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Exercise 10.1.1. It is a fact that if F and G are disjoint finite nonempty subsets of R and F U G is linearly independent over Q, then naEF Ua fl ftEG Da 96 0. Use this fact, together with Theorem 10.8 to show that (fiN, +) has a collection of 2` pairwise disjoint right ideals. (This fact also follows from Theorem 6.44.)
Exercise 10.1.2. Let a > 0 and let p, q E fN. Prove that ga(p + q) E (ga(p) + ga (q), ga (P) + ga (q) - 1, ga (P) + $a (q) + 1).
Exercise 10.1.3. Let a > 0 be irrational, let k E N with k > a, and let 8 = k - a. Prove that
(i) for all p E fN, ha (P) _ -ha (P),
(ii) Z. = Za, (iii) X. = Ys, and (iv) Y. = X8.
10.2 Homomorphisms from #T into S* We know that topological copies of fiN are abundant in N*. Indeed, every infinite compact subset of N* contains a copy of this kind, by Theorem 3.59. However, as we shall show in this section, there are no copies of the right topological semigroup fiN in N*. In fact, if 0 : fiN N* is a continuous homomorphism, then 0[#N] is finite and IOfN*ll = 1. We shall show that, if S and T are countable, commutative, cancellative semigroups
and if S can be embedded in the unit circle, then we can often assert that 4[T*] is a finite group whenever ¢ : fiT -> S* is a continuous homomorphism.
Lemma 10.13. Let S be an infinite discrete semigroup which can be algebraically embedded in T and let V be an infinite subsemigroup of S. Then, for every x and p in
S*, there exists y E V* for which x+ y+ p 0 y+ p+ x. Proof. Let h : S --). T be an injective homomorphism, with h : fl S -* T denoting its continuous extension. Let U and D be defined as in Lemma 10.1. By this lemma,
applied to V instead of S, if U' = {y E V* : IS E V : h(y) -< h(s)} E y} and D' = {y E V* : IS E V : h(s) -< h(y)} E y}, then U' and D' are non-empty. Let x, p E S*. If X E U, we can choose y E D. Since D' C D and U and D are
right ideals offS,x+y+pEUand y+p+xEDand so x+y+p#y+p+x. Similarly, if x E D, we can choose y E U' and deduce that x + y + p 0 y + p + x. Lemma 10.14. Suppose that S is a countably infinite discrete semigroup which can be algebraically embedded in T and that T is a countably infinite commutative weakly left cancellative semigroup. Let 0 : 0T -a S* be a continuous homomorphism. Then
K(O[T*]) is a finite group. Furthermore, for every c E T, {b E T : O(b + c) E K(O[T *])) is infinite.
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proof Observe that by Theorem 4.23, if a E S, then a commutes with every member
of ,BS. Also, if t E T, then t commutes with every member of fT and hence ¢(t) commutes with every member of 0[p T].
We shall show that if p is any minimal idempotent in O[T*] and q E T* with 4,(q) = p,then (1) there exists t E T such that 0(t) E ¢[T *] + p and
(2) given any c E T and any B E q, there exists t E T such that 0(b) + 95(c) E K(O[T*]) To see this, let such p and q be given, let c E T, and let B E q. Let V = (a E S : a + O (q) E ¢[f T]}. We claim that V is finite. Suppose instead that V is infinite, and notice that V is a subsemigroup of S. (Given a, b E V, a + b + 0 (q) = a + b + 0 (q) +
¢(q)=a+0(q)+b+0(q).) For every Y E V*, wehave y+0(q) E O[PT], by the continuity Ofpoigi. Then foreveryy E V*, wehave ¢(c)+y+O(q) = y+O(q)+O(c) which contradicts Lemma 10.13.
Choose A E 0 (c) for which A fl v = 0. Now 0 (c) + 0 (q) _ (q) + 0 (c), 0(c) + 0(q) E C1 (A + 0(q)), and 0(q) + 0(c) E ct(O[B] + 0(b)). It follows from Theorem 3.40 that one of the two following possibilities must hold:
(i) a +-0 (q) = -0 (v) +0(c) for some a E A and some v E fl T; (ii) u + 0(q) = 0(b) + 0(c) for some b E B and some u E fl S.
The first possibility is ruled out by the assumption that A fl v = 0. So we may suppose that (ii) holds. Let t = b + c. Then 0(t) E P S + p and so ¢(t) + p = 0(t). Since 0(t) = 0(t+q)+4'(q) and t+q E T* (by Theorem 4.31), 0(t) E 0[T*]+ p c K(0[T*]). Thus (1) and (2) hold.
Now let p be a minimal idempotent in 0[T*] and pick by (1) t E T such that 0(t) E 0[T*]+ p. Let F = 0[T*] +0(t). Then, since t E T, F = 4(t) +0[T*] so F is both a minimal left ideal and a minimal right ideal of 0[T*], and is therefore a group (by Theorem 1.61). Since F is compact, it must be finite, because an infinite compact subset of fS cannot be homogeneous by 6.38. If p' is any other minimal idempotent in 0[T*], there exists t' E T such that 0(t') E
0[T*] + p'. Now 0(t + t') = 0(t'+ t), and so the minimal left ideals 0[T*] + p' = 0[T*] + 0(t') and 0[T*] + p = 0[T*] + 0(t) of 0[f T] intersect. They are therefore equal. This shows that F is the unique minimal left ideal of ¢[T *] and so F = K (0[T *]) (by Theorem 1.64).
The fact that {b E T : 0(b + c) E K(¢[T*])} is infinite for each c E T follows immediately from (2).
Theorem 10.15. Suppose that S is a countably infinite discrete semigroup which can be algebraically embedded in the unit circle. Let T be a countably infinite commutative weakly left cancellative semigroup with the property that any ideal of T has finite complement and let 0 : PT -+ S* be a continuous homomorphism. Then 0[ f T] is finite and 0[T*] is a finite group.
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Proof. By Lemma 10.14, if F = K(" [T*]), then F is a finite group. Let I =-1[F]. We claim that I n T is an ideal of T. It follows from Lemma 10.14 that I n T 0 0. If c E I n T, then ¢ (c) = 0 (c) + p for some minimal idempotent p of cb [T * ]. There exists q E T* for which O (q) = p. For every b E T, we have O (b + c) = O (b) + O (c)
4(b)+cb(c)+¢(q) = O(b+q+c) = $(b+q)+O(c) E K(4[T*]). Sob+c E If1T and we have shown that I fl T is an ideal of T.
Thus T\I is finite. Given q E T*, I E q and so O[T*] c F. Since ¢[fT] C 4[T\I] U F, ¢[flT] is finite. Since t[T*] is a finite cancellative semigroup, it is a group by Exercise 1.3.1. p Notice that the hypotheses of Theorem 10.15 are satisfied in each of the following three cases.
(i) T is a countably infinite commutative group;
(ii) T = (N, +); or (iii) T = (N, v). The following lemma is simple and well known.
Lemma 10.16. Let G be a commutative group with identity 0 which contains no elec, then G can be embedded in (R, +).
ments of finite order apart from 0. If I G I
Proof. Let IGI = K. Suppose that G\{0} has been arranged as a K-sequence (aa)a
whenever a < a' < . We shall show that we can define fp so that these properties hold with 6 replaced by 0 + 1.
Let H = Ua R by stating that fp (b +ma) = f (b) + ms for every b E H and every m E Z. Then fp is well defined because Zap fl H = {0} and is easily seen to be an injective homomorphism. Thus we may assume that kap = c for some k E Z\{0} and some c E H. We define
fp : H + Zap
JR by stating that f# (b + map) = f (b) + k f (c) for every b E H
and every m E Z. To see that fp is well defined, suppose that b + map = b+ m'a p, where b, b' E H
andm,m' E Z. Then k(b'-b) =k(m-m')ap = (m-m')candsok(f(b')- f(b)) _ (m - m') f(c) and .f (b) + k f (c) = f (b') + k f (c) It is easy to see that fp is a homomorphism. To see that it is injective, suppose that
fp (b + map) = 0 for some b E H and some m E Z. Then f (b) + f (c) = 0 and so f (kb + mc) = 0. Since f is injective, kb + me = 0. That is to say,k k(b + map) = 0 and so b + map is of finite order in G and hence b + map = 0.
