We claim that I n T is an ideal of T. It follows from Lemma 10.14 that I n T 0 0. If c E I n T, then ¢ (c) = 0 (c) + p for some minimal idempotent p of cb [T * ]. There exists q E T* for which O (q) = p. For every b E T, we have O (b + c) = O (b) + O (c)
4(b)+cb(c)+¢(q) = O(b+q+c) = $(b+q)+O(c) E K(4[T*]). Sob+c E If1T and we have shown that I fl T is an ideal of T.
Thus T\I is finite. Given q E T*, I E q and so O[T*] c F. Since ¢[fT] C 4[T\I] U F, ¢[flT] is finite. Since t[T*] is a finite cancellative semigroup, it is a group by Exercise 1.3.1. p Notice that the hypotheses of Theorem 10.15 are satisfied in each of the following three cases.
(i) T is a countably infinite commutative group;
(ii) T = (N, +); or (iii) T = (N, v). The following lemma is simple and well known.
Lemma 10.16. Let G be a commutative group with identity 0 which contains no elec, then G can be embedded in (R, +).
ments of finite order apart from 0. If I G I
Proof. Let IGI = K. Suppose that G\{0} has been arranged as a Ksequence (aa)a
whenever a < a' < . We shall show that we can define fp so that these properties hold with 6 replaced by 0 + 1.
Let H = Ua
R by stating that fp (b +ma) = f (b) + ms for every b E H and every m E Z. Then fp is well defined because Zap fl H = {0} and is easily seen to be an injective homomorphism. Thus we may assume that kap = c for some k E Z\{0} and some c E H. We define
fp : H + Zap
JR by stating that f# (b + map) = f (b) + k f (c) for every b E H
and every m E Z. To see that fp is well defined, suppose that b + map = b+ m'a p, where b, b' E H
andm,m' E Z. Then k(b'b) =k(mm')ap = (mm')candsok(f(b') f(b)) _ (m  m') f(c) and .f (b) + k f (c) = f (b') + k f (c) It is easy to see that fp is a homomorphism. To see that it is injective, suppose that
fp (b + map) = 0 for some b E H and some m E Z. Then f (b) + f (c) = 0 and so f (kb + mc) = 0. Since f is injective, kb + me = 0. That is to say,k k(b + map) = 0 and so b + map is of finite order in G and hence b + map = 0.
10.2 Homomorphisms from ,BT into S*
213
Thus we can define an injective homomorphism fs : G,6 + JR for every f < K,
so that f,6 c ff, whenever f < #' < K. Then Ufi
Theorem 10.17. Let G be a countably infinite commutative group which contains no nontrivial finite subgroups, and let T be a countably infinite commutative weakly left cancellative semigroup with the property that any ideal of T has finite complement. If 0: PT + G* is a continuous homomorphism, then cb[19T] is finite and I4[T*]I = 1. Proof. By Lemma 10.16, G can be embedded in JR and therefore in T. (If S is a countable
subsemigroup of 1R, we can choose Of 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[jT] is finite and O[T*] is a group. By Theorem 7.17, G* contains no nontrivial finite subgroups and so I0[T*]1 = 1.
Theorem 10.18. If ¢ : ON + N* is a continuous homomorphism, 0[,8N] is finite and I0[N*]1 = 1.
Proof. This follows from Theorem 10.17 with T = N and G = Z. Question 10.19. Is there any continuous homomorphism i¢ : ON > N* for which
I0[fN31 : 1? We note that, in view of Theorem 10.18, there are two simple equivalent versions of Question 10.19
Corollary 10.20. The following statements are equivalent. (a) There is a nontrivial continuous homomorphism from fN to N*.
(b) Thereexistp
gins`Y*suchthatp+p=q+p=p+q=q+q=q.
(c) There is a finite subsemigroup of N* whose elements are not all idempotent.
Proof. (a) implies (b). Let 0 : ON
N* be a nontrivial continuous homomorphism. Then by Theorem 10.18, ¢[fN] is finite and there is some q E N* such that O[N*] = (q). There is a largest x E N such that 0 (x) # q. (Since 0 is nontrivial, there is some such x, and if infinitely many x have 0(x) # q and r E ce{x E N : 0(x) q} fl N*, then
0(r) # q.) Let p = O(x). (b) implies (c). The set {p, q} is such a subsemigroup. (c) implies (a). Let S be a finite subsemigroup of N* whose elements are not all
idempotent and pick p E S which is not an idempotent. Define a homomorphism f from N to S by stating that f (n) is the sum of p with itself n times and let 0 be the continuous extension of f to fN. Then 0 is a homomorphism by Corollary 4.22 and is nontrivial because {p} is not a semigroup.
Question 10.21. Which discrete semigroups S have the property that any continuous homomorphism from 0 S to S* must have a finite image?
214
10 Homomorphisms
10.3 Homomorphisms from T* into S* A good deal of specialized notation will apply throughout this section. S and T will denote countably infinite discrete semigroups which are commutative and cancellative. G will denote the group generated by S and 0 will denote the identity of G. (So G is a countably infinite commutative group in which every element can be expressed as the difference of two elements of S). We shall regard S as embedded in G and PS as embedded in PG. We shall assume that T has an identity. (This is no real restriction, because an identity can be adjoined to any semigroup). We shall assume that we have chosen sequential orderings for G and T. For each m E N, G,,, will denote the set whose elements are the first m elements of G, and T. will denote the set whose elements are the first m elements of T. We shall suppose that 0 : T* * S* is a continuous injective homomorphism. We choose q to be a nonminimal idempotent in T * and we put p = 0 (q). (The existence of a nonminimal idempotent in T* follows from Corollary 6.34.)
Lemma 10.22. cb[K(T*)] = K(,[T*]). Proof. The mapping q5 defines an isomorphism from T* onto O[T *].
Lemma 10.23. For every s E S and every x E ,3S, $ + x E K(f S) implies that x E K(fiS). Proof Suppose that s + x E K(fiS). Since K(PS) is a union of groups (by Theorem 1.64), there is an idempotent u E K (fS) for which s + x + u = s + x. By Lemma 8.1 this implies that x + u = x and hence that x E K(flS). Lemma 10.24. There is a subset Y = {tn : n E N} of T with the following properties: (i)For every y E Y* and every b, c E T, O (b + q) O f G + O (c + y) + p. (ii) For every m and n in N, every a E G,,,, and every b, c E Tm, if m < n, then
cb(b+tn +q) # a+q5(c+tn +q). Proof. We can choose a minimal idempotent u E T* satisfying u = q + u = u + q (by Theorem 1.60). We note that O(u) is in the smallest ideal of q5[T*] by Lemma 10.22. We claim that, for every b E T, ¢(b + q) 0 PG + 0(u). To see this, we note that
¢(b+q) E fiG+O(u) implies that 4(b+q) =b(b+q)+O(u) E K(O[T*]). By Lemma 10.22, this would imply that b + q E K(T*) and hence that q E K(T*) (by Lemma 10.23). This contradicts our choice of q. So for each b E T, O (b + q) has a clopen neighborhood Wb in PG disjoint from the compact subset PG + O (u) of f G. For each a E G and each b, c E T, let
Ea,b,c = {x E T* : a +t(c+x) + p E Wb}. By the continuity of the map x hi a + O (c + x) + p, Ea,b,c is a clopen subset of T*.
We note that u 0 Ea,b,c,because a+O(c+u)+p=a+O(c+u+u)+O(q)=
10.3 Homomorphisms from T* into S*
215
a+O(c+u)+cb(u+q). Sinceu+q =u,a+cb(c+u)+p E fG+q5(u) e fG\Wb. Thus UaeG UbET UcET Ea,b,c i6 T*, because u does not belong to this union. It follows from Exercise 3.4.5 that there is a nonempty open subset U of T* disjoint from UaEG UbeT UcET Ea.b.c We can choose an infinite subset V of T for which V* C U.
We claim that,ify E V* and b, c E T,then 0(b+q) gl #G+o(c+y)+p. Suppose, on the contrary, that q, (b + q) E ,0G + 0(c + y) + p. Since Wb is a neighborhood of 0(b + q) in PG and we are assuming that 0(b + q) E cf(G + 0(c + y) + p), we must have a + O(c + y) + p E Wb for some a E G. This implies that y E Ea,b,c and contradicts our assumptions that y E U and U fl Ea,b,c = 0. To obtain property (ii), we shall enumerate V as {vn : n E N) and then replace (vn)n__i by a subsequence (tn)n° It defined inductively.
W e put tj = v1. W e then assume that we have chosen elements r1, t2, ... , tk in { vn : n E N) with the following property: if i, j E ( 1 , 2, ... , k), a E G; , b, c E Ti, and
i < j, then O(b+t; +q) # a+O(c+tj +q). Suppose that i E {1, 2, ... , k). If a E G and b, c E T, we claim there are at most a finite number of values of n for which 0 (b + t, + q) = a + 0 (c + vn + q). To see this, suppose that this set of values of n is infinite. Then the corresponding elements vn have
a limit pointy in V* satisfying 0(b+t, +q) = a+0(c+y+q) = a+0(c+y)+ p. We have seen that this cannot occur. It follows that we can choosevn with the property that 0(b+t;+q) ,A a+¢ (c+vn+q) whenever i E { 1 , 2, ... , k}, a E Gk and b, c E Tk. We put tk+1 = v, . This shows that we can define a sequence (tn)n° , with the property specified in (ii).
We then put Y = {tn : n E N}. The fact that Y* a V* implies that property (i) holds as well.
Lemma 10.25. Let Y denote the set guaranteed by Lemma 10.24, let y E Y* and let t E T. Then 0(t + y) + p is right cancelable in P G.
Proof. If we suppose the contrary, then 0(t + y) + p = x + 0(t + y) + p for some x E G* (by Theorem 8.18). Now O (t + y) + p E cf ({¢ (t + to + q) : n E N}), because
0 (t + y) + p = 0 (t + y + q) and t + y + q Ecf({t+tn+q:nEN}). We also have x + 0(t + y) + p E cE((G\{0)) + ¢(t + y) + p). It follows from Theorem 3.40 that 0(t + to + q) = x' + 0(t + y) + p for some n E N and some x' E P G, or else
O(t + y'+ q) = a + O(t + y) + p forsomey' E Y* andsomea 54 0inG. The first 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 t E Tm. Since 0(t + y' +q) E
cf({0(t+tn+q):n>m})and a+¢(t+y)+pEcf({a+0(t+t, +q):n>m}), a second application of Theorem 3.40 allows us to deduce that 0(t + t + q) = a +
O(t+y"+q)or4'(t+tn+q)=a+O(t+y"+q) forsomen > m inN and
some y" E Cf((tn : n > m)). Now, by condition (i) of Lemma 10.24, neither of these equations can hold if y" E Y*. Soy" = t, for some r > m. By condition (ii) of Lemma 10.24, neither of these equations can hold if r 0 n. However, they cannot hold if r = n, by Lemma 6.28. Thus 0 (t + y) + p is right cancelable in PG.
216
10 Homomorphisms
Lemma 10.26. Let Y = {tn : n E N} be the set guaranteed by Lemma 10.24. Suppose
that o(b + y +q) E OG + O(c + y + q) for some b, c E T and some y E Y*. Then
O(c+q) E G+O(b+q). Proof. Sinceo(b+y+q) E ct(O[b+Y+q])ando(b+y+q) E Cf(G+r(c+y+q)), it follows from Theorem 3.40 that o (b + t +q) E fG + ¢ (c + y +q) for some n E N, or else O (b + z + q) = a + O(c + y + q) for some z E Y* and some a E G. The first possibility is ruled out by condition (i) of Lemma 10.24,, and so we assume the second.
We choose m E N such that a, a E G,n and b, c E T,,,. Now 0(b + z + q) E cf({q5(b + t + q) : n > m}) anda+O(c+y+q) E cf(a + {0(c + t + q) : n > m}). So a second application of Theorem 3.40 shows that there exists n > m in N and V E c8 ({tn : n > m)) satisfying O (b + to + q) = a + O (c + v + q) or O (b + v + q) _ a + 0 (c + to + q). By condition (i) of Lemma 10.24, neither of these equations can hold if v E Y*. So v = tr for some r > m. However, by condition (ii) of Lemma 10.24, neither of these equations can hold if r 0 n. Thus r = n.
So 0(b+tn+q) = a'+O (c + tn +q), where a' E (a, a}. By adding 0(y+q)onthe right of this equation, we see that 0 (b+q)+O (tn +y+q) = a'+ ¢ (c+q)+O (tn +y+q ). By Lemma 10.25, q5 (tn + y +q) is right cancelable in fi G. So 0 (b + q) = a'+ 0 (c +q)
andO(c+q) E G+g5(b+q). Lemma 10.27. For every c E T, ¢ (c + q) E G + p. Proof. Let Y be the set guaranteed by Lemma 10.24 and let y E Y*. For every b E T, we have 0 (c + q) +0 (b + y + q) = 0 (b +q) + ¢ (c + y +q). It follows from Corollary
6.20 that 0(b+y+q) E P
or0(c+y+q) E P G
In either case, Lemma 10.26 implies that 4i(c + q) E G + 0 (b + q). By choosing b to be the identity of T, we deduce that '(c + q) E G + p.
In the statement of the following theorem, we remind the reader of the standing hypotheses that have been applying throughout this section.
Theorem 10.28. Let S and T be countably infinite, commutative, and cancellative semigroups and assume that T has an identity. Let G be the group generated by S and assume that ¢ : T* * S* is a continuous injective homomorphism. Then there is an
injective homomorphism f : T  G such that f (x) = ¢ (x) for every x E T*. Proof. By Lemma 10.27, for each t E T, there exists a E G for which 0 (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 (t) + p. It is easy to check that f is an injective homomorphism, and so f is also an injective homomorphism (by Exercise 3.4.1 and Corollary 4.22). Since O (t +q) = f (t) + p for every t E T, it follows by continuity that O (x +q) _ T (x) + p for every x E PT. Let Y be the set guaranteed by Lemma 10.24 and let y E Y*. For every x E T*,
wehave ¢(x)+¢(y+q)=O(x+y+q)= f(x+y)+P= f(x)+f(y)+P=
.L(x) + t(y + q). Now O(y + q) is right cancelable in $G, by Lemma 10.25. So f(x) = O(x).
10.3 Homomorphisms from T* into S*
217
Corollary 10.29. Let f denote the mapping defined in Theorem 10.28 and let T' = f1 [S]. Then T' is a subsemigroup of T for which T \T' is finite. Proof. It is immediate that T' is a subsemigroup of T. By Theorem 10.28, T (X) =
¢(x) E S* for every x E T*. Thus, if x E T*, then S E T (x) so T' = f '[S] E x by Lemma 3.30. Consequently T \T' is finite.
Theorem 10.30. Let S and T be countably infinite, commutative, and cancellative semigroups and assume that T has an identity and that 0 : T *  S* is a continuous injective homomorphism. There is a subsemigroup T' of T for which T \T' is finite, and an injective homomorphism h : T' > S for which h(x) = 0(x) for every x E T*.
Proof. Let f denote the mapping defined in Theorem 10.28. We put T' = f 1 [S] and let h = fIT'. Our claim then follows from Corollary 10.29 and Theorem 10.28. The necessity of introducing T' in Theorem 10.30, is illustrated by choosing T = cv and S = N. Then T* and S* are isomorphic, but there are no injective homomorphisms from T to S. In this example, T' = T\{0}.
In the case in which S is a group we have S = G and hence T' = f1 [S]
f1[G] = T. Theorem 10.31. Let S and T be countably infinite, commutative, and cancellative semigroups and assume that T has an identity. Suppose that 8 : PT > 8S is a continuous injective homomorphism. Then there is an injective homomorphism f
T  S for which 8 = f . Proof. Let 0 = 81T*. We observe that O[T*] C S*, because x E T* implies that O(x) is not isolated in PS and hence that O(x) S. By Theorem 10.28 there is an injective homomorphism f : T G such that ¢(x) = T (x) for every x E T*. Let Y denote the set guaranteed by Lemma 10.24 and let y E Y*. For each t E T,
w_ehave 8(t+y+q)=0(t)+6(y+q)=e(t)+0(y+q)andalso 6(t+y+q)= f(t+y+q)= f(t)+f(y+q)= f(t)+0(y+q). By Lemma 10.25, 0 (y + q) is right cancelable in G. So f (t) = 0(t). Now f (t) E G and 0(t) E 18S so f (t) E S.
Theorem 10.32. Let S and T be countably infinite, commutative, and cancellative semigroups and assume that T has an identity. Then S* does not contain any topological and algebraic copies of PT.
Proof. This is an immediate consequence of Theorem 10.31. We do not know the answer to the following question. (Although, of course, algebraic copies of Z are plentiful in N*, as are topological ones.)
Question 10.33. Does N* contain an algebraic and topological copy of Z? Exercise 10.3.1. Show that the only topological and algebraic copies of N* in N* are the sets of the form (kN)*, where k E N.
10 Homomorphisms
218
Exercise 10.3.2. Show that the only topological and algebraic copies of Z* in V. are the sets of the form (kZ)*, where k E Z\{0}.
Exercise 10.3.3. Show that the only topological and algebraic copies of (N*, +) in (N*, ) are those induced by mappings of the form n r+ kn from N to itself, where k E N\{1}.
Exercise 10.3.4. Let S be an infinite discrete cancellative semigroup. Show that S* contains an algebraic and topological copy of N. (Hint: By Theorem 8.10, there is an element p E S* which is right cancelable in $S. Define pn for each n E N by stating that pt = p and p.+1 = pn + p for every n E N. Then {pn : n E N} is an algebraic and topological copy of N.)
Exercise 10.3.5. Show that N* contains an algebraic and topological copy of w. (Hint: Let p be an idempotent in N* for which FS((3n )n° ,) E p and let q E ce({2.3" n E NJ) rl N*. Define xn for every n E co by stating that xo = p, x1 = p + q + p and xn+1 = xn + x1 for every n E N. Then {xn : n E w} is an algebraic and topological copy of w. To see that (xn : n E w) is discrete, consider the alterations of l's and 2's in the ternary expansion of integers. For example, {EtEF, 3' + E,EF2 2. 3' + EtEF3 3'
max F1 < min F2 and max F2 <minF3}Ex1.)
10.4 Isomorphisms Defined on Principal Left and Right Ideals In this section, we shall discuss continuous isomorphisms between principal left ideals or principal right ideals defined by idempotents. Throughout section, S and T will denote countably infinite discrete semigroups, which are commutative and cancellative, G will denote the group generated by S and H will denote the group generated by T.
Lemma 10.34. Suppose that p and q are nonminimal idempotents in S* and T* respectively, and that i/r : q +P1 + q * p + CBS + p is a continuous isomorphism. Let
S'={a EG:a+pE.$S}and T'={bE H:b+q E$T}. Then 3fr(q)= pand there is an isomorphism f : T'
S' such that* (t + q) = f (t) + p for every t E V.
Proof. Since q is in the center of q + f3T + q, *(q) is in the center of p + $S + p. So i,1r(q) = a + p for some a E G (by Theorem 6.63). Since VI(q) is idempotent, it follows from Lemma 6.28 that a is the identity of G and hence that p = 3lr(q).
Foreveryt E T', t + q = q+ (t + q) + q is in q + fT + q and is in the center of this semigroup. So ilr (t + q) is in the center of p + PS + p. It follows from Theorem 6.63
that *fr(t + 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 *(t + q) = f (t) + p for every t E T'.
10.4 Isomorphisms on Principal Ideals
219
It is easy to see that f is injective and a homomorphism. To see that f is surjective,
lets E S'. Since s + p is in the center of p + fiS + p, s + p = * (x) for some x in the center of q + fT + q. We must have x = t + q for some t E T' by Theorem 6.63, and this implies that f (t) = s. Lemma 10.35. Let p and q be nonminimal idempotents in S* and T* respectively. Let
S'=(aEG:a+pEfiS}andletT'=(bEH:b+gEfT).Ifi/r:fT+q). fS+p is a continuous isomorphism, then if (q) is a nonminimal idempotent, fS + p = PS +
i/,(q), and there is an isomorphism f : T'  S' for which *(t + q) = f (t) + >/r(q) for every t E V.
Proof. We first show that 14S + p = fS + *(q). On the one hand, *(q) E /3S + p
and so 1S + *(q) c f S + p. On the other hand, *'(p) = *'(p) +q and so p = i/r(*'(p)+q) = p+*(q) and therefore ,BS+p S fS+*(q). It follows that *(q) is a nonminimal idempotent (by Theorem 1.59).
We also observe that (a E G : a+p E,PS} = {a E G : a+ *(q) E PS}. To see, for example, that {a E G : a+ p E 1PS} C (a E G : a+i,r(q) EPS), let a E G and assume
that a+p E 14S. Then a+*(q) = a+*(q)+p = +/r(q)+a+p E fS+fS C fS. The result now follows from Lemma 10.34 and the observation that * defines a
continuous isomorphism from q + fiT + q onto *(q) + 1S + *(q). (To see that
ifr(q)+fS+ilr(q) c ilr(q+16T+q),letr E ifr(q)+fS+ilr(q). Thenr =r+*(q)so r = *(v)forsomev E fiT+q. Thenq+v E q+fT+gandi/r(q+v) = *(q)+r = r.) 11
We omit the proof of the following lemma, since it is essentially similar to the preceding proof.
Lemma 10.36. Let p and q be nonminimal idempotents in S* and T* respectively. Let
S' = (aEG:a+pE,6S)andletT'={bEH:b+qE fT}.If*:q+$T  p+,BS is a continuous isomorphism, then ifr(q) is a nonminimal idempotent, p+fS = *(q)+
PS, and there is an isomorphism f : T' * S' such that ifr(t +q) = f (t) + *(q) for every t E T'.
Lemma 10.37. Suppose that y E G*\K(f G). Let f : H ) G be an isomorphism. Then there is an element x E T* for which T (x) + y is right cancelable in fl G.
Proof We note that f : PH + fl G is an isomorphism and so ILK (f H)] = K(6G).
By Lemma 6.65, T fl K(fH) i6 0, and so f [T] fl K(fG) = f [T fl K(fH)] # 0. It follows from Theorem 6.56 (with S = G) that there is an element v E G* such that f [T] E v and v + y is right cancelable in fiG. Since V E 7(T1, v = f(x) for some
xET*. Theorem 10.38. Suppose that G and H are countable discrete commutative groups and that L and M are principal left ideals in /3G and ,3H respectively, defined by nonminimal idempotents. If ilr : M . L_ is a continuous isomorphism, there is an isomorphism f : H G for which i' = fM. 
220
10 Homomorphisms
Proof Let L = 0 G+p and M = t3 H+q, where p and q are nonminimal idempotents in G* and H* respectively. By applying Lemma 10.35 with G = S = S' and H = T = T',
we see that 1A (q) is nonminimal and there is an isomorphism f : H  G such that T/r(t + q) = f (t) + i/r(q) for every t E H. By continuity, this implies that *(x + q) = f (x) + (q) for every x E fiH. Forx E fH,wehave >/r(q+x+q) = f (q+x)+i/r(q) = f(g) + f(x)+i/i(q) and
alsoi/r(q+x+q) = i(q)+i/r(x+q) =''lr(q)+f(x)+*(q). So f(q)+f(x)+i/r(q) = i/r(q) + L (x) + i/r(q) for every x E /3H. By Lemma 10.37, we can choose x E $H such that f (x) + ilr(q) is right cancelable in 0 G. Thus ilr(q) = f (q). This implies that i/i (x + q) = T (x + q) for every x E ,0H.
Theorem 10.39. Suppose that L and M are principal left ideals in )4N defined by M > L is a continuous isomorphism then M = L nonminimal idempotents. If and i/' is the identity map.
Proof. Let p and q be nonminimal idempotents in )4N for which L = fiN + p and M = fN + q. Then L = fi7L + p and M = fi7L + q, since, given r E fi7L, r + p = r + p + p E fiN + p because N* is a left ideal in f7L (by Exercise 4.3.5). So L and M are principal left ideals in fi7L defined by nonminimal idempotents in 7L*. It follows
7L for which i/r = fIM. from Theorem 10.38 that there is an isomorphism f : 7L Now there are precisely two isomorphisms from 7L to itself: the identity map t r t and the map t r> t. We can rule out the possibility that f is the second of these maps, because this would imply that f (q) fN. Hence f is the identity map and so is i/r.
Theorem 10.40. Let G and H be countable commutative groups and let L and M be principal right ideals in G* and H* respectively, defined by nonminimal idempotents. Suppose that there is a continuous isomorphism i/r : M + L. Then there is an
isomorphism f : H a G for which f [M] = L. Proof. Let p and q be nonminimal idempotents in G* and H* respectively for which
L = p + fiG and M = q +,8H. By applying Lemma 10.36 with G = S = S' and H = T = T', we see that i/i(q) is nonminimal and there is an isomorphism f : H
G
such thati/r(t+q) = f (t)+(q). It follows, by continuity, that*(x+q) = f (x)+*(q) for every x E M. Putting x = q shows that i/r(q) = 7(q) + (q). On the other hand, we also have, for every x E P H, that ilr(q ± x + q) = f (q) +
f(x)+*(q)and*(q+x+q) _(q)+(q+x+q) _(q)+f(q)+f(x)+(q)
So f(q)+f(x)+ '(q) =/r(q)+f(q)+f(x)+* (q)forevery x r= '6 H. We can choose x E H* for which T (x) + *(q) is right cancelable in fiG (by Lemma 10.37)
and so T (q) _(q) + F (q). It follows easily that * (q)+/3G= f(q)+/3G. Wealso know that i/r(q)+/3G=L (by Lemma 10.36). So T [M] = 7(q) + fiG = *(q) + f G = L. The preceding results in this section tell us nothing about principal left or right ideals defined by minimal idempotents.
Notes
221
if S is any discrete semigroup, we know that any two minimal left ideals in 9S are both homeomorphic and isomorphic by Theorems 1.64 and 2.11. However, the maps which usually define homeomorphisms between minimal left 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 left ideals in ,BN with the property that one cannot be mapped onto the other by a continuous homomorphism? Question 10.42. Are there two distinct minimal left ideals in ,8N with the property that one can be mapped onto the other by a continuous homomorphism?
Notes Most of the result of Section 10.1 are from [33], results of collaboration with V. Bergelson and B. Kra. Theorem 10.8 is due to J. Baker and P. Milnes in [10]. Theorem 10.18 is from [228] and answers a question of E. van Douwen in [80].
Most of the results in Section 10.3 were proved in collaboration with A. Maleki in [1801.
Chapter 11
The RudinKeisler Order
Recall that if S is a discrete space, then the RudinKeisler order
for which f (q) = p. Also recall that if p, q E fS, we write p
reflexive and transitive, it is not antisymmetric. The relation
^RK The RudinKeisler order has proved to be an important tool in analyzing spaces of ultrafilters. It shows how one ultrafilter can be essentially different from another. There are deep and difficult theorems about the RudinKeisler order which we shall
not attempt to prove in this chapter. Our aim is to indicate some of the connections between the RudinKeisler order and the algebraic structure of flS. If S is a discrete semigroup, one would expect that the RudinKeisler order on fi S would have little relation to the algebra of P S, because any bijective mapping from one subset of S to another induces a RudinKeisler equivalence between ultrafilters, and a bijection may have no respect for algebraic structure. Nevertheless, there are algebraic properties which are related to the RudinKeisler order. For example, if S is countable and cancellative, an
element q E S* is right cancelable in f3S if and only if p
Exercise 11.0.2. Let S be a countable discrete space and let p E
S. Show that
{q E PS : q
Exercise 11.0.3. Let S be a discrete semigroup and let a E S and P E PS. Show that
ap
Exercise 11.0.4. Let S be a countable discrete space and let p be a Ppoint in S*. If q E S* satisfies q
11.1 Connections with Right Cancelability
223
11.1 Connections with Right Cancelability In this section we shall explore the connections between the relation
Definition 11.1. Let S be a discrete space and let p, q E PS. The tensor product p® q of p and q is defined by
p®q={AcSxS:{s:{t:(s,t)EA}Eq}E p}. The reader is asked to show in Exercise 11.1.1 that p ® q is an ultrafilter on S x S. If A C S x S, then A E p ®q if and only if A contains a set of the form ((s, t) : s E P and t E Qs }, where P E p and Q,t E q for each s E P. Note that, if s, t E S, we have s ® t = (s, t) where, as usual, we identify the point (s, t) with the principal ultrafilter it generates. If we choose any bijection 4) : S x S + S, we can regard p ®q as being an ultrafilter
on S by identifying it with 4)(p (9 q), where 4 : fi(S x S)  CBS is the continuous extension of 0. In particular, if s, t E S, then s 0 t can be identified with the element 4)(s, t) of S. In this section, we shall sometimes think of ® as being a binary operation on S, assuming that a bijection 0 has been chosen. Of course, different bijections will result in p ® q being identified with different elements of PS. These will, however, all be
RudinKeisler equivalent. So it does not matter which one we choose for studying properties of the relation
Lemma 11.2. Let S be a discrete space and let p, q E S*. Then p
q
fft(P (9 q) = P and ti2(P ®q) = q. Sop
Lemma 11.3. Let S be a discrete space and let p, q E fiS. Then
p ®q = lim lim (s, t) s* p t.q
where s and t denote elements of S and the limits are taken to be in fl(S x S). Proof. This is Exercise 11.1.2. Recall that in Section 4.1, we extended an arbitrary binary operation on S to 14 S.
Lemma 11.4. Let S be a discrete space and let * be a binary operation defined on S. Let p, q, x, y E PS. If X
11 The RudinKeisler Order
224
Proof Let f : S , S and g : S + S be Functions for w hick f(p) = x and g(q) = y. We define h : S x S ). S by h(s, t) = f (s) * g(t). Then
h(p (& q) = lim lim h(s, t) = lim lim f (s) * g(t) = T (p) * g(q) = x * y. s*pt>q s>pt q Corollary 11.5. Let S be a discrete space and let * be a binary operation defined on S. For every p, q E 8S, we have p * q
defined by h(s,t)=s*t,then h(p(& q)=p*q. Proof. The proof is the same as the proof of Lemma 11.4, with f and g taken to be the identity maps.
Corollary 11.5 shows that, for any p, q E fS, p ® q
(1) p * q RK p ®q (2) There is a set D E p for which D * q is strongly discrete and s * q # t * q whenever s and t are distinct members of D. Proof. (1) implies (2). If p*q " RK p®q, there is a set A E p®q on which the mapping (s, t) r). s * t from S x S to S is injective (by Corollary 11.5 and Theorem 8.17.) We
may suppose thatA = WAS) xES),where D E pandEs E gforeach s E D. Then for each s E S, cfps(s * Es) is a neighborhood of s * q in S. If s and s' are distinct elements of D, s * Es and s' * Es, are disjoint, and so cfps(s * Es) and cPSs(s' * Es,) 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 Us of s * q in OS such that Us fl Us, = 0 whenever s and s' are distinct elements of D. For each s E D, we can choose ES E q for which s * Es C U.
If A = s
x Es), then A E p ® q and, since s * Es fl s' * Esc = 0 when
s' and s * t # s * t' when t 0 t', the mapping (s, t) H s * t is injective on A. So
p*q ~RK p®q Corollary 11.7. Let S be a countable left cancellative discrete semigroup and let q be a right cancelable element of PS. Then pq ~RK p ® q for every p E fiS.
Proof. By Theorem 8.11, Sq is strongly discrete in ,S and sq 0 tq when s ¢ tin S, and so our claim follows from Theorem 11.6.
11.1 Connections with Right Cancelability `We are now able to
.
225
establish the equivalence of statements (1), (4) and (5) of Theorem.
8.18 under weaker hypotheses.
Theorem 11.8. Suppose that S is a countable cancellative discrete semigroup and that q e S*. Then the following statements are equivalent.
(1) q is right cancelable in 8S. (2) p
(3) q
there exist a E S and p E CBS\{a} such that aq = pq. By Lemma 6.28, p 0 S. We have aq "RK q by Exercise 11.0.3. So q pq, a contradiction. Theorem 11.9. If S is an infinite discrete space, there are no elements of S* which are 22K maximal in the RudinKeisler order. In fact, if ISI = K, every element of S* has distinct RudinKeisler successors.
Proof. We may suppose that S is a group, because there are groups of cardinality K. (The direct sum of K copies of Z2 provides an example.) It then follows from Theorem 6.30 that there is a subset P of S* such that I P I = 22K,
ap # by when a i4 b in S and Sp is strongly discrete in 1S for every p E P, and (jS)p fl (l4S)q = 0 whenever p and q are distinct elements of P. By Lemma 11.2 and Theorem 11.6, x
Theorem 11.10. Let S be an infinite discrete cancellative semigroup. There is no element of S* whose RudinKeisler predecessors form a subsemigroup of 8S.
Proof. Let p E S* and let L = {x E ,BS : x
Theorem 11.11. Let S be an infinite discrete left cancellative semigroup. Let D C S and let q E S*. Suppose that Dq is strongly discrete in PS and that sq # tq whenever s and t are distinct elements of D. Suppose also that x, y E S*, P E D fl S*, x
11 The RudinKeisler Order
226
Corollary 11.12. Let S be a discrete countable left cancellative semigroup and let q be a right cancelable element of 6S. Then, for every x, p E S*, X
Proof By Theorem 8.11(1), Theorem 11.11 applies with y replaced by q. We now prove a converse to Corollary 11.12.
Theorem 11.13. Let S be a countable cancellative semigroup. Let q be a right cancelable element of fiS and let p E fl S satisfy p
Proof Let f : S * S be a function for which f (yp) = xq. If q E S, then p E S and sox RK xq _
Let B = (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 Lemma 6.28. So we can define a function g : S * S by putting g(b) = c if b E B, and extending g arbitrarily to S\B. We may suppose that g(y) # x,
since otherwise x
Bay.
We now claim that
(*) for each X E x and each Y E y, if X C S\g[V] and Y C V, then aq = f (zp) for some a E X and some z E Y.
To see this, suppose instead that (*) fails and pick X E x and Y E y witnessing this failure. Now xq E ct(Xq) fl cZ f [Yp] so by Theorem 3.40 and the assumption that (*) fails, we must have wq = f (bp) for some w E X and some b E Y. Then wq
Let A = {a E S : aq E f [(fS)p]}. If A 0 x, then letting X = S\(A U g[V]) and Y = V one obtains a contradiction to (*), and so A E x. For each a E A, let Ca = {z E f8S : aq = f(zp)}. Then C. = (f o pp)1[{aq}] so Ca is closed. We claim that for each a E A, Ca f1ci (UbEA\(a) Cb) = 0. Indeed, by Theorem 8.11, aq 0 ct{bq : b E A\{a}} so pick a neighborhood W of aq which misses {bq : b E A\(a)). Given Z E Ca, pick D E z such that f [Dq] C W. Then D fl UbEA\(a) Cb = 0.
Thus one can pick for each a E A a clopen set Va such that Ca C Va and Va fl (UbEA\(a) Cb) = 0. Let A be sequentially ordered by <. For the first element ao of A, let Uao = V. For later elements, let Ua = Va\ Ub
clopen,CaCU,,,and UaflUb=0if a#b. Define h : S ). S by h(s) = a ifs E Ua, defining h arbitrarily on S\ UaEA Ua. We claim that h(y) = x. Suppose instead that h(y) # x and pick X E x such that h(y) 0 X and pick Y E y such that h[Y] fl X = 0. We may assume that X C S\g[V]
11.1 Connections with Right Cancelability
227
and Y C V. Then by (*), aq = f (zp) for some a E X and some z E Y. So Z E Ua and thus h(z) = a, contradicting the fact that h[Y] f1 X = 0. 
Corollary 11.14. Let S be a countable cancellative semigroup and let p E 8S. The following statements are equivalent.
(1) p is right cancelable in PS. (2) For every x, y E ,BS, xp
Theorem 11.15. Let S be a countable discrete space and let * be a left cancellative binary operation defined on S. Suppose that p, q E S and that A E p has the property that A * q is discrete in fi S and that a * q # b * q whenever a and b are distinct elements
of A. Then p * q
Let it, and n2 denote the projection maps of S x S onto S. Then >zi (x (9 q) = x and n2 (x (9 q) = q for every x E 14 S.
Let f : S  S x S be a function for which f (q * p) = p ® q. If P E p and Q E q, p ® q belongs to each of the sets Theorem 3.40 that either
0 q) and cf(f [Q * p]). It follows from
(i) T (b * p) = x ® q for some b E Q and some x E P or (ii) a ® q = T (z * p) for some a E P and some z E Q. Now if (i) holds for some P and Q, then n2 o 7o Ab is a continuous extension of the function s i 7r2 (f (b * s)) taking p to q so that q
228
11 The RudinKeisler Order
Let B =_{b E S : )Fj( (b * p)) E S) and define.g : S  S by stating that g(b) = n, ( f (b * p)) if b E B and defining g arbitrarily on S\B. We claim that W(q) = p, so suppose instead that W(q) # p. Pick P E p and Q E q such that W[U] f1 P = 0. By (ii) pick some a E P and some z E Q such that a ®q = f (z * p), Thus a = ii (a (9 q) = n j (f (z * p)) and a is isolated so there is some R E z with n, [ f [R * p]] = {a}. Pick b E R fl Q. Then ail (7(b * p)) = a sob E B and thus g(b) = a, a contradiction. Corollary 11.16. Let S be a countable discrete space and let p, q E S*. If p ® q
Proof The hypotheses of Theorem 11.15 are clearly satisfied if ® is regarded as a binary operation on S via a fixed bijection between S x S ;nd S. Corollary 11.17. Let S be a discrete countable cancellative semigroup and let q be a right cancelable element of S*. If p E S* and pq <_RK qp, then p and q are RudinKeisler comparable. Proof. Theorem 11.15 applies by Theorem 8.11(1). If S is an infinite discrete semigroup, we know that $S is rarely commutative. In the following corollary, we show something stronger than noncommutativity 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 are not RudinKeisler 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 RudinKeisler comparable. Proof By Theorem 6.36, there are elements p and q of S* which are not RudinKeisler comparable. We may suppose that p and q are right cancelable in $S, because there is an infinite subset T of S with the property that every element of T * is right cancelable in $S (by Theorem 8.10). Since S can be mapped bijectively onto T, p and q are RudinKeisler equivalent to elements in T*. Our claim then follows from Corollary 11.17.
Exercise 11.1.1. Let S be a discrete space and let p, q E CBS. Show that p ® q E
#(S x S). Exercise 11.1.2. Prove Lemma 11.3. Exercise 11.1.3. We cannot normally use the most obvious function to establish a re
lation of the form p ®q
p ® q.
Exercise 11.1.4. Let S be an infinite cancellative semigroup. Show that there is no element p in S* for which {q E $S : q ^ RK p} is a subsemigroup of $S. (Hint: Consider the proof of Theorem 11.10.)
11.2 Connections with Left Cancelability in N*
229
11.2 Connections with Left Cancelability in N* Because PS is a right topological semigroup, left cancelability in ,6S is more difficult to handle than right cancelability. As we have seen, it is easy to derive connections between the RudinKeisler order and right cancelability. However, whether the analogous results hold for left cancelability remains a difficult open question, even in flN. For example, we do not know whether q
Recall that, for each k E N, the natural homomorphism hk : Z + 7Gk has a continuous extension Wk : fl7G + Zk which is also a homomorphism (by Theorem 4.8).
If x, y E 6N and k E N, we may write x = y (mod. k) if hk(x) = hk(y).
Theorem 11.19. Suppose that p E N* and that there exists A E p with the property
that x = p (mod k) for every x E A* and every k E N. Then p
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 0 b + q whenever a b in A and A + q is strongly discrete in fN. 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 Theorem 11.6 and Lemma 11.2. Comment 11.20. The set of ultrafilters pin N* which satisfy the hypotheses of Theorem 11.19 contains a dense open subset of N*. To see this, let U be any nonempty clopen
subset of N* and let q E U. For each k E N, let Vk = {x E ON : x = q (mod k)}. Then Vk is a clopen subset of fN. Since q E U fl n i Vk, this set is a nonempty Gssubset of N* and therefore has a nonempty interior in N* (by Theorem 3.36). Any ultrafilter p in the interior of U fl nk__, Vk satisfies the hypotheses of Theorem 11.19.
Theorem 11.21. Let (an)n°, be an infinite increasing sequence in N. Suppose that either of the two following conditions is satisfied:
(i) For every n E N, an+i is a multiple of an. (ii) For every m, n E N with m # n, an is not a multiple of an,.
If A = {an : n E N} and P E A*, then p
Proof. We first consider the case in which q E nn° , nN. For each n E N, let Bn = nk±i akN so that Bn E q. If condition (i) is satisfied, we define f : N  N by stating that
f(n) = max{a, : an,In)
if an, In for some m,
11 The RudinKeisler Order
230
defining f arbitrarily otherwise. For every n E N and every b E B, f (an + b) = a,,. So am + B. and a + B. are disjoint if 'm i4 n. Since am + Bm E am + q, the set A + q is strongly discrete in flN and our claim follows from Theorem 11.6 and Lemma 11.2. If condition (ii) is satisfied, we define g : N * N by stating that g(n) = min{am : am I n}
if am n for some m,
defining g arbitrarily otherwise. For every n E N and every b E Bn, g(an + b) = a,,, So we can again deduce that A + q is discrete and that Theorem 11.6 and Lemma 11.2 apply.
Now let q be an arbitrary element of N*. For each n E N, we can choose bn E N satisfying bn + q = 0 (mod k) f o r every k E { 1, 2, ... , n). To see this, let n E N be given. Then {c E N : c  q (mod k) for every k E (1, 2, ... , n} } E q so pick c in this set. Pick M E N such that n!m > c and let bn = n!m  c. Let B = {bn : n E N) and let r E B*. Then q + r E nn t nN since for each k, {b : b + q = 0 (mod k)} E r. By what we have already proved, with q + r in place of q, we know that a + q + r qE
b + q + r when a and b are distinct members of A (by Lemma 6.28) and A + q + r is discrete in fN. This is equivalent to stating that a + q + r zA x + q + r if a E A and X E A\{a}, by Theorem 8.11. This implies that a + q 0 x + q for every a E A and every x E A\{a} and hence that A + q is strongly discrete. So Theorem 11.6 and Lemma 11.2 apply once again.
Comment 11.22. By Ramsey's Theorem (Theorem 5.6), every infinite sequence in N contains an infinite subsequence satisfying the hypotheses of Theorem 11.21. So the set of ultrafilters p E N* to which this theorem applies also contains a dense open subset of N*.
Theorem 11.23. Let S be a discrete countable cancellative semigroup and let p be a Ppoint in S*. Then p
by Lemma 6.28. Let S' = (a E S : aq # pq). So S' = S or IS\S'I = 1. For each b E S', let Cb = {x E S* : xq = bq }. Then Cb is a compact subset of S*. Since p 0 Cb and since p is a Ppoint in S*, p ¢ UbES' Cb. It follows that there is a member V of p such that V c S' and v fl UbEs, Cb = 0. We claim that for any b E V and any x E V\{b}, we have bq 0 xq. To see this, note that the equation bq = xq cannot hold if x E S by Lemma 6.28. Also if x E S*, then x 0 Cb so bq # xq. It follows from Theorem 8.11 that Vq is strongly discrete in fS. So Theorem 11.6 and Lemma 11.2 apply.
Comment 11.24. We observe that the ultrafilters p E N* to which Theorems 11.19, 11.21, and 11.23 apply are left cancelable in 6Z. This is a consequence of Corollaries 8.36, 8.37, and 8.38 and Theorem 8.39.
Question 11.25. If p is a left cancelable ultrafilter in N*, must it be true that p
p+gandq
11.3 Further Connections with the Algebra of 8S
231
11.3 Further Connections with the Algebra of fiS We remind the reader that a family F' of subsets of a given set is said to be almost disjoint if the intersection of any two distinct members of F is finite. Theorem 11.26. Let A be a subset of N*. If IAA < c, the elements of A have a common RK successor in IIII.
Proof. We index A as (px)xEtt, allowing repetitions if JAI < c. Let (Ex)XER 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 qx E Ex fl N* such that qx ARK Px. For each F E J'f (H), where .Pf (H) denotes the set of nonempty finite subsets of ]R, we put SF = EXEF qx, with the terms in the sum occurring in the order of increasing indices. We order .Pf (R) by set inclusion and choose q to any limit point of the net (SF) FE.Pf(R) in ON, noting that q E H. Recall that supp(n) is the binary support of n. For each x E H, we define fx : N * N by stating that
f, (n) = min(Ex fl {2' m E supp(n)}) if Ex fl (2m : m E supp(n)) * 0 and defining f, (n) arbitrarily otherwise. We shall show that fx (q) = qx for each x E R. We suppose that this equation does
not hold. Then we can choose B E qx such that fx [B] ¢ q, and we can choose Q E q such that fx  t [B] fl Q = 0. Since q is a limit point of the net (SF)FE.Pf(R), there exists F E .Pf(R) such that x E F and Q E SF. We can choose a disjoint family (By)yEF of subsets of N satisfying
Bx C Ex fl B and, for every y r= F, By c Ev, By e qy, and By fl Ex = 0 if y # x. Let M denote the set of all integers m of the form m = EYEF ny, where ny E By for each y E F. We observe that this expression for m is unique, since each ny is a power of 2, and that fx(m) = nx E B. Suppose that F is arranged in increasing order. By Exercise 4.1.6, M E E EF qy = SF. Thus we can choose n E M fl Q. Since fx (n) E B, it follows that n E fx 1[B] fl Q,
contradicting our assumption that this set is empty. So fx (q) = qx and we have Px "RK qx ARK q for every x E R. Corollary 11.27. Let A be a subset of N* with I A 1 < c. The elements of A have a common
Proof. By Theorem 11.26, the elements of A have a common
pE{n!:nEN}f1N*such that p is right cancelable in ON, by Corollary 8.38. So P
u E N*, by Theorem 11.8. We also have p
232
11 The RudinKeisler Order
Theorem 11.21. We can choose u to lie in any given left ideal or in airy given right ideal of ON. Let L= ON + p and R= p+ ON. Then x
Corollary 11.28. Every minimal left ideal of ON contains an increasing
Theorem 11.29. There are at most c elements of N* whose <_ R K successors form a subsemigroup of ON.
Proof. Suppose that p E N* has the property that its
thus M + u = L. Consequently p
11.4 The RudinFroilk Order There are other useful topological order relations on spaces of ultrafilters, one of these being the RudinFroltlc order.
Definition 11.30. Let S be a countable discrete space. The RudinFrolik order C on OS is defined by stating that, if p, q E ,BS, p C_q if there is an injective function f : S $S for which f [S] is discrete in ,BS and f (p) = q.
Ifp,q E PS, we may write pCgif pCgand q ¢ p. It is immediate that the relation C is reflexive, and we shall see in the next lemma that it is transitive. It is not, however, antisymmetric. Lemma 11.31. Let S be a countable discrete space. The relation C on 8S is transitive.
Proof. Let g : S OS be an injective function for which g[S] is discrete in fiS, and let X be a countable discrete subset of PS. Then g[X] is also discrete, because it follows from Exercise 11.4.1 that g defines a homeomorphism from 8S onto cP,8s g[S].
Suppose now that p, q, r E ,S and that p C q and q C r. So there are injective functions f and g from S to 16S for which f [S] and g[S] are discrete, T (p) = q,
11.4 The RudinFrolffc Order
233
and g(q) = r. Then g o f is injective by Exercise 11.4.1, g[ f [S]Ii is discrete, and
(8 f)(P) = g(.f(P)) = r. So p r r. We now observe that the assertion that p C q is stronger than the assertion that
p:5RKq. Lemma 11.32. Let S be a countable discrete space and suppose that p, q E /3S satisfy
pCq.Then p
We now see that there are elements q E S* which are not right cancelable, 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 11.33. Let G be a discrete countable group and let q be a right maximal idempotent in G*. Let C = {x E G* : xq = q}. Then, for every p E PG, either p E (fG)C or there exists P E p such that Pq is discrete in ,PG. Proof. Suppose that p V (/3G)C. Since C is finite (by Theorem 9.4), (,G)C is compact. So we can choose P E p for which P fl (/3G)C = 0. We claim that Pq is discrete. If we assume the contrary, then aq E c2((P\{a})q)
for some a E P. This implies that aq = xq for some x E P\{a}. Then ax E C so x E aC, contradicting our choice of P.
Theorem 11.34. Let G be a discrete countable group and let q be a strongly right maximal idempotent in G*. Then p C pq for every p E G*. In fact, for every p E G*,
either p C pq and q
Exercise 11.4.1. Let S be a countable discrete space and let f : S > /3S be an irective function for which f [S] is discrete in 14 S. Show that f is also injective, where
f : PS  /3S denotes the continuous extension of f. (Hint: let x and y be distinct elements of PS and let X and Y be disjoint subsets of S for which X E x and Y E Y. Observe that f [X ] fl T [Y1 = f[X] n f [Y] = 0 and apply Theorem 3.40.)
Exercise 11.4.2. Let S be a countable discrete space and let p E /3S. Show that
{qE/3S:gCp}l
234
11 The RudinKeisler Order
Exercise 11.4.3. Let p, q E #N. If p EZ q and q
p, we shall say that p and q are
RudinFrolic equivalent. Show that p and q are RudinFroh1c equivalent if and only if they are RudinKeisler equivalent. (Hint: Use Theorem 8.17 and Lemma 11.32)
Exercise 11.4.4. Let p be a weak Ppoint in N*. Show that, for any q E N*, q C p implies that p
Exercise 11.4.5. Let S be a discrete countable semigroup and let p, q E S*. Suppose
that there is a set P E p such that Pq is discrete in /BS and aq 0 bq whenever a and b are distinct elements of P. Show that p C pq. (Hint: Once you have shown that p C pq, apply Lemma 11.2, Theorem 11.6, and Lemma 11.32 to deduce that one cannot have pq C p.)
Exercise 11.4.6. Let S be a discrete countable semigroup and let q E S* be right cancelable in fl S. Show that p C pq for every p E S*. (Hint: Use Exercise 11.4.5 and Theorem 8.11.) Exercise 11.4.7. Show that every element of N* has 2` distinct RudinFroltlc successors. (Hint: use Exercise 11.4.6, Theorem 6.30, and Theorem 8.11.)
Exercise 11.4.8. Let S be a discrete countable cancellative semigroup and let p E S*.
Show that p is right cancelable in /3S if and only if there is an injective function f : S /3S for which f [S] is discrete in /3S and f (q) = qp for every q E S*. (Hint: Use Theorem 8.11. For the sufficiency consider the proof of Lemma 6.37. For the necessity consider pp.)
Notes Some of the results of Sections 11.1 and 11.2 are from [103], obtained in collaboration with S. GarciaFerreira. Theorem 11.10 contrasts strongly with the situation that holds for the Comfort order of ultrafilters. Given an ultrafilter p on a set S, a space X is said to be pcompact if and only if whenever f : S * X, p lim f (s) exists in X. One then S
defines p
Notes
235
The RudinKeisler order was introduced and studied by M. Rudin in [2111 and [212] and independently by H. Keisler in lectures at UCLA in 1967. The RudinFrolik order was implicitly introduced by Z. Froltlc in [95] and some of its basic properties were established by M. Rudin in [212]. There are several very interesting properties of the RudinKeisler and RudinFrohlc
orders that we have not presented because they do not relate to the algebra of OS. Included among these is the equivalence of the following three statements for p E fS: (a) p is minimal in the RudinKeisler order. (b) p is a selective ultrafilter. (That is, given any f : S S, there is some A E p such that the restriction of f to A is either onetoone or constant.) (c) p is a Ramsey ultrafilter. (That is, for every n E N and every partition { P0, Pi } of [S]", there exist some A E p and some i E (0, 1) such that [A]' c Pi.) Another interesting result is that the RudinKeisler equivalence classes (which are the
same as the RudinFrolic equivalence classes by Exercise 11.4.3) of RudinFrolic predecessors of a given element of fiN are linearly ordered by
Chapter 12
Ultraffiters Generated by Finite Sums
We have seen in Theorem 5.8 that if p is an idempotent in (ON, +) and A E p, then A in N. However, it is not necessarily contains FS((x,,)n° 1) for some sequence 1) E p. We investigate in this chapter the existence and properties true that of ultrafilters satisfying the requirement that each A E p contains some 1) with 1) E p. 1
12.1 Martin's Axiom We shall have several proofs 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 Axiom, although some of it may seem strange at first glance.
Definition 12.1. A partially ordered set is a pair (P, <) where P is a nonempty set and < is a transitive and reflexive relation, that is a partial order. Notice that a partial order is not required here to be antisymmetric. (However, the partial orders that we will use are antisymmetric.)
Definition 12.2. Let (P, <) be a partially ordered set. A filter in P is a subset G of P such that
(1) G 0 0, (2) for every a, b E G, there is some c E G such that c < a and c < b, and (3) for every a E G and b E P, if a < b, then b E G. Thus, if S is a set, a filter on S is a filter in the partially ordered set (,P(S)\{0}, g).
Definition 12.3. Let (P, <) 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 E D such that b < a.
12.1 Martin's Axiom
237
Thus, in more usual terminology, a subset of P is dense if and only if it is cofinal downward.
Definition 12.4. Let (P, <) be a partially ordered set. (a) Elements a and b of P are compatible if and only if there is some c E P such that c < a and c < b. (Otherwise they are incompatible.) (b) (P, <) is a c.c.c. partially ordered set 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 actually asserts that all antichains are countable.
Definition 12.5. (a) Let K be an infinite cardinal. Then MA(K) is the assertion that whenever (P, <) is a c.c.c. partially ordered set and D is a collection of at most K dense subsets of P, there is some filter G in P such that for all D E £, G fl D # 0. (b) Martin's Axiom is the assertion that for all cardinals K, if w < K < c, then MA (K ).
We first note that Martin's Axiom follows from the continuum hypothesis.
Theorem 12.6. MA(co) is true.
Proof. Let (P, <) be a partially ordered set and let {An : n E N} be a set of dense subsets of P. Choose a, E Ai. Inductively, having chosen an E An, pick an+1 E An+I such that an+t < an. Let G = {b E P : there is some n E N with an < b}.
Notice that, for the truth of MA(w) one does not need that (P, <) 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 ordered set (P, <) and a collection £ of w, dense subsets of P such that no filter in P meets every member of D.
Proof. Let P = if : there exists A E J" f(N) such that f : A  wi ). For f, g E P agree that f < g if and only if g C f. For each a < w1, let Da = (f E P : a is in the range of f) and let `a) = {Da : a < wi }. Trivially each Da is dense in P. Suppose now that we have a filter G such that G fl Da 0 0 for each a < a),. Let g = U G. 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 that k < f and k < h. Then (a, b) E k and (a, c) E k so b = c. But now g is a function whose domain is contained in N and whose range is all of wi, which is impossible. 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, IA fl BI < w.
238
12 Ultrafilters Generated by Finite Sums
Definition 12.8. Let S be a set. A set A of subsets of S. is a maximal almost disjoint family if and only if it is an almost disjoint family which is not properly contained in any other almost disjoint family. Theorem 12.9. Let A be a maximal almost disjoint family of infinite subsets of a set S. Then A is not countably infinite.
Proof This is Exercise 12.1.2. As a consequence of Theorem 12.9, one sees immediately that the continuum hypothesis implies that any infinite maximal almost disjoint family of subsets of N must have cardinality c. We shall see that the same result follows from Martin's Axiom.
