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CR and let V = -1 [U]. Then 7r-'[V] _ *-1 [U] so V is open. is one-to-one. 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(x-1) 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 =7r-1[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 (xa-1)aEI converges to x-1. Let z be any cluster point of (xa-1)aEl in S and choose a subnet (xs-1)6EJ which converges to z. Then by the continuity of at (x, z), (xsxs-1)3EJ converges to xz. Therefore z = Proof.
x-1.
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 Stone-tech Compactification of a Discrete Space
There are many different constructions of the Stone-tech compactification of a topological space X. In the case in which the space is discrete, the Stone-tech 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 non-empty 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).
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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 Zermelo-Fraenkel 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 K-uniform ultrafilter
on D is an ultrafilter U on D such that 119111 ? K. The set UK(D) = {91 : 91 is a K-uniform ultrafilter on D}. A uniform ultrafilter on D is a K-uniform 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 K-uniform 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 one-to-one 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 one-to-one 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 non-empty 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 = e-1 [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 Stone-tech 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 Stone-tech 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 a-1[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 a-1 [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 Stone-tech Compactification In this section we show that 9D is the Stone-tech 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 Stone-tech compactification.
Definition 3.25. Let X be a completely regular topological space. A Stone-Cech 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 Stone-Cech compactification of a space X rather than a Stone-Cech 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 Stone-Cech 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 Stone-Cech 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 Stone--tech 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 Stone-tech 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 Stone-Cech 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 (Stone-tech 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 : f-1[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 g-1 [A] = f-1 [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 non-empty, 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 = g-1 [{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 non-empty Gs-subset 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 non-empty 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 P-point 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 P-points 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 P-points 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 nn-1 B I = w so we may choose a one-to-one sequence (x n )°O n-1
witheachxn E DflnmBn,. LetA = (xn : n E N) and let A, = (A, D) UUa 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 one-to-one 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 one-to-one and so f [14N] is homeomorphic to ^ (The proof that f is one-to-one 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 x-uniform ultrafilters.
Definition 3.60. Let A be a set of sets and let K be an infinite cardinal. Then A has the K-uniform 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 "w-uniform 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 K-uniform 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 K-uniform 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 x-sequence (Ao )a 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
a--Wpb-.q c-->r
a--Wpb-q
= lim a-p(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-.pb-gc-sr 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 p-limit. 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,q-limxsrEU.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(p-limq-limst)
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 one-to-one.
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 s-1 in A, then JAS' [A] = {s-'a : a E A}. However, even if S can be embedded in a group G (so that s-1 is defined in G), s-1A need not equal {s-'a : a E A}. For example, in (N, ), let
A=N2+1={2n+1:n EN}.Then 2-1A={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)A-s={tES:t+sEA}.
Theorem 4.12. Let (S, ) be a semigroup and let A c S.
(a) For any sESand
ifs-AEq.
(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 s-1 A, we are done. ,
Sufficiency. Assume s-IA 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 s-1 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), s-t (A*(P)) E p.
Proof Lets E A"(p), and let B = s-i A. Then B E p and, since p is an idempotent, B*(p) E p. We claim that B*(p) C s-1 (A*(p)). So let t E B*(p). Then t E B so st E A. Also t-t 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 x-1A Eq. Thus A C {x E S : x'1A E q} so {x E S : x-tA 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 y-q(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 x-pp (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, ... , k-2) and every a = (al, a2, ... , ai) E D,
a E Aa implies that (a1, a2, ... , ai, a) E D.
ak:aI EA0and(a1,a2,...,ai-1)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 x-1A 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 : x-1A E q} E p} (the product p q computed in PT). (The notation x-IA 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 t-q 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
p-q= 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 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 : s-1 Jul E p} E q so choose a one-to-one 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 xn-1 (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 py-t[{a}] is infinite. Choose -a one-to-one sequence (xn)n_I in py-I [{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 finite-to-one, 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 one-to-one 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 {a-1(UIEG t-1 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 =
t-1A" in the definition of syndetic by "S = U:EG At-1". We also
would have replaced "a-' (UIEG t-1 A)" in the definition of piecewise syndetic by "(UIEG At-1)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 : x-1 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 t-1A. Thus the sets of the form t-tA cover the compact set L and hence L C UIEG t-t A for some finite subset G of S. To see that S c UIEG t-1B, 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 = a-1(t-'A) E p and so to E B and thus a E t-1 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 : x-1A E p} and pick G E ?j(S) such that S = UrEG t-1 B. Pick t E G such that t-1 B E q. Then B E tq. That is, {x E S : x-1A 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 t-1 B for some G E .?f(S). For each a E S, a E t-1 B for some t E G and so a-1(t-'A) = (ta)-'A E P. It follows t-1A) : a E S} has the finite that a'1(U,EG t-I A) E p and hence that {a-1(UfEG intersection property.
Sufficiency. Assume that A is piecewise syndetic and pick G E J'f (S) such that (a-1 ( UrEG t_'A) : a E S) has the finite intersection property. Pick q E /3S such
that {a-1(UrEG t-' A) : a E S} c q. Then S q c UIEG t-1 A. This implies that (/3S) -q S UrEG t-1 A. We can choosey E K(,BS) fl (,lS q). We then have y E t-1 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 : x-1 A is central} is syndetic. (c) There is some x E S such that x-1 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 $S with p E L and pick an idempotent e E L. Then p = p e so pick y E S such that y'1A E e. Now by Theorem 4.39 B = (z E S : z-t(y-'A) E e} is syndetic, so pick finite G C S such that S = UtEG t-t B. Let D = {x E S : x-1A is central}. We claim that S= U,E(y.G)t-t D. Indeed,letx E Sbegivenandpickt E Gsuch that E B. Then
(t . x)-(y-1 A) E e so (t x)'t (y't A) is central. But (t x)-t (y-t A) = (y t x)-t A. Thus y- t- x E D so x E (y . t)-t D as required. (b) implies (c). This is trivial. (c) implies (a). Pick X E S such that x-'A is central and pick an idempotent p E K ($ 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 £NC-compactification 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 r-colored", 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 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 cxm-1 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 a-l A* E p and hence n0EE a-1 A* E p. Pick xn+l E A* fl naEE a-t 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 a-IA'. By Lemma 4.14 a-I 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)n-1 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 Ui-1 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 2n-1.)
Exercise 5.2.4. Prove Corollary 5.15.
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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. Then-FS((yn )n° ,) C A so by Lemma 5.19, q p E r.
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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 = ut-k+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)k-1) _ { 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.
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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 = {M-1 [X] fl m-1 [X] : X E x}. We claim that A has the finite intersection property, for which it suffices to show that M-1 [X] fl m-t [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 G-1 V = (F E Yf(N) : G U F E V} and let U* = (G E 2l : G-1 U E p}. Then by Lemma 4.14, U* E p and for each G E 2(*, G-1 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(
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(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 r-coloring of X is a function (p : X -). {1, 2, ... , r}. (b) Given an r-coloring 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 r-colored, 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 r-coloring 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 r-coloring r of S by r(se) = a(f). Then S = Ui=1 r-1 [{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 r-1[{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 r-1 [{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 r-colored, 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
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Proof. Let r, m E N be given and suppose the conclusion fails. For each n E RI
choose an r-coloring 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
iftn.
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 r-1[{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 r-1[{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 j7-1 [{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")) m-sp 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-
to-one, 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 one-to-one, 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 one-to-one, 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) = p-zl XEN
YEN
q-yliEmrp(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 : x-1A 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 : X-1 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 = s-I xq. In the second, q = t-I 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 = s-txf in,6G, and hence that of = s-'xf f = s-txf = 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 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 one-to-one 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 = h-1(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 h-1 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\h-1 [A]. 1 if n E Then p = h-' (q) E h-' [c2 A] = ct h-1 [A] and f and h agree on h-1 [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 one-to-one function f : N -> 1`Y*
such that f (n) = h(n) for all n E h-1 [B] and f [N] is discrete. Then p = h-' (q) E h-' [c8 B] = cf h-' [B] and f and h agree on h-1 [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 x-1 and y-1 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 K-sequence (sI )a 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 non-empty 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.
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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
$ScPG.
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6 Ideals and Commutativity in ,5S
Theorem 6.55. Let S be a countably infinite discrete semigroup embedded in a count-
able discrete group G. Let q E K($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
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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, a-tbm # 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 d-iz' E fiG\{1} in the first case, and (d')-iw' E fG\{1} in the second case. (If Z' E G, then d-iz' 1 because z 0 W. If z' E G*, then d-iz' 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.
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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 = b-tzx 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 = a-Jr'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.)
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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 well-ordered chains of semiprincipal left ideals. It is not known whether every totally ordered chain of semiprincipal left ideals of N* is reverse well-ordered. 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.
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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 = s-tt 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 s-lq E fiS and so rq = s-'tq = is-tq 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*.
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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 n+l+m.
Also,
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132
Lemma 6.70. Ifaa,b' E ®°_1 Z, E,°°1aizi = E,°=-1 b:zi,andforeachi E N, JaiI bn. Then E" 1 aizi = E°=1 bizi so
n- 2i+ 11 Zi < Zn, n-I (bi - aizi < Ei=1 Zn 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 non-empty 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).
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138
(3) X (t"v) U X (t" 1) = X (t) and X(t"0) n X (t" 1) = 0.
(4) x(t-0) = 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 vice-versa.
(10) If S, t E F ands _ 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(tk-1), and (iii) if h(ak) = gp(tk-1), 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)
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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 = t-1 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')).
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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(tn-l). 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(tn-1) + 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+1-1) + (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,U-Egyp (ii) {U" : U E cp} is a base for the filter rp-, and
(iii) for all U E cp and all a E U', a-1 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.
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To verify (iii), let U E V and a E U" be giver, and suppose that a-1 U" V V-. Then C"\a-1 U. Then there is an element y E
G\a-1U"={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 a-lc E U- so by (iii), c-1aU" 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 a-1 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\a-1 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
Px-1 [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 : a-I 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 : a-1Uyz E Z) E y. Thus nzEC {a E G: a-1Uyz 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.
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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. a-1Ayz 0 z so that G\a-1 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 non-empty 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:a-1An+iEY1} so that An+t E ylny1 = Y1n}1. So let a E An and pick s1 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 = 0-1 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 on-the 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 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\0-1 [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\0-1[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 0-1[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 = p-lim(x * y) = p-lim(y x) = p x yES
yES
and, similarly, p * x = x p. Consequently, the left-right 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 Rudin-Keisler 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: x-1 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:X-1BE p}. Proof. The sufficiency is easy. Given r
Af1T={xET:x-1BEp}.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:x-iBEp}. 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
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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 x-1B 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 : s-1 C E p).
(Ifa E A, then a-1Ba C a-1C andifa E T\A, then a-1Ban a-1C=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 non-empty open subset of UK (S) contains a non-empty 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 s-1 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:s-1CEp}.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.
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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 P-point 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 P-point in S* is right cancelable in PS.
Proof. Suppose that p is a weak P-point 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 P-point 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 P-point 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 Rudin-Keisler order 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 P-point 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 P-point of 7L*. Then there is a set A E p such that every
q E A fl r is cancelable in $Z. Proof. For each n E N let Bn = {m r= N : hn(m) = hn(p)}. Then each Bn E p so nn_i Bn is a neighborhood of p so pick A E p such that A fl Z* c no,, i Bn. Then Corollary 8.36 applies to A and any q E A fl V.
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176
Corollary 8.38. Let (xn)n° 1 be an infinite sequence in N with the property that xn+l is a multiple of xn for every n E N. Then every ultrafilter in N* fl {x : n E N) is cancelable in (f7G, +). Proof. This is Exercise 8.4.1.
By Ramsey's Theorem (Theorem 5.6), every infinite sequence in N contains an infinite subsequence which satisfies the hypothesis of Corollary 8.38 or the hypothesis of the following theorem. So these two results taken together provide a rich set of left cancelable elements of flN. Theorem 8.39. Let (x,1)n-_l be an infinite sequence in N with the property that, for every m 0 n in N, x" is not a multiple of xm. Then everyaltrafrlter in (xn : n E N) is left cancelable in ($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+3e-E,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,...,j-1},aj=2,a,E{1,2}iftE{j+l,j+2,...,f-1}, 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 = Ek-e+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 P-point 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 P-point 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 $S. -
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8 Cancellation
The reader may have noticed that all the criteria for left cancelability give- u. ;is section, apart from that of being a P-point in an arbitrary countable cancellative semi. group, describe a clopen subset of S* all of whose elements are left cancelable. Thus these criteria cannot be used to find left cancelable elements in K($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 t-1 H. We can choose P E p with the property that e 0 P and F-1 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 ={u-1v:U,VEFP({aEG:a mk_i. We then have bn, = u-1 Vbn,k, where u = bn,bn2 ... bn,_, and v = abn,,b,n2 . . .b,nk_,. This implies that u = v, because otherwise we should have b, E F711 P, contradicting our assumption that bn, E P,,-,. However, u = v implies that b,nk = bni . Similarly, if mk_1 > ni_1, the equation b,nk = v-1ubn, implies that bmk = bn,. So bn,k = bn, and these terms can be cancelled from the equation and the argument repeated until we have a = bn, b12 ... bn, if i = l - k > 0, or else abn,, bn,2 ... bn,j = e
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 a-1 = b,n,, contradicting our assumption that a-1 < bn,,. Lemma 8.49. The expression for an element of T as a P-product is unique.
Proof. We apply Lemma 8.48 with a = e. We observe that we cannot express e as a P-product bn, bn2 ... b, with i > 0. To see this, note that otherwise, if i = 1, we
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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 P-product. Thus, if bn,, bn2 ... bn,k = bn, bn2 ... bn, , where these are P-products, 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 P-product. 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 P-product: 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,/ 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 P-product and bn,, E Pk, let r = max{n, m; + 1}. (iii) If a = xbj, b12 ... b4 = y where b1, b12 ... b11 is a P-products and b1, E Pn let
r=max{Ij+1,k}.
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(iv) If a = xbt, b12 ... big = ybm, bm2 ... bm; where bl, b12 . . . big and b,,,, b,,,, ... b,,,. are P-products, 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 P-product 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 P-product. 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 P-product, 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 1 + 2 + 3 + + bn;_, . We shall refer to a sum of this kind as a P-sum. We define Tn to be the set of sums of all P-sums 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 P-sum 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. El-Mabhouh 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 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 a-i 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)
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198
Now let b E G. We claim that Ab is r-dense 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 $G. Let V be a nonempty r-open subset of G and pick a E V. Then a-t V is a r-neighborhood of e so a-' V E po(°,b). Since also Ab E apo(a,b), we have
VflAb360.
D
Corollary 9.20. Let G be a group with a left invariant topology r with respect to which G has no isolated points. If there is a countable basis for r then G can be partitioned into co disjoint subsets which are all r-dense in G.
Proof Let {V,, : n E N} be a basis for the neighborhoods of the identity e of G. By Theorem 3.36 the interior in G* of nt1 is nonempty and thus by Theorem 8.10 there is some p E nn_t which is right cancelable in (3G (where PG is the Stone-tech compactification of the discrete space G). Thus Theorem 9.19 applies. E3
Exercise 9.2.1. Let S be any discrete semigroup and let p be any idempotent in S*. Prove that the left ideal (fS)p is extremally disconnected. (Hint: Consider Lemma 7.41.)
Exercise 9.2.2. Let G be an infinite discrete group and let p and q be idempotents in G*. Show that the following statements are equivalent.
(a) q :5R P. (b) The function Gp -). Gq defined by /r(ap) = aq is continuous. (c) The topology induced on G by (Pp)IG is finer than or equal to the one induced by (Pq)IG.