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Thus we can define an injective homomorphism fs : G,6 -+ JR for every f < K,
so that f,6 c ff, whenever f < #' < K. Then Ufi
Theorem 10.17. Let G be a countably infinite commutative group which contains no non-trivial finite subgroups, and let T be a countably infinite commutative weakly left cancellative semigroup with the property that any ideal of T has finite complement. If 0: PT -+ G* is a continuous homomorphism, then cb[19T] is finite and I4[T*]I = 1. Proof. By Lemma 10.16, G can be embedded in JR and therefore in T. (If S is a countable
subsemigroup of 1R, we can choose Of E R such that a is irrational with respect to S. The mapping ha then embeds S in T). Thus it follows from Theorem 10.15 that q[jT] is finite and O[T*] is a group. By Theorem 7.17, G* contains no non-trivial finite subgroups and so I0[T*]1 = 1.
Theorem 10.18. If ¢ : ON -+ N* is a continuous homomorphism, 0[,8N] is finite and I0[N*]1 = 1.
Proof. This follows from Theorem 10.17 with T = N and G = Z. Question 10.19. Is there any continuous homomorphism i¢ : ON -> N* for which
I0[fN31 : 1? We note that, in view of Theorem 10.18, there are two simple equivalent versions of Question 10.19
Corollary 10.20. The following statements are equivalent. (a) There is a nontrivial continuous homomorphism from fN to N*.
(b) Thereexistp
gins`Y*suchthatp+p=q+p=p+q=q+q=q.
(c) There is a finite subsemigroup of N* whose elements are not all idempotent.
Proof. (a) implies (b). Let 0 : ON
N* be a nontrivial continuous homomorphism. Then by Theorem 10.18, ¢[fN] is finite and there is some q E N* such that O[N*] = (q). There is a largest x E N such that 0 (x) # q. (Since 0 is nontrivial, there is some such x, and if infinitely many x have 0(x) # q and r E ce{x E N : 0(x) q} fl N*, then
0(r) # q.) Let p = O(x). (b) implies (c). The set {p, q} is such a subsemigroup. (c) implies (a). Let S be a finite subsemigroup of N* whose elements are not all
idempotent and pick p E S which is not an idempotent. Define a homomorphism f from N to S by stating that f (n) is the sum of p with itself n times and let 0 be the continuous extension of f to fN. Then 0 is a homomorphism by Corollary 4.22 and is nontrivial because {p} is not a semigroup.
Question 10.21. Which discrete semigroups S have the property that any continuous homomorphism from 0 S to S* must have a finite image?
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10.3 Homomorphisms from T* into S* A good deal of specialized notation will apply throughout this section. S and T will denote countably infinite discrete semigroups which are commutative and cancellative. G will denote the group generated by S and 0 will denote the identity of G. (So G is a countably infinite commutative group in which every element can be expressed as the difference of two elements of S). We shall regard S as embedded in G and PS as embedded in PG. We shall assume that T has an identity. (This is no real restriction, because an identity can be adjoined to any semigroup). We shall assume that we have chosen sequential orderings for G and T. For each m E N, G,,, will denote the set whose elements are the first m elements of G, and T. will denote the set whose elements are the first m elements of T. We shall suppose that 0 : T* -* S* is a continuous injective homomorphism. We choose q to be a nonminimal idempotent in T * and we put p = 0 (q). (The existence of a nonminimal idempotent in T* follows from Corollary 6.34.)
Lemma 10.22. cb[K(T*)] = K(,[T*]). Proof. The mapping q5 defines an isomorphism from T* onto O[T *].
Lemma 10.23. For every s E S and every x E ,3S, $ + x E K(f S) implies that x E K(fiS). Proof Suppose that s + x E K(fiS). Since K(PS) is a union of groups (by Theorem 1.64), there is an idempotent u E K (fS) for which s + x + u = s + x. By Lemma 8.1 this implies that x + u = x and hence that x E K(flS). Lemma 10.24. There is a subset Y = {tn : n E N} of T with the following properties: (i)For every y E Y* and every b, c E T, O (b + q) O f G + O (c + y) + p. (ii) For every m and n in N, every a E G,,,, and every b, c E Tm, if m < n, then
cb(b+tn +q) # a+q5(c+tn +q). Proof. We can choose a minimal idempotent u E T* satisfying u = q + u = u + q (by Theorem 1.60). We note that O(u) is in the smallest ideal of q5[T*] by Lemma 10.22. We claim that, for every b E T, ¢(b + q) 0 PG + 0(u). To see this, we note that
¢(b+q) E fiG+O(u) implies that 4(b+q) =b(b+q)+O(u) E K(O[T*]). By Lemma 10.22, this would imply that b + q E K(T*) and hence that q E K(T*) (by Lemma 10.23). This contradicts our choice of q. So for each b E T, O (b + q) has a clopen neighborhood Wb in PG disjoint from the compact subset PG + O (u) of f G. For each a E G and each b, c E T, let
Ea,b,c = {x E T* : a +t(c+x) + p E Wb}. By the continuity of the map x hi a + O (c + x) + p, Ea,b,c is a clopen subset of T*.
We note that u 0 Ea,b,c,because a+O(c+u)+p=a+O(c+u+u)+O(q)=
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a+O(c+u)+cb(u+q). Sinceu+q =u,a+cb(c+u)+p E fG+q5(u) e fG\Wb. Thus UaeG UbET UcET Ea,b,c i6 T*, because u does not belong to this union. It follows from Exercise 3.4.5 that there is a non-empty open subset U of T* disjoint from UaEG UbeT UcET Ea.b.c We can choose an infinite subset V of T for which V* C U.
We claim that,ify E V* and b, c E T,then 0(b+q) gl #G+o(c+y)+p. Suppose, on the contrary, that q, (b + q) E ,0G + 0(c + y) + p. Since Wb is a neighborhood of 0(b + q) in PG and we are assuming that 0(b + q) E cf(G + 0(c + y) + p), we must have a + O(c + y) + p E Wb for some a E G. This implies that y E Ea,b,c and contradicts our assumptions that y E U and U fl Ea,b,c = 0. To obtain property (ii), we shall enumerate V as {vn : n E N) and then replace (vn)n__i by a subsequence (tn)n° It defined inductively.
W e put tj = v1. W e then assume that we have chosen elements r1, t2, ... , tk in { vn : n E N) with the following property: if i, j E ( 1 , 2, ... , k), a E G; , b, c E Ti, and
i < j, then O(b+t; +q) # a+O(c+tj +q). Suppose that i E {1, 2, ... , k). If a E G and b, c E T, we claim there are at most a finite number of values of n for which 0 (b + t, + q) = a + 0 (c + vn + q). To see this, suppose that this set of values of n is infinite. Then the corresponding elements vn have
a limit pointy in V* satisfying 0(b+t, +q) = a+0(c+y+q) = a+0(c+y)+ p. We have seen that this cannot occur. It follows that we can choosevn with the property that 0(b+t;+q) ,A a+¢ (c+vn+q) whenever i E { 1 , 2, ... , k}, a E Gk and b, c E Tk. We put tk+1 = v, . This shows that we can define a sequence (tn)n° , with the property specified in (ii).
We then put Y = {tn : n E N}. The fact that Y* a V* implies that property (i) holds as well.
Lemma 10.25. Let Y denote the set guaranteed by Lemma 10.24, let y E Y* and let t E T. Then 0(t + y) + p is right cancelable in P G.
Proof. If we suppose the contrary, then 0(t + y) + p = x + 0(t + y) + p for some x E G* (by Theorem 8.18). Now O (t + y) + p E cf ({¢ (t + to + q) : n E N}), because
0 (t + y) + p = 0 (t + y + q) and t + y + q Ecf({t+tn+q:nEN}). We also have x + 0(t + y) + p E cE((G\{0)) + ¢(t + y) + p). It follows from Theorem 3.40 that 0(t + to + q) = x' + 0(t + y) + p for some n E N and some x' E P G, or else
O(t + y'+ q) = a + O(t + y) + p forsomey' E Y* andsomea 54 0inG. The first possibility contradicts condition (i) of Lemma 10.24, and so we shall assume the second. Let M E N satisfy a, -a E G, and t E Tm. Since 0(t + y' +q) E
cf({0(t+tn+q):n>m})and a+¢(t+y)+pEcf({a+0(t+t, +q):n>m}), a second application of Theorem 3.40 allows us to deduce that 0(t + t + q) = a +
O(t+y"+q)or4'(t+tn+q)=-a+O(t+y"+q) forsomen > m inN and
some y" E Cf((tn : n > m)). Now, by condition (i) of Lemma 10.24, neither of these equations can hold if y" E Y*. Soy" = t, for some r > m. By condition (ii) of Lemma 10.24, neither of these equations can hold if r 0- n. However, they cannot hold if r = n, by Lemma 6.28. Thus 0 (t + y) + p is right cancelable in PG.