Lemma 12.10. Let A be a family of infinite subsets of N with the property that, for every F E pf(A), N\ U F is infinite. Let CAI = K. If MA(K), then there is an infinite subset B of N such that A n B is finite for every A E A.
Proof Let Q = {(H, F) : H E 9f(N) and F E ,Pf(A)}. Order Q by agreeing that (H', .T'') < (H, F) if and only if
(1) HCH', (2) F S .T'', and
(3) (H'\H) n OF = 0. It is routine to verify that < is a partial order on Q (which is in fact antisymmetric).
Next notice that if (H, F) and (H', F') are incompatible elements of Q, then
H 56 H' (since otherwise (H, F U F) < (H, F) and (H, F U F') < (H', F') ). Thus, since J.Af(N)I = co, it follows that (Q, <) is a c.c.c. partially ordered set.
Now, for each A E A, let D(A) = ((H, F) E Q : A E F). Then given any (H, F) E Q, one has (H, F U (A)) E D(A) and (H, F U (A)) < (H, F). So D(A) is dense.
For each n E N, let E(n) = ((H, F) E Q : H\{1, 2, ... , n} # 0). Given any (H, F) E Q, we can choose m > n such that m U F. Then (H U {m}, F) E E(n) and (H U {m}, F) < (H, F). So E(n) is dense in Q By MA(K), there is a filter G in Q such that G n D(A) 96 0 for each A E A and G n E(n) 0 0 for each n E N. Let B = U{H : there is some .T" with (H, .T') E G}. Since G n E(n) 0 for each n E N, it follows that B is infinite. We shall show that AnB is finite forevery A E A. To this end, let A E A and pick (H, F) E GnD(A). We claim that A n B C H. Let X E A n B and pick some (H', F') E G such that x E H'.
Since (H, F) E G and (H', F') E G, pick (H", F") E G such that (H", F") _< (H, F) and (H", F") < (H', F'). Then X E H' C H", (H"\H) n U F = 0, and A E F. Sox E Has required.
12.1 Martin's Axiom
239
Corollary 12.11. Let K < c and let (Fa)a
Assuming MA(K), if Ua
Proof. Let x E N*\ Ua
of N* such that F. a D. and x 0 D. We have Da = A* for some Aa C N by Theorem 3.23. Now Ua
Corollary 12.12. Let K < c and let (Ga)a
Proof. This follows immediately from Lemma 12.10.
Corollary 12.14. Let A be an infinite maximal almost disjoint family of subsets of N and assume Martin's Axiom. Then CAI = c. Proof By Theorem 12.13 one cannot have IAA < c and, since A S; J" (N), I A 1 < c. We also see as a consequence of the next corollary that Martin's axiom is the strongest assertion of its kind which is not simply false.
Corollary 12.15. MA(c) is false.
Proof. Let £ 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 S C A. Necessarily CAI < c so Theorem 12.13 together with the assumption of MA(c) yield a contradiction.
Exercise 12.1.1. Prove that the partially ordered set in Theorem 12.7 is not a c.c.c. partially ordered set.
Exercise 12.1.2. Prove Theorem 12.9.
Exercise 12.1.3. Recall that a Ppoint in a topological space is a point x with the property that any countable intersection 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 Ppoints in N*. Provide a direct proof of this fact. (Hint: Let be an enumeration of all countable families of clopen subsets of N*. Use Corollary 12.11 to construct an increasing family (Da)a« of proper clopen subsets of N* with the property that, for every a < c, either n F. a Da or N*\ n F. C Da.)
12 Ultrafilters Generated by Finite Sums
240
Exercise 12.1.4. Show that Martin's Axiom implies that N* satisfies the following strong form of the Baire Category Theorem: If F is a family of nowhere dense subsets of N* with IFI < c, then U F is nowhere dense in N*. (Hint: Let K < c and let (Fa)a
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.
Definition 12.16. Let P E ON. (a) The ultrafilter p is strongly summable if and only if for every A E p, there is a sequence (xn)' 1 in N such that FS((xn)n° 1) C A and FS((xn)n° 1) E p. (b) The ultrafilter p is weakly summable if and only if for every A E p, there is a sequence (xn)n° 1 in N such that FS((xn)n° 1) C A.
Recall that r =c2{pc $N:p+p=p}. Theorem 12.17. The set r = {p EON : p is weakly summable}. In particular, every idempotent is weakly summable.
Proof. This is an immediate consequence of Theorem 5.12. We set out to show that every strongly summable ultrafilter is an idempotent.
Lemma 12.18. Let p be a strongly summable ultrafilter. For each A E p there is an increasing sequence (xn)n in N such that FS((xn)n_1) C A and for each m E N, 1
FS((xn)n°_m) E P.
Proof. For i E [1, 2}, let Ci = (3n (3k +i) : n, k E (0}. (That is, Ci is the set of positive integers whose rightmost nonzero ternary digit is i.) Pick i E (1, 2) such that Ci E P.
Define f : N + a) by f(3n(3k+j)) =nforalln,k E wand j E 11, 2). Notice that if x, y E Ci and f (x) = f (y), then x + y 0 Ci. Consequently, if {x, y, x + y} C Ci, then f (x + y) = min{ f (x), f (y)}. Let A E p and pick a sequence (x)1 such that FS ((xn )n_ 1) E p and FS((xn)O 1) C A n Ci. By the observation above, the sequence (xn)n1 is one to one, so we may presume that it is increasing. Let m E N and suppose that 0 p. Then m > 1. Now
FS((xn)n'1) = FS((xn)n 1) U UFc{1.2,
m
11(EneFxn +FS((xn)
m))
12.2 Strongly Summable Ultrafilters  Existence
241
where EnEO xn = 0. Since FS((xn)n_11) is finite and, by assumption, EnEO xn +
p, pick F with 0 0 F C (1, 2, ..., m  1) such that EnEF Xn +
FS((xn)n_m)
FS((xn)n°_m) E p. Given any y E EnEFxn + FS((xn)rt°_,n), one has that for some G E J'f({m, m +I, m +2, ...}), y = EnEFxn + EnEG Xn, where {EnEFxn, EnEG xn, EnEF xn + EnEG xn} C Ci
so, as observed above, f (y) < f (EnEF xn)
Pick a sequence (yn)n° with FS((yn)R_1) C EnEFxn + FS((Xn)n_m) and FS((yn)n_1) E p. Then for each k E N, f (yk) < f(EnEFXn) so pick some k 1
such that f (yk) = f (ye). Then yk + ye 0 Ci, a contradiction.
Theorem 12.19. Let p be a strongly summable ultrafilter. Then p = p + p. Proof. Let A E p and pick by Lemma 12.18 a sequence (xn)n
in N such that FS((xn)n° 1) c A and for each m E N, FS((xn)nm) E p. It suffices to show that 1
FS((xn)n° 1) C {y E N : y + A E p}, so let y E FS((xn)' 1) and pick F E J'f(N) such that y = EnEFxn. Let m = max F + 1. Then FS((xn)n m) E p and FS((Xn)n°_m) C y + A.
Lemma 12.20. Let p be a strongly summable ultrafilter. For every A E p there is a sequence (xn)n° 1 in N such that (1) FS((xn)noo=1) C A,
(2) for each m E N, FS((xn)n°_m) E p, and (3) for each n E N, Xn+1 > 4 En=1 xt.
Proof. Let A E p. For i E {0, 1, 2} let 23n+i + 1, 23n+i + 2 ..
Bi = U 0 0
23n+i+l
 1}
and pick i E {0, 1, 2} such that Bi E P. Pick by Lemma 12.18 an increasing sequence (xn)n° 1 in N such that FS((xn)n 1) c A f1 Bi and for each m E N, FS((Xn)nm) E P. We claim that (xn)n° is as required, so suppose not and pick the first n such that xn+1 < 4 Et =1 Xt. Pick k E w such that 23k+i < Xn < 23k+i+l Then in fact Eni_1 Xt < 23k+i+1. (If n = 1 this is immediate. Otherwise, x,, > 4 En=,' X, so En1 SO En __ rX < 23k+i+l + 23k+i1 < 23k+i+2 and, since En __ xt E t=I Xt < 1
23k+i1
1
1
Bi, Ei_1 Xt <
23k+i+l.)
Now Xn+l < 4 Ei_1 xt < 23k+i+3 and Xn+l E Bi so
Xn+1 < 23k+i+1 Since also x,, < Xnt1 we have that 23k+i+l < Xn + Xn+1 < 23k+i+2
so xn + xn+1 0 Bi, a contradiction.
The following easy theorem shows that not nearly all idempotents are strongly summable.
Theorem 12.21. Let P E cB K (,8N, +). Then p is not strongly summable.
12 Ultrafilters Generated by Finite Sums
242
Proof. Suppose that p is strongly summable and pick by Lemma 12.20 a sequence (xn)n_ 1 such that E p, and for each n E N, xn+t > 4 Ef 1 xt. But then FS((xn)0 1) is not piecewise syndetic, contradicting Corollary 4.41. Theorem 12.22. There is a weakly summable ultrafilter which is not an idempotent.
Proof This was shown in Corollary 8.24. Theorem 12.23. The set r is not a semigroup. In fact, there exist idempotents p and q in fN such that p + q 1t F. Proof. This was shown in Exercise 6.1.4.
We now introduce a special kind of strongly summable ultrafilter. Its definition is somewhat complicated, but we shall prove in Section 12.4 that these idempotents can only be written in a trivial way as sums of other ultrafilters. Recall that for y E N we define the binary support of y by y = Enesapp(y) 2n. In a similar fashion, for elements y of FS((t!)°O t) we define the factorial support of y. Definition 12.24. (a) For y E FS((t!)°O 1), fsupp(y) is defined by y = EfEfS pp(y) 0(b) The ultrafilter p E fiN is a special strongly summable ultrafilter if and only if (1) for each m E N, FS((n!)n_m) E p,
(2) for each A E p, there is an increasing sequence (xn)' such that 1
FS((xn)n__t) E p, FS((xn)n° 1) C A, and for each n E N, x,, divides xn+t, and (3) for each infinite L C ICY, there is a sequence (xn )n_ such that FS ((xn )n _ 1) C 1
FS((t!)°O 1), FS((xn)o 1) E p, and L\ U{fsupp(y) : y E FS((xn)n_ 1)} is infinite.
We shall show that Martin's Axiom implies that special strongly summable ultrafilters exist. Indeed, we shall show that Martin's Axiom implies that any family of subsets of FS((n!)n_1) which is contained in an idempotent and has cardinality less than c, is contained in a special strongly summable idempotent.
Definition 12.25. (a) 8 denotes the set of infinite subsets S of N with the property that
m divides n whenever m, n E S and m < n. .8f denotes the set of nonempty finite subsets of N which have the same property.
(b) If X is a subset of N which can be arranged as an increasing sequence (xn)M 1, we put FSm(X) = FS((xn)n_m) for each m E N. We put FS,,,(X) _ nmEN Cf fN(FSm(X)).
Lemma 12.26. Let F denote a family of subsets of N with the finite intersection prop
erty, such that mN E F for every m E N. Suppose that B E F and that B contains
an idempotent p. Suppose also that B* = (b E B : b + B E p) E Y' and that b + B* E .T' for every b E B*. Let w < K < c and assume that ITI < K. Then it
12.2 Strongly Summable Ultrafilters  Existence
243
follows from MA(K) that there exists X E & such that FS(X) S; B and x n A $ O for every A E F. Proof. We may suppose that F is closed under finite intersections. Let Q = {F E 8f : FS(F) c B*}. We define a partial order on Q by stating that
F'
Since Q is countable, it is trivial that it satisfies the c.c.c. condition. For each A E 3', let D(A) = {F E Q : F f1 A # 0}. To see that D(A) is dense in Q,
let F E Q and let m = fI F. We can choose a E A n B* n fbeFS(F) (b + B*) n mN.
Then FU{a}
0forevery A E F. LetX=U(D. We shall show that X has the properties required. To see that X is infinite, let m E N and pick F E (D n D(mN). Then x n mN # 0. Since CD is a filter, any two elements x and y of X belong to a set F E (D. So, if x < y, x divides y. Thus X E 44. Furthermore,
if H is any finite subset of X, H C F for some F E CD and so FS(H) c B. Thus FS(X) C B. Finally, for any A E F, there exists F E CD n D(A) and so x n A # 0.
Lemma 12.27. Let p be an idempotent in N*, let F c p, and let to < K < c. If ITI < K, then MA(K) implies that there exists X E 8 such that FS(,.(X) C nAE. A.
Proof Let g= 'U{mN:mEN}U{B*:BEF}U(b+B*: BeFandbEB*) and note that by Lemmas 4.14 and 6.6, g c p. Let 3e be the set of finite intersections of members of 9 and note that 3e c p and IJeI < K. Well order F as (Ax)a
(a) FS(Xa) S; Aa and Xa n A # 0 for every A E 3e and (b) if S < a, then IXa\XsI < w. Using condition (b) and the facts that R is closed under finite intersections and that mN E Je for every m E N one sees that
(*) for any finite nonempty ^a) c 3e U {Xa : a < P}, there exist Ac 3e and a <
with Af1XaC(l;D. In particular, by condition (a), {A n Xa : A E 3e and a < #} has the finite intersection property.
We apply Lemma 12.26, with {A n Xa : A E 3e and a < l3} in place of F and Aj in place of B. Pick WW E 44 such that FS(W,8) C Ap and WW n A n Xa 0 for every A E 3e and every a < P. By the observation (*), (l{ W, n A n Xa : A E N and a < #} # 0 so by Corollary 12.12, there exists an infinite subset Xf of N such that X0* 9 WW n A n Xa for every A E 3e and every a < 0. We may suppose that
244
12 Ultrafilters Generated by Finite Sums
X# C W. We may also suppose that Xp E 3, because x,6 fl mN 0 for every m E N, and so X# contains a set in 4. It is clear that conditions (a) and (b) are satisfied with P in place of a.
We put X = XK. If a < K, X\Xa is finite and so there is some n E N such that FSn(X) C FS(Xa) and thus FS... (X) C FS(Xa) C Aa. Corollary 12.28. Let F be a family of subsets of N contained in an idempotent p for
which FS((n!)n° t) E p. Let L be any infinite subset of N and let co < K < c. If MA(K) and IFI < K, then there exists X E 3 with FS(X) C FS((n!)n° 1) such that FSOO(X) C A for every A E .F and L\ UxEFS(x) fsupp(x) is infinite.
Proof. We may suppose that F is closed under finite intersections. By Exercise 12.2.1, we may suppose that FS ((n !) n m) E F for every m E N.. _ By Lemma 12.27, there exists Y E 8 such that FSQO(Y) C A for every A E Y.
We claim that Y fl FS((n!)' m) # 0 for every m E N. To see this, let m E N. Since FS0O(Y) = nkENFSk(Y) C (FS(n!)n°_m), there exists k E N for which FSk(Y) C FS((n!)n_m). Thus we can inductively choose a sequence (xn)n°1 in Y fl FS((n!)n° 1) with the property that, for each n E N, max fsupp(xn) < in <min fsupp(xn+1) for some in E L. If X = {xn : n E N}, then X has the properties required.
Notice that, while the previous results only use MA(K) for a fixed K < c, the following theorem uses the full strength of Martin's Axiom. Theorem 12.29. Let F be a family of subsets of N contained in some idempotent q E N*
for which FS((n!)n° 1) E q. If IFS < c, Martin's Axiom implies that there is a special strongly summable ultrafilter p for which F C p. Proof. We assume Martin's Axiom. Let (Yu)a« be an enumeration of JP (N) and let (La)a« 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 X0 E i with FS(Xo) C FS((n!)n° 1) such that FSC,(Xo) C A fl Zo for every A E F and Lo\ UxCFS(xo)fsupp(x) is infinite. We then make the inductive assumption that 0 < P < c and that Xa E 3 with FS(Xa) C_ FS((n!)n 1) has been defined for every a < B so that the following conditions hold: (a) FSOO(Xa) C T,,, or FS,,, (X,,) C N\ Y,,;
(b) if 8 < a, then FSOO(Xa) C FS,,,(XS); and (c) La\ UxEFS(xa)fsupp(x) is infinite. By Lemma 5.11, for each a < ,B, FS,,.(Xa) is a compact subsemigroup of N* and hence so is na <,6 FS,, (Xa) which therefore contains an idempotent r. Since FS (Xo) C FS((n!)n° 1), FS((n!)n° 1) E r. We can choose Zf E {YS, N\ YO) satisfying ZS E r. By Corollary 12.28 (applied to {FSm(Xa) : a < 6 and m E N} U {ZS) in place of F) we
12.3 Strongly Summable Ultrafilters  Independence
245
can choose X, E 3 with FS(Xa) C FS((n!)O 1) such that Lp\ UXEFS(xs)fsupp(x) is infinite, FS00(Xp) c Zp, and FS,,0, (Xp) C FSm(Xa) for every a < 0 and every m E N. Thus FS00(X,6) c FS.(Xa) for every a < P. So conditions (a)(c) are satisfied with f in place of a. This shows that we can define Xa E 3 for every a < c so that conditions (a)(c) are satisfied. We put p = (B C N : FSOO(Xa) C_ B for some a < c). It is clear that
p is a filter. For every Y C N, Y E p or N\Y E p, and so p is an ultrafilter. Since FSoo (Xo) c fAE.J1 A, we have .F C p.
Since FS(Xo) C FS((n!)' 1), we have by Exercise 12.2.1 that for each m E N, FS((n!)n_m) E p. Given A E p, A = Ya for some a and so FSOO(Xa) c A and thus for some m, FSm(Xa) C A. Condition (3) of the definition holds directly, and so we have that p is a special strongly summable ultrafilter.
Exercise 12.2.1. Show that if p is an idempotent in N* and 1) E p, then for all m E N, FS((n!)n_m) 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 strongly summable ultrafilters exist. In this section we show that their existence cannot be deduced in ZFC.
Definition 12.30. U is a union ultrafilter if and only if 'U is an ultrafilter on Jr'f (N) and for every A E U there is a sequence (FF)n 1 of pairwise disjoint members of 9'f (N) such that FU((Fn)n° 1) E U and FU((Fn)n° 1) C A.
It would appear that union ultrafilters and strongly summable ultrafilters are essentially equivalent notions. This is indeed true, as we shall shortly see. The correspondence in one direction is in fact immediate.
Theorem 12.31. Let 2( be a union ultrafilter. Let
P={(ErEF2r1:FE.A):AEU}. Then p is a strongly summable ultrafilter.
Proof. The function f : J'f(N) >. N defined by f (F) = E4EF 2t1 is a bijection, so p is trivially an ultrafilter. To see that p is strongly summable, let A E p. Let A = (F E p1(N) : EtEF 2r1 E A). Then A E U so pick a sequence (F,,)' 1 of pairwise disjoint members of J'f(N) such that FU((F,,)R° 1) E 'U and FU((Fn)n° 1) C_ A. For E p and FS((xn)n° 1) C A. each n, let xn = EIE F 2t1. Then To establish the correspondence in the reverse direction, we need some purely arithmetic lemmas. They are somewhat stronger than needed here, but they will be used in their full strength in the next section.
12 Ultrafilters Generated by Finite Sums
246
Lemma 1232. Let (xn)n° be a sequence in N such that, for each n E N, 4 E" t=1 xt.' Then each member of N can be written in at most one way as a linear 1
combination of xn 's with coefficients from (1, 2, 3, 4}.
Proof Suppose not and let z be the smallest counterexample. Then (taking as usual EtEO xt = 0) one has Z = EtEFI Xt + E/EF2 2xt + EtEF3 3Xt + EIEF4 4xt = EtEGI Xt + EtEG2 2x1 + EtEG3 3xt + EtEG4 4Xt
where (F1, F2, F3, F4) and {G1, G2, G3, G4} are each pairwise disjoint and not each
F; = G, . Let n = max U4 1(F; U G,) and assume without loss of generality that n E U41 F,. Then n iE U41 G,, since if it were we would have z  xn as a smaller counterexample. Thus Z = EtEFI XI + EtEF2 2xt + EtEF3 3xt + E:EF4 4Xt xn
>4En_;xt EtEGI XI + EZEG2 2x1 + EIEG3 3xt + EtEG4 4xt = Z,
a contradiction.
Lemma 1233. Let (xn)R° 1 be a sequence in N such that, for each n E N, xn+1 > 4 En 1 xt. Then each member of Z can be written in at most one way as a linear combination of xn's with coefficients from {2, 1, 1, 21. Proof. Assume that we have EtEFI 2x1 + EtEF2 Xl  EfEF3 XI  EtEF4 2xt = EIEGI 2xt + EtEG2 Xt  EtEG3 Xt  EtEG4 2xt
where {F1, F2, F3, F4) and {G1, G2, G3, G4} are each pairwise disjoint. Then EtEFInG4 4Xt + EtE(FInG3)u(FZnG4) 3x1 + EtE(F,\(G3UG4))U(G4\(F1uF2))U(F2nG3) 2xt + EtE(F2\(G3UG4))U(G3\(F1UF2)) Xt
= EtEGinF4 4xt + EtE(G1nF3)U(G2nF4) 3x, + EtE(Gi\(F3UF4))u(F4\(GtuG2))U(G2nF3) 2xt + EtE(G2\(F3UF4))u(F3\(G1uG2)) X1.
Thus, by Lemma 12.32, one has F1 n G4 = G 1 n F4,
(F1nG3)U(F2nG4) _ (G1nF3)U(G2nF4), (F1\(G3UGa))U(G4\(FluF2))u(F2nG3) _ (G1\(F3UF4))U(F4\(GiUG2)) U (G2 n F3), (F2\(G3 U G4)) U (G3\(F1 U F2)) _ (G2\(F3 U F4)) u (F3\(G1 U G2)).
12.3 Strongly Summable Ultrafilters  Independence
247
It is routine to establish from these equations that F, = G1 for each i E (1, 2, 3, 4).13
Lemma 12.34. Let (xn)n° 1 be a sequence in N such that, for each n E N, xn+1 > 4 Et xt. If {a, b, a + b} a FS((xn),°_1), a = E:EF xt, and b = FrteG xr, then
FnG=0.
Proof. Pick H E Pf(N) such that a + b = E:EH xt. Then EtEH xt = E!EF1G xt + EtEFnG 2xt so by Lemma 12.32, F n G = 0. Theorem 12.35. Let p be a strongly summable ultrafilter and pick by Lemma 12.20 a sequence (xn)n° 1 in N such that FS((xn)' 1) E p and for each n E N, xn+1 > 4 E1_1 x,. Define tp : FS((xn)'1) + Pf(N) by (P(ErEFxt) _ {x! : t E F). Let
U = {A C Pf(N) : tp1 [A] E p). Then U is a union ultrafilter. Proof Notice first that by Lemma 12.32, the function tp is well defined. One has then immediately that U is an ultrafilter. To see that U is a union ultrafilter, let .4 E U. Let A = to1 [A]. Then A E p so pick a sequence (yn)n° 1 such that FS((yn)n° 1) C A
and FS((yn)n° 1) E p. For n E N, let F,, = to(yn). By Lemma 12.34, one has for each H E Pf(N) that co(EnEH Yn) = UnEH Fn. Consequently, one has that FU((Fn)n°_1) C A and FU((F,)n° 1) E U.
Theorem 12.36. Let 2( be a union ultrafilter and let q = mak(U), where max 18 (Pf (N)) + fl N is the continuous extension of max : Pf (N) + N. Then q is a Ppoint of N*.
Proof Notice that q is a nonprincipal ultrafilter because U is and the function max is finite to one on Pf (N). To see that q is a Ppoint, let (An )n°_ be a sequence of members of q. We need to show that there is some B E q such that B* C fn A,,*. We may 1
1
presume that Al = N and that An+1 C An and n 0 An+1 for all n. (In particular nn1 An = 0.) Define f : N 3. N by f (x) = max{n E N : x E An). Let
Po=(F E Pf(N): f(maxF) <minF} and
°81=(FEPf(N): f(maxF)>minF} and pick i E {0, 11 such that Si E U. Pick a pairwise disjoint sequence (Fn)n_1 in Pf(N) such that FU((F,)n° 1) C Si and FU((Fn)n° 1) E U. We may presume that min Fn < min Fn+1 for all n. Since FU((F,)n° 1) E `U, we have that (max G : G E FU((Fn)n 1)} E q. Since
(maxF,:nEN}=(maxG:GEFU((F,,)n_1)}, we have {max Fn : n E N) E q.
12 Ultrafilters Generated by Finite Sums
248
We show first that i 0 0, so suppose instead that i = 0. Pick k E N such that for all
n > k, max F > max Ft. Since q is nonprincipal {max F : n E N and n > k} E q. Let f = min Ft. Now Ae+t E q so pick n ? k such that max F E At+1. Then
f+1 < f(maxFn)=
<min(FnUF1)=minF1 =8,
a contradiction.
Thus i = 1. Let B = {max F : n E N}. Now, if n, k E N and max Fn E B\Ak, then n < min Fn < f (max Fn) < k so B\Ak I < k. Thus B* cnn° 1 A,* as required. 13
We have need of the following fact which it would take us too far afield to prove.
Theorem 12.37. The existence of Ppoints in N* cannot be established in ZFC. Proof. [222, VI, §4].
0
Corollary 12.38. The existence of strongly summable ultrafilters can not be established in ZFC.
Proof. By Theorem 12.37 it is consistent relative to ZFC that there are no Ppoints in N*. Theorem 12.35 shows that if strongly summable ultrafilters exist, so do union ultrafilters, while Theorem 12.36 shows that if union ultrafilters exist so do Ppoints in N*.
Exercise 12.3.1. As in Theorem 12.31, let U be a union ultrafilter and let p = {{ErcF 2t1 : F E Al : A E U}. Define f : 30f (N) N by f(F) = EtEF 2Prove that 7(u) = p. Exercise 12.3.2. As in Theorem 12.35, let p be a strongly summable ultrafilter and pick a sequence (xn)n° 1 in N such that FS((xn)n 1) E p and for each n E N, x,,+, > 4 En xt. Define tp : FS((xn)n° 1) ' J'f(N) by tp(E,EFxt) = {xt : t E F} and i .1 extend tp arbitrarily to the rest of N. Let 2( = {A c_ 5'f (N) : rp1 [A] E p). Prove that AP(P) = U.
12.4 Algebraic Properties of Strongly Summable Ultrafilters Recall that for any discrete semigroup (S, +), an idempotent p in 16S is strongly right maximal in PS if and only if {q E 8S : q + p = p} = {p}. We saw in Theorem 9.10 that strongly right maximal idempotents in PN exist.
Theorem 12.39. Let P E ,BN be a strongly summable ultrafilter. Then p is a strongly right maximal idempotent.
12.4 Algebraic Properties
249
Proof. By Theorem 12.19, p is an idempotent. Let q # p be given. We show that q + p 0 p. Pick A E p\q and pick by Lemma 12.20 a sequence (xn)n° 1 in N such that FS((xn)n° 1) C A, for each m E N, FS((xn)n__m) E p, and xn+l > 4 E 1 xt for all n. It suffices to show that
{y E N : y +FS((xn)n° 1) E p} C FS((xn)n_1).
(In fact equality holds, but we do not care about that.) Indeed, one has then that {y E N : y + FS((xn) n_ 1) E p} 0 q so FS ((xn )n° 1) 0 q + p and thus p 0 q + p. To this end, let a E N such that a + FS((xn)n°1) E p. Then
a+FS((xn)n° 1))
fFS((xn)n__1) E P.
such that
So pick a sequence (yn)n
FS((yn)n° 1) C (a +FS((xn)n__1)) f FS((xn)n 1). Pick F and G in JP1(N) such that y1 = EtEF xt and Y2 = EtEG xt. By Lemma 12.34, F fl G = 0. Also, a + yt and a + y2 are in FS((xn)n_1). So pick H and K in ?f (N)
such that a+yl = EtEHX1 anda+y2 = EtEKxt. Then a = EtEH Xt  EtEF Xt = EtEK Xt  EtEG Xt. SO EtEGOH Xt + EtEGnH 2xt = EtEFOK Xt + EtEFnK 2xt. Thus, by Lemma 12.32
GAH = FAK and G fl H = F fl K. Since F fl G = 0, one concludes from these equations that F c H. So a = EtEH\F Xt and hence a E FS((xn)n°1) as required. We saw in Corollary 7.36 that maximal groups in K(PN) are as large as possible (and highly noncommutative). That is, they all contain a copy of the free group on 21 generators. We set out to show now that if p is strongly summable, then H(p) is as small (and commutative) as possible, namely a copy of Z. Lemma 12.40. Let (x n )n_ be a sequence in N such that for each n, xn+1 > 4 E° .1 Xt. Let F, G 1, and G2 be finite subsets of N such that G 1 fl G2 = 0 and F U G 1 U G2 0 0. 1
Leta =min(FUG1 UG2) and let t = EnEFXn  EnEGI Xn  EnEG2 2xn.
Then t=0orItl > Xa Proof. Assume that t 0. Let F' = F\(G1 U G2), let G1' = (G1\F) U (G 2 fl F), and let G2' = G2\F. Then F', G1', and G2' are pairwise disjoint and t = EnEF' Xn  EfEG,' Xn  EnEG2' 2xn.
250
12 Ultrafilters Generated by Finite Sums
Since t # 0, F' U G1' U G2' 0 0. Let b = max(F' U G1' U.G2') andnotice that b > a. Then if b E F', we have t > Xb  EnEG, Xn  EnEGZ 2Xn
Xb  Ebn=12x. Xb
Xb
2=2
>Xb
>
Xa
2.
Otherwise b E (G1' U G2') so that
t < EnEF'Xn Xb b1
< En=1 Xn  Xb 3xb
4
3xa
4
The following lemma is exceedingly technical and we apologize in advance for that fact. It will be invoked twice in the proof of the main result of this section.
Lemma 12.41. Let (xn)° 1, (yn)n1, and (zn)n__1 be sequences in N and assume that for each n, xn+1 > 4 E 1 xt. Let 2 E N and assume that whenever k, j ? f, one has some Hk,j E Pf(N) such that Zk + Yj = EnEHk,j xn. Let b = max He,t and assume that whenever min{k, j} > f one has min Hk, j > b. Let t = (EnEHH,1\He.e+1 xn)  YE
and fork > t, let Fk = Hk,t n Hk,e+l and Gk = He,k n He+l,k Then Iti < xb+l and for each k > 2, Zk = (EnEFk xn) + t and Yk = (EnEGk Xn)  t
4
Proof We firstshowthatforanyk > f,Hk,t\Hk,e+1 = He,e\Ht,e+1 andHt,k\He+l,k = He,e\Ht+l,e Indeed, consider the equations Ze +Ye = EnEHe.e Xn zt + Ye+1 = EnEHe,e+1 Xn
Zk + Ye = EnEHk,e X. and
Zk +Ye+1 = EnEHk.e+l Xn
From the first two equations we have Ye+1  Ye = EnEHe,e+i\He.e Xn  EnEHe.e\He.e+i Xn
while the last two equations yield Ye+1  Ye =`'JnEHk,e+i\Hk,e Xn  EnEHk,e\Hk,e+i X.
Thus by Lemma 12.33, one has that Hk,t\Hk,e+1 = He,e\Ht,e+1 Likewise, working with zt+l  zt, one gets that He,k\He+1,k = Ht,e\Ht+l,e
12.4 Algebraic Properties
251
Now let s = (EnEHu,e\Ht+1,e Xn)  ze Then Xb+1
4
b <  En=1 Xn
 EnEHe,I X.
_ (ze + ye) < Ye EnEHt.e\Ht.t+1 Xn  Ye = t < EnEHt,e Xn
< Eb n=1
Xb+1
X n

X6+1
4
Xb+t
. Similarly, Isi < and thus Is + tl 4 4 Now let k > f be given. Then Zk + ye = EnEHk,e Xn SO
and so Iti <
Xb+1
<2
Zk = EnEHk,eflHk,e+l Xn + EnEHk,e\Hk,e+l Xn  Ye = EnEFk Xn + EREHe,t\He,e+1 Xn  Ye = EnEFk X. + C.
Also zt + Yk =
EnEHI,k Xn SO
Yk = EnEHe,knHe+l,k Xn + EnEHe.k\Ht+l,k Xn  Ze
= EnEGk Xn + EnEHH.e\He+l.e Xn  Ze = EnEGk xn +s.
To complete the proof, it therefore suffices to show that s = t. From the derivations above we have that Zk + Yk = EnEFk Xn + EnEGk xn + t + s and we also know that Zk + Yk = EnEHk,k Xn. Thus EnEHk.k Xn  EnEFk xn  EnEGk Xn = s + t. Let K1 = (Fk\Gk) U (Gk\Fk) and let K2 = Fk fl Gk. Then S + t = EnEHk k Xn  EnEK1 Xn  EnEK2 2xn
and min(Hk,k U K1 U K2) = min(Hk,k U Fk U Gk) ? b + 1 because Fk c Hk,e+l and Gk c He+1,k. Since Is + t l < XX , Lemma 12.40 tells us that s + t = 0 as required. II
As a consequence of the next theorem (and Theorem 12.29) it is consistent with ZFC
that there are maximal groups of (,BIN +) 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. Let p be a strongly summable ultrafilter. If q E N* and there exists
r E N* such that q + r = r + q = p, then q E Z + p. In particular, H(p) = Z + p. Proof, Let us notice first that since, by Theorem 4.23, Z is contained in the center
of (fl7L, +), Z + p is a subgroup of (fl7L, +). And, by Exercise 4.3.5, 7G + p c ON. Consequently Z + p S H(p). The "in particular" conclusion then follows from the
12 Ultrafilters Generated by Finite Sums
252
first assertion immediately, because any member of H (p) is in N* and is invertible with respect to p.
So assume we have q and r in N* such that q + r = r + q = p and suppose that q 0 Z + p. Then likewise, p ¢ 7G + q. Since p 0 { 1 + q, q, 1 + q), pick A l E p such
that 1 + Al 0 q, Al 0 q, and 1 + Al ¢ q. By Lemma 12.20 we may presume (by passing to a subset) that Al = FS((xi,n)n°1) where for each n, x1,n+1 > 4 EJ=1 x1, j,
and for each m, FS((xl,n)n_m) E p. Let B1 = {x E N : x + Al E q}. Since
p =r+q, BiEr.PickyiEB1.Let C1 = (yl + A1) n {x E N : x + Al E r} n ni__1(N\(i + A1)). Since yi E B1 and p = q +r, C1 E q. Pick zl > Yl with z1 E C1. Since yi +zl E Al, pick J1,1,1 E Rf(N) such that Y1 + zl = EnEJ1,1,1 xl,n Let bi = max J1,1,1 and let ai = b1 + 1. Note that al is the smallest member of N with xl,a1 > EnEJ1,1,1 X1,n Assume now that k > 1 and we have chosen Ak_I = FS((xk_ l,n )n_1) E p, BkI E r, Ck1 E q, Yk1 E Bk1, Zk1 E Ck1 C {x E N : x + Ak_1 E r}, bk_i, and ak_i such that for each n, Xk1,n+1 > 4 xk_l, j, and for each m, FS((Xk_i,n)n_m) E P. Then FS((xk_l,n)n° bk_1+1 E p and FS((xi,n)n° ak__,+2) E p. Also, p ¢ {t +q : t E 7L and Iti < k}. Thus we may choose by Lemma 12.20, Ak = FS((xk,n)n°1) E p where for each n, Xk,n+l > 4 Ey=1 xk,j, for each m, FS((xk,n)n°Om) E p, Ak 1* t + q for any
t E Z with Itl < k, and Ak C FS((xk1,n)n°_bk1+1) n FS((x1,n)n_ak_1+2).
Note that Ak c Ak_i.
NowAkEp=r+q. Let Bk=(Zk1+Ak_I)n{X EN: x+Ak E q}nBk1. Then Bk E r. Pick yk E Bk with yk > ZkI. Let
Ck = Ck1 n (Yk + Ak) n {x E N : x + Ak E r} n nk _k (N\(i + Ak)). Then Ck E q. Pick zk E Ck with Zk > yk. Let ak be the smallest member of N with x1,,,, > Yk + Zk. Pick Jk,k,k E J" f(N) With Yk + Zk = EnEJk,k,k Xk,n and let bk = maX Jk,k,k.
The inductive construction being complete, let Q, k, j E N be given with £ min{ j, k}.
Ifk> jwehave ZkECkcCj c y/+Aj soyj +zkEAj. Ifk < j we have y j E B j C Bk+1 C Zk + Ak SO Yj + Zk E Ak.
Consequently yj + zk E At. Further, if i > f, then Ai C At+1 C FS((xe,n)n° b,+l). Consequently, we may choose j E f(N) such that yj + zk = EnEJI,k j xe,n and, if f < min{ j, k}, then min Jt,k,j > be. Note that, if f = k = j, then the definition of Je,k, j agrees with the one given earlier.
12.4 Algebraic Properties
253
Now let t = (EnEJ1,l,1\J,,I,2 xl,n)  Yi. Fork > 1, let Fk = J1,k,I n J1,k,2 and 1
Gk = Jl, I,k n Jl,2,k. By Lemma 12.41 we have that It I< 1 Xl,bl+I and for each k > 1, Zk = EnEFk X1,n + t and yk = EnEGk X1,n
t
Next let e = Itl (or e = 2 if Itl < 1). Let t' _ (EnEJe,e,e\Je,z,e+, xt,n)  yt. For k > f, let Fk' = Jt,k,e n Je,k,t+1 and Gk' = J&,e,k n e,e+1,k. Then again by Lemma 12.41 we have that for each k > f, zk = EnEFk'Xt,n + t' and Yk = EnEGk' Xe,n  t'. To complete the proof we show that t = t'. Indeed, once we have done this we will have ze+I = EnEFe+I' Xe,n + t E t + At while ze+1 E Ce+l S Ce c (N\(t + At)), a contradiction. Now, given n, xt,n E At 9 FS((xl,m so pick T,, such that xe,n = EmET" Xi,m Also, if n k, then xe n + Xt,k E FS((xl,n),°,° 1) so by Lemma 12.34, Tn n Tk = 0. Let k = f + 1. We show that if n E Jt,k,k, then min Tn > at + 2. Indeed, zk + yk E
Ak c FS((xl,m)m ae+2) so pick H with Zk + Yk = EmEH Xl,m and min H > at + 2. On the other hand, using at the second equality the fact that the Tn's are pairwise disjoint, we have Zk + Yk = EJEJJ,k,k Xe,n = EmEL X1,m where L = UnE Je k k T. Thus L = H
and in particular each min Tn > min H > at + 2. Now let Fk" = UnEFk' Tn and let Gk" = UnEGk' Tn. Since Fk' c Jt,k,e+1 = Jt.k,k and Gk' c Je,e+l,k = e,k,k, we have min Fk" > at + 2 and min Gk" > at + 2. Note also that zk = EnEFk" X1,n + t' and Yk = EnEGk" X1,n  t'. Now let
LI={nEFk:n>at+2}, L2 = In E Fk : n < at + 2},
Ml = {nEGk:n>at+2}, and
M2={nEGk :n
EnEFk" X1,n  EnEL1 X1,n = EnEL2 X1,n + t  t'.
Likewise, from the expressions for yk we have EnEGk" X1,n  EnEMI X1,n = EnEM2 XI,n + t'  t.
Since min(Fk" U L 1) > at + 2 and min(Gk" U Ml) > at + 2, we have by Lemma 12.40 that
(1) EnEL2 X1,n + t  t' = 0 or I EnEL2 Xl.n + t  t'I > ZXl,ae+2 and (2) EnEM2 X1,n + t'  t = 0 Or I EnEM2 X1,1 + t'  tI > ?Xi,ae+2 Now 0 < EnEL2 X1,n < Enel XI,n < 4X1,ae+2 We have already seen that
12 Ultrafilters Generated by Finite Sums
254
Itl <
1
1
1
1
1
16Xl,ae+1 < XI,at+2,
4Xl,a,
t = (EnElt,t.e\Je,e.x XP,n)  YP < EnEJe,e.e XP,n = YP + Ze
t=(
< xl,at, and
nEJt.e,e\Je.t.k XP.n)  YP 2: yP > (ZP + yP) > xl.at
Thus I EnEL2 XI,. + t  t'I < g4XI,at+2 < jXl.at+2 SO It'I < X1,ae < 16x1 SO EnELZ Xl,n + t  t' = 0. Likewise, EnEMZ xl,n + t'  t = 0. Thus EnEL2 Xl,n + EnEMZ xl,n = 0 so that L2 = M2 = 0 and hence t = t' as required. 'ae+2
We show in the remainder of this section that it is consistent with the usual axioms of set theory that there exist idempotents in (#N, +) which can only be written as a sum of elements of IBN in a trivial fashion.
Lemma 12.43. Let (xn)n°_1 be an increasing sequence in N with the property that for each n E N, xn divides xn+1. If a, b, m E N, a < xn xn,+l divides b, and a + b E FS((xn)n° 1), then a E FS((xn)n° 1) and b E FS((xn)' 1). P r o o f . Pick F E .Pf(N) such that a + b = EtEF x,. Let G = F fl {1, 2, ... , m} and
let H = F\{1, 2, ... , m}. Then a  E,EG Xt = EtEH xt  b (where EIEO X, = 0). Then xn,+l divides E,EH xt  b while Ia  EIEG XII < xtn+l, so a  EIEG Xt =
ErEHxtb=0. Lemma 12.44. Let p be a special strongly summable ultrafilter, let q E fN, and let r E nc,'I 1 cf (Nn). If p = q + r, then q = r = p. Proof By Theorem 12.39 it suffices to show that r = p. So suppose not and pick A E p\r. Pick an increasing sequence (xn)n°_1 such that FS((xn)n° 1) e A, FS((xn)n_1) E p, and for each n E N, xn divides xn+1. Then FS((xn)n° I) E q + r so pick a E N such that a + FS((xn )n° 1) E r. Pick m E N such that xn a. Pick b E (a +FS((xn)n°O__I)) fl Nxn+1 fl (N\A).
Then a + b E FS((xn)n° 1), a < xn,, and x,n+l divides b so by Lemma 12.43, b E FS((xn )n° 1) c A, a contradiction.
Of course, any idempotent p in fiN 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.
Theorem 12.45. Let p be a special strongly summable ultrafilter and let q, r E fiN. If
p=q+r,then gandrarein7L+p. Proof. We claim that it suffices to show that r E 7L + nrt_1 c2(Nn). For assume we
have k E Z such that k + r E nn ct(Nn). Then (k + q) + (k + r) = p so by Lemma 12.44, k + q = k + r ='p. I
12.4 Algebraic Properties
Thus we suppose that r o Z + f n° 1 ct(Nn), so that also q o z + I
255 no=1 cZ(Nn).
Define a : N * X00110, 1,...,t}
by the equation x = E°_ 1 a (x) (t) t ! (It is easy to see that any positive integer has a unique such factorial expansion.) Let a be the continuous extension of a to #N.
Let M = it E N : a(q)(t) 0 t} and let L = It E N : a(q)(t) 0 0}. We claim that both M and L are infinite. To see that M is infinite, suppose instead we have some k E N such that f o r all t > k, a(q)(t) = t. Let z = (k + 1)!  Ei1 a(q)(t) t!. We claim that q + Z E fn_1 ct(Nn), a contradiction. To see this, let n E N be given with
n > k. We show that z + N(n + 1)! E q. Now, by the continuity of a, X
foralltE{1,2,...,n},
a(x)(t) = a(q)(t)) E q. We claim that
N(n + 1)! + En=1 a(q)(t) t! C z + N(n + 1)!. To see this, let w E N(n + 1)! + E°=1 a(q)(t) P. Then
so
as required.
Similarly, if L is finite so that for some k, one has a(q)(t) = 0 for all t > k, and
z = Ek,=t a(q)(t) fl, then q  z E fn__1 ct(Nn). Since p is a special strongly summable ultrafilter, pick a sequence (zn) 1 such that FS((zn)n_1) S FS((t!)°O1), FS((zn)O_1) E p and M\ U(fsupp(y) : y E FS((zn)n_1)} is infinite. Let K = U{fsupp(y) : y E FS((zn)001)}.
Now FS((zn)n_1) E q + r so pick a E N such that a + FS((zn)n_1) E r. Pick m E N such that a < m! and pick f> s > m such that f E M\K and S E L. Now FS((t!)°Ot+1) E p so
{x E N : x +FS((t!)i__e+1) E r} fl (N(f + 1)! + Ei_1 a(q)(t) t!) E q so pick b E N(f + 1)! + Ei=1 a(q)(t) t! such that b + FS((t!)°Ot+1) E r. Pick x E (b + FS((t!)°_e+1)) fl (a + FS((Zn)n__1)). Then b + x = EIEG t! where min G > f + 1 and for some d E N,
b = d (2+1)!+EI_1a(q)(t)t!.
256
12 Ultrafilters Generated by Finite Sums
Since s E L, a(q)(s)
0 so
and
Ei_1 a(q)(t) t!  a < Ei_1 a(q)(t) t! < (s + 1)! so E1
a(q)(t) t!  a = Es1 =1 ct t! where each c1 E {0, 1, ... , t} and not all ct's
are 0.
Now b + x E FS((Zn)n° 1) + (b  a) so for some H E
ErEGt! = b+x EnEH Zn + (b  a)
= EnEH Zn +d (t + 1)! + EI=5+1 &(q)(t) t! + Es 1 a(q)(t) t!  a
= E JP f ( N) such that EnEH Zn = EtEJ t ! and let 11 = I fl { 1 , 2, ... , t} and 12 = I\{1,2,...,t}. Then EtEG t!  d (t + 1)!  EtE12 t! = EtE1, t! + Ets+1 a(q)(t) t! + Et=1 ct t!. Now (t + 1)! divides the left hand side of this equation. On the other hand, t 0 K so
f f I, and f EM so a(q)(t)
El=zt!+t < (t + l)!
a contradiction.
Notes The presentation of Martin's Axiom is based on that by K. Kunen in [171]. See the book by D. Fremlim [94] for substantial information about Martin's Axiom. It was shown in 1972 in [ 117] that the (then unproved) Finite Sums Theorem together with the continuum hypothesis implied the existence of an ultrafilter 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 ($N, +). (See the notes to Chapter 5 for the history of the discovery of this fact.) With this knowledge it appeared that nothing interesting remained in [117]. Then, in 1985, E. van Douwen noted in
Notes
257
conversation that in fact the ultrafilter produced there had the stronger property that it was generated by sets of the form FS((x,,) 1), and wondered if the existence of such an ultrafilter could be proved in ZFC. This question led to the investigation of strongly summable ultrafilters, beginning with [131]. Union ultrafilters were introduced by A. Blass in [44]. Theorem 12.37, one of three
nonelementary results used in this book that we do not prove, is due to S. Shelah. Corollary 12.38 is due to P. Matet [181]. The proof of Corollary 12.38 presented here is from [47], a result of collaboration with A. Blass. Theorem 12.39 is from [46], a result of collaboration with A. Blass. Theorem 12.42 is from [136]. Some of the results of this chapter have been generalised in [151] (a result of collaboration with I. Protasov) to countably infinite abelian groups. If G is a group of this kind, the existence of strongly summable nonprincipal ultrafilters on G follows from Martin'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 equation
p + x = p has the unique solution x = p in G*, and so 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 = p implies that x and y are both in 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 nondiscrete 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 fS
We deal in this chapter with the relationships among different operations on the same set S, as for example the semigroups (N, +) and (N, ). We also include a section considering the relationships between the left continuous and right continuous extensions of an operation on S.
13.1 Sums Equal to Products in flZ The earliest applications of the theory of compact right topological semigroups to Ram
sey Theory involved the semigroups (N, +) and (N, ). (See Chapters 5 and 17 for several examples.) In investigating these applications, the question arose whether the equation q + p = s r has any solutions in N*, or indeed in V. These questions remain open, but we present 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 XP 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) = q p and 2p(q) = p q.
IfpEfZ, p denotes Lemma 13.1. Let n E 7G and let p, q E PZ. Then n
(p +q)
(p + q) = n p + n q and
n.
Proof. Since, by Theorem 4.23, 7L is contained in the center of (fZ, ), the second
assertion follows from the first.
Now n (p + q) = (en o pq) (p) and n p + n . q = (pn.q o en) (p) so it suffices to show that the functions in o Pq and pn.q o in agree on Z. So let m E Z be given. Then
and
(Pn.goen)(m)=nm+nq=()nmofn)(q)
13.1 Sums Equal to Products in fiZ
Since the distributive law holds in Z, the functions t, o 1. and
259
o en agree on Z.
We have occasionally used the canonical mapping from Z to Z before. In this chapter it will be used in several proofs, so we fix some notation for it.
Definition 13.2. Let n e N. Then hn : Z + Z denotes the canonical mapping. Notice that the continuous extension W. : fiZ + Zn of hn is a homomorphism both from OZ, +) to (Zn, +) and from (f Z, ) to (Zn, ) (by Lemma 2.14).
Definition 133. Let b be a real number satisfying b > 1. We define 4b : N  co by putting 'b(n) = [logb(n)J. As usual, b : fiN o fim denotes the continuous extension of ¢b.
L e m m a 13.4. Let p E N* and let b > 1.
(i)b(q+P) EOb(P)+{1,0, 1}forevery q E fZ; (ii)
E b(q) + Ab(p) + (0, 1} for every q E #N.
Proof. (i) Choose E > 0 with the property that I logb(1 + t) I < 1 if t E (E, e). If m E 7G, n E N, and Imp < En, we have logb(m +n) = logb n +logb(1 + n) and hence 4Pb(m + n) E Ob(n) + 11, 0, 1).
For each i E (1, 0, 1), we define M, C Z by putting in E M; if in E N Ob(m+n) _ ob(n)+i} E P. SincethesetsM,partition Z,wecanchoose j E (1,0, 1) such that Mj E q. For each in E Mj, let N. = {n E N : ob(m + n) _ ¢b (n) + j) E p. We have:
Ob(q + p) = q lim p lim (Pb(m + n) mEMj
nEN,,,
= q lint p lim (Ob(n) + j) mEMj
nEN,,,
= ob(P) + j.
(ii) This follows in the same way from the observation that, for every m, n E N, cbb(mn) E vPb(m) + Ob(n) + (0, 11.
Theorem 13.5. Let P E Z*, let q E f87G, let n E Z and assume that n  p = q + p. Then n = 1 and thus p = q + p. If q E Z, then q = 0. Proof. We may assume that p E N*, since otherwise, by Lemma 13.1, we could replace
p by p and q by q. Now by Exercise 4.3.5, q + p E N*. And trivially 0 p = 0, while n E N implies that n p E N*. Thus n E N. Suppose that n > 1. We can then choose b E 1 satisfying 1 < b2 < n. So ¢b(n) > 2. It follows from Lemma 13.4 that.b(n  p) = Ab(p) + k for some k > 2 in N. On the other hand, ¢b(q + p) E ob(p) + {1, 0, 1). By Lemma 6.28, this is a contradiction.
It is easy to find solutions to the equation q + p = n r where n E Z. The next two results are concerned with multiple solutions to such an equation for fixed p and r. Such multiple solutions are used in Theorem 13.9 to characterize the existence of q and
sinZ*with q+p=sr.
13 Multiple Structures in 8S
260
Lemma 13.6. Let p, r E 7G*, let s, t E ,8Z, and let m, n E Z. Ifs + p = m r and
t+p= n r, then m=n. Proof. We may suppose that p E N*, since otherwise, by Lemma 13.1, we could replace
p, s, t, and r by p, s, t, and r respectively. This implies that s + p and t + p are in N*, by Exercise 4.3.5, and hence that in ¢ 0 and n 0 0. Therefore r, m, and n are all in fl N or all in ,6N. We may suppose that r, m, n E fiN, since otherwise we could replace r, 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 1l satisfying b > 1 and mb4 < n. So ¢b(n) > cbb(m) + 4. By Lemma 13.4, ¢b (n r) = ob (m r) + k for some k > _3 in N. We also know that ob (s + p) and b (t + p) are both in,b (p) + {1, 0, 1). So ¢b (t + p) = Ob (s + p) + f for some f E {2, 1, 0, 1, 2). Since f ¢ k, this contradicts Lemma 6.28. This establishes that n = m. Lemma 13.7. Let p, r E 7G* and letm, n, m', n' E Z. Ifm+p = n r and m'+p = n'r, then (m, n) = (m', n'). Proof. By Lemma 13.6, n = n'. It then follows from Lemma 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 V. Now, of course, an affirmative answer to the first 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 E Z* and there exists q E Z* such that q + p = s r, then there exists q' E Z* such that q' + I pI = IsI Ir1, where I p I 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 q + p = s  r but not have any q' E N* such that q' + p = s r. Accordingly, we address both questions.
Lemma 13.8. Let p, q, r E 7L* and let H = {m E 7Z : there exists t E i7G such that in + p = t r}.
There exists s E Z* such that q + p = s  r if and only if H E q.
Proof. Necessity. Picks E Z* such that q + p = s  r. We note that, by Lemma 13.6, there is at most one a E 7L for which a r E f 7L+ p. Thus, if S = in E Z : n r 0 ,67L+p},
we have S E S. Suppose that H 0 q and let B = Z\H E q. Since q + p E cf(B + p) ands r E c2 (S r), it follows from Theorem 3.40, that m + p E #Z r for some m E B, or else n r E f87G + p for some n E S. The first possibility can be ruled out because it implies that m E B fl 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 E 7L : m + p c Z*  r). Then H' E q and hence
q + p E cPSz(H' + p) 9 cfpz(Z* r) = Z*  r.
13.1 Sums Equal to Products in PZ
261
Theorem 13.9. Let p, r E Z*, let G = {q E Z* : there exists s E Z*such that q + p = s r} and let
H = {m E 7L : there exists t E ,8Z such that m + p = t r}. Then the following statements are equivalent.
(a) G#0. (b) IHI = co.
(c) IGI = 2` Proof. (a) implies (b). Pick q E G. By Lemma 13.8, H E q so, since q E Z*, I H I = Co. (b) implies (c). By Theorem 3.58, I{q E 7Z* : H E q}I = 2` so by Lemma 13.8, IGI = 2c. That (c) implies (a) is trivial.
Theorem 13.10. Let p, r E N*, let G = {q E N* : there exists s E N* such that q + p = s r} and let
H = {m E N : there exists t E ON such that m + p = t r}. Then the following statements are equivalent.
(a)G00. (b) IHI = w. (c) IGI = 2`. Proof. (a) implies (b). Pick q E G. By Lemma 13.8, {m E 7L : there exists t E fl Z such that m + p = t r} E q. Since also N E q, one has
{m E N : there exists t E ,87Z such that m + p = t r} E q. Now, given m E N and t E OZ such that m + p = t r, one has m + p E N*, so one also must have t E ON. (Otherwise t r E N* U {0}.) Thus H E q and therefore I H I= co. (b) implies (c). By Theorem 3.58, 1{q E N* : H E q}I = 2`. Further,
H C {m E 7G : there exists t E PZ such that 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. Since r and q + p are in N*, s EN*, sothatq E G. To put Theorems 13.9 and 13.10 in context, notice that we do not know whether the set H ever has two members, let alone infinitely many members. Notice also that if H,
p, and r are as in Theorem 13.10, then H + p SON r. Theorem 13.12 characterizes the existence of such H.