(Hint: Consider the proof of Theorem 9.13.) Exercise 9.2.3. Let G be a group with identity e and let r be a non-discrete left invariant
topology on G. Let Xr = n{cI$G(U\{e}) : U is a r-neighborhood 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 non-discrete 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
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199
p is an idempotent in G* such that the T-neighborhoods 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 r-dense 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 r-dense subsets, but cannot be partitioned into n + 1 disjoint z-dense 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 non-minimal 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` non-minimal idempotents, in
the sense that there are 21 non-minimal idempotents q E Z* satisfying q < p, which are maximal subject to this condition. This will allow us to construct w1-sequences of idempotents in Z*, with the property that each idempotent in the sequence corresponding
to a non-limit 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 non-minimal
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.
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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 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.
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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' = f-1 [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' = f-1 [S]
f-1[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.
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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. -
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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 Rudin-Keisler Order
Recall that if S is a discrete space, then the Rudin-Keisler order 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 En-1 SO En __ rX < 23k+i+l + 23k+i-1 < 23k+i+2 and, since En __ xt E t=I Xt < 1
23k+i-1
1
1
Bi, Ei_1 Xt
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 non-empty 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'. N defined by f (F) = E4EF 2t-1 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 2r-1 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 2t-1. 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) : tp-1 [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 = to-1 [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 P-point 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 P-point, 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 P-points 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 P-points 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 P-points in N*.
Exercise 12.3.1. As in Theorem 12.31, let U be a union ultrafilter and let p = {{ErcF 2t-1 : 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) : rp-1 [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') and-notice 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 b-1
< 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
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, Bk-I E r, Ck-1 E q, Yk-1 E Bk-1, Zk-1 E Ck-1 C {x E N : -x + Ak_1 E r}, bk_i, and ak_i such that for each n, Xk-1,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°O--m) E p, Ak 1* t + q for any
t E Z with Itl < k, and Ak C FS((xk-1,n)n°_bk-1+1) n FS((x1,n)n_-ak_1+2).
Note that Ak c Ak_i.
NowAkEp=r+q. Let Bk=(-Zk-1+Ak_I)n{X EN: -x+Ak E q}nBk-1. Then Bk E r. Pick yk E Bk with yk > Zk-I. Let
Ck = Ck-1 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 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! + Et-s+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) 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) + 1-1, 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.
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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.
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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, x-1A 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 =s-r- 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.
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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
IPIp-p 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,...,r-1},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') 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 : Ax-1 E P) E q, where Ax-I = {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 : X-1Mk 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 $S. For each k E N, let Rk = { yk } U Syk U Yk S U Syk S = {X E S : yk occurs in x}.
Then Rk is an ideal of S so by (the left-right switch of) Corollary 4.18, Rk is an ideal of
($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 Mix-i\ Ui=1 R, E p so pick z E Mix-1 \ Ur-i R1 and pick j E N such that y1 occurs in z. Then j > k so zx 0 Mi, a contradiction.
For the left-right 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 only-if 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 At-i.
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 t-1A. Letm = max{1(t) : t E G}+1
and let F = (a'}. Pick X E S such that amx E U,EG t-tA 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 Aa-t 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 left-right 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)=p-s. Thus
poq =
fp(q-liEms)
= q-limpos SES = q-limp 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. El-Mabhouh 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 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,...,m-1},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 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)))j-i) 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 Hales-Jewett 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 2e-1 < n + 1 so 2e-1 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 -
2rk-P
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
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)t-1(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 fin-1 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 f--X C x-1 B j.
(d) Let (S, ) be a semigroup and let A C S. Then T is an FP-tree in A if and only if T is a tree in A and for all f E T, 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 FP-tree is a tree of finite products. It is this notion which provides the most fundamental combinatorial characterization of the notion of "central". A *-tree arises more directly from the proof outlined above. Lemma 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 FP-tree 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 x-IA*, 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 n-1) 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 FP-tree 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 x-I 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 FP-tree 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 FP-tree. Then given f E T and X E B f, we claim that
Bf-C 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 x-1 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 x-1 Bfk = x-1 Cn. We close this section by pointing out a Ramsey-theoretic 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 FP-tree 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(.)Y-I t-1 (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 Hales-Jewett 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 Hales-Jewett 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 non-negative 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 non-zero. (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.
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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 - ZJ-l,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 Zj-l,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 Zj-l,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 Zj-1,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 Zj-l,t
EnEF dxl,n + EnEF EtEH, Ej=2 ai,jzj-l,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 r-colored 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 non-zero 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 non-zero. 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 r-colored, 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 C-0
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 = C-1. 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
(ki-t 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,,kl-t
= EI=s ci > kj-S
-
ikj-t lci,,Ikj-i
Ei=s+1
> kj-s - 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
Ujt-t
Ij.
We observe that B' satisfies the first entries condition, with the first non-zero 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 = C-z 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 = C-x'. (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 = C-B (and all entries of C+B and of C-B are non-negative) so C+x = CBE = C-BE = 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' = C-BE = C-x'. 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' to-mean 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}k-1 be a finite partition of N. Then {0-1 [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 (0-1[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 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
II-el - E2 Si
II 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,...,p-1}x{0,1,...,p-1}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 pY-2 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 Uj-2 Dj, we have y(p, yi) < y - 2; let Ai = yi/pY and note that 0 < Xi < 1/p. Now Ax' =y so
Ei=]xi-4=Y=Ei-]Yi'di =EiED,Yi'di+Ej2EiEDjYi-diThus 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.
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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.
k-1
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 py-1 + zi ' pY-2 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 v-tuples) 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 v-tuples 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 v-tuples of n-tuples, 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 n-dimensional 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 n-dimensional 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 (rpn-1[{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 At-1 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 (rp-1[{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; s-1 A so s-1 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 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, t-1 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
_ F-1 ), 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
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 E-j+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 a-1 B E p and pick (xt ).1 such that FS ((xt )m 1) C a-1 B. Then
FS((axt)m1) C B.
Toseethatp - gET,letBEp - q andletmEN.LetC={a EN: a-1BEq}. 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 a-1 B E q so pick b E naEE a-1 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 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 n-1(nA) = A. Show that in (N, +) one need not have n(n-1 A) = A. Show in fact that n(n-1 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 a-1(Ab-1) or (a-1A)b-1. In either case, y E a-1 Ab-1 if and only if ayb E A. a-1Ab-1
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 Ab-1 is an IP* set in (S, +). (c) If (S, +, ) is a weak ring and A is an IP* set in (S, +), then a Ab-1 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 left-right 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 a-1 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)i-1, (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{a-1A : a E AP((xt)r=i)}.
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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, a-t 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{a-lA : 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 a-lA 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 {a-lA : a E AP((x,)i_',)} n n {Ab'1 : b E AP((x,)'_',)} n n {a-tAb-1 : a, b E AP((x,)'=',)}. Then replace the definition of Bn+1 with
Bn+1 = FS((Ys)tmk) n n {a-1A : a E AP((x,)i_',)} n n
{Ab-1
: b E AP((x,)'=j)}
n n {a- Ab-t : a, b E AP((x,)t-i)} 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((22t-t)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, j-1-/; (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 g-j-/«(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
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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 y-nonhomogenous 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, -).
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Recall the notion of FP-tree introduced in Definition 14.23. We call the corresponding additive notion an FS-tree. Since any member of a combinatorially rich ultrafilter is additively central, it must contain by Theorem 14.25 an FS-tree 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 FS-tree T in C such that for each f E T, d (B f) > 0. (f) There is an FP-tree 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 FS-tree 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.
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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.
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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 one-to-one 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(2e-3 + u
v + (u + v)2e-2 +
uv2e-1)
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 one-to-one 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.
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Lemma 17.13. Let (xn)r°i° be a one-to-one 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 one-to-one 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 one-to-one 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 one-to-one 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(xn)+I (1 - 2-k) z and x,. > 2a(xr)+1(1
- 2-k).
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 - 2-k)2 and xr < 2a(x,)+1(1 - 2-k)2. We have a(xr + Xn) - C(Xr + xn) = a(xr) - C(Xr) = k. Since XrXn < 2a(xrxn)+l(1 - 2-k) = 2a(xrxn)+1 - 2a(x,xn)+I-k 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(1-2-k)2 andxr > 2a(xr)+1(1-2-k)2. Now
a(xr+Xn) =a(xr)SOXr+xn > 2a(xr+xn)+l(1-2-k)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 - 2-k) = 2a(xrxn)+1 - 2a(xrxn)+1-k so C(XrXn)
1. Let k = min Ft and let j = max Fe-1. 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 + (- Ee-i 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 one-to-one sequences (x1)°O 1 and (yt)°° 1 with SPm((x1)°O1) C A andSP1((Y1)O0,) C A.
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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 one-to-one 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 one-to-one 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')
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352
that for each t > m, xt > ma(xt)(1 +m-d(xl )-1). Now 1 +m-d(xl>-1)e+1 > m and hence f, (1
1m+t X > m t=m
+m-d(xl)-1
> 1 so for some
^'e a(x,)+l
t
Pick the first k such that
Since Ek-I
r1k-1
t=m Xt < m
t_, a(x,)+1 and xk < ma(xk)+l
we have that r1kt=m x
t
mEk
a(xt), we have that for each s E {m, m + 1, ..., k}, a (x1 + X2 + + xm _ l + xs) = a (xs) and, since +X2+...+Xm-1
X1
a (xs) is even. Also a (X l +X2 +
+Xs E SPm((Xt)°=1) C A4,
+ Xm -1 + 1-1 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 = +Xm-1 +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 + .. + xm-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 one-to-one 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 - m-d(xl)-I). Then for some 1,
(1 -
m-d(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 1hk-1
t=m xt
k
Xt
t=m
nt Et=m
t
a(xt)+k-m
k-1
> mE=ma(x,)+k-I-m andxk >
so
m xt
>m
V=m
Ma(sk),
a(xt)+k-1-m
and hence
a(xl +x2 + ... +xm_1 + fk=m xt) = a(nt=m xt) = Ee=m a(xt) +k - 1 - in. + xm-1 + xt) = a(xt) so each Also for each t E {m, m + 1, ..., k}, a(xl + x2 + a(xt) as well as Ek-m 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 = +xm-1 +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 one-to-one sequence (x1) J ° 1 in N with SPm ((xt) °O 1) C B, and (4) there is no one-to-one 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 one-to-one 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 one-to-one 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 "Milliken-Taylor" systems because of their relation to the Milliken-Taylor 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 non-zero, (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,...,m-1),Fl,F2,...,FeE,P1((1,2,...,r)), and F1 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, 1/2n-1) 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 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
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 xm-1Am 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 one-to-one 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 Milliken-Taylor Theorem. The main feature of the adaptation is that we utilize ultrafilters with special properties. (For example, to obtain the Milliken-Taylor Theorem, we use an idempotent in place of the 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]t-1, 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,...,k-l}andEE[S]t-1,
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]t-1. 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,...,k-1}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 N and assume that there is some C E p on which rp is finite-to-one. 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,...,k-1)andeachE E [(DjnH,n_,)
1',rp]t 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] 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
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 (Xe-1 FP((xi,n)n 1)) x FP((xe,n)nnOO--1) 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, ... , xe-1, j) = {Y E Se : (XI, x2, ... , xe-1, 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 t-I Si and each j E (1, 2, ... , r}. Given r-1
(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, ... , xt-1, j) = {y E St : Bt+1(x1, x2, ..., xt-1, 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,...,xt-1) E X;=,' Si,S, =Uj=1 Bt(xl,x2.....xt-1,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, ... , xt-1, 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,...,xt-l,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.
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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 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 (*).
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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
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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 Milliken-Taylor 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 (Milliken-Taylor 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 Ui-1 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+j-k. 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 Milliken-Taylor 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 d-dimensional 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 d-dimensional subspaces of W(kP) are partitioned into finitely
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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 Stone-tech 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 (p-limT.(.f))(s) =I UES q p-lim(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)=P-simsq=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
ifj-10AnB.
0
Let
C={9L+t:
2 -a _ 0. Then for each k E N there exists H E R f(N) with min H > k such that A fl TH-t [A]) > 0.
Proof. Pick M E N such that g(A) > g(X)/m. For n E {1, 2, ... , m}, let Fn = g(TF,-1 [A]) = g(A) > g(X)/m so {k + n, k + n + 1, ... , k + m}. Then each [A] fl TFe-1 [A]) > 0. Let H = pick n, f E {1, 2, ..., m} such that n < f and [TH-1 [A]] {k+n, k+n+ 1, ... , k+2-1}. Then TH oTFe = TF, so TF -1 [A] = TF,-1 so
g(TH-1[A] fl A) =
1-t(TF,-1[TH-1[AJ
fl A])
fl TF,-'[Al) = g(T F-1[A] fl TFe-'[A]) > 0.
=
lt(TFF-1 [TH-1[Al]
0
Lemma 19.32. Let (X, B, A, (Ts),,Es) be a measure preserving system and let A E B satisfy g (A) > 0. For every sequence (sn )n° 1 and everyk E N, there exists F E J" f (N)
such that k < min F and g(A fl T,-'[Al) > 0, where s = nnEF Sn Proof. For F E . f(N), let RF = TT.EFsn. Then (RF)FE.mf(N) is a monotone action of J" f(N) on X. So we can choose a set F with the required properties by Lemma 19.31. 13
Theorem 19.33. Let (X, B, A, (Ts)sEs) be a measurepreserving system and let C C S. If there is some A E B with g(A) > 0 such that IS E S : g(A fl TS 1 [A]) > 0} C C, then C is an IP* set. Proof. This follows immediately from Lemma 19.32.
Definition 19.34. Let S be a semigroup. A subset C of S is a dynamical IP* set if and only if there exist a measure preserving system (X, B, A, (Ts)sEs) and an A E 2 with g(A) > 0 such that IS E S : g(A fl T, 1 [A]) > 0) C C.
Recall that by Theorem 16.32, there is an IP* set B in (N, +) such that for each n E N, neither n + B nor -n + B is an EP* set. Consequently, the following simple result shows that not every 1P* set is a dynamical IP* set.
Theorem 19.35. Let B be a dynamical IP* set in (N, +). There is a dynamical IP* set
C C B such that for each n E C, -n + C is a dynamical JP* set (and hence -n + B is a dynamical IP* set).
19.4 Dynamically Generated IP* Sets
409
Proof. Pick a measure space (X, 2, µ), a measure preserving action (Tn)fE11 of N on X, and a set A E 2 such that µ(A) > 0 and {n E N : p (A n Tn-1 [A]) > 0) C B. Let
C = {n E N : µ(A n Tn-1[A]) > 0}. To see that C is as required, let n E C and let D = A n Tn-1 [A]. We claim that (m EN: µ(D n T,n-1 [D]) > 01 C -n + C. To this end, let m E N such that µ(D n Tn,-1 [D]) > 0. Then D n T , n 1 [D] = A n T,- 1 [A] n T 1 [A n T,,-'[A]l
C AnTm1[T,i 1[A]]
=
An(TnoTm)-1[A]
= A n T,,+m [A]
p
so n+In E C.