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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).
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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. -
220
10 Homomorphisms
Proof Let L = 0 G+p and M = t3 H+q, where p and q are nonminimal idempotents in G* and H* respectively. By applying Lemma 10.35 with G = S = S' and H = T = T',
we see that 1A (q) is nonminimal and there is an isomorphism f : H - G such that T/r(t + q) = f (t) + i/r(q) for every t E H. By continuity, this implies that *(x + q) = f (x) + (q) for every x E fiH. Forx E fH,wehave >/r(q+x+q) = f (q+x)+i/r(q) = f(g) + f(x)+i/i(q) and
alsoi/r(q+x+q) = i(q)+i/r(x+q) =''lr(q)+f(x)+*(q). So f(q)+f(x)+i/r(q) = i/r(q) + L (x) + i/r(q) for every x E /3H. By Lemma 10.37, we can choose x E $H such that f (x) + ilr(q) is right cancelable in 0 G. Thus ilr(q) = f (q). This implies that i/i (x + q) = T (x + q) for every x E ,0H.
Theorem 10.39. Suppose that L and M are principal left ideals in )4N defined by M -> L is a continuous isomorphism then M = L nonminimal idempotents. If and i/' is the identity map.
Proof. Let p and q be nonminimal idempotents in )4N for which L = fiN + p and M = fN + q. Then L = fi7L + p and M = fi7L + q, since, given r E fi7L, r + p = r + p + p E fiN + p because N* is a left ideal in f7L (by Exercise 4.3.5). So L and M are principal left ideals in fi7L defined by nonminimal idempotents in 7L*. It follows
7L for which i/r = fIM. from Theorem 10.38 that there is an isomorphism f : 7L Now there are precisely two isomorphisms from 7L to itself: the identity map t r t and the map t r-> -t. We can rule out the possibility that f is the second of these maps, because this would imply that f (q) fN. Hence f is the identity map and so is i/r.
Theorem 10.40. Let G and H be countable commutative groups and let L and M be principal right ideals in G* and H* respectively, defined by nonminimal idempotents. Suppose that there is a continuous isomorphism i/r : M -+ L. Then there is an
isomorphism f : H -a G for which f [M] = L. Proof. Let p and q be nonminimal idempotents in G* and H* respectively for which
L = p + fiG and M = q +,8H. By applying Lemma 10.36 with G = S = S' and H = T = T', we see that i/i(q) is nonminimal and there is an isomorphism f : H
G
such thati/r(t+q) = f (t)+(q). It follows, by continuity, that*(x+q) = f (x)+*(q) for every x E M. Putting x = q shows that i/r(q) = 7(q) + (q). On the other hand, we also have, for every x E P H, that ilr(q ± x + q) = f (q) +
f(x)+*(q)and*(q+x+q) _(q)+(q+x+q) _(q)+f(q)+f(x)+(q)
So f(q)+f(x)+ '(q) =/r(q)+f(q)+f(x)+* (q)forevery x r= '6 H. We can choose x E H* for which T (x) + *(q) is right cancelable in fiG (by Lemma 10.37)
and so T (q) _(q) + F (q). It follows easily that * (q)+/3G= f(q)+/3G. Wealso know that i/r(q)+/3G=L (by Lemma 10.36). So T [M] = 7(q) + fiG = *(q) + f G = L. The preceding results in this section tell us nothing about principal left or right ideals defined by minimal idempotents.
Notes
221
if S is any discrete semigroup, we know that any two minimal left ideals in 9S are both homeomorphic and isomorphic by Theorems 1.64 and 2.11. However, the maps which usually define homeomorphisms between minimal left ideals are different from those which define isomorphisms. It may be the case that one minimal left ideal cannot be mapped onto another by a continuous isomorphism. It is tantalizing that we do not know the answer to either of the following questions.
Question 10.41. Are there two minimal left ideals in ,BN with the property that one cannot be mapped onto the other by a continuous homomorphism? Question 10.42. Are there two distinct minimal left ideals in ,8N with the property that one can be mapped onto the other by a continuous homomorphism?
Notes Most of the result of Section 10.1 are from [33], results of collaboration with V. Bergelson and B. Kra. Theorem 10.8 is due to J. Baker and P. Milnes in [10]. Theorem 10.18 is from [228] and answers a question of E. van Douwen in [80].
Most of the results in Section 10.3 were proved in collaboration with A. Maleki in [1801.
Chapter 11
The Rudin-Keisler Order
Recall that if S is a discrete space, then the Rudin-Keisler order
for which f (q) = p. Also recall that if p, q E fS, we write p
reflexive and transitive, it is not anti-symmetric. The relation
^RK The Rudin-Keisler order has proved to be an important tool in analyzing spaces of ultrafilters. It shows how one ultrafilter can be essentially different from another. There are deep and difficult theorems about the Rudin-Keisler order which we shall
not attempt to prove in this chapter. Our aim is to indicate some of the connections between the Rudin-Keisler order and the algebraic structure of flS. If S is a discrete semigroup, one would expect that the Rudin-Keisler order on fi S would have little relation to the algebra of P S, because any bijective mapping from one subset of S to another induces a Rudin-Keisler equivalence between ultrafilters, and a bijection may have no respect for algebraic structure. Nevertheless, there are algebraic properties which are related to the Rudin-Keisler order. For example, if S is countable and cancellative, an
element q E S* is right cancelable in f3S if and only if p
Exercise 11.0.2. Let S be a countable discrete space and let p E
S. Show that
{q E PS : q
Exercise 11.0.3. Let S be a discrete semigroup and let a E S and P E PS. Show that
ap
Exercise 11.0.4. Let S be a countable discrete space and let p be a P-point in S*. If q E S* satisfies q
11.1 Connections with Right Cancelability
223
11.1 Connections with Right Cancelability In this section we shall explore the connections between the relation
Definition 11.1. Let S be a discrete space and let p, q E PS. The tensor product p® q of p and q is defined by
p®q={AcSxS:{s:{t:(s,t)EA}Eq}E p}. The reader is asked to show in Exercise 11.1.1 that p ® q is an ultrafilter on S x S. If A C S x S, then A E p ®q if and only if A contains a set of the form ((s, t) : s E P and t E Qs }, where P E p and Q,t E q for each s E P. Note that, if s, t E S, we have s ® t = (s, t) where, as usual, we identify the point (s, t) with the principal ultrafilter it generates. If we choose any bijection 4) : S x S -+ S, we can regard p ®q as being an ultrafilter
on S by identifying it with 4)(p (9 q), where 4 : fi(S x S) - CBS is the continuous extension of 0. In particular, if s, t E S, then s 0 t can be identified with the element 4)(s, t) of S. In this section, we shall sometimes think of ® as being a binary operation on S, assuming that a bijection 0 has been chosen. Of course, different bijections will result in p ® q being identified with different elements of PS. These will, however, all be
Rudin-Keisler equivalent. So it does not matter which one we choose for studying properties of the relation
Lemma 11.2. Let S be a discrete space and let p, q E S*. Then p
q
fft(P (9 q) = P and ti2(P ®q) = q. Sop
Lemma 11.3. Let S be a discrete space and let p, q E fiS. Then
p ®q = lim lim (s, t) s-* p t-.q
where s and t denote elements of S and the limits are taken to be in fl(S x S). Proof. This is Exercise 11.1.2. Recall that in Section 4.1, we extended an arbitrary binary operation on S to 14 S.
Lemma 11.4. Let S be a discrete space and let * be a binary operation defined on S. Let p, q, x, y E PS. If X
11 The Rudin-Keisler Order
224
Proof Let f : S , S and g : S -+ S be Functions for w hick f(p) = x and g(q) = y. We define h : S x S -). S by h(s, t) = f (s) * g(t). Then
h(p (& q) = lim lim h(s, t) = lim lim f (s) * g(t) = T (p) * g(q) = x * y. s-*pt->q s->pt- q Corollary 11.5. Let S be a discrete space and let * be a binary operation defined on S. For every p, q E 8S, we have p * q
defined by h(s,t)=s*t,then h(p(& q)=p*q. Proof. The proof is the same as the proof of Lemma 11.4, with f and g taken to be the identity maps.