13 Multiple Structures in fiS
262
Definition 13.11. Let 0 0 H c N. Then
TH = (A c N : there exists F E [H]`w such that N = UnEF(n + A)). Theorem 13.12. Let H C N with CHI > 1. Then there exist p, r e N* such that H + p S ON r if and only if there is a choice of xA E N for each A E TH such that (XA' A : A E TH ) has the infinite finite intersection property.
Proof Sufficiency. Pick by Corollary 3.14 r E N* such that {xA' A : A E TH } C ,Recall that C(r) = {A C N : for all x E N, x1A E r} and that, by Theorem 6.18,
ON  r = (p E ON : C(r) C p). We claim that {n + B : B E C(r) and n E H) has the finite intersection property. Since C(r) is a filter, it suffices to let B E C(r) and F E .Pf (H) and show that lfEF (n + B) # 0. So suppose instead that InEF (n + B) = 0. Let A = N\B. Then A E TH so xA' A E r so B 0 C(r), a contradiction. Pick p E ON such that (n + B : B E C(r) and n E H) C p. Then for each n E H, C(r) C n + p so by Theorem 6.18, n + p E ON r. Since, by Corollary 4.33, I
ON  r c N*, it follows that p E N*. Necessity. Pick p and r in N* such that H + p S ON  r. Given A E TH, it suffices to show that there is some xA E N such that XA ' A E r. So let A E TH and pick
F E [H]`w such that N = UnEF (n + A). Pick n E F such that n + A E p and picks E fNsuchthatn+ p =sr Then A E n+ p = s  r so {x E N : x'A E r} E s so pick xA E N such that xA' A E r.
By Lemma 13.8 and Theorem 13.10, if there exist p, q, r, and s in N* with q + p = s r, then (q E N* : q + p = s r for some p, s, r E N*) has nonempty interior in N*. By way of contrast, the set of points occupying any of the other positions of such an equation must be topologically small.
Theorem 13.13. Let
q,s,rEN*), and Then each of P, S, and R is nowhere dense in N*.
Proof. To see that P is nowhere dense in N*, we show that P C Z + N* N*. (We have that N*  N* is nowhere dense in N* by Theorem 6.35 and so 7G + N* N* is nowhere dense in N* by Corollary 3.37.) Let P E P and pick q, s, and r in N* such that q + p = s r. By Lemma 13.8, {m E N : m + p E ON r) E q and by Lemma 13.7 at most one m E N has m + p E N r. Thus P E Z + N* N* as claimed. To see that S is nowhere dense in N*, suppose instead that there is an infinite subset B of N such that B* C c1pu S and choose x E B*. Choose a sequence (bn),°,° in B with the property that h. (bn) = h , , , (x) f o r every n E N and every m E 11, 2, ... , n). Let A = (bn : n E N). Then for every s E At and every m E N, h,n (s) = h,n (x).
263
13.1 Sums Equal to Products in 147E
Pick S E A* n S and pick q, p, and r in N* such that q + p = s  r. By Lemma 13.6, there is at most one n E N for which n  r E ON + p. If such an n exists, let C = A\{n}, and if not let C = A. Let D = {m E N : m + p E C* r}. We claim that D E q so suppose instead that N\ D E q. Then q + p E (N\D) + p and s r E C rr so by Theorem 3.40 either n  r E (N\D) + p for some n E C or m + p = s'  r for some m E N\D and some s' E C*. Both possibilities are excluded and so D E q as claimed. Choose m, k E D with k < m. Now hm is constant on C* r, so hm (m +p) = hm (k+ p) and so hm (m) = hm(k), a contradiction. To see that R is nowhere dense in N* we show first that R C Un EN en  t [7G+N*  N* ]
Let r E R and let q + p = s  r, where q, p, s E N*. By Lemma 13.8, we can choose m < m' in N such that m + p = x  r and m'+ p = y r for some x, y c= ON. Putting
a =m'm, wehave
y  r. Nowa+x  r E a+N  randy  r E N.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 or y E N. By Lemma 13.7, this equation cannot hold with x and y both in N. (Let So
there exists n E N for which n r E Z + N* N*. Thus R S
ins [Z + N* N*]
as claimed. As we have observed, Z + N* N* is nowhere dense in N*. Now if M is a nowhere dense subset of N* and n E N, then enI [M] is also nowhere dense in N*. Otherwise,
if fn 1 [M] were dense in A* for some infinite subset A of N, M would be dense in f.
[A*] = (n A)*. So UnEN in 1 [7G+N* N*] is nowhere dense in N* by Corollary 3.37. 13
The following result establishes that one cannot find p, q, r, and s in Z* with q + p = s  r in the most familiar places.
Theorem 13.14. Letp,q,r,sE7G*. Ifs{aEN:aZEr}I=ce,then q+p# s r. Proof. Suppose instead that q + p = s r. By Theorem 13.9, pick m ZA n in Z such that
m+p E
Pick a > max{Imp,In k}such that aZEr. Then ha (r) = 0 and so ha (m + p) = ha (n + p) = 0. This implies that ha (m) = ha (n), a contradiction.
Corollary 13.15. K(flN, ) n K(flN, +) = 0.
Proof Suppose one has some r E K(fN, ) n K(6N, +). Pick a minimal left ideal L of (ON, ) such that r E L. Then by Lemma 1.52, L = Lr so r = s  r for some s E L C N*. Similarly r = q + r for some q E N*. But nn° 1 nN is an ideal of (ON, ). Thus K(fN, ) e nn° , nN and consequently r E n00 nN. But then by 1
Theorem 13.14, q + r
s r, a contradiction.
By way of contrast with Corollary 13.15, we shall see in Corollary 16.25 that K(flN, ) n ce K(14N, +) 96 0. We conclude this section with a result in a somewhat different vein, showing that a certain linear equation cannot be solved.
13 Multiple Structures in fl S
264
Lemma 13.16. Let u # v in 7L\{0}. There do not exist p E I that
nN and q E N* such
uq + p = vq + p. Proof. Suppose we have such elements p and q and pick a prime r > lu I + I v I. For i E (1,2..... r  11, let C1 = {r' (rk + i) : n, k E w}. Then C1 is the set of x E N whose rightmost nonzero digit in the base r expansion is i. Pick i such that C, E q. For each m E N, let f (m) denote the largest integer in co for which rf (m) is a factor
of m. Since rf (')+1 N E p, it follows from Theorem 4.15 that {um + s : m E C, and
s E r f(m)+1N} E uq + p and {vn + t : n E C, and t E
rf(n)+1 N}
E vq + p. Thus
there exist m,n E Ci,s E rf(m)+1 N, and t E rf(n)+1N such that um+s = vn + t. We observe that f (um + s) = f (m), because rf(m) is a factor of um + s, but rf(m)+1 is not. Similarly, f(vn + t) = f (n). So f (m) = f (n) = k, say.
We have m = irk (mod rk+1) and n = irk (mod rk+1) So um + s = uirk This implies that u = v (mod r), a (mod rk+1) and vn + t = virk (mod rk+1).
contradiction.
Theorem 13.17. Let u # v in 7L\{O}. There do not exist p, q E N* such that
uq + p = vq + p. Proof. Suppose we have such elements p and q. Pick r E K (MN, +). Then p + r E K (8N, +) so by Theorem 1.64 there is some maximal group G in K (f N, +) 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 + p = vq + p, one obtains uq + e = vq + e. This contradicts Lemma 13.16, because hn (e) = 0 for every n E N, and so e E I InEN nN. Exercise 13.1.1. Given p E ,BZ, define
IPIpp if ifpEfiN p E f7L\j9N. Prove that if p, r, s E Z* and there exists q E 7L* such that q + p = s r, then there exists q' E Z* such that q' + ; p I = Is I
 IrI
13.2 The Distributive Laws in flZ We saw in Lemma 13.1 that if n E Z and p, q E flZ, then n (p +q) = n p +n . q. This is the only nontrivial instance of the distributive law known to hold involving members of Z*. We establish in this section that any other instances, if they exist, are rare indeed.
Theorem 13.18. Let n E 7L\{0} and let k, m E Z. The following statements are equivalent.
13.2 The Distributive Laws in flZ
265
(a) There exists p E 7L* such that m p + n p = k p. (b) k = n and either m = 0 or m = n. Proof. (a) implies (b). Assume that we have p E Z* 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 N* and n, k E N. We show first that k = n so suppose instead that k n. Choose a real n_umberb > 1 forwhich nb3< korkb3 < n. So Icbb(n)Ob(k)I > 3. ByLemma 13.4, cbb(m p+n p) E 4)b(n)+b(p)+{1, 0, 1, 2} and, b(k p) E Ob(k)+cb(p)+{0, 1}. It follows from Lemma 6.28, that 10b(M)  Ib (n) I < 2. This contradiction shows that
k = n. Suppose now that m 0. Let r E N be a prime satis Ding r > Imp + n. For any a E N, we have hra (m)hra (p) + hra (n)hra (p) = hra (n)hra (p). Since r is not a factor of Im I, it follows that hra (p) = 0. That is, r' N E p for each a E N.
Fori E {1,2,...,r1},let A; = {rt(tr + i) : 1, t E w}. (Thus A, is the set of positive integers whose rightmost nonzero digit in the base r expansion is i.) Pick i E { 1 , 2, ... , r  1} for which A, E p. For each x E N, let f (x) denote the largest integer in w for which rf (x) is a factor of x. We have {mx + s : x E A, ands E rf (x)+1 N} E m p + n p (by Theorem 4.15) and nA; E n p. Thus there exist x, y E A; and S E rf (x)+1 N for which mx + s = ny.
We note that f (mx + s) = f (x) and f (ny) = f (y). So f (x) = f (y) = f, say. Now x = ire (mod re+t) and y  in (mod re+1). So mx + s  mire (mod re+t) and ny  nir1 (mod re+1). It follows that m  n (mod r) and hence that m = n. (b) implies (a). Assume that k = n. If in = 0, then m p+n p = k p for all p E PZ. If m = n and p + p = p, then by Lemma 13.1,n p+n p
p
( p) + p.
Proof. This is immediate from Theorem 13.18.
Corollary 13.20. Let p E Z* and let n, m E Z\{0}. Then (n + m) p # n p + m p Proof Since, by Theorem 4.23, 7G is contained in the center of (f67Z, ), it suffices to
show that (n +m) p
n p +m p. Suppose instead that (n + m) p = n p +m p.
Then by Theorem 13.18, it + m = m, son = 0, a contradiction. Theorem 13.18 has special consequences for the semigroup H.
Theorem 13.21. Let q, r E Handler p E V. Then p (q + r) 0 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 H.
13 Multiple Structures in ,8S
266
Proof. By Lemma 6.8, q + r E IIii, so by Theorem 13.14, p (q + r) f Z* + 7C. Since (Z\{0}, ) is cancellative, we have by Corollary 4.33 that Z is a subsemigroup
Similarly (p + q) r 0 Z* +Z* and p r + q r E Z* +Z*.
13
We now set out to establish the topological rarity of instances of a distributive law in N*, if indeed any such instances exist at all.
Definition 13.22. Let A C N. A is a doubly thin set if and only if A is infinite and whenever (m, n) and (k, 1) are distinct elements of N x co, one has I (n+mA)fl (l+kA) I < w.
Lemma 13.23. Let B be an infinite subset of N. There is a doubly thin set A c B.
Proof Enumerate N x co as ((m(k), n(k)))k°1 and pick s( E B. Inductively let k E N and assume that sI , sZ, ... , sk have been chosen. Pick sk+t > sk such that sk+1 E B
jt
st m(l)+n(l)n(i) i,l,t E (1,2,...,k) }. m(i)
LetA=(sk:kEN}. Then A is infinite. Suppose that one has some 10 i such that
1(n(l) +m(1)A) fl (n(i) +m(i)A)I = w. Now, since (m(l), n(1)) gE (m(i), n(i)), there is at most one x E A such that n(l) + m(l)x = n(i) + m(i)x. If there is such an x = s,,, let b = max{v, i, l}, and otherwise let b = max(i, l).
Pick Z E (n(l) + m(1)A) fl (n(i) + m(i)A) with z > n(l) + m(l)sb and z > n (i) + m (i )sb and pick t, k E N such that z = n (l) +m (1)s, = n (i) + m (i )sk and notice
that t > b and k > b. Then s, # sk so assume without loss of generality that t < k. But then sk
n(l)+m(l)st  n(i) =
m(i)
a contradiction.
Lemma 13.24. Let A be a doubly thin subset of N, let p E A*, and let B C N x w. Then cE{n + m p : (m, n) E B} fl cE{n + in p : (m, n) E (N x w)\B} = 0.
Proof Supposethatq E
(m, n) E
(m, n) E (Nx(0)\B}.
Order N x w by < in order type a). (That is, < linearly orders N x co so that each element has only finitely many predecessors.) Let
C = U(m,n)EB ((n + mA) \ U(m',n')<(m,n) (n' + m'A)) and let
/
D = U(m,n)E(Nxw)\B 1(n + mA) \ U(m',n')<(m,n) (n' + m'A)).
13.2 The Distributive Laws in,6Z
267
Notice that c fl D = 0. Thus without loss of generality we have N\C E q. Pick
(m,n)EBsuch that and for every (n', m') < (n, m), I(n' + m'A) fl (n + mA)I < w since A is doubly thin. Thus
(n' + m'A) E n +m p
(n +mA) \ so C E n+ m p, a contradiction. We have one final preliminary.
Lemma 13.25. Let A be a doubly thin subset of N and let p E A*. Then N* N* + N* p, and N* (N* + p) have pairwise disjoint closures.
p,
Proof. By Lemma 13.24, it suffices to observe that
and
n>m}.
C c2in
The first observation is immediate. For the second, just notice that if q, r E N*, then nEN
For the last observation, let q, r E N*, let B E q (r + p), and pick m E N such
that m1 B E r + p. Then {n E N : n + m1 B Ep} E r so pick n such that
n+m'BEp.Then
Theorem 13.26. The set (P E N* : N* p, N* + N* p, and N* (N* + p) have pairwise disjoint closures} has dense interior in N*.
Proof. Let B be an infinite subset of N. Pick by Lemma 13.23 a doubly thin subset A of B. Then apply Lemma 13.25.
Corollary 13.27. The set (P E N* : for all q, r E N*, (q + r)  p
r (q + p) # r q + r p} has dense interior in N*. Proof. Given p, q, r E N*,
(q +
E N* E
so Theorem 13.26 applies.

q p + r p and
13 Multiple Structures in PS
268
13.3 Ultrafilters on }I8 Near 0 We have dealt throughout this book with the Stonetech compactification of a discrete space, especially a discrete semigroup. One may naturally wonder why we restrict our attention to discrete spaces, especially given that the Stonetech compactification exists for a much larger class of spaces. We shall see in Theorem 21.47 that for many
nondiscrete semigroups S, including (R, +), the Stonetech 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 proceed as we always have. It would seem that by doing that, one would lose any information about the original topological semigroup. We shall see however in Section 17.5 that we can get information about partitions of the real interval (0, oo), every member of which is measurable, or every member of which is a Baire set, by first giving (0, oo) the discrete topology and passing to its Stonetech compactification. Because of these interesting applications, we shall investigate the algebra of (,BIIld, +) and (8Rd, ), where Rd is the set IIl with the discrete topology. We shall be primarily concerned with (,BIIld, +).
Definition 13.28. (a) Given a topological space X, Xd is the set X endowed with the discrete topology.
(b) Let a : ,JRd
[oo, oc] be the continuous extension of the identity function.
(c) B(R) _ {p E find : a(p) 0 {oc, oo}}. (c) Given X E IIl,
x+= {pEB(JR)a(p)=xand(x,00)Ep}and x = {p E B(R) : a(p) = x and (oo, x) E p}. (d) U = UXER x+ and D = UXEIIB XThe set B(R) is the set of "bounded" ultrafilters on R. That is, an ultrafilter p E 6 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 13.29. (a) Let X E IIl and let p E 14II8d. Then p E x+ if and only if for every E > 0, (X, X + E) E p. Also, p E X if and only if for every E > 0, (x  E, X) E p.
(b) Let p, q E B(R), let x = a(p), and let y = a(q). If P E x+, then p + q E (x + y)+. If p E x, then p + q E (x + y). (c) B(IR)\IR = U U D and U and D are disjoint right ideals of (B(IR), +). In particular, B(R) is not commutative. (d) The set 0+ is a compact subsemigroup of (B(R), +).
(e)Ifx E Rand p E,BIIBd, then x+p = p+x. (f) The set 0+ is a two sided ideal of the semigroup ($ (0, 1)d, ). Proof. The proof of (a) is a routine exercise, (e) is a consequence of Theorem 4.23, and (c) and (d) follow from (b). We establish (b) and (t).
To verify (b) assume first that p E x+. To see that p + q E (x + y)+, let E > 0 be given and let A = (x + y, x + y + E). To see that A E p + q, we show that
13.3 Ultrafilters on R Near 0
269
(X,x+E) c {z ER : z+A E q}. Soletz E (x, x+e). LetS = min{zx, x+ez}. Since a (q) =y, we have (y  S, y + S) E q and (y  S, y + S) C z+ A so z+ A E q as required. The proof that p + q E (x + y) if p E x is nearly identical. To verify (f), let p E 0+ and let q E 0 (0, 1)d. To see that p q E 0+ and q p E 0+,
let E > 0 be given. Since (0 E) (0, 1) = (0, E), we have (0, E) C (x E (0, 1) : x1(0, E) E q} (so that (0, E) E p q) and (0, 1) C {x E (0, 1) : x' 1(0, E) E p} (so that (0, E) E q p). We shall restrict our attention to 0+, and it is natural to ask why. On the one hand, it is a subsemigroup of (8Rd, +) and for any other x E R, x+ and x  are not semigroups. On the other hand we see in the following theorem that 0+ holds all of the algebraic structure of (B(R), +) not already revealed by R.
Theorem 13.30. (a) (0+, +) and (0, +) are isomorphic. (b) The function rp : Rd x ({0} U 0+ U 0) + B(R) defined by cp(x, p) = x + p is a continuous isomorphism onto (B(R), +). Proof. (a) Define r
: 0+ + 0 by r(p) = p, where p = {A : A E p}.
It
is routine to verify that r takes 0+ onetoone onto 0. Let p, q E 0+. To see that r (p +q) = r (p) + r (q) it suffices, since r (p +q) and r (p) + r (q) are both ultrafilters,
to show that r(p+q) c r(p) + r(q). So let A E r(p + q). Then A E p + q so
B={x ER:x+AEq}Ep and hence B E r(p). Then B C {x E R : x + A E r(q)} so A E r(p) + r(q) as required. (b) To see that q is a homomorphism, let (x, p) and (y, q) be in Rd x ({0} U0+ U0).
Since p + y =y+p,wehave rp(x, p)+'p(y,q) ='P(x+y, p + q). To see that cp is onetoone, assume we have 'p (x, p) = V (y, q). By Lemma 13.29(b)
wehavex=a(x+p)=a(y+q)=y.Thenx+p=x+q sop=x+x+q=q. To see that 'p is onto B(R), let q E B(R) and let x = a(q). Let p = x + q. Then
q=x+p='P(x, p).
To see that rp is continuous, let (x, p) E Rd x ({0} U 0+ U 0) and let A E x + p. Then x + A E p so {x} x (cf#Rd (x + A)) is a neighborhood of (x, p) contained in c1 [cf $RdA].
As we have remarked, 0+ has an interesting and useful multiplicative structure. But much is known of this structure because, by Lemma 13.29(t), 0+ is a two sided
ideal of (f(0, 1)d, ), so K(f(0, 1)d, ) C 0+ and hence by Theorem 1.65, K(0+,) _ K(48(0, 1)d, ')On the other hand, 0+ is far from being an ideal of (B(S), +), so no general results apply to (0+, +) beyond those that apply to any compact right topological semigroup. We first give an easy characterization of idempotents in (0+, +), whose proof we leave as an exercise.
Theorem 13.31. There exists p = p + p in 0+ with A E p if and only if there is some 9 A. 1 in (0, 1) such that ER_I x converges and
sequence
270
13 Multiple Structures in SS
Proof. This is Exercise 13.3.1.
As a compact right topological semigroup, 0+ has a smallest two sided ideal by Theorem 2.8. We now turn our attention to characterizing the smallest ideal of 0+ and its closure. If (S, +) is a discrete semigroup we know from Theorem 4.39 that a point p E 16S
is in the smallest ideal of fS if and only if, for each A E p, (x E S : x + A E p) is syndetic (and also if and only if for all q E fS, p E PS + q + p). We obtain in Theorem 13.33 a nearly identical characterization of K(0+, +).
Definition 13.32. A subset B of (0, 1) is syndetic near 0 if and only if for every e > 0
there exist some F E ?f((0, e)) and some 8 > 0 such that (0, 8) C U:EF(t + B). Theorem 1333. Let p E 0+. The following statements are equivalent.
(a) p E K(0+,+). (b) For all A E p, {x E (0, 1) : x + A E p) is syndetic near 0.
(c) For allrE0+, pE0++r+p. Proof (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 e > 0 such that for all F E .Pf ((0, e)) and all 8 > 0,
(0, 8)\ UIEF(t + B) 96 0. Let = ((0, 8)\ U,EF(t + B) : F E R f ((0, e)) and 8 > 0}. Then g has the finite intersection property so pick r E P(0, 1)d with g e r. Since {(0, 8) : 8 > 015; r we have r E 0+. Pick a minimal left ideal L of 0+ with L C 0+ + r, by Corollary 2.6. Since K(0+) is the union of all of the minimal right ideals of 0+ by Theorem 2.8, pick a minimal right ideal R of 0+ with p E R. Then L n R is a group by Theorem 2.7, so let q be the identity
of L n R. Then R= q +O+, so p E q +O+ sop = q+ p so B Eq. Also q E 0++r so
pick wE0+such that q=w+r.Then (0,e)Ewand{tE(0,1):t+BEr)Ew so pick t E (0, e) such that t + B E r. But (0, 1)\(t + B) E g C r, a contradiction. (b) implies (c). Let r E 0+. For each A E p, let B(A) = {x E (0, 1) : x + A E p} and let C(A) = It E (0, 1) : t + B(A) E r}. Observe that for any A1, A2 E p, one has B(A1 n A2) = B(A1) n B(A2) and C(A1 n A2) = C(Ai) n C(A2).
We claim that for every A E p and every e > 0, C(A) n (0, e) # 0. To see this, let A E p and e > 0 be given and pick F E Pj((0, E)) and 8 > 0 such that (0, 8) C U,EF(t + B(A)). Since (0, 8) E r we have U,EF(t + B(A)) E r and hence there is some t E F with t + B(A) E r. Then t E C(A) n (0, e). Thus ((0, e) n C(A) : E > 0 and A E p} has the finite intersection property so pick q E f(0, 1)d with {(0, e) n C(A) : E > 0 and A E p} C q. Then q E 0+. We claim that p = q + r + p for which it suffices (since both are ultrafilters) to show that
p e q + r + p. Let A E p be given. Then {t E (0, 1) : t + B(A) E r} =C(A)E q so B(A) E q + r, so A E q + r + p as required. (c) implies (a). Pick r E K(0+). 
13.3 Ulirafilters on R Near 0
271
If (S, +) is any discrete semigroup, we know from Corollary 4.41 that, given any discrete semigroup S and any p E 13 S, p is in the closure of the smallest ideal of PS if and
only if each A E p is piecewise syndetic. We obtain a similar result in 0+. Modifying the definition of "piecewise syndetic" to apply to 0+ is somewhat less straightforward than was the case with "syndetic". Definition 13.34. A subset A of (0, 1) is piecewise syndetic near 0 if and only if there exist sequences (F,,) n° t and (Sn)n° t such that
(1) for each n E N, Fn E Pf ((0, 1/n)) and S E (0, 1/n), and (2) for all G E Pf((0, 1)) and all µ. > 0 there is some x E (0, µ) such that for all n E N, (G n (0, s,)) +x C U:EFF(t + A).
Theorem 13.35. Let A C (0, 1). Then K(0+) n C 0(0,lld A 0 0 if and only if A is piecewise syndetic near 0.
Proof. Necessity. Pick p E K(0+)nct$(o,1)d A and let B = {x E (0, 1) : x+A E p}. By Theorem 13.33, B is syndetic near 0. Inductively for n E ICY pick Fn E Pf ((O, 1 /n))
and Sn E (0,11n) (with S < 6n_1 if n > 1) such that (0, Sn) c UIEF (t + B). Let G E Pf ((0, 1)) be given. If G n (0, 31) = 0, the conclusion is trivial, so assume
Gn(0,Si)#0andletH=Gn(0,S1).Foreach yEH,let m(y)=max{n EN:y <Sn}. For each y E H and each n E {1, 2, ... , m(y)}, we have y E
B) so pick
t(y,n) E Fn such that y E t(y,n)+B. Then given y E H andn E {1,2,...,m(y)}, one has (t(y, n) + y) + A E P. Now let µ > 0 be given. Then (0, µ) E p so pick
x E (0, µ) n nyER nn (i)((t(y, n) + y) + A). Thengivenn E Nandy E Gn(O,Sn),onehasy E Handn < m(y)sot(y,n)+y+x E A so
y+x E t(y,n)+A c Sufficiency. Pick (F,,)', and (Sn)n_i satisfying (1) and (2) of Definition 13.34. Given G E Pf((0, 1)) and µ > 0, let
C(G, µ) = {x E (0, IL) : for all n E N, (G n (0, Sn)) + x c U:EF, (t + A)).
By assumption each C(G, µ) 96 0. Further, given Gt and G2 in Rf ((0, 1)) and µl,µ2 > 0, one has C(Gi U G2, min{µl, µ2}) c C(GI, µi) n C(G2, µ2) so {C(G, µ) : G E Pf((0, 1)) and p, > 0) has the finite intersection property so pick p E 6(0, 1)d with {C(G, µ) : G E Pf ((0, 1)) and 1L > 0) c p. Note that since each C(G, A) C (0, µ), one has p E 0+.
272
13 Multiple Structures in PS
Now we claim that for each n E N, 0+ + p S A)), so let n E N and let q E 0+. To show that UtEF (t + A) E q + p, we show that (0, Sn) C (y E (0, 1) : y + UtEF (t + A) E p}.
So let y E (0, &). Then C({y}, Sn) E p and C({y}, SO c y + UtEF, (t + A).
Now pick r E (0+ + p) fl K(0+) (since 0+ + p is a left ideal of 0+). Given n E N, (t + A) E r so pick t E Fn such that tn + A E r. Now each to E Fn C (0, 1/n) so nlimntn = 0 so pick q E0+ f1 ctf(o,1)d{tn : n E N}. Then
q+rEK(0+)and {to:nENJ C(tE(0,1):t+AEr)so AEq+r. Since (0+, +) is a compact right topological semigroup, the closure of any right ideal is again a right ideal. Consequently cfo+ K(0+) = c2B(o,I)d K(0+) is a right ideal of 0+. On the other hand, if S is any discrete semigroup, we know from Theorem 4.44 that the closure of K(PS) is a two sided ideal of OS. We do not know whether cto+ K(0+) is a left ideal of 0+, but would conjecture that it is not.
Exercise 13.3.1. Prove Theorem 13.31. (Hint: Consider Lemma 5.11 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 (flS, ) 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 S and P E PS we can define pox = p lim y x (so yES
that if p E S, then pox = p x) and given p, q E PS,
Then (PS, o) is a compact left topological semigroup and for each x E S, the function rx : ,BS fS defined by rx (p) = pox is continuous. Further, Theorems 2.7 through 2.11 apply to (OS, o) with "left" and "right" interchanged. Notice that, given p, q E 46S and A C S,
A E p o q if and only if {x E S : Ax1 E P) E q, where AxI = {y E S : yx E Al. Definition 13.36. Let (S, ) be a semigroup. Then o is the extension of to 48S with the property that for each p E ,6S, the function £ is continuous, and for each x E S, the function rx is continuous, where for q E 46S, fp(q) = p o q and rx(q) = q o x.
13.4 Left and Right Continuous Extensions
273
We see that if S is commutative, then there is no substantive difference between
($S, ) and ($S, o). Theorem 13.37. Let (S, ) be a commutative semigroup. Then for all p, q E i9S, p o q = q p. In particular K($BS, ) = K($S, o) and, if H is a subset of 48S, then H is a subsemigroup of ($S, ) if and only if H is a subsemigroup of ($S, o). Proof. This is Exercise 13.4.1. We show now that if S is not commutative, then both of the "in particular" conclusions may fail.
Theorem 13.38. Let S be the free semigroup on countably many generators. Then there is a subset H of $ S such that
(1) H is a subsemigroup of ($S, ) and (2) for every p, q E fS, p o g 0 H. Proof. Let the generators of S be { yn : n E N}. For each k E N, let Mk = {Yn(1)Yn(2) . . .Yn(I) :
I E N and k < n(l) < n(2) < . . . < n(l)},
and let H=fk_1c2Mk. To see that H is a subsemigroup of (,8S, ), let p, q E H and let k E N. We need to show that Mk E p q. To see this, we show that Mk C {x E S : X1Mk E q}, so let X E Mk and pick l E N and n(1), n(2), ... , n(l) E N such that k < n(l) < n(2) < < n(l) and x = Yn(1)Yn(2) ... Y,(1) Then Mn(1)+1 E q and Mn(q+1 S X' Mk To verify conclusion (2), let p, q E $S. For each k E N, let Rk = { yk } U Syk U Yk S U Syk S = {X E S : yk occurs in x}.
Then Rk is an ideal of S so by (the leftright switch of) Corollary 4.18, Rk is an ideal of
($S, o). Since Rk fl Mk+i = 0 for each k, one has that p o q 0 H if p E U0 i Rk. So assume that p Uk° 1 Rk. We claim that M1 0 p o q, so suppose instead that
M1 E p o q and let B = {x E S : M1x'1 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 Mixi\ Ui=1 R, E p so pick z E Mix1 \ Uri R1 and pick j E N such that y1 occurs in z. Then j > k so zx 0 Mi, a contradiction.
For the leftright switch of Theorem 13.38, see Exercise 13.4.2.
We see now that we can have K($S, o) 0 K($S, ). Recall that after Definition 4.38 we noted that there are really two notions of "piecewise syndetic", namely the notion of right piecewise syndetic which was defined in Definition 4.38, and the notion of left piecewise syndetic. A subset A of the semigroup S is left piecewise syndetic if and onlyif there is some G E JPf (S) such that for every F E J" f (S) there is some x E S
with x F c UIEG Ati.
13 Multiple Structures in fiS
274
Lemma 13.39. Let S be the free semigroup on two generators. There is a subset. A of S which is left piecewise syndetic but is not right piecewise syndetic. Proof Let the generators of S be a and b. For each n E N, let Wn = (x E S : 1(x) < n), where 1(x) is the length of x. Let A = UR__1 bnWna. Suppose that A is right piecewise syndetic and pick G E .Pf(S) such that for every F E P1(S)thereissomex E S; U:EG t1A. Letm = max{1(t) : t E G}+1
and let F = (a'}. Pick X E S such that amx E U,EG ttA and pick t E G such that tamx E A. Pick n E N such that tamx E bn Wa. Then t = bnv for some v E S U (0) son < 1(t). But now, the length of tamx is at least n + m + 1 while the length of any element of bn Wna is at most 2n+1. Since m > 1(t)+1 > n+ 1, this is a contradiction. To see that A is left piecewise syndetic, let G = (a) and let F E 7'f(S) be given.
Pick n E N such that F c Wn and let x = bn. Then x F a C bn Wna c A so x F C Aat as required. For a contrast to Lemma 13.39, see Exercise 13.4.3
Theorem 13.40. Let S be the free semigroup on two generators. Then
¢0.
00 and
Proof. We establish the first statement, leaving the second as an exercise. Pick by Lemma 13.39 a set A c S which is left piecewise syndetic but not right piecewise syndetic. Then by Theorem 4.40, A fl K(fS, ) = 0 so A fl c2 K(fS, ) = 0. On the other hand, by the leftright switched version of Theorem 4.40, A fl K(f S, o) $ 0. We do not know whether K(PS, o) fl K(9BS, ) = 0 for some semigroup S or even whether this is true for the free semigroup on two generators. The following theorem tells us that in any case the two smallest ideals are not too far apart.
Theorem 13.41. Let S be any semigroup. Then K(fS, o) fl ct K(fS, ) # 0 and 0.
Proof. We establish the first statement only, the other proof being nearly identical.
Let P E K(,3S, ) and let q E K(flS, o). Then p o q E K($S, o). We show that p o q E ce K (PS, ). Given S E S, the continuous functions ps and rs agree on S, hence
onPS,sopos=rs(p)=ps(p)=ps. Thus
poq =
fp(qliEms)
= qlimpos SES = qlimp s. sES
For each s E S, p s E K(j8S, ) c ct K(fiS, ), so p o q E ct K(fiS, ). Exercise 13.4.1. Prove Theorem 13.37.
Notes
275
Exercise 13.4.2. Let S be the free semigroup on countably many generators. Prove that there is a subset H of fl S such that
(1) H is a subsemigroup of (,6S, o) and
(2) for every p, q E ,BS, p q f H. Exercise 13.4.3. Let S be any semigroup and assume that r E N and S = U,=i A;. Prove that there is some i E (1, 2, ..., r} such that A; is both left piecewise syndetic and right piecewise syndetic. (Hint: Use Theorems 4.40 and 13.41.)
Notes Most of the material in Section 13.1 is based on results from [ 124], except for Theorem 13.17 which is based on a result of B. Balcar and P. Kalasek in [13] and on a personal communication from A. Maleki. In [124] it was also proved that the equation q + p = q p has no solutions in N*. That proof is quite complicated so is not included here. Theorem 13.18, Corollary 13.20, and Theorem 13.21 are from [138] and Corollary 13.19 is from [117]. Theorem 13.26 and Corollary 13.27 are due to E. van Douwen in [80].
Most of the material in Section 13.3 is from [143], results obtained in collaboration with I. Leader.
Theorem 13.38 is from [87], a result of collaboration with A. ElMabhouh and J. 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 the powerful Central Sets Theorem and some of its consequences. 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 Our first introduction 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 discrete semigroup S, A n K(fS) 0 0 if and only if A is piecewise syndetic. Thus, in particular, any central set is piecewise syndetic. Theorem 14.1. Let A be a piecewise syndetic set in the semigroup (N, +) and let e E N.
There exista,d ENsuch that {a,a+d,a+2d,...,a+Id} C A. Proof. Let Y = X t_ofBN. Then by Theorem 2.22, (Y, +) is a compact right topological semigroup and if! E X, =0N, then Xx is continuous. Consequently Y is a semigroup compactification of X r=0N. Let
E°={(a,a+d,a+2d,...,a+ed):aENanddENU{0}} and let
I° _ ((a,a+d,a+2d,...,a+ed) : a,d E N). Let E=ctyE
Thus
280
14 The Central Sets Theorem
By Theorem 2123, K(Y) = Xi_0K(,6N). We now claim that E n K(Y) # 0 so that by Theorem 1.65, K(E) = E n K(Y). In fact we show that
(*) if p E K(#N) and P = (p, p...., p), then p E E. To establish (*), let U be a neighborhood of p and fort E {0, 1, ..., t} pick At E p
such that X 0Ar c U. Then ne=0 At EP so nr0 At 0 0 so pick a En't=O A,. Then (a,a,...,a) E Ufl E. Since K(E) = E fl K(Y) = E fl XOK(PN) and I is an ideal of E, we have En X,=0K(,BN) C 1. Now since A is piecewise syndetic, pick by Theorem 4.40 some p E A fl K(,BN) and let (p, p, ... , p). Then by (*) p E E fl X i=OK(fN) C I.
Let U = X ,=OA. Then U is a neighborhood of p so u n 1°
0 so pick a, d E N
such that (a,a+d,a+2d,...,a+td) E U. The above 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 E° contains the diagonal of X0N. Then one concludes immediately that p 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 = Ui=1 A1, 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 from our point of view is to note that fiN = U;=1 Ti so some Ti fl K(fiN) 0 0 so Theorem 4.40 applies. Corollary 14.2 (van der Waerden [238]). Let r E N and let N = U;=1 A. For each
t E N there exist i E {1,2,...,r}anda,d E Nsuch that {a,a+d,...,a+td} C A;. Proof. Some Ai is piecewise syndetic so Theorem 14.1 applies.
In Exercise 5.1.1 the reader was asked to show that the above version of van der Waerden's Theorem implies the apparently stronger version wherein one i is chosen which "works" for all t. In fact the current method of proof yields the stronger version directly.
Corollary 14.3 (van der Waerden [238]). Let r E N and let N = U1=1 Ai. There e x i s t s i E 11, 2, ... , r} such that for each t E N there exists a, d E N such that
{a,a+d,...,a+td} C A;.
Proof. Some A; is piecewise syndetic so Theorem 14.1 applies. One may make the aesthetic objection to the proof of Theorem 14.1 that one requires a different space Y for each length of arithmetic progression. In fact, one may do the proof once for all values oft 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.
14.2 The HalesJewett Theorem
281
As a consequence of van der Waerden's Theorem, we establish the existence of a combinatorially interesting ideal of (ON, +) and of (14N, ), namely the set of ultrafilters every member of which contains arbitrarily long arithmetic progressions.
Definition 14.4. AY = {p E ON : for every A E p and every f E N, there exist
a,d E Nsuch that (a,a+d,a+2d,...,a+id) c A}. Theorem 14.5. 4
is an ideal of (1BIN +) and of (ON, ).
Proof. By Theorems 4.40 and 14.1 K(fiN, +) C AR so AeP 0. To complete the proof, let p E A5 and let q E fiN. Given A E q + p and f E N, pick x E N such
that x+A E P. Pick a, d E Nsuch that {a, a+d,a+2d,...,a+Id} c x + A. Then {x + a,x + a + d,x + a + 2d,...,x + a + Id} C A. Given A E p + q and f E N, pick a, d E Nsuch that (a,a+d,a+2d,...,a+Id) C {x E N: x + A E
q}. Then n1o((a + td) + A) E q so pick x E nf.0((a + td) + A). Then
{x+a,x+a+d,x+a+2d,...,x+a+Id} C_ A.
Given A E qp and f E N, pick x E N such that x1 A E p. Pick a, d E N such that
(a, a+d, a+2d,... , a+Id} C x1 A. Then {xa, xa+xd, xa+2xd, ... , xa+Ixd} C_ A Given A E pq and f E N, pick a, d E N such that (a, a + d, a + 2d, ... , a + Id ) C
{x E N : X1 A E q}. Then n1p(a + td)'A E q so pick x E n1=0(a + td) Then {xa, xa + xd, xa + 2xd,... , xa + Ixd} C A.
A.
Exercise 14.1.1. Using compactness (see Section 5.5) and Corollary 14.2 prove that given any k, I E N there exists n E N such that whenever 11, 2, ... , n} is rcolored, there must be a length f monochrome arithmetic progression.
Exercise 14.1.2. Prove Theorem 14.1 as follows.
Let Y = X 000N. Let E'
{(a, a+d, a+2d.... ) : a E N and d E N U {0} } and let I° _ {(a, a+d, a+2d, ...) : a, d E N}. Let E = cIY E° and let I = cIY I Prove that E is a subsemigroup of Y and that I is an ideal of E. Then prove
(*) if pEK(fN)and p=(p,p,p.... ),then pEE. Pick P E K(ON) fl A, let p = (p, p, p, ...), and show that p E I. Given I E N let
U={q"EY:foralliE{0,1,...,I},q; EA}andshow that UflI'
0.
14.2 The HalesJewett 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.)
14 The Central Sets Theorem
282
Definition 14.6. Let S be the free 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 free 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 member a of A, w(a) is the word in which each occurrence of v is replaced by an occurrence of a. Recall that the free semigroup on the alphabet A was formally defined (Definition 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 free 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 formally that w(a) is the function with the same domain as w(v) such that for i in this domain, if W(V); v w(a), = j w(v)i a if W(V); = U.
Informally, if A = (1,2,3,4,5) and w(v) = M0251 v4, then w(3) = 1333325134 and w(2) = 1322325124. Theorem 14.7. Let A be a finite nonempty alphabet, let S be the free semigroup over A, and let B be a piecewise syndetic subset of S. Then there is a variable word w(v) such that (w (a) : a E A) C B. Proof.
Let f _ IAI and write A = (al, a2, ... , at). Let Y = X f=148S. Denote the
operation of Y (as well as that of S) by juxtaposition. Then by Theorem 2.22, Y is a compact right topological semigroup and if x" E X,=1 S, then Xi is continuous. Thus Y is a semigroup compactification of X e1 S. Let 1° = ((w(ai), W 0 2 ) , w(at)) : w(v) E S(v)} and let
E
I'U{(w,w,...,w): w r= S}.
Let E = cfy E° and let I = cdy V. Observe that E° is a subsenugroup of XI 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. We now claim that EflK(Y) 00 By Theorem 2.23 we have K(Y) = so that by Theorem 1.65, K(E) = E fl K(Y). In fact we show that
(*) if p E K(PS) and p = (p, p, ..., p), then p E E. To establish (*), let U be a neighborhood of p and fort E 11, 2, ... , e} pick Bt E p
such that Xt_IBt c_ U. Then nr=1 Bt E p so nt=1 Bt # 0 so pick w E n,t_1 Bt. Then (w, w, ... , w) E U fl E. Since K(E) = E fl K(Y) = E n X f_1 K(48S) and I is an ideal of E, we have En X ,=i K(48S) C I. Now since B is piecewise syndetic, pick by Theorem 4.40 some
14.3 The Commutative Central Sets Theorem
283
p E B f1 K(fS) and let p = (p, p, ... , p). Then by (*) p E E fl Xi_1 K($S) _c I. Let U = X e1 B. Then U is a neighborhood of p so u fl I` # 0 so pick w(v) E S(v) such that (w(al), w(a2), ... , w(at)) E U. Corollary 14.8 (HalesJewett [113]). Let A be a finite nonempty alphabet, let S be the free semigroup over A, let r E N and let S = U, =1 B; . Then there exist i E (1, 2, ... , r) and a variable word w(v) such that (w(a) : a E A) C B,. Proof. Some B; is piecewise syndetic so Theorem 14.7 applies.
Exercise 14.2.1. Using compactness (see Section 5.5) and Corollary 14.8 prove that given any finite nonempty alphabet A and any r E N there is some n E N such that whenever the length n words over A are rcolored there must be a variable word w(v) such that {w (a) : a E A) is monochrome. Exercise 14.2.2. Derive van der Waerden's Theorem (Corollary 14.2) from the HalesJewett Theorem (Corollary 14.8). (Hint: Consider for example the alphabet (0, 1, 2, 3, 4) and the following length 5 arithmetic progression written in base 5: 20100010134, 20111011134,20122012134,20133013134,20144014134.)
14.3 The Commutative Central Sets 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 first 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 HalesJewett 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 x for a function constantly equal to x.)
Lemma 14.9. Let J be a set, let (D, <) be a directed set, and let (S, ) be a semigroup. Let be a decreasing family of nonempty subsets of S such that for each i E D and each x E Ti there is some j E D such that x Tj C Ti. Let Q = niED cfss Ti. Then Q is a compact subsemigroup of PS. Let and (I,)iED be decreasing families of nonempty subsets of X r EJ S with the following properties:
(a) foreachi E D, I; C E; C X,EJT,, (b) for each i E D and each x' E I, there exists j E D such that x Ej C 1,, and (c) for each i E D and each x E E, \I, there exists j E D such that x . Ej C E, and
 .!j CI,. Let Y = X,EJJ S, let E= n1ED c y E,, and let I= f;ED cfY I,. Then E is a subsemigroup of X,Ej Q and I is an ideal of E. If in addition, either
14 The Central Sets Theorem
284
(d) for each i E D, Ti = S and {a E S : a 0 E; } is not piecewise syndetic, or (e) for each i E D and each a E Ti, a E E,,
then given anypEK(Q),onehaspEEnK(XtEj Q)=K(E)CI. Proof. It follows from Theorem 4.20 that Q is a subsemigroup of 0 S.
By condition (a) and the fact that (I,)iED is decreasing we have 0 0 1 c E C X tEJ Q. To complete the proof that E is a subsemigroup of X (EJ Q and I is an ideal of E, we let p, q" E E and show that p q E E and if either p' E I or q" E I, then p" q" E I. To this end, let U be an open neighborhood of p" .4 and let i E D be given. We show that U fl E; # 0 and if p E I or q E I, then U fl I; 96 0. Pick a neighborhood V of p" such that V q" c U and pick x E E; n V with x E 1; if p E I. If! E I; pick
j E D such that z Ej c I. If x' E E; \I,, pick j E D such that x EJ C_ E; and 2 . I j C Ii. Now 2 g E U so pick a neighborhood W of q such that x W C U and
picky E W n Ej with y E Ij if q E I. Then x" y E U n E, and if either p E I or
4EI,then . 5 Ufll,.
To complete the proof, assume that (d) or (e) holds. It suffices to establish
(*) if p E K(Q), then p E E. Indeed, assume we have established (*). Then p E E f1 X t ,j K (Q) and X tE K (Q) _ K (X tEJ Q) by Theorem 2.23. Then by Theorem 1.65, K (E) = E n K (X tE.t Q) and,
since I is an ideal of E, K(E) C 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 T. Pick F E P" f(J) and for each t E F pick some At E p such that ntEF rri 1 [ceps At] C U. Assume now that (d) holds. Since P E K(88S) and {a E S : a 0 E,} is not piecewise syndetic, we have by Theorem 4.40 that {a E S : a 0 E, } 0 p and hence
{aES:REE,}Ep.Pick aE(n,EFAt)n{aES:dEE,}.Then ii EUnE,. Finally assume that (e) holds. Since p E ce T,, pick a E (n,1F At) fl Ti. Then
aEUfE,.
Definition 14.10. 4) is the set of all functions f : N n E N.
N for which f (n) < n for all
We are now ready to prove the Central Sets Theorem (for commutative semigroups).
Theorem 14.11 (Central Sets Theorem). Let S be a commutative semigroup, let A be a central subset of S, and for each e E N, let (yy,n )n_ 1 be a sequence in S. There exist a sequence (an) in S and a sequence (Hn )n° in .% f (N) such that max H,, < min Hn+1 1
1
for each n E N_and such that for each f E 4', FP((an ntEH Yf(n),t)a1) c A. Proof. Let Y = X o_=, PS. By Theorem 2.22, Y is a compact right topological semigroup and X e01 S C A(Y). Given a E S and H E R f (ICY) let
2(a, H) = (a i 1tEH Yl,t , a M EH Y2,t , a  111EH Y3,t ,
)
285
14.3 The Commutative Central Sets Theorem
(so that x!(a, H) E X e01 S C Y). For each i E N, let T, = S, let I; = {x2(a, H) : a E S, H E Yf (N), and min H > i },
and let E; = I; U {a : a E S}. We claim that Ti, E,, and I, satisfy hypotheses (a), (b), (c), and (d) of Lemma 14.9. Indeed, (a) and (d) hold trivially. To verify (b), let i E N and X E I; be given. Pick a E S and H E Yf (N) such that min H > i and X = !(a, H). Let j = max H + 1. Then, since S is commutative, one sees easily that Ej C Ii. To verify (c), let i E N and z r= E; \I, be given. Pick a E S such that x"
cl;.
x" =i.Then 1E;
Since A is central in S, pick an idempotent p E K (fl S) such that A E p. By Lemma
14.9pEI=niENcayII. We now construct the sequences (an)n°O_1 and (H,,)n° inductively. The construction
_
1
is reminiscent of the proof of Theorem 5.8.
Let A* = A*(p) = {x E A : x1A E p} and let U = {q E Y : ql E A*}. By Theorem 4.12 A* E p so U is a neighborhood of p. Since also p E I, pick z' E U n I,.
Pick al E S and H1 E Pf(N) such that z' = z(al, Hl). Since x(al, Hl) E U we have that al n > 1 and assume that we have chosen (am)m 11 and (Hm)m=1 so that, if m E { 1, 2, ... , n  2}, then max Hm < min H,,,+ 1, and so that for any f E 4), FP((am nrEHm Yf(m),r)11) = c A*. Let k = max Hn_1 + 1. Let
L = UJE1 FP((am
n/EH, Yf(m).t)m11).
By Lemma 4.14 we have for each b E L that b1 A* E p. Let B = A* n nbEL b1 A*.
Since L is finite (although (D isn't) we have that B E p. Let U = {q" E Y : for all e E { 1 , 2, ... , n}, qe E B). Then U is a neighborhood of p so pick i E U n Ik. Pick an E S and Hn E J" f(N) with min Hn > k such that z = x' (an, Hn). To complete the proof we need to show that for any f E 4), FP((am
ntEHm Yf(m),t)n1=1) C A*.
So let f E 4) and let 0 0 F C (1, 2, ... , n). If n V F, then nmEF(am ' ntEH,,, Yf(m),t) E FP((am . ntEHm Yf(m),t)m 11) C
A*
so assume that n E F. Now an
n:EH,, Yf(n),t E B since z(an, Hn) = i E U. If F = {n}, then we have IImEF(am ' nzEH,,, Yf(m),t) =an ' ntEH Yf(n),t E B C_ A*. Otherwise, let G = F\{n} and let b = fImEG(am ' T7 111EH,,, Yf(m),t) Then b E L so an' fItEH Yf(n),t E B C b' A* SO f ImEF(am' n1EHH Yf (m),t) = b'an ntEH Yf(n),t E A*.
Of course we have the corresponding partition result.
Corollary 14.12. Let S be a commutative semigroup and for each f E N, let (Ye,n)n°
1
be a sequence in S. Let r E N and let S = U;=1 A. There exist i E 11, 2, ... , r}, a sequence (an )n° 1 in S and a sequence (H)1 in J'f (N) such that max Hn < min Hn+1 for each n E N and for each f E 4), FP((an I1rEH Yf (n),t)n' 1) 9 A;
14 The Central Sets Theorem
286
Proof. Some Al is central. A special case of the Central Sets Theorem is adequate for many applications. See Exercise 14.3.2.
It is not surprising, since the proof uses an idempotent, that one gets the Finite Products Theorem (Corollary 5.9) as a corollary to the Central Sets Theorem (in the case S is commutative). For a still simple, but more interesting, corollary consider the following extension of van der Waerden's Theorem, in which the increment can be chosen from the finite sums of any prespecified sequence. Corollary 14.13. Let (xn)n° be a sequence in N, let r E N, and let N = U;=1 A. T h e n t h e r e is some i E {1, 2, ... , r} such that for all f E N there exist a E N and d E FS((xn )n° 1) with {a, a + d, a + 2d, ..., a + 1d) C A1. 1
Proof. Pick i E (1, 2, ..., r) such that A; is central in (w, +). For each k, n E N, let 1 as guaranteed by Theorem Yk,n = (k  1) xn. Pick sequences (an)n° 1 and 14.11. Given f E N, pick any m > f and let a = a. and let d = EtEH,,, xt. Now, given k E (0, 1, ... , P}, pick any f E (D such that f (m) = k + 1. Then
a + kd = an +
E FS((an . EtEHn yf(n).t)
1) S A1. 13
Exercise 14.3.1. Let (xn)no l be a sequence in N, let r E N, and let N = Ui=1 A. Prove that there is some i E { 1, 2, ... , r) such that for all e E N there exist a E N and d E FS((xn)n° 1) and c r= FS((xn2)° 1) with {a, a + d, a + 2d, ..., a + id) U {a +
c,a+2c,...,a+tc} c A;.
Exercise 14.3.2. Let S be a commutative semigroup, let A be central in S, let k e N and for each P E {1, 2, ... , k}, let (ye,n)n__1 be a sequence in S. Prove (as a corollary to Theorem 14.11) that there exist a sequence (a,,)n° 1 and for each f E {1, 2, ..., k} a product subsystem (ze,n)n__1 of (ye,n)n_1 such that for each f E (1, 2, ... , k}, FP((an ze.n)n° t) S A. Show in fact that the product subsystems can be chosen in a uniform fashion. That is, there is a sequence (Hn)n_1 in Pf(N) with max H,, < min Hn+1 for
each n such that for each f E (1, 2, ... , k) and each n E N, ze,n = n,,H ye.t
14.4 The Noncommutative Central Sets Theorem In this section we generalize the Central Sets Theorem (Theorem 14.11) to arbitrary semigroups. The statement of the Central Sets Theorem in this generality is considerably more complicated, because the "a" in the conclusion must be split up into many parts.
(In the event that the semigroup S is commutative, the conclusion of Theorem 14.15 reduces to that of Theorem 14.11.)
14.4 The Noncommutative Central Sets Theorem
287
Definition 14.14. Let M E N. Then
1m = {(H1, H2, ... , Hm) E d'f(N)m : if m > l and
t E {1,2,...,m1},then maxHr < min H,+,). Theorem 14.15. Let S be a semigroup, let A be a central subset of S, and for each I e N, let (ye,,1)n°_1 be a sequence in S. Given 1, m E N, a E Sm+l, and H E 1m, let
w(a, H, e) _ (nm 1(ai . ntEtj ye,,)) . am+1 There exist sequences (m(n))R° 1,
(a"n)rto 1,
(1) for each n E N, m(n) E N, an E
and (Hn)n° such that Hn E 1m(n), and max Hn,m(n) < Sm(n)+l.
min Hn+t, t, and
(2) for each f E
z'(a, H) = (w(a, H, 1), w(a", H, 2), w(a, H, 3),...).
(Observe that if S is commutative, a = fl' a,, H = Um1 Hi, and s(a, H) is as defined in the proof of Theorem 14.11, then x' (a, H) = z(a, H).)
For each i E N, let T, = S, let I; = {i(aa, H) : there exists m E N such that a E S'n}1, H E 1m, and min H1 > i } and let E; = I, U {a : a E S}. We claim that Ti, E,, and I, satisfy hypotheses (a), (b), (c), and (d) of Lemma 14.9. Indeed, (a) and (d) hold trivially.
To verify (b), let i E N ands E Ii be given. Pick M E N, a E 5m+1, and H E 1m i and x' = i(a, H). Let j = max Hm + 1. To see that x" Ei C I,, let E Ej. If = s f o r some s E S, let b ' = (a1, a2, ...,am, am+1 s). Then such that min Hl
= i(bb, if) E I,. So assume y E Ij, pick k E N, E E Sk+l, and G E 1k such that min G1 > j and y" = i(b, G). Let c = (a1, a2, ... , am, am+1 b1. b2, ... , bk+l ) and let K = (H1, H2,..., Hm, G1, G2,..., Gk). Then c' E Sk+m+t, K E 1m+k and x"
To verify (c), let i e N and l E E, \I, be given. Pick a E S such that s = a. Then as above one easily sees that x" E, c E, and x" I, C I,. Since A is central in S, pick an idempotent p E K (PS) such that A E p. By Lemma
14.915 EI=fiElyctyIi. Now we construct the sequences (m(n))n° t, (a""n)n° and (l~In)n_ inductively. Write A* = A*(p). Let U = {q" E Y q1 E A*). By Theorem 4.12 A* E P SO U is a neighborhood of T. Since also 15 E I, pick y" E U fl 11. Pick m(1) E N, a"1 E Sm(1)+l, and H1 E 1m(1) such that y' = z(a1, H1). Since i(a"1, H1) E U we have that w(a1, H1, 1) E A*. Inductively, let n > 1 and assume that we have chosen (m(j))l=', and so that, if j E { 1, 2, ... , n  2}, then max Hj,m(j) < min Hj+1,1, and so that
for any f E 4',FP((w(aj, Hj, f(j)))j=1) c A.