Recall from Theorem 18.15 that an IP* set in N x N need not contain {w} x FS((xn),°,°_l) for any sequence (xn)n° 1. In the final result of this section we show that a dynamical IP* set in the product of two senigroups with identities must contain sets of the form FP((xn)n° ,) x FP((yn )n , ). Indeed, it has a much stronger property: given any sequences (w,) , and (Zn)' , one can choose infinite product subsystems (xn)' of (wn)' 1 and (yn)n_1 of (Zn)n° with FP((xn)n_1) x FP((yn),,° ,) contained in the given dynamical IP* set. More than this, they can be chosen in a parallel fashion. 1
That is if xn = rl,EH wt, then yn = fltEH Z1. We only discuss the product of two semigroups for simplicity and because the generalization to arbitrary finite products is straightforward. In the proof of Theorem 19.36 we utilize product measure spaces. In doing so we
shall use the customary notation. If (X, 2, ji) is a measure space and m E N, then (X', 2m, /1') denotes the measure space defined as follows: X' is the usual Cartesian product, 2m denotes the a-algebra generated by the sets X m, A; where A; E `£ for each i E { 1 , 2, ... , m}, and µ'n denotes the countably additive measure on 21 such that µn' (X m 1 A;) = fl"', A (A;) for every sequence (Ai )"' , in 2. Any transformation T : X -+ X defines a transformation flm T : Xm Xm for which flm T (x1, x2, ... , Xm) = (T(xI ), T (x2), ... , T (x n)). If T is measure preserving, so is flm T, because (flm T)-1[Xm,A,] = Xm1(T-1[A,]) for every
A1,A2,...,AmCX. Theorem 19.36. Let S1 and S2 be semigroups with identities and let C be a dynamical IP* set in S1 x S2. Let (wn)n°, be a sequence in S1 and let (Zn)n_1 be a sequence in S2. There exists a sequence (Hn)R° in J" f(N) such that
(a) for each n, max Hn < min H,,+ I and (b) if for each n, xn = fItEH wt and yn = n1EH Zt, then FP((xn),°,°_,) x FP((yn)n° 1) C C.
Proof. Let S = S1 x S2. Pick a measure space (X, 2, µ), a measure preserving action (T(s,t))(s,t)Esj x52 of S on X, and B E £ with µ(B) > 0 such that,
D = {(s, t) E S : ji(B n
T(s,t)-1 [B])
> 0) 9 C.
410
19 Relations With Topological Dynamics
Let e and f denote the identities of S1 and S2 respectively. For each a = (s, t) E S, we define UQ : X3 -a X3 by putting Ua (xi, x2, x3) = (T(s, f) (xi ), T(e,t) (X2), T(s,t) (x3)).
Since UUl[A1xA2xA3]=T(s f)[AI]xT(af)[A2]xT(sf)[A3]forevery A1,A2,A3 E 2, it follows that (X3, 23 µ3 (UQ)oes) is a measure preserving system. For each n E N, we define an E S by an = (Win, Z.O. We shall inductively choose a sequence (Hn)R° 1 satisfying (a) and the following condition, which is clearly stronger than (b):
(c) if, = nkEH, wkandyi = nkEH; zk,then{(x, f), (e, y), (x, y)} a Dwhenever (x, y) E FP((xi)n 1) x FP((Yi)n 1). We first apply Lemma 19.32 to the measure preserving system (X3, 23, µ3, (U, ),,S) and the set A = B3. This allows us to choose Hl E P f (1Y) such that I.c3 (An Ua '[A]) >
0, where a = niEH1 ai = (xi, Yt) This means that n (T(x,1,f)[B] x 7 '),1)[B] x T(x,,y1)[B])) > 0. So Hl satisfies (c). We now suppose that H1, H2,..., Hn have been chosen. Let m = J FP((x; )" 1) x FP((yi );`_ 1) j. We apply Lemma 19.32 again, this time to the measure preserving system ((X3)3m, (23)3m (.3)3m, M3. U-11),IES), and the set A= X { (B n T(x f)[B])3 x (B n T(e Y) [B])3 lx (B n TAX y)[B])3) (x, Y) E FP((x1)j 1) X FP((y,)l 1)J
This allows us to choose Hn+l E ?f(N) such that max(Hn) < min(HH+i) and (A3)3m(A n (n3m UQ)-I [A]) > 0, where a= niEHH+, ai = (xn+i, Yn+i) This implies that, for every (x, y) E FP((xi)n_I) x FP((y; )" 1), we have
,u (BnT,, I[B]nTi-1[BnT. 1[B]])=A(BnT,, 1[B]nTj1[B]nT,,-,1[B])>0 whenever a E {(x, f), (e, y), (x, y)} and fl E {(xn+l , f), (e, Yn+i), (xn+l, Yn+i)). We can therefore deduce each of the following statements:
(1) (xn+i, f) E D and (xxn+I, f) E D (putting a = (x, f) and fl = (xn+i, f)). (1') (e, yn+i) E D and (e, YYn+1)) E D (putting a = (e, y) and ,8 = (e, Yn+I) ) (2) (xn+I, Y) E D (putting a = (e, y) and P = (xn+1, f))(2) (x, Yn+i) E D (putting a = (x, f) and ,8 = (e, yn+1) ) (3) (xxn+i , y) E D (putting a = (x, y) and _ (xn+i , f) ). (3') (x, YYn+i) E D (putting a = (x, y) and fl = (e, y,,+1)). (4) (xn+l , YYn+I) E D (putting a = (e, y) and = (xn+l , Yn+i) ) (4') (xxn+I, Y) E D (putting a = (x, f) and fi = (xn+1, yn+1) ) (5) (Xn+I, Yn+I) E D and (xxn+l , YYn+I) E D (putting a = (x, y) and ,
(xn+1, Yn+l) ) It is now easy to check that our inductive assumption extends to H1, H2, ..., H"+ I. So the sequence (Hn)n_1 can be chosen inductively.
Notes
411
Notes The notion of "dynamical system" is often defined only for compact metric spaces. The greater generality that we have chosen (which is essential if one is going to take fiS as the phase space of a dynamical system) is also common in the literature of dynamical
systems. General references for topological dynamics include the books by R. Ellis [86] and J. Auslander [5]. Theorem 19.8 and Lemma 19.9 are due to B. Balcar and F. Franek in [12] as is the proof that we give of Theorem 19.10. Theorem 19.10 is proved by R. Ellis [86] in the case that S is a group.
Theorem 19.15 is due to S. Glasner in [105], where it is stated in the case that S is a countable abelian group. Theorems 19.17 and 19.18 are from [141], a result of collaboration with J. Lawson and A. Lisan. Theorem 19.19 is due to W. Ruppert in [219]. S. Glasner has recently published [106] a proof that the enveloping semigroup of a minimal left ideal in fiZ is not topologically and algebraically isomorphic to $Z via a map taking n to A. As a consequence, the condition of Theorem 19.18 does not hold in flZ. The equivalence of the notions of "central" and "dynamically central" was established in the case in which S is countable and the phase space X is metric in [27], a result of collaboration with V. Bergelson with the assistance of B. Weiss. The equivalence of these notions in the general case is a result of S. Hong-ting and Y. Hong-wei in [164]. Theorem 19.24 is due to J. Auslander [4] and R. Ellis [85]. Most of the results of Section 19.4 are from an early draft of [32], results obtained in collaboration with V. Bergelson. Lemma 19.31 is a modification of standard results about Poincard recurrence. Theorem 19.33 for the case S = Z is due to H. Furstenberg [98].
Chapter 20
Density - Connections with Ergodic Theory
As we have seen, many results in Ramsey Theory assert that, given a finite partition of some set, one cell of the partition must contain a specified kind of structure. One may ask instead that such a structure be found in any "large" set. For example, van der Waerden's Theorem (Corollary 14.3) says that whenever N is divided into finitely many classes, one of these contains arbitrarily long arithmetic progressions. Szemerddi's Theorem says that whenever A is a subset of N with positive upper density, A contains arbitrarily long progressions.
Szemerddi's original proof of this theorem [232] was elementary, but very long and complicated. Subsequently, using ergodic theory, H. Furstenberg [97] provided a shorter proof of this result. This proof used his "correspondence principle" which can be viewed as a device for translating some problems involving sets of positive density in N into problems involving measure preserving systems, the primary object of study in ergodic theory. We present in Section 20.2 a proof of Furstenberg's Correspondence Principle using the notion of p-limit and in Section 20.3 a strong density version of the Finite Sums Theorem obtained using the algebraic structure of (i3N, +).
20.1 Upper Density and Banach Density We have already dealt with the notion of ordinary upper density of a subset A of N, which was defined by
d(A)=lim sup
IA fl {1, 2,...,n}l
n-oo
n
Another notion of density that is more useful in the context of ergodic theory is that of Banach density which we define here in a quite general context.
Definition 20.1. Let (S, -) be a countable semigroup which has been enumerated as (sn)n°l and let A C S. The right Banach density of A is
dr (A) = sup{a E R: for all n E N there exist m> n and x E S such that
IA fl {st, s2, ... , sm } - XI m
> a}
20.1 Upper Density and Banach Density
413
and the left Banach density of A is
dj (A) = sup{a E R: for all n E N there exist m> n and x E S such that
>a}.
m
If S is commutative, one has d, = dl and we simply write d*(A) for the Banach density of A. Notice that the Banach density of a subset A of S depends on the fixed enumeration of S. For example, in the semigroup (N, +), if N is enumerated as 1, 2, 3, 4, 6, 5, 8, 10, 12, 14, 7, 16, 18, 20, 22, 24, 26, 28, 30, 9, 32, .. .
then d* (2N) = 1, while with respect to the usual enumeration of N, d* (2N) = 12' When working with the semigroup (N, +) we shall always assume that it has its usual enumeration so that
d*(A) = sup{a E R: for all n E N there exist m> n and x E N such that
IAl{x+1,x+2,...,x+m)I m
> a}.
Lemma 20.2. Let (S, ) be a countable semigroup which has been enumerated as (sn)' t and let A, B c S. Then d, (A U B) < d7 (A) + d, (B) and d, (A U B) di (A) + dl (B). If S is right cancellative, then d; (S) = 1 and if S is left cancellative, then dl (S) = 1 Proof. This is Exercise 20.1.1.
Recall that we have defined A = {p E fiN : for all A E p, d(A) > 0}. We introduce in a more general context the similar sets defined in terms of d; and d,*.
Definition 20.3. Let (S, ) be a countable semigroup which has been enumerated as
(sn)°.Then
for all Ac p,
0S: for all A Ep, d;(A)>0}. Again, if S is commutative we write A*(S, ) for A* (S, ) = A (S, ).
Lemma 20.4. Let (S, ) be a countable semigroup which has been enumerated as (sn)n° , and let A C S. (a) If dr (A) > 0, then A n
* (S, )
(b) If dl (A) > 0, then A fl A (S, )
0. 0.
Proof. We establish (a) only. By Theorem 3.11 it suffices to show that if.Y is a finite nonempty subset of .P (S) and d; (U .T') > 0, then for some B E .T", d, (B) > 0. This follows from Lemma 20.2.
414
20 Density - Connections with Ergodic Theory
Recall from Theorems 6.79 and 6.80 that both A and N* \ A are left ideals of (flN, +)
and of (fN, ). By way of contrast, we see in the following theorems that S*\A; is far from being a left ideal of PS and S*\A* is far from being a right ideal of fiS. Theorem 20.5. Let (S, ) be a countable right cancellative semigroup which has been enumerated as (sn)n° I. Then A* (S, ) is a right ideal of CBS.
Proof One has that A* (S, ) # 0 by Lemma 20.4 and the fact from Lemma 20.2 that
d; (S) > 0. Let P E A*(S, ), let q E j 6S, and let A E p q. We need to show that
d,(A)>0.Let B=(yES:y-1AEq}.ThenBEpsod,(B)>0.Picka>Osuch that f o r all n E N t h e r e exist m > n and x E S such that
J B n (sl , s2, ... , sm ) x i >
of.
m
To see that d; (A) > a, let n r= N and pick m > n and x E S such that JBn(sl,s2,...,sm) x1 > a. m Let C = B n (s1,s2 ...,sm ) x andpickz E nyEC y-1 A. Then, since S is right cancellative, pzlc is a one-to-one function from C to A n (s1, S2, ... , sm} xz so a. M
Theorem 20.6. Let (S, ) be a countable left cancellative semigroup which has been enumerated as (sn
I. Then A* (S, ) is a left ideal of 14S.
Let P E A7 (S, ), let q E fS, and let A E q p. We need to show that d* (A) > 0. Now (yES : y-1 A E P) E q so picky E S such that y-IA E p. Then Proof. d1
(y-1 A)
> 0 so pick a > 0 such that for all n E N there exist m > n and x E S such
Iy-'Anx {sl,s2,...,sm}I that m
> a. Now given m and x, since S is left cancellative,
Ay is a one-to-one function from y-1 A n x (sl, s2, ..., sm ) to A n yx (sl, s2, ... , sm }, and consequently, d! (A) > a. As a consequence of Theorems 20.5 and 20.6, if S is commutative and cancellative, then A*(S) is an ideal of fiS.
Theorem 20.7. Let (S, ) be a countable right cancellative semigroup which has been enumerated as (sn)n° 1. For any A C S, if A is piecewise syndetic, then there is some
G E Pf(S) such that d; (U(EG t-1 A) = 1. If (S, ) _ (N, +), then the converse holds. Consequently,
ci K (f N, +) = (p E f N :for all A E p there exists G E P f (N)
such that d*(UIEG _t +A)= U. Proof Let A C S, assume that A is piecewise syndetic, and pick G E P f (S) such
that for each F E .I f(S) there exists x E S with F x C UtEG t- IA. To see that dr ( U I E G t- A) = 1, let n E N be given and pick x E S such that (sl , s2, ... , sn } x C UIEG t-1 A.
20.1 Upper Density and Banach Density
415
Then
I UfEG t-'A fl {st,s2,...,sn} x] =n.
Now assume that (S, ) _ (N, +) and we have G E .P1(N) such that d*( UtEG -t + A) = 1. Let F E ,Pf(N) be given and pick k E N such that F C (1,2,...,k}. We claim that there issomex E Nsuchthat{x+ 1,x+2,...,x+k} C (UtEG -t + A). To see this, pick some m > 2k2 and some y E N such that
{y+1,y+2,...,y+m}fl(UfEG -t+A)I> (I --L)m and pick v E N such that vk < m < (v + 1)k, noting that v > 2k. If for some i c- {0, 1, ... , v-1} onehas {y+ik+1, y+ik+2,... , y+(i + l)k} e UtEG -t+A, then we a r e done, so suppose instead that f o r each i E (0, 1, ... , v - 1),
I{y+ik+1,y+ik+2,...,y+(i+1)k}fl (UtEG -t+A)I < k-1. Then
(1-yf)m
vk, one has that (1- ) vk < v (k -1) +k so that 2k > v, a contradiction. The final conclusion now follows from Corollary 4.41. The following result reflects the interaction of addition and multiplication in PN which was treated in Chapter 13.
Theorem 20.8. The set A*(N, +) is a left ideal of (fN, ).
Proof. Let P E A*(N, +) and let q E #N. To see that q- p E A*(N, +), let A E q
p. Then {x E N : x-1A E p} E q so pick x such that x-'A E p. Then
d* (x-' A) > 0 (where here, of course, we are referring to the additive version of d*) so pick a > 0 such that for all n E N there exist m > n and y E N such that IX-1 A fl {y + 1, y + 2, .... y + m) l m
> a. Then given such n, m, and y, one has that
IAfl(xy+1,xy+2,...,xy+xm)I ?a m so that d*(A) >
«. X
0
Recall that, given a subset A of a semigroup (S, ) that we have defined a tree T in A whose nodes are functions. Recall further that given a node f of T, the set of successors to f is denoted by B f and that T is an FP-tree provided that for each node
416
20 Density - Connections with Ergodic Theory
f of T, Bf consists of all finite products of entries on paths extending f which occur after f . (By a path in T we mean a function g : co -a A such that for each n E w, the restriction of g to (0, 1, ... , n - 1) is in T.) Theorem 20.9. Let (S, ) be a countable cancellative semigroup which has been enumerated as (sn)n° 1. Let r E N and let S = U;=1 A;. There exist i E {1, 2, ..., r} and a tree T in A, such that for each path g in T, FP((g(n))n° O) C A1, and for each node
f of T,d;(Bf)>0and d1(Bf)>0. Proof. By Theorems 20.5 and 20.6, A* (S, ) is a right ideal of ,S and A* (S, ) is a left ideal of PS. By Corollary 2.6 and Theorem 2.7 pick an idempotent p E A* (S, .) n A* (S, .) and pick i E { 1 , 2, ... , r} such that A, E p. Then by Lemma 14.24 there is an FP-tree T in A, such that for each f E T, Bf E p. Thus, in particular, for each f E T,
d,(Bf)>0and d7(Bf)>0. The following theorem is not a corollary to Theorem 20.9 because d*(A) > 0 does not imply that d(A) > 0. The proof is essentially the same, however, so we leave that proof as an exercise. Theorem 20.10 is an interesting example of the strength of combinatorial results obtainable using idempotents in ,8N. Although many of the Ramsey-theoretic results which we have presented were first obtained by elementary methods or have subsequently been given elementary proofs, we doubt that an elementary proof of Theorem 20.10 will be found in the near future.