Corollary 11.5 shows that, for any p, q E fS, p ® q
(1) p * q -RK p ®q (2) There is a set D E p for which D * q is strongly discrete and s * q # t * q whenever s and t are distinct members of D. Proof. (1) implies (2). If p*q " RK p®q, there is a set A E p®q on which the mapping (s, t) r-). s * t from S x S to S is injective (by Corollary 11.5 and Theorem 8.17.) We
may suppose thatA = WAS) xES),where D E pandEs E gforeach s E D. Then for each s E S, cfps(s * Es) is a neighborhood of s * q in S. If s and s' are distinct elements of D, s * Es and s' * Es, are disjoint, and so cfps(s * Es) and cPSs(s' * Es,) are disjoint. (2) implies (1). Suppose that for some D E p, D * q is strongly discrete and s * q ,{ t * q whenever s and t are distinct members of D. Then, for each s E D, there
is a neighborhood Us of s * q in OS such that Us fl Us, = 0 whenever s and s' are distinct elements of D. For each s E D, we can choose ES E q for which s * Es C U.
If A = s
x Es), then A E p ® q and, since s * Es fl s' * Esc = 0 when
s' and s * t # s * t' when t 0 t', the mapping (s, t) H s * t is injective on A. So
p*q ~RK p®q Corollary 11.7. Let S be a countable left cancellative discrete semigroup and let q be a right cancelable element of PS. Then pq ~RK p ® q for every p E fiS.
Proof. By Theorem 8.11, Sq is strongly discrete in ,S and sq 0 tq when s ¢ tin S, and so our claim follows from Theorem 11.6.
11.1 Connections with Right Cancelability `We are now able to
.
225
establish the equivalence of statements (1), (4) and (5) of Theorem.
8.18 under weaker hypotheses.
Theorem 11.8. Suppose that S is a countable cancellative discrete semigroup and that q e S*. Then the following statements are equivalent.
(1) q is right cancelable in 8S. (2) p
(3) q
there exist a E S and p E CBS\{a} such that aq = pq. By Lemma 6.28, p 0 S. We have aq "RK q by Exercise 11.0.3. So q pq, a contradiction. Theorem 11.9. If S is an infinite discrete space, there are no elements of S* which are 22K maximal in the Rudin-Keisler order. In fact, if ISI = K, every element of S* has distinct Rudin-Keisler successors.
Proof. We may suppose that S is a group, because there are groups of cardinality K. (The direct sum of K copies of Z2 provides an example.) It then follows from Theorem 6.30 that there is a subset P of S* such that I P I = 22K,
ap # by when a i4 b in S and Sp is strongly discrete in 1S for every p E P, and (jS)p fl (l4S)q = 0 whenever p and q are distinct elements of P. By Lemma 11.2 and Theorem 11.6, x
Theorem 11.10. Let S be an infinite discrete cancellative semigroup. There is no element of S* whose Rudin-Keisler predecessors form a subsemigroup of 8S.
Proof. Let p E S* and let L = {x E ,BS : x
Theorem 11.11. Let S be an infinite discrete left cancellative semigroup. Let D C S and let q E S*. Suppose that Dq is strongly discrete in PS and that sq # tq whenever s and t are distinct elements of D. Suppose also that x, y E S*, P E D fl S*, x
11 The Rudin-Keisler Order
226
Corollary 11.12. Let S be a discrete countable left cancellative semigroup and let q be a right cancelable element of 6S. Then, for every x, p E S*, X
Proof By Theorem 8.11(1), Theorem 11.11 applies with y replaced by q. We now prove a converse to Corollary 11.12.
Theorem 11.13. Let S be a countable cancellative semigroup. Let q be a right cancelable element of fiS and let p E fl S satisfy p
Proof Let f : S -* S be a function for which f (yp) = xq. If q E S, then p E S and sox --RK xq _
Let B = (b E S : f (bp) E Sq). For each b E B, there is a unique c E S for which f (bp) = cq, by Lemma 6.28. So we can define a function g : S -* S by putting g(b) = c if b E B, and extending g arbitrarily to S\B. We may suppose that g(y) # x,
since otherwise x
Bay.
We now claim that
(*) for each X E x and each Y E y, if X C S\g[V] and Y C V, then aq = f (zp) for some a E X and some z E Y.
To see this, suppose instead that (*) fails and pick X E x and Y E y witnessing this failure. Now xq E ct(Xq) fl cZ f [Yp] so by Theorem 3.40 and the assumption that (*) fails, we must have wq = f (bp) for some w E X and some b E Y. Then wq
Let A = {a E S : aq E f [(fS)p]}. If A 0 x, then letting X = S\(A U g[V]) and Y = V one obtains a contradiction to (*), and so A E x. For each a E A, let Ca = {z E f8S : aq = f(zp)}. Then C. = (f o pp)-1[{aq}] so Ca is closed. We claim that for each a E A, Ca f1ci (UbEA\(a) Cb) = 0. Indeed, by Theorem 8.11, aq 0 ct{bq : b E A\{a}} so pick a neighborhood W of aq which misses {bq : b E A\(a)). Given Z E Ca, pick D E z such that f [Dq] C W. Then D fl UbEA\(a) Cb = 0.
Thus one can pick for each a E A a clopen set Va such that Ca C Va and Va fl (UbEA\(a) Cb) = 0. Let A be sequentially ordered by <. For the first element ao of A, let Uao = V. For later elements, let Ua = Va\ Ub
clopen,CaCU,,,and UaflUb=0if a#b. Define h : S -). S by h(s) = a ifs E Ua, defining h arbitrarily on S\ UaEA Ua. We claim that h(y) = x. Suppose instead that h(y) # x and pick X E x such that h(y) 0 X and pick Y E y such that h[Y] fl X = 0. We may assume that X C S\g[V]
11.1 Connections with Right Cancelability
227
and Y C V. Then by (*), aq = f (zp) for some a E X and some z E Y. So Z E Ua and thus h(z) = a, contradicting the fact that h[Y] f1 X = 0. -
Corollary 11.14. Let S be a countable cancellative semigroup and let p E 8S. The following statements are equivalent.
(1) p is right cancelable in PS. (2) For every x, y E ,BS, xp
Theorem 11.15. Let S be a countable discrete space and let * be a left cancellative binary operation defined on S. Suppose that p, q E S and that A E p has the property that A * q is discrete in fi S and that a * q # b * q whenever a and b are distinct elements
of A. Then p * q
Let it, and n2 denote the projection maps of S x S onto S. Then >zi (x (9 q) = x and n2 (x (9 q) = q for every x E 14 S.
Let f : S - S x S be a function for which f (q * p) = p ® q. If P E p and Q E q, p ® q belongs to each of the sets Theorem 3.40 that either
0 q) and cf(f [Q * p]). It follows from
(i) T (b * p) = x ® q for some b E Q and some x E P or (ii) a ® q = T (z * p) for some a E P and some z E Q. Now if (i) holds for some P and Q, then n2 o 7o Ab is a continuous extension of the function s i- 7r2 (f (b * s)) taking p to q so that q
228
11 The Rudin-Keisler Order
Let B =_{b E S : )Fj( (b * p)) E S) and define.g : S - S by stating that g(b) = n, ( f (b * p)) if b E B and defining g arbitrarily on S\B. We claim that W(q) = p, so suppose instead that W(q) # p. Pick P E p and Q E q such that W[U] f1 P = 0. By (ii) pick some a E P and some z E Q such that a ®q = f (z * p), Thus a = ii (a (9 q) = n j (f (z * p)) and a is isolated so there is some R E z with n, [ f [R * p]] = {a}. Pick b E R fl Q. Then ail (7(b * p)) = a sob E B and thus g(b) = a, a contradiction. Corollary 11.16. Let S be a countable discrete space and let p, q E S*. If p ® q
Proof The hypotheses of Theorem 11.15 are clearly satisfied if ® is regarded as a binary operation on S via a fixed bijection between S x S ;nd S. Corollary 11.17. Let S be a discrete countable cancellative semigroup and let q be a right cancelable element of S*. If p E S* and pq <_RK qp, then p and q are Rudin-Keisler comparable. Proof. Theorem 11.15 applies by Theorem 8.11(1). If S is an infinite discrete semigroup, we know that $S is rarely commutative. In the following corollary, we show something stronger than non-commutativity in the case in which S is countable and cancellative. In this case, there are elements p and q of S* for which pq and qp are not Rudin-Keisler comparable.