14 The Central Sets Theorem
288
Let k =
Let E = UfE.v FP(.(w(dj, flj, By Lemma 4.14 we have for each b E E that bIA* E p. Let B = A* nnb,E b 'A*. Since E is finite we h a v e that B E p. Let U = { q ' E Y : f o r all f E {1, 2, ... , n}, qe E B}. Then U is a neighborhood of p so pick y' E U n Ik. Pick m(n) E N, 1.
do E Sm(n)+1 and Hn E .tm(n) with min Hn,I > k such that Y = Z(dn, Hn). To complete the proof we need to show that for any f E CD,
FP((w(aj, Hj, f(j)));_1) C A'. (1,2,
. . .
, nn}. ).
If n 0 F, then
njEF w(dj, Hj, f(j)) E FP((w(dj, Hj, f(j)))ji) c A*, so assume that n E F. Now w(a"n, fin, f (n)) E B since zz(a'n, H,,) = y" E U. If F = {n}, then we have f1jEF w(dj, Hj, f (j)) = w(dn, fin, f (n)) E B C A'. Otherwise, let G = F\{n} and let b = 11j EG w(dj , Hj, f (j)). Then b E E so w(dn, fin, f (n)) E
B C b_IA* so njEF w(dj, Hj, f(j)) = b w(dn, fn, .f (n)) E A*. Exercise 14.4.1. Derive the HalesJewett Theorem (Theorem 14.7) as a corollary to the noncommutative Central Sets Theorem (Theorem 14.15).
14.5 A Combinatorial Characterization of Central Sets Recall from Theorem 5.12 that members of idempotents in P S are completely characterized by the fact that they contain FP((xn) n° ,) for some sequence (xn) n° 1. Accordingly, it is natural to ask whether the Central Sets Theorem characterizes members of minimal idempotents, that is, central sets. We see now that it does not.
Lemma 14.16. Let n, m, k E N and for each i E {1, 2, ..., n}, let (yi,t)°O 1 be a sequence in N. Then there exists H E Pf (N) with min H > m such that for each i E {l,2,...,n}, E,EHYi.t E N2k Proof. Choose an infinite set G I C N such that for all t, s E G 1, y,j  yi,,. (mod 2k). Inductively, given i E {1, 2, ..., n  1} and Gi, choose an infinite subset Gi+i of Gi such that for all t, S E Gi+i, Yi+1,: = Yi+i,s (mod 2k). Then for all i E (1, 2, ..., n) and all t, S E Gn one has yi,,  yi,s (mod 2k). Now pick H C_ Gn with min H > in and IHI = 2k.
Recall from Definition 6.2 that given n E N, supp(n) E JPf(co) is defined by n = Ei EsuPP(n) 2i'
Lemma 14.17. There is a set A C N such that (a) A is not_piecewise syndetic in (N, +).
289
14.5 Combinatorial Characterization
(b) For all x E A there exists n E N such that 0 # A n N2n C x + A. (c) Foreachn E N and any n sequences (yj,t)°. 1 , (y2,t)00 1, . , (yn,t)°°t in N there exista E N and H E J P f ( N ) such thatfor all i E (1, 2, ... , n), a+ElEH yi,t E AnN2n.
Proof. For each k E N let Bk = (2k, 2k + 1, 2k + 2, ... , 2k+1 1 ) and let A= {n E N : for each k E N, Bk\ supp(n) # 0}. Then one recognizes that n E A by looking at the binary expansion of n and noting that there is at least one 0 between positions 2k and 2k+1 for each k E N. To show that A is not piecewise syndetic we need to show that for each g E N there is some b E N such that f o r any x E N there is some y E {x + 1 , x + 2, ... , x + b} with {y + 1, y + 2, ... , y + g j n A = 0. To this end let g E N be given and pick k E N such that 22k > g. Let b = 22k+t . Let X E N be given and pick the least a E N such 2k . Then x < y < x + b and for each that a . 22k+1  22k > x and let y = a . 22k+1 2k+1  1) C supp(y + t) so t E { 1, 2, ... , 22k  11 one has {2k 2k + 1, 2k + 2
2
y+t 0 A. To verify conclusion (b), let x E A and pick k E N such that
22' 1 > x. Let n = 2k.
Then 0 96 A n N2n C x + A. Finally, to verify (c) let n E N and let sequences (yl,t)°O1, (y2,t)°O1, ... , (yn,t) ij be given. We first observe that by Lemma 14.16 we can choose H E P f (N) such that for each i E {1, 2, ... ,}, E:EH Yi,t E N2n+1 Next we observe that given any z 1, z2, ... , zn in N and any k with 2k > n, there 0 f o r each i E (1, 2, ... , n). Indeed, if exists r E Bk such that Bk \ supp(2r + zi) r E Bk and Bk C supp(2' + z) then supp(z) n Bk = Bk\{r}. Consequently I{r E Bk there is some i E { 1, 2, ... , n) with Bk g supp(2' + zi) } I < n.
For i E {1, 2, . . , n}, let zo,i = EtEH Yi,t Pick the least t such that 2e > n. Now given i we have 2n+1 Izo,i and 2e1 < n + 1 so 2e1 E BQ_1\ supp(zo,i) Pick
ro E Be such that Be\ supp(2r0 + zo,i) : 0 for each i E 11, 2, ..., n) and let zl,i = zo,i + 2r0. Inductively choose rj E Be+j such that Be+/ \ supp(2ri + z j,i) 0 0 for each
i E {1, 2, ..., n} and let zj+1,i = zi,i + 2ri. Continue the induction until f + j = k where 22k > EtEH At f o r each i E { 1, 2, ... , n} and let a = 2r0 + 2rl +  .+ 11 
2rkP
Theorem 14.18. There is a set A C N such that A satisfies the conclusion of the Central Sets Theorem (Theorem 14.11) for (N, +) but K(,BN, +) n of A = 0.
Proof. Let A be as in Lemma 14.17. By conclusion (a) of Lemma 14.17 and Theorem 4.40 we have that K(fiN, +) n ce A = 0.
For each f E N, let (Ye,n)n° 1 be a sequence in N. Choose by condition (c) of Lemma 14.17 some at and H1 with at + EIEH, Yl,t E A. Inductively, given an and Hn, let f = max H n and pick m > n in N such that for all i E { 1 , 2, ... , n}, an + EtEH Yi,t < 22'". Pick by condition (c) of Lemma 14.17 (applied to the sequences (Yi,e+t)
1, (Y2,e+1)°0_1, ... , (Yn+1.e+t)°0_1, so that
min Hn+1 > f = max Hn) some
an+1 and Hn+1 such that for all i E (1, 2, ..., n + 1), an+1 + Et
E A n N22'.
290
14 The Central Sets Theorem
Observe that if x, y E A and for some k, x <
22k
and 22k ly, then x + y E A.
Consequently for f E 0 one has FS((an + EfEH Yf(n),t)°O_1) c A. We now proceed to derive a combinatorial characterization of central sets. This characterization involves the following generalization of the notion of a piecewise syndetic set.
Definition 14.19. Let (S, .) be a semigroup and let A c P(S). Then A is collectionwise piecewise syndetic if and only if there exist functions G : Pf(A) * Pf(S) and x : Pf(A) x Pf(S) a S such that for all F E Pf(S) and all .F and 3e in Pf(A) with
F c 3e one has F x(3e, F) c UZEG(.)t' (n P) Note that a subset A is piecewise syndetic if and only if {A} is collectionwise piecewise syndetic. An alternate characterization is: A is collectionwise piecewise syndetic if and only if there exists a function G :.Pf(A) > Pf(S) such that {y1(G(P))'(n F) : y E Sand' E Pf(A)}
has the finite intersection property.
(Here y' (G(.F))' ( n F) = UIEG(.F)Y I t' ( n F), see Exercise 14.5.1.) 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 semigroup S = (an : n E N} and {An : n E NJ C P(S). The family {An : n E N} is collectionwise piecewise syndetic if and only if there exist a sequence (g(n))n° 1 in N and a sequence (yn)n° 1 in S such that for all n, m E N with
m>n, {alym,a2Ym,...,amym} c U8(l a!'(nn IAi). Proof For the necessity, pick functions G and x as guaranteed by Definition 14.19. Given n E N, let
g(n) = min{k E N : G({A1, A2, ... , An)) c {al, a2, ..., ak}} and y n = x({A1, A2, ... , An}, {al, a2, ..., an}). Then if m > n, letting
F _ {A1, A2, ... , An},
3e = {A1, A2, ... , Am},
and F = {al, a2, ... , am},
one has F C Re and
F x(Je, F) c UrEG(.y)t' (n r)
{alym, a2Ym,... , amym} =
s(n)
c U,=1
ai_
1
(n,=1
A,).
For the sufficiency, let (g(n))n° and (yn)R I be as in the statement of the lemma. I
Given F E Pf((An : n E N}), let m = max{k E N : Ak E F} and let G(F) _
291
14.5 Combinatorial Characterization
{a1, a2,.  ., ag(m)). Given 3e E ,?'f((An : n E N)) and F E 3'f (S), let m = max(k E N : Ak E 3e or ak E F) and let x(3e, F) = ym. Then if .F, 3e E 3'f({An : n E N)),
FEpf(S), "cJe,m=max{kEN:AkE3eorakEF},andn=max{kEN: Ak E F), then n<mso F x(3e, F) C {a1Ym,a2Ym,
 , amYm}
c Ui=t aj (nf=t A;) 8(n)
n
1
c utEG(y)t_'(nF)
11
Theorem 14.21. Let (S, ) be an infinite semigroup and let A C .P (S). There exists p E K(f S) with A c p if and only if A is collectionwise piecewise syndetic.
Proof Necessity. Pick P E K ($ S) such that A c p. For each F E 3'f (.A), n F E p,
so by Theorem 4.39, B(F) = It E S : t' (n .F) E p} is syndetic. Pick G(F) E UZEG(F)t' B(.F). For each y E S and each F E ?f (A), pick ,Pf(S) such that S = F))1 B(F) (so that (t (y, .F) y)' (n .F) E P). t(y, F) E G(F) such that y E (t(y, Given 3e E R f (A) and F E ,Pf (S), { (t (y, F) y)' (n 3:') : y E F and 0 # F c 3e}
is a finite subset of p so pick
x(3e, F) E n{(t(Y, F)  Y)1 (n F) : y E F and O :A F C 3e}.
To see that the functions G and x are as required, let F E ?(S) and let F, 3e E UZEG(.n)t' (n F), let y E F. Then P f (,A) with .F c 3e. To see that F x (3e, F) c x(3e, F) E (t (y, .F) Y)1(n .F), so
y . x(3e, F) E (t(y, F))1(n F) c UtEG(T)t`(n F). Sufficiency. Pick G : .i'1(A) > Pf(S) andx : J'f(A) x pf(S) > S as guaranteed by the definition. For each F E ?f (.A) and each y E S, let
D(F,y)={x(3e,F):3eEJ"f(A), FEJ"f(S), .FcJ?, andyEF). Then {D(F, y) :.F E ?1(A) and y E S) has the finite intersection property so pick u E PS such that {D(.F, y) :.F E 5'f (,A) and y E S) c u. We claim that (*) for each F E .P f(,A), S S. u C U,EG(.x)t1(n F)
To this end, let .F E ,Pf(A) and let y E S. Then D(Y, y) c {x t (n F)} so USEG(F)t (n F) E y u as required.
:
y
xE
UfEG(.F)t ,8S
UtEG(x)t' (n .F). Since By (*) we have that for each F E Rf (A), @S u c u is a left ideal of $S, pick a minimal left ideal L of j 8S such that L c PS u
and pick q E L. Then for each 3e E .P f(ib) we have
t(3e) E G(3e) such that (t(3e))1(n 3e) Eq.

UIEG(,t)t' (n 3e) E q so pick
292
14 The Central Sets Theorem
For each F E ?f (A), let E(F) = {t (3e) : 3e E .Pf(A) and F C 3e}. Then {E(F) : F E .P1 (A)} has the finite intersection property. So pick w E PS such that
{E(F) F E ?f (A)} C w. Let p = w q. Then p E L C K(P S). To see that A C p, let A E A. Since E({A}) E w, it suffices to show that E({A}) S= It E S :
t'1A E q}, so let R E Pf(A) with {A} C 3e. Then (t(3e))1(n) E q and (t(3f))1(n
3e) C (t(3e))'A, so (t(3e))'A Eq.
Our combinatorial characterization of central is based on an analysis of the proofs of Theorem 5.8. The important 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, so that any path through that tree yields a sequence (xn)0° 1 with FP((xn) n° 1) C A. (Recall that in FP((xn) n° 1), the products are taken in increasing order of indices.) We formalize the notion of "tree" below. We recall that each ordinal is the set of its predecessors. (So 3 = {0, 1, 2} and 0 = 0 and, if f is the function {(0, 3), (1, 5), (2, 9),
(3, 7), (4, 5)}, then f13 = {(0, 3), (1, 5), (2, 9)).)
Definition 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 co and range(f) c A and if domain(f) = n > 0, then fin1 E T. T is a tree if and only if for some A, T is a tree in A.
The last requirement in the definition is not essential. Any set of functions with domains in co can be converted to a tree by adding in all restrictions 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. Definition 14.23. (a) Let f be a function with domain(f) = n E (o and let x be given.
Then f x = f U {(n, x)}.
(b) Given a tree Tand f ET,Bf=Bf(T)={x: f"xET}. (c) Let (S, .) be a semigroup and let A C S. Then T is a *tree in A if and only if T is a tree in A and for all f E T and all x E B f, B fX C x1 B j.
(d) Let (S, ) be a semigroup and let A C S. Then T is an FPtree in A if and only if T is a tree in A and for all f E T, Bf = g(t) : g E T and f Z g and
0 0 F C dom(g)\dom(f)}. The idea of the terminology is that an FPtree 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 14.24. Let (S, ) be an infinite semigroup and let A C S. Let p be an idempotent in fiS with A E p. There is an FPtree T in A such that for each f E T,
BfEp. Proof. We shall define the initial segments Tn = If E T : dom(f) = n} inductively.
Let To = {0} (of course) and let Co = A*. Then Co E p by Theorem 4.12. Let
T1={{(0,x)}:xE Co}.
14.5 Combinatorial Characterization
293
Inductively assume that we have n E N and have defined T so that for each f E Tn
one has FP((f (t))t=o) c A*. Given f E T, write Pf = FP((f (t)) r=o ), let Cf = A*nfXEPf xIA*, andnotethatbyLemma4.14, Cf E p. LetTn+I = if Y : f E Tn and y E Cf1. Then given g E T,+ 1, one has FP((g(t))% o) c A*.
The induction being complete, let T = URo T, Then T is a tree in A. One sees immediately from the construction that for each f E T, Bf = Cf. We need to show that for each f E T one has Bf = {{TIFF 9W : g E T and f Z g and
0# F c 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 = fI(EF g(t). For the other inclusion we first observe that if f, h E T with f C h then Pf c Ph so Bh C Bf. Let f E Tn and let x E {{TIFF g(t) : g E T and f Z g and 0 0 F c_ dom(g)\ dom(f )}. Pickg E T with f Z g and pick F with 0 of F C dom(g)\ dom(f ) such that x = f11EF g(t). First assume F = {m}. Then m >_ n. Let h = gym. Then f c h and h"x = glIn+I E T. Hence X E Bh Bf as required. Now assume IFl > 1, let m = max F, and let G = F\{m}. Let h = glIn, let w = ntEG g(t), and let n1) and Ph = FP((h(t))m ol). We need y = g(m). Then y E Bh. Let Pf = FP((f (t))1=0 E A*andforallz E Pf, E z'A*. E Bf. That is, weneed to show that and yEBhSO
so
Theorem 14.25. Let (S, ) be an infinite semigroup and let A C S. Statements (a), (b), (c), and (d) are equivalent and are implied by statement (e). If S is countable, then all five statements are equivalent.
(a) A is central. (b) There is a FPtree T in A such that (B1 : f E T) is collectionwise piecewise syndetic.
(c) There is a *tree T in A such that { B f
f E T j is collectionwise piecewise
syndetic. (d) There is a downward directed family (CF) FEI of subsets of A such that
(i) for each F E I and each x E CF there exists G E I with CG c xI CF and (ii) {CF : F E I } is collectionwise piecewise syndetic. (e) There is a decreasing sequence (Cn)n_I of subsets of A such that (i) for each n E N and each x E Cn, there exists m E N with CIn C x ' Cn and (ii) (Cn : n E N) is collectionwise piecewise syndetic.
(a) implies (b). Pick an idempotent p e K($S) with A E p. By Lemma 14.24 pick an FPtree with (Bf : f E T) e p. By Theorem 14.21 {Bf : f E T} is
Proof.
collectionwise piecewise syndetic. (b) implies (c). Let T be an FPtree. Then given f E T and X E B f, we claim that
BfC x 'Bf. Tothisendlety E Bf,randpickg E TandF C_ dom(g)\dom(f"x) such that f"x g and y = f1IEF g(t). Let n = dom(f) and let G = F U (n). Then x Y = fl(EG g(t) and G C dom(g)\dom(f), sox y E Bf as required.
294
14 The Central Sets Theorem
(c) implies (d). Let T be the given *tree. Since {Bf : f E T} is collectionwise piecewise syndetic, so is {nfEF Bf : F E Rf(T)}. (This can be seen directly or by invoking Theorem 14.21.) Let I = Pf(T) and for each F E I, let CF = nfEF Bf. Then (CF : F E 1) is collectionwise piecewise syndetic, so (ii) holds. Let F E I and
let x E CF. Let G = if x : f E F). Then G E I. Now for each f E F we have Bf^X C x'Bf so CG c x1 CF. (d) implies (a). Let M= nFE( ci CF. By Theorem 4.20 M is a subsemigroup of PS. Since (CF : F E 1) is collectionwise piecewise syndetic, we have by Theorem 14.21 that Mf1K(8S) 0 so we may picka minimal left ideal L of flSwith LnM 96 0. Then L fl m is a compact subsemigroup of, S 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 that (c) implies (e). So let T be the
given *tree in A. Then T is countable so enumerate T as (f)1. For each n E N, let C,, = nk=, Bfk. Then [Cn : n E N) is collectionwise piecewise syndetic. Let n E N and let x E Cn. Pick M E N such that { fk"x : k E {1, 2, ... , n}} c If, : t E (1, 2, ... , m)). Then C, = nm 1 Bf, c I kn=e Bfk"x c fk=1 x1 Bfk = x1 Cn. We close this section by pointing out a Ramseytheoretic consequence of the characterization.
Corollary 14.26. Let S bean infinite semigroup, let r E N, and let S = U;=t A; . There e x i s t i E [1, 2, ... , r} and an FPtree Tin Ai such that {B f : f E T j is collectionwise piecewise syndetic.
Proof Pick an idempotent p E K(9S) and pick i E (1, 2, ... , r) such that Ai E p. Apply Theorem 14.25. Exercise 14.5.1. Let S be a semigroup. Prove that a family A C ,P (S) is collectionwise piecewise syndetic if and only if there exists a function G :.Pf (A) ). Rf (S) such that
{y' (G(F))' ((l jr) : y E S and F E .Pf(A)) has the finite intersection property.
(Where y' (G(F))' (n F) =
UtEG(.)YI t1 (n F).)
Notes The original Central Sets Theorem is due to Furstenberg [98, Proposition 8.21] and applied to the semigroup (N, +). It used a different but equivalent definition of central. See Theorem 19.27 for a proof of the equivalence of the notions of central. This original version also allowed the (finitely many) sequences (Ye,n)n° i to take values in Z. Since any idempotent minimal in (#N, +) is also minimal in (flZ, +), and hence any central set in (N, +) is central in (Z, +), the original version follows from Theorem 14.11.
Notes
295
The idea for the proof of the Central Sets Theorem (as well as the proofs in this chapter of van der Waerden's Theorem and the HalesJewett Theorem) is due to Furstenberg and Katznelson in [99], where it was developed in the context of enveloping semigroups. The idea to convert this proof into a proof in PS is due to V. Bergelson, and the construction in this context first appeared in [27]. Corollary 14.13 can in fact be derived from the HalesJewett Theorem (Corollary 14.8). See for example [28, p. 434]. The material in Section 14.5 is from [148], a result of collaboration with A. Maleki, except for Theorem 14.21 which is from [147], a result of collaboration with A. Lisan.
Chapter 1 5
Partition Regularity of Matrices
In this chapter we present several applications of the Central Sets Theorem (Theorem 14.11) and of its proof.
15.1 Image Partition Regular Matrices Many of the classical results of Ramsey Theory are naturally stated as instances of the following problem. Given u, u E N and a u x v matrix A with nonnegative integer entries, is it true that whenever N is finitely colored there must exist some z E N° such that the entries of Ax are monochrome? Consider for example van der Waerden's Theorem (Corollary 14.2). The arithmetic progression {a, a + d, a + 2d, a + 3d} is precisely the set of entries of 0 1
1
1
2
1
3
(a) d
Also Schur's Theorem (Theorem 5.3) and the case m = 3 of Hilbert's Theorem (Theorem 5.2) guarantee an affirmative answer in the case of the following two matrices: 1
1
0 0
1
0
1
1
0
1
1
1
0 0
0
1
1
0 0
1
1
1
1
1
0
1
1
0
1
1
1
1
1
1
This suggests the following natural definition. We remind the reader that for n E N and x in a semigroup (S, +), nx means the sum of x with itself n times, i.e. the additive version of x". We use additive notation here because it is most convenient for the matrix manipulations. Note that the requirement that S have an identity is not substantive since
15.1 Image Partition Regular Matrices
297
one may be added to any semigroup. We add the requirement so that 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.) Definition 15.1. Let (S, +) be a semigroup with identity 0, let u, v E N, and let A be a u x v matrix with entries from co. Then A is image partition regular over S if and only if whenever r E N and S = U,_f Ei, there exist i E (1, 2, ... , r) and x E (S\{0})° such that Az E Ei'1. It is obvious that one must require 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 nonzero. (Consider van der Waerden's Theorem with increment 0.) We have restricted our matrix A to have nonnegative entries, since in an arbitrary semigroup x may not mean anything. In Section 15.4 where we shall deal with image partition regularity over N we shall extend the definition of image partition regularity to allow entries from Q.
Definition 15.2. Let u, v E N and let A be a u x v matrix with entries from Q. Then A satisfies the first 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 11, 2, ... , v} : ai,, 96 0} = min{t E (1, 2, ... , v) : aj,, 0 0}, then ai,k = aj,k > 0. An element b of Q is a first entry of A if and only if 0}. there is some row i of A such that b = ai,k where k = mint E (1, 2, ..., v} : ai,, Given any family R of subsets of a set S, one can define the set 92* of all sets that meet every member of R. We shall investigate some of these in Chapter 16. For now we need the notion of central* sets.
Definition 15.3. Let S be a semigroup and let A C S. Then A is a central* set of S if and only if A fl c 0 0 for every central set C of S. Lemma 15.4. Let S be a semigroup and let A C S. Then the following statements are equivalent. (a) A is a central* set. (b) A is a member of every minimal idempotent in 18 S.
(c) A fl c is a central set for every central set C of S.
Proof. (a) implies (b). Let p be a minimal idempotent in 9S. If A ¢ p then S\A is a central set of S which misses A. (b) implies (c). Let C be a central set of S and pick a minimal idempotent p in ,BS such that C E p. Then also A E p so A fl C E P. That (c) implies (a) is trivial.
In the following theorem we get a conclusion far stronger than the assertion that matrices satisfying the first entries condition are image partition regular. The stronger conclusion is of some interest in its own right. More importantly, the stronger conclusion is needed as an induction hypothesis in the proof.
298
15 Partition Regularity of Matrices
Theorem 15.5. Let (S, +) be an infinite commutative semigroup with identity 0, let u, v E N, and let A beau x v matrix with entries from w which satisfies the first entries condition. Let C be central in S. If for every first entry c of A, cS is a central* set, then there exist sequences (xl,n)n° 1, (x2,n)n__l, ..., (x,,,n)n°_1 in S such that for every F E ?f (N), xF E (S\{0})v and AIF E Cu, where EnEF xl.n
XF=
EPEF x2.n
EnEF xv,n
Proof S\{0} is an ideal of S so by Corollary 4.18, 0 is not a minimal idempotent, and thus we may presume that 0 f C. We proceed by induction on v. Assume first v = 1. We can assume A has no repeated rows, so in this case we have A = (c) for some c E N such that cS is central*. Then C fl cS is a central set so pick by Theorem 14.11 (with the sequence yl,n = 0 for each n and the function f constantly equal to 1) some sequence (kn)n°_1 with FS((kn)n°l) c C fl cS. (In fact here we could get by with an appeal to Theorem 5.8.) For each n E N pick some xl,n E S such that k,, = cxThe sequence (xl,,,)n° l is as required. (Given F E 9f(N), EnEF kn 0 0 so EnEF X1,n 0 Now let v E N and assume the theorem is true for v. Let A beau x (v + 1) matrix with entries from co which satisfies the first entries condition, and assume that whenever
c is a first entry of A, cS is a central* set. By rearranging the rows of A and adding additional rows to A if need be, we may assume that we h a v e some r E 11, 2, ... , u and some d E N such that 
_
0
1}
ifi E (1,2,...,r}
a`'l  d ifi E {r+1,r+2,...,u}. Let B be the r x v matrix with entries bij = aij+l. Pick sequences (zl,n)n=1, (z2,n)n_l,
, (zv,n)n_l in S as guaranteed by the induction hypothesis for the matrix
B. For each i E {r+1,r+2,...,u} andeachn E N, let u+l
Yi,n = EJ=2 aij  ZJl,n
and let yr,n = 0 for all n E N. (For other values of i, one may let yi,n take on any value
inSatall.) Now Cfl d S is central, so pick by Theorem 14.11 sequences (kn )n° i in S and (Hn) n_
in pf (N) such that max H,, < min Hn+l for each n and for each i E {r, r + 1, ... , u},
FS((kn + EtEH Yi,t)n°_l) S C n dS. (If (k,,)n° l and (H,, )n l areas given by Theorem 14.11, let k,, = kn+u and H,, = H,+u
for every n, and for i E jr, r + 1, ... , u), let fi (n) = n if n < i and let fi (n) = i if
n>i.)
15.1 Image Partition Regular Matrices
299
Note in particular that each kn = kn + EtEH, Yr,t E C fl dS, so pick x1,n E S such
that kn = dxi,n. For j'E (2,3,...,v+ 1}, let xj,n = EtEHn Zjl,t We claim that the sequences (xj,n)n° 1 are as required. To see this, let F E show that for each j E { 1 , 2 , . . . , V
be given. We need to u),
Ej+1 a,j EnEFXj,n E C.
For the first assertion note that if j > 1, then EnEFxj,n = EtEG Zjl,t Where G = UnEFHn. If j = 1, then d EnEF xl,n = EnEF(kn + EtEH1 Yr.t) E C. To establish the second assertion, let i E (1, 2, ... , u) be given. Case 1. i < r. Then u+1
v+1
Ej=1 a,,i EnEF xi n = Ej_=2 ai,1 EnEF EtEHn Zj1,t Ej=1 bi,1 EtEG Zj,t E C
where G = UnEFHn. Case 2. i > r. Then
Ej+l a1,j EnEFxj,n = d EnEFX1,n +
ai,j EnEFXj,n = EnEF dxi,n + EnEF Ej+2 a1,J E(EHn Zjl,t
EnEF dxl,n + EnEF EtEH, Ej=2 ai,jzjl,t = EnEF(kn + EtEH, Yi,t) E C. We now present the obvious partition regularity corollary. But note that with a little effort a stronger result (Theorem 15.10) follows.
Corollary 15.6. Let (S, +) be an infinite commutative semigroup and let A be a finite matrix with entries from to which satisfies the first entries condition. If for each first entry c of A, cS is a central* set, then A is image partition regular over S.
Proof. Let S =
Ei. Then some Ei is central in S. If S has a two sided identity, then Theorem 15.5 applies directly. Otherwise, adjoin an identity 0. Then, by Exercise 15.1.1, Ei is central in SU {0}. Further, given any first entry c of A, c(SU {0}) is central* in S U {0} so again Theorem 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 N} (the multiplicative analogue of 2S) is not even a central set. (See Exercise 15.1.2.) Consequently B = N\{x2 : x E N) is a central* set and A = (2) is a first entries matrix. But there does not exist a E N with a2 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 every idempotent in (fiN, +) by Lemma 6.6.) Corollary 15.7. Any finite matrix with entries from to which satisfies the first entries condition is image partition regular over N.
300
15 Partition Regularity of Matrices
Corollary 15.8. Let A be a finite matrix with entries from.w which satisfies the first entries condition and let r E N. There exists k E N such that whenever { 1, 2, ..., k} is rcolored there exists z E (1, 2, ..., k)° such that the entries of Al are monochrome. Proof. This is a standard compactness argument using Corollary 15.7. (See Section 5.5 or the proof that (b) implies (a) in Theorem 15.30.) See also Exercise 15.1.3. As we remarked after Definition 15.1, in the definition of image partition regularity
we demand that the entries of z are all nonzero because that is the natural version for the classical applications. However, one may reasonably ask what happens if one weakens the conclusion. (As we shall see, the answer is "nothing".) Likewise, one may strengthen the definition by requiring that all entries of Ax" be nonzero. Again, we shall see that we get the same answer.
Definition 15.9. Let (S, +) be a semigroup with identity 0, let u, v E N, and let A be a u x v matrix with entries from w. (a) The matrix A is weakly image partition regular over S if and only if whenever r E N and S = Ur=1 E;, there exist i E 11, 2, ... , r} and . E (S°\{0}) such that Az E Ej u (b) The matrix A is strongly image partition regular over S if and only if whenever
r E N and S\{0} = Ui_1 E;, there exist i E { 1 , 2, ... , r} and l E (S\{0})1 such that Ax" E E;u
Theorem 15.10. Let (S, +) be a commutative semigroup with identity 0. The following statements are equivalent. (a) Whenever A is a finite matrix with entries from to which
satisfies the first entries condition, A is strongly image partition regular over S. (b) Whenever A is a finite matrix with entries from co which satisfies the first entries condition, A is image partition regular over S. (c) Whenever A is a finite matrix with entries from w which satisfies the first entries condition, A is weakly image partition regular over S. (d) For each n E N, n S : (0}. Proof. That (a) implies (b) and (b) implies (c) is trivial. (c) implies (d). Let n E N and assume that nS = {0}. Let 1
A=
1
n
0
n
.
Then A satisfies the first entries condition. To see that A is not weakly image partition
regular over S, let E1 = {0} and let E2 = S\{0}. Suppose we have some 1 =
\ / XZ
E
15.2 Kernel Partition Regular Matrices
301
S2\{0} and some i E 11, 2} such that A. E (E;)3. Now 0 = nx2 so i = 1. Thus xl = xl + nx2 = 0 so that x2 = XI + x2 = 0 and hence x = 0, a contradiction. (d) implies (a). Let A be a finite matrix with entries from w which satisfies the first entries condition. To see that A is strongly image partition regular over S, let r E N and let S\{0} = U;=1 E;. Pick by Corollary 15.8 some k E N such that whenever (1, 2, ... , k) is rcolored, there exists . E ((I, 2, .... k})° such that the entries of Az are monochrome.
There exists z E S such that {z, 2z, 3z, ... , kz} fl {0} = 0. Indeed, otherwise one would have k!S = (0). So pick such z and for i E {1, 2, ..., r}, let B; = It E ( 1 , 2, ... , k) : tz E E; } . Pick i E 11, 2, ... , r} and y E ({1, 2, ... , k})° such that Ay E (Bi)". Let x = 5z. Then Ax E (Ei)". 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 prove that C is central in S U (0). Exercise 15.1.2. Prove that {x2 : x E N} is not piecewise syndetic (and hence not central) in (N, .). Exercise 15.1.3. Prove Corollary 15.8. Exercise 15.1.4. Let u, u E N and let A be a u x v matrix with entries from co which is image partition regular over N. Let T denote the set of ultrafilters p E ON with the property that, for every E E p, there exists x' E N' for which all the entries of Ax are in E. Prove that T is a closed subsemigroup of ON.
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 kernel partition regular over G if and only if whenever G\{0} is finitely colored there will exist z with monochrome entries such that Cx" = 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. Definition 15.11. Let u, v 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 cv defined by cI'J _ (Ici, I + c,)/2 and c,,i = (I ci.i I  ci,i )/2. Thus, for example, if
C0
2 2
then
Note that C = C+  C.
C+_(0
0
0)
and C0
2
0
302
15 Partition Regularity of Matrices
Definition 15.12. Let (S, +) be a semigroup with identity 0, let u, v E N and let C be a u x v matrix with entries from Z. Then C is kernel partition regular over S if and only Di, there exist i E (1, 2, ... , r) and x" E (Di)° if whenever r E N and S\{0} =
such that C+x = C1. The condition which guarantees kernel partition regularity over most semigroups is known as the columns condition. It says that the columns of the matrix can be gathered
into groups so that the sum of each group is a linear combination of columns from preceding groups (and in particular the sum of the first group is 0).
Definition 15.13. Let u, v E N, let C be a u x v matrix with entries from Q, and let ci, c2, ... , c 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 lY and Ii , 12, ... ,1such that
(1) {11,12,...,1,,,}is a partition of 11, 2,..., v). (2) EiE1, 4= 6 . (3) If m > 1 and t E (2, 3, ... , m), let Jt = U'1 I. Then there exist (fir,; )ZEJ, in R such that EiE/, Ci = EiEJ, st,i . Observe that one can effectively check whether a given matrix satisfies the columns
condition. (The 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. Suppose that C is a u x m matrix with entries from Z which satisfies the f i r s t entries condition. Let k = max { Ici, j I : (i, j) E (1, 2, ... , u } x ( 1, 2, ... , m) } + 1,
and let E be the m x m matrix whose entry in row t and column j is
(kit ift<j
ei.i = {l
ift>j.
0
Then CE is a matrix with entries from co which also satisfies the first entries condition.
Proof. Let D = CE. To see that D satisfies the first entries condition and has entries from w, let i E (1, 2, ... , u) and let s = min(t E (1, 2, ... , m) : ci,, 01. Then for
j < s, di, j = 0 and di,s = ci,s. If j > s, then
di,j =
ci,,klt
= EI=s ci > kjS

ikjt lci,,Ikji
Ei=s+1
> kjs  Et_s+1(k  1)k'1 = 1.
A connection between matrices satisfying the columns condition and those satisfying the first entries condition is provided by the following lemma.
15.2 Kernel Partition Regular Matrices
303
Lemma 15.15. Let u, V E N and let C be a u x v matrix with entries from Q which satisfies the columns condition overQ. There exist In E { 1, 2, ..., v} and a v x m matrix B with entries from co that satisfies the first entries conditions such that CB = 0, where 0 is the u x m matrix with all zero entries. If C satisfies the columns condition over 7L, then the matrix B can be chosen so that its only first entry is 1.
Proof Pick m E N,
(Ji)m 2, and {(St,i)iEi,) 2 as guaranteed by the columns condition for C. Let B' be the v x m matrix whose entry in row i and column t is given (It)mt,
by
St,i if i E Jt b,
1
if iEIt
0
if i 0
Ujtt
Ij.
We observe that B' satisfies the first entries condition, with the first nonzero entry in each row being 1 . W e also observe that CB' = 0. (Indeed, let j E {1, 2, ... , u} and
t E {1, 2, ... , m}. If t = 1, then E° 1 cjj b,,t = EIE/, cjj = 0 and if t > 1, then Z"=1 cjj ' bi,t = EiEJ, St,i ' cjj + EiE4 cjj = 0.) We can choose a positive integer d for which dB' has entries in Z. Then dB' also satisfies the first entries condition and the equation C(dB') = 0. If all of the numbers
Si,j arein7L,letd= 1. Let k=max{Idb!1I : (i, j) E (1,2,...,v) x (1,2,...,m)}+1, andletEbethe m x m matrix whose entry in row i and column j is
e''
kj' jl 0
ifi< j
ifi > j.
By Lemma 15.14, B = dB'E has entries in w and satisfies the first entries condition.
Clearly, CB = 0. Now, given i E 11 , 2, ... , v}, chooses E {1, 2, ... , m} such that i E IS. If t < s, then b t = 0 while if t > j, then et, j = 0. Thus if j < s, then bi, j = 0 while if j > s, then bi, j = E'=S dbI tki t. In particular the first nonzero entry of row i is d. We see that, not only is the columns condition over Q sufficient for kernel partition
regularity over most semigroups, but also that in many cases we can guarantee that solutions to the equations C+i = Cz can be found in any central set. Theorem 15.16. Let (S, +) be an infinite commutative semigroup with identity 0, let u, v E N, and let C beau x v matrix with entries from Z. (a) If C satisfies the columns condition over 7L, then for any central subset D of S, there exists x E D° such that C+i = Cx'. (b) If C satisfies the columns condition over Q and for each d E N, d S is a central*set in S, then for any central subset D of S, there exists x" E D° such that C+x = C. (c) If C satisfies the columns condition over Q and for each d E N, d S 0 {0}, then C is kernel partition regular over S.
304
15 Partition Regularity of Matrices
Proof. (a) Pick a v x m matrix B as guaranteed for C by Lemma 15.15. Then the only first entry of B is 1 (and IS = S is a central* set) so by Theorem 15.5 we may pick
some E E S' such that Bzz E D". Let x = B. Now CB = 0 so C+B = CB (and all entries of C+B and of CB are nonnegative) so C+x = CBE = CBE = C. (b) This proof is nearly identical to the proof of (a). (c) Pick a matrix B as guaranteed for C by Lemma 15.15. Let r E N and let S\{0} _ Ui=1 Di. By Theorem 15.10 choose a vector E E (S\{0})'° and i E { 1, 2, ... , r} such
that BE E (Di)", and let x' = B. As above, we conclude that C+x = C+Bz' = CBE = Cx'. 13 Exercise 15.2.1. Note that the matrix (2 2 1) satisfies the columns condition. Show that there is a set which is central in (N, ) but contains no solution to the equation x12 X3 = x22. (Hint: Consider Exercise 15.1.2.)
15.3 Kernel Partition Regularity Over N Rado's Theorem In this section we show that matrices satisfying the columns condition over Q are precisely those which are kernel partition regular over (N, +) (which is Rado's Theorem) and are also precisely those which are kernel partition regular over (N, ). Given a u x v matrix C with integer entries and given i E N" and y' E Nu we write
xC = y' tomean that for i E {l, 2, ... , u}, IIjv_l xj`i,i = yi.
Theorem 15.17. Let u, v E N and let C be a u x v matrix with entries from Z. The following statements are equivalent.
(a) The matrix C is kernel partition regular over (N, +). (b) The matrix C is kernel partition regular over (N, ). That is, whenever r E N and N\{1} = U;=1 Di, there exist i E 11, 2, ... , r} and x E (Di)" such that xxc = 1.
Proof. (a) implies (b). Let r E N and let N\ (11 = U,, Di. For each i E (1, 2, ... , r), let Ai = In E N : 2" E Di}. Pick i E 11, 2, ... , r} and y E (Ai)" such that Cy; = 0. For each j E { 1, 2, ... , u }, let xj = 2yi. Then xc = 1. (b) implies (a). Let (pi ) °O l denote the sequence of prime numbers. Each q E Q\ {0}
can be expressed uniquely as q = II°, p7 , where ei E Z for every i. We define Q\{0} > Z by putting 4b(q) = °__I ei. (Thus if q E N, r0 (q) is the length of the prime factorization of q.) We extend 0 to a function 1r : (Q\{0})" * 7G" by putting t / r (x)i = / (xi) for each x E (Q\{0})" and each i E { 1, 2, ... , v}. Since 0 is a homomorphism from (Q\{0}, ) to (Z, +), it follows easily that ilr(x" C) = CVr(x) for every! E (Q\{0})". Let {Ei}k1 be a finite partition of N. Then {01 [Ei] Cl N}k, is a finite partition of N\11), because 0 [N\{ 1 }] C N. So pick i E { 1, 2, ... , k} and
15.3 Kernel Partition Regularity Over N
305
x E (01[Ei] fl N)° such that z c = 1. Then *(x) E (Ei)° and CG(i) _ *(x c) = Vr(1) = 0.
Definition 15.18. Let u, V E N, let C be a u x v matrix with entries from Z, and let J and I be disjoint nonempty subsets of 11, 2, ... , v}. Denote the columns of C as
C1,C2,...,cv. (a) If there exist (zj)jEJ in Q such that EjEt c'j = EjEJ zjc'j, then E(C, J, I) = 0. (b) If there do not exist (zj)jEJ in Q such that E1E1 cj = EjEJ zjc'j, then
E(C, J, I) = (q : q is a prime and there exist (zj) jEJ in co, a E {1,2,...,q  1}, d E w, and y' E7G" such that EjEJ zjc'j +aqd EJEI cj = qd+1 y' } Lemma 15.19. Let u, v E N, let C beau x v matrix with entries from Z, and let J and I be disjoint nonempty subsets o f { 1 , 2, ... , v). Then E (C, J, I) is finite. Proof. If (a) of Definition 15.18 applies, the result is trivial so we assume that (a) does not apply. Let b = E1 1 c'j. Then b is not in the vector subspace of QU spanned by (c'j )j E J so pick some w E QU such that w c'j = 0 for each j E J and w bb # 0. By multiplying by a suitable member of Z we may assume that all entries of w are integers
and that w b > 0. Let s = w b. We show now that if q E E(C, J, I), then q divides s.
Let q E E (C, J, I) and pick (z j) j EJ in co, a E [1, 2, ... , q 1 }, d E co, and y E 7Gu
such that EJEJ zjcj +aqd E,11 cj = qd+lY Then
zjc'j +agdb =
qd+1
so
EjEJ zj(w jj) +agd(w b) = qd+1(w . Y) Since w c'j = 0 for each j E J we then have that agds = qd+t (iv y"") and hence
as = q (w . 5 ). Since a E (1, 2, ... , q  1), it follows that q divides s as claimed. The equivalence of (a) and (c) in the following theorem is Rado's Theorem. Note that Rado's Theorem provides an effective computation to determine whether a given system of linear homogeneous equations is kernel partition regular.
Theorem 15.20. Let u, v E N and let C be a u x v matrix with entries from Z. The following statements are equivalent.
(a) The matrix C is kernel partition regular over (N, +). (b) The matrix C is kernel partition regular over (N, ). That is, whenever r E N and
N\{1}_U=1Di, thereexisti E{l,2,...,r}andxE(Di)°such thatx =1. (c) The matrix C satisfies the columns condition over Q. Proof. The equivalence of (a) and (b) is Theorem 15.17.
306
15 Partition Regularity of Matrices
(a) implies (c). By Lemma 15.19, whenever J and I are disjoint nonempty subsets of { 1 , 2, ... , v}, E(C, J, I) is finite. Consequently, we may pick a prime q such that
q>max{IEjEJ ci,jl :i E{1,2,...,u}and0:0J C(1,2,...,v)} and whenever j a n d I are disjoint nonempty subsets of (1, 2, ... , v), q 0 E(C, J, 1).
Given x E N, pick a(x) E (1, 2, ... , q  1) and I
and b(x) in w such that
x = a(x) qe(x) + b(x) ql(x)+1. That is, in the base q expansion of x, E(x) is the number of rightmost zeros and a(x) is the rightmost nonzero digit.
For each a E (1, 2, ... , q  1) let Aa = {x E N\{1j : a(x) = a}. Pick a E
(1,2,...,q 1)andxl,x2,...,x E Aa such that C! =0. Partition 11, 2, ..., v) according to 1(x1). That is, pick m E N, sets 11, 12.... , Im
and numbers II <£2 <... <1msuchthat{11,12,..., Im}isapartitionof(1,2,...,v) and for each t E {1, 2, ... , m} and each j E I,, I(xj) = t,. For n E {2, 3, ... , m}, if any, let Jn = Ui=i We now show that
(1) EjEl, c = 6 and (2) given n E (2, 3, ... , m), E jE,, cj is a linear combination over Q of (c"j) jEjq, so that C satisfies the columns condition over Q.
To establish (1), let d = 11. Then for any j E 11 let ej = b(xj) and note that xj = a qd + ej qd+1. For j E (1, 2, ..., v)\Il we have f(xj) > d so xj = ej qd+l for some e j E N. Then
0 = Ej_l xj
c"j
EJEI, a . qd , Cj + EJ_l ej qd+l , cj.
Suppose that EjEl, c'j Then
0 and pick some i E 11, 2, ... , u) such that EJEl, ci, j 96 0.
so, since q Xa we have qI EjEI, ci, j, contradicting our choice of q.
Now consider (2). Let n E (2, 3, ... , m) be given and let d = E,,. For j E In, let ej = b(xj) and note that xj = a qd +ej . qd+l. For j E U7 n+l Ii, if any, pick ej E N
such that xj = ej qd+t Thus
6 =Ev Let
UT
EjEU.iint_(ej)  cj. Then

qd+I
i = EjEJn xj
j + EjEln a qd
.j
Since q 0 E(C, Jn, 1n) we have that EjEln c'j is a linear combination of required. (c) implies (a). This follows from Theorem 15.16(c).
as
15.3 Kernel Partition Regularity Over N
307
To illustrate the use of Theorem 15.20 we establish the case t = 4 of van der Waerden's Theorem. That is, we show that whenever N is finitely colored we can get
some a, d E N with a, a + d, a + 2d, a + 3d monochrome. So one lets xt = a,
x2 =a+d,x3 =a+2d,andx4 = a+3d. Then x2xi =d SO X3 =xi+2(x2xi) and x4 = X1 + 3(x2  x i ). That is we have the equations
xi 2x2+x3 = 0 2xi  3x2 + X4 = 0 so we are asking to show that the matrix 1
2
1
2 3 0
0 1
is kernel partition regular. Indeed it satisfies the columns condition because the sum of its columns is 0.
But there is a problem! The assignment xi = x2 = X3 = 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, x2 = d, x3 = a +d, x4 = a + 2d, and x5 = a + 3d. Then we get the equations
X3 = xl +X2 X4 = xi + 2x2 X5 = xi + 3x2 so we need to show that the matrix 1
1
1
2
1
3
1
0
0 0
1
0 0
0
1
is kernel partition regular. For this, let Ii = 11, 3, 4, 5}, 12 = 12}, 52,1 = 0, 82,3 = 1,
52,4 = 2, and 52,5 = 3. Exercise 15.3.1. Derive the finite version of the Finite Sums Theorem as a consequence
of Rado's Theorem. That is, show that for each n, r E N, whenever N = Ui=1 Aj, there exist i E 11, 2, ..., r) and (xt)" 1 such that FS((xt)" 1) c A. For example, the case n = 3 requires that one show that the matrix
is kernel partition regular.
1
1
0 1
1
0
1
0
1
1
1
1
1
0 0 0
0
1 0 0
0 0
1
0 0 0
0
1
15 Partition Regularity of Matrices
308
15.4 Image Partition Regularity Over N We concluded the last section with a verification of the length 4 version of van der 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 problem. On the other hand, we have already seen that the original problem is naturally seen as one of determining that the matrix 1
0
1
1
1
2
1
3
is image partition regular. (And the strengthened problem clearly asks that the matrix 0
1
1
0
1
1
1
2
1
3
be image partition 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.
Definition 15.21. Let u, v E N, and let A beau x v matrix with entries from Q. Then A is image partition regular over N if and only if whenever r E N and N = Uii=1 Ei, there exist i E { 1 , 2, ... , r} and x E N' such that Ax E E, u. We need a preliminary lemma which deals with a specialized notion from linear algebra.
Definition 15.22. Let u, v E N. let cl, c2, ... , c" be in Q", and let I c {1, 2, ... , v}. The Irestricted span of (c'1, c'2, ... , {E
1 ai ci
:
is
each a1 E 1R and if i E I, then ai > 0}.
Lemma 15.23. Let u, v E N, let c ' 1 , c ' 2 , ... , c " be in Qu, and let 1 c { 1, 2, ... , v}. Let
S be the Irestricted span of (cl, c2, ... , c'v). (a) S is closed in R".
(b) If y E S rl Q", then there exist 81, 32, ... , S in Q such that y = E Si > O for each i E I.
Si
c'i and
Proof. (a) We proceed by induction on I I I (for all v). If I = 0, this is simply the assertion
that any vector subspace of R' is closed. So we assume 10 0 and assume without loss
309
15.4 Image Partition Regularity Over N
of generality that 1 E I. Let T be the (I\(1))restricted span of (c2, c3, ... ,
By the induction hypothesis, T is closed. To see that S is closed, let b E C1 S. We show b E S. If cj = 0, then b E cZ S = such that c2 T = T = S, so assume that cl 0 0. For each n E N, pick (ai (n)) ai (n) > 0 for each i E 1 and 11b  E I a; (n) c"; II < 1/n. Assume first that {aI (n) : n E N} is bounded. Pick a limit point S of the sequence
(ai(n))n° t and note that S > 0. We claim that b  8 cl E T. To see this we show that b  S c"i E ct T. So let E > 0 be given and pick n > 2/c such that jai (n)  SI < E/(21Ict II). Then IIb  S c1  Ei_2 ai (n)  ci 11
llb  Ei=t a; (n) iF; Il + llai (n)cl  Scj 11 < E.
Sinceb
E T, we have Now assume that {al (n) : n E N) is unbounded. To see that S is closed, it suffices to
show that cj E T. For then S is in fact the (I\(l))restricted span of (cl, c'2, ... , which is closed by the induction hypothesis.
To see that ci E T, we show that c1 E ct T. Let E > 0 be given and pick n E N such that al (n) > (1 + IIbII)/E. For i E {2, 3, ... , v} let Si = a; (n)/c i (n) and note that for i E I \(1), S; > 0. Then
IIel  E2 Si
II
(b) Again we proceed by induction on III. The case I = 0 is immediate, being merely 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).
So assume 1 # 0. Let X = {x E ]E8°
:
Ziv= I xi
c;
Thus X is an affine
subspace of Rv, and we are given (by the fact that y E S) that there is some x' E X such that xi > 0 for all i E 1. Also (by the case I = 0) there is some 'z E X such that z; E Q f o r all i E 11, 2, ... , v). If z; > 0 for all i E I, then we are done, so assume there is some i E I such that zi < 0. Given i E I, {t E [0, 11 : (1  t)xi + tzi > 0} is [0, 1] if
zi > 0 and is [0, xi/(xi  zi)J if zi < 0. Let t be the largest member of [0, 1] such that the vector w = (I  t) x + t  z satisfies wi > 0 for all i E I. Then for some i E I we have that wi = 0. We assume without loss of generality that 1 E I and wI = 0. Then y = Ei_2 wi c ; so y ' is in the (I\{I})restricted span of (c2, c3, ... , c) so by the induction hypothesis we may pick S2, 33, ..., S in Q such that y' = Ei=2 Si ci and Si > 0 for each i E I\[ 1). Letting S i = 0, we are done.
Theorem 15.24. Let u, v E N and let A be a u x v matrix with entries from Q. The following statements are equivalent.
(a) A is image partition regular over N.
15 Partition Regularity of Matrices
310
(b) Let c"t, cZ...... be the columns of A. There exist st, s2, ... , sv E {x E Q : x > 0} such that the matrix
1
0
0
M=
1
... ...
0 0
Sv'Cv 0
0
... 1
is kernel partition regular over N.
(c) There exist m E N and a u x m matrix B with entries from Q which satisfies the first entries condition such that given any 3 E N' there is some x E Nv with
Al =By. (d) There exist m E N and a u x m matrix C with entries from Z which satisfies the first entries condition such that given any y' E N' there is some x" E N° with Ax" = Cy. (e) There exist m e N and a u x m matrix D with entries from w which satisfies the first entries condition such that given any y' E N' there is some x" E N" with
Ax"=Dy'. (f) There exist m E N, a u x m matrix E with entries from w, and c E N such that E satisfies the first entries condition, c is the only first entry of E, and given any y E N' there is some x' E N° with A. = Ey. Proof. (a) implies (b). Given any p E N\{1}, we let the start base p coloring of N be the function
op:N . {1,2,...,p1}x{0,1,...,p1}x{0,1} defined as follows: given y E N, write y = E°o at p, where each atE {0, 1, ... , p 
1) and an # 0; if n > 0, ap(y) = (a,,, a,, I, i) where i  n (mod 2); if n = 0, ap(y) = (ap, 0, 0). (For example, given p > 8, if x = 8320100, y = 503011, z = 834, and w = 834012, all written in base p, then ap(x) = vp(z) _ (8, 3, 0), ap(w) = (8, 3, 1), and ap (y) = (5, 0, 1). Let c"t, c"2, ... , c be the columns of A and let dl, d2, ... , du denote the columns of the u x u identity matrix. Let B be the matrix t St ct
S2
C2
..:
Sv
Cv
dt d2
... d
where st, s2, ... , sv are as yet unspecified positive rationale. Denote the columns of B , bu+v Then by bt, b2,
_ b`
St ct if t < v d,_v if t > v.
Given any p E N\{1} and any x E N, let y(p,x) = max{n E w : p" < x}. Now temporarily fix some p E N\{1}. We obtain m = m(p) and an ordered partition
15.4 Image Partition Regularity Over N
311
(D] (p), D2(p), ... , Dm(p)) of 11,2,...,u I as follows. Pick! E N° such that Az is monochrome with respect to the start base p coloring and let y" = Al. Now divide up 11, 2, . .. , u } according to which of the yi's start furthest to the left in their base p representation. That is, we get D, (p), D2 (p), ..., Dm (p) so that
(1) if k E (1, 2, ... , m} and i, j E Dk(p), then y(p, yi) = y(p, yj), and (2) if k E 12, 3, ... , m), i E Dk(p), and j E Dk_I (p), then y(P, Yj) > y(P, y). We also observe that since ap(yi) = ap(yj), we have that y(p, yj)  y(p, yi) (mod 2) and hence y(p, yj) > y(p, yi) + 2. There a r e only finitely many ordered partitions of 11, 2, ... , u }. Therefore we may pick
an infinite subset P of N\{1}, M E N, and an ordered partition (D], D2, ..., Dm) of
(1,2,...,u)sothatforallpE P,m(p)=mand (D] (P), D2(P), ... , Dm (P)) = (DI, D2, ... , Dm). W e shall utilize (D], D2, ... , Dm) to find s ] , s2, ... , s° and a partition of {1, 2, ... , u + v} as required for the columns condition. W e proceed by induction. First we shall find El e { 1 , 2, ... , v), specify si E Q+ _
(s EQ:s> 0)foreach i E E], letIl =El U(v+D1),andshow that EiEJ,bi =0. That is, we shall show that EiEE, Si * ji = EiED, d,. In order to do this, we show that EiED, ji is in the ( 1 , 2, ... , v)restricted span of (c], c2, ... . c°). For then by Lemma
15.23(b) one has EiED, ji = Ej ] ai c'i, where each ai E Q and each ai > 0. Let El = {i E (1, 2.... , v) : ai > 0} and for i E Ei, let si = ai. Let S be the 11, 2, ... , v}restricted span of (c'], c2, ... , c"°). In order to show that
EiED, ji is in S it suffices, by Lemma 15.23(a), to show that EiED, ji is in cf S.