Theorem 20.10. Let r E N and let N = U;=1 A1. There exist i E 11, 2, ..., r} and a tree T in A, such that for each path g in T, FS((g(n))n° o) e A1, and for each node f of T, d(Bf) > 0. Proof. This is Exercise 20.1.2. The following lemma will be needed in the next section.
Lemma 20.11. Let A C_ N such that d* (A) = a > 0. Then there exists some sequence (Ik)k° 1 of intervals in N such that lim IikI = oo and
k-oo
lim
k-oo
IAnikI
= a.
Ilkl
Proof. This is Exercise 20.1.3.
Exercise 20.1.1. Prove Lemma 20.2. Exercise 20.1.2. Using the fact that A is a left ideal of (flN, +) (Theorem 6.79), prove Theorem 20.10.
Exercise 20.1.3. Prove Lemma 20.11.
417
20.2 The Correspondence Principle
20.2 The Correspondence Principle Recall that a measure preserving system is a quadruple (X, 2, {.s, (Ts)SES) where (X, 2, µ) is a measure space and (Ts)sEs is a measure preserving action of a semigroup S on X. If the semigroup is (N, +), the action (Tn)fEN is generated by a single function T1 and we put T = Tl and refer to the measure preserving system (X, 2, µ, T).
Theorem 20.12 (Furstenberg's Correspondence Principle). Let A C N with d * (A) > 0. There exist a measure preserving system (X, 2, N T) (in which X is a compact metric space and T is a homeomorphism from X onto X) and a set A' E B such that
(1) µ(A') = d*(A) and (2) for every F E J"f(N), d*(A fl nfEF (-n + A)) > /2(A' n nnEF T -n[A'])
Proof. Let 12 = Z{0, 1) with the product topology and let T : 0 0 be the shift defined by T(x)(n) = x(n + 1). (By Exercise 20.2.1 one has in fact that SZ is a metric space.) Let 1; be the characteristic function of A (viewed as a subset of Z). Let n E Z), the orbit closure of . Then X is a compact metric space and X= (the restriction of) T is a homeomorphism from X onto X.
For each n E 7G, let Dn = X fl irn-'[{l}] = {S E X : S(n) = 1} and notice that Dn is clopen in X. Let A be the Boolean algebra of sets generated by {Dn : n E Z} and let 2 be the a-algebra generated by {Dn : n E 7G}. Notice that for each n E Z,
T[Dn] = Dn_I and T-1 [D,,] = Dn+I. Consequently, if B E 2, then T-1 [B] E B (and T[B] E 2). Define cp : R(X) - Y(N) by W(B) = {n E N : Tn(l) E B}. Let a = d*(A) and pick by Lemma 20.11 a sequence (Ik)k I of intervals in N such that lim IIkI = no and k-aoo
lim
I A fl IkI
k-*oo
= a.
IIkI
Pick any p E N* and define v : J" (X) --> [0, 1) by
v(B) = p-lim
Icp(B) fl IkI IlkI
Notice that v(X) = 1. Further, given any B, C C X, one has cp(B fl c) = cp(B) fl cp(C) (so that, in particular, if Bf1C = 0, then W(B)f)cp(C) = 0) and cp(BUC) = rp(B)Uq (C). Thus, if B and C are disjoint subsets of X, then
v(B U C) = P-lim I
(_(B) U y(C)) n IkI
kEN
IIkI
Icp(B)nIkI+Ico(C)flIkI
-pkEN = v(B) + v(C).
IIkI
418
20 Density - Connections with Ergodic Theory
Next we claim that for any B c X, v(T-'[BI) = v(B). Indeed, cp(T-'[B]) _ (-1 + w(B)) n N so
v(T-'[BI) =P- k
I(-1 +(p(B)) n lkl =
_
iw(B) n lkI
P_ khm
Ilkl
_ -"(B)
IIkI
because for any k E N, I (-1 + rp(B)) n Ik I and IV(B) n Ik I differ by at most 1. Thus v is finitely additive and T-invariant on ?(X) and hence in particular on A. Further all members of A are clopen in X, so if one has a sequence (Bn)n° I in A such that each Bn+ t c Bn and nn° t Bn = 0, then since X is compact there is some k such
that for all n > k, Bn = 0 and hence lim v(B,,) = 0. n-a o0 Therefore, we are in a position to apply Hopf's Extension Theorem. (This is a standard result in first year analysis courses. See for example [115, Exercise 10.371 or [14, Satz 3.2].) The finitely additive measure v on A can be extended to a countably additive measure Ec on 2 where for each B E 2,
µ(B) = inf{EcEg v(C) : a c A, Igi < w, and B c U g}. Using this description of A (B) and the fact that for C E A,v(T-'[C]) = v(C),onesees immediately that for all B E B, µ(T -' [Bj) =µ(B). Thus we have that (X, B, z, T) is a measure preserving system.
Let A' = Do = {S E X : 8(0) = 1} and observe that (p (A') = A and for each n E N, T'n[A'] = Dn and cp(Dn) = (-n + A) n N. Thus A (A') = v(A') = P-lim
n ikI Ilki
_
P
IA n Ikl
_
IIkI
_- «
and given any F E Rf(N),
µ(A' n nnEF T-n[A']) = l2(Do n n
= =
Dn)
n nnEF co(Dn) n Ikl kEN
P_ him
IIkI
IAnnnEF (-n+A)nikI Ilki d*(nnEFAn(-n+A)).
0
When proving Szemeredi's Theorem, one wants to show that, given a set A with d*(A) > 0 and f E N, there is some d E N such that {a E N :{a,a + d,a + 2d,...,a + ed} c A} # 0.
As is typical of ergodic theoretic proofs of combinatorial facts, one establishes this fact by showing that the set is in fact large. The proof of the following theorem requires extensive background development, so we do not present it here.
20.3 A Density Version of the Finite Sums Theorem
419
Theorem 20.13. Let (X, 2, µ) be a measure space, let T1, T2,..., Tt be commuting measure preserving transformations of (X, 2, N,), and let BE B with u(B) > 0. Then there exists d E N such'that µ(B n n,=I T;-d[B]) > 0. Proof. [98, Theorem 7.15].
Corollary 20.14 (Szemeredi's Theorem). Let A C N with d*(A) > 0. Then A contains arbitrarily long arithmetic progressions.
Proof. Pick by Theorem 20.12 a measure preserving system (X, 2, Fs, T) and a set A' E `£ such that
(1) µ(A) = d* (A) and (2) for every F E Rf (N),
d*(A n nnEF (-n + A)) > µ(A' n nnEF T-"[A']) Let f E N be given and for each i E [1, 2, ... , 8}, let T; = T'. Notice that T1, T2.... , Tt are measure preserving transformations of (X, 2, A) that commute with each other. Pick by Theorem 20.13 some d E N such that A(A' n Hl T -d[A']) > 0 and notice that for each i, T,-d = T-'d. Let F = {d, 2d, 3d, ... , 8d}. Then
d*(A n nREF (-n + A)) > js(A' n nnEF T-" [A']) > 0
so AnnnEF (-n +A) 2d,...,a + id} C A.
0sopicka E AnnnEF (-n + A). Then {a, a + d, a +
r in n, define p (l; , rl) = k+ where k = min{ItI : t E Z and l;(t) 0 q(t)) (and of course p Exercise 20.2.1. Let Q = Z [O, 1) and for
0
metric
on f2 agree.
20.3 A Density Version of the Finite Sums Theorem The straightforward density version of the Finite Sums Theorem (which would assert
that any set A C N such that d(A) > 0 - or perhaps d*(A) > 0 - would contain for some sequence (x,,)',) is obviously false. Consider A = 2N + 1. However, a consideration of the proof of Szemeredi's Theorem via ergodic theory reminds us that the proof was obtained by showing that a set which was only required to be nonempty was in fact large. Viewed in this way, the Finite Sums Theorem says that whenever r E N and N =
U r A;, there exist i E (1, 2, ..., r) and a sequence (x")R_, such that for each n, xn+l E A, n n{- EtEF x: + A; : 0 0 F C (1, 2, ... , n) }. In this section we show
that not only this set, but indeed many of its subsets, can be made to have positive upper density and specify how big these sets can be made to be. -
20 Density - Connections with Ergodic Theory
420
Lemma 20.15. Let A C N such that a (A) > 0 and let B be an infinite subset of N. For
every c > 0 there exist x < y in B such that d (A n (-(y - x) + A)) > d(A)2 - E. Proof. Let a = W (A) and pick a sequence (x,) ' 1 in N such that lim x, = co and n-> oo
lim
n-+oo
Notice that for every t E N, lira
IAn(1,2,...,Xn)1 =a. Xn
1(-t + A) n (1, 2, ... , xn } I = a
n-aoo
Xn
Enumerate B in increasing order as (yi )°O ,. Let E > 0 be given. If e > a2, the conclusion is trivial, so assume that E < a2 and let b = /a2 - E/2. Pick k E N such 4a that k > and pick f E N such that for all i E {1, 2, ..., k} and all n > f, b
vi,1 I(-yi+A)n(-yj+A)n{1,2,...,x,}I xn
321FI and
(2) for each f E T, d(B f) > 0.
Proof Given n E w, g : {0, 1, ... , n - 1 }
and F C {0, 1, ... , n - 1 },let
D(F, g) = A fl n{-y + A : y E FS((g(t))tEF)}, C(F, g) = {x E D(F, g) : -x + D(F, g) E p and d (D(F, g) fl (-x + D(F, g))) >
321Fi+i
),
and
B(g) _ (l{C(F, g) : F C (0, 1, ..., n - 1)}. Notice that, following our usual convention that n 0 is whatever semigroup we are concerned with at the time, one has D(0, g) = A. Let To = {0} and inductively for n E w, let
Tn+1 = {g U {(n, x)} : g E T,,, X E B(g), and if n > 0, x > g(n - 1)).
Let T = Uno T. Then trivially T is a tree in A. (Given X E B(g), X E C(0, g) c D(0, g) = A.) Notice that for n E N and g E Tn, Bg = {x E B(g) : x > g(n - 1)). We show first by induction on n that for all g E Tn and all F C (0, 1, ... , n - 1), C(F, g) E p and d (D(F, g)) > S21FI. Observe that
(*) ifk=maxFandhistherestrictionofgto{0, 1,...,k},thenD(F,g) = D(F,h) and C(F, g) = C(F, h). To ground the induction assume that n = 0. Then g = F = 0 and D(F, g) = A so that d (D(F, g)) > S = 820. Let e = j (A)2 - 32. Then
C(F, g) = {x E A : -x + A E pandd(Afl(-x+A))-> d(A)2-E}
20.3 A Density Version of the Finite Sums Theorem
423
so that by Lemma 20.16, C(F, g) E P.
Now let n E w and assume that f o r all g E T and all F c (0, 1, ... , n -
1),
C(F, g) E p and d(D(F, g)) > 821FI. Let g E T,,+1 and F c (0, 1, ... , n) be given. By the observation (*), we may assume that n E F. Let h be the restriction of g to (0, 1, ... , n - 1), let H = F\{n}, and let x = g(n). Then h E T and X E B(h). We now claim that
(**) D(F, g) = D(H, h) n (-x + D(H, h)).
To see that D(H, h) n (-x + D(H, h)) c D(F, g), let z E D(H, h) n (-x + D(H, h)). Then z E D(H, h) c A. Let y E FS((g(t))tEF) and pick G S; F such that Y = E:EG g(t). We show that z E -y + A. If n f G, then y E FS((h(t)),EH) and hence, since z E D(H, h), z E -y + A. We thus assume that n E G. If G = {n},
then z E -x + D(H, h) C -x + A = -y + A as required. We therefore assume that K = G\{n} 54 0. Then K C H. Let V = EtEH g(t) = E FS((h(t))ZEH) SOX +z E D(H,h) c -v +A and thus v +x +z = Y +Z E A as required.
To see that D(F, g) C D(H, h) n (-x + D(H, h)), let z E D(F, g). Since FS((h(t))tEH) S; FS((g(r))tEH), Z E D(H, h). Since x = g(n) E FS((g(t))tEF), x + z E A. To see that x + z 'E D(H, h), let y E FS((h(t)):EH) Then y + x E FS((g(t))tEF) so z E -(y + x) + A so x + z r= -y + A. Thus (**) is established. Since X E B(h) c C(H, h), we have -x + D(H, h) E p. Thus
D(H, h) n (-x + D(H, h)) E p.
That is D(F, g) E p. Now since x E C(H, h) and D(F, g) = D(H, h) n (-x + D(H, h)), d(D(F, g)) > 321H1+1 = S21FI. Let y = d(D(F, g)) - 321" and let e _ 2y321 F1 + y2. Let
E = {z E D(F, g) : -z + D(F, g) E p and d(D(F, g) n (-z + D(F, g))) > d(D(F, g))2 - E = S2iF1+1' we By Lemma 20.16, E E P. Since d(D(F, g))2 _'E = have that E = C(F, g) so that C(F, g) E p. The induction is complete. (32iFi + y)2
Now let f be any path of T. We show by induction on BFI, using essentially the first proof of Theorem 5.8, that if F E R f (w), n = min F, and h is the restriction of
f to {0, 1, ... , n - 1}, then ErEF f (t) E D({0, 1, ... , n - 1}, h). First assume that F = {n}, let g be the restriction off to (0, 1, ... , n}, and let h be the restriction of f
to (0, 1,...,n- 1). Theng =hU((n, f(n))) with f (n) E B(h) C C({0, I, ... , n - 1}, h) c D({0, 1, ... , n - 1}, h) as required. Now assume that I F1 > 1, let G = F\ {n }, and let m = min G. Let h, g, and k be the restrictions of f to {0, 1 , ... , n - 1 }, {0, 1, ... , n}, and (0, 1, ... , in - 1) respectively. Then
EtEG f (t) E D({0, 1, ... , m - 1), k) 9 D({0, 1, ... , n), g)
424
20 Density -Connections with Ergodic Theory
and D({0, 1..... n}, g) = D({0, 1,..., n-1}, h)fl(- f (n)+D({0, 1,...,n- I I, h)) by (**). Thus E,EG f (t) + f (n) E D({0, 1, ... , n - 1}, h) as required. In particular, conclusion (1)(a) holds.
To establish conclusion (1)(b), let F E 9'f (w), pick n > max F, and let h be the restriction off to 10, 1, ... , n -1}. The conclusion of (1)(b) is precisely the statement, 321Fi proved above, that d(D(F, h)) >
As we observed above, for n E N and g E T, Bg = {x E B(g) : x > g(n - 1)} and thus Bg E p so, since p E A, d(Bg) > 0.
Notes The notion which we have called "Banach density" should perhaps be called "Polya density" since it appears (with reference to N) explicitly in [196]. To avoid proliferation of terminology, we have gone along with Furstenberg [98] who says the notion is of the kind appearing in early works of Banach. The definition of Banach density in the generality that we use in Section 20.1 is based on the definition of "maximal density" by H. Umoh in [237].
It is easy to construct subsets A of N such that for all n E N, -t + A) _ d(A) > 0. On the other hand, it was shown in [126] that if d*(A) > 0, then for each e > 0 there is some n E N such that d* (U' -t + A) > 1 - e. Consequently, in view of Theorem 20.7, it would seem that A*(N, +) is not much larger than cf K(,N, +). On the other hand it was shown in [73], a result of collaboration with D. Davenport, that
cP K(8N, +) is the intersection of closed ideals of the form cf(N* + p) lying strictly between cE K(PN, +) and A*(N, +). Theorem 20.13, one of three nonelementary results results used in this book that we do not prove, is due to H. Furstenberg. Lemma 20.15 is due to V. Bergelson in [18]. Theorem 20.17 is from [24], a result of collaboration with V. Bergelson.