Corollary 11.18. Let S be a countable infinite discrete cancellative semigroup. There are elements p, q E S* for which pq and qp are not Rudin-Keisler comparable. Proof By Theorem 6.36, there are elements p and q of S* which are not Rudin-Keisler comparable. We may suppose that p and q are right cancelable in $S, because there is an infinite subset T of S with the property that every element of T * is right cancelable in $S (by Theorem 8.10). Since S can be mapped bijectively onto T, p and q are RudinKeisler equivalent to elements in T*. Our claim then follows from Corollary 11.17.
Exercise 11.1.1. Let S be a discrete space and let p, q E CBS. Show that p ® q E
#(S x S). Exercise 11.1.2. Prove Lemma 11.3. Exercise 11.1.3. We cannot normally use the most obvious function to establish a re-
lation of the form p ®q
p ® q.
Exercise 11.1.4. Let S be an infinite cancellative semigroup. Show that there is no element p in S* for which {q E $S : q ^ RK p} is a subsemigroup of $S. (Hint: Consider the proof of Theorem 11.10.)
11.2 Connections with Left Cancelability in N*
229
11.2 Connections with Left Cancelability in N* Because PS is a right topological semigroup, left cancelability in ,6S is more difficult to handle than right cancelability. As we have seen, it is easy to derive connections between the Rudin-Keisler order and right cancelability. However, whether the analogous results hold for left cancelability remains a difficult open question, even in flN. For example, we do not know whether q
Recall that, for each k E N, the natural homomorphism hk : Z -+ 7Gk has a continuous extension Wk : fl7G -+ Zk which is also a homomorphism (by Theorem 4.8).
If x, y E 6N and k E N, we may write x = y (mod. k) if hk(x) = hk(y).
Theorem 11.19. Suppose that p E N* and that there exists A E p with the property
that x = p (mod k) for every x E A* and every k E N. Then p
hk(a) = hk(b) for every k E N and so a = b. We may replace A by A\{a}, if such an element a exists. So we may suppose that A + q is disjoint from A* + q. We claim that a + q 0 b + q whenever a b in A and A + q is strongly discrete in fN. Otherwise we should have b + q = x + q for some b E A and some x E A\{b}, by Theorem 8.11. Now this cannot hold if x E A*; neither can it hold if x E A (by Lemma 6.28). Thus our theorem follows from Theorem 11.6 and Lemma 11.2. Comment 11.20. The set of ultrafilters pin N* which satisfy the hypotheses of Theorem 11.19 contains a dense open subset of N*. To see this, let U be any non-empty clopen
subset of N* and let q E U. For each k E N, let Vk = {x E ON : x = q (mod k)}. Then Vk is a clopen subset of fN. Since q E U fl n i Vk, this set is a non-empty Gs-subset of N* and therefore has a non-empty interior in N* (by Theorem 3.36). Any ultrafilter p in the interior of U fl nk__, Vk satisfies the hypotheses of Theorem 11.19.
Theorem 11.21. Let (an)n°, be an infinite increasing sequence in N. Suppose that either of the two following conditions is satisfied:
(i) For every n E N, an+i is a multiple of an. (ii) For every m, n E N with m # n, an is not a multiple of an,.
If A = {an : n E N} and P E A*, then p
Proof. We first consider the case in which q E nn° , nN. For each n E N, let Bn = nk±i akN so that Bn E q. If condition (i) is satisfied, we define f : N - N by stating that
f(n) = max{a, : an,In)
if an, In for some m,
11 The Rudin-Keisler Order
230
defining f arbitrarily otherwise. For every n E N and every b E B, f (an + b) = a,,. So am + B. and a + B. are disjoint if 'm i4 n. Since am + Bm E am + q, the set A + q is strongly discrete in flN and our claim follows from Theorem 11.6 and Lemma 11.2. If condition (ii) is satisfied, we define g : N -* N by stating that g(n) = min{am : am I n}
if am n for some m,
defining g arbitrarily otherwise. For every n E N and every b E Bn, g(an + b) = a,,, So we can again deduce that A + q is discrete and that Theorem 11.6 and Lemma 11.2 apply.
Now let q be an arbitrary element of N*. For each n E N, we can choose bn E N satisfying bn + q = 0 (mod k) f o r every k E { 1, 2, ... , n). To see this, let n E N be given. Then {c E N : c - q (mod k) for every k E (1, 2, ... , n} } E q so pick c in this set. Pick M E N such that n!m > c and let bn = n!m - c. Let B = {bn : n E N) and let r E B*. Then q + r E nn t nN since for each k, {b : b + q = 0 (mod k)} E r. By what we have already proved, with q + r in place of q, we know that a + q + r qE
b + q + r when a and b are distinct members of A (by Lemma 6.28) and A + q + r is discrete in fN. This is equivalent to stating that a + q + r zA x + q + r if a E A and X E A\{a}, by Theorem 8.11. This implies that a + q 0 x + q for every a E A and every x E A\{a} and hence that A + q is strongly discrete. So Theorem 11.6 and Lemma 11.2 apply once again.
Comment 11.22. By Ramsey's Theorem (Theorem 5.6), every infinite sequence in N contains an infinite subsequence satisfying the hypotheses of Theorem 11.21. So the set of ultrafilters p E N* to which this theorem applies also contains a dense open subset of N*.
Theorem 11.23. Let S be a discrete countable cancellative semigroup and let p be a P-point in S*. Then p
by Lemma 6.28. Let S' = (a E S : aq # pq). So S' = S or IS\S'I = 1. For each b E S', let Cb = {x E S* : xq = bq }. Then Cb is a compact subset of S*. Since p 0 Cb and since p is a P-point in S*, p ¢ UbES' Cb. It follows that there is a member V of p such that V c S' and v fl UbEs, Cb = 0. We claim that for any b E V and any x E V\{b}, we have bq 0 xq. To see this, note that the equation bq = xq cannot hold if x E S by Lemma 6.28. Also if x E S*, then x 0 Cb so bq # xq. It follows from Theorem 8.11 that Vq is strongly discrete in fS. So Theorem 11.6 and Lemma 11.2 apply.
Comment 11.24. We observe that the ultrafilters p E N* to which Theorems 11.19, 11.21, and 11.23 apply are left cancelable in 6Z. This is a consequence of Corollaries 8.36, 8.37, and 8.38 and Theorem 8.39.
Question 11.25. If p is a left cancelable ultrafilter in N*, must it be true that p
p+gandq
11.3 Further Connections with the Algebra of 8S
231
11.3 Further Connections with the Algebra of fiS We remind the reader that a family F' of subsets of a given set is said to be almost disjoint if the intersection of any two distinct members of F is finite. Theorem 11.26. Let A be a subset of N*. If IAA < c, the elements of A have a common RK -successor in IIII.