To this end, let c > 0 be given and pick p E P with p > u/E. Pick the l E N° and y E Nu that we used to get (D] (p), D2 (P), ... , D. (p)). That is, Ax" = y, y is monochrome with respect to the start base p coloring, and (D], D2,..., Dm) = (DI (p), D2 (p), ... , D. (p)) is the ordered partition of (1, 2, ... , u) induced by the starting positions of the yi's. Pick y so that for all i E D1, y(p, yi) = y. Pick (a, b, c) E (1, 2, ... , p  1) x 10, 1, ... , p  1) x (0, 11 such that ap (yi) _ (a, b, c)
for all i E (1,2,...,u). LetZ =a+b/p and observe that 1 < f < p. For i E D], yi = a  pY +b  pY ] + zi pY2 where 0 < zi < p, and hence yi /pY = f +z, /p2; let Xi = zi/P2 and note that 0 < Xi < 1/p. For i E Uj2 Dj, we have y(p, yi) < y  2; let Ai = yi/pY and note that 0 < Xi < 1/p. Now Ax' =y so
Ei=]xi4=Y=Ei]Yi'di =EiED,Yi'di+Ej2EiEDjYidiThus E;=] (xi / pY) c; = EiED, f  di + Eu=, xi di and consequently
11 EiED, di  E
]
(xi /(tPY))  ei If = II Eu_] (Ai/e) . di ll < E°=] lei /el < u/p < OF.
Since E°=] (xi/(ZpY)) 6 is in S, we have that EiED, di E ct S as required.
312
15 Partition Regularity of Matrices
Now let k E {2, 3, ... , m) and assume we have chosen E1, E2, ... , Ek_ I e {1, 2, ... , v}, Si E Q+ for i E U;=1 Ei, and Ii = Ei U (v + Di) as required for the columns condition. Let Lk = U Ei and let Mk = U;=, Di and enumerate Mk in order as q(1), q(2), ..., q(r). We claim that it suffices to show that EiEDk di is in the { 1 , 2, ... , v)restricted span of (c1, c " 2 , ... , IF,,, dq(1), dq(2), ... , dq(r)), which
we will again denote by S. Indeed, assume we have done this and pick by Lemma 15.23(b) a1, a2, ... , a in {x E Q : x > 01 and 6. (1), 8 q ( 2 ) . . . . . Sq(r) in Q such that EiEDk di = Ei I ai ci + Ei=1 Sq(i) dq(i). Let Ek = (i E {1, 2, ... , v}\Lk : ai > 0) and for i E Ek, let si = ai. Let Ik = Ek U (v + Dk). Then + EiEDk di EiEIk bi = EiEEk ai = EiEEk of 61 + Ei=1 ai ' ,F; + Ei=1 Sq(i) dq(i) Now, if i E (1, 2, ... , v) and ai # 0, then i E Lk U Ek so
EiEEk ai'ci+E° 1 a, ci+Ei=1 Sq(i)' dq(i) = EiELk
ci+Ei=1 Sq(i)' dq(i)
and
EiELk ai ' 4 + E1=1 Sq(i) Let jai = ai /si if i E Lk and let E.
k1
EiELk a, Ci + EiEMk Si . di EiELk (ai/Si) ' bi + EiEM, Si ' by+i Si if i E Mk. Then we have
bi = EiELk (a4/SO ' bi + EiEMk Si ' by+i = E,Elk bi
as as required for the columns condition.
In order to show that EiEDk di is in S, it suffices, by Lemma 15.23(a) to show that EiEDk di is in ci S. To this end, let c > 0 be given and pick p E P with p > u/E. Pick
the l E N" and y E N" that we used to get (D1(p), D2(p), ..., D(p)). Pick y so that f o r all i E Dk, y(p, Yi) = y. Pick (a, b, c) E ( 1 , 2, ... , p  1 } x {0, 1, ... , p 1 } x {0, 1 } such that a (yi) _ (a, b, c) for all i E { 1, 2, ... , u). Let t = a + b/ p. For i E Dk, Yi = a pY + b py1 + zi ' pY2 where 0 < zi < p, and hence
yi/py = f + zi/p2; let ? = zi/p2 and note that 0 < Xi < 1/p. For i E U7k+1 Di, we have y (p, yi) < y  2; let .li = yi / pY and note that 0 < A, < I /p. (Of course we have no control on the size of yi/pY for i E Mk.) Now Ax = y so EVi=1 xi ci
u = Ei=1 yi
' di
EiEMk Yi ' di + EiEDk Yi ' di + Ei k+1 EiEDi Yi ' di,
where EJk+I EiEDi yi d; = 0 if k = m.
15.4 Image Partition Regularity Over N
313
Consequently,
II EiEDk di  (E(_1(xi/(epY)) ' Ci + EiEMk (yil (epY))
di) II
= IIj Ejk EiEDj Pi/fl < u/p < E,
which suffices, since xi /(epY) > 0 f o r i E (1, 2, ... , v}.
Having chosen 11, 12, ... , I , , if (1, 2, ... , u + v) = U; 1 Ij, we are done. So
assume (1,2,...,u+v)
U7='i. Let 1,,,+l =(1,2,...,u+v)\U; 1Ij. Now
11, 2, ... , u} = Uj 1 Dj, so {d1, d2, ... ,
i
c {bi : i E U; Ij} and hence we
can write Ei Et,,,+i bi as a linear combination of {bi : i E U, (b) implies (c). By Theorem 15.20 M satisfies the columns condition over Q. Thus
1 j).
by Lemma 15.15, there exist some m E {1, 2, ... , u + v} and a (u + v) x m matrix F with entries from w which satisfies the first entries condition such that MF = 0, the u x m matrix whose entries are all zero. Let S denote the diagonal v x v matrix whose diagonal entries are Si, s2, ... , s,,. Then Ml can be written in block form as (AS 1) and F can be written in block form as ( H I, where I denotes the u x u identity matrix and G and H denote v x m and u x m\\matrices respectively. We observe that G and H are first entries matrices and that ASG = H, because MF = O. We can choose d E N such that all the entries of d SG are in w. Let B = dH. Then B is a first entries matrix. Let y E N°' be given and let x = dSGy. Then Al = By as required. (c) implies (d). Given B, let d E N be a common multiple of the denominators in entries of B and let C = Bd. Given y, let z' = dyy and pick z such that Ax = BE = C y". (d) implies (e). Let C be given as guaranteed by (d). By Lemma 15.14, there is an m x m matrix E for which D = CE has entries from co and also satisfies the first entries condition. Given y E N'", we define z E N' by z = Ey. Pick x E N° such that
Ax= CE. Then CE=CEy"=Dy'. (e) implies (f). Let D be given as guaranteed by (e), and f o r each j E ( 1 , 2, ...
pickwj E Nsuchthatforanyi E {1, 2, ... , v} if j = min ft E {1, 2, ... , m} : di,, r 0}, then di, j = wj. (That is, wj is the first entry associated with column j, if there are any first entries in column j.) Let c be a common multiple of {wt, W2, ... , in }. Define the u x m matrix E by, for (i, j) E (1, 2, ... , u) x (1, 2, ..., m}, ei, j = (c/wj)di, j. Now, given y E N'", we define z' E N' by, for j E (1, 2, ... , m), zj = (c/wj)yj. Then Dz' = Ey. (f) implies (a). For each c E N, cN is a member of every idempotent in iBN by Lemma 6.6, and is in particular a central* set. Thus by Corollary 15.6 E is image partition regular over N. To see that A is image partition regular over N, let r E N and
let N = Ui'=1 Fi. Pick i E 11, 2, ... , r} and y E N' such that E3 E F,u. Then pick z E N' such that Ax = Ey.
314
15 Partition Regularity of Matrices
Statement (b) of Theorem 15.24 is a computable condition. We illustrate its use by determining whether the matrices and
are image partition regular over N. Consider the matrix
(
3si 2si
1
1
3
2
4
6
152
1
0
3s2
0
1
where sI and s2 are positive rationals. One quickly sees that the only possible choice 1, 2, 3, 4) and then, solving the equations for a set Il of columns summing to 0 is 11
3sl + 52 = 1 2st + 3S2 = 1
one gets sl = 2/7 and s2 = 1/7 and one has established that 3
1
(2
3
is image partition regular. Now consider the matrix
lsl 3st 4sI
1s2 1
0
0 0
252 652
0
1
0
0
1
where again si and s2 are positive rationals. By a laborious consideration of cases one sees that the only non zero choices for si and s2 which make this matrix satisfy the columns condition are Si = 3/5 and s2 = 2/5. Consequently,
is not image partition regular.
Exercise 15.4.1. Prove that the matrix 2 4
0
0
1
9
2 2
3
is image partition regular.
Exercise 15.4.2. Let A be a u x v matrix with entries from w. Prove that A is image partition regular over (N, +) if and only if A is image partition regular over (N, ). (Hint: Consider the proof of Theorem 15.17.)
315
15.5 Matrices with Entries from Fields
15.5 Matrices with Entries from Fields In the general situation where one is dealing with arbitrary commutative semigroups, we restricted our coefficients to have entries from w. In the case of (N, +), we allowed the entries of matrices to come from Q. There is another natural setting in which the entries of a coefficient matrix can be allowed to come from other sets. This is the case in which the semigroup is a vector space over a field. We show here that the appropriate analogue of the first entries condition is sufficient for 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 vector space over a finite field, the columns condition is also necessary for kernel partition regularity. We begin by generalizing the notion of the first entries condition from Definition 15.2.
Definition 15.25. Let F be a field, let u, v E N, and let A beau x v matrix with entries from F. Then A satisfies the first entries condition over F 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} : ai,t # 0} = min{t E {1, 2, ..., v} : aj,t
0},
then ai k = aj,k. An element b of F is a first entry of A if and only if there is some row i of A such that b = ai,k where k = min{t E {1, 2, ..., v} : ai,t 0}, in which case b is the first entry of A from column k. Notice that the notion of "first entries condition" from Definition 15.2 and the special
case of Definition 15.25 in which F = Q are not exactly the same since there is no requirement in Definition 15.25 that the first entries 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 vtuples) each of whose entries is a vector (meaning a member of V). We shall use bold face for the members of V and continue to represent vtuples by an arrow above the letter.
The 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\{0}, one has dV = V so that dV is automatically central* in (V, +).) Accordingly, we leave the proof as an exercise.
Theorem 15.26. Let F be a field and let V be an infinite vector space over F. Let u, v E N and let A be a u x v matrix with entries from F which satisfies the first entries
condition over F. Let C be central in (V, +). Then there exist sequences (xI,n)n0 II (x2,n)n°j, ...,(x,,,n)n° j in V such that for every G E J'f(N), XG E (V\{0})° and AxG E Cu, where
xG =
316
15 Partition Regularity of Matrices
Proof. This is Exercise 15.5.1.
The assertion that V is an infinite vector space over F reduces, of course, to the assertion that either F is infinite and V is nontrivial or V is infinite dimensional over F.
Corollary 15.27. Let F be a field and let V be an infinite vector space over F. Let u, v E N and let A be a u x v matrix with entries from F which satisfies the first entries condition over F. Then A is strongly image partition regular over V. That is, whenever
r E N and V\{0} = U;=1 E;, there exist i E (1,2, ..., r) and! E (V\{0})° such that A. E E;". Proof. We first observe that (0) is not central in (V, +). To see this, note that by Corollary 4.33, V* is an ideal of (fV, +) so that all minimal idempotents are in V*. Consequently one may choose i E {1, 2, ... , r} such that E; is central in (V, +). Pick, by Theorem 15.26 some x' E (V\{O})° with Al E E;". Now we turn our attention to kernel partition regularity. We extend the definition of the columns condition to apply to matrices with entries from an arbitrary field.
Definition 15.28. Let F be a field, let u, v E N, let C be a u x v matrix with entries f r o m F, and let c1, c2, ... , c be the columns of C. The matrix C satisfies the columns condition over F if and only if there exist m E N and I, , 12, ..., Im such that
(1) {I1, 12, ... , Im} is a partition of {1, 2, ... , v}.
(2) EiEJ, c; = 6(3) If m > I and t E {2, 3, ... , m}, let Jt = U;, Ij. Then there exist (&, i ); F such that EiEJ, c; = E;EJ, 5,,; c";.
in
Note that Definitions 15.13 and 15.28 agree in the case that F = Q. Theorem 15.29. Let F be afield, let V be an infinite vector space over F, let u, v E N, and let C be a u x v matrix with entries from F that satisfies the columns condition
over F. Then C is kernel partition regular over V. That is, whenever r E N and V\{0} = U;=1 E;, there exist i E {1, 2, ..., r} and l E E° such that Ci = 0.
Proof Pick m E N, 11, 12, ..., Im, and for t E {2, 3, ... , m}, J, and (St,i
as
guaranteed by the definition of the columns condition.
Define the v x m matrix B by, for (i, t) E {1,2,...,v} x (1,2,...,m}, b;,, =
St,i if i E J, 1 if i E It 0
ifi
Uj_1 Ij.
We observe that B satisfies the first entries condition. We also observe that CB = 0.
(Indeed, let j E ( 1 , 2, ... , u} and t E 11, 2, ... , m}. If t = 1, then Ei_1 cjj b;,, =
E;El, cjj = 0 and if t > 1 then Ei_1 cjj . bi,t = E, .,, St,i . cj,i + EiEt, cjj = 0.)
15.5 Matrices with Entries from Fields
317
Now let r E N and let V\{0} = U* =1 E,. Pick by Corollary 15.27 some i E { 1 , 2, ... , r} and y E (V\{0})' such that By" E E; °. Let . = B$ . Then Cx" _
CBy=0y=0.
We see that in the case that F is a finite field, we in fact have a characterization of kernel partition regularity.
Theorem 15.30. Let F be a finite field, let u, v E N, and let C be a u x v matrix with entries from F. The following statements are equivalent.
(a) For each r E N there is some n E N such that whenever V is a vector space of dimension at least n over F and V\{0} = Ui Ei there exist some i E (1, 2, ... , r} and some x E Ei o such that Cz = 0. (b) The matrix C satisfies the columns condition over F.
P r o o f (a) implies (b). Let r = I F I  I and pick n E N such that whenever V is a vector space of dimension at least n over F and V\{0} = U;=1 Ei there exist some i E (1, 2, ... , r) and some x' E Ei u such that Cx = 0. Let V = X n F. Since in this case we will be working with vtuples of ntuples, let us establish our notation. Given I E V°, we write 1
x and given i E ( 1 , 2, ... , v), xi = (xi (1), xi (2), ... , xi (n)). We color V according to the value of its first nonzero coordinate. For each x E
V\{0}, let y(x) = min{i E {1, 2, ..., n} : x(i) # 0}. For each a E F\{0}, let Ea = {x E V\(01 x(y(x)) = a}. Pick some a E F\{0} and some l E Eau such that Cs = 0. Let D = { y (xi) : i E (1, 2, ... , v) } and let m = I D 1. Enumerate D in increasing
order as (d1,d2,...,dm). For eacht E (1,2,...,m),letIt = {i E (1,2,...,v} y(xi) = d,}. Fort E (2,3..... m), let J, = U'; II and for i E Jt, let Sr,i =
xi(dt) a1 To see that these are as required for the columns condition, we first show that EiEI, ci = 0. To this end, let j E {1, 2..... u} be given. We show that EiEI, cj,i = 0. Now C. = 0 so E 1 cj,i  xi = 0 so in particular, E 1 cj,i  xi (d1) = 0. Now if
i E (1,2,...,v}\Il,then d1 < y(xi)soxi(dl)=0. Thus 0= EiE11 Cj,i xi(dl) _ a . EiEI, Cj,i SO EiE/, Cj,i = 0.
Nowassumem> 1and t E {2,3,...,m}. To see thatEiEl, ci again let j E (1, 2, ..., u) be given and show that EiE/, Cj,i = E cj,i  xi = 0 so in particular, E°_1 cj,i  xi (dt) = 0. Ifi E {1, 2, ..., v}\(It U J,), thenxi(dt) = OsoO = EiE1, c Cj,i'xi(dt) = EiEI, Cj
318
15 Partition Regularity of Matrices
(St j a). Thus a E(EI, Cj,j = «' EiEJ, cjj 8t j and so EiEI, Cj.i = F'iEJ, cj,j ' st,j as required.
(b) implies (a). We use a standard compactness argument. Let r E N and suppose the conclusion fails. For each n E N let Vn = {x E X °_1 F :for each i > n, x (i) = 0}. Then for each n, V,, is an ndimensional vector space over F and V,, c Vn+1
Now given any n E N, there is a vector space V of dimension at least n over F for which there exist E1, E2,..., Er with V\(0) = U,=1 Ej such that for each i E 11, 2, ... , r} and each x E Ej ', Cxi # 0. The same assertion holds for any ndimensional subspace of V as well. Thus we can assume that we have some tpn
:
Vn\{0)) {1,2,...,r}such that foranyi E {1,2,...,r}andanyx" E (rpn1[{i}])°, Ci 96 0. Choose an infinite subset A I ofNsothatforn, m E Al onehasrpnly,\{0} = WmIV,\{0} Inductively, given A,_1, choose an infinite subset At of At1 with min A, > t such that for n, m E At one has Wnly( \101 = `'mlV,\10). For each t pick n(t) E At. Let V = Un__1 Vn. Then V is an infinite vector space over F. For X E V\{0} pick the first t such that x E Vt. Then X E Vt S; Vn(t) because n(t) E At and min At > t. Define v(x) = rpn(t)(x). By Theorem 15.29 pick i E (1, 2, ... , r) and x' E (rp1[{i}])° such that CX = 0. Pick t E N such that {xt , x2, ... , xv} c Vt.
We claim that for each j E {1, 2, ..., v} one has rpn(t)(xj) = i. To see this, let j E { 1, 2, ... , v} be given and pick the least s such that xj E VS. Then n(s), n(t) E A,
SO rpn(t)(xj) = rpn(s)(xj) = rp(xj) = i. Thus x E (rpn(t)1[{i}])° and Ci = 0, a contradiction.
Of course any field is a vector space over itself. Thus Corollary 15.27 implies that, 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 over F.
Exercise 15.5.1. Prove Theorem 15.26 by suitably modifying the proof of Theorem 15.5.
Notes The terminology "image partition 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 [76]. The columns condition was introduced by Rado [206] and he showed there that a matrix is kernel partition regular over (N, +) if and only if it satisfies the columns condition over Q. Other generalizations of this result were obtained in [206] and [207]. The proof that (a) and (c) are equivalent in Theorem 15.20 is based on Rado's original arguments [206]. It is shown in [ 162], a result of collaboration with W. Woan, that there
Notes
319
are solutions to the system of equations x'c = 1 in any central set in (N, ) if and only if C satisfies the columns condition over Z. The proof given here of the sufficiency of the columns condition using the image partition regularity of matrices satisfying the first entries condition is based on Deuber's proof that the set of subsets of N containing solutions to all kernel partition regular matrices is partition regular [76]. The material from Section 15.4 is taken from [142], a result of collaboration with I. Leader, where several other characterizations of image partition regular matrices are 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 partition regular matrices was not obtained until 1993 [142].
Theorems 15.26, 15.29, and 15.30 are from [211, written in collaboration with V. Bergelson and W. Deuber, except that there the field F was required to be countable and the vector space V was required to be of countable dimension. (At the time [211 was written, 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 combinatorial content. And our introduction to the combinatorial applications of the algebraic structure
of ,S came through the Finite Products Theorem (Corollary 5.9). We shall see in this chapter that sets which intersect FP((x )n° I) for every sequence (xn )n° , (that is, the IP*
sets) have very rich combinatorial structure, especially in the semigroups (N, +) and (N, ). Further, by means of the old and often studied combinatorial notion of spectra of numbers, we exhibit a large class of examples of these special sets in (N, +).
16.1 IP, IP*, Central, and Central* Sets in Arbitrary Semigroups Recall that we have defined a central 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 are central in an arbitrary semigroup.
Theorem 16.1. Let S be a semigroup and for each F E J" f(S), let XF E S. Then XF) is central in S.
Proof. Let A = UFEPf(s)(F xF). For each F E Rf (S), let
BF ={XH:HE?f(S)and FCH}. Then (BF : F E ,Pf(S)) is a set of subsets of S with the finite intersection property so choose (by Theorem 3.8) some p E ,9S such that {BF : F E , f(S)} C p. We claim that PS p C A for which it suffices, since pp is continuous, to show that S p C A. To this end, lets E S. Then B(,) S; s1 A so s1 A E p so by Theorem 4.12 A E sp. By Corollary 2.6, there is a minimal idempotent q E fS p and so A E q.
In the event that S is countable, the description given by Theorem 16.1 can be simplified.
16.1 Sets in Arbitrary Semigroups
321
Corollary 16.2. Let S = (an : n E N) be a countable semigroup and let (xn)n° 1 be a sequence in S. Then {an xm n, m E N and n < m} is central in S. Proof. This is Exercise 16.1.1. We now introduce some terminology which is due to Furstenberg [98] and is commonly used in Topological Dynamics circles.
Definition 16.3. Let S be a semigroup. A subset A of S is an IP set if and only if there is a sequence (xn) n° 1 in S such that FP ((xn) n° 1) c A.
Actually, as defined by Furstenberg, an IP set is a set which can be written as FP((xn)n° 1) for some sequence (xn)n° 1. We have modified the definition because we already have a notation for FP((xn)n 1) and because of the nice characterization of IP set obtained in Theorem 16.4 below. The terminology may be remembered because of the intimate relationship between
IP sets and idempotents. However, the origin as described in [98] is as an "infinite dimensional parallelepiped". To see the idea behind that term, consider the elements of FP( (xi )3_ t) which we have placed at seven of the vertices of a cube (adding an identity e at the origin). x1x2x3
x2x3
XIX2
X1
Theorem 16.4. Let S be a semigroup and let A be a subset of S. Then A is an IP set if and only if there is some idempotent p E 0S such that A E P. Proof. This is a reformulation of Theorem 5.12. In general, given a class ,'R of subsets of a 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.
Definition 16.5. Let S be a semigroup and let A C S. Then A is an IP* set if and only if for every IP set B C S, A n B # 0. 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. Let S be a semigroup and let A C S. The following statements are equivalent.
322
16 EP, EP*, Central, and Central* Sets
(a) A is an JP* set. (b) A is a member of every idempotent of 14S.
(c) A fl B is an IP set for every IP set B of S.
Proof. (a) implies (b). Let p be an idempotent of CBS and suppose that A f p. Then in S with FP((xn )n° t) C S\A. S\A E p so by Theorem 5.8 there is a sequence (xn That is, S\A is an IP set which misses A, a contradiction. (b) implies (c). Let B be an IP set and pick by Lemma 5.11 an idempotent p of fS such that B E p. Then A fl B E p so A fl B is an IP set by Theorem 5.8. 1
That (c) implies (a) is trivial. As a trivial consequence of Lemma 15.4, Theorems 16.4 and 16.6, and the definition of central, one has that IP*
central* = central
IP.
One also sees immediately the following:
Remark 16.7. Let S be a semigroup and let A and B be subsets of S. (a) If A and B are IP* sets, then A fl B is an IP* set. (b) If A and B are central* sets, then A fl B is a central* set.
We see now that in most reasonable semigroups, it is easy to produce a specific central* set which is not an IP* set. Theorem 16.8. Let S be a semigroup and assume that (x n ) is a sequence in S such that FP((xn )n°_ 1) is not piecewise syndetic. Then S\ FP((xn) n 1) is a central* set which is not an IP* set.
Proof. Trivially S\FP((xn)n° 1) is not an IP* set. Since FP((xn)n° 1) is not piecewise syndetic, we have by Theorem 4.40 FP((xn)n_1) fl K($S) = 0 so for every minimal idempotent p, S\ FP((xn )n_1) E P. Of course, since the notions of central and IP are characterized by membership in an idempotent, they are partition regular notions. In trivial 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 a semigroup. (a) The notion of IP* is partition regular in S if and only if fS has a unique idempotent. (b) The notion of central* is partition regular in S if and only if K(fS) has a unique idempotent.
Proof We establish (a) only, the other proof being very similar. Necessity. Suppose that fS has two idempotents p and q and pick A E p\q. Then A U (S\A) is an IP* set while neither A nor S\A is an IP* set.
16.1 Sets in Arbitrary Semigroups
323
Sufficiency. Let p be the unique idempotent of fS. Then a subset A of S is an IP* set if and only if A E p. As a consequence of Lemma 16.9 we have the following.
Remark 16.10. Let S be a semigroup. If the notion of IP* is partition regular in S, then so is the notion of central*.
Exercise 16.1.3 shows that one may have the notion of central* partition regular when the notion of IP* is not. We see now that in more civilized semigroups, the notions of EP* and central* are not partition regular. (Theorem 16.11 is in fact a corollary to Corollary 6.43 and Lemma 16.9, but it has a simple self contained proof, so we present it.) Theorem 16.11. Let S be an infinite weakly left cancellative semigroup. There exist disjoint central subsets of S. Consequently, neither the notions of central* nor IP* are partition regular in S.
Proof Let K = IS1 and enumerate Pf(S) as (Fa)a
(Uo
xa E Ssuch that
0
andchoose ya ESsuch that By Theorem 16.1 we have that ( Jo
ya) are disjoint
In any semigroup, we see that IP* sets satisfy a significantly stronger combinatorial conclusion than that given by their definition.
Theorem 16.12. Let S be a semigroup, let A be an IP* set in S, and let (xn)n° 1 be a sequence in S. There is a product subsystem (yn)n°1 of (xn)n 1 such that FP((Yn)n°_1) C A.
Proof. Pick by Lemma 5.11 an idempotent p E fS such that FP((xn)n°_m) E p for every m E N. Then A E p by Theorem 16.6, so Theorem 5.14 applies.
Exercise 16.1.1. Prove Corollary 16.2. Exercise 16.1.2. Let S be a set and let a E S. Define xy = a for all x and y in S. Show that a subset A of S is an IP set if and only if a E A and consequently that the notions of IP* and central* are partition regular for this semigroup.
324
16 IP, IP*, Central, and Central* Sets
Exercise 16.1.3. Recall the semigroup (N, A) where n Am = min(n, m). (See Exercise 4.1.11.) Verify each of the following assertions. (a) In (flN, A) every element is idempotent and K(f N, A) _ {1}. (b) The IP sets 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 (N, A) is N, while the central* sets in (N, 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. Recall the semigroup (N, v) where nvm = max{n, m}. (See Exercise 4.1.11.) Verify each of the following assertions. (a) In (fN, v) every element is idempotent and K(fN, v) = N*. (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* are partition regular in (N, v).
16.2 IP* and Central Sets in N In this section we compare the additive and multiplicative structures of N. We begin with a trivial observation.
Lemma 16.13. Let n E N. Then Nn is an IP* set in (N, +). Proof. This is an immediate consequence of Theorem 16.6 and Lemma 6.6.
0
Now we establish that central sets in (N, +) have a richer structure than that guaranteed to an arbitrary commutative semigroup (and in fact richer than that possessed by central sets in (N, ) ).
Theorem 16.14. Let B be central in (N, +), let A be a u x v matrix with entries from Q. (a) If A is image partition regular, then there exists y E N' such that all entries of Ay are in B. (b) If A is kernel partition regular, then there exists y" E B° such that Ay' = 0. Proof. (a) Pick by Theorem 15.24 some m E N and a u x m matrix D satisfying the first
entries condition with entries from w such that for any i E N' there is some y E N" with Ayy = Di. By Lemma 16.13, for each n E N, Nn is an IP* set and hence a central* set. Consequently, by Theorem 15.5, pick 'z E N' such that all entries of Dz are in B and pick y' E N' such that Ay" = Dzz. (b) Let d E N be a common multiple of all of the denominators of entries of A. Then dA is a matrix with entries from 7G which is kernel partition regular. (If N = u =1 C,,
325
16.2 IP* and Central Sets in N
pick i E (1, 2, ... , r) and x' E C`;' such that Ax' = 0. Then (dA)x'. = 0.) Thus by Rado's Theorem (Theorem 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 the first entries condition such that d AC = 0. As in part (a), pick x E NM such that all entries of Cxi are in B and let j = Cx'. Then dAy = dAC! = 01 = 0 so Ay" = 0. In fact, central sets in (N, +) 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. be a sequence of subsets Definition 16.15. Let (S, +) be a semigroup and let of S. Then FS((Y,,)n° 1) = {EfEF an : F E .Pf(N) and for all n E F, an E 1
We define FS((Y,,)n 1) and
analogously.
Theorem 16.16. Let B be central in (N, +). Let (A(n))n° enumerate the image partition regular matrices with entries from Q and for each n, let m(n) be the number of columns of A(n). There exists for each n r= N a choice of z(n) E N,(n) such that, if Yn is the set of entries of A(n)x(n), then FS((Yn)' 1) c B. 1
Proof. Pick a minimal idempotent in ($N, +) such that B E p. Pick by Theorem 16.14 some 7C(I) E N'(1) such that all entries of A(I)i(l) are in B* = {a E B : a + B E P). Let Y1 be the set of entries of A(1)x"(1).
Inductively, let n E N and assume that we have chosen x(k) E N, (k) for each k E {1, 2, ... , n} so that, with Yk as the set of entries of A(k)x'(k), one has FS((Yk)k=1) c
B*. By Lemma 4.14, for each a E B*, a + B* E p so
B* fl n{a + B* : a E FS((Yk)k_1)} E p so pick x"(n + 1) E Nm(n+1) such that all entries of A(n + 1)x"(n + 1) are in
B* fl n{a + B* : a E FS((Yk)k=1)}. Letting Yn+l be the set of entries of A(n + 1). (n + 1) one has FS((Yk)k±i) C B. A similar result applies to kernel partition regular matrices.
Theorem 16.17. Let B be central in (N, +). Let (A(n))n° 1 enumerate the kernel partition regular matrices with entries from Q and for each n, let m(n) be the number of columns of A(n). There exists for each n E N a choice of x' (n) E N'(') such that A(n)z(n) = 0 and, if Yn is the set of entries of. (n), then FS((Yn)' 1) C B. Proof. This can be copied nearly verbatim from the proof of Theorem 16.16.
The contrast with Theorems 16.14, 16.16, and 16.17 in the case of sets central, in fact central* in (N, ) is striking. Recall from Section 15.3 the notation za = 1 which represents the multiplicative analogue of the equation Az = 0.
326
16 IP, IP*, Central, and Central* Sets
Theorem 16.18. Let C = N\(x2 : x E N), let A = (2) and let B = (2 2 1). Then C is central* in (N, ), A is image partition regular in (N, ), and B is kernel partition regular in (N, ). However, for no x E N is X2 E C and for no z E C3 is
xe=1. Proof The reader was asked to show in Exercise 15.1.2 that (x2 : x E N) is not central in (N, ). Consequently, its complement must be central*. Trivially A is image partition regular in (N, ) and B satisfies the columns condition over Q, so is kernel partition regular in (N, ) by Theorem 15.20. The following lemma establishes that multiplication preserves IP* sets in (N, +).
Lemma 16.19. Let A C N and let n E N. The following statements are equivalent.
(a) A is an IP* set in (N, +). (b) n1A is an IP* set in (N, +). (c) nA is an IP* set in (N, +). Proof. (a) implies (b). Assume that A is an lP* set, and let B be an IP set. Pick a sequence
(xt)°_, such that FS((xt)t°O1) c B. Then FS((nxt)°1) fl A 96 0 so n1A n B # 0. (b) implies (a). Assume that n1A is an IP* set, and let B be an EP set. Pick
a sequence (x,)', such that FS((xt)°°,) C B. By Lemma 16.13, Nn is an IP* set in (N, +) so pick by Theorem 16.12 a sum subsystem (yt)°°, of (x,)°_1 such that FS((yt)°_°,) C Nn. In particular n divides yt for each t. Consequently
FS((A )°_1) fl n1A # 0 so that FS((yt)°O,) fl A 54 0. Since FS((yt)°_,) S; FS((xt)°__i) c B, we are done. Since n1(nA) = A, the equivalence of (a) with (c) follows from the equivalence of (a) with (b). We see now that IP* sets in (N, +) are guaranteed to have substantial multiplicative structure. Theorem 16.20. Let 3 be the set of finite sequences in N (including the empty sequence)
and let f : 8 s N. Let (yn )n° , be a sequence in N and let A be an IP* set in (N, +). Then there is a sum subsystem (xn)n° , of (yn)n°_, such that whenever F E J'f(N), Z = f ((xk)k 1 F1), and t E (1, 2, ... , 8), one has t EnEF Xn E A. Proof. Pick by Lemma 5.11 some p = p + p in fiN such that FS((yn)n_m) E p for each m E N. Then by Lemma 16.19, we have for each t E N that t1 A is an lP* set in (N, +) and hence is in p. Let B1 = FS((yn)n° 1) fl n` 10) tA and note that B1 E P. Pick X1 E B1* and pick H1 E .% f(N) such that x1 = EtEH, Yt Inductively, let n E N and assume that we have chosen (x i) l , , (H,)=,, and (B,)"_, such that for each i E { 1, 2, ... , n}:
16.2 IP* and Central Sets in N
327
(1) Xi = ErEH; yr, (2) if i > 1, then min Hi > max H1 _ 1,
(3) Bi E p, (4) if 0 F c {1, 2, ... , i} and m = min F, then EjEF xj E Bm*, and
(5) if i > 1, then Bi C n;((Xj)'') CIA. Only hypotheses (1), (3), and (4) apply at n = I and they hold trivially. Let k = max Hn + 1. By assumption FS((yt)°__k) E p. Again we have by Lemma 16.19 that for each t E N, t1 A is an IP* set in (N, +) and is thus in p. For each m E (1, 2, ... , n} let
E. = {EjEFXi : 00 F C {1,2,.._,n} andm =minF}. By hypothesis (4) we have for each m E (1, 2, ... , n) and each a E Em that a E Bm* and hence, by Lemma 4.14, a + Bm* E p. Let Bn+1 = FS((yt)°O_k) n
nr(IXJ)J_i)
t' A n nm_1 naEE,, (a + Bm*).
Then Bn+l E P. Choose E Bn+l*. Since xn+l E FS((yt)°_k), choose k and xn+i = EtEHn+, yt. Then hypotheses (1), Hn+l E Rf (N) such that min (2), (3), and (5) are satisfied directly. To verify hypothesis (4), let 0 0 F C 11, 2, ... , n + 1} and let m = min F. If n + 10 F, the conclusion holds by hypothesis, so assume that n+ I E F. If F = In + 1), then EjEF Xj = Xn+1 E Bn+1*, so assume F 54 In + 1} and let G = F\{n + 1}. Let
a = EjEGxj. Thena E E. SOXn+1 E a + Bm* SO EjEFXj =a+Xn+l E Bm* as required. We thus have that (xn )n° is a sum subsystem of (yn ) 1
i
To complete the proof, let
_ F1 ), and let t E 11, 2, ... , e}. Let m =min F. Then F E PPf (N), let e= f ((xk)in by hypotheses (4) and (5), EjEF xj E Bm g tA sot EnEF Xn E A. Corollary 16.21. Let A be an IP* set in (N, +) and let (yn)no_1 be a sequence in N. There is a sum subsystem (xn)n° 1 of (yn)n_1 such that
FS((xn)n_1) U FP((xn)n° 1) C A.
Proof Let 3 be the set of finite sequences in N. Define f(0) = 1 and given (xj)ixi. Choose (xn)n_1 as guaranteed by Theorem 16.20. define f ((xj)nj_I) = fl Letting t = 1, one sees that FS((xn)n° 1) C A. To see that FP((xn)n 1) c A, let F E .% f (N) and let n = max F. If I F = 1, then f I j E F xj = xn E A, so assume I F l > 1
andletG = F\{n}. Lett = fljEGxj. Thent <
E A.
Theorem 16.20 and Corollary 16.21 establish that an IP* set in (N, +) 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
328
16 IF, IP*, Central, and Central* Sets
which are central* in (N, +) have significant multiplicativestructure, in fact are central in (N, ). On the other hand, central sets in (N, +) need have no multiplicative structure at all (Theorem 16.27) while central sets in (N, ) must have a significant amount of additive structure (Theorem 16.28).
Definition 16.22. M = {p E ON: for each A E p, A is central in (N, +)}. As an immediate consequence of the definition of central we have the following.
Remark 16.23. M = cf E(K(6N, +)). Theorem 16.24. M[ is a left ideal of (13N, ). Proof. Let P E M[ and let q E flN. To see that qp E MI, let B E qp and pick a E ICY such that a1 BE p. Then a1 B is central in (N, +) so pick, by Theorem 14.25 a decreasing
sequence (C,)', such that (i) for each n E N and each x E Cn, there is some m E N such that Cm g x + Cn and (ii) {Cn : n E NJ is collectionwise piecewise syndetic.
Then (aCn)n 1 is a decreasing sequence of subsets of B and immediately one has that for each n E ICY and each x E a Cn, there is some m E N such that aCm c x +a Cn .
It thus suffices by Theorem 14.25 to show that {aCn : n E N} is collectionwise piecewise syndetic. That is, using Lemma 14.20, that there exist sequences (g(n))n° and (xn)' 1 in N such that for all n, m E N with m > n, 1
{Xm+1,xm+2,...,Xm+m} c US(n1(.1+ni=l aCi) If a = 1, the conclusion is immediate from the fact that {Cn : n E N } is collectionwise piecewise syndetic so assume that a > 1. Since {Cn : n E N} is collectionwise piecewise syndetic, choose sequences (h (n))' n=1 and (yn )n° 1 in N such that for all n, m E N with m > n,
For each n E N, let xn = ayn and let g(n) = ah(n) + a. Given n, m E N with m > n and given k E [1, 2, ... , m }, pick r E {0, 1, ..., a  11 such that k + r = al for some
I E {1,2,...,m}. Then for some j E {1,2,...,h(n)}wehaveym+I Ej+ni=1 Ci so that aym + al E aj + ni=1 aCi so xm + k E (aj + r) + n,1 aCi. Since aj +r < g(n) we are done. We saw in Corollary 13.15 that K(PN, ) fl K(PN, +) = 0. We pause to observe now that these objects are nonetheless close.
Corollary 16.25. K($N, ) fl ct K(#N, +) 0 0.
16.2 IP* and Central Sets in N
329
Proof. By Theorem 16.24,1 is a left ideal of (ON, ) and consequently has nonempty intersection with K(8N, ) while by Remark 16.23, M C ci K(18N, +).
Corollary 16.26. Let A be central* in (N, +). Then A is central in (N, .).
Proof. By Lemma 15.4, E(K (ON, +)) C A so M C A. Since M is a left ideal of (ON, ), M n E(K(fN, )) 0 0 by Corollary 2.6 so A is central in (N, ). Theorem 16.27. There is a set A C N which is central in (N, +) such that for no y and z in N is {y, z, yz} C A. In particular for no y E N is {y, y2} C A. Proof. Let x1 = 2 and inductively for n E N choose
(xn + n)2. Notice
in particular that for each n, x > n. Let A = {xn + k : n, k E N and k < n}. By Corollary 16.2 A is central in (N, +). Suppose now we have y < z in N with {y, z, yz} C A and pick n E N such that z E {Xn + 1, xn + 2, ... , xn + n}. Then y > 2 so yz > 2z > 2xn > xn + n so yz > Xn+1 + 1 > (xn + n)2 > z2 > yz, a contradiction. We see now that sets central in (N; ) must contain large finite additive structure.
Theorem 16.28. Let A C N be central in (N, ). For each m E N there exists a finite sequence (xt )m 1 such that FS ((xt )m 1) C A.
Proof Let
T={p EON: for all BEpand all In ENthere exists (xt)m1 with FS((xt)m1) C B}. By Theorem 5.8, all idempotents in (ON, +) are in T so T 0 0. We claim that T is a two sided ideal of (ON, ) so that K(fN, ) C T.
Let P E T and let q E ON. To see that q p E T, let B E q p and let m E N. Pick a E N such that a1 B E p and pick (xt ).1 such that FS ((xt )m 1) C a1 B. Then
FS((axt)m1) C B.
Toseethatp  gET,letBEp  q andletmEN.LetC={a EN: a1BEq}. Then C E p so pick (xt)m1 such that FS((xt)m1) C C. Let E = FS((xt)m1). Then E is finite, so naEE a1 B E q so pick b E naEE a1 B. Then FS((bxt )7 1) C B. Since A is central in (N, ), pick a minimal idempotent p in (ON, ) such that A E P.
Then p E K(i4N, ) c T so for each m E N there exists a finite sequence (xt)m1 such that FS((xt)m1) 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 C_ N which is central in (N, ) such that for no sequence (yn)n
1
does one have FS((yn)n 1) C A.
16 IP, IP*, Central, and Central* Sets
330
Proof Let xt = 1 and for n E N pick some xn+1 > nxn. Let
A={kxn:n,kENandk
system of (yn )'1 can be chosen. Pick m, n, and r in N such that n < r, Ym E {xn, 2xn, 3xn, ... , nxn}, and ym+t E {xr, 2Xr, 3Xr, ... , rxr}. Pick k E (1, 2, ... , r} such that ym+t = kxr. Then
kxr < ym + ym+t < kxr + nxn < kxr + xr = (k + 1)Xr
sOym+Ym+i 0A. Given any central set A in (N, +), one has by definition that there is a minimal idempotent p in (fiN, +) such that A E p. Since p = p + p, one has that
{xEN:x+AEp}Ep and so, in particular, (x E N : x + A is central) is central. That is, every central set often translates down to a central set.
Recall that for any p E ON, p = (1) p E fiZ. Since by Exercise 4.3.5, N* is a left ideal of (f Z, +), one cannot have p = p + (p) for any p E N*. If one had p = (p) + p for some minimal idempotent pin (fiN, +) one would have as above that for any A E p, {x E N : x + A E p) E p and hence {x E N :_x +A is central} is central. However, according to Corollary 13.19 such an equation cannot hold. Nonetheless, we are able to establish that for any central set A in (N, +), {x E N : x + A is central) is central.
Theorem 1630. Let P E K(fiN, +) and let L be a minimal left ideal of (ON, +). There is an idempotent q E L such that p = (q) + p.
Proof. Let T = {q E L : (q)+p = p}. Givenq, r E ON, (q+r) = (q)+(r) by Lemma 13.1, soifq,r E T,thenq+r E T. Define : f7G + fiZbye(p) = p. (That is,inthesemigroup(f7G, ),2 =.1._1.) ThentiscontinuousandT = Lfl(ppot)1[{p}] so T is compact. Therefore, it suffices to show that T # 0. For then, T is a compact subsemigroup of ON, hence contains an idempotent by Theorem 2.5.
Since P E K(fN, +), pick a minimal left ideal L' of ON such that p E L'. Now N* is a left ideal of f7L by Exercise 4.3.5 so by Lemma 1.43(c), L and L' are left ideals
of fZ. In particular L + p S L. We claim that L + p is a left ideal of ON so that
L+p = L'. Toseethis,letq E Landletr E ON. Weshowthatr+q+p E L+p. Now q E L so r + q E L. Since r + q = (r + q) by Lemma 13.1 we have
r + q E L so r + q + p E L + p. Since L+p=L'andpEL'wehave T#0 as required.
16.2 IP* and Central Sets in N
331
Corollary 16.31. Let A C N be central in (N, +). Then (x E N : x + A is central) is central:
Proof. Pick a minimal idempotent p in (fN, +) such that A E p and pick, by The
orem 16.30, a minimal idempotent q in (,ON, +) such that p = (q) + p. Then
{xEN:x+AEp}Egand Ix EN:x+AEp}C{xEN:x+A is central}. We do not know whether every central* set often translates either up or down to another central* set. However we do have the following strong contrast with Corollary 16.31 for IP* sets. Theorem 16.32. There is an JP* set A of (N, +) such that for all n E N, neither n + A nor n + A is an JP* set.
Proof. Let (Dn)nEZ be a sequence of pairwise disjoint infinite sets of positive even integers such that for each n, min D,, > 12n 1. For each n E Z enumerate D. in 2a(n,2k1) increasing order as (a(n, k))k_1 and for each k E N, let Yn,k = For each n E 76\{0}, let Bn = FS((yn,k)k° 1). Let C = {n +z : n E 76\{0} and z E Bn}.
Notice that if n E Z\{0} and Z E Bn, then Ini < z so C c N. Let A = N\C. Then given any n E 7G\(0), (A + n) fl B_n = 0, so A + n is not an IP* set. We now claim that A is an EP* set, so suppose instead we have a sequence (xn)n_1 with FS((xn)n° 1) fl A = 0. That is, FS((xn)n° 1) c C. By passing to a suitable sum subsystem we may presume that the sequence (xn)n° 1 is increasing. We first observe that if n E Z\{O}, U E n + Bn, and u = EtEF 2', then max F E D.
Indeed u = n + z where z E Bn and z = EtEH 2' where H C Dn and H has at least two members (because each yn.k has two binary digits). Thus, if n < 0, borrowing will not reach max H. Since then min H > 12n I one has max F = max H. We now show that one cannot haven E 7.\{0} and i < j with {xi, x!) C n + Bn, so
suppose instead that we do. Now xj = E,EF 2' where max F E D. Let k = max F. Since xi < xj we have that xj + xi = E,EH 2` where either max H = k or max H = k + 1. But as we have just seen, given any member of C, the largest element of its binary support is even, so the latter case is impossible. The former case tells us that xj + xi E n + B. But now {xi  n, xj  n, xi + xj  n) C Bn, so if P = min Dn we have that 2e divides each of xi  n, xj  n, and xi + xj  n, and hence 21 divides n, a contradiction.
Consequently, we may choose for each i E N. some n(i) E Z\{0} such that xi E
n(i) + Bn(,) and n(i) 96 n(j) for i # j. Choose i such that Jn(i)i > x1. Then xi = E,EF 2t and xi + xl = E,EG 21 where max F = max G and consequently, Xi + X1 E n (i) + Bn(i ). But now x1 = (xi + x1)  xi is a difference of two members of B,,(i) and is hence divisible by 2e, where f = min D(1). This is a contradiction because
f > 12n(i)I > xi. Exercise 16.2.1. In the proof of Lemma 16.19 we used the fact that in the semigroup (N, +), one must have n1(nA) = A. Show that in (N, +) one need not have n(n1 A) = A. Show in fact that n(n1 A) = A if and only if A c Nn.
16 UP, IP*, Central, and Central* Sets
332
Exercise 16.2.2. Prove that if r E N and N =
C and for only one i
E
{1, 2, ... , r} is there a sequence (yn)n° 1 in N with FS((yn)n° 1) c C,, then in fact for this i, C, is an IP* set (so that there is a sequence (xn)n° 1 in N such that FS((xn)n° 1) U FP((xn)n° 1) C C1),
16.3 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 16.33. (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 E S one has 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 E S one has (x + 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 regarding order of oper
ations. That is x y + x z = (x y) + (x z). Notice that neither of the semigroups (S, +) nor (S, ) is assumed to be commutative. 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 right zero semigroup. Then (S, +, ) is left weak ring. If (S, +) has at least one element which is not idempotent, then (S, +, ) is not a right weak ring. Proof. This is Exercise 16.3.2. Analogously to Lemma 16.19, we have the following. Notice that it does not matter whether we define to be a1(Ab1) or (a1A)b1. In either case, y E a1 Ab1 if and only if ayb E A. a1Ab1
Lemma 16.35. Let S be a set, let A C S, and let a, b E S. (a) If (S, +, ) is a left weak ring and A is an IP* set in (S, +), then a A is an 1P* set in (S, +). (b) If (S, +, ) is a right weak ring and A is an IP* set in (S, +), then Ab1 is an IP* set in (S, +). (c) If (S, +, ) is a weak ring and A is an IP* set in (S, +), then a Ab1 is an IP* set in (S, +).
16.3 IP* Sets in Weak Rings
333
Proof It suffices to establish (a) since then (b) follows from a leftright switch and (c) follows from (a) and (b). So, let (xn)n° be a sequence in S. Then 1
FS((a xn)n°O__1) n A # 0
so pick F E J"f (ICY) such that EnE F a x E A. Then, using the left distributive law we have that E a1 A. Recall that in
1), the products are taken in increasing order of indices.
Definition 16.36. Let (S, ) be a semigroup, let (xn)n° l be a sequence in S, and let k E N. Then AP((xn)n_1) is the set of all products of terms of (xn)n_1 in any order with no repetitions. Similarly AP((xn)n_ in any order with no repetitions.
is the set of all products of terms of (xn)n1
For example, with k = 3, we obtain the following: AP((xn)n=1) = (xl, X2, X3, Xlx2, XIx3, x2x3, X2Xl, X3X2, X3x1, XiX2x3, x1 X3X2, X2XIX3, X2x3x1, X3XIX2, X3X2X1 ).
Theorem 16.37. Let (S, +, ) be a left weak ring, let A be an IP* set in (S, +), and let (yn)' be any sequence in S. Then there exists a sum subsystem (xn)n° 1 of (yn)n 1 1
such that if m > 2, F E Rf(N) with min F > m, and b E AP((xn)n=11), then b  EtEF Xt E A. In particular FS((xn)n_1) U {b xm : m > 2 and b E AP((xn)n_11)} C A.
Proof. Pick by Lemma 5.11 some idempotent p of ($S, +) with
OEnm 1ceFS((xn)nom) Then by Lemma 16.35, we have for each a E S that a A is an IP* set in (S, +) and hence is in p. Let B1 = FS((yn)n 1) and note that B1 E p. Pick xl E B1* and pick H1 E J" f(1`N) such that xl = EtEHI Y,Inductively, let n E Nandassume that wehave chosen (xi)i1, (Hip"=1,and(B,)"_l such that for each i E (1, 2, ... , n): (1) Xi = EIEH, Yt, (2) if i > 1, then min Hi > max Hi _ 1,
(3) Bi E p,
(4) if 0¢ FC (1,2,...,i) andm=minF,then EjEFxj E B.*, and (5) if i > 1, then Big n{a1A : a E AP((xt)r=i)}.
334
16 EP, IP*, Central, and Central* Sets
Only hypotheses (1), (3), and (4) apply at n = 1 and they hold trivially. Let k = max Hn + 1. By assumption FS((yt)°_k) E p. Again we have by Lemma 16.35 that for each a E S, at A is an EP* set in (S, +) and is thus in p.
Foreach m E (1, 2,..., n) let
Em={EjEFXi:0#FC{1,2,...,n}and m=min F}. By hypothesis (4) we have f o r each m E { 1, 2, ... , m} and each a E Em that a E Bm* and hence, by Lemma 4.14, a + Bm* E p. Let
Bn+1 = FS((Yz)°O_k) n n{alA : a E AP((x,)i 1)} n nm=1 I IaEEm(a + Bm*).
Then Bn+l E p. Choose xn+l E Bn+l*. Since Xn+l E FS((yt)O0_k), choose Hn+l E ? f(N) such that min Hn+t > k and xn+l =
Yt Then hypotheses (1), (2), (3), and (5) are satisfied directly. T o v e r i f y hypothesis (4), let 0 # F C { 1, 2, ... , n + 1) and let m = min F. If n+ 10 F, the conclusion holds by hypothesis, so assume that n + 1 E F. If F = {n+ 1),
then EjEF Xj = Xn+t E Bn+1*, so assume F # {n + 11 and let G = F\{n + 1}. Let a = EEEF xj. Then a E E. SO Xn+l E a + Bm* SO EjEF Xj = a +xn+l E Bm* as required. We thus have that (xn )n° 1 is a sum subsystem of (yn) ' 1. To complete the proof, let m > 2, let F E 9f (N) with min F > m and let a E AP((xn )n=11). Then by hypotheses
(4) and (5), EjEF xj E Bm c alA so that a EtEF xt E A. Observe that {b xm : m > 2 and b E AP((xn)n_1t)} is the set of all products (without repetition) from (xn)n° 1 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 stronger conclusion in the event that one has a weak ring rather than just a left weak ring.
Theorem 16.38. Let (S, +, ) be a weak ring, let A be an IP* set in (S, +), and let be any sequence in X. Then there exists a sum subsystem (xn)n° of (y, )R° 1 in S such that FS((xn)R° 1) U AP((xn)n° 1) C A. 1
Proof Modify the proof of Theorem 16.37 by replacing induction hypothesis (5) with: (5) if i > 1, then
Big n {alA : a E AP((x,)i_',)} n n {Ab'1 : b E AP((x,)'_',)} n n {atAb1 : a, b E AP((x,)'=',)}. Then replace the definition of Bn+1 with
Bn+1 = FS((Ys)tmk) n n {a1A : a E AP((x,)i_',)} n n
{Ab1
: b E AP((x,)'=j)}
n n {a Abt : a, b E AP((x,)ti)} n nm=1 naEE, (a + Bm*).
16.3 IP* Sets in Weak Rings
335
Theorems 16.37 and 16.38 raise the natural question of whether the stronger con
clusion in fact holds in any left weak ring. The example of Lemma 16.34 is not a counterexample to this question because, in this left weak ring, given any sequence (xn)n° i, one has AP((xn)°_t) = {xn : n E N). Theorem 16.39. Let S be the free semigroup on the two distinct letters a and b and let hom(S, S) be the set of homomorphisms from S to S. Let o be the usual composition of functions and define an operation ® on hom(S, S) as follows. Given f, g E hom(S, S) and u i, u2, ..., ut E {a, b},
(f (D g)(uiu2...u,) = f(uI) g(ul) 'f(u2) g(u2) ...f(ut)g(ut) Then (hom(S, S), ®, o) is a left weak ring and there exist an IP* set A in (hom(S, S), ®) and a sequence (fn)n° 1 in hom(S, S) such that no sum subsystem (gn)n°t of
has AP((gn)n°1) C A.
Proof The verification that (hom(S, S), (D, 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. Define ft E hom(S, S) by fi(a) = ab and fi(b) = b and define inductively for n E N, fn+i = fn ® fl. Notice that for each n E N, fn (a) = (ab)n and fn (b) = bn. Let A = hom(S, S)\{ fr o fs : r, s E N and r > s). Notice that, given r, s E N,
fr(fs(a)) = fr((ab)s)) = (fr(a)fr(b))s = ((ab)rbr)S. (We have used the fact that fr is a homomorphism.) In particular, notice that if fr o fs =
fm o f, then (r, s) = (m, n).
Weclaimthatifm,n,r,s,f,t E N,h = fno ,,k= fro fs,andh®k= ftoft, then t =m = r and t =n +s. Indeed, ((ab)tbt)t = (h (9 k)(a) = h(a)k(a) =
((ab)mbm)n((ab)rbr)s
Suppose now that (hn)t°t°_1 is a sequence in hom(S, S) with
FS((hn)n°_1)C{fro fs:r,sENandr>s}. Then, using the fact just established, there is some r and for each n some s(n) such that
hn = fro fs(n). Then hi
where t = En
®h2®...®hr =froft
s(n) > r, a contradiction. Thus A is an IP* set in (hom(S, S), +). ° with AP((gn )n° t) c A.
Finally, suppose that (gn )n°i is a sum subsystem of (f,
1
Then gi = EnEH fn for some H E pf(N) so gi = fk where k = E H. Then gk+t = ft for some I > k and thus gk+t o gt = ft o fk ¢ A; a contradiction. Exercise 16.3.1. Show that if (S, +, ) is a weak ring in which (S, +) is commutative, n E N and the operations on the n x n matrices with entries from S are defined as usual (so, for example, the entry in row i and column j of A B is Ek=1 ai,k bk,j), then these matrices form a weak ring. Give an example of a left weak ring (S, +, ) for which (S, +) is commutative, but the 2 x 2 matrices over S do not form a left weak ring.
16 IF, IP*, Central, and Central* Sets
336
Exercise 16.3.2. Prove Lemma 16.34.