Chapter 21
Other Semigroup Compactifications
Throughout this book we have investigated the structure of OS for a discrete semigroup S. According to Theorem 4.8, 9S is a maximal right topological semigroup containing S within its topological center.
In this chapter we consider semigroup compactifications that are maximal right topological, semitopological, or topological semigroups, or topological groups, defined not only for discrete semigroups, but in fact for any semigroup which is also a topological space.
21.1 The J NCO, WAR, AR, and 8AR Compactifications Let S be a semigroup which is also a topological space. We shall describe a method of associating S with a compact right topological semigroup defined by a universal property. The names for the compactifications that we introduce in this section are taken from [39] and [40]. These names come from those of the complex valued functions on S that extend continuously to the specified compactification. The four classes of spaces in which we are interested are defined by the following statements.
Definition 21.1. (a) 'Y, (C) is the statement "C is a compact right topological semigroup". (b) tP2(C) is the statement "C is a compact semitopological semigroup". (c) %Y3 (C) is the statement "C is a compact topological semigroup". (d) q/4(C) is the statement "C is a compact topological group".
Lemma 21.2. Let S be a set with ISI = K > 22x w, let X be a compact space, and let f : S -' X. If f [S] is dense in X, then IX I < Proof. Give S the discrete topology. Then the continuous extension f off to PS takes 22k fS onto X so that IXI < I,SI. By Theorem 3.58, I14Si = In the following lemma we need to be concerned with the technicalities of set theory
more than is our custom. One would like to define F; = {(f, C) : %Pi (C), f is a
426
21 Other Semigroup Compactifications
continuous homomorphism from S to C, and f [S] c A(C)}. However, there is no such set. (Its existence would lead quickly to Russell's Paradox.) Notice that the requirements in (1) and (2) concerning the topological center are redundant if i gf 1.
Lemma 21.3. Let S be an infinite semigroup which is also a topological space and let i E {1, 2, 3, 41. There is a set F; of ordered pairs (f, C) such that
(1) if (f, C) E F then'Pi (C), f is a continuous homomorphism from S to C, and f [S] S A(C), and (2) given any D such that %Pi (D) and given any continuous homomorphism g : S ->
D with g[S] c A(D), there exist (f, C) E F, and a continuous one-to-one homomorphism rp : C -> D such that (p o f = g. Proof. Let K = ISI and fix a set X with I X I = 22". Let
9 _ {(f, (C, T, )) : C C X, T is a topology on C, is an associative binary operation on C, and f : S -> C). Note that if C C X, -T is a topology on C,
is a binary operation on C, and f : S -> C,
thenf CSxCCSXX,CcX,TS;R(C)C,I(X),and :CxC->Cso that C(CxC)xCC(XxX)xX.Thus A C P (S X X) X (J'(X) x .P(.P (X)) X ,P((X X X) X X)) so 9 is a set. (More formally, the axiom schema of separation applied to the set
.P(S X X) x (?(X) x ,P(.P(X)) X .P((X X X) X X)) and the statement "C S; X, 7 is a topology on C, is an associative binary operation on C, and f : S --> C" guarantees the existence of a set a as we have defined it. In defining 9 we have, out of necessity, departed from the custom of not specifically mentioning the topology or the operation when talking about a semigroup with a topology. In the definition of the set F, we return to that custom, writing C instead of
(C, 7, ). Let F, _ { (f, C) E 9.: '17, (C), f is a continuous homomorphism from S to C, and f [S] c A(C)). (Notice again that the requirement that f [S] c A(C) is redundant if i 0 1.) Trivially F, satisfies conclusion (1). Now let g and D be given such that 'P (D) and g is a continuous homomorphism from S to D. Let B = cf g[S]. By Exercise 2.3.2, since g[S] C A(D), B is a subsemigroup of D. Further, if D is a topological group (as it is if i = 4), then by Exercise 2.2.3 B is a topological group. Since the continuity requirements of 'Y, are hereditary, we have 4, (B). 22K Now by Lemma 21.2, I B I < = I X I so pick a one-to-one function r : B -> X and let C = r[B]. Give C the topology and operation making r an isomorphism and a
21.1 The £.4 C, WAY, AR, and .BAY Compactifications
427
homeomorphism from B onto C. Let f = r o g and let rp = r-1. Then (f, C) E F V is a continuous one-to-one homomorphism from C to D, and ri o f = g. Notice that one may have f and distinct Ci and C2 with (f, CI) E F; and (f, C2) E F,. We can now establish the existence of universal semigroup compactifications with respect to each of the statements %i.
Theorem 21.4. Let S be an infinite semigroup which is also a topological space and let i E [1, 2, 3, 4). There exist a pair (ri,, yiS) such that (1) 'Pi (Yi S),
(2) rii is a continuous homomorphism from S to y, S, (3) ri, [S] is dense in yi S, (4) ni [S] S A (Yi S), and
(5) given any D such that Wi (D) and any continuous homomorphism g : S ->. D with g[S] c A(D), there exists a continuous homomorphism s : y,S -+ D such
thatµorii = g. So the following diagram commutes: yi S
S
g
D.
Proof. Pick a set F, as guaranteed by Lemma 21.3 and let T = X (fc)EF; C. Define i7i : S -> T by i , (s) (f, C) = f (s). Let y, S = rii [S]. By Theorem 2.22 T (with the product topology and coordinatewise operations) is a compact right topological semigroup and r)i [S] c A(T). If i E (2, 3, 4) one easily verifies the remaining requirements needed to establish that 1i (T). By Exercise 2.3.2 y, S is a subsemigroup of T and, if i = 4, by Exercise 2.2.3 y, S is a topological group. Consequently'Yi (yi S).
For each (f, C) E Fi, 1r(f,C) o rii = f so rii is continuous. Given s, t E S and
(f,C)EF1, (in(s)n,(t))(f, C) = (rni(s)(f, C))(r7i(t)(f, C)) = f(s)f(t) = f(st) = ni(st)(f,C) so rii is a homomorphism. Trivially rii [S] is dense in y, S and we have already seen that i7i [S]
A (T) so that
?7i[S] c A(YiS).
Finally, let D be given such that 1, (D) and let g : S -), D be a continuous homomorphism. Pick by Lemma 21.3 some (f, C) E F, and a continuous one-to-one homomorphism cp : C -> D such that cp o f = g. Let u = (p o 1r(fc). Then, given
s E S, (t o r!i)(s) = ((P o n(fC) a ni)(s) = (p(ni(s)(.f, C)) = q,(f(S)) = g(s) We now observe that the compactifications whose existence is guaranteed by Theorem 21.4 are essentially unique.
428
21 Other Semigroup Compactifications
Theorem 21.5. Let S be an infinite semigroup which is also a topological space and let i E (1, 2, 3, 4}. Let (77i, yi S) be as guaranteed by Theorem 21.4. Assume that also the pair ((p, T) satisfies (1) 'Pi (T), (2) 7 is a continuous homomorphism from S to T, (3) 7p[S] is dense in T, (4) (p[S] C A(T), and
(5) given any D such that Ti (D) and any continuous homomorphism g S D with g[S] C A(D), there exists a continuous homomorphism µ : T -+ D such that it o tp = g. Then there is a function S : yi S --)- Y which is both an isomorphism and a homeomorphism such that S takes Tb [S] onto (p[S] and S o rli = cp. Proof. Let S : yi S - T be as guaranteed by conclusion (5) of Theorem 21.4 for T and rp and let lc : T yi S be as guaranteed by conclusion (5) above for yi S and iii. Then S and A are continuous homomorphisms, S o ni = rp, and A o cp = Ti. Now 3[yiS] is a compact set containing (p [S] which is dense in T so 3[yiS] = T: Also A o S o ni = it o V = 77i so A o S agrees with the identity on the dense set 77i [S] and thus A o S is the identity on yi S. Consequently j = S-I so 8 is a homeomorphism and an isomorphism.
As a consequence of Theorem 21.5 it is reasonable to speak of "the" £.MCcompactification, and so forth.
Definition 21.6. Let S be an infinite semigroup which is also a topological space and for each i E (1, 2, 3, 4} let (?7i, yi S) be as guaranteed by Theorem 21.4. (a) (711, yI S) is the £.M C-compactification of S and £ obi C (S) = yI S.
(b) (712, y2S) is the 'WA.P-compactification of S and WAY(S) Y2S. (c) 03, Y3S) is the A P-compactification of S and A (S) = Y3S. (d) 014, y4S) is the -SA.P-compactification of S and -SAY(S) = y4S. Notice that by Theorem 4.8, if S is a discrete semigroup, then (c, ! S) is another candidate to be called "the" £.MC-compactification of S. We remind the reader that semigroup compactifications need not be topological compactifications because the functions (in this case 77i) are not required to be embeddings. In Exercise 21.1.1, the reader is asked to show that if it is possible for ni to be an embedding, or just one-to-one, then it is. Exercise 21.1.1. Let S be an infinite semigroup which is also a topological space and let i E (1, 2, 3, 4). Let Y be a semigroup with topology such that 'IJ1(Y) Let r : S -* Y be a continuous homomorphism with r[S] C A(Y). (a) Prove that if r is one-to-one, then so is ni. (b) Prove that if r is an embedding, then so is 77i .
21.2 Right Topological Compactifications
429
Exercise 21.1.2. If S is a group (which is also a topological space); show that AR (S) --3A.P(S). (Hint: Prove that A31'(S) is a group and apply Ellis' Theorem (Corollary 2.39).)
Exercise 21.1.3. Let C denote any of the semigroups 'WA,P(N), AY (N), or .3AJ?(N). Let X E C and let m and n be distinct positive integers. Show that
m+x n+x.
Exercise 21.1.4. Show that each of the compactifications WA.P(N), AR(N) and ,SAP(N) has 2` elements. (Hint: By Kronecker's Theorem, N can be densely and homomorphically embedded in a product of c copies of the unit circle.)
21.2 Right Topological Compactifications The requirements in conclusions (4) and (5) of Theorem 21.4 referring to the topological center seem awkward. They are redundant except when i = 1, when the property being
considered is that of being a right topological semigroup. We show in this section that without the requirement that g[S] c A(D), there is no maximal right topological compactification of any infinite discrete weakly right cancellative semigroup in the sense of Theorem 21.4.
Theorem 21.7. Let S be an infinite discrete weakly right cancellative semigroup and let K be-an infinite cardinal. There exist a compact right topological semigroup T and an injective homomorphism r : S -> T such that any closed subsemigroup of T containing r[S] has cardinality at least K. Proof. Let oo be a point not in K x S and let T = (K X S) U {oo}. (Recall that the cardinal
K is an ordinal, so that K = (a : a is an ordinal and a < K).) Define an operation on T as follows.
(0, s) . (t', s') = (t', ss'), (t, s) (t', s') = (t' + t, s) if t 36 0, and where t' + t is ordinal addition. We leave it as an exercise (Exercise 21.2.1) to verify that the operation is associative. Now fix an element a E S and define a topology on T as follows.
(1) Each point of (K x (S\{a})) U {(0, a)) is isolated. (2) Basic open neighborhoods of oo are of the form {00} U ({t" : t' < t" < K) X S) U ({t'} x (S\F))
where t' < K and F is a finite subset of S with a E F.
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21 Other Semigroup Compactifications
(3) If t = t' + 1, then basic open neighborhoods of (t, a) are of the form
{(t, a)} U ((t') x (S\F)) where F is a finite subset of S with a E F. (4) If t is a nonzero limit ordinal, then basic open neighborhoods of (t, a) are of the form
{(t, a)} U ({t" : t' < t" < t} x S) U ({t'} x (S\F)) where t' < t and F is a finite subset of S with a E F. That is to say, a basis for the topology on T is
B = {{(t, s)} : t < K ands E S\{a}} U {{(0, a))} U{{oo} U ({t" : t' < t" < K} X S) U ({t'} x (S\F)) t' < K, F E 2f (S), and or E F}
U{{(t' + 1, a)} U ({t'} x (S\F)) : t' < K, F E Pf(S), and a E F}
U{{(t, a)} U ({t" : t' < t" < t} x S) U ({t'} x (S\F)) : t is a limit, t' < t < K, F E P1(S), and or E F). The verification that .8 is a basis for a Hausdorff topology on T is Exercise 21.2.2.
To see that, with this topology, T is compact, let U be an open cover of T. Pick U E U such that oo E U and pick ti < K such that {t" : ti < t" < K} X S c U. If a finite subfamily of U covers It : t < tt } x S, then we are done. So assume that no finite subfamily of U covers It : t < ti } x S and pick the least to such that no finite subfamily of U covers It : t < to) x S. Pick V E U such that (to + 1, a) E V and pick F E 3"f(S) such that {to) x (S\F) C V. Pick finite g c U such that (to) x F C U g. Assume first that to = t' + 1 for some t' and pick finite F C U such that It : t < t'} x S C U F. Then F U g U {V) covers It : t< to) x S, a contradiction. Thus to is a limit ordinal. If to = 0, then 9 U { V) covers {0) x S. Thus to 0 0. Pick W E U such that (to, a) E W and pick t' < to such that (t" : t' < t" < to) x S c W. Pick finite F C U such that It : t < t'} x S U F. Then F U g U {V, W) covers It : t < to) x S, a contradiction. Thus T is compact as claimed. We verify now that T is a right topological semigroup. Since p. is constant it is continuous. Let (t', s') E K x S. Trivially p(N,s') is continuous at each isolated point of T, so we only need to show that p(t',s') is continuous at oo and at each point of (K\{0}) x {a}. To see that p(r'.s') is continuous at oo, let U be a neighborhood of oo and pick u < K
such that {t":u I X I and let T and r be as guaranteed by Theorem 21.7 for S and K. Suppose one has a continuous homomorphism rl : X -- T such that ri o rp = r. Then q[X) is a closed subsemigroup of T containing r[S] and thus K > l X i > II[X] > K, a contradiction.
Exercise 21.2.1. Prove that the operation on T defined in the proof of Theorem 21.7 is associative.
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21 Other Semigroup Compactifications
Exercise 21.2.2. Let B and T be as in the proof of Theorem 21.7. Prove that B is a basis for a Hausdorff topology on T.
21.3 Periodic Compactifications as Quotients Given a semigroup S with topology, the WAR-, AR-, and -S.4.9 -compactifications are all compact right topological semigroups that are equal to their topological centers and thus, by Theorem 4.8 each is a quotient offSd, where Sd denotes S with the discrete topology. In this section we shall identify equivalence relations on ,6Sd that yield the 'WARand ,A, -compactifications. We shall use the explicit descriptions of these compactifications as quotients to characterize the continuous functions from a semitopological semigroup S to a compact semitopological or topological semigroup that extend to the
WAR- or .4.?-compactifications of S. We shall also show that, if S is commutative, -SAP(S) is embedded in 'WAP(S) as the smallest ideal of W,4.P(S). We first give some simple well-known properties of quotient spaces defined by families of functions. Definition 21.9. Let X be a compact space and let 0 (f, Y) be a statement which implies
that f is a continuous function mapping X to a space Y. We define an equivalence relation on X by stating that x = y if and only if f (x) = f (y) for every f for which 4(f, Y) holds for some Y. Let X/- denote the quotient of X defined by this relation, and let rr : X - X/= denote the canonical mapping. We give X/= the quotient topology, in which a set U is open in X/- if and only if a-' [U] is open in X. Of course, the space X/= depends on the statement 0, although we have not indicated this in the notation.