Proof. We index A as (px)xEtt, allowing repetitions if JAI < c. Let (Ex)XER be an almost disjoint family of infinite subsets of {2" : n E N}. (Such a family exists by Theorem 3.56.) For each x E R, we choose qx E Ex fl N* such that qx ARK Px. For each F E J'f (H), where .Pf (H) denotes the set of non-empty finite subsets of ]R, we put SF = EXEF qx, with the terms in the sum occurring in the order of increasing indices. We order .Pf (R) by set inclusion and choose q to any limit point of the net (SF) FE.Pf(R) in ON, noting that q E H. Recall that supp(n) is the binary support of n. For each x E H, we define fx : N -* N by stating that
f, (n) = min(Ex fl {2' m E supp(n)}) if Ex fl (2m : m E supp(n)) * 0 and defining f, (n) arbitrarily otherwise. We shall show that fx (q) = qx for each x E R. We suppose that this equation does
not hold. Then we can choose B E qx such that fx [B] ¢ q, and we can choose Q E q such that fx - t [B] fl Q = 0. Since q is a limit point of the net (SF)FE.Pf(R), there exists F E .Pf(R) such that x E F and Q E SF. We can choose a disjoint family (By)yEF of subsets of N satisfying
Bx C Ex fl B and, for every y r= F, By c Ev, By e qy, and By fl Ex = 0 if y # x. Let M denote the set of all integers m of the form m = EYEF ny, where ny E By for each y E F. We observe that this expression for m is unique, since each ny is a power of 2, and that fx(m) = nx E B. Suppose that F is arranged in increasing order. By Exercise 4.1.6, M E E EF qy = SF. Thus we can choose n E M fl Q. Since fx (n) E B, it follows that n E fx 1[B] fl Q,
contradicting our assumption that this set is empty. So fx (q) = qx and we have Px "RK qx ARK q for every x E R. Corollary 11.27. Let A be a subset of N* with I A 1 < c. The elements of A have a common
Proof. By Theorem 11.26, the elements of A have a common
pE{n!:nEN}f1N*such that p is right cancelable in ON, by Corollary 8.38. So P
u E N*, by Theorem 11.8. We also have p
232
11 The Rudin-Keisler Order
Theorem 11.21. We can choose u to lie in any given left ideal or in airy given right ideal of ON. Let L= ON + p and R= p+ ON. Then x
Corollary 11.28. Every minimal left ideal of ON contains an increasing
Theorem 11.29. There are at most c elements of N* whose <_ R K -successors form a subsemigroup of ON.
Proof. Suppose that p E N* has the property that its
thus M + u = L. Consequently p
11.4 The Rudin-Froilk Order There are other useful topological order relations on spaces of ultrafilters, one of these being the Rudin-Froltlc order.
Definition 11.30. Let S be a countable discrete space. The Rudin-Frolik order C on OS is defined by stating that, if p, q E ,BS, p C_q if there is an injective function f : S $S for which f [S] is discrete in ,BS and f (p) = q.
Ifp,q E PS, we may write pCgif pCgand q ¢ p. It is immediate that the relation C is reflexive, and we shall see in the next lemma that it is transitive. It is not, however, anti-symmetric. Lemma 11.31. Let S be a countable discrete space. The relation C on 8S is transitive.
Proof. Let g : S OS be an injective function for which g[S] is discrete in fiS, and let X be a countable discrete subset of PS. Then g[X] is also discrete, because it follows from Exercise 11.4.1 that g defines a homeomorphism from 8S onto cP,8s g[S].
Suppose now that p, q, r E ,S and that p C q and q C r. So there are injective functions f and g from S to 16S for which f [S] and g[S] are discrete, T (p) = q,
11.4 The Rudin-Frolffc Order
233
and g(q) = r. Then g o f is injective by Exercise 11.4.1, g[ f [S]Ii is discrete, and
(8 f)(P) = g(.f(P)) = r. So p r- r. We now observe that the assertion that p C q is stronger than the assertion that
p:5RKq. Lemma 11.32. Let S be a countable discrete space and suppose that p, q E /3S satisfy
pCq.Then p
We now see that there are elements q E S* which are not right cancelable, but have the property that p C pq for every p E S*. We remind the reader that an idempotent q E S* is strongly right maximal if the equation xq = q has the unique solution x = q in S*. (And strongly right maximal idempotents exist by Theorem 9.10.)
Lemma 11.33. Let G be a discrete countable group and let q be a right maximal idempotent in G*. Let C = {x E G* : xq = q}. Then, for every p E PG, either p E (fG)C or there exists P E p such that Pq is discrete in ,PG. Proof. Suppose that p V (/3G)C. Since C is finite (by Theorem 9.4), (,G)C is compact. So we can choose P E p for which P fl (/3G)C = 0. We claim that Pq is discrete. If we assume the contrary, then aq E c2((P\{a})q)
for some a E P. This implies that aq = xq for some x E P\{a}. Then ax E C so x E aC, contradicting our choice of P.
Theorem 11.34. Let G be a discrete countable group and let q be a strongly right maximal idempotent in G*. Then p C pq for every p E G*. In fact, for every p E G*,
either p C pq and q
Exercise 11.4.1. Let S be a countable discrete space and let f : S -> /3S be an irective function for which f [S] is discrete in 14 S. Show that f is also injective, where
f : PS -- /3S denotes the continuous extension of f. (Hint: let x and y be distinct elements of PS and let X and Y be disjoint subsets of S for which X E x and Y E Y. Observe that f [X ] fl T [-Y1 = f[X] n f [Y] = 0 and apply Theorem 3.40.)
Exercise 11.4.2. Let S be a countable discrete space and let p E /3S. Show that
{qE/3S:gCp}l
234
11 The Rudin-Keisler Order
Exercise 11.4.3. Let p, q E #N. If p EZ q and q
p, we shall say that p and q are
Rudin-Frolic equivalent. Show that p and q are Rudin-Froh1c equivalent if and only if they are Rudin-Keisler equivalent. (Hint: Use Theorem 8.17 and Lemma 11.32)
Exercise 11.4.4. Let p be a weak P-point in N*. Show that, for any q E N*, q C p implies that p
Exercise 11.4.5. Let S be a discrete countable semigroup and let p, q E S*. Suppose
that there is a set P E p such that Pq is discrete in /BS and aq 0 bq whenever a and b are distinct elements of P. Show that p C pq. (Hint: Once you have shown that p C pq, apply Lemma 11.2, Theorem 11.6, and Lemma 11.32 to deduce that one cannot have pq C p.)
Exercise 11.4.6. Let S be a discrete countable semigroup and let q E S* be right cancelable in fl S. Show that p C pq for every p E S*. (Hint: Use Exercise 11.4.5 and Theorem 8.11.) Exercise 11.4.7. Show that every element of N* has 2` distinct Rudin-Froltlc successors. (Hint: use Exercise 11.4.6, Theorem 6.30, and Theorem 8.11.)
Exercise 11.4.8. Let S be a discrete countable cancellative semigroup and let p E S*.
Show that p is right cancelable in /3S if and only if there is an injective function f : S /3S for which f [S] is discrete in /3S and f (q) = qp for every q E S*. (Hint: Use Theorem 8.11. For the sufficiency consider the proof of Lemma 6.37. For the necessity consider pp.)
Notes Some of the results of Sections 11.1 and 11.2 are from [103], obtained in collaboration with S. Garcia-Ferreira. Theorem 11.10 contrasts strongly with the situation that holds for the Comfort order of ultrafilters. Given an ultrafilter p on a set S, a space X is said to be p-compact if and only if whenever f : S -* X, p- lim f (s) exists in X. One then -S
defines p
Notes
235
The Rudin-Keisler order was introduced and studied by M. Rudin in [2111 and [212] and independently by H. Keisler in lectures at UCLA in 1967. The Rudin-Frolik order was implicitly introduced by Z. Froltlc in [95] and some of its basic properties were established by M. Rudin in [212]. There are several very interesting properties of the Rudin-Keisler and Rudin-Frohlc
orders that we have not presented because they do not relate to the algebra of OS. Included among these is the equivalence of the following three statements for p E fS: (a) p is minimal in the Rudin-Keisler order. (b) p is a selective ultrafilter. (That is, given any f : S S, there is some A E p such that the restriction of f to A is either one-to-one or constant.) (c) p is a Ramsey ultrafilter. (That is, for every n E N and every partition { P0, Pi } of [S]", there exist some A E p and some i E (0, 1) such that [A]' c Pi.) Another interesting result is that the Rudin-Keisler equivalence classes (which are the
same as the Rudin-Frolic equivalence classes by Exercise 11.4.3) of Rudin-Frolic predecessors of a given element of fiN are linearly ordered by
Chapter 12
Ultraffiters Generated by Finite Sums
We have seen in Theorem 5.8 that if p is an idempotent in (ON, +) and A E p, then A in N. However, it is not necessarily contains FS((x,,)n° 1) for some sequence 1) E p. We investigate in this chapter the existence and properties true that of ultrafilters satisfying the requirement that each A E p contains some 1) with 1) E p. 1
12.1 Martin's Axiom We shall have several proofs in this chapter for which Martin's Axiom is a hypothesis. Since we expect that Martin's Axiom may be unfamiliar to many of our readers, we include in this section an elementary introduction. We use the terminology which is standard when dealing with Martin's Axiom, although some of it may seem strange at first glance.
Definition 12.1. A partially ordered set is a pair (P, <) where P is a nonempty set and < is a transitive and reflexive relation, that is a partial order. Notice that a partial order is not required here to be anti-symmetric. (However, the partial orders that we will use are anti-symmetric.)