Exercise 16.3.3. Prove that (hom(S, S), ®, o) as described in Theorem 16.39 is a left weak ring.
Exercise 16.3.4. We know that the left weak ring (hom(S, S), ®, o) of Theorem 16.39 is not a right weak ring because it does not satisfy the conclusion of Theorem 16.38. Establish this fact directly by producing f, g, h E hom(S, S) such that (f (D g) o h 0
f oh®goh.
16.4 Spectra and Iterated Spectra Spectra of numbers are sets of the form ([na j : n E N} or { [nct + y J : n E N} where a and y are positive reals. Sets of this form or of the form { [naj : n E A} or { [na + 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 central* sets is considerably different however. We know from Theorem 16.8 that whenever (x,,)n_, is a sequence in N such that FS ((x,,)n° ,) is not piecewise
syndetic, one has N\FS((xn)n° t) is a central* set which is not IP*. So, for example, {FIEF 21 F E ?f (N) and some t E F is even} is central* but not IP* because it is N\FS((22tt)oo')
We also know from Lemma 16.13 that for each n E N, 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.) In this section we shall be utilizing some information obtained in Section 10.1. Recall
that we defined there for any subsemigroup S of (R, +) and any positive real number a, functions ga : S > 7L, fa : S  [2, 2), and ha S + T by g,,, (x) [xa + 1J, fa (x) = xa  ga (x), and ha (x) = 7r (fa (x)) where the circle group T = R/7L and n is the projection of R onto T.
Definition 16.40. Let a > 0 and let 0 < y < 1. The function ga,y : N > co is defined
byga,y(n) = [na+yJ. We denote by g;,y, the continuous extension of gay from #N to jaw. Notice that ga = ga, z Recall that for a > 0 we have defined Za = (p E CBS : fa(p) = 0).
Lemma 16.41. Let a > 0, let 0 < y < 1, and let p E Za. Then F.,y (p) = ga (p).
16.4 Spectra and Iterated Spectra
337
Proof. Let c = min(y, 1  y} and let A = (n E N : E < fa (n) < e). Then A E p so it suffices to show that ga,y and ga agree on A. So let n E A and let m = g,, (n). Then
m  E < na <m+6so
m <m  E+y < na+y < m + E + y <m+1 so m = ga,y (n). As regards spectra, we see that "as you sew, so shall you reap".
Theorem 16.42. Let a > 0, let 0 < y < 1, and let A C N. (a) If A is an IP* set, then ga,y[A] is an IP* set. (b) If A is a central* set, then ga,y[A] is a central* set. (c) If A is a central set, then ga,y[AI is a central set. (d) If A is an IP set, then ga, y [A] is an IP set. Proof. (a) By Theorem 16.6 we need to show that ga.y[A] is a member of every
idempotent in (flN, +). So let p be an idempotent in ($N, +). Since Zi1a is the kernel of a homomorphism by Lemma 10.3 we have that p E Z1la. Since, by Theorem 10.12, gi/« is an isomorphism from ZI/a to Za, j1/; (p) is an idempotent and so A E gj/«(p). By Lemma 16.41 ga,y(gi/«(p)) = g«(j (p)) which is p by Theorem 10.12. Since (p) and g., y (ji7 (p)) = p we have that g«,y [A] E p as required. AE (b) Let p be a minimal idempotent of (fiN, +). We need to show that ga,y[A] E P. Now P E K(flN) fl Z1 and K(,BN) fl Zi1« = K(Zila) by Theorem 1.65. Thus p is a minimal idempotent in Zl/a so by Theorem 10.12 gj/«(p) is a minimal idempotent in Za. That is, g11.(p) E K(Z«) = K(66N) fl Z so gig«(p) is a minimal idempotent in (,8N, +). Consequently, A E gji«(p) so as in part (a), ga,y[A] E p.
(c) Since A is central, pick a minimal idempotent p with A E p. Then P E K(8BN) fl Z« = K(Z,,) so by Theorem 10.12, g«(p) E K(Zil«) = K(jN) fl Z11« so g«(p) is a minimal idempotent and g«,y[A] E ga y(P) = (p) (d) Since A is an IP set, pick an idempotent p with A E p. Then P E Z« so by Theorem 10.12, ga (p) is an idempotent and ga, y [A] E g«, y (p) = 8a (P).
In the event that a > 1 we get an even stronger correspondence, because then ga,y is one to one.
Corollary 16.43. Let a > 1, let 0 < y < 1, and let A C N. (a) A is an IP* set if and only if ga, y [A] is an IP* set. (b) A is a central* set if and only if ga, y [A] is a central* set. (c) A is a central set if and only if ga,y [A] is a central set. (d) A is an IP set if and only if ga, y [A] is an IP set.
Proof. 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 ga.y[A] is an IP* set and suppose that A is not an IP* set. Pick an IP set B such that A fl B = 0. Since ga.y is
338
16 IP, EP*, Central, and Central* Sets
one to one, ga,y[A] fl ga,y[B] = 0 while by Theorem 16.42(d), ga,y[B] is an IP set, a contradiction. (c) The necessity is Theorem 16.42(c). Assume that ga,y[A] is a central set and suppose that A is not a central set. Then N\A is a central* set so by Theorem 16.42(b), ga,y[N\A] is a central* set. But this is a contradiction because ga,y[A] fl ga,y[N\A] _ 0. Because for a > 1 the output of ga,y is the same kind of set as the input, one may iterate the functions 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 {L[n
+.2Jrr+ 1i:nEA} n
is a central* set which is not IP*. Notice also that the description of a set ga, y [N] is effective. That is, one only needs a sufficiently precise decimal approximation to a and y in order to determine whether a specified number is a member of ga, y [N].
Notes The notions of IP, IP*, central, and central* are due to H. Furstenberg in [98] where central sets were defined in terms of notions from topological dynamics. See Chapter 19 for a proof of the equivalence of the notions of central. Theorem 16.20 is from [ 120] where it had a purely combinatorial proof. Most of the remaining results of Section 16.2 are from [31 ], obtained in collaboration with V. Bergel
son, except for Theorem 16.16 (which is from [78] and was obtained in collaboration with W. Deuber) 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 (hom(S, S), ®, o) is due to J. Clay in (65] where it is given as an example of a left nearring 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 are from [33], results of collaboration with V. Bergelson and B. Kra. H. Furstenberg [98, Proposition 9.4] gives an explicit description of certain IP* sets in terms of topological dynamics, and other examples can be deduced from his paper [100] with B. Weiss. Spectra of the form ([na + y j : n E N) were introduced by T. Skolem [224] and given the name the ynonhomogenous spectrum of a by R. Graham, S. Lin, and C. Lin in [ 109]. See the introduction to [33] for a brief history of the spectra ([na J : n E N) and further references.
Chapter 17
Sums and Products
We saw in Chapter 16 that any IP* set in (N, +) has extensive multiplicative structure in addition to the additive structure that one would expect. In particular, by Corollary 16.21, if A is an IP* set in (N, +), then there is some sequence (xn)n° 1 in N such that FS((xn)n° 1) UFP((xn)n 1) C A. On the other hand,1P* sets in (N, +) are not partition regular by Theorem 16.11 so this fact does not yield any results about finite partitions of N. In fact, we shall show in Theorem 17.16 that there is a finite partition of N such that no cell contains all pairwise sums and products from the same sequence. We saw in Corollary 5.22 that given any finite partition of N, there must be one cell A and sequences and (yn)n° 1 with FS((x,)°O 1) U FP((yn)R° 1) c A. We are concerned in this chapter with extensions of this result in various different directions.
17.1 Ultrafilters with Rich Additive and Multiplicative Structure Recall that we have defined M _ {p E flN : for all A E p, A is central in (N, +)} and
A=(pc fN:forallAEp,d(A)>0). Definition 17.1. A combinatorially rich ultrafilter is any p E M n A fl K (ON, ) such
thatp  p=p. We shall see the reason for the name "combinatorially rich" in Theorem 17.3. First we observe that they exist.
Lemma 17.2. There exists a combinatorially rich ultrafilter. Proof. By Theorem 6.79 we have that A is a left ideal of (16N, +) and so, by Corollary 2.6 contains an additive idempotent r which is minimal in (fN, +). By Remark 16.23 we have r E M and consequently M fl A # 0. Thus by Theorems 6.79 and 16.24, M n A is a left ideal of (#N, ) and hence contains a multiplicative idempotent which is minimal in (,8N, ).
17 Sums and Products
340
Recall the notion of FPtree introduced in Definition 14.23. We call the corresponding additive notion an FStree. Since any member of a combinatorially rich ultrafilter is additively central, it must contain by Theorem 14.25 an FStree T such that { B f : f E T) is collectionwise piecewise syndetic, and in particular each B f is piecewise syndetic. (Recall that Bf is the set of successors 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, recall from Theorem 6.80 that N*\A is a left ideal of (flN, +) and hence there are central sets in (N, +) with zero density.)
Theorem 17.3. Let p be a combinatorially rich ultrafilter and let C E p. (a) C is central in (N, +). (b) C is central in (N, ). (c) Let (A(n))' enumerate the image partition regular matrices with entries from Q and for each n, let m(n) be the number of columns of A(n). There exists for each n E N a choice of i(n) E N,(n) such that, if Y is the set of entries of A(n)x"(n), then 1
FS((YY)n° 1) C C.
(d) Let (A(n))' 1 enumerate the kernel partition regular matrices with entries from Q and for each n, let m(n) be the number of columns of A(n). There exists for each n E N a choice of x"(n) E N" `W such that A(n)x(n) = 0 and, if Yn is the set of entries of x(n), thenFS((Y.)n°_1) C C. (e) There is an FStree T in C such that for each f E T, d (B f) > 0. (f) There is an FPtree T in C such that for each f E T, d(B1) > 0.
Proof. Conclusion (a) holds because p E M while conclusion (b) holds because p p =
p E K(#N, .)Conclusions (c) and (d) follow from Theorems 16.16 and 16.17 respectively. To verify conclusion (e), notice that by Lemma 14.24 there is an FStree T in A such
that for each f E T, Bf E P. Since each member of p has positive upper density the conclusion follows. Conclusion (f) follows in the same way.
As a consequence of conclusions (a) and (b) of Theorem 17.3 we have in particular that any member of a combinatorially rich ultrafilter satisfies the conclusions of the Central Sets Theorem (Theorem 14.11), phrased both additively and multiplicatively. There is naturally a corresponding partition result. Notice that Corollary 17.4 applies in particular when C = N.
Corollary 17.4. Let C be a central* set in (N, +), let r E N, and let C = U;_1 Ci. T h e r e is some i E (1 , 2, ... , r } such that each of the conclusions of Theorem 17.3 hold
with C; replacing C.
Proof. By Lemma 15.4 we have {p E K(14N, +) : p = p + p} C C so M C C. Pick a combinatorially rich ultrafilter p. Then C E p so some C; E p.
17.2 Pairwise Sums and Products
341
17.2 Pairwise Sums and Products We know from Corollary 16.21 that any IP* set in (N, +) contains FS((xn)°° 1) U FP((xn )°° 1) for some sequence in N. It is natural to ask whether there is some partition analogue of this result. We give a strong negative answer to this question in this section. 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*. Definition 17.5. Let (xn)°° 1 be a sequence in N. (a) PS((xn)n1) = {xn +xm : n, m E N and n 96 m}. n
m}.
The partition we use is based on the binary representation of an integer. Recall that for any x E N, X = EIESUpp(X) 2.
Definition 17.6. Let X E N. Then
(a) a(x) = max supp(x). (b) If x 0 {2` : t E co), then b(x) = max(supp(x)\{a(x)}). (c) c(x) = max({1, 0, 1, ... , a(x)}\ supp(x)). (d) d(x) = min supp(x). (e) If x 0 {2` : t E 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 e(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. : t E N) and let k E w. k if and only if x > 2aw + 2k
Remark 17.7. Let X E ICY\(2`
(a) b(x) (b) b(x) (c) c(x) (d) c(x)
k if and only if x < 2a(X) + 2k+1 k if and only if x < 2a(X)+1  2k. k if and only if x > 2a(X)+1  2k+1
(e) e(x) = k if and only if k > d(x) and there is some m E w such that x = 2k+1 m + 2k + 2d(X).
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 (x) is even and 2a(X) < x < 2a(X)+ } A 1 = {x E N : a (x) is even and 2a(X)+z < x < 2a(x)+l } A2 = {x E N : a(x) is odd and 2a(X) < x < 2a(X)+ } A3 = {x E N : a(x) is odd and 2a(x)+Z < x < 2a(X)+1 }
A4={2f:tECol.
17 Sums and Products
342
(b) For i E {0, 1, 2)
B; = {x E N\A4 : x < 21(x)+1(1  2`(x)Q(x))I and a(x)  c(x) = i (mod 3)) U{x E N\A4 : x > 2°(x)+1(1 2`(x)°(x))7 and a(x)  c(x) as i + 1 (mod 3)). (c) (Co, Cl) is any partition of N such that for all k E N\{ 1), k + 1 E Co if and only if 2k E C1.
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 does not contain k + 1.
Remark 17.9. (a) If x, y E A0 U A2, then a(xy) = a(x) + a(y). (b) If x, y E A 1 U A3, then a(xy) = a(x) + a(y) + 1.
Lemma 17.10. (a) If x, y E A0 U A2, then a(xy)  b(xy) < a(x)  b(x). (b) If x, y E Al U A3, then a(xy)  c(xy) < a(x)  c(x). Proof We use Remarks 17.7 and 17.9. (a) We have x > 2°(x) + 2b(x) and y > 2°(y) so that xy > 2a(x)+a(y) + 2b(x)+a(y) = 2a(xy) + 2a(xy)a(x)+b(x)
so b(xy) > a(xy)  a(x) + b(x). (b) We have x < 2°(x)+1  2c(x) and y < 2°(y)+l so that xy < 2°(x)+a(y)+2  2a(y)+c(x)+l = 2a(xy)+l  2a(xy)a(x)+c(x)
so c(xy) > a(xy)  a(x) + c(x). We are now ready to define a partition of N. When we write x y (mod R) we mean, of course, that x and y are elements of the same member of R. Definition 17.11. Define a partition JZ of N by specifying that A4 and 2N + 1 are cells
of ,R and for any x, y E N\((2N Ir 1) U A4), x  y (mod ,R) if and only if each of the following statements holds.
(1) For i E {0, 1, 2}, x E B; if and only if y E Bi. (2) For i E {0, 1), d (x) E C1 if and only if d (Y) E Ci.
(3) a(x)  b(x) < d(x) if and only if a(y)  b(y) < d (y). (4) a(x)  c(x) d (x) if and only if a(y)  c(y) < d(y). (5) a(x)  b(x) = a(y)  b(y) (mod 3). (6) a(x) =a (y) (mod 2). (7) e(x) as e(y) (mod 2). (8) x as y (mod 1)6.
343
17.2 Pairwise Sums and Products
C A4 or with
Notice that there is no sequence (xn)n_1 with PS((xn )n°_1) C 2N + 1.
Lemma 17.12. Let (xn)R°1 be a onetoone sequence in N. If PS((xn)°O 1) U PP((xn)n° 1) is contained in one cell of the partition JR, then (d(xn) : n E N} is unbounded.
Proof. Suppose instead that (d(xn)_ : n E N} is bounded and pick k E N such that for infinitely many n, d(xn) = k. If k = 0, then we would have PP((xn)n° 1) C 2N + 1 so PS((xn)n° 1) C 2N + 1, which is impossible. Thus we may assume that k > 1. If k > 1, then pick n < r such that d (xn) = d(Xr) = k and either k + 1 E supp(xn)flsupp(Xr)
or
k + 10 SUPP(Xn) U supp(xr).
Then d(xn + xr) = k + 1 and d(xnxr) = 2k so one can't have i E (0, 1) with d(xn +xr), d(Xnxr) E Ci. Thus we must have that k = 1. Suppose first that for infinitely many n one has
d (xn) = 1 and e(xn) = 2. Pick n < r and u, v E w such that u = v (mod 2) and
xn = 2+4+8u andxr = 2+4+8v. Thenxn+xr = 12+16 ("2°) m 12 (mod 1)6 while xnxr = 36 + 48(u + v) + 64uv = 4 (mod 1)6. Consequently one has infinitely many n with d(xn) = 1 and e(xn) > 2. Pick n < r and u, v E co such that d(xn) = d(xr) = 1, 3 < e(xn) < e(xr), u = v (mod 2), and
xn = 2 + 2e(xa) + u
.
2e(xn)+1
and
Xr = 2 + 2e(xr) + v , 2e(xr)+1
Suppose first that e(xn) < e(xr). Then 2e(xr)e(x.))
u+v
xn + Xr = 4 + and
XnXr = 4 +
2e(xr)2
+2 e(x,)+2 (u +
+ u 2e(xr)1 + v
2e(xr)e(x,,) + v .
2e(xr)1
+ uv 2e(xr))
so e(xn + Xr) = e(xn) while e(xnxr) = e(xn) + 1, a contradiction. Consequently we have some f > 2 such that e(xn) = e(xr) = f. Then
Xn+xr =4+2e+1
+2e+2(u+v\
2
and
XnXr = 4 + 2t+2 + 2t+3(2e3 + u
v + (u + v)2e2 +
uv2e1)
so that e(xn + Xr) = f + 1 while e(xnxr) = f + 2 2, again a contradiction. As a consequence of Lemma 17.12, we know that if (xn )n° 1 is a onetoone sequence
with PS((xn)R° 1) U PP((xn)n 1) contained in one cell of the partition ,R, then we can assume that for each n, a(xn) < d(xn+1) and consequently, there is no mixing of the bits of x, and xr when they are added in binary.
17 Sums and Products
344
Lemma 17.13. Let (xn)r°i° be a onetoone sequence in N. If PS((xn)n° 1) U PP((xn),,_1) is contained in one cell of the partition .R, then in E N : xn E Ao} is infinite or {n E N : xn E A3} is infinite 1
Proof. One cannot have PS ((x,,) n° 1) c A4 and one cannot have PS ((xn) n° 1) C 2N+ 1
so one has a(xn + xr)  a(xnxr) (mod 2) whenever n and r are distinct members of
N. By the pigeon hole principle we may presume that we have some i E (0, 1, 2, 3,4) such that (xn : n E N) C A; . If one had i = 4, then one would have PP((xn )n° 1) C A4 and hence that PS((xn)l1) C A4, which we have already noted is impossible. By Lemma 17.12 we have that {d(xn) : n E NJ is unbounded so pick n E N such
thatd(xn) > a(xl). Thena(xn +xl) =a(xn). Suppose i = 1, that is (xn : n E N) C A1. Then by Remark 17.9, a(xnxl) _ a(xn) +a(xl) + 1 so a(xnxl) is odd while a(xn +xl) is even. Similarly, if i = 2, then a(xnxl) is even while a(xn + xl) is odd. Lemma 17.14. Let (xn)R° be a onetoone sequence in N. If d(xn+l) > a(xn) for each n, PS((xn)n1) U PP((xn)n 1) is contained in one cell of the partition .R, and {xn : n E N) C A0, then {a(xn)  b(xn) : n E N) is bounded. 1
Proof. Suppose instead that {a(xn)  b(xn) : n E N) is unbounded. Pick n E N such that a (xn)  b (xn) > d (x 1). Then, using the fact that d (xn) > a (x 1) we have that a(xn + xl)  b(xn + x1) = a(xn)  b(xn) > d(xl) = d(xn + xt)
while by Lemma 17.10
a(xnxl)  b(xnxl) < a(xl)  b(x1) < d(xn) < d(xnxl).
0
Lemma 17.15. Let (xn )n_ be a onetoone sequence in N. If d (xn+l) > a (xn) for each n, PS((xn)°i°r1) U PP((xn)r°i 1) is contained in one cell of the partition R, and {xn : n E N) C A3, then {a(xn)  c(xn) : n E N} is bounded. 1
Proof. This is nearly identical to the proof of Lemma 17.14.
Theorem 17.16. There is no onetoone sequence (xn) ' in N such that PS ((xn) n° 1) U PP((xn)r°i°_I) is contained in one cell of the partition R. 1
Proof. Suppose instead we have such a sequence. By Lemmas 17.12, 17.13, 17.14, and 17.15 we may presume that for each n E ITV, d (xn+1) > a (xn) + 1 and either
(i) {xn : n E N) C A0 and {a(xn)  b(xn) : n E N) is bounded, or (ii) {xn : n E N) C A3 and {a(xn)  c(xn) : n E N} is bounded. Assume first that (xn : n E N) C A0 and {a(xn)  b(xn) : n E N) is bounded.
Pick some k E N and n < r in N such that a (xn)  b(xn) = a (xr)  b(Xr) = k. Then a(Xr + Xn)  b(Xr + Xn) = a(Xr)  b(Xr) = k.
17.2 Pairwise Sums and Products
345
Also 2a(xn) + 2a(xn)k < xn and 2a(xr) + 2a(xr)k < xr, SO 2a(x,xn) + 2a(xrx')k+l = 2a(xr)+a(xn) +2 a(xr)+a(xn)k+1 < X x rn
Sob(XrXn) > a(xrxn)k+1. Andxn < so
2a(xn)+2a(xn)k+t andXr
<
2a(r)+2a(x,)k+l
XrXn < 2a(x,)+a(xn) + 2a(x,)+a(xn)k+2 + 2a(xr)+a(xn)2k+2 = 2a(x,xn) + 2a(xrxn)k+2 + 2a(x,xn)2k+2 < 2a(xrxn) + 2a(xrxn)k+3
SOb(xrXn) 5 a(xrx,)k+2. Thusa(xrxn)b(xrxn) E {k1,k2}so a(xrxn)b(xrxn) # a(Xr +xn)  b(Xr +xn) (mod 3), a contradiction. Finally assume that {xn : n E N} c A3 and {a(xn)  c(xn) : n E N) is bounded. We may presume that we have some k E N such that a(xn)  c(xn) = k for all n E N. By the pigeon hole principle, pick n < r in N such that either (a) x, < 2a(xn)+1(1
 2k) 2 and xr < 2a(xr)+t (1  2k)2 or
(b) x, > 2a(xn)+I (1  2k) z and x,. > 2a(xr)+1(1
 2k).
In either case we have a(xrxn) = a(xr) + a(xn) + 1. Also in either case we have xn > 2a(xn)+1  2a(xn)k+l and xr > 2a(xr)+1  2a(x,)k+l So that Xrxn > 2a(xrxn)+1  2a(x,xn)k+2 + 2a(xrxn)2k+1 > 2a(xrxn)+1  2a(xrxn)k+2 and consequently C(xrxn) < a(xrxn)  k + 1.
Assume that zn < 2a(xn)+t (1  2k)2 and xr < 2a(x,)+1(1  2k)2. We have a(xr + Xn)  C(Xr + xn) = a(xr)  C(Xr) = k. Since XrXn < 2a(xrxn)+l(1  2k) = 2a(xrxn)+1  2a(x,xn)+Ik and hence c(xrxn) > a(xrxn)  k + 1, one has C(Xrxn) = a(XrXn)  k + 1 . B u t a(XrXn)  C(XrXn) a(xr + Xn)  C(Xr + Xn) (mod 2),a contradiction.
Thus wemusthavethatxn > 2a(xn)+1(12k)2 andxr > 2a(xr)+1(12k)2. Now
a(xr+Xn) =a(xr)SOXr+xn > 2a(xr+xn)+l(12k)2 anda(Xr+Xn)C(xr+Xn) _ a(xr)  c(xr) = k, so picking i E {0, 1, 2} such that i + 1  k (mod 3) we have that Xr + Xn E Bi and consequently XrXn E B,. Now XrXn > 2a(xrxn)+1(1  2k) = 2a(xrxn)+1  2a(xrxn)+1k so C(XrXn) <
a(XrXn)  k. By Lemma 17.10 a(xrXn)  C(XrXn) < k so a(xrxn)  c(xrxn) = k.
Notice that (1  2k1  22k2) < (1  2k)2, a fact that may be verified by squaring both sides. Since xn < 2a(xn)+1  2 a(x.)k and xr < 2a(xr)+l _ 2a(xr)k we have
Xrxn < 2a(xrxn)+1 _2a(x,xn)k+1
+2a(x,xn)2k1

< 2a(xrxn)+1  2a(x,xn)k 2k1 22k2) = 2a(xrxn)+1(1

< 2a(xrxn)+I (1  2k) 2 = 2a(xrxn)+1 (I  2r(xrxn)a(xrx0 )2
2a(x,xn)2k1
346
17 Sums and Products
Thus since XrX,, E Bi we have that k = a(xrxn)  c(xrxn)  i = k  1 (mod 3), a contradiction.
Recall from Theorem 13.14 that if p E ON and Nn E p for infinitely many n, then there do not exist q, r, and s in N* such that q p = r + s. The following corollary has a much weaker conclusion, but applies to any p e N*.
Corollary 17.17. Let p E N*. Then p + p 0 p p. Proof. Suppose instead that p + p = p p and pick some A E R such that A E p p. Let B = {x E N : x + A E p} f1 {x E N : xt A E p} and pick xl E B. Inductively, let n E N and assume we have chosen
Pick
Xn+1 E (B l (l; 1(x, + A) fl (x,1 A))\{xt, x2, ... , X,,)Then PS((xn)n_1) U PP((xn)' 1) C A, contradicting Theorem 17.16. Theorem 17.16 establishes that one cannot expect any sort of combined additive and multiplicative results from an infinite sequence in an arbitrary finite partition of N. On the other hand, the following question remains wide open.
Question 17.18. Let r, n E N. If N = U;=1 A;, must there exist i E {1, 2, ... , r} and 1 such that FS ((x, )1 1) U FP((xl) l 1) c A; ?
a onetoone sequence (x1)
We would conjecture strongly that the answer to Question 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 Finite Sums Theorem and its corollary, the Finite Products Theorem, Erdo"s asked [89] 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 "multilinear combinations", the answer is "no". (Theorem 17.16.) We see in this section, however, that a certain regularity can be imposed on sums of products of a sequence.
Recall that, if n E N and p E ON, then n p is the product of n and p in the semigroup (ON, ) which need not be the same as the sum of p with itself n times. (We
already know from Theorem 13.18 that if p E N*, 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 ON. Then al (p) = p and, given n E N, an+1 (p) _ an(P) + p.
17.3 Sums of Products
347
Of course, if p E N, then for all n, n p = an (p). The question naturally arises as to whether it is possible to have n p = Q, (p) for some n E N\{ 1) and p E N*. We shall see in Corollary 17.22 that it is not possible.
Lemma 17.20. Let n E N, let p E N*, let A E p and let B E on(p). There is a onetoone sequence (x,)001 in A such that, for each F E [N]", E:EF Xt E B. Proof. This is Exercise 17.3.1.
Theorem 17.21. Let n E N\ { 11. There is a finite partition ,R of N such that there do not exist A E JR and a onetoone sequence (x`)001 such that
(1) for each t E N, n x, E A, and (2) whenever F E [N]", FIEF Xt E A. Proof For i E (0, 1, 2, 31, let A = Uk=O °O {x E N : n2k+i/2 < x < n 2k+(i+l)/2 JI
and let R = {A0, Al, A2, A3). Suppose that one has a onetoone sequence (x,)0_1 and i E 10, 1, 2, 3) such that
(1) foreacht E N,n  x, E A,,and (2) whenever F E [N]", EiEF Xt E A. By the pigeon hole principle, we may presume that we have some j E (0, 1, 2,3) such
that (x,:tEN)cAj. Now if n2k+j/2 < x, < n2k+(j+1)12 then n2k+(j+2)/2 < n . x, < n2k+(j+3)/2 so that i = j + 2 (mod 4). On the other hand, for sufficiently large t, if n2k+j/2 < X, < n2k+(j+l)/2 then n2k+j/2 < X1 + X2 +
+Xn_l +xt < n2k+(j+2)/2
so that i # j + 2 (mod 4), a contradiction. Corollary 17.22. Let P E N* and let n E N\( 11. Then n p # a" (p). Proof. Suppose that n p = an (p) and pick A E ,R such that A E n p. Then n' A E p so, by Lemma 17.20, choose a sequence (xt)001 in n1 A such that for each F E [N]", E,EF Xt E A. This contradicts Theorem 17.21.
Recall that, given F, G E Pf (N) we write F < G if and only if max F < min G. Definition 17.23. Let (x,)001 be a sequence in N and let m E N. Then I
F m E 2f(N) and Fl < F2 <.
< Fm}.
17 Sums and Products
348
Theorem 17.24. Let p p = p E flN, let m E N, and letA E am (p). Then there is a sequence (xt) °O 1 such that SPm ((xt) t0O 1) C A.
Proof If m = 1, this is just the finite products theorem (Theorem 5.8). In the proof, we are dealing with two semigroups, so the notation B* is ambiguous. We shall use it to
refer to the semigroup (fN, ), so that B* = {x E B : x B E p}. We do the m = 2 case separately because it lacks some of the complexity of the
general theorem, so assume A E p + p. Let B1 = {x E N : x + A E p). Then B1 E p so B1* E p. Pick xl E Bl*. Inductively, let n E N and assume that we have chosen (xt) in N and B1 ? B2 ? D Bn in p such that, if 0 0 F c {1, 2, ... , n}, 1
k = min F, andf =maxF,then (1) ITtEFXt E Bk* and
(2) if f < n, then Be+1 C ( IItEF Xt + A).
Fork E{1,2,...,n},letEk={IItEFxt::F(z {1,2,...,n}and k=min F}.Let Bn+1 = Bn nn Ik=1 laEEk ((a + A) fl al Bk*).
Now, given k E {1, 2, ... , n} and a E Ek, a E Bk* c_ B1 so a + A E p and a1 Bk* E p so Bn+1 E p. Choose Xn+l E Bn+1* T o verify the induction hypotheses, let 0 F C { 1 , 2, ... , n + 1), k = min F, and
f = max F. If e < n, then both hypotheses hold by assumption. Assume that f = n. Then (1) holds by assumption and I1tEF xt E Ek so (2) holds. Finally assume that f = n + 1. Then (2) is vacuous. Ilk = n + 1, then IItEF X. = Xn+1 E Bn+l*. If k < n, then let G = F\{k}. Then fIfEG xt E Ek SO Xn+l E Bn+l C A EG X0 ' Bk* and thus (1) holds.
The construction being complete, let F1, F2 E . f (N) with F1 < F2. Let k = min F2 and let e = max F1. Then
11tEF2 Xt E Bk* C Be+1 C ( ftEF, xt + A). This completes the proof in the case m = 2.
Now assume that m > 3. Let B1 = {x E N : x + A E a,,, I (p)} and choose x1 E B1*. Inductively, let n E N and assume that we have chosen (Xk)k=1 in N and (Bk)k=1 in
p so that for each r E {1, 2, ... , n} each of the following statements holds.
(I) If 0 F C (1, 2, ... , r} and k = min F, then IIrEF Xt E Bk*. (II) If r < n, then Br+1 C Br
(III) Iff E {1,2,...,m1},F1,F2,...,Fe E JPf({1,2,...,r}), and F1 < F2 < < Ft, then  El _1'1teF Xt + A E am1(p).
(IV) If F1,F2,...,Fm_1 EPf({l,2,...,r}), F1 < F2 < ...
17.3 Sums of Products
349
(V) If f E (1,2. , m  2), F1, F2, ... , Ft E .Pf({12. , r}) with F1 < F2 <
At n = 1, hypothesis (I) says that x1 E Bl*. Hypotheses (II), (IV), and (V) are vacuous, and hypothesis (III) says that x1 + A E am1(p).
Fort E{1,2,...,m1},let Y = {(Fl, F2, ... , Ft) : F1, F2, ... , Ft E ,Pf({1, 2, ... , n})
and F1
F C {1,2,...,n} and minF = k}.
Ek = {Il EFX, : 0
Given a E Ek, we have that a E Bk* by hypothesis (I) and so a1Bk* E p by Lemma 4.14. If (F1, F2, ... , Fm_1) E Fm 1, then by (III) we have
Ei=, 1 REF Xt + A E P. If f E ( 1 , 2, ... , m  2} and (F1, F2, ... , Ft) E Fe we have by hypothesis (III) that
{X E N : x + ( Eti=1 REF, Xt + A) E ama1(P)} E P. So we let
Bn+1 = B. fl
n
Ik=1n
'
I
nn(FI,F2,,
IaEEk a1 Bk*
. ,FF_,)EF,_,
1
Em=1
(
n I Ie=12I I(F,, F2, ...Fe)EFe {X E ITV
fltEF xt + A)
: x +
fltEF Xt + A) E ama1(P)}
and notice that Bn+1 E p. For simplicity of notation we are taking n 0 = N in the above. Thus, for example,
ifn =m3,then .Fm2=pm1 = 0 and so n
Bn+1 = B,n nn IInk=1 nIaEE,p a
*
1
Bk
nI 1,=13l(F1,F2,.. ,Ft)E.F, {X E N : x + ( Ef=1 IEEE xt + A) E ame1(P)}.
Choose xn+1 E Bn+I*. T o verify hypothesis (I), let 0 # F C 11, 2, ... , n + 1} and let
k = min F. If n + 1 0 F, then (I) holds by assumption, so assume that n + I E F. If k = n + 1, we have xn+1 E Bn+1* directly, so assume that k < n + 1 and let G = F\{n + 11. Then rltEG xt E Ek so xn+1 E xt)1 Bk* Hypothesis (II) holds trivially and hypotheses (IV) and (V) hold directly.
T o verify hypothesis (III), let f E ( 1 , 2, ... , m  1 ) and let F1, F2, ... , Fe E
Yf({1, 2, ... , n + 1}) with F1 < F2 < ... < Fe. If f = 1, then by hypothesis
17 Sums and Products
350
(I) and (U), fIrEF1 X, E Bt' C B1 SO  f1EEF, Xt +A E am1(P). So assume that
I > 1. Let k = min Ft and let j = max Fe1. Then by hypotheses (I) and (II), fIIEFE Xt E Bk* c Bj+1 and by hypothesis (V) at r = j,
Bj+1 C {x E N : x + ( Eei f11EF, Xt + A) E am_e(p)}. The induction being complete, we have that whenever F1, F2, ..., Fm E .Pf (N) with F1 < F2 < . . . < Fm, if k = min Fm and r = max F,,, 1, then by (I), (II), and (IV), Xt E
Bk*
C Br+1 C  Em11 f11EF, Xt + A
and thus Em l Il(EF; X, E A.
Corollary 17.25. Let r, in E N and let N = u;=, A j. Then there exist j E ( 1, 2, ... , r) and a sequence (xt)°O 1 such that SPm((xt)°° 1) C Aj.
Proof. Pick any idempotent p in (fN, ) and pick j E (1, 2, ..., r) such that Aj E am (P)
There is a partial converse to Theorem 17.24. In this converse, the meaning of SP, ((xt)1'k) should be obvious. Notice that one does not require that p p = p. Theorem 17.26. Let (x1)°O, be a sequence in N. If P E nk°°_, FP((xt)0Ok), then for all n and k in N, SPn((xr)Q0k) E an(P)
Proof. We proceed by induction on n, the case n = 1 holding by assumption. So let n E N and assume that for each k E N, SPn ((xt)O0k) E an (p). Let k E N. We claim that
SPn((X,)tmk) C (a E N : a+SPn+I((xr)°_k) E P}
so that SPn+1((xr)°Ok) E an(P) + p = an+1(P) So let a E SPn((x1)t k) and pick F1 < F2 < < Fn in Pf(N) such that min F1 > k and a = E;_, n,EFF x1 and let
f=maxFt, +1.Then FP((x1)°Oe) C a + SPn+1((xt)0Ok)
so that a + SPn+I ((XI)t'k) E p. We see now that one cannot necessarily expect to find SPm ((xt)°O,) and SP1((x1)°O, ) (= FP((xt )O01)) in one cell of a partition, indeed not even SPm ((xt )°°,) U SP1((y1)°O_1) with possibly different sequences (x1) °O, and (,, ) °O , . In fact much stronger conclusions
are known  see the notes to this chapter.
Theorem 17.27. Let m E N\{I). There is a finite partition .R of N such that, given any A E JR, there do not exist onetoone sequences (x1)°O 1 and (yt)°° 1 with SPm((x1)°O1) C A andSP1((Y1)O0,) C A.
17.3 Sums of Products
351
Proof. For X E N, define a(x) and d(x) in w by ma(x) < x < ma(x)+1 and d(x) = max{t E w : m'jx}. Thus when x is written in base in, a(x) is the position of the leftmost nonzero digit and d(x) is the position of the rightmost nonzero digit. Let
Ao =N\Nm={xEN:d(x)=0} A1= {xEN:d(x)=1}
A2={mt:rENandt>1)
A3 = {x E N : a(x) is even, d(x) > 1 and ma(x) < x < maw + ma(x)d(x)} ma(x)d(x) < x} A4 = {x E N : a(x) is even, d(x) > 1 and ma(x) + A5 = {x E N : a(x) is odd, d(x) > 1 and ma(x)+1  ma(x)d(x) < x} A6 = {x E N : a(x) is odd, d(x) > 1 and maW < x < ma(x)+1 _ ma(x)d(x)} Then {Ao, Al, A2, A3, A4, A5, A6} is a partition of N. Trivially, neither Ao nor A2 and Al does not contain SPi((xt)°_1) = FP((xt)' 1) for any contains sequence (xt) °O 1 in N.
We now claim that A3 does not contain any FP((xt)°O_1), so suppose instead we have some onetoone sequence (x,)°O1 with FP((xt)0O 1) C A3. Then for each t E N, d (xt) > 1 so pick a product subsystem (yt) of (xt) °_ such that for each t, a (yt) < d(xt+1). Then ma(Y') < Yl < ma(Y1)+1 and ma(Y2) < Y2 < ma(Y2)+1 so ma(tt)+a(Y2) < YlY2 <ma(YI)+a(Y2)+2 and consequently,a(yjY2) E {a(yl)+a(y2),a(yl)+a(Y2)+1}. 1
Since a(yl), a(Y2), and a(yly2) are even, a(yly2) = a(yl) +a(Y2). Now d(ylY2) ? d(y1) +d(y2) so
a(y1y2)a(yl)+d(yl) > a(yly2)a(y) > a(Yly2)d(Y2) > a(Yly2)d(y1y2). Also, Yl > ma(Yt) + md(Yt) and y2 > ma(Y2) so Yl Y2 > ma(Y1)+a(Y2) + ma(Y2)+d(Y1) = ma(YIY2) +ma(Yty2)a(Yt)+d(Y1)
>
ma(Y1Y2) +ma(Y1Y2)d(Y1Y2)
a contradiction. Next we claim that A4 does not contain any SPm ((xt)°O 1), so suppose that we have a onetoone sequence (xt) °O 1 with SPm ((xt) 0O 1) C A4. If for infinitely many is we
have d (xt) = 0, we may choose F E [N]' such that for any t, s E F one has
xt=x5.# m (modm2). Then d(EtEF x,) = ISO EtEF Xt E SPm((xt)°O1)\A4
Thus we may assume that each d(xt) > 0 and hence, by passing to a product subsystem as we did when discussing A3, we may presume that for each t, a(xt) < d(xt+1), so that there is no carrying when xt and xt+1 are added in base m arithmetic. (Notice that by Exercise 17.3.2, if (yt)°°1 is a product subsystem of (xt)°O1, then SPm((Yt)t1) C SPm((Xt)
00
We may further assume that for each t > m, a (xt) > a (xm 1) + d (x 1) + 2. We +ma(x,)d(x,)1. Suppose instead claim that there is some t > m such that x, < ma(x')
17 Sums and Products
352
that for each t > m, xt > ma(xt)(1 +md(xl )1). Now 1 +md(xl>1)e+1 > m and hence f, (1
1m+t X > m t=m
+md(xl)1
> 1 so for some
^'e a(x,)+l
t
Pick the first k such that
Since EkI
r1k1
t=m Xt < m
t_, a(x,)+1 and xk < ma(xk)+l
we have that r1kt=m x
t
mEk
<
=_a(x,)+2
and consequently a(r1k=m xt) = Ek m a(xt) + 1. Since, for each t, d(xt+1) > a(xt), we have that for each s E {m, m + 1, ..., k}, a (x1 + X2 + + xm _ l + xs) = a (xs) and, since +X2+...+Xm1
X1
a (xs) is even. Also a (X l +X2 +
+Xs E SPm((Xt)°=1) C A4,
+ Xm 1 + 11 k=m Xt) = a (IIk=m xt ), so a (r1 km Xt )
is even, a contradiction.
Thus we have some t > m such that xt < ma(xi) + ma(xi)d(xl)1. Let y = +Xm1 +xt. Then, a(y) = a(xt) and d(y) = d(x1). And because there
X1 + X2 +
is no carrying when these numbers are added in base m, we have
Xl + X2 + .. + xm1 <
ma(xmI)+1
< ma(xi)d(xl)1 so
y < ma(xt) < ma(xi)
+ma(x,)d(Xi)1
+ma(xt)d(xl)1
+ma(xt)d(xl)
= Ma(y) + ma(y)d(y)
contradicting the fact that y E A4. Now we claim that A5 does not contain any FP((xt)°Ol). So suppose instead that
we have a onetoone sequence (xt)°°1 such that FP((xt)°Oi) c A5. As before, we may presume that for each t, d(xt+1) > a(xt). Now x1 < ma(xi)+l  md(x,) as can be seen by considering the base m expansion of x1. Also, X2 < ma(x2)+l so that X1X2 < ma(xl)+a(x2)+2  ma(x2)+d(x,)+1
Since also XJX2 > ma(xi)+a(x2) and a(x1),
a(x2), and a(xlx2) are all odd, we have a(xlx2) = a(xl) + a(X2) + 1. Also
a(x2) +d(x1) + 1 = a(X1x2) a(XI)+d(xl) > a(xlx2)  d(xIX2) because d(x1x2) > d(x1) + d(X2) > a(x1)  d(xl). Thus
XIX2 < ma(XIX2)+1  ma(xIx2)d(xlx2)
173 Sums of Products
353
contradicting the fact that xlx2 E A5. Finally we show that A6 does not contain any SPm ((xt) ° t ), so suppose instead that we have a onetoone sequence (xt) °_ such that SPm ((xt) °O 1) c A. As in the consideration of A4, we see that we cannot have d (xt) = 0 for infinitely many is and 1
hence we can assume that for all t, d(xt+I) > a(xt) We claim that there is some t > in such that xt > ma(st)+I  ma(rt)d(x1). For suppose instead that for each t > m, xt < ma(st)+t (1  md(xl)I). Then for some 1,
(1 
md(xi)1)Q+I
< 1/m so nm+Z t=m
n+e
rF_ a(xt)+f
X tt
Pick the first k such that
Ilk Then 1hk1
t=m xt
k
Xt
t=m
nt Et=m
t
a(xt)+km
k1
> mE=ma(x,)+kIm andxk >
so
m xt
>m
V=m
Ma(sk),
a(xt)+k1m
and hence
a(xl +x2 + ... +xm_1 + fk=m xt) = a(nt=m xt) = Ee=m a(xt) +k  1  in. + xm1 + xt) = a(xt) so each Also for each t E {m, m + 1, ..., k}, a(xl + x2 + a(xt) as well as Ekm a(xt) + k  1  in are odd, which is impossible.
Thus we have some t > m such that xt > ma(st)+t  ma(st)d(xl)Let y = +xm1 +xt. Then a(y) = a(xt) and d(y) = d(x1) and y > ma(st)+1 XI +X2 + ma(xt)d(xl) = ma(y)+1  ma(y)d(y) a contradiction. 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\ 11). There is a set B C N such that (1) whenever E am(p), p=pE (3) there is no onetoone sequence (x1) J ° 1 in N with SPm ((xt) °O 1) C B, and (4) there is no onetoone sequence (x,)°D 1 in N with SP1((xt)°O 1) C N\B. Proof. Let 52 be the partition guaranteed by Theorem 17.27 and let
B = U{A E 5: there exists (x1)°01 with FP((xt)°_1) C A}.
To verify conclusion (1), let p p = p E $N and pick A E .'R such that A E P. Then by Theorem 5.8 there is a sequence (xt) °O 1 such that FP((xt) °° 1) C A, so A C B
so B E p.
17 Sums and Products
354
To verify conclusion (2), let p p = p E fH and pick A E R such that A E om (p). Then by Theorem 17.24 pick a sequence (yt)0O t with SPm((yt)O01) S A. By Theorem 17.27 there does not exist a sequence (xt)t= 1 with FP((xt)0° 1) C A, so A n B = 0. To verify conclusion (3), suppose that one had a onetoone sequence (xt)°°1 with
SPm((xt)0O1) c B. By Lemma 5.11, pick p p = p E f°k°__1 FP((xt)t k). Then by Theorem 17.26 SPm ((xt) o01) E am (p) so that B E om (p), contradicting conclusion (2).
To verify conclusion (4), suppose that one had a onetoone sequence (xt)0O with 1
FP((xt)001) C N\B. Again using Lemma 5.11, pick p p = p E fk__1 FP((xt)0Ok). Then FP((xt)0°1) E p so N\B E p, contradicting conclusion (1). Exercise 17.3.1. Prove Lemma 17.20. (Hint: See the proofs of Corollary 17.17 and Theorem 17.24.)
Exercise 17.3.2. Prove that, if m E N and (yt) °O is a product subsystem of (xt) 00 1
then SPm((yt)°O_1) C SPm((xr)0_°1)
Exercise 17.3.3. Let (y,111. Modify the proof of Theorem 17.24 to show that if p p = p E nk 1 FP((yt)0°k) and A E om(p), then there is a product subsystem (x).1 of (yt) t' k such that SPm ((xt) 0O 1) c A. (Hint: See the proof of Theorem 17.31)
Exercise 17.3.4. Let n, r E N and let N = U =1 A;. Prove that there is a function f : 11, 2, ... , n) s { 1, 2, ... , r) and a sequence (xt) 0 ° 1 such that for each m E 11, 2, ..., n), SPm((xt)0_1) c Af(m). (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 further that the cell can depend on the choice of coefficients.
In many respects, the results of this section are similar to those of Section 17.3. However, the motivation is significantly different. In Section 15.4, we characterized the (finite) image partition regular matrices with entries from Q One may naturally extend the notion of image partition regularity to infinite dimensional matrices.
Definition 17.29. Let A be an to x co matrix with entries from Q such that each row of A has only finitely many nonzero entries. Then A is image partition regular over N if and only if whenever r E N and N = U; _ 1 C; , there exist i E { 1, 2, ..., r} and x' E N' such that all entries of Ax are in C; .
17.4 Linear Combinations of Sums
355
A simple example of an infinite partition regular matrix is
A=
1
0
0
0
1
0
1
1
0
0
0
1
1
0
1
0
1
1
1
1
1
0 0 0 0 0 0 0
... ... ... ... ... ... ...
Here A is a finite sums matrix. That is, the assertion that all entries of Ax" are in Ci is the same as the assertion that FS((z )n° 1) C Ci. We establish here the image partition regularity of certain infinite matrices and provide a strong contrast to the situation with respect to finite image partition regular matrices. For example, any central set C in (N, +) has the property that, given any finite image partition regular matrix A, there must exist x' with all entries of A. in C. (Theorems 15.5, 15.24 and Lemma 16.13.) Consequently, given any finite partition of N, some one cell must contain the entries of Al for every finite image partition regular matrix A. By way of contrast, we shall establish here that a certain class of infinite matrices are image partition regular, and then show that there are two of these matrices, A and B, and a two cell partition of N. neither cell of which contains all entries of Ax" and B5 for any x and y".
We call the systems we are studying "MillikenTaylor" systems because of their relation to the MillikenTaylor Theorem, which we shall prove in Chapter 18.
Definition 1730. (a) A = {(a, )"' 1 E Nt : m E N and for all i E (1, 2, ... , m  1), ai
ai+1 }.
(b) Given a" E A with length m and a sequence (xt)°O 1 in N,
[MT(a, (xt)'21) = {E;"_1 ai EtEF; xt : Fl, F2, ... , Fm E ,7'f (N) and
The reason for requiring that ai ,A a;+1 in the definition of A is of course that a EtEF xt + a EtEG xt = a EIEFUG xt when F < G. As a consequence of Corollary 17.33 below, whenever as E A and N is partitioned into finitely many cells, one cell must contain MT(a, (xt)°O1) for some sequence (xt)O1. This is the same as the assertion that a particular infinite matrix is image partition regular.
17 Sums and Products
356
For example, if 2 0
0
0
1
2
0
1
2
1
1
2
1
2
2
0 0 0 0
1
0
0
2
0
1
2
1
1
0 0
1
A=
2
... ... ... ... ... ... ... ...
andx= (xt)OO1,then MT((1,2), (xr)0O1)isthesetofentries ofAx. (The rows of A are all rows with entries from 10, 1, 2} such that (1) only finitely many entries are nonzero, (2) at least one entry is 1, (3) at least one entry is 2, and (4) all occurrences of 1 come before any occurrences of 2.)
(al, a2, ... , am) E A, let (y1)0° 1 be a sequence in N, let Theorem 17.31. Let P + P = P E nk l FS((yt)rO0_k), and let A E al p + a2 P + + am p. There is a sum such thatMT(a, (x1)°_1) c A. subsystem (xr)°°1 Proof. Assume first that m = 1. Then a1
A E p so by Theorem 5.14 there is a sum subsystem (x1)0 .1 of (yt)001 such that FS((xt)001) al A. This says precisely that MT(a, (x,)001) C A. Assume now that m > 2 and notice that
{xEN:x+
Ealp
so that
{X EN:alx+AEa2P+a3p+ +amp) Ep. Let
B1 ={x EN:
(1FS((yt)001).
Pick X1 E Bl' and pick H1 E pf(N) such that xl = EtEHI yt. (Here we are using the additive version of B1'. That is, Bl. = (x E B1 : x + Bi E P).) Inductively, let n E N and assume that we have chosen (xk)k=1 in N, (Bk)k=1 in p,
and (Hk)k_1 in J f(N) so that for each r E (1, 2, ..., n):
(I) If00 F C {1,2,...,r}andk=min F, then EtEFx, E Bk*. (H) If r < n, then Br+1 C Br and Hr < Hr+l.
(III)IffE(1,2,...,m1),Fl,F2,...,FeE,P1((1,2,...,r)), and F1
Vj_laiErEF;xr+AEae+1P+ae+2P++amp.
17.4 Linear Combinations of Sums
357
(IV) If F1, F2, ... , Fm1 E Pf({1, 2, ... , r}), F1 < F2 < : < Fm_1, and r < n, then
Br+l Cam
1
(
Emi=1Iai
EtEF xt + A).
(V) IftE{1,2,...,m2},F1,F2,...,FtE?f({1,2,...,r}),FI Fe, and r
Br+1C {xEN: ae ai EtEF Xt+A) E ae+2P+ae+3P+" '+amP}. a+Ix+(Ei_I e
(VI) Xr = EtEH, Yr.
At n = 1, hypotheses (I) and (VI) hold directly, hypotheses (II), (IV), and (V) are vacuous, and hypothesis (III) says that a 1x1 + A E a2 P + a3P + +amp which is true because x1 E B1.
Fore E {l,2,...,m 11, let Fe = {(F1, F2,..., Fe) : F1, F2,..., Fe E Pf({1,2,...,n}) and F1 < F2 < . < Fe} and for k E {1, 2, ... , n}, let
Ek={EtEFXt:0#FC {l,2,...,n}and minF=k}. Given b E Ek, we have b E Bk* by hypothesis (I), and so b + Bk* E p. If (FI, F2, ... , Fm_l) E Fmj, then by (III) we have ( Em1l ai EtEF xt + A) E amp so that am1( Em11 ai EtEF xt + A) E p. If e E {1, 2, ... , m  2} and (F1, F2, ..., Fe) E Fe we have by (III) that Ej=1 ai ErEF xr + A E ae+1 P + ae+2P +
+ am p
so that {
e xEN:ae+lx + ( Ei=1 ai ErEF xt + A) E ae+2P +ae+3P + ....{_ amp} E p.
Let s = max Hn + 1. Then we have that Bn+1 E p, where
Bn+1 = B,, n FS((yt) i) n I
n I'(F,,F2,
Ik=l I
mi)E m_I am
IbEEk(b + Bk*) n I( Em1Iai EtEF, Xr + A) n
n f_1 n(F,,F2,...,Ft)EY'e{X E ICY : ae+lx + ( Eje=1 ai ErEF,, xt + A)
E ae+2P+ae+3P++amp}. Here again we use the convention that n 0 = N so, for example, if m = 2, then
Bn+l = Bn n FS((yt)OQ,) n nk=1 nbEEk(b + Bk*) n nO#Fcj1,2,....n) a21(a1 EIEF Xt + A).
17 Sums and Products
358
and pick E a (s, s + 1, s + 2, ...) such that x,,+1 = Yt Hypothesis (I) can be verified as in the second proof of Theorem 5.8 and hypothesis
Choose
(II) holds trivially. Hypotheses (IV), (V) and (VI) hold directly. T o v e r i f y hypothesis (III), let t E { 1, 2, ... , m  1) and let Ft < F2 < ... < Fe in
d'f((1, 2, ... , n + 1)). If f = 1, we have by hypotheses (I) and (II) that EtEF, xt E Bt so that al EtE F, xt + A E a2 P + a3 P + amp as required. So assume that f > 1 andletk=min Fe and j =maxFe_1. Then EtEF, Xt E Bk* a Bj+1
a {x E N : aex + ( Ef_j a; EtEFF xt + A)
E ae+lp+at+2P+...+amp} (by hypothesis (V) at r = j) so
(E;_1a; EteF,Xr)+AEae+1P+ae+2P +
a, p
as required.
The induction being complete, let Fl < F2 < ... < F. in Pf(N) and let k = min Fm and j = max F,,, 1. Then EtEF,R Xt E Bk* C Bj+1 C am1(
Emil a; EtEF;
xt + A)
So Em1 a; EtEF; Xt E A.
Just as Theorem 17.26 was a partial converse to Theorem 17.24, we now obtain a partial converse to Theorem 17.31. Notice that p is not required to be an idempotent.
Theorem 17.32. Let a = (a,, a2, ... , am) E A, let (x,)' 1 be a sequence in N, and let
PEn
1
FS((xr)°°_k). Then MT(a, (Xr)°__1) E al p + a2P { ...
amp,
P r o o f . W e show, by downward induction on f E ( 1 , 2, ... , m) that for each k E N,
{Emea;EtEF,xt:Fe,Fe+1,...,FmERf({k,k+1.... })and Fe
E aeP+ae+tP+...+amp. For E = m, we have FS((xt)°Ok) E p so that FS((amxt)°°k) E amp. So let f E { 1, 2, ... , m  1) and assume that the assertion is true fore + 1. Let k E N and let
A = (Eat EtEF; x, : Fe, Fe+t, ... , Fm E d'f({k, k + 1, ...}) and Fe
FS((atxt)°_k) 9 {x E N : x + A E ae+lp +ae+2p +
+ amp).
17.4 Linear Combinations of Sums
359
So let b E FS((aext)O°k) and pick Fe E 1f ({k, k + 1, ...}) such that b = at EtEFL xt
Let r = max Ft + 1. Then
{E;"a+t ai EtEF; xt : Fe+1, Fe+2.... , F. E .Pf({r, r + 1, ...})
andFt+i
+ amp as required.
13
We see now that MillikenTaylor systems are themselves partition regular.