Lemma 21.10. Let X be a compact space and let 0 (f, Y) be a statement which implies that f is a continuous function mapping X to a space Y. Then X/- has the following properties:
(1) If g : X - Z is a continuous function and if g : Xl
Z is a function for
which g = g o r, then i is continuous. (2) X/- is a compact Hausdorff space. (3) A net (zr(x,)),E j converges to 7r(x) in Xl= if and only if (f (z,)),E1 converges to f (x) in Y for every (f, Y) for which i(f, Y) holds.
Proof. (1) If V is an open subset of Z, then 7r-'[j -'[VII = g-' [V ], which is open in
X and so[V] is open in X/=. (2) Since X/= is the continuous image of a compact space, it is compact. To see that it is Hausdorff, suppose that x, y E X and that 7r (x) 7r (y). Then there is a pair
21.3 Periodic Compactifications as Quotients
433
(f, Y) such that 0(f, Y) holds and f (x) f (y). For every u, v E X, 7r(u) = 7r (V) implies that f (u) = f (v). So there is a function f : X/- -- Y for which f = f o 7r. By (1), f is continuous. If G and H are disjoint subsets of Y which are neighborhoods
of f (x) and f (y) respectively, then f -[G] and f -'[H] are disjoint subsets of X/= which are neighborhoods of r(x) and 7r(y) respectively. So X/- is Hausdorff. (3) As in (2), for each (f, Y) for which 0(f, Y) holds, let f : X/- -+ Y denote the continuous function for which f = f o 7r. The set {f1[U] : 0(f, Y) and U is open in Y) is a subbase for a topology r on X/- which is coarser than the quotient topology and is therefore compact. However, we saw in (2) that, for every x, y E X, 7r(x) # 7r(y) implied that 7r(x) and 7r(y) had disjoint r-neighborhoods. So r is Hausdorff and is therefore equal to the quotient topology on X/-. Let (XI)LEI be a net in X and let x E X. If (f (x,)),E1 converges to f (x) in Y whenever Off, Y) holds, then (7r(x,)),EI converges to 7r(x) in the topology r. So (7r(x,)),EI converges to 7r(x) in the quotient topology.
Now assume that (7r(x,)),EI converges to 7r(x) and let (f, Y) be given such that Off, Y) holds. If U is a neighborhood of f (x), then f -'[U] is a neighborhood of 7r(x). Thus (7r(x,)),EI is eventually in f -1 [U] and so (f (x,)),E1 is eventually in U.
We recall that we defined in Section 13.4 a binary operation o on ,9Sd by putting x o y = lim lim st, where s and t denote elements of S. Note that, in defining x o y,
ty s-+x
we have reversed the order of the limits used in our definition of the semigroup (P Sd, -).
The semigroup (,BSd, o) is a left topological semigroup while (fSd, ), as we defined it, is a right topological semigroup. The following lemma summarizes the basic information that will be used in our construction of the W A,P- and .A,P-compactifications of a semitopological semigroup.
Lemma 21.11. Let S be a semitopological semigroup and let 9(f, C) be a statement which implies that f is a continuous function f mapping S to a compact semitopological
semigroupC. Let ¢(g, C) be the statement that there exists f for which 9(f, C) holds and g = f : 18Sd --> C. Suppose that, for every (f, C) for which 0(f, C) is true, we have:
(1) 6(f o Aa, C) and 8(f o pa, C) hold for every a E S (where Xa, A, : S -+ S) and (2) for all x, y E 6Sd, T (X - y) = T (X o y).
Let 9Sd/- and r be defined by the statement 0 as described in Definition 21.9. We define on 8Sd/- by putting 7r(x) - 7r(y) = 7r (x - y). (a) The operation - on 8Sd/- is well defined. (b) With this operation, ,BSd/- is a compact semitopological semigroup and 7r is a continuous homomorphism from (,BSd, ) and from (OSd, o). (c) The restriction 7rIs is continuous.
Proof. Notice that for x, y E 18Sd, we have x - y if and only if for every pair (f, C) satisfying 0 (f, C), one has T (X) = 'T(Y) -
21 Other Semigroup Compactifications
434
(a) Given a E S and x, x' E fSd with x - x', we claim that x a - x' a and
N
ax-a
N
x'. To see this, let (f, C) be given satisfying 8(f, Q. Then 8(f o pa, C) and 8(f o Xa, C) hold. Since f o pa = f o pa and f o Xa = f o ,la we have that f (x a) = T (x' a) and T (a x) = f (a x') as required. Now assume that x x' and y E fiSd. We claim that x y x' y. Recall that we denote by ix the continuous function from f Sd to, Sd defined by Zx(z) = x o z. Now for each a E S and each (f, C) for which 8(f, C) holds, .To
and so f o f, and f o ex, are continuous functions agreeing on S, hence on fSd. Thus
f (x y) = f (x' y). Similarly (using px(z) = z x), y x - y x'. Now let x-x'and y-y'.Then (b) By Lemma 21.10, fSd/= is compact and Hausdorff. By (a), tr is a homomorphism from (flSd, ), and by (a) and (2), Jr is a homomorphism from (flSd, o). By Exercise 2.2.2, since ,BSd/=_ is the continuous homomorphic image of ($Sd, ), it is a right topological semigroup, and since it is the continuous homomorphic image of (fSd, o), it is a left topological semigroup. (c) To see that 7rjs : S --* fSd/- is continuous, lets E S and let (s,),EI be a net in S converging to s in S. To see that (7r(s,))IEI converges to 7r(s) it suffices by Lemma 21.10 to let (g, C) be given such that 0(g, C) holds and show that (g(s,)),EI converges
to g(s). Pick f such that 8(f, C) holds and g = T. Then f : S -* C is continuous so (f (s,)),EI converges to f (s). That is, (g(s,)),EI converges to g(s) as required.
o
The numbering 82 and 02 in the following definition is intended to correspond to the function r12 : S --0- WAR (S). The statements 82 and 02 depend on the semigroup S,
but the notation does not reflect this dependence.
Definition 21.12. Let S be a semitopological semigroup. 62(f, C) is the statement "C is a compact semitopological semigroup, f is a continuous function from S to C, and f (x o y) = T (x y) for every x, y E flSd". ¢2(g, C) is the statement "there exists
(f, C) for which 82(f, C) holds and g = f : fSd -+ C". Lemma 21.13. Let S be a semitopological semigroup. If 02 (f, C) holds and a E S, then 82(f 0 ),a, C) and 02(f o pa, C) hold as well (where Xa, Pa : S --- S). Proof. Since Xa is continuous (because S is a semitopological semigroup), we have that
f o Xa is continuous. Notice also that f o Xa = f o ,la. Now, let x, y E fSd. Then, since
foAa(xoy) =f_(ao(xoy))=_f((aox)oy)=
f
o
0
21.3 Periodic Compactifications as Quotients
435
The reader may wonder why we now require S to be a semitopological semigroup, after pointing out in Section 21.1 that we were not demanding this. -The reason is that we need this fact for the validity of Lemma 21.13.
Theorem 21.14. Let S be a semitopological semigroup, let ¢ = ¢2 and let (8Shc - and
n : flSd - $Sd/- be defined by Definition 21.9. Then (1rls, fSd/=) is a WA.Pcompacti(Ication of S. That is:
(a) 'l/2(fSd/=), (b) nIs is a continuous homomorphism from S to fSd/-, (c) 7r [S] is dense in QSd/=, and (d) given any D such that W2 (D) and any continuous homomorphism g : S -> D, there exists a continuous homomorphism µ : fSd/m -+ D such that Fs o nIs = g. Proof. Conclusions (a), (b), and (c) are an immediate consequence of Lemmas 21.11 and 21.13. Let g and D be given such that %P2 (D) and g is a continuous homomorphism from S to
D. We claim that 92(g, D) holds. So let x, y E f3Sd. Notice that g is a homomorphism from (flSd, ) to D by Corollary 4.22. Thus
g(x o y) = g(y-limx-lims t) = y-limx-limg(s) g(t) = g(x) g(y) = g(x y). tES
tES
SES
sEs
Let µ Sd /- -+ D denote the function for which Ec o n = W. Then ,u is continuous, by Lemma 21.10, and is easily seen to be a homomorphism. We now turn our attention to characterizations of functions that extend to W A P (S).
Theorem 21.15. Let S be a semitopological semigroup, let C be a compact semitopological semigroup, and let f be a continuous function from S to C. There is a continuous function g : "W A P (S) -* C for which f = g o r12 if and only if .7(x y) = T (x o y)
forevery x, y E ISd. Proof. Necessity. Pick a continuous function g : WAR(S) Then
C such that f = g o n2.
AX-Y) = x-limy- lim f (s t) = x-limy- lim g (112 (s t)) sES (ES SES tES = g(x-limy-lim>)2(s) 172(t)) = g(i 2(x) i2(Y)) SES tES = g(Y- lim(ESx- hmSES 772(S)
772(t))
= y-limx-limg(r12(s t)) = y-limx-lim f(s t) = f(x oy). tES
SES
tES
SES
Sufficiency. Define - as in Theorem 21.14. Then by Theorems 21.5 and 21.14, it suffices to show that there exists a continuous function g : 6Sd/- -+ C for which
f = g o rris. Since 2(f, C) holds, there is a function g : fiSd/- -+ C for which f = g o n. By Lemma 21.10, g is continuous. If S is commutative, we obtain a characterization in terms of (flSd, ) alone.
21 Other Semigroup Compactifications
436
Corollary 21.16. Let S be a commutative semitopological semigroup, let C be a compact semitopological semigroup, and let f be a continuous function from S to C. There
is a continuous function g : 'W A.P (S) -* C for which f = g o 112 if and only if f (x y) = f (y x) for every x, y E flSd. Proof. This is an immediate consequence of Theorems 13.37 and 21.15. Recall that C(S) denotes the algebra of bounded continuous complex-valued functions defined on S.
Definition 21.17. Let S be a semitopological semigroup and let f E C(S). (a) The function f is weakly almost periodic if and only if there is a continuous g : WAR(S) -- C such that g o 112 = f .
(b) The function f is almost periodic if and only if there is a continuous g ,4.'P(S) - C such that g o 113 = f. Theorem 21.18. Let S be an infinite semitopological semigroup and let f E C(S). The following statements are equivalent: (1) The function f is weakly almost periodic.
(2) Whenever x, y E 6Sd, T (X y) = f (x o y). (3) Whenever (an)n° and (bn)n° 1 are sequences in S and all indicated limits exist, lim lim f (am bn) = lim lim f (am bn). 1
m-icon-soo
n-room-moo
Proof. By Theorem 21.15, (1) and (2) are equivalent. We shall show that (2) and (3) are equivalent.
(2) implies (3). Let (an)n° and (bn)n_1 be sequences in S for which all the limits indicated in the expressions lim lim f (am bn) and lim lim f (am bn) exist. Let 1
m-*oon-goo
n-+oom-+oo
x and y be limit points in #Sd of the sequences (an)n° 1 and (bn)n 1 respectively. Then lim lim f (am bn) = 7(x o y) and lim lim f (am bn) = f (x y). So (2) implies n-Boom--soo
m--soon--oo
that these double limits are equal.
(3) implies (2). Let X,2 E /Sd, let k = f (x y), and let £ = L (x o y). For each
n, then If (am bn) - 7(x bn) I
m,then
m, and
C. We show next that 93(f, C) implies 92(f, C).
Lemma 21.25. Let S be a semitopological semigroup. If 93(f, C) holds and a E S, then 93 (f o pa , C) and 93 (f 0 Xa, C) hold as well (where pa, ka : S -+ S). Furthermore,
if f * is as guaranteed by this statement, then for all x, y E 8Sd, T (X y) = f *(x, y) _ .f (x o Y)
Proof Suppose that 93(f, C) holds and that a E S. Let g : f Sd x fiSd --> C be defined by g(x, y) = f *(ax, y). Then g is continuous and g(s, t) = (f o ,1a)(s t) for every s, t E S. So 93(f o la, C) holds. Similarly, 03 (f o pa, C) holds.
21.3 Periodic Compactifications as Quotients
439
Let x, y E fSd. Then
f (x y) = x- lim y- lim f (s t) = x- lim y- lim f * (s, t) = f* (x, y) SES
tES
SES
tES
= y- lim x- lim f * (s, t) = y- lim x- lim f (s t) = T (X o y). tES
5ES
tES
sES
0
For the remainder of this section, we reuse the notations - and 7r for a different quotient.
Theorem 21.26. Let S be a semitopological semigroup. Let l4Sd/- and n be defined by Definition 21.9 with 0 = 03. For each x, y E f Sd, let 7r(x) .7r(y) = n(x y). Then (7r1s, f Sd/=_) is an M -compactification of S. That is, (a) 'P3(8Sd/=), (b) Iris is a continuous homomorphism from S to fiSd/=,
(c) 7r [S] is dense in fSd/-, and (d) given any D such that 4'3(D) and any continuous homomorphism g : S -). D, there exists a continuous homomorphism µ : ,Sd/- --> D such that /2 o 7r1s = g. Proof. It follows immediately from Lemmas 21.11 and 21.25 that 7r is well-defined. Conclusions (b) and (c) are also immediate from these lemmas, as is the fact that (P Sd/-, ) is a compact semitopological semigroup. To complete the verification of (a), we need to show that multiplication in flSd/- is jointly continuous.
Suppose that x, y E flSd and that (x,),EI and (y,),E/ are nets in #Sd for which (n(x,))ZE1 and (n(Yj))IEI converge to 7r (x) and n(y) respectively in fSd/-. We claim that (7r (x,) n(y,)),EI converges to 7r (x) 7r(y), that is that (7r (x, y,)),EI converges to 7r(x y). By passing to a subnet, we may presume that (x1)1Ej and (y,),EJ converge in /3Sd to limits u and v respectively. Then 7r (u) = 7r (x) and n(v) = n(y). To see that (7r(x, ,y,)),EI converges to 7r (x y) = 7r (u v), it suffices by Lemma
21.10 to show that (f (x, . y,)),EI converges to f (u v) for every (f, C) for which 03(f, C) holds. So assume that 03(f, C) holds and let f* be the function guaranteed by this statement. Then (f* (x y,))IEI converges to f* (u, v) and so by Lemma 21.25, (f(x, . y,)),EI converges to f (u v) as required. To verify (d), let g and D be given such that %P3(D) and g is a continuous homomorphism from S to D. We claim that 93(g, D) holds. To see this, define g* : fiSd x fiSd -- D by g*(x, y) = g(x) g(y). Then g* is continuous and, given s, t E S, one has g*(s, t) = g(s) g(t) = g(s t). Let µ : OSd/- ->. D be the function for which µ. o n = W. Then µ is continuous (by Lemma 21.10) and is easily seen to be a homomorphism. Theorem 21.27. Let S be a semitopological semigroup, let C be a compact topological
semigroup, and let f be a continuous function from S to C. There is a continuous function g : AR(S) -+ C for which f = g 0 773 if and only if there is a continuous function f * : iSd x 0Sd -+ C such that f *(s, t) = f (s t) for all s, t E S.
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21 Other Semigroup Compactifications
Proof. Necessity. Pick a continuous function g : A5'(S) -* C such that f = g o p3. Let r13 : 8Sd - AY (S) denote the continuous extension of 713. Define f*(x, y) _ gS713(x) n3 (y)) Then given s, t E S one has g(r13(s) 7j3(t)) = g(173(s) 173(t)) _
To see that f * is continuous, let x, y E fSd and let W be a neighborhood of g(n3(x) r13(y)). Pick neighborhoods U and V of j3(x) and i3(y) in AY (S) such that U. V C g-1 [W]. Then f*[713-'[U] x r13-1[V]] c W. Sufficiency. Since 03(f, C) holds, there is a function g Sd/- - C such that f = g o 7r. By Lemma 21.10, g is continuous. Theorem 21.28. Let S be a semitopological semigroup and let f E C(S). The following statements are equivalent:
(1) The function f is almost periodic. (2) There is a continuous function f #Sd x i8Sd -- C such that for all s, t E S,
f*(s,t)=
(3) For every e > 0, there is an equivalence relation on S with finitely many equivalence classes such that If (s t) - f (s' t') I < e whenever s s' and
tit'.