Definition 12.2. Let (P, <) be a partially ordered set. A filter in P is a subset G of P such that
(1) G 0 0, (2) for every a, b E G, there is some c E G such that c < a and c < b, and (3) for every a E G and b E P, if a < b, then b E G. Thus, if S is a set, a filter on S is a filter in the partially ordered set (,P(S)\{0}, g).
Definition 12.3. Let (P, <) be a partially ordered set. A subset D of P is dense in P if and only if for every a E P there is some b E D such that b < a.
12.1 Martin's Axiom
237
Thus, in more usual terminology, a subset of P is dense if and only if it is cofinal downward.
Definition 12.4. Let (P, <) be a partially ordered set. (a) Elements a and b of P are compatible if and only if there is some c E P such that c < a and c < b. (Otherwise they are incompatible.) (b) (P, <) is a c.c.c. partially ordered set if and only if whenever D is a subset of P consisting of pairwise incompatible elements, one has that D is countable. In the above definition, "c.c.c." stands for "countable chain condition". However, it actually asserts that all antichains are countable.
Definition 12.5. (a) Let K be an infinite cardinal. Then MA(K) is the assertion that whenever (P, <) is a c.c.c. partially ordered set and D is a collection of at most K dense subsets of P, there is some filter G in P such that for all D E £, G fl D # 0. (b) Martin's Axiom is the assertion that for all cardinals K, if w < K < c, then MA (K ).
We first note that Martin's Axiom follows from the continuum hypothesis.
Theorem 12.6. MA(co) is true.
Proof. Let (P, <) be a partially ordered set and let {An : n E N} be a set of dense subsets of P. Choose a, E Ai. Inductively, having chosen an E An, pick an+1 E An+I such that an+t < an. Let G = {b E P : there is some n E N with an < b}.
Notice that, for the truth of MA(w) one does not need that (P, <) is a c.c.c. partial order. However, the next theorem (together with Exercise 12.1.1) shows that the c.c.c. requirement cannot be deleted.
Theorem 12.7. There exist a partially ordered set (P, <) and a collection £ of w, dense subsets of P such that no filter in P meets every member of D.
Proof. Let P = if : there exists A E J" f(N) such that f : A -- wi ). For f, g E P agree that f < g if and only if g C f. For each a < w1, let Da = (f E P : a is in the range of f) and let `a) = {Da : a < wi }. Trivially each Da is dense in P. Suppose now that we have a filter G such that G fl Da 0 0 for each a < a),. Let g = U G. We claim that g is a function. To this end, let a, b, c be given with (a, b) E g and (a, c) E g. Pick f, h E G such that (a, b) E f and (a, c) E h and pick k E G such that k < f and k < h. Then (a, b) E k and (a, c) E k so b = c. But now g is a function whose domain is contained in N and whose range is all of wi, which is impossible. We now illustrate a typical use of Martin's Axiom. The partially ordered set used is similar to those used in succeeding sections. Recall that a set ,A of subsets of a set S is an almost disjoint family if and only if whenever A and B are distinct members of A, IA fl BI < w.
238
12 Ultrafilters Generated by Finite Sums
Definition 12.8. Let S be a set. A set A of subsets of S. is a maximal almost disjoint family if and only if it is an almost disjoint family which is not properly contained in any other almost disjoint family. Theorem 12.9. Let A be a maximal almost disjoint family of infinite subsets of a set S. Then A is not countably infinite.
Proof This is Exercise 12.1.2. As a consequence of Theorem 12.9, one sees immediately that the continuum hypothesis implies that any infinite maximal almost disjoint family of subsets of N must have cardinality c. We shall see that the same result follows from Martin's Axiom.
Lemma 12.10. Let A be a family of infinite subsets of N with the property that, for every F E pf(A), N\ U F is infinite. Let CAI = K. If MA(K), then there is an infinite subset B of N such that A n B is finite for every A E A.
Proof Let Q = {(H, F) : H E 9f(N) and F E ,Pf(A)}. Order Q by agreeing that (H', .T'') < (H, F) if and only if
(1) HCH', (2) F S .T'', and
(3) (H'\H) n OF = 0. It is routine to verify that < is a partial order on Q (which is in fact antisymmetric).
Next notice that if (H, F) and (H', F') are incompatible elements of Q, then
H 56 H' (since otherwise (H, F U F) < (H, F) and (H, F U F') < (H', F') ). Thus, since J.Af(N)I = co, it follows that (Q, <) is a c.c.c. partially ordered set.
Now, for each A E A, let D(A) = ((H, F) E Q : A E F). Then given any (H, F) E Q, one has (H, F U (A)) E D(A) and (H, F U (A)) < (H, F). So D(A) is dense.
For each n E N, let E(n) = ((H, F) E Q : H\{1, 2, ... , n} # 0). Given any (H, F) E Q, we can choose m > n such that m U F. Then (H U {m}, F) E E(n) and (H U {m}, F) < (H, F). So E(n) is dense in Q By MA(K), there is a filter G in Q such that G n D(A) 96 0 for each A E A and G n E(n) 0 0 for each n E N. Let B = U{H : there is some .T" with (H, .T') E G}. Since G n E(n) 0 for each n E N, it follows that B is infinite. We shall show that AnB is finite forevery A E A. To this end, let A E A and pick (H, F) E GnD(A). We claim that A n B C H. Let X E A n B and pick some (H', F') E G such that x E H'.
Since (H, F) E G and (H', F') E G, pick (H", F") E G such that (H", F") _< (H, F) and (H", F") < (H', F'). Then X E H' C H", (H"\H) n U F = 0, and A E F. Sox E Has required.
12.1 Martin's Axiom
239
Corollary 12.11. Let K < c and let (Fa)a
Assuming MA(K), if Ua
Proof. Let x E N*\ Ua
of N* such that F. a D. and x 0 D. We have Da = A* for some Aa C N by Theorem 3.23. Now Ua
Corollary 12.12. Let K < c and let (Ga)a
Proof. This follows immediately from Lemma 12.10.
Corollary 12.14. Let A be an infinite maximal almost disjoint family of subsets of N and assume Martin's Axiom. Then CAI = c. Proof By Theorem 12.13 one cannot have IAA < c and, since A S; J" (N), I A 1 < c. We also see as a consequence of the next corollary that Martin's axiom is the strongest assertion of its kind which is not simply false.
Corollary 12.15. MA(c) is false.
Proof. Let £ be an infinite set of pairwise disjoint infinite subsets of N. A routine application of Zom's Lemma yields a maximal almost disjoint family A with S C A. Necessarily CAI < c so Theorem 12.13 together with the assumption of MA(c) yield a contradiction.
Exercise 12.1.1. Prove that the partially ordered set in Theorem 12.7 is not a c.c.c. partially ordered set.
Exercise 12.1.2. Prove Theorem 12.9.
Exercise 12.1.3. Recall that a P-point in a topological space is a point x with the property that any countable intersection of neighborhoods of x is again a neighborhood
of x. It is a consequence of Theorems 12.29, 12.35, and 12.36, that Martin's Axiom implies that there are P-points in N*. Provide a direct proof of this fact. (Hint: Let be an enumeration of all countable families of clopen subsets of N*. Use Corollary 12.11 to construct an increasing family (Da)a« of proper clopen subsets of N* with the property that, for every a < c, either n F. a Da or N*\ n F. C Da.)
12 Ultrafilters Generated by Finite Sums
240
Exercise 12.1.4. Show that Martin's Axiom implies that N* satisfies the following strong form of the Baire Category Theorem: If F is a family of nowhere dense subsets of N* with IFI < c, then U F is nowhere dense in N*. (Hint: Let K < c and let (Fa)a
12.2 Strongly Summable Ultrafilters - Existence We deal in this section with the relationships among three classes of ultrafilters on N. One of these classes, namely the idempotents, we have already investigated extensively.
Definition 12.16. Let P E ON. (a) The ultrafilter p is strongly summable if and only if for every A E p, there is a sequence (xn)' 1 in N such that FS((xn)n° 1) C A and FS((xn)n° 1) E p. (b) The ultrafilter p is weakly summable if and only if for every A E p, there is a sequence (xn)n° 1 in N such that FS((xn)n° 1) C A.