Corollary 17.33. Let a E A, let B C N, let (x1)0°_1 be a sequence in N, and assume that MT(d, (xt)O° 1 ) C B. I f r E N and B.= B;, then t h e r e e x i s t i E {1, 2, ... , r} and a sum subsystem (y,)1 of (x1) 001 such that MT (a, (y)1) a Bi.
Proof. Pick by Lemma 5.11 some p + p = p E nk FS((xt)O°k). By Theorem 1
17.32, MT(a, (x,),) E al p + a2 P + + am p, where as = (a1, a2, ..., am ). Pick i E (1, 2, ... , r) such that B, E al p + a2 P + + am p, and apply Theorem 17.31. We observe that certain different members of A always have MillikenTaylor systems in the same cell of a partition.
Theorem 17.34. Let a, b E A, and assume that there is a positive rational s such that a = sb. T h e n whenever r E N and N = U;=1 Ai, t h e r e e x i st i E (1, 2, ... , r) and sequences (xt)O°1 and (yt)OD1 such that MT(a, (x,)0°1) C A; and MT(b, (yt)O°1) e Ai.
Proof. Assume that s = n where m, n E N and let c = mb. Pick by Corollary 17.33
some i E (1,2,...,r) and a sequence (zt)O°1 such that MT(F, (zt)O1) e Ai. For each t E N, let x, = nzt and yt = mz,. Then MT(a", (x1)001) = MT (b, (y,)°01) _ MT(F, (zt)O°1) We show now, as we promised earlier, that we can separate certain MillikenTaylor systems.
Theorem 17.35. There is a set A C N such that for any sequence (x1)O° 1 in N, MT((1,2), (x,)001) A and MT((l, 4), (x1)0°1) N\A. Proof. For X E N, let supp(x) be the binary support of x so that x = E,ESUPP(X) 21. 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) counts the number of blocks of zeros of odd length interior to the binary expansion of x, that is between a(x) and d(x). Thus, for example, if in
binary, x = 1011001010001100000, then z(x) = 3. (The blocks of length 1 and 3 were counted, but the block of length 5 is not counted because it is not interior to the expansion of x.)
17 Sums and Products
360
Let A = {x E N : z(x) = 0 (mod 4) pr z(x)  3 (mod .4)}. Suppose that we have a sequence (X: ) °O I such that either
MT((1, 2), (x1)001) C A
or
MT((1, 4), (xt)O01) C N\A.
Pick a sum subsystem (yt)°_I of (xt)I01 such that for each t, d(yt+i) > a(yt). (See Exercise 17.4.1.) For i E (0, 1, 2, 3) let
Bi={xEN:z(x)i (mod4)}. By Corollary 17.33 we may presume that if
MT((1, 2), (xt)101) c A, then
MT((1, 2), (Yt)`O1) c Bp
or
MT((1, 2), (yt)°O1) S B3
and, if
MT((1,4), (x,)°01) c N\A, then
MT((1, 4), (yt)JO1) S B1
or
MT((1, 4), (Yt)°O1) c B2.
By the pigeon hole principle we may presume that for t, k E N,
a(yt) = a(Yk) (mod 2),
d(Yt) = d(Yk) (mod 2), and z(yt) = z(Yk) (mod 4).
Assume first that fort E N, a(yt)  d(yt) (mod 2). Now if MT((1, 2), (yt)to 1) c A, we have y1 + Y2 + 2y3 and Y2 + 2y3 are in the same B; so
z(YI + Y2 + 2Y3) = z(Y2 + 2y3) (mod 4).
Since a(yi) = d(y2) (mod 2) there is an odd block of zeros between yt and Y2 in YI+Y2+2Y3,soz(Y1+Y2+2Y3) = z(yI)+1+z(y2+2Y3) and consequentlyz(yt) = 3 (mod 4). Butd(2Y3) = d(y3)+ 1 #a (y2) (mod 2) so z(y2+2y3) = z(y2)+z(y3) 3 + 3 = 2 (mod 4), a contradiction to the fact that MT((1, 2), (yt)°O 1) S A. If MT((1, 4), (yt)°O 1) c N\A, we have as above that
z(yt + y2 + 4y3) = z(y2 + 4Y3) (mod 4)
and z(yt + y2 + 4y3) = z(yt) + 1 + z(y2 + 4y3) so that z(yt) = 3 (mod 4). Since d(4y3) = d(y3) +2 = a(y2) (mod 2) we have z(y2 +4y3) = z(y2) + I + z(y3) = 3 (mod 4), a contradiction to the fact that MT ((1, 4), (yt)O° 1) c N\ A.
Now assume that fort E N, a(yt) 0 d(yt) (mod 2). If MT((1, 2), (yt)°O1) c A we have z(yt + y2 + 2Y3) = z(y2 + 2Y3) (mod 4) and, since d (y2) # a(yt) (mod 2), z(yt + y2 + 2y3) = z(yt) + z(y2 + 2y3) so that z(y1) = 0 (mod 4). Since d(2Y3) = d(Y3) + 1 = a(y2) (mod 2)
17.5 Sums and Products in (0, 1)
361
we have that z(y2 + 2y3) = z(y2) + 1 + z(y3) = 1 (mod 4), a contradiction. If MT((1, 4), (y,) f ° t) C N\A, we have as above that z(yt) = 0 (mod 4) and since d(4y3) = d(y3) + 2 0 a(y2) (mod 2), z(y2 + 4Y3) = z(Y2) + z(Y3) = 0 (mod 4), a contradiction. As a consequence of Theorem 17.35, we have the following contrast between infinite and finite image partition regular matrices.
Corollary 17.36. There exist infinite image partition regular matrices B and C and a two cell partition of N, neither cell of which contains all of the entries of Bs and of C y for any 1, y E N°'. Proof. The matrices for the MillikenTaylor systems MT((1, 2), (xt)°O1) and MT((1, 4), (yt) °O 1) are image partition regular.
In Section 15.4, we found several characterizations of finite image partition regular matrices.
Question 17.37. Can one find a reasonable combinatorial characterization of infinite image partition regular matrices? Exercise 17.4.1. In Exercise 5.2.2, the reader was asked to show combinatorially that given any sequence (xt)1°1 there is a sum subsystem (yt)°_1 of such that whenever 2k < yt, one has 2Jc+1IYt+1. Use the fact that, by Lemma 6.8, given any
p + p = p E I lkk,1 FS((xt)O0k), one has p E IHII, to show that there is a sum subsystem (yt)°° 1 of (xt)°O 1 such that for each t E N, a(yt) + 2 < d(yt+1).
Exercise 17.4.2. Prove that if H is a finite subset of A, r E N, and N = A then there exist a function f : H > {1, 2, ... , r} and a sequence (xt)°° 1 such that for each a E H, MT(a, (xt)O01) C Af(a). (Hint: Use Corollary 17.33.)
17.5 Sums and Products in (0,1) 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 set, then one cell will contain 1) for some sequence in 1) U (0, 1). (The terminology "Baire set" has different meanings in the literature. We take it to mean members of the smallest aalgebra containing the open sets and the nowhere dense sets.) In fact, we show that certain sums of products and products of sums can also be found in the specified cell. We use the algebraic structure of the subset 0+ of 0(0, 1)d which we discussed in Section 13.3. The results that we obtain are topological in nature, so it is natural to 1
17 Sums and Products
362
wonder how one can obtain them starting with the discrete topology on (0, 1). The answer is that we use certain topologically characterized subsets of 0+. Recall that a subset A of R is meager (or first category) if and only if A is the countable union of nowhere dense sets.
Definition 17.38. (a) A set A e (0, 1) is Baire large if and only if for every E > 0, A n (0, e) is not meager.
(b) B = (p E 0+ : for all A E p, A is Baire large).
Lemma 17.39. B is a left ideal of (0+, ). Proof 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 theorem 5.7, it follows that {p E fl(0, 1)d : for all A E p, A is Baire large) # 0. On the other hand, if p E P (O, 1)d\0+ one has some e > 0 such that (e, 1) E p and (e, 1) is not Baire large. Thus B # 0 . To see that B is a left ideal of
0+, let p E 2 and q E 0+ and let A E q p. Pick x E (0, 1) such that xtA E p. Given e > 0, xtA n (0, e) is not meager so x(xA n (0, e)) is not meager and x(x'An ( O, E)) c A n (0, E). The following result, and its measurable analogue Theorem 17.46, are the ones that allow us to obtain the combined additive and multiplicative results. Recall that a subset of R is a Baire set if and only if it can be written as the symmetric difference of an open set and a meager set. (Or do Exercise 17.5.1.)
Theorem 17.40. Let p be a multiplicative idempotent in B and let A be a Baire set which is a member of p. Then {x E A : x' A E p and x + A E p} E p.
Proof. Let B = {x E A: x1A E p}. Then since p = p p, B E p. Also A is a Baire set so pick an open set U and a meager set M such that A = UOM. Now M\U is meager so M\U 0 p so U\M E p. We claim that
(U\M)nBc{xEA:x'AEpand x+AEp}. So let x E (U\M) n B and pick E > 0 such that (x, x + e) e_ U. To see that x + A E p, we observe that (0, 1)\(x + A) is not Baire large. Indeed one has ((0, 1)\(x + A)) n (0, E) c x + M, a meager set. (Given y E (0, E), x + y E U so, if x+ y 0 A, then x+ y E M.) Now we turn our attention to deriving the measurable analogue of Theorem 17.40. We denote by µ(A) the Lebesgue measure of the measurable set A, and by µ*(B) the outer Lebesgue measure of an arbitrary set B. We assume familiarity with the basic facts of Lebesgue measure as encountered in any standard introductory analysis course. 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 Sums and Products in (0, 1)
363
Definition 17A1. Let A e (0, 1). (a) The upper density near 0 of A, d (A) is defined by
a (A) = lim sup
µ* (A n (0, h)) h
hlo
(b) A point x E (0, 1) is a density point of A if and only if Urn
µ* (A n (x  h, x + h))  1 2h
hyo
The following result, while quite well known, is not a part of standard introductory analysis courses so we include a proof.
Theorem 17.42 (Lebesgue Density Theorem). Let A be a measurable subset of (0, 1). Then Et, ({x E A : x is not a density point of A}) = 0.
Proof. If x is not a density point of A, then there is some e > 0 such that lim
µ(A n (x  h, x + h)) ao
2h
<
Since the union of countably many sets of measure 0 is again of measure 0, it suffices to let e > 0 be given, let
B={xEA:limihnfo
µ(A n (x  h, x + h)) 2h
<1E}
and show that µ(B) = 0. Suppose instead that IL*(B) > 0. Pick an open set U such that B C U and µ(U) < ui tee We first claim that
(t) if (If)nEF is a (finite or countably infinite) indexed family of pairwise disjoint
intervals contained in U and for each n E F, p(A n In) < (1  e)p(In), then µ*(B\ UnEF 1n) > 0. To see this, since in general µ*(C U D) < µ*(C) +µ*(D), it suffices to show that µ*(B n UnEF In) < µ*(B). And indeed
µ.*(B n UnEF 1n) .5 µ(A n UfEF In) = EnEF P(A n in) < FinEF(1  E)Fl(In) < (1  011 (U)
< A*(B). Now choose XI E B and h 1> 0 such that [x 1 h 1, x, + hl ] e U and
12(An(xi h1,x1 + hi)) < (1 e)2h1.
17 Sums and Products
364
Inductively, let n E N and assume we have chosen x1, x2, ... , xn in B and positive h 1, h2..... hn such that each [xi  h1, xi + hi ] c U, each u, (A n (xi  h i , xi + hi )) <
(1E)2hi,and[xihi,xi+hi]n[xjhj,xj+hj]=0fori 96 j. Let
do = sup{h : there exist x E B such that [x  h, x + h] c U,
[xh,x+h]nU"=1[xihi,xi+hi]=0,andµ(An(xh,x+h)) <(1E)2h}. Notice that do > 0. Indeed, by (t) one may pick x E B\ U"=1 [xi  hi, xi + hi] and then pick 6 > 0 such that (x  d, x + S) n Un 1 [xi  hi, xi + hi ] = 0 and (x  6, x + 3) c U. Then, since x E B, one may pick positive h < 6 such that µ(A n (x  h, x + h)) < (1  E)2h. Pick xn+1 E B and hn+1 >
such that [xn+i  hn+1, xn+1 + hn+1]
2 [xn+l  hn+t , xn+1 + hn+1 ] n U"=1 [xi  hi, xi + hi] = 0,
U,
and µ(A n (xn+l  hn+1, xn+1 + hn+1)) < (1  E)2hn+1 The inductive construction being complete, let C = B\ Un° [xn  hn, x, + hn]. Then by (]') µ*(C) > 0. Also, since ([xn  hn, xn + hn})n°1 is a pairwise disjoint 1
collection contained in (0, 1), En° 1 hn converges, so pick m E N such that
zoo Then µ(Un=m+1 [xn 3hn, xn + 3hn]).
m+l
hn < g*(C)/6.
 3h, xn + 3hn]) < jA,* (C) so pick x E C\(Un
m+1 [xn 
Then X E B\ Un t [xn  hn, xn + hn] so pick h > 0 such that [x  h, x + h] c U,
[xh,x+h]nUn
1[xnhn,xn+hn]=0,andis(An(xh,x+h)) <(1E)2h.
Now there is some ksuch that [x  h, x + h] n Un=t [xn  hn, xn + hn] # 0 (since otherwise each dk > h and consequently En_ 1 hn diverges) so pick the first k such that
[x  h, x + h] n [xk  hk, xk + hk] # 0 and notice that k > m. Then hk ? dk2' > 2
so that Ix  xk I < h + hk 3hk and hence x E [xk  3hk, xk + 3hk], contradicting the fact that x ¢ Un°_m+t [xn  3hn, xn + 3hn]. We also need another basic result about measurable sets.
Lemma 17.43. Let A be a measurable subset of (0, 1) such that d (A) > 0. There exists
B c A such that B U (0) is compact and d(A\B) = 0.
Proof. Foreachn E N, let An = An(1/2n, 1/2n1) and let T = {n E N : s(An) > 0}. As is well known, given any bounded measurable set C and any E > 0 there is a compact
subset D of C with µ(D) > µ(C)  E. Thus for each n E T, pick compact Bn S; An
with A(Bn) > lt(An)  4^1 . For n E N\T, if any, let Bn = 0. Let B = U (0) is compact.
Suppose now that d(A\B) = a > 0. Pick M E N such that s m < a. Pick x < 1/2m such that 1i((A\B)n(0, x))/x > 1 2' . Pickn E Nwith 1/2n <_ x < 1/2n1 and note that n > m. Then
1
A((A\B) n (0, x)) < Ek_n s(An\Bn) < Eko n 4k+t = 1 3 .4n
17.5 Sums and Products in (0, 1)
and x?

365
so
p ((A\B) n (0, x))/x <
2n < 3 2m'
a contradiction.
Definition 17.44. £ = (p E 0+ : for all A E p, d(A) > 0}. Lemma 17.45. £ is a left ideal of (0+, ). Proof. It is an easy exercise to show that if d(A U B) > 0 then either d(A) > 0 or
d(B) > 0. Consequently, by Theorem 3.11, it follows that {p E 18(0, 1)d : for all A E p, d(A) > 0) 0 0. On the other hand, if p E $(0, 1)d\0+ one has some e > 0 such that (e, 1) E p and d((e, 1)) = 0. Thus £ 0. Let p E £ and let q E 0+. To see that q p E £, let A E q p and pick x such that x1 A E p. Another easy exercise establishes thatd(A) = d(x1 A) > 0. Theorem 17.46. Let p be a multiplicative idempotent in £ and let A be a measurable member of p. Then {x E A : x1 A E p and x + A E P) E P.
{yEA:y is not a density point of Al. By Theorem 17.42, A(C) = 0. Consequently since p E X,
C gE p so B\C E P. We claim that B\C C {x E A x1 A E p and x + A E p}. Indeed, given x IE C one has 0 is a density point of x + A so by an easy computation,
a ((O, 1)\(x + A)) = 0 so (0, 1)\(x + A) 0 p so x + A E P. Now let us define the kind of combined additive and multiplicative structures we obtain. Definition 17.47. Let (xn)n° be a sequence in (0, 1). We define FSP((xn)n° ) and a : FSP((xn )n° 1)  * Y (5f (N)) inductively to consist of only those objects obtainable by iteration of the following: 1
(1) If m E N, then xm E FSP((xn)n° 1) and (m) E a(xm).
(2) Ifx E FSP((xn)n° 1),m E N, F E a(x),andminF > m, then{xm x,xm+x} C FSP((xn)n° 1), F U {m} E a(xm x), and F U {m} E a(xm +x). For example, if z = X3 + X5 X7 (x8 +x10 x11), then z E FSP((xrz)n 1) and 13, 5, 7, 8, 10, 111 E a (z). (Of course, it is also possible that z = x4 +x12 x 13, in which case also 14, 12, 13) E a(z).) Note also that (x3 +x5) x7 is not, on its face, a member 1) UFP((xn)n° 1) C FSP((xn)n 1). of FSP((xn)n°1). Notice that trivially The proof of the following theorem is reminiscent of the first proof of Theorem 5.8.
Theorem 17.48. Let p be a multiplicative idempotent in 2 and let A be a Baire set which is a member of p. Then there is a sequence (xn)n' 1 in (0, 1) such that FSP((xn)n° 1) C A
17 Sums and Products
366
Proof. Let Al = A. By Theorem 17.40, {x E Al : x1A1 E. p and x +At E p) E p so pick x, E Al such that x1 ' A 1 E p and x, + A, E p. Let
A2 = A, nx,'A, fl (x, +A,). Then A2 E p and, since multiplication by x,
and addition of x, are homeomor
phisms, A2 is a Baire set. Inductively, let n E N\{1} and assume that we have chosen (x,)°_1 and (At)' 1 so
that An is a Baire set which is a member of p. Again invoking Theorem 17.40, one
has that {x E An : x' An E p and x + An E p} E p so pick xn E An such that xn' An E p and xn + A. E p. Let An+1 =An n xn'Ann (xn + An). Then An+1 is a Baire set which is a member of p. The induction being complete, we show that FSP((xn )n_1) c A by establishing the following stronger assertion: If
z E FSP((xn)n 1), F E o(z), and m = min F, then Z E Am. Suppose instead that this conclusion fails, and choose
Z E FSP((xn)n° 1), F E o(z), and m= min F such that z 0 Am and IF I is as small as possible among all such counterexamples. Now
if F = {m}, then z = X. E Am, so we must have BFI > 1. Let G = F\{m} and pick y E FSP((xn)n° 1) such that G E a(y) and either z = xm + y or z = xm y. Let r = min G. Since IGI < BFI, we have that
y E Ar C Am+1 C xm1Am n(xm + Am) and hence xm + y E Am and xm y E Am, contradicting the choice of z.
Corollary 17.49. Let r E N and let (0, 1) = Ui=1 At. If each Ai is a Baire set, then there exist i E {1, 2, ... , r} and a sequence (x,)' 1 in (0, 1) such thatFSP((xn)n° 1) C A, .
Proof By Lemma 17.39, 2 is a left ideal of (0+, ), which is a compact right topological semigroup by Lemma 13.29. Thus, by Corollary 2.6 we may pick an idempotent p E S. Pick i E {1, 2, ... , r} such that A; E p and apply Theorem 17.48. We have similar, but stronger, results for measurable sets.
Theorem 17.50. Let p be a multiplicative idempotent in £ and let A be a measurable set which is a member of p. Then there is a sequence (xn)n°1 in (0, 1) such that c1FSP((xn)n° 1) C A U {0}
Proof Since A E p, d(A) > 0. By Lemma 17.43, pick B C A such that B U {0) is compact and d(A\B) = 0. Then A\B ¢ p so B E p.
Notes
367
Now, proceeding exactly as in the proof of Theorem 17.48, invoking Theorem 17.46 instead of Theorem 17.40, one obtains a sequence (xn) O such that FSP((xn) ' 1) :B and consequently ct FSP((xn )n0 1) C B U {0) C A U {0}. 1
Notice that ifctFSP((xr)°O 1) c AU(0), then inparticular ctFS((xn)n01) c AU{0} and hence, if F is any subset of N. finite or infinite, one has EnEF Xn E A.
Corollary 17.51. Let r E N and let (0, 1) = U =1 A. If each A, is a measurable set, then t h e r e e x i s t i E (1, 2, ... , r) and a sequence (x)01 in (0, 1) such that CBFSP((xn)n01) C A, U (0). Proof By Lemma 17.45, ..C is a left ideal of (0+, ), which is a compact right topological semigroup by Lemma 13.29. Thus, by Corollary 2.6 we may pick an idempotent p E X. Pick i E 11, 2, ... , r} such that Ai e p and apply Theorem 17.50.
Exercise 175.1. Let X be a topological space and let .8 = { U A M : U is open in X and M is meager in X}. Show that B is the set of Baire sets in X.
Notes The material in Section 17.1 is from [25], [27], and [31] (results obtained in collaboration with V. Bergelson).
The material in Section 17.2 is from [127]. (In [127], 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 a result of R. Graham (presented in [ 120, Theorem 4.3]) that whenever 11, 2, ... , 252) = A 1 UA2, there must be some x 0 y and some i E 11, 2} with {x, y, x+y, x y} C Ai, and consequently the answer to Question 17.18 is "yes" for n = r = 2. Theorem 17.21 is from [119] in the case n = 2. It is shown in [119] that the partition :R can in fact have 3 cells. It is a still unsolved problem of J. Owings [ 188] as to whether there is a two cell partition of N such that neither cell contains (xn +x,n : n, m E N) for any onetoone sequence (x,) 0 0 a (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 m i6 n in N, there is a two cell partition of N neither cell of which contains both SPn((x,)°_°_1) and SPm((yr)°O_1) for any sequences (x,)00_1 and (y,)°°_1. The results of Section 17.4 are from [79], which is a result of collaboration with W. Deuber, H. Lefmann, and I. Leader. In fact a result much stronger than Theorem 17.35 is proved in [79]. That is, if a" and b are elements of A, neither of which is a rational multiple of the other, then there is a subset A of N such that for no sequence
(x,)0°1 inN, is eitherMT(aa, (x,)01) c AorMT(b, (x,)0°1) c N\A. Most of the material in Section 17.5 is from [35], results obtained in collaboration with V. Bergelson and I. Leader. The proof of the Lebesgue Density Theorem (Theorem
368
17 Sums and Products
17.42) is froth [189]. The idea of considering Baire or measurable partitions arises from research of Plewik, Promel, and Voigt: Given a sequence (tn)n° 1 in (0, 1], such that En__i to converges, define the set of all sums of the sequence by AS((tn)n__t) F C N). In [197], Promel and Voigt considered the question: If (EneF to : 0 (0, 1] Ai, must there exist i E {1, 2, ..., r} and a sequence (tn)n° 1 in (0, 1]
with AS((tn)n_1) e A;? 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 A; has the property of Baire, then the answer becomes "yes". In [ 195] Plewik and Voigt reached the same conclusion in the event that each A; is assumed to be Lebesgue measurable. A unified and simplified proof of both results was presented in [36], a result of collaboration with V. Bergelson and B. Weiss.
Chapter 18
Multidimensional Ramsey Theory
Several results in Ramsey Theory, including Ramsey's Theorem itself, naturally apply to more than one "dimension", suitably interpreted. That is, while van der Waerden's Theorem and the Finite Sums Theorem, for example, deal with colorings of the elements
of N, Ramsey's Theorem deals with colorings of finite subsets of N, which can be identified with points in Cartesian products.
18.1 Ramsey's Theorem and Generalizations In this section we present a proof of Ramsey's Theorem which utilizes an arbitrary nonprincipal ultrafilter. Then we adapt that proof to obtain some generalizations, including the MillikenTaylor Theorem. The main feature of the adaptation is that we utilize ultrafilters with special properties. (For example, to obtain the MillikenTaylor Theorem, we use an idempotent in place of the arbitrary 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 [S]k = U,=1 A;. For each i E {1, 2, ..., r}, each t E {1, 2, ..., k}, and each E E [S]t1, define Bt(E, i) by downward induction on t:
(1) For E E [S]' , Bk(E, i) = {y E S\E : E U {y} E A;). (2) Fort E{1,2,...,kl}andEE[S]t1,
Bt (E, i) = {y E S\E : Bt+l(E U {y}, i) E p}. Then for each t E {1, 2, ... , k} and each E E [S]!1, S\E =
Bt(E,1).
Proof. We proceed by downward induction on t. If t = k, then for each y E S\E, E U {y} E A; for some i.
So let t E {1, 2, ... , k  1} and let E E [S]t1. Then given y E S\E, one has by the induction hypothesis that S\(E U {y}) = U;=1 Bt+1(E U {y}, i) so for some i, Bt+1(E U {y}, i) E p.
18 Multidimensional Ramsey Theory
370
Theorem 18.2 (Ramsey's Theorem). Let S be an infinite set and let k, r E N. If [S]k = U,=1 A,, then there exist i E {1, 2, ..., r} and an infinite subset C of S with [C]k C A.
Proof. If k = 1, this is just the pigeon hole principle, so assume that k > 2. Let p be any nonprincipal ultrafilter on S. Define Bt (E, i) as in the statement of Lemma 18.1. Then S = U;=1 Bt (0, i) so pick i E { 1, 2, ... , r} such that B, (0, i) E p. Pick x1 E B, (0, i) (so that B2({xl }, i) E p. Inductively, let n E N and assume that (xm)',_1 has been chosen so that whenever tE{1,2,...,k1}and has Bt+l ({Xm,, Xm...... Xmj }, i) E P.
Choose
xn+1 E (Bl(0,i)\{xl,x2,...,xn}) n n(Bt+l ({xm,, xm2, ... , xmj}, i) : t E {1, 2, ... , k  11
and m, <m2 <... <mt
m, < m2 <
< mt < n + 1, let such t and ml, m2, ... , m, be given. If m, < n, then the induction hypothesis applies, so assume that mt = n + 1. If t = 1, the conclusion holds because xn+, E B, (0, i). If t > 1, the conclusion holds because Xn+1 E Bt ({Xm1, Xm2, ... , Xmj_, }, 0
The sequence (xn)n° having been chosen, let ml < m2 < 1
< Mk Then
Xmk E Bk ({Xm, , Xm2 , ... , Xmk_ i }, I) SO {X.,, X.,_., Xmk) E Ai.
In order to establish some generalizations of Ramsey's Theorem, we introduce some special notation.
Definition 18.3. Let S be a set, let k E N, and let tp : S + N. (a) If (Dn)n° 1 is a sequence of subsets of S, then
[(D.)1,tp]k
(xi,x2,...,xk):xi 0xjfori
jin{1,2,...,k}
and there exist m, < m2 <
< ink in N such that f o r each j E {1, 2, ... , k}, Xj E Dmi and for each
j E {2, 3, ...,k}, mj > tp(xj_l) (b) If t E N and (D.)n [(Dn)n_1, tp]<
is a sequence of subsets of S, then
(xl, x2, ... , xk} : xi # xj for i # jin {1, 2, ..., k} and there exist m, < m2 < < ink in (1, 2, ..., t} such that for each j E {1,2..... k}, xj E Dmi and f o r each j E 12, 3, ... , k}, mj > tp(xj_1)}.
For example, if S = N, tp(n) = n, and Dn = (m E N : m > n) for each n E N, then
[(Dn)n°=1,tp]<={{x,,x2,...,xk}9N:x,
<x2<<xk}=[N]k.
18.1 Ramsey's Theorem and Generalizations
371
Theorem 18.4. Let S be an infinite set, let p E S*, let k, r E N, and assume that [S]k = Ut=1 A;. Let rp : S > N and assume that there is some C E p on which rp is finitetoone. Then there exist i E 11, 2, ... , r} and a sequence (Dn )n° 1 of members of p such that Dn+1 c D for each n and [(Dn)n° 1, rp]k c A;. P r o o f . First assume that k = 1 . Pick i E { 1, 2, ... , r) such that
E={XES:{x}EA;}Ep and let Dn = E for each n. Then [(D,,)',, rp] = A;. Now assume that k > 1 and let B, (E, i) be defined f o r each i E (1, 2, ... , r), each t E (1, 2, ... , k), and each E E [S]t'' as in Lemma 18.1. Then by Lemma 18.1,
S = U, =i B1(0, i ), so pick i E 11, 2, ... , r} such that Bt (0, i) E P. For each 2 E N, let He = {x E S : rp(x) < B}. Then by assumption, for each f, He n C is finite. Let D, = C fl B, (0, i). Inductively let n E N and assume that we have chosen (Dm)m=1 in p such that for each m E (1, 2, ... , n}:
(1) If m < n, then D,n+1 C D,n. (2) I f > 1, thenforeach t E {1,2,...,k1)andeachE E [(DjnH,n_,)
1',rp]t<,
_
Dm C Bt+1(E,i) Notice that both hypotheses are vacuous when n = 1.
W e now claim that if t E (1, 2, ... , k  1) and E E [(Dj n Hn)j=1,
then
Bt+1(E, i) E p. So let such t and E be given. Then there exist x,, x2, ... , xt in S and m, < m2 < . < mt in {1, 2, ... , n} such that E = {xi, x2, ... , xt}, Xj E Dmj n Hn for each j E {1, 2, ... , t}, and for each j E {2, 3, ..., t}, mj > rp(xj_,). If t = 1, we have x1 E Dm, c D, S; B1(0, i), so B2({x11, i) E p. So now assume
thatt > landletF = {x,,x2, ...,xt_,}. WeclaimthatF E
1.
Indeed, one only needs to verify that f o r each j E { 1, 2, ... , t  1), x j E Hm,_, . To see this, notice that rp(xj) < mj+1 < mt. Since F E [(Dj n V]' 1, we have by hypothesis (2) that Xt E Dm, S B,(F, i)
so that Bt+1(E, i) E pas required. Now f o r each t E 11, 2, ... , k  1}, [(Dj n
rp] is finite (because C n H,,
is finite), so we may choose Dn+i E p such that D,+1 c D. and for each t E (1, 2, ..., k  1) and for each E E [ (D j n Hn
rp] <, Dn+1 C Bt+1(E, i ). The construction being complete, we now only need to show that [(Dn)n° 1, (p]< c A. So let E E [(Dn)n° 1, rp]k and pick xl, x2, ... , xk, and m 1 < m2 < < Ink such that for each j E ( 1, 2, ..., k}, X j E Dmj , and for each j E {2, 3, ..., k}, mj > rp (x j _, ),
and E = {x,, x2, ... , xk}. Let F = (x,, x2, ... , xk_, }. Then for each j E (1, 2, ... , k  1), cp (x j) < m j+1 <
Mk so F E [(DinH,nk_i)k1',rp]<' SO Xk E Dmk C Bk(F, i), so E E A, as required. 0
18 Multidimensional Ramsey Theory
372
As the first application of Theorem 18.4, consider the following theorem which is a common generalization of Ramsey's Theorem and van der 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.
Theorem 18.5. Let k, r E N and let [N]k = Ui_1 At. Then there exist sequences (an)n° 1 and (dn)n in N such that whenever nl < n2 < < nk and for each 1
j E{1,2,.. ,k},tj E{0,1,.... nj }, one has {ant +tido,,ant+t2dn2,...,ank+tkdnk} E A;. Proof. Let cy : N ) N be the identity map and pick p E A.P (which is nonempty by
Theorem 14.5). Pick i E (1, 2, ... , r} and a sequence (D.)1 °11 of members of p as guaranteed by Theorem 18.4. Pick ai and dl such that (a1, ai+di} C DI andlete(1) = 1. Inductively assume that (a.)m_1, (dm)m, and (2(m))m_1 have been chosen so that for each m E (1, 2, ... , n),
{am,am+dm,am+2dm,...,am+mdm} C_ Dt(m). Lete(n+1) =max{e(n)+l,an+ ndn + 1) and pick an+i and do+i such that {an+1, an+1 + do+1, an+l + 2dn+1,
Assume now that n j 10, 1, ..., nj}. Then
< n2 <
, an+1 + (n + 1)dn+i } C Dt(n+i )
< nk and for each j E (1, 2, ... , k) tj
{any + tidn,, ant + t2dn2, .... ank +tkdnk} E [(Dn)
1, (p]< C A1.
E
o
The case k = 2 of Theorem 18.5 is illustrated 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 generalization of Ramsey's Theorem, which has often been applied to obtain other results, is the MillikenTaylor Theorem. To state it, we introduce some notation similar to that introduced in Definition 18.3. We state it in three forms, depending on whether the operation of the semigroup is written as "+", " ", or "U".
Definition 18.6. (a) Let (S, +) be a semigroup, let (xn)n° 1 be a sequence in S. and let k E N. Then [FS((Xn)nn i)]<
{EtEHI Xt, `tEH2 Xt, ... , EtEHk xt } f o r each j E { 1 , 2, ... , k}, Ht E P'f(N)
and if j < k, then max Hj < min Hj+i }. (b) Let (S, ) be a semigroup, let (x)1 be a sequence in S, and let k E N. Then 7
[FP((Xn)oo 1)]< = { {11tEH1 Xt, iltEH2 Xt, ... , 171tEHk Xt}
for each j e {1,2,...,k}, Ht E.l f(N) and if j < k, then max Hj < min Hj+i }.
18.1 Ramsey's Theorem and Generalizations
373
Figure 18.1: Monochrome Products of Arithmetic Progressions
(c) Let (Fn)n1 be a sequence in the semigroup (PPf(N), U) and let k E N. Then
[FU((Fn)n 1)]<
{UtEH, Ft, UtEH2 Ft, ... , UtEHk Ft} : for each j E (1, 2, ... , k), Ht E J" f(N) and if j < k, then max H1 < min Hj+1 }.
The MillikenTaylor Theorem is proved by replacing the arbitrary nonprincipal ultrafilter used in the proof of Theorem 18.2 by an idempotent. Theorem 18.7 (MillikenTaylor Theorem  Version 1). Let k, r E N, let (xn) n° be a sequence in N, and assume that [N]k = U =1 A. Then there exist i E (1, 2, ... , r) and a sum subsystem (yn)n°1 of (xn)n° such that 1
1
[FS((Yn)n 1)]< C A;. Proof. By passing to a suitable sum subsystem if necessary, we may assume that (xn) n° 1 satisfies uniqueness of finite sums. (That is, if F, G E Rf(N) and EnEF Xn = EnEG Xn,
then F = G.) Define tp : N > N so that for each F E J"f(N), cp(EnEF Xn) = max F, defining tp arbitrarily on N\ FS ((xn )n 1). Notice that (p is finitetoone on FS( (xn )n° 1).
Pick by Lemma 5.11 an idempotent p EON such that FS((X)°,°=n) E p for every
n E N. Pick by Theorem 18.4 some i E (1, 2, ... , r) and a sequence (Dn)n° 1 of members of p such that Dn+1 c Dn for each n and [(Dn)n° 1, tp]' c A.
18 Multidimensional Ramsey Theory
374
Let B1 = D1 f1FS((xm)m 1) and pick y1 E Bj*. Inductively, let n E N and assume that we have chosen (yt)r in N and (B,)'=, in p. Let 1
B.+1 = Bn n (Yn + Bn) n
fl FS((xm)m (p(Y,)+1)
and pick y.+1 E B.+1*.
Since, for each n E N, yn+l E FS((xm)m we have immediately that (yn)n° is a sum subsystem of (x,,)' 1. One can verify as in the first proof of Theorem 5.8 that whenever F E ,Pf(N) and in = min F, one has EnEF yn E Bm. To complete the proof, it suffices to show that 1
[FS((Yn) 1)]< c [(Dn) 1,V]< So, let a E [FS((yn)n°_1)]< and pick F1, F2, ... , Fk E ,f f(N) such that
maxFj <minFj+1 foreach j E {1,2,...,k 1} and
a = {EtEF, Yt, EIEF2 Y1,
,
EtEFk Yd
For each j E (1, 2, ..., k), let tj = min Fj, let ej = max Fj, and let mj = tp(Ytj1) + 1. Then EZEF; Yt E Btj C Dmj. Further, if j E 12, 3, ... , k), then tp(E,EFj_I Yr) = tp(Ytj_I)
tp(Ytj1) < mj
Corollary 18.8 (MillikenTaylor Theorem  Version 2). Let k, r E N and let [Pf(N)]k = Ui=1 A,. Then there exist i E {1, 2, ... , r} and a sequence (Fn)n_1 in ?f(N) such that max Fn < min Fn+1 forall n and [FU((Fn)n° 1)]< C A,. P r o o f For each i E 11, 2, ... , r), let
Bi = {(EIEHI 21t, E,EH2
2t1, . , , EtEHk
2r1) : {H1, H2,..., Hk) E A, ).
Then [N]k = Ui=1 Bi so pick by Theorem 18.7, i E (1, 2, ... , r) and a sum subsystem (yn)n of(2n'1)n° 1 suchthat[FS((yn)n° 1)]< C B. Since(y,,)n°1 isasumsubsystem 1
of (2n1)n° 1, pick for each n, some F,, E J f(N) such that y,, = E1EF 21'1 and maxFn < min F,,+,. To see that [FU((Fn)°1)]< C Ai, let H1, H2,..., Hk E ?f(N) with max Hj < min Hj+1 f o r each j E { 1, 2, ... , k  1). Then (EtEHI Yt, EIEH2 y , . . . . . EIEHk YI) E [FS((Yn)n°_1)]< C Bi
so there exist (K1, K2, ... , Kk) E Ai such that {E/EHI Yt, EIEH2 Yt,
,
EtEHk YI} _ {E1EKI
2t1
EIEK2 2
t1
.. , EIEKk
2`1}.
18.1 Ramsey's Theorem and Generalizations
375
Now, given j E (1, 2,.., k}, let Gj = UnEHJ F. Then EIEHj Yt = EIEG; 2t1 and if j < k, then max Gj < min G1+1. Thus (EtEK, 2f1, EtEK22`1,
, EtEKk 2t1}
= {E1EG, 2`1, E,EG2211, .., EIEGk 2r1}
so (G 1 , G2,..., Gk) = (K1, K2, ... , Kk ) (although, we don't know for example that G1 = Kt) so that {G1, G2, ... , Gk} E Al as required. It is interesting to note that we obtain the following version of the MillikenTaylor Theorem (which trivially implies the first two versions) as a corollary to Corollary 18.8, which was in turn a corollary to Theorem 18.7, but cannot prove it using the straight
forward modification of the proof of Theorem 18.7 which simply replaces sums by products. The reason is that the function tp may not be definable if the sequence (x n )n_t
in the semigroup (S, ) does not have a product subsystem which satisfies uniqueness of finite products.
Corollary 18.9 (MillikenTaylor Theorem  Version 3). Let k, r E N, let (S, ) be a semigroup, let 1 be a sequence in S, and assume that [S]k = U;=1 A. Then there exist i E (1, 2, ..., r) and a product subsystem (yn)R° of (xn)°O 1 such that 1
[FP((Yn)n° 1)]< C Ai. P r o o f . F o r each i E [1, 2, ... , r}, let
Bi = {(Fl, F2, ... , Fk} E [pf (N)]k : (REF, Xt, 11lEF2 Xt, ... , 11tEFk Xt } E Ai }.
Then [ 7'f (N)]k = Ui=1 Bi so pick by Corollary 18.8 i E 11, 2, ... , r) and a sequence (Fn)n°1 in .9f (N) such that max Fn < min Fn+1 for all n and [FU((FF,)n 1)]< C Bi. For each n E N, let yn = IItEF x,. Then [FP((yn)n° 1)]< C A. (The verification of this assertion is Exercise 18.1.2.)
Exercise 18.1.1. Let
Al={ (X, y)E[N]2:x
A2 = {{x, y} E [N]2:y>2x}. Then [N]2 = Al U A2. Prove that there do not exist i E 11, 2) and sequences (an)n°1 and (dn)n° 1 in N such that (1) whenever n t < n2 and f o r each j E 11 , 2) and t j E {1, 2, ..., nj) one has {ant +tldn,,an2 +t2dn2} E Ai and (2) whenever n E N and 1 < t < s < n one has {an + tdn , an + sdn } E A; . Exercise 18.1.2. Complete the proof of Corollary 18.9 by verifying that
[FP((Y.)n 1)]< C A1.
376
18 Multidimensional Ramsey Theory
Exercise 18.1.3. Apply Theorem 18.4 using a minimal idempotent in a countable semigroup S. See what kind of statements you can come up with. Do the same thing using
a combinatorially rich ultrafilter in fiN. (See Section 17.1.) (This is an open ended exercise.)
18.2 IP* Sets in Product Spaces We show in this section that IP* sets in any finite product of commutative semigroups contain products of finite subsystems of sequences in the coordinates. We first extend the notion of product subsystems to finite sequences.
Definition 18.10. Let (yn)' 1 be a sequence in a semigroup S and let k, m E N. Then (xn)n I is a product subsystem of FP((yn)n°k) if and only if there exists a sequence (Hn)n I in .I f(N) such that
(a) min HI > k, (b) max Hn < min Hn+1 for each n E { 1, 2, ... , m  1), and
(c) Xn = n,,H Yr for each n E (1,2..... m}. The proof of the following theorem is analogous to the proof of Theorem 18.4.
Theorem 18.11. Let f E N\{1} and for each i E {1, 2, ... , e} let Si be a semigroup and let (yi,n)no 1 be a sequence in Si. Let m, r E N and let X; _1S; = U;=1 CJ. There e x i s t s j E ( 1 , 2, ... , r) and f o r each i E (1, 2, ... , f  1), there exists a product subsystem (xi,n )' of FP((y,,, )' I) and there exists a product subsystem (xe,n )R° 1 of FP((ye,n) 1) such that (Xe1 FP((xi,n)n 1)) x FP((xe,n)nnOO1) C C3.
Proof. Given i E (1, 2, ... , e}, pick by Lemma 5.11 an idempotent pi with pi E I lk,.1 CE,BSi FP((Yi,n)n° k).
For(XI,x2,...,xe_I) E Xe=ISi and j E 11, 2,..., rl, let Be(xi, x2, ... , xe1, j) = {Y E Se : (XI, x2, ... , xe1, Y) E Cj}.
Now given t E {2, 3, ... , f  1}, assume that Br+1(x1, x2, ... , xr, j) has been defined for each (x1, x2, ... , x,) E X tI Si and each j E (1, 2, ... , r}. Given r1
(XI,x2,.. ,xr_I) E X1=1S,
and
j E {1,2,.. ,r},
377
18.2 IP* Sets in Product Spaces
let
Bt (xi, x2, ... , xt1, j) = {y E St : Bt+1(x1, x2, ..., xt1, Y, j) E Pt+1 }. Finally, given that B2(x, j) has been defined for each x E S1 and each j E (1, 2, ... , r},
letB1(j)={xES!:B2(x,j)EP2}. We show by downward induction on t that for each t E {2, 3, ... , e} and each
(xl,x2,...,xt1) E X;=,' Si,S, =Uj=1 Bt(xl,x2.....xt1,j). This is trivially true for t = f. Assume t E {2, 3, ..., e  1) and the statement is true fort + 1. Let (xl, x2, ... , xt_1) E X;=iSi. Given y E St, one has that St+1 = Uj=1 Bt+1(xi, x2, ... , xt1, y, j)
so one may pick j E { 1 , 2, ..., r} such that Bt+i (xl, X2.... , xt_1, y, j) E pt+1 Then
y E Bt(xl,x2,...,xtl,j) Sinceforeachx E S1, S2=Uj=i B2(x, j), one sees similarly that S1 =Uj=1 B1(j) Pick j E { 1, 2, ... , r) such that BI (j) E P1. Pick by Theorem 5.14 a product subsystem (xl,n)n° of FP((yi,n)n° 1) such 1
that FP((x1,n)n°1) c Bi(j). Let D2 = n(B2(a, j) : a E FP((xi,n)n 1)}. Since FP((xl,n)n 1) is finite, we have D2 E P2 so pick by Theorem 5.14 a product subsystem (x2,n)n' 1 of FP((y2,n)n° 1) such that FP((x2,n)n° 1 c D2. Let t E 12, 3, ... , f  1} and assume (xt,n)n° has been chosen. Let 1
Dt+i=1 (Bt+l(al,a2,..., at, l):(al,a2,...,a1)E Xi.lFP((xi,n)n 1)}. Then D1+1 E Pt+l so pick by Theorem 5.14 a product subsystem (xt+i,, )n_1 of FP((Yt+l,n)n° 1) such that FP((xt+1,n)n° 1) C Dt+1 Then
(XP=1 FP((xi,n)n 1)) X FP((x2.n)no 1) C Cj
as required. In contrast with Theorem 18.11, where the partition conclusion applied to products of arbitrary semigroups, we now restrict ourselves to 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 (yn )n° 1 be a sequence in S. and let m E N. The sequence (xn)n is a weak product subsystem of (yn)n° if and only if there exists a sequence (Hn )' 1 in J" f (N) such that Hn fl Hk = 0 when n # kin { 1, 2, ... , m } and 1
xn ='ItEH yt for each n E (1, 2, ..., m). Recall by way of contrast, that in a product subsystem one requires that max Hn < min Hn+1.
378
18 Multidimensional Ramsey Theory
Lemma 18.13. Let f E N and f o r each i E { 1, 2, ... , fl, let Si be a commutative semigroup and let (yi,n)n° i be a sequence in Si. Let
aC = (p E i8(X'_1Si) : for each A E p and each m, k E N there exist for each i E 11, 2, ..., f) a weak product subsystem (xi,,,)' n=1
of FS((yi n)) ksuch that Xe_i FP((xi,n)n 1) c A}. Then X is a compact subsemigroup of 6 (X
Si ).
Proof Since any product subsystem is also a weak product subsystem, we have by Theorems 18.11 and 5.7 that X 96 0. Since X is defined as the set of ultrafilters all of whose members satisfy a given property, X is closed, hence compact. To see that £ is a semigroup, let p, q E £, let A E p q and let m, k E N. Then {a E X i=1 Si : as ' A E q} E p so choose for each i E (1, 2, ... , f) a weak product subsystem (xi,n)n of FP((yi,n)n_ k) such that 1
Xi_1 FP((Xi,n)m=1) C {a E Xe_1Si : a'A E q}.
Given i E { 1 , 2, ... , f) and n E (1, 2, ... , m), pick Hi,n E Pf (N) with min Hi,n > k such that xi,n = nIEH,,,, yi,t and if 1 < n < s < m, then H, ,n fl Hi,5 = 0. Let r = max(U;_1 Un Hi,n) + 1 and let 1
B= n{a'A : a E Xe_1 FP((xi,n)rt 1)). Then B E q so choose for each i E 11, 2, ... , f) a weak product subsystem (zi,n)n 1
of FP((yi,n)n_,) such that X e_1 FP((zi,n)n 1) S B. Given i E (1, 2, ..., e} and n E {1, 2, ... , m}, pick K,,n E JPf(N) with min Ki,n > r such that zi,n = I1IEx,,,, yi,t
andif1
=0.
For i E {1, 2,_. , e} and n E {1, 2, ..., m}, let Li,n = Hi,n U Ki,n. Then i itELi,n yi,t = ftEH,,,, yi,f . flfEK,,,, yi,[ = Xi,f ' zi,t
and if 1 < n < s < m, then Li,n fl Li,s = 0. Thus for each i E (1, 2, ... , e}, (xi,t zi,t) 1 is a weak product subsystem of FP((yi,n)n' k).
Finally we claim that X E_1 FP((xi,n zi,n)n 1) S A. To this end let c E X;_1 FP((xi,n zi,n)n 1) be given. For each i E (1, 2, ... , e}, pick F, c 11,2,..., m} such that ci = IIIEFi (Xi,t Zia) and let ai = fItEF Xi,t and bi = f1:EFj zi,t Then b E XE_1 FP((zi,n)n 1) so b E B. Since a E Xe_1 FP((Xi,n)"'= 1) one has that b E as ' A so that a b E A. Since each Si is commutative, we have for each i E (1, 2, ..., f) that ntEFi (Xi.t zi,t) = (fl,EF xi,t) . (f1,EF zi,t) so that c" = a b as required.
13
379
18.2 IP* Sets in Product Spaces
Theorem 18.14. Let I E N and for each i E ( 1, 2, ..., e), let Si be a commutative e semigroup and let (y;,,,)' i be a sequence in Si. Let C be an IP* set in Xi_1S, and let m E N. Then for each i E (1, 2, ..., t) there is a weak product subsystem (xi,,,)n 1 of FP((Yi,n)n°_I) such that Xc_1 FP((xi,n)n 1) C C. Proof. Let X be as in Lemma 18.13. Then X is a compact subsemigroup of X i=1 Si so by Theorem 2.5 there is an idempotent p E X. By Theorem 16.6, C E p. Thus by the definition of X for each i E {1, 2, ..., B} there is a weak product subsystem (xi,n)n1
of FP((yi,n)n_1) such that XFP((xi,n)n 1) C C. Three natural questions are raised by Theorem 18.14. (1) Can one obtain infinite weak product subsystems (defined in the obvious fashion) L such that X i_ 1 FP((xi n ) 1) _ C? (2) Can one replace "weak product subsystems" with "product subsystems"? (3) Can one omit the requirement that the semigroups Si be commutative? We answer all three of these questions in the negative. The first two questions are answered in Theorem 18.15 using the semigroup (N x N, +). Since the operation is addition we refer to "sum subsystems" rather than "product subsystems".
Theorem 18.15. There is an IP* set C in N x N such that.
(a) There do not exist z E N and a sequence (x)1 in N such that either {z} x FS((xn)n_1) C C or FS((xn)n° 1) x {z} C C. (In particular there do not exist infinite weak sum subsystems (xl,n)n° 1 and (X2,n)n°_1 of FS((2n)n° 1) with FS((xl,n)' 1) x FS((x2,n)n° 1) S C.) (b) There do not exist sum subsystems (xn)n=1 and (yn)rs_1 of FS((2n)n° 1) with {xl, x2} x {yl, y2} c C. Proof. Let
C = (N X N)\
2n, EnEG 2n) : F, G E ?(co) and max F < min G or max G < min F}.
To see that C is an IP* set in N x N, suppose instead that we have a sequence ((xn , yn)) n_
in N x N with
FS(((xn, Yn))' 1) C {(EnEF 2n, EnEG 2n) : F, G E ?(w) and max F < min G or max G < min F}. Pick Fl and G I in R(w) such thatxl = E,EF, 2' and yl =
2'. Letk = max(F1 U
G 1). Choose, by Lemma 14.16, H E R f (N) such that min H > 1, 2k+1 I EnE H xn, and
2k+1I EnEH Yn Pick F', G' E P(w) such that
xn = EnEF 2' and EnEH Yn =
380
18 Multidimensional Ramsey Theory
EnEG' 2". Then k + 1 < min(F' U G'), so xl + EnEH xn = EnEF, UF' 2" and Yl + EnEH Yn = EnEG,UG' 2". Also EnEl1)UH(xn, Yn) = (xl + EnEH Xn, Y1 + EnEH Yn) E {(EnEF 2", EnEG 2") : F, G E
`J'and (co)
max F < min G or max G < min F}.
Thus k + 1 < max(F1 U F') < min(G1 U G') < k or k + 1 < max(G1 U G') < min(F1 U F') < k, a contradiction. To establish (a), suppose that one has z E N and a sequence (x n )n° in N such that either (z) x FS((xn)n° 1) c C or FS((xn)n° 1) x {z} c C and assume without loss of
generality that {z} x FS((xn)n° 1) c C. Pick F E P(w) such that z = EtEF 2' and let k = max F. Pick H E pf(N) such that 2k+1 I EnEH xn. Pick G E J.P(w) such that EnEH xn = E,EG 2'. Then max F < min G so (z, EnEH xn) 0 C. To establish (b) suppose that one has sum subsystems (xn)2of n  and (yn )2= n FS((2n)n° 1) with {xl, x2} x {y1, y2} c C. Pick F1, G1, F2, G2 E .P(w) such that 2", yl = EnEG, 2", y2 = EnEG2 2", max Fl < min F2, xl = EnEF, 2", X2 = and max G l < min G2. Without loss of generality, max F1 > max G 1. But then we have max GI < max Fl < min F2 so (x2, yt) 0 C, a contradiction. 1
A striking contrast is provided by Theorem 18.15(b) and the case t = 2 of Theorem 18.11. That is IP* sets are not in general partition regular. It is easy to divide most semigroups into two classes, neither of which is an IP* set. Consequently it is not too surprising when one finds a property that must be satisfied by an IP* set in a semigroup S (such as containing a sequence with all of its sums and products when S = N) which
need not be satisfied by any cell of a partition of S. In this case we have a property, namely containing FS((x1,n)n 1) x FS((x2,n)n° 1) for some sum subsystems of any given sequences, which must be satisfied by some cell of a partition of N x N, but need not be satisfied by IP* sets in N x N. The third question raised by Theorem 18.14 is answered with an example very similar to that used in the proof of Theorem 18.15.
Theorem 18.16. Let S be the free semigroup on the alphabet {yl, Y2, Y3, ...}. There is an 1P* set C in S x S such that there do not exist weak product subsystems (x")n1 and (wn)n1 of FP((yn)n° 1) such that {x1, x2} x {wl, w2} c C.
Proof Let
C=(SXS)\{(nnEFYn,nnEGYn):F,GCE ,%f(N) and max F < min G or max G < min F). To see that C is an IP* set, suppose one has a sequence ((xn, wn))n°l with FP(((xn, wn))
1) C {(nnEF Yn, nnEG Yn) : F, G E Jnf(N) and max F < min G or max G < min F}.
18.3 Spaces of Variable Words
381
Givenanyi < j inN,pick Fi, Fj, Gi, Gj, Fi,j, Gi, j E J'f(N) such thatxi = fIfEFF yn, Xi = nnEFj Yn, Wi = nnnEG, Yn, Wj = nnEGj yn, xixj = nnEFi,j yn, and wi wj = flnEGi,j Yn. Sincex1xj = nnEFi,j yn, we have thatmax F1 < min Fj andFF,j = FFUFJ and similarly max Gi < min Gj and Gi, j = Gi U Gj. Thus we may pick j E N such that max(F1 U G1) < min(Fj U Gj). Then
(Xixj, wiwj) 0 {(fIfEF Yn, nnEG Yn) : F, G E .! f(ly)
and maxF <minGor maxG <minF}, a contradiction.
Now suppose we have weak product subsystems (xn)n1 and (wn)n=1 of FP((yn)'1) such that {xl, x2} x (wl, w2) C. Pick F1, F2, G1, G2 E J" f(N) such that xl = I1nEF, Yn, X2 = nnEF2 Yn, WI = nfEG, Yn, and w2 = nnEG2 Yn Since x1x2 E FP((yn )n° 1), we must have max F1 < min F2 and similarly max G 1 < min G2. Without loss of generality, max F1 > max G 1. But then we have max G 1 < max F1 < min F2 so (x2, WO 0 C, a contradiction. 0
18.3 Spaces of Variable Words In this section we obtain some infinite dimensional (or infinite parameter) extensions of the HalesJewett Theorem (Corollary 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 %P to be a word over %P U {v} in which v actually occurs (where v is a "variable" not in 'Y). We introduce new notation (differing from that previously used) for variable words and nonvariable words, since we shall be dealing simultaneously with both.
Definition 18.17. Let W be a (possibly empty) set (alphabet). (a) W (T; v) is the set of variable words over 'P. (b) W (%P) is the set of (nonvariable) words over %P.