Proof. The equivalence of (1) and (2) is a special case of Theorem 21.27. (2) implies (3). Let e > 0 be given. Each element of p Sd x 0 Sd has a neighborhood of the form U x V, where U and V are clopen subsets of,BSd and I f * (x, y) - f * (u, v) I
91
defined by g(p) C p is a continuous surjection.
Proof. That 91 is Hausdorff is an immediate consequence of (b). That 91 is compact will follow from the fact that 91 is a continuous image of fl S.
Now let p E ,BS and suppose first that for no A E 91 is A c p. Then for each f (A). E
A E 91 pick f (A) E A\p. By (a) pick some tj E J" f(91) such that S = 1.1. Then S E p so pick some A E 3 such that f (A) E p, a contradiction.
Next suppose that we have distinct A and 2 in 91 with A c p and 2 c p. Pick
A E A and B E 2 as guaranteed by (b). Then S\A E 2 c p and A E A c p, a contradiction.
The function g is trivially onto 91. To see that g is continuous, let p E fS and let
A c Swithg(p) E At. Foreach2 E 91\{g(p)}pickby(b)some f(9) E S and some D(2) E g(p) so that for all e E'R, S\ f (2) E G or S\D(2) E C. Let f (g(p)) = A. Pick by (a) some a E P f(91) such that S = USEa f (2) and let 0 = \{g(p)}. Then
S\A C U2EO f(2). Let D= n.,Ee D(2). Then D E g(p) C p. We claim that g[D] c At. So let q E D and suppose that A 0 g(q). Pick by Corollary 3.9 some r E PS such that g(q) U {S\A} c r and pick 2 E 6 such that f (2) E r. Then one cannot have S\ f (2) E g(r) = g(q) so S\D(B) E g(q) c q, a contradiction. o Now we bring the algebra of S into play. The operation * on filters is a natural generalization of the operation on ultrafilters. Definition 21.33. Let S be a semigroup and let A and 9 be filters on S. Then
A*2=(ACS: {xES:x-'AES}EA}. Remark 21.34. Let S be a semigroup and let A and 2 be filters on S. Then A * 2 is a filter on S.
Definition 21.35. Let S be a semigroup and let 91 be a set of filters on S. If for all A and £ in 91, there is a unique C E 91 such that C' C A * 2, then define an operation
C A*S.
Theorem 21.36. Let S be a discrete semigroup and let ((p, Y) be a semigroup compactification of S. Let 91 = in (Q' 1 [{x}] : x E Y). Then 931 is a set of filters on S satisfying
(a) given any choice function f for 91 there exists a E J 'Of (91) such that S = U.,eE3 f (A),
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21 Other Semigroup Compactifications
(b) given distinct A and 2 in 91, there exist A E A and B E 2 such that whenever C E 9t, either S\A E C or S\B E C, and (c) given any A and 2 in 91 there is a unique C E 91 such that C C A * B. Further, with the quotient topology and the operation on 91, the function h : Y --* 9t defined by h (x) = n v1 [{x }] is an isomorphism and a homeomorphism. Proof. Conclusions (a) and (b) follow from Theorem 21.31. To establish (c), let A, 2 E
91. We note that it is enough to show that there exists e E 91 such that e c: A* S. (For then, if dJ E 91 and l) C pick by (b) some C E C such that S\C E D. Then
C E A * B so S\C E £\(A * 2).) Pick x, y E Y such that A= n iy-1 [{x}] and s = n v1 [{y}] and let e= n v1 [{xy}]. To see that e C A * S, let A E e. Let U = Y\rp[S\A]. Then U is open in Y and xy E U. (For if xy 0 U pick some p E S\A such that i p ( p ) = xy. Then A 0 p so A o n v1 [{xy}], a contradiction.) Since xy E U and Y is right topological, pick a neighborhood V of x such that
Vy C U. Then by Remark 21.29, rp-1[V] = rp-1[V] fl S E A. We claim that rp-1[V] C_ {s E S : s-1A E 2} so that A E A * B as required. To this end, let s E rp-1 [V]. Then q (s)y E U. Since A,(S) is continuous pick a neighborhood W of y such that rp(s)W C U. Then by Remark 21.29, rp-1 [W] E 2 so it suffices to show that
W-1[W] C s-1A Let t E V-1[W]. Then re(st) = rp(s)rp(t) E U so St 0 S\A, i.e., St E A as required. 91 defined by h(x) = n ip-1 [{x}] is a By Theorem 21.31, the function h : Y homeomorphism from Y onto 91. To see that it is a homomorphism, let x, y E Y. We
have just shown that h(xy) = nip-1 [{xy}] C h(x)*h(y)sothat
h(xy).
The following theorem tells us that the description of semigroup compactifications provided by (a), (b), and a weakening of (c) of Theorem 21.36 in fact characterizes semigroup compactifications. Theorem 21.37. Let S be a discrete semigroup and let 91 be a set of filters on S satisfying
(a) given any choice function f for 91 there exists E 9'f(91) such that S = UAEa f (A), (b) given distinct A and 2 in 91, there exist A E A and B E 2 such that whenever C E 9t, either S\A E C or S\B E C, and (c) given any A and £ in 9t there exists C E 91 such that e C A * 2. Then in fact for any A and 2 in 91 there exists a unique C E 91 such that C C A * 2. Also, with the quotient topology and the operation , 91 is a compact right topological
semigroup and the function g : 6S -+ 91 defined by g(p) _C p is a continuous homomorphism from $S onto 91 with g[S] c A(9t) and g[S] dense in 91. In particular, (gds, 91) is a semigroup compactification of S.
Proof. As in the proof of Theorem 21.36 we see that statements (b) and (c) imply
that for any A and 2 in 9t there exists a unique e E 91 such that C C A * 2, so the operation is well defined. By Theorem 21.32, with the quotient topology 91 is a
21.5 Uniform Compactifications
445
compact Hausdorff space and g is a (well defined) continuous function from $ S onto
We show next that g is a homomorphism. To this end, let p, q E S. To see that
g(p - q) = g(p) g(q), we show that g(p) - g(q) c p q. Let A E g(p) g(q). Then A E g(p) * g(q) so B = {s E S : s-1A E g(q)} E g(p) c p and ifs E B, then s-1A E g(q) c q so {s E S: s-1A E q} E p. That is, A E p- g as required. Since g is a homomorphism onto 9i, we have immediately that the operation on 91 is associative. By Exercise 2.2.2, 91 is a right topological semigroup.
Let s E S. To see that kg(,) is continuous, note that since g is a homomorphism g o )'S = AAg(5) o g. Let U be open in 91 and let V = kg(s) -' [U]. Then g-' [V ] _ (g o A,)-1 [U] is open in S. Then V = 9i\g[,aS\g-1 [V]] is open in 91. Since S is dense in 3S and g is continuous, g[S] is dense in 9t.
21.5 Uniform Compactifications Every uniform space X has a compactification (0, y,, X) defined by its uniform structure. For many uniform spaces X which are also semigroups, (q5, y,, X) is a semigroup compactification of X. It has the property that the map (a, x) H 0 (a)x is a continuous map from X x yu X to yu X, and yu X is the maximal semigroup compactification of X for which this holds.
This is true for all discrete semigroups and for all topological groups. If X is a discrete semigroup, then yuX 0X and so the material in this section generalizes our definition of fX as a semigroup. We first remind the reader of the definition of a uniform structure.
Definition 21.38. Let X be any set. If U, V c X x X, we define U-' and UV by U-' = {(y, x) : (x, y) E U} and UV = {(x, y) : (x, z) E V and (z, y) E U for some z E X}. We may use U2 to denote UU. A will denote the diagonal {(x, x) : x E X}. A uniform structure (or uniformity) `U on X is a filter of subsets of X x X with the following properties:
(1) AcUfor every UEU; (2) for every U E 2(, U-1 E U; and (3) for every U E 'U, there exists V E U for which V2 C U.
Let 2( be a uniform structure on X and let U E U. For each x E X, we put U(x) = (y E X : (x, y) E U), and, for each Y C X, we put U[Y] = UyEY U(y). U generates a topology on X in which a base for the neighborhoods of the point x E X are the sets of the form U(x), where U E 'U. If X has this topology, (X, 21) is called a uniform space and X is called a uniformizable space.
If (X, U) and (Y, V) are uniform spaces, a function f : X -> Y is said to be uniformly continuous if, for each V E V. there exists U E U such that (f (XI), f (x2)) E V whenever (X I, x2) E U.
446
21 Other Semigroup Compactifications
Metric spaces and topological groups provide important examples of uniformizable spaces. If (X, d) is a metric space, the filter which has as base the sets of the form {(x, y) E X x X : d (x, y) < r}, where r > 0, is a uniform structure on X. This example includes
all discrete spaces. If X is discrete, it has the trivial uniform structure U = (U C
XxX:ACU). If G is a topological group, its topology is defined by the right uniform structure which has as base the sets {(x, y) E G x G : xy-1 E V}, where V denotes a neighborhood of the identity. In this section, we shall assume that we have assigned this uniform structure to any topological group to which we refer. It is precisely the completely regular topological spaces which are uniformizable. Suppose that X is a space and that CR(X) denotes the subalgebra of C(X) consisting of the real-valued functions in C(X). X is said to be completely regular if, for every closed subset E of X and every x E X \ E, there is a function f E CR(X) for which
f(x) = 0 and f[E] = {1}. For each f E CR(X) and each e > 0, we put UfE = {(x, y) E X X X : I f (x) - f (y)I < e). The finite intersections of the sets of the form Uf,E then provide a base for a uniform structure on X. In particular, every compact space X is uniformizable. In fact, X has a unique uniform structure given by the filter of neighborhoods of the diagonal in X x X (see Exercise 21.5.1). In the next lemma we establish a relation between compactifications of X and subalgebras of CR(X). Lemma 21.39. Let X be any topological space and let A be a norm closed subalgebra of CR(X) which contains the constant functions. There is a compact space Y and
a continuous function 0 : X -+ Y with the property that 0[X] is dense in Y and A = If E CR(X) : f = g o 0 for some g E CR(Y)). The mapping 0 is an embedding if, for every closed subset E of X and every x E X\E, there exists f E A such that
f(x)=0and f[E]={1}. Proof For each f E A, let I f = It E R : It l < II f II } Let C = X fEA I f and let X --> C be the evaluation map defined by 0(x)(f) = f (x). Let Y = ctc 0[X].
For each f E A, let trf : C -> If be the projection map. If f E A, we have f = trfIy o 0. Conversely, let G = {g E CR(Y) : g o 0 E A). By Exercise 21.5.2, G is a closed subalgebra of CR(Y) which contains the constant functions. We claim that G also separates the points of Y. To see this, let y' and z' be distinct points in Y and choose f E A such that y f 0 z f. Since 7r f I y o 95 = f, we have n f,, E G and n fI Y (y) 0 n1i y (i). So G separates the points of Y and it follows from the StoneWeierstrass Theorem that G = CR (Y). Thus A = If E CR (X) : f = g o 0 for some g E CR(Y)}. Suppose now that, for every closed subset E of X and every x E X \E, there exists
f E A such that f (x) = 0 and f [E] = {1}. Then 0 is clearly injective. To see that 0 is an embedding, let E be closed in X and let y E 4)[X]\¢[E]. Pick X E X such that y = 0(x) and pick f E A such that f (x) = 0 and f [E] = {1}. Pick g E C(Y) such that f = g o 0. Then g(y) = g(O(x)) = 0 and g[O[E]] = {1} so y V c2 4)[E].
21.5 Uniform Compactifications
447
The following result establishes the existence of the Stone-tech compactification for any completely regular space. (Of course, since any subset of a compact space is completely regular, this is the greatest possible generality for such a result.)
Theorem 21.40. Let X be a completely regular space. Then X can be embedded in a compact space fX which has the following universal property: whenever g : X -> C is a continuous function from X to a compact space, there is a continuous function g : f3X -+ C which is an extension of g. Proof. Let A = CR(X). By Lemma 21.39, X can be densely embedded in a compact space fX with the property that every function in A extends to a continuous function in Cn(fX). We shall regard X as being a subspace of fX. Let C be a compact space and let F denote the set of all continuous functions from C to [0,1 ]. We define a natural embedding i : C -+ F [0, 1 ] by putting i (u) (f) = f (u) for every u E C and every f E F.
Suppose that g : X - C is continuous. Then, for every f E F, f o g has a continuous extension (f o g) : OX -). [0, 1]. We define h : fiX - F[0, 1] by h(y)(f) = (f o g) (y) for every y E 8X and every f E F. We observe that h(x) = i(g(x)) whenever X E X. Since X is dense in P X, h[X] is dense in h[flX]. So h[fX] c i[C], because h[X] c i[C] and i[C] is compact. We can therefore define g
by8=i-' oh. Recall that a topological compactification (rp, Y) of a space X has the property that p is an embedding of X into Y.
Theorem 21.41. Let (X, U) be a uniform space. There is a topological compactification (0, yuX) of X such that it is precisely the uniformly continuous functions in C(X) which have continuous extensions to Y. X. (That is, { f E CR(X) : f = g o 0 for some g E Ca(yuX)} = If E CR(X) : f is uniformly continuous).) Proof. It is a routine matter to prove that the uniformly continuous functions in CR(X) are a norm closed subalgebra of CR(X). The conclusion then follows from Exercise 21.5.3 and Lemma 21.39.
Since 0 is an embedding, we shall regard X as being a subspace of yuX. The compactification yuX will be called the uniform compactification of X. It can be shown that all possible topological compactifications of a completely regular space X arise in this way as uniform compactifications. We now see that yu X may be very large.
Lemma 21.42. Let (X, U) be a uniform space. Suppose that there exist a sequence in X and a set U E U such that xm 0 U (x,,) whenever m # n. Let f : N - X (xn) R be defined by f (n) = xn. Then f : ON --), yuX is an embedding.
Proof. We claim that, for any two disjoint subsets A and B of N,
f [Al fl
f[B] = 0. To see this, we observe that U[{x : n E A)] and {x : n E B)
448
21 Other Semigroup Compactifications
are disjoint, and hence that there is a uniformly continuous function g : X -> [0, 1] for which g[{xn : n E A}] = {0} and g[{xn : n E B)] = (1) (by Exercise 21.5.3). By Theorem 21.41, g can be extended to a continuous function g : yuX -+ R. Since g[ctyux f[A]] = {0} f[B]] = {1}, we havec1y.x f[A]flcty.xf[B] = 0.
Now suppose that u and v are distinct elements of N. We can choose disjoint subsets A and B of N such that A E u and B E v. Since f (u) E [A]) and .T(V) E c1yu x (f [B]), it follows that f (u) 0 T (v). So f is injective and is therefore an
embedding.
We now assume that X is a semigroup and that (X, U) is a uniform space. We shall give sufficient conditions for (0, yuX) to be a semigroup compactification of X. We observe that these conditions are satisfied in each of of the following cases:
(1) X is discrete (with the trivial uniformity);
(2) (X, d) is a metric space with a metric d satisfying d(xz, yz) < d(x, y) and d(zx, zy) < d(x, y) for every x, y, z E X; (3) X is a topological group. Notice that requirement (ii) below is stronger than the assertion that pa is uniformly continuous for each a E X.
Theorem 21.43. Suppose that X is a semigroup and that (X, U) is a uniform space. Suppose that the two following conditions are satisfied:
(i) For each a E X, Aa : X -- X is uniformly continuous and (ii) for each U E U, there exists V E U such that for every (s, t) E V and every a E X, one has (sa, ta) E U. Then we can define a semigroup operation on yuX for which (0, yuX) is a semigroup compactification of X.
Proof. For each a E X, there is a continuous extension
yuX -a yuX of Aa (by
Exercise 21.5.4). We put ay = ,La (y) for each y E yuX X.
Given Y E yuX, we define ry : X -a yuX by ry(s) = sy. We shall show that ry is uniformly continuous. We have observed that the (unique) uniformity on yuX is generated by {Ug,E : g E CR(yuX) and e > 0}, where Ug,E = {(u, v) E yuX x yuX g(u) - g(v)I < e}. So let g E CLR(yuX) and e > 0 be given.
Now gix is uniformly continuous by Theorem 21.41 so pick W E U such that Jg(s) - g(t)J < i whenever (s, t) E W. Pick by condition (ii) some V E 2( such that for all a E X and all (s, t) E V, one has (sa, ta) E W. Then, given (s, t) E V, one has E g(sy) - g(ty)I = Q m Ig(sa) - g(ta)I < 2, where a denotes an element of X, and so (ry(s), ry(t)) E Ug,E. It follows from Exercise 21.5.4 that ry can be extended to a continuous function
ry : yuX - Y" X.
21.5 Uniform Compactifications
449
We now define a binary operation on yuX by putting xy = Ty(x) for every x, y E
yuX. We observe that, for every a E X, the mapping y H ay from yuX to itself is continuous; and, for every y E yuX, the mapping x i-+ xy from yuX to itself is continuous, because these are the mappings Aa and Ty respectively. It follows that the operation defined is associative because, for every x, y, z E yuX ,
x(yz) = lim lim lim s(tu) and S-x t->y u-.z
(xy)z = lim lim lim(st)u, t-sy u-*z
where s, t, and u denote elements of X. So x(yz) = (xy)z. Thus yuX is a semigroup compactification of X. We now show that the semigroup operation on yu X is jointly continuous on X x yu X.
Theorem 21.44. Let (X, U) satisfy the hypotheses of Theorem 21.43. Then the map (s, x) r+ sx is a continuous map from X x yuX to yuX.
Proof. Lets E X and X E yuX. Choose any f E Ct (Y. X) and any e > 0. Since fix is uniformly continuous, there exists U E U such that If (s) - f (s') I
I whenever m # n. Let be a binary operation on /BS which extends the semigroup operation of S and has the
property that the mapping H x - from 0S to itself is continuous for every x E S. Then there is a point p E fS such that the mapping x i--+ x - p is discontinuous at every non-isolated point x of S. Proof Let p be any limit point of (yn) n° 1 and let x E S be the limit of a sequence (xn) R
of distinct elements of S. We may suppose that 3(x,n, xn) < 1 for every m, n E N. For every m, m', n, n' E N with n # n', we have S(xmYn, xm'yn') ? 3(xmYn, xmYn') - 3(XmYn', Xm'Yn') = S(Yn, Yn') - 8(xm, xm') > 12
Let A = {xmyn : m < n and m is even) and B = {xm yn : m < n and m is odd). We claim that A and B have no points of accumulation in S. This can be seen from the fact that, for any s E S, It E S : S(s, t) < 1) cannot contain two points of the form X,nyn and xn'yn' with n # n', and can therefore contain only a finite number of points
inAUB. Thus cis A = A, cis B = B and cis An cfsB = 0.
21.5 Uniform Compactifications
451
It follows from Urysohn's Lemma that there is a continuous function f : S -). [0, 11
for which f [A] = (0) and f [B] = [1). Let f : f S --* [0, 1] denote the continuous extension of f. Then f (xn p) = 0 if m is even and f (x,n p) = 1 if m is odd. So the sequence (x,n p)m cannot converge to x p. We shall now look at some of the properties of y l8, where ]R denotes the topological group formed by the real numbers under addition. We first note that P Z can be embedded topologically and algebraically in y,,lt. By Theorem 21.45, the inclusion map i : Z --> IR can be extended to a continuous homomorphism i : P7L --* y R. We can show that i is injective and therefore an embedding, by essentially the same argument as the one used in the proof of Lemma 21.42.
We shall therefore assume that fZ c y .R, by identifying $Z with ii[fZ]. This because i [f Z] = [7d]). identifies PZ with The following theorem gives an expression for an element of analogous to the expression of a real number as the sum of its fractional and integral parts.
Theorem 21.48. Every X E y. R can be expressed uniquely as x = t + z, where
tE(0,1)andzEPZ. Proof. Let jr : ]R -* T = R/Z denote the canonical map. Since jr is uniformly continuous, jr can be extended to a continuous homomorphism n : yu1R --> T.
We claim that n(x) = 0 if and only if x E f7L. On the one hand, J (x) = 0 if x E (3Z, because n [7G] = (0). On the other hand, suppose that x 0 #7G. Then there is a
continuous function f : y l8 -> [0, 1] for which f (x) = 1 and f [7G] = (0). Since fez is uniformly continuous, there exists S > 0 such that f [UnEZ[n - S, n + S]] c [0, (UfEz[n - S, n + S] and therefore x E ctyua(Un,z(n + S, n + 1 - S)) So x ¢ and n(x) 96 0, because 0 0 cta n [UfEZ(n + S, n + 1 - S)].
2].
Let t denote the unique number in [0, 1) for which ir(t) = ir(x). If z = -t + x, then rr(z) = 0 and so z E PZ. Thus we have an expression for x of the type required. To prove uniqueness, suppose that x = t' + z', where t' E [0, 1) and z' E P Z. Then
3f(x) = tr(t') and so t' = t and therefore z' = -t + x = z. Theorem 21.48 allows us to analyze the algebraic structure of y III in terms of the algebraic structure of #Z. Corollary 21.49. The left ideals of y,,l8 have the form l8 + L, where L is a left ideal of PZ; and the right ideals of yyR have the form l8 + R, where R is a right ideal of 16Z. Proof. This follows easily from Theorem 21.48 and the observation that l8 is contained in the center of y. R.
We omit the proofs of the following corollaries, as they are easy consequences of Theorem 21.48. Corollary 21.50. K(ynl$) = ]R + K(#7G).
21 Other Semigroup Compactifications
452
Corollary 21.51. Every idempotent of
is in #Z.
Corollary 21.52. y,,lR has 2` minimal left ideals and 2` minimal right ideals, each containing 2` idempotents.
Proof. This follows from Corollary 21.49 and Theorem 6.9.
Exercise 21.5.1. Let (X, U) be a uniform space. Show that every U E U is a neighborhood of the diagonal A in X x X. If X is compact, show that 2( is the filter of all neighborhoods of A in X x X. Exercise 21.5.2. Let G and Y be as in the proof of Lemma 21.39. Prove that G is a closed subalgebra of CR(Y) which contains the constant functions.
Exercise 21.5.3. Let (X, U) be a uniform space. Suppose that E and F are subsets of X and that U[E] n F = 0 for some U E U. Show that there is a uniformly continuous function f : X -+ [0, 1] for which f [E] = (0) and f [F] = {1}. (Hint: Choose a sequence (U")"EW in 'U by putting Uo = U n U-1 and choosing for every n > 0. Then define subsets E, of X U to satisfy U = U,,- 1 and U,2 c for each dyadic rational r E [0, 1] with the following properties: Eo = E,
E1 = X\F, and for each n E N and each k E {0, 1 , 2, ... , 2" - 1}, Un[E k ] C Ek+I 2n
.
2"
This can be done inductively by putting E0 = E and E1 = X\F, and then assuming that E k has been defined for every k E 10, 1, 2, ... , 2'). If k = 2m + 1 form E 2^ 10,1,...,2 n - 1), E k can be defined as U, +I [Em ]. Once the sets Er have been Fn -+I
constructed, define f : X
2n
[0, 1] by putting f (x) = 1 if x E F and f (x) = inf {r :
X E Er} otherwise.)
Exercise 21.5.4. Suppose that (X, 'U) and (Y, V) are uniform spaces and that 9 : X Y is uniformly continuous. Show that 9 has a continuous extension 9 : y"X y,, Y. (Hint: Let A = CR(y,, Y). The mapping 0 : Alb defined by q5 (y)(f) = f (y) is then an embedding. If f E A, fly is uniformly continuous and so f o 9 is uniformly continuous and has a continuous extension f : R. Let i/r : y"X -+ 0[y,, YJ be
defined by i/r (x)(f) = j (x). For every x c X, i/r(x) _ 0(9(x)).) Exercise 21.5.5. A uniform space (X, U) is said to be totally bounded if, for every U E U, there exists a finite subset F of X such that X = UXEF U(x). If (X, 'U) is not totally bounded, show that y"X contains a topological copy of ,BN . (Hint: Apply Lemma 21.42.)
Exercise 21.5.6. Let X be a topological group. Show that f E CR(X) is uniformly continuous if and only if the map x H f o AX from X to CR(X) is continuous, where the topology of CR (X) is that defined by its norm. (For this reason, y, X may be called the left uniformly continuous compactification of X and denoted by X'UC(X).)
Notes
453
Notes It is customary to define the £,MC, WAR, AR, and -3A.% compactifications only for semitopological semigroups S. Of course, if S is not semitopological, then the map nI cannot be an embedding (because 711 [S] is semitopological). It was shown in [150] (a result of collaboration with P. Milnes) that there exists a completely regular semitopological semigroup S such that ill is neither one-to-one nor open and that there exist completely regular semigroups which are neither left nor right topological for which ril is one-to-one and other such semigroups for which 171 is open as a map to 11 [S].
In keeping with our standard practice, we have assumed that all hypothesized topological spaces are Hausdorff. However, the results of Section 21.1 remain valid if S is any semigroup with topology, without any separation axioms assumed. We defined weakly almost periodic and almost periodic functions in terms of extendibility to the 'WAR- and A P-compactifications. It is common to define them in terms of topologies on C(S). One then has that a function f E C(S) is weakly almost periodic if cf{ f o ps : s E S} is compact in the weak topology on C(S) and is almost periodic provided cP{ f o ps : S E S} is compact in the norm topology on C(S). For the equivalence of these characterizations with our definitions see [40]. Theorem 21.7 and Corollary 21.8 are due to J. Berglund, H. Junghenn, and P. Milnes in [39], where they credit the "main idea" to J. Baker. The equivalence of statements (1) and (3) in Theorem 21.18 is due to A. Grothendieck [1121.
Theorem 21.22 is due to W. Ruppert [217]. The algebraic properties of 'W A J' (N) are much harder to analyze than those of fi (ICY).
It is difficult to prove that 'WA31'(N) contains more than one idempotent. T. West was the first to prove that it contains at least two [245]. It has since been shown, in the
work of G. Brown and W. Moran [54], W. Ruppert [220] and B. Bordbar [51], that 'WAR(N) has 2` idempotents. Whether the set of idempotents in WAD' (N) is closed, was an open question for some time. It has recently been answered in the negative by B. Bordbar and J. S. Pym [53]. The results of Section 21.4 are from [37] and were obtained in collaboration with
J. Berglund. In [37] a characterization of the WASP-compactification as a space of filters was also obtained. Theorem 21.40 is due (independently) to E. tech [60] and M. Stone [227]. See the notes to Chapter 3 for more information about the origins of the Stone-tech compactification. Theorem 21.48 is due to M. Filali [90], who proved it for the more general case of Yu
n -
Suppose that S is a semitopological semigroup and that f E C(S). It is fairly easy
to prove that f is a 'WAR function if and only if {(f o As)) : s E S) is relatively compact in {g : g E C(S)) for the topology of pointwise convergence on C(8Sd). It follows from [81, Theorem IV.6.14] that this is equivalent to if o Xs : s E S} being weakly relatively compact in C(S).
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[2] P. Anthony, The smallest ideals in the two natural products on $S, Semigroup Forum 48 (1994), 363-367. [3] R. Arens, The adjoint of a bilinear operation, Proc. Amer. Math. Soc. 2 (1951), 839-848.
[4] J. Auslander, On the proximal relation in topological dynamics, Proc. Amer. Math. Soc. 11 (1960), 890-895. [5] J. Auslander, Minimal flows and their extensions, North Holland, Amsterdam, 1988.
[6] J. Baker and R. Butcher, The Stone-tech compactification of a topological semigroup, Math. Proc. Cambridge Philos. Soc. 80 (1976), 103-107. [7] J. Baker, N. Hindman, and J. Pym, K-topologiesfor right topological semigroups,
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List of Symbols
A - s 76
As-' 76 A* 55 A3 441 A* 76, 93
A*(p) 76, 77, 93 [A]" 91
[A]"` 91 A 53
A *,B 443 AR 281, 372 an ideal, 281 compactification, 428
As(S) 428 as quotient, 439
A 54 C(p) 113,262 a filter, 114
C(S) 436 C(X) 70 C*-embedded 155 C+ 301 C- 301 C,, 179
chains of idempotents, 181 contains free group on 21 generators, 181
does not meet K(fG), 181
in K(fG), 182 minimal left and right ideals, 180 Cp(,6G) copy of Cq(fN), 183 CP(flN) 185 CR(X) 71, 446 cZ K (H) 172 c 66
D* 55,61 clopen subsets, 55 Da 207 d(A) 363 e 56, 57 e(a) 53 identified with a, 57
E(C, J, I) 305 E(S) 8, 33 e(x) 55 e[D] identified with D, 57 (e, BBD) 56
FP-trees and Banach density, 416
FP((xn)n t) 90 FP((xn)n°_1) 90 FP((xn)n°_m) 93
FS((xn)n ,) 90 90 FS((YY)n° l) 325 FSm 242 FSO2242 FU((Fn)n°_1) 95
f [A] 2
f-'[A] 2 f, 207 Fa subset 62 fsupp 242 f 58, 59, 62
g 207 as mapping on Za, 209 GS subsets of D* 60 H(e) 8, 9, 20, 24, 35
H(p) isomorphic to Z, 251 hn 259
List of Symbols
472
ha 207 H-map 148 H 107, 150, 151, 361 as anoid, 111 closure of K(IHI), 171 contains all idempotents of (P N, +), 109
contains infinite chains of idempotents, 110 copies of, 115, 117, 148, 150, 151, 152, 183
homomorphic images of H, 108 homomorphisms, 108 IP set 321 if and only if, 321 IP* set 321, 339, 376 and measure preserving systems, 408 dynamical, 408 examples, 337 if and only if, 321, 326 may never translate to IP*, 331 not a dynamical IP* set, 408 sums and products in, 327 K (S)
a group if commutative, 23 a group, 41 elements not cancelable, 30 elements of, 30 equal to SeS, 24 homomorphic image, 30 union of minimal left ideals, 21
K(T) equal to K(S) n T, 29 K (H) closure not a left ideal of 1H!, 171
K(OS) 85, 291 and piecewise syndetic sets, 86 and syndetic sets, 85 closure as ideal, 87 closure, 86 equal to K(,BG) n OS, 130 not closed, 169
K(,BS, ) . and K(8S, o), 274 K(,BN, +) 87 closure, and Banach density, 414
K(fN, ) K(,BN, ) n K(f N, +) = 0, 263 hits cB K($N, +), 328 misses K(8N, +), 263 K(,8S) contains right cancelable elements, 169
K(PN)
and sums of elements outside (K(,6N), 172
L(S) 7 2p 258, 272
lim 63
a-*x
£.MC compactification, 428
£ MC(S) 428 £`UC 453 (m, p, c) sets 318, 319 M 328,339
N2 N* 69 contains copy of co, 218 copies of N* in N*, 217 not a copy of IHI, 152
topological center empty, 170
(N*, +) and (-N*, +) as left ideals of (OZ, +), 85 (N, +) center, 109, 134 topological center, 110
(N, ) center, 109, 134 topological center, 110 (N, v) 3, 80, 324 weakly left cancellative, 82 (N, n) 3, 80, 324
(N*, ) copies of (N*, +) in (ICY*, ), 218
-N* 330 naFU 98
473
List of Symbols
p-limit 63, 74
W(ql; v) 381
p+q
WAR compactification as quotient, 435 compactification, 428
P