Notice that W(4'; v) U W('Y) = W('Y U {v}). Definition 18.18. Let %P be a set and let (sn (v)) n° 1 be a sequence in W('P; v). (a) The sequence (t, (v))n° 1 is a variable reduction of (sn (v))n° 1 if and only if there exist an increasing sequence of integers (ni )°O 1 with n j = 1 and a sequence (a1)O° in 'P U {v} such that for each i, 1
v E {ani , ani+1 , ... , ani+1 t } and
ti (v) = Sni (ani)^Sni+l (ani+l) ... ^Sni+I 1 (ani+i1).
18 Multidimensional Ramsey Theory
382
(b) The sequence (tn(v))n 1 is a variable extraction of (sn(v))n°1 if and only if (tn(v))n°1 is a variable reduction of some subsequence of (sn(v))n° 1. Thus, a variable reduction of (sn(v))n_1 is any infinite sequence of variable words
obtained from (sn(v))n° 1 by replacing each s,(v) by one of its instances, dividing the resulting sequence (of words over W U {v}) 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. For example, if (sn(v))n were the sequence (avbv, v, bav, abvb, vavv, ...) then (tn(v))n° 1 might 1
be (aabavbab, abvbbabb, ... Definition 18.19. Let W be a set and let (sn(v))n° 1 be a sequence in W(W; v). (a) The word w is a reduced word of (sn 1 if and only if there exist k E N and
a1, a2, ... , ak in W U (v) such that w = s, (al)'s2(a2) `'sk(ak) (b) A variable reduced word of (sn(v)) ° 1 is a variable word which is a reduced word of (sn(v))°O 1.
(c) A constant reduced word of (sn(u))n° 1 is a reduced word of (sn(v))n° 1 which is not a variable reduced word.
Reduced words, variable reduced words and constant reduced words of a finite sequence (sn (v))n=1 are defined analogously. Notice that a word is a variable reduced word of (sn (v))n° 1 if and only if it occurs as the first term in some variable reduction of (sn (v)) n° 1.
Definition 18.20. Let q, be a set and let (sn(v))n° 1 be a sequence in W(W; v). (a) The word w is an extracted word of (sn (v))n°1 if and only if w is a reduced word of some subsequence of (sn (v))n' 1. (b) A variable extracted word of (sn (v))n° 1 is a variable word which is an extracted word of (sn(v))n' 1. (c) A constant extracted word of (sn (v))R° 1 is an extracted word of (s n (v))n° which is not a variable extracted word. 1
Notice that w is an extracted word of (sn (v))n°1 if and only if there exist k E N, < nk in N, and al, a2, ..., a/, E W U {v) such that
n1 < n2 <
w = Sni (a1)sn2(a2)' ...^Snk(ak) Extracted words are defined in the analogous way for a finite sequence (sn (v))n=1 We now introduce some special notation which will be used in the proof of the main 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 W be a set, let k E N U loo) and let (sn(v))n_1 be a sequence in W (W ; v). Let S be the semigroup (W (W U {v}), "). (a) Foreach M E N, A((sn(V)),k,=,n) is the set of extracted words of (sn(v))n=m
(b) For each m E N, V((sn(v))R=m) is the set of variable extracted words of (Sn(v))n=m
18.3 Spaces of Variable Words
383
(c) For each m E N, C((sn(v))k=m) is the set of constant extracted words of (sn (v))n=m A((sn(v))n_m).
(d) A((sn(v))n_1) = nm=1 ctps (e) V((sn(v))' 1) = nm=1 cE,6s V((sn(v))n°=m). (0 C((sn(v))n= 1) = nm=1 CIfS C((sn(v))n m).
Notice that 4((sn(v))n° 1) = 'V((sn(v))' 1) U G((sn(v))R° 1). Notice also that if tY = 0, then each C((sn(v))k=m) = 0. Lemma 18.22. Let %P be a set, let (sn(v))rs_1 be a sequence in W(41; v), and let S be
the semigroup (W(W U {v}), "). Then A((sn(v))' 1) is a compact subsemigroup of 18S, and V((sn(v))n° 1) is an ideal of A((sn(v))n° 1). If 9l 0, then C((sn(v))°O 1) is a compact subsemigroup of PS.
Proof. We use Lemma 14.9 with J = (11, D = N, and Tm = S for each m E N. To see that ,it((sn (v))n° 1) is a subsemigroup of PS and V((sn (v))n°1) is an ideal of ,4((sn(v))n°_1), for each m E N let Em = A((sn(v))n°_m) and let I. = V ((sn(v))n°=m). In order to invoke Lemma 14.9 we need to show that
(1) for each m E N, Im c Em c Tm, (2) for each m E N and each W E Im, there is some k E N such that w"Ek C Im, and (3) for each m E N and each w E Em \ Im , there is some k E N such that w Ek S Em
and w"'Ik c Im. Statement (1) is trivial. To verify statement (2), let m E N and W E Im be given. Nwith nl>m,and al,a2,...,a,E'IU(v)(with Pick at least one a; = v) such that W = sn, (a1)^snz(a2)^ ... sn,(at).
Let k = nt + 1. Then w"Ek c Im as required. The verification of statement (3) is identical except that a1, a2, ... , at E 4'. Now assume that 41 0 0. To show that C ((sn (v))n°1) is a subsemigroup it suffices by Theorem 4.20 to show that for each m E N and each w E C((sn(V))°n°__m) there is some k E N such that w"'C((sn(v))n° k) c C((sn(v))°_m). The verification of this assertion is nearly identical to the verification of statement (2) above. o By now we are accustomed to using idempotents to obtain Ramseytheoretic results. In the following theorem, we utilize two related idempotents in the case 9! 0.
Theorem 18.23. Let 41 be a finite set and let (sn (v))n° 1 be a sequence in W(4'; v). Let
Y and g be finite partitions of W(W; v) and W(4') respectively. There exist V E ,F and C E $, and a variable extraction (wn(v))n_1 of (sn(v))' such that all variable 1
extracted words of (wn(v))n° 1 are in V and all constant extracted words of (wn(v))n° are in C.
18 Multidimensional Ramsey Theory
384
Proof Assume first that 'Y = 0. Then W(i/r; v) = {vk : k E N} and W(AY) = {0}. Given n E N, pick k(n) E N such that sn(v) = vkini. For each V E F let B(V) = {k E N : vk E V}. Pick by Corollary 5.15 V E F and a sum subsystem (f(n))n° of (k(n))n° t such that FS((2(n))n° r) C B(V). For each n E N let wn(v) = Onl. Then (wn(v))n is a variable extraction of (sn(v))n° and the variable extracted words of 1
1
1
(wn(v))n° 1 correspond to the elements of FS((e(n))n° 1) so that all variable extracted
words of (wn(v))' 1 are in V. The conclusion about constant extracted words of (wn(v))n° 1 holds vacuously. Now assume that 'I 0. By Lemma 18.22, C ((sn(v))n° 1) is a compact subsemi
group of ,BS, where S is the semigroup (W ('Y U {v}), " ). Pick by Theorem 1.60 an idempotent p which is minimal in C((sn(v))n° 1). Then p E A((sn(v))n° 1) so pick, again by Theorem 1.60, an idempotent q which is minimal in As n (v))n° 1) such that q < p. Since, by Lemma 18.22, V((sn(V)),° 1) is an ideal of A((sn(v))n° 1), we have that q E V ((sn(v))n1). Now, for each a E 'Y, define the evaluation map Ea : W(' U {v}) W(4<) as follows. Given W E W (tP U {v}), Ca (w) is w with all occurrences of v (if any) replaced by a. Thus, if w = w(v) E W ('Y; v), then Ea (w (v)) = w(a), while if w E W (%P), then Ea(W) = w. Notice that Ca is a homomorphism from W (4Y U {v}) to W (%P) and hence,
by Corollary 4.22, Ea is a homomorphism from,(W(q, U {v})) to,n(W(q')). We claim that for each a E 'If, Ea(q) = p. Since EQ is a homomorphism, we have that Z. (q) < EQ (p). Since ca is equal to the identity on W (*), we have EQ (p) = p and consequently EQ (q) < p. Since p is minimal in C ((sn (v))00 1), to verify that Ea (q) = p
we need only show that a(q) E e((sn(u))n° 1). To see that Ea(q) E C((sn(v))n 1), let m E N be given. Now V((sn(V));_,,,) E q SO Ca IV ((Sn (V))n° m)] E Ea (q) by Lemma 3.30. It is routine to verify that
Ea[V((Sn(V))nm)] C C((Sn(V))nm) m) E Q(q) as required. We shall, according to our usual custom, use " for the binary operation on PS which extends the operation " on S. SO C((sn(V))
Let V E F n q and let C E gn p. We put V a= {y E V : Y P E V and y"q E V) and CO = {x E C : x"p E C and x"q E V}. Since q"'q = p"q = q"p = q and p p = p, it follows that V G E q and Co E P.
We claim that for each y E Va and each x E Ca, y'(Vt) E q, y'(Vt) E p, x' (Vtt) E q, and X' (CO) E p. To see this, let y E Vt and X E Ct be given. To see that Y1 (0) E q, we observe that ky' pP' IV] is a neighborhood of q in PS, and so is x 'p1 Y
4
[V]. Also y'V Eq. Since y1 (0) = Y'V n,l Y 'p'{V] n xY p` {V], P
it it follows that y1(Vt) E q. Likewise, since p E Ay'pp'[V] n xyIp4'[V], and
y' V E p, y1(Vt) E P. Similarly x1(Vt) = x'VnAX' pP I[V]nAx I py l[V] E q andx1(Ct)=x1CnAXlpp'[C]nAX1pq'[V]
E P.
If x, y E A((sn(u))n° 1), we shall write x < < y if there exists m E N for which x E A((sn(v))n 1) and y E A((sn(v))n°m+i). We observe that, for any given x E A((Sn(V))n°1), {Y E A((s(v))n° 1) : x << y} E q.
18.3 Spaces of Variable Words
385
We now inductively construct a variable extraction (wn(v))n° 1 of (sn(v))n° 1 for which V((wn(v))nm 1) c Vt and C((wn(v))n° 1) c C. We first choose w1 (v) E Va ifl naEV1 e;' [CU]. We note that this is possible because
ea1[Ct] E q for every a E'Y. W e then let n E N and assume that we have chosen w; (v) f o r each i E { 1, 2, ... , n) so that wi(v) < < w;+1(v) for every i E (1,2,...,n  11, V((w;(v))" 1) e Va and
C((w; (v))i 1) C C. We can then choose wn+1 (v) satisfying the following conditions.
(1) wn+1(v) E Va n naEgv ea1[Ct], (ii) wn (v) < < wn+1(v), (iii) wn+1(v) E y1(Va) for every y E V((wi(v))n_1), (iv) wn+1(v) E ea1[y1(Vt)] for every y E and every a E'Y, (v) wn+1(v) E x1(Vt) for every x E C((w; (v))" 1), and (vi) wn+l (v) E ea1 [x1(0)] for every x E C((wi (v))"=1) and every a E %P. This choice is possible, because each of these conditions for wn+l (v) is satisfied by all the elements of some member of q.
To see that V((wi(v))n+i) C VU, let u E and pick k and ml < < Mk in N and al, a2, ..., ak E 'Y U (v) such that some ai = v and
m2 <
u = wm i (a l) ^ wm2 (a2)^ ... ^ wmk (ak)
If mk < n + 1, then u E Vt by the induction hypothesis so assume that mk = n + 1. If k = 1, then u E V t by condition (i), so assume that k > 1. If ak E W, then u E V O by condition (iv). If ak = v and {a1, a2, ... , ak_ 1 } c 'P, then u E V t by condition (v). If ak = v and v E jai, a2, ... , ak _ 1), then u E V1 by condition (iii). That C((w; (v))n+11) c Ct follows from the induction hypotheses and conditions (i) and (vi). Finally, notice that (wn(v))n 1 is a variable extraction of (sn(v))n° 1 by conditions (i) and (ii).
The following corollary includes the "main lemma" to the proof of Carlson's Theorem. Corollary 18.24. Let kY be a finite set and let (sn (v)) ' 1 be a sequence in W (%P; v). Let
F and g be finite partitions of W (4Y; v) and W ('Y) respectively. There exist V E F and C E 9 and a variable reduction (tn(v))n° 1 of (sn(v))n° 1 such that all variable reduced words of (tn (v))n° 1 are in V and all constant reduced words of (tn (v))' 1 are in C. Proof. Define cp : W(W U {v}) > W
U {v}) by
rp(a1a2 ... ak) = sI (a1)^s2(a2)^ ...
sk(ak)
where a 1,a2,...,akare in xPU{v}. Note that cp[W(%Y; v)] c W(W; v)andrp[W('P)] C
W(W). Let 3f = IV' [B] : B E F) and 3C = {rp1[B]
B E }. Then 3f and X
386
18 Multidimensional Ramsey Theory
are finite partitions of W(%P; v) and W(%P) respectively. Pick V' E X, C' E X, and a variable extraction (wn(v))'1 of the sequence (v, v, v, ...) as guaranteed by Theorem 18.23. (That is, (wn (v)) °_ 1 is simply a sequence of variable words, all of whose variable
extracted words are in V' and all of whose constant extracted words are in C'.) Pick V E F and C E a such that V' = q [V ] and C' = (p1 [C]. For each n E N pick kn e N and an,1, an,2, ... , an,kn E'P U {v) such that wn = an, I an,2 ... an,kn .
Let 81 = 0 and for each n > 1, let In = E7. k; . For each n E N let to (v) = St.+I
(an.2)
. st.+kn (an.kn )
and notice that (tn(v))°_1 is a variable reduction of (sn(v))n° 1. Notice further that if
b1,b2,...,bn E WU{v},then op(wl (bt)_w2(b2)_ ... wn (bn)) = t1(b1)t2(b2) ... tn(bn) so if b1b2 ... bn E W (%P; v), then t1(b1)"t2(b2)"
tn (bn) E V and if b1b2 ... bn E
W M, then t100t202)_ ... tn (bn) E C. Theorem 18.23 produces a sequence (wn (v))n_ 1 with a stronger homogeneity prop
erty than that in Corollary 18.24, but to obtain such a (wn(v))n° 1 we must take an extraction of (sn(v))°_j; a reduction won't do. (To see why it won't do, consider (sn(v))n'1 = (av, by, by, by, ...)and partition the variable words according to whether the first letter is a or not.) Exercise 183.1. Derive the Finite Sums Theorem (Corollary 5.10) as a consequence of Corollary 18.24. (Hint: Consider the function f : W(W; v) * N defined by the condition that f (w) is the number of occurrences of v in w.)
18.4 Carlson's Theorem In 1988 T. Carlson published [58] a theorem which has as corollaries many earlier Ramseytheoretic results. We prove this theorem in this section. Definition 18.25. Let %P be a (possibly empty) finite set and let 3 be the set of all infinite sequences in W (W ; v). Denote the sequence (sn )' 1 by i. For s" E 8, let
B0(s) = (7 E 8 : t is a variable reduction of s), and, for each n E N, let Bn ( s ' ) = (d E B o ( s ' ) : for each i E { 1, 2, ... , n}, t; = si }.
18.4 Carlson's Theorem
387
Remark 18.26. For every ss", t E s and m, n E w, if tt E Bm(i) and if m < n, then Bn (F) C B. (i)

The following lemma is an easy consequence of Remark 18.26. Lemma 18.27. A topology can be defined on s by choosing {Bn(ss) as a basis for the open sets.
E sand n E w}
Proof. Let s, t E s, let m, n E w, and let i E Bn (s) n Bn (t'). Let k = maxim, n}. Then by Remark 18.26 Bk (i) C Bn (s) n Bn (t). Definition 18.28. The Ellentuck topology on ,& is the topology with basis {Bn(i) : i E s and n E W). For the remainder of this section, we shall assume that ,& has the Ellentuck topology.
Definition 18.29. A set X c s is said to be completely Ramsey if and only if for every n E w and every s" E s, there is some i E B. (s) such that either Bn (t') c X or
Bn(t')nx=0. Recall that we take the Baire sets in a topological space to be the aalgebra generated by the open sets and the nowhere dense sets. (Other meanings for the term Baire sets also exist in the literature.) 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 meager (or first category) sets are those sets that are the countable union of nowhere dense sets. Carlson's Theorem (Theorem 18.44) is the assertion that a subset of s is completely Ramsey if and only if it is a Baire set (with respect to the Ellentuck topology). The proof in one direction is simple.
Theorem 18.30. If X is a completely Ramsey subset of s, then X is a Baire set. In fact X is the union of an open set and a nowhere dense set. Proof Let X° denote the interior of X. We shall show that X is a Baire set by showing that X \X' is nowhere dense so that X is the union of an open set and a nowhere dense set.
If we assume the contrary, there exists i E s and n E co for which Bn (s') c cfs (X \X °). Since X is completely Ramsey, there exists t' E B. (s') such that Bn (t) c X
or Bn(t) c &\X. However, if Bn(t) X, then Bn(t) c X° and hence B,,(F) n cfs(X\X°) = 0. This is a contradiction, because Bn (t) c Bn (i ). If B,, (t) c 3 \X, then Bn (t) n ces(X) = 0. This is again a contradiction, because
Bn(t) c B. (g) c cfs(X). Lemma 18.31. The completely Ramsey subsets of s form a subalgebra of .l (s).
18 Multidimensional Ramsey Theory
388
Proof It is immediate from the definition of a completely Ramsey set that a subset of 3 is completely Ramsey if and only if its complement is. So it is sufficient to show that the completely Ramsey subsets of 3 are closed under finite unions. To this end let X and Y be completely Ramsey subsets of 3. To see that X U Y is completely Ramsey, lets E 8 and n E w be given. Pick t E Bn (s") such that B (t') C X
or B.(1) c .8\X. If B.(!) C X, then Bn(i) c X U Y and we are done, so assume that B1(!) C 3 \X. Since Y is completely Ramsey, pick u" E B, (7) such that Bn(ua) C Y or Bn(u) C 45\Y. Now Bn(u') C B,, (F) and so if Bn(u') C Y, then Bn(ua) C X U Y and if
B. (a) C 3\Y, then Bn(i) C 3\(X U Y). Remark 18.32. Let (7m )m l be a sequence in 3 with the property that t"'+1
E
Bm+l(?n) for every m E N, and let t = (t11, t22, t33, ...). Then t E Bm(t'") for every m E Co.
We now need to introduce some more notation.
Definition 18.33. Let n E w. (a) Ifs E 45 and n > 0, then sin = ( 5 1 , 5 2 . . . . . sn) and s'1o = 0. (b) If s' E 8, then tails (s) = (sn+1, sn+2, ) (c) If s' E W (%P; v)", then Is ] = n. (d) Ifs is a finite sequence and t is a finite or infinite sequence, then (s, t) will denote the sequence in which s' is followed by F. (Thus, if s = (sl , s2, ... , sn,) and
t = (ti, t2, ...), then (s", t) = (s1, s2, ... , sm, ti, t2....).) (e) A subset X of 8 is said to be almost dense in Bn (s") if X n Bn (t') t E Bn (s').
0 whenever
Notice that the notation iijn differs 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 function s. Definition 18.34. We define a condition (*) on a sequence (T")nE,,, of subsets of 43 as follows: (*) For every n E w and every s' E 3
(a) B1(s) n T" 0 and (b) if s' E T", then B1(s) C T". We make some simple observations which will be useful in the proof of Carlson's Theorem.
Remark 18.35. Lets E 45. (a) If X is almost dense in Bn (i), then X is almost dense in Bn (I)for every t' E Bn (s).
(b) If X n Bn (s) = 0, then X n Bn (t) = O for every t' E Bn (s). (c) For any t r= 43, t E Bn (s') if and only if t1n = Sin and tails (t) E Bo(tailn (s )).
18.4 Carlson's Theorem
389
Lemma 18.36. Let (T")nE(o be a sequence of subsets of .8 which satisfies (*) and let s' E 3. Then for each k E to, there exists t E Bk (s') such that, for every n > k in N and every variable reduction r' of tin, we have (F, tailn(t)) E TM. Proof. We first show that
(t) for every t E & and every n E N, there exists a E Bn(t) such that (r, tailn(a)) E T1'1 for every variable reduction r" of in.
We enumerate the finite set of variable reductions of in as {r' 1, r' 2, We put r" ° = tails (t) and then inductively choose i 1, i 2, E .8 so that the following properties hold f o r each i E { 1 , 2, ...
(i) i' E Bo(?i') and (ii) (F', V) E TI't1. Suppose that i E (0, 1, 2, ... , m  1} and that we have chosen V. Let I = Ir'i+1I. By condition (a) of (*), we can choose u E B1((i'+1, i')) n TI. Let V+' = taill(u'). r+1) = u' Then V+' E Bo(i `) and (i'+, V+')
E TI' i+' I.
Having chosen z "', we put o = (tin, T "`). Since a""` E Bo(tailn(t)), we have E Bn(t). Now let r be a variable reduction of tin. Then r" = r"' for some i E {1, 2, ... , m}.
We note that s"m E Bo(i'), and hence that (r', ?m) E BI;i ((r"', i')). So (r`, i") _ (r"', tailn(a)) E TI'`1, by condition (b) of (*). Thus (t) is established. Now let k E co be given. By (f) we can inductively define elements TO, 71, t'2, of 8 with the following properties:
. .
(i)t°=t'=.. =tk=S, (ii) t " E Bn (t'"') for every n > k, and (iii) for every n > k and every variable reduction r of t n1 In, (r', tail,, (in )) E T1'I.
Let t = (t11, t22, t3, ...). Then t E Bk(s'). Let n > k be given and let r' be a variable reduction of tin. Now t1n =
t"1
,, so r" is a variable reduction of t n1In and
thus (r', tailn(i")) E TI'I. Also (r, tailn(t)) E Blr, ((r, tailn(t"))) and so by condition (b) of (*), (r', tails (t)) E T171.
Corollary 18.37. Let s' E 3, k E w, and X c 3. Then there exists t e Bk(s') such that, for every n > k and every variable reduction r' of in, either X is almost dense in Bl;, ((r', tailn(t))) or else X fl Bl;l ((r", tailn(t))) = 0. Proof. For each n E w we define T" C 8 by stating that i E T" if and only if either X is almost dense in Bn(r') or X fl Bn(?) = 0. We shall show that (T"),E,,, satisfies (*), and our claim will then follow from Lemma 18.36. Let n E co and i E 3. If X is almost dense in Bn (r'), then i E Bn (r") fl T" . Otherwise there exists i E Bn (i) for which Bn (u') fl X = 0 and soil E Bn (r') fl T". Thus (T" )nE,,, satisfies condition (a) of (*).
390
18 Multidimensional Ramsey Theory
To see that (T")fEW satisfies condition (b) of (*), let n .E CO, let i E T" and let u" E Bn (s). Then B. (1) C Bn (r"). Consequently; if X is almost dense in Bn (t'), then X is almost dense in Bn (u") and if x fl Bn (i) = 0, then x fl B. (u') = 0. Definition 1838. (a) Let F E 44. then R(t) = {w E W('1'; v) : w is a variable reduced word of 71.
(b) Let t E d and let w E R(?). Then N(w, t) is that 1 E N for which there exist al, a2,..., al E W U {v} such that w = tt(al)"'t2(a2) "'ti(al)
Remark 1839. Let s" E 8, t E Bo(s") and W E R(7). Then (w, tai1N(w,i)(t)) E BO ((w, tailN(w,i)(s)))
Lemma 18.40. Lets E 44, k E w and X C J. Then there exists t E Bk(s) such that, for every n E N satisfying n > k and every variable reduction F of tln, either (1) forevery w E R(tailn(t)), X isalmostdenseinB1;I}1((r', w, ta11N(w,taz1n(F))+n(t )) or
(2) for every w E R(tailn(t")), x fl Bjrj+1((r, w, tallN(w,W"(i))+n(t))) = 0. Proof We define a sequence (T")fE,, of subsets of S by stating that r' E T" if and only
if either x fl Bn+i((iln, w,
0 for every w E R(tailn(r')), or
else X is almost dense in Bn+1Win, w, taiIN(w,tai1n (T))+n (i)) for every w E R (tail, (r")).
We shall show that (T")nEW satisfies (*). The claim then follows from Lemma 18.36. Let n E co and i E 44. We shall show that T" and r' satisfy conditions (a) and (b) of
(*). By Corollary 18.37, there exists u E Bn (?) such that, for every m > n in N and every variable reduction F of u'I,n, X f) BI;I ((F, tail,, (ii))) = 0 or else X is almost dense in BIF, ((F, tail, (9))). Let
V0 = {w E R(tailn(a)) : X n Bn+1((uln, w, tailN(w,tatln(ti))+n(u))) = 0), V1 = R (tails Q) \ Vo, and
V2 = W('1'; v)\R(tailn(i )). We note that, if w E R(tail,, (u')), then w E V1 if and only if X is almost dense in Bn+1((uln, W, ta11N(w,tai1n(i))+n(u)))
By Corollary 18.24, there exists i E (0, 1, 2) and Q E Bo (tails (ii)) such that R (o) c
V,. Since R(v) C R(tailn(u)), i i6 2. Let p' = (u"In, a). Then p' E Bn(r') fl T". So T" and i satisfy condition (a) of (*). To verify that T" and i satisfy condition (b) of (*), assume that i E T" and let 'v E Bn(r"). Then tail,("v) E Bo(tailn(i)) and R(tailn(i)) c R(tailn(s)). If W E R (tails 0)), then (vin, w, ta11N(w.taitn(v))+n(v)) E B+1 Win, w, tailN(w,tailn(f))+n(r)))
so that Bn+1((i ,, w, tailN(w,tailn(v))+n(v))) C Bn+l ((=In, w,
It follows easily that 'v E T".
18.4 Carlson's Theorem
391
Lemma 18.41. Every closed subset X of 3 is completely Ramsey.
Proof Let s" E 3 and k E W. We need to show that there exists a E Bk(s) for which Bk(a) c X or Bk(a) fl x = 0. We suppose, on the contrary, that no such element exists. So X is almost dense in Bk (s). Let tt E Bk (s) be the element guaranteed by Lemma 18.40. We shall show that, for
every ii E Bk (t) and every n > k in w, X is almost dense in B (u). This is true if n = k. We shall assume that it is true for n and deduce that it is also true for n + 1. Suppose then that u E Bk (t) and let m E N be the integer for which u 1n is a variable reduction of By our inductive assumption, there exists z E X fl Bn (ii). Let w = xn+1 E R(tail,n(t)). Since l E Bn+t ((uiIn, w, taIIN(w.tailm(i ))+m(t ))) n X, it follows from our choice of t that X is almost dense in Bn+t ((uln, w', tailNw.taitm(i))+m (N) for every w' E R(tailm (t )). In particular, this holds if we put w' = un+, . We then have u" E Bn+t ((i 1,,, w', tallN(w'.taam(?))+m (t ))), and so X is almost dense in Bn+i (u). We have thus shown that, for every u E Bk (t) and every n > k in w, Bn (u) fl X # 0.
Since X is closed, this implies that i E X. Thus Bk (t') c X, a contradiction. Lemma 18.42. Let (F n )n° o be an increasing sequence of closed nowhere dense subsets of 3 and let N = Un°_a Fn. Then N is nowhere dense in 3.
Proof. For each n E w, let Tn = (i E 3 : Bn(i) fl Fn = 0). Since each Fn is completely Ramsey (by Lemma 18.41) and nowhere dense, (T")nEa, satisfies condition (a) of (*). It clearly satisfies condition (b) of (*).
To see that N is nowhere dense in 3, suppose instead that we have some s in the interior of cE N. Pick k E w such that Bk (s) c cP N. By Lemma 18.36, there exists t E Bk (s) such that, for every m > k in co and every variable reduction r of tam, we have
(F, tail. (F)) E T T.
Then t E ce N so pick u E Bk (t) n N and pick n > k such that u E Fn. Pick m > k such that urn is a variable reduction of
Then ii E B,((u1n, tailm(t"))) and
(u' I,, tail. (t )) E Tn so u iE Fn, a contradiction.
Corollary 18.43. If N is a meager subset of 3, then N is completely Ramsey.
Proof We have that N is the union of an increasing sequence (X,)' o of nowhere dense subsets of 3. Let X = cB ( U°_o Xn. Then X is nowhere dense, by Lemma 18.42, and X is completely Ramsey, by Lemma 18.41. It follows that, for each i E 3 and n E w, there exists t E Bn (s) for which B,, (t) fl x = 0. Since N c X, this implies
that Bn(t)f1N=0. Theorem 18.44 (Carlson's Theorem). A subset of 3 is completely Ramsey if and only if it is Baire.
Proof. By Theorem 18.30, every completely Ramsey subset of 3 is Baire. Now assume that X is a Baire 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
18 Multidimensional Ramsey Theory
is completely Ramsey. Applying Lemma 18.31 again, one has that X is completely Ramsey.
As an amusing consequence of Theorems 18.30 and 18.44 one sees that in 8, every Baire set is the union of an open set and a nowhere dense set. Notice that one has an immediate partition corollary to Carlson's Theorem.
Corollary 18.45. Let r E N and assume that = f r_, X. If each X; is a Baire set, then for every n E w and every i E 3, there exist some i E {1, 2, ... , r} and some t E B,, (9) such that Bn(t) C X.
Let n E w ands E 8 and suppose that f o r each i E {1, 2, ... , r} and each
Proof.
t E Bn (s') one does not have Bn (t) C Xi. Pick P E Bn (s") as guaranteed by the fact that X, is completely Ramsey. (So B n (t'') n X 1 = 0.) For i E {1,2,...,r I }, assume that t i has been chosen and pick F'+' E Bn (P) as guaranteed by the fact that Xi+1 is completely Ramsey. (So Bn (ti+') n Xi+t = 0.) Then B (t r) c 0, a contradiction. An extensive array of theorems in Ramsey Theory are a consequence of Carlson's
Theorem. For example, Ellentuck's Theorem [82] is the special case of Carlson's Theorem which has the alphabet iY = 0. We have already seen in Exercise 18.3.1 that the Finite Sums Theorem is derivable from the "main lemma" to Carlson's Theorem. To illustrate a more typical application, we shall show how the MillikenTaylor Theorem (Theorem 18.7) is derivable from Carlson's Theorem. As is common in such applications, one only uses the fact that open sets are completely Ramsey.
Corollary 18.46 (MillikenTaylor Theorem). Let k, r E N, let (xn)n° 1 be a sequence in N, and assume that [N]k = U =1 Ai. Then there exist i E {1, 2, .... r} and a sum subsystem (yn)I, of (xn)n_, such that [FS((Yn)n 1)]k C Ai.
Proof. Let 'Y = (a) and define the function f : W(4J; v) > N by letting f (w) be the number of occurrences of v in w.
For each i E (1,2,...,r) let Xi = {S E
3:
W (S,), SO, P S2), ... , f (Sk)} E Ai )
and let Xr+1 = {S E I S : I { f (Sl ), f(S2), ... , .f (Sk))I < k}.
Then 3 = Ur+I Xi
Now, given i E 11, 2, ... , r + 1) ands E Xi, one has that
Bk(s) C Xi so Xi is open. For each n E N, let s, = vX^, so that f (s,) = xn. Pick by Corollary 18.45 some
i E (1, 2, ..., r + 1) and some t E B0(s') such that BO(F) C Xi. Since B0(t) C Xi, i # r + 1. For each n E N, let yn = f Then (yn)no 1 is a sum subsystem of (xn
I
393
Notes
To see that [FS((yn)n° 1)]< c Ai, let H1, H2, ..., Hk E J'f(IY) be given such that max H. < min Hj+1 f o r each j E { 1 , 2, ... , k  1). For each j E (1, 2, ... , k), let Ej = max H j and let to = 0. For each n E { 1 , 2, ... , fk), let
a
bn
{v
k
if n 0 Ui1 Hi
ifnEUk_1Hi
and for j E { 1, 2, ... , k}, let wj = tej_,+1(bej_,+1)`'tei, +2(bej,+2) ...  te, (bej).
For j > k, let wj = tlk+jk. Then t E Bp(i) so W WI), (wl), P W2), ... , f (wk)} E Ai. Since
{f(wl), f(w2), ... , f(wk)} = {EnEH, Yn, InEH2 Yn,
,
EnEHk Yn},
we are done.
Notes The ultrafilter proof of Ramsey's Theorem (Theorem 18.2) is by now classical. See [67, p. 39] for a discussion of its origins. Version 1 of the MillikenTaylor Theorem is essentially the version proved by K. Milliken [183] while Version 2 is that proved by A. Taylor [233]. The rest of the results of Section 18.1 are from [26] and 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 Corollary 18.24 are from [20], a result of collaboration with V. Bergelson and A. Blass. The part of Corollary 18.24 that corresponds to W(ql) is [28, Corollary 3.7], a result of collaboration with V. Bergelson. It is a modification of a result of T. Carlson and S. Simpson [59, Theorem 6.3]. The part of Corollary 18.24 that refers to W(W; 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 introduced by T. Carlson and S. Simpson in [59, Section 6]. It is analagous to a topology on the set of infinite subsets of w which was introduced by E. Ellentuck in [82]. Given a finite alphabet T, call L an infinite dimensional subspace of W (%P) if and only if there is a sequence (sn (v))n 1 in W (%P; v) such that L is the set of all constant reduced words of (sn (v))1. Similarly, given a finite sequence (wn (v))d=1 in W (4'; v), call L a ddimensional subspace of W (4') if and only if
L = {wl(al)w2(a2) ... wd(ad) : al, a2, ..., ad E XP}. It is a result of H. Furstenberg and Y. Katznelson [99, Theorem 3.1 ] that whenever the collection of all ddimensional subspaces of W(kP) are partitioned into finitely
394
18 Multidimensional Ramsey Theory
many pieces, there exists an infinite dimensional subspace of W('Y) all of whose ddimensional subspaces lie in the same cell of the partition. This result can be established by methods similar to those of Section 18.3. (See [20].)
Part IV Connections With Other Structures
Chapter 19
Relations With Topological Dynamics
We have already seen that the notions of syndetic and piecewise syndetic, which have their origins in topological dynamics, are important in the theory of OS for a discrete semigroup S. We have also remarked that the notion of central, which is very important in the theory of f3S 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 reader that we take all hypothesized topological spaces to be Hausdorff.
Definition 19.1. A dynamical system is a pair (X, (TS)SEs) 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, T5 is a continuous function from X to X; and (4) for all s, t E S, TS o Tr = TSr If (X, (TS)SEs) is a dynamical system, one says that the semigroup S acts on X via (TT)SES.
We observe that we are indeed familiar with certain dynamical systems.
Remark 19.2. Let S be a discrete semigroup. Then (,OS, (X )IEs) is a dynamical system.
Definition 19.3. Let (X, (TS)SEs) be a dynamical system. A subset Y of X is invariant if and only if for every s E S, TS [Y] c Y.
19 Relations With Topological Dynamics
398
Lemma 19.4. Let S be a semigroup. The closed invariant subsets of the dynamical system (VS, (A5)SEs) are precisely the closed left ideals of $S.
Proof Given a left ideal L of,8S, p E L, ands E S, one has XS(P) = sp E L so left ideals are invariant.
Given a closed invariant subset Y of fS and p E Y one has that (PS) p = ct(Sp) S
cl'Y=Y.
t]
Definition 19.5. The dynamical system (X, (TS)SEs) 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, (TS)SEs), X contains a minimal nonempty closed invariant subset Y, and consequently (Y, (T5)sE5) is a minimal dynamical system, where T,s is the restriction of TS to Y.
Lemma 19.6. Let S be a semigroup. The minimal closed invariant subsets of the dynamical system (,BS, (AS)SES) are precisely the minimal left ideals of PS.
Proof. This is an immediate consequence of Lemma 19.4 and the fact that minimal left ideals of )SS are closed (Corollary 2.6). o
Of course, a given semigroup can act on many different topological spaces. For example, if X is any topological space, f is any continuous function from X to X, and is a dynamical system. And, since any for each n E N, T = f", then (X, dynamical system contains a minimal dynamical system, a given semigroup may act minimally on many different spaces. Definition 19.7. Let S be a semigroup. Then (X, (Ts)sEs) is a universal minimal dynamical system for S if and only if (X, (Ts)ses) is a minimal dynamical system and, whenever (Y, (Rs )SEs) is a minimal dynamical system, there is a continuous function (p
from X onto Y such that for each s E S one has that RS o v = w o TS. (That is, given any s E S, the diagram TS X (P
RS
Y
commutes.)
Theorem 19.8. Let S be a semigroup, let L be a minimal left ideal of,BS, and for each s E S, let )ls be the restriction of As to L. Then (L, (As)SEs) is a universal minimal dynamical system for S.
19.1 Minimal Dynamical Systems
399
Proof Let (Y, (Rs )sEs) be a minimal dynamical system and fix y E Y. Define g : S > Y by g(s) = RA(y) and let V = gL. Then immediately we have that (p is a continuous function from L to Y. To complete the proof it suffices to show that for all s E S and all p E L, ((,ls(p)) = Rs ((p(p)). (For then R.[(p[L]] C W[L] so (p[L] is invariant and thus by minimality (p[L] = Y.) To see that tp(X's(p)) = Rs(rp(p)) for all s E S and all p E L, it suffices to show
that for each s E S, g o A = Rs o g on PS for which it in turn suffices to show that g o As = Rs o g on S. So let t E S be given. Then g(As(t)) = g(st) = Rsr(y) = Rs(Rt(y)) = Rs(g(t)). We now wish to show that (L, (As)SEs) is the unique universal minimal dynamical
system for S. For this we need the following lemma which is interesting in its own right.
Lemma 19.9. Let S be a semigroup and let L and L' be minimal left ideals of fS. If tp is a continuous function from L to L' such that ,ls o rp = cp o As for all s E S (where),' is the restriction ofAs to L), then there is some p E L' such that tp is the restriction of pp to L.
Proof. Pick by Corollary 2.6 some idempotent q E L and let p = ((q). Then pp and tP o Pq are continuous functions from fiS to L'. Further, given any s E S,
pp(s)=sp=stp(q)=(),sotp)(q)=(tpoA')(q)=W(sq)=(tpopq)(s) and thus pp and tp o pq are continuous functions agreeing on S and are therefore equal. In particular, ppjL = (tp o pq)IL. Now, by Lemma 1.30 and Theorem 1.59, q is a right identity for L, so (tp o pq)IL = tp and thus tp = PpIL as required. The following theorem is in many respects similar to Remark 3.26 which asserted the uniqueness of the Stonetech compactification. However, unlike that remark, the proof of Theorem 19.10 is not trivial. That is, given another universal minimal dynamical system (X, (T5)3E5) and given s E S, one obtains the following diagram L
XS'
V
X
Ts
r
L
Asp
400
19 Relations With Topological Dynamics
and would like to use it to show that cp is one to one. However, if for example S is a left zero semigroup, so that 14S is also a left zero semigroup, then X (p) = As (q) for all p and q in PS, and the above diagram is of no use in showing that cp is one to one.
Theorem 19.10. Let S be a semigroup and let L be a minimal left ideal of CBS. Then, up to a homeomorphism respecting the action of S, (L, ()).f )SES) is the unique universal minimal dynamical system for S. That is, given any universal minimal dynamical system (X, (TS)SES)for S there is a homeomorphism (p : L ) X such that cp o)4 = TS o cp for every s E S. Proof. Pick cp : L + X as guaranteed by the fact that (L, (,4)SEs) is a universal
minimal dynamical system for S and pick r : X
L as guaranteed by the fact that (X, (Ts)SEs) is a universal minimal dynamical system for S.
Then one has immediately that cp o)4 = Ts o cp 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 cp is one to one. Now r o ' : L * L and for each
5ES, As o(r o(p) =)4 o(rocp)=(7o(p)0X, 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 Theorem 2.11(c), the restriction of pp to L is one to one, we are done. 0
19.2 Enveloping Semigroups We saw in Theorem 2.29 that if X is a topological space and 7 is a semigroup contained in XX (with the product topology) and each member of T is continuous, then the closure of T in XX is a semigroup under o, called the enveloping semigroup of T. In particular
(and this is the origin of the notion of enveloping semigroup), if (X, (Ts)sEs) is a dynamical system, then the closure of {Ts : s E S) in XX is a semigroup which is referred to as the enveloping semigroup of the dynamical system.
Theorem 19.11. Let (X, (TS)SES) be a dynamical system and define cp : S * XX by p(s) = Ts. Then ip is a continuous homomorphism from fS onto the enveloping semigroup of (X, (TS)sEs) Proof. Immediately one has that cp is continuous and that p[j6S] = cl'{Ts : s E S}. Also by Theorem 2.2, each TX is in the topological center of XX so by Corollary 4.22, cp is a homomorphism. o The following notation will be convenient in the next section.
Definition 19.12. Let (X, (TS)IES) be a dynamical system and define (p : S + XX by cp(s) = Ts. For each p E ,BS, let Tp = (p(p).
19.2 Enveloping Semigroups
401
As an immediate consequence of Theorem 19.11 we have the following. Be cautioned however that Tp is usually not continuous, so (X, (TP)PEPS) is usually not a dynamical system.
Remark 19.13. Let (X, (Ts)SES) be a dynamical system and let p, q E 13S. Then Tp o Tq = Tpq and for each x E X, Tp (x) = p lms TS (x).
How close the enveloping semigroup comes to being a copy of 8S can be viewed as a measure of the complexity of the action (Ts)sEs of S on X. With certain weak cancellation requirements on S, we see that there is one dynamical system for which 6S is guaranteed to be the enveloping semigroup. Lemma 19.14. Let Q be a semigroup, let S be a subsemigroup of Q, and let 0 = Q{0, 1 } with the product topology. For each s E S, define TS : n + n by TS (f) = f o ps. Then (S2, (Ts)SES) is a dynamical system. Furthermore, for each p E S. f E S2, and
5ES, (Tp (f)) (s) = 1 q {tES: f(t)=1}Esp. Proof. Let S E S. To see that TS is continuous it is enough to show that 7rt o TS is continuous for each t E Q. So, let t E Q and let f E Q. Then
(nt 0 T,) (f) = irt(.f ° Ps) = (f ° P,) (t) = .f (ts) = nts(.f) That is to say that jr. o TS = irts and is therefore continuous. Now lets, t E S. To see that TS o Tt = Tst, let f E 22. Then (Ts ° Tt) (.f) = TS (Tt (.f)) = Ts (.f ° Pt) = f ° Pt 0 Ps = f ° Pst = Tst (.f)
Finally, let p E 0S, f E S2 and S E S. Then
(TT(f))(S) = 1 q (plimT.(.f))(s) =I UES q plim(TT(f))(s) = 1 uES
p lim f (S U) = 1 UES
=1 It E S : f (t) = 1}Esp. T (SP)
.
0
Notice that by Lemma 8.1, the hypotheses of the following theorem hold whenever S has any left cancelable element. In particular, they hold in the important cases in which S is N or Z.
Theorem 19.15. Let S be a semigroup, let 0 = S{0, 1), and for each s E S, define TS : S2  0 by TS (f) = f o ps. If for every pair of distinct elements p and q of 0S there is some s E S such that sp ¢ sq, then fiS is topologically and algebraically isomorphic to the enveloping semigroup of (0, (T5)sEs)
402
19 Relations With Topological Dynamics
Proof. By Lemma 19.14, (S2, (Ts)SE5) is a dynamical system. Define W : S + 12 52 by
rp(s) = Ts. Then by Theorem 19.11, iP is a continuous homomorphism from $S onto the enveloping semigroup of (0, (TS)SEs). Thus we need only show that is one to one.
So let p and q be distinct members of $S and picks E S such that sp g0 sq. Pick A E sp\sq and let XA be the characteristic function of A. Then by Lemma 19.14, (TP(XA))(s) = 1 and (Tq(XA))(s) = 0. Therefore i(P) = TP # Tq = Rq) We have seen that if L is a closed left ideal of $S and Xs is the restriction of 1s to L, then (L, is a dynamical system. We know further (by Theorems 19.8 and 19.10) that if L is a minimal left ideal of fS, then (L, (As)$Es) is the universal minimal dynamical system for S. It is a natural question as to whether $ S can be the enveloping semigroup of (L, y )sEs).
Lemma 19.16. Let S be a semigroup and let L be a closed left ideal of $S and for each s E S, let Xs be the restriction of h s to L. Define W : S  LL by sp(s) = .)4. Then for all p E $S, W(p) = XPIL.
Proof. Let p E CBS, let q E L, and let TP  ip(p). Then by Remark 19.13, ip(p)(q) _
TP(q)=Psimsq=Pq Notice that if L contains any element which is right cancelable in iS then the hypotheses of the following theorem are satisfied. (See Chapter 8 for characterizations of right cancelability.) If S is a countable semigroup which can be embedded in a group and if L ¢ K(flS), then L does contain an element which is right cancelable in PS (by Theorem 6.56). So, in this case, the enveloping semigroup of (L, (As)SEs) is topologically and algebraically isomorphic to flS.
Theorem 19.17. Let S be a semigroup, let L be a closed left ideal of PS, and for each s E S, let Xs be the restriction of A5 to L. Then f S is topologically and algebraically isomorphic to the enveloping semigmup of (L, (As)sEs) via a map which takes s to Xs if and only if whenever q and r are distinct elements of $ S, there is some p E L such that qp 96 rp. Proof. Define (p : S + LL by V(s) = ,1;.. By Theorem 19.11 j5 is a continuous homomorphism onto the enveloping semigroup of (L, (,ls)5Es) (and is the only continuous function extending rp). The conclusion now follows from Lemma 19.16. If L is a minimal left ideal of $S, we have the following superficially weaker condition. Theorem 19.18. Let S be a semigroup, let L be a minimal left ideal of p S, and for each s E S, let Ls be the restriction of As to L. Then PS is topologically and algebraically isomorphic to the enveloping semigroup of (L, (AS)SES) via a map which takes s to Xs if and only if whenever q and r are distinct elements of fiS, there is some p E K(fS)
such that qp 0 rp.
403
19.2 Enveloping Semigroups
Proof Since L c K(fS), the necessity follows immediately from Theorem 19.17. For the sufficiency, let q and r be distinct members of 5S and pick some p E K(flS) such that qp 96 rp. Pick by Theorem 2.8 some minimal left ideal L' of PS such that p E L' and pick some z E L. By Theorem 2.11(c), the restriction of pz to L' is one to one and qp and rp are in L' so qpz rpz. Since pz E L, Theorem 19.17 applies. We conclude this section by showing that ON is the enveloping semigroup of certain natural actions of N on the circle group T, which we take here to be the quotient IR/Z under addition.
T * T by
Theorem 19.19. Let a E (2, 3, 4, ...) and for each n E N define Tn
Tn(7G + x) = Z + a"x. Then (T, (Tf)nEN) is a dynamical system. If rp N + TT is is a continuous isomorphism from ON onto the enveloping
defined by V(n) = Tn, then semigroup of (T, (Tn)nEN)
Proof Since a E N, the functions Tn are well defined. Let n E N, let E : R + R be multiplication by a", and let it : 111 + T be the projection map. Then Tn o it = it o 2 so Tn is continuous.
Trivially, if n, m E N, then Tn o Tn = Tn+m Thus (T, (Tn)fEN) is a dynamical system.
To complete the proof it suffices, by Theorem 19.11, to show that is one to one. To this end, let p and q be distinct members of fN. Pick B E {2N, 2N  1} such that B E p and pick A E p\q. Define x E (0, 1) by x = Ej°O1 = aja
j where aj=
ifj  IEAnB
1
ifj10AnB.
0
Let
C={9L+t:
2 a _
1
1
}
Notice that (since a2 +2 < a3), C n D = 0. We claim that (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 E N, if n E A n B, then TT(x) E C, and if n f A n B, then Tn(x) E D. So let n E N be given. Then
Tn (x) = Z + E! ° 1 a ja"J = Z + Ej n+l ajanj since aja"J E Z if j < n. Now assume that n E A n B. Then an+1 = 1 and (because aja"j = a1 + E n+3 aja"3 So
n + 10 B) an+2 = 0. Thus E n+1
U
< Ej=n+1
and hence Tn (x) E C as claimed.
ajanj
<
a
+
a3
404
19 Relations With Topological Dynamics
Next assume that n 0 A fl B. Then an+, = 0 and either an+2 = 0 or an+3 = 0. Consequently 0 < Ej n+la%anJ
T22 + a4
and hence T,, (x) E D as claimed.
If one views the circle group as {z E C : Izi = 11, then Theorem 19.19 says that the enveloping semigroup of the system (T, (f°)fEN) is isomorphic to $N, where
f(z)=z°.
19.3 Dynamically Central Sets The term "central set" was originally defined by Furstenberg [98, Definition 8.3] for subsets of N. Sets defined algebraically as elements of minimal idempotents were called "central" because many of the same theorems could be proved about them. The original definition (restricted to the semigroup N and applied only to metric dynamical 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. Definition 19.20. Let S be a semigroup. A set C c S is dynamically central if and only if there exist a dynamical system (X, (TS)SES), points x and y in X, and a neighborhood U of y such that
(1) y is a uniformly recurrent point of X, (2) x and y are proximal, and (3) C= Is E S: T,.(x) E U}. The terms "uniformly recurrent" and "proximal" are notions from topological dynamics, defined as follows. Definition 19.21. Let (X, (TS)SES) be a dynamical system. (a) A point y E X is uniformly recurrent if and only if for every neighborhood U of y, IS E S : TS(y) E U} is syndetic. (b) Points x and y of X are proximal if and only if there is a net (s,)1Et in S such that the nets (TS, (x)),EI and (TS, (y)),EI converge to the same point of X.
The definition of "proximal" as standardly given for a dynamical system with a metric phase space (X, d) says that there is some sequence (s n )n° , such that lim d (T,,, (x), TS (y)) = 0. It is easy to see that this definition is equivalent to the n
00
one given above. The characterization of proximality provided by the following lemma will be very convenient for us.
19.3 Dynamically Central Sets
405
Lemma 19.22. Let (X, (TS)SEs) be a dynamical system and let x, y E S. Then x and y are proximal if and only if there exists a point p E CBS such that Tp(x) = Tp(y).
Proof Notice that if p E V, (S,61 is a net in S which converges to p, and z E X, then Tp(z) = limLEJ TS,(z). Thus if p E 6S and Tp(x) = Tp(y), then picking any net (s,),EJ in S which converges to p, one has limLE! Ts, (x) = lim,EI Ts, (y).
Conversely, assume that one has a net (s,),EJ in S such that limLEJTs,(x) _ lim,EJ Ts, (y). Let p be any limit point of (s,),E, in fS. By passing to a subnet we may presume that (S,),EJ converges to p. Then Tp(x) = lim,Ej Ts, (x) = lim,EI Ts, (y) _ Tp (Y).
In order to establish the equivalence of the notions of "central" we investigate relationships between some algebraic and dynamical notions. Theorem 19.23. Let (X, (Ts)SES) be a dynamical system, let L be a minimal left ideal of ,6S, and let x E X. The following statements are equivalent. (a) The point x is a uniformly recurrent point of (X, (Ts)SES) (b) There exists u E L such that Tu(x) = x. (c) There exists an idempotent e E L such that Te(x) = x. (d) There exists y E X and an idempotent e E L such that Te(y) = x. Proof (a) implies (b). Choose any v E L. Let .A be the set of neighborhoods of x in X. For each U E dV let BU = IS E S : Ts(x) E U). Since x is uniformly recurrent, each
BU is syndetic and so there exists a finite set FU c S such that S = U,EFU(t'Bu). Pick to E FU such that tU  t BU E v so that tU v E BU. Let u be a limit point of the net (tUV)UE.N in , S and notice that u E L. We claim that T. (x) = x. By Remark 19.13, it suffices to show that B v E u for each V E X. So let V E X and suppose that By 0 u.
Pick U C V such that tuv E S\By. Then tUV E BU C By, 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) Te (x) = Te (Tu (x)) _ Teu(x) = Tu(x) = X. That (c) implies (d) is trivial.
(d) implies (a). We note that Te(x) = Te(Te(y)) = Tee(y) = Te(y) = x. Let U be a neighborhood of x. We need to show that IS E S : Ts(x) E U} is syndetic. Pick a neighborhood V of x such that ce V C_ U and let A = Is E S : Ts(x) E V}. Since
x = elim Ts(x) by Remark 19.13, A E e. Let B = Is E S : se E A}. Then by sEs Theorem 4.39, B is syndetic. We claim that B C Is E S : Ts(x) E U}. Indeed, if
SEB,then T5(x)=Ts(Te(x))=TSe(x)E(T,(x):tEA)CVCU. Theorem 19.24. Let (X, (Ts)SEs) be a dynamical system and let x E X. Then there is a uniformly recurrent point y E ce{Ts (x) : s E S} such that x and y are proximal.
Proof. Let L be any minimal left ideal of CBS and pick an idempotent u E L. Let y = Tu(x). Then trivially y E ce{Ts(x) : s E S}. By Theorem 19.23 y is a uniformly
406
19 Relations With Topological Dynamics
recurrent point of (X, (Ts)SEs). By Remark 19.13 we have that T. (y) = Tu(Tu(x)) = y are proximal. Tuu (x) = T. (x) so by Lemma 19.22
Theorem 19.25. Let (X, (Ts)sEs) be a dynamical system and let x, y E X. If x and y are proximal, then there is a minimal left ideal L of $S such that Tu (x) = Tu (y) for all
uEL. Proof. By Lemma 19.22, {p E fS : Tp(x) = Tp(y)} # 0. It is a left ideal in $S, because, for every p, q E AS, Tp(x) = Tp(y) implies that Tqp(x) = Tq(Tp(x)) = Tq (Tp(Y)) = Tgp(Y)
Theorem 19.26. Let (X, (Ts)5Es) be a dynamical system and let x, y E X. There is a minimal idempotent u in fS such that Tu(x) = y if and only if x and y are proximal and y is uniformly recurrent. Proof. Necessity. Since u is minimal, there is a minimal left ideal L of $S such
that u E L. Thus by Theorem 19.23 y is uniformly recurrent. By Remark 19.13 Tu(y) = Tu(Tu(x)) = Tuu(x) = Tu(x) so by Lemma 19.22 x and y are proximal. Sufficiency. Pick by Theorem 19.25 a minimal left ideal L of 1S such that Tu (x) _ Tu (y) for all u E L. Pick by Theorem 19.23 an idempotent u E L such that Tu (y) = y. We are now prepared to establish the equivalence of the notions of central.
Theorem 19.27. Let S be a semigroup and let B C S. Then B is central if and only if B is dynamically central.
Proof Necessity. Let Q = S U {e} where e is a new identity adjoined to S (even if S already has an identity). Let S2 = Q{0, 1} and for s E S define Ts : 12 + f2 by TS(f) = fops. Then by Lemma 19.14, (f2, (T5)SEs) is a dynamical system. Let x = X B, the characteristic function of B g Q. Pick a minimal idempotent u in $ S such that B E u and let y = Tu (x). Then by Theorem 19.26 y is uniformly recurrent and x and y are proximal. Now let U = {z E f2 : z(e) = y(e)}. Then U is a neighborhood of y in Q. We note
that y(e) = 1. Indeed, y = Tu(x) so {s E S : Ts(x) E U} E u so choose some s E B such that T,(X) E U. Then y(e) = TS(x)(e) = x(es) = 1. Thus given any s E S,
5EB
.
x(s)=I Ts(x)(e) = 1
q TS(x)EU. Sufficiency. Choose a dynamical system (X, (Ts)SEs), points x, y E X, and a neighborhood U of y such that x and y are proximal, y is uni