Recall that r =c2{pc $N:p+p=p}. Theorem 12.17. The set r = {p EON : p is weakly summable}. In particular, every idempotent is weakly summable.
Proof. This is an immediate consequence of Theorem 5.12. We set out to show that every strongly summable ultrafilter is an idempotent.
Lemma 12.18. Let p be a strongly summable ultrafilter. For each A E p there is an increasing sequence (xn)n in N such that FS((xn)n_1) C A and for each m E N, 1
FS((xn)n°_m) E P.
Proof. For i E [1, 2}, let Ci = (3n (3k +i) : n, k E (0}. (That is, Ci is the set of positive integers whose rightmost nonzero ternary digit is i.) Pick i E (1, 2) such that Ci E P.
Define f : N -+ a) by f(3n(3k+j)) =nforalln,k E wand j E 11, 2). Notice that if x, y E Ci and f (x) = f (y), then x + y 0 Ci. Consequently, if {x, y, x + y} C Ci, then f (x + y) = min{ f (x), f (y)}. Let A E p and pick a sequence (x)1 such that FS ((xn )n_ 1) E p and FS((xn)O 1) C A n Ci. By the observation above, the sequence (xn)n1 is one to one, so we may presume that it is increasing. Let m E N and suppose that 0 p. Then m > 1. Now
FS((xn)n'1) = FS((xn)n -1) U UFc{1.2,
m
11(EneFxn +FS((xn)
m))
12.2 Strongly Summable Ultrafilters - Existence
241
where EnEO xn = 0. Since FS((xn)n_11) is finite and, by assumption, EnEO xn +
p, pick F with 0 0- F C (1, 2, ..., m - 1) such that EnEF Xn +
FS((xn)n_m)
FS((xn)n°_m) E p. Given any y E EnEFxn + FS((xn)rt°_,n), one has that for some G E J'f({m, m +I, m +2, ...}), y = EnEFxn + EnEG Xn, where {EnEFxn, EnEG xn, EnEF xn + EnEG xn} C Ci
so, as observed above, f (y) < f (EnEF xn)-
Pick a sequence (yn)n° with FS((yn)R_1) C EnEFxn + FS((Xn)n_m) and FS((yn)n_1) E p. Then for each k E N, f (yk) < f(EnEFXn) so pick some k 1
such that f (yk) = f (ye). Then yk + ye 0 Ci, a contradiction.
Theorem 12.19. Let p be a strongly summable ultrafilter. Then p = p + p. Proof. Let A E p and pick by Lemma 12.18 a sequence (xn)n
in N such that FS((xn)n° 1) c A and for each m E N, FS((xn)nm) E p. It suffices to show that 1
FS((xn)n° 1) C {y E N : -y + A E p}, so let y E FS((xn)' 1) and pick F E J'f(N) such that y = EnEFxn. Let m = max F + 1. Then FS((xn)n m) E p and FS((Xn)n°_m) C -y + A.
Lemma 12.20. Let p be a strongly summable ultrafilter. For every A E p there is a sequence (xn)n° 1 in N such that (1) FS((xn)noo=1) C A,
(2) for each m E N, FS((xn)n°_m) E p, and (3) for each n E N, Xn+1 > 4 En=1 xt.
Proof. Let A E p. For i E {0, 1, 2} let 23n+i + 1, 23n+i + 2 ..
Bi = U 0 0
23n+i+l
- 1}
and pick i E {0, 1, 2} such that Bi E P. Pick by Lemma 12.18 an increasing sequence (xn)n° 1 in N such that FS((xn)n 1) c A f1 Bi and for each m E N, FS((Xn)nm) E P. We claim that (xn)n° is as required, so suppose not and pick the first n such that xn+1 < 4 Et =1 Xt. Pick k E w such that 23k+i < Xn < 23k+i+l Then in fact Eni_1 Xt < 23k+i+1. (If n = 1 this is immediate. Otherwise, x,, > 4 En=,' X, so 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 <
23k+i+l.)
Now Xn+l < 4 Ei_1 xt < 23k+i+3 and Xn+l E Bi so
Xn+1 < 23k+i+1 Since also x,, < Xnt1 we have that 23k+i+l < Xn + Xn+1 < 23k+i+2
so xn + xn+1 0 Bi, a contradiction.
The following easy theorem shows that not nearly all idempotents are strongly summable.
Theorem 12.21. Let P E cB K (,8N, +). Then p is not strongly summable.
12 Ultrafilters Generated by Finite Sums
242
Proof. Suppose that p is strongly summable and pick by Lemma 12.20 a sequence (xn)n_ 1 such that E p, and for each n E N, xn+t > 4 Ef 1 xt. But then FS((xn)0 1) is not piecewise syndetic, contradicting Corollary 4.41. Theorem 12.22. There is a weakly summable ultrafilter which is not an idempotent.
Proof This was shown in Corollary 8.24. Theorem 12.23. The set r is not a semigroup. In fact, there exist idempotents p and q in fN such that p + q 1t F. Proof. This was shown in Exercise 6.1.4.
We now introduce a special kind of strongly summable ultrafilter. Its definition is somewhat complicated, but we shall prove in Section 12.4 that these idempotents can only be written in a trivial way as sums of other ultrafilters. Recall that for y E N we define the binary support of y by y = Enesapp(y) 2n. In a similar fashion, for elements y of FS((t!)°O t) we define the factorial support of y. Definition 12.24. (a) For y E FS((t!)°O 1), fsupp(y) is defined by y = EfEfS pp(y) 0(b) The ultrafilter p E fiN is a special strongly summable ultrafilter if and only if (1) for each m E N, FS((n!)n_m) E p,
(2) for each A E p, there is an increasing sequence (xn)' such that 1
FS((xn)n__t) E p, FS((xn)n° 1) C A, and for each n E N, x,, divides xn+t, and (3) for each infinite L C ICY, there is a sequence (xn )n_ such that FS ((xn )n _ 1) C 1
FS((t!)°O 1), FS((xn)o 1) E p, and L\ U{fsupp(y) : y E FS((xn)n_ 1)} is infinite.
We shall show that Martin's Axiom implies that special strongly summable ultrafilters exist. Indeed, we shall show that Martin's Axiom implies that any family of subsets of FS((n!)n_1) which is contained in an idempotent and has cardinality less than c, is contained in a special strongly summable idempotent.
Definition 12.25. (a) -8 denotes the set of infinite subsets S of N with the property that
m divides n whenever m, n E S and m < n. .8f denotes the set of 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'
Since Q is countable, it is trivial that it satisfies the c.c.c. condition. For each A E 3', let D(A) = {F E Q : F f1 A # 0}. To see that D(A) is dense in Q,
let F E Q and let m = fI F. We can choose a E A n B* n fbeFS(F) (-b + B*) n mN.
Then FU{a}
0forevery A E F. LetX=U(D. We shall show that X has the properties required. To see that X is infinite, let m E N and pick F E (D n D(mN). Then x n mN # 0. Since CD is a filter, any two elements x and y of X belong to a set F E (D. So, if x < y, x divides y. Thus X E 44. Furthermore,
if H is any finite subset of X, H C F for some F E CD and so FS(H) c B. Thus FS(X) C B. Finally, for any A E F, there exists F E CD n D(A) and so x n A # 0.
Lemma 12.27. Let p be an idempotent in N*, let F c p, and let to < K < c. If ITI < K, then MA(K) implies that there exists X E 8 such that FS(,.(X) C nAE. A.
Proof Let g= 'U{mN:mEN}U{B*:BEF}U(-b+B*: BeFandbEB*) and note that by Lemmas 4.14 and 6.6, g c p. Let 3e be the set of finite intersections of members of 9 and note that 3e c p and IJeI < K. Well order F as (Ax)a
(a) FS(Xa) S; Aa and Xa n A # 0 for every A E 3e and (b) if S < a, then IXa\XsI < w. Using condition (b) and the facts that R is closed under finite intersections and that mN E Je for every m E N one sees that
(*) for any finite nonempty ^a) c 3e U {Xa : a < P}, there exist Ac- 3e and a <
with Af1XaC(l;D. In particular, by condition (a), {A n Xa : A E 3e and a < #} has the finite intersection property.
We apply Lemma 12.26, with {A n Xa : A E 3e and a < l3} in place of F and Aj in place of B. Pick WW E 44 such that FS(W,8) C Ap and WW n A n Xa 0 for every A E 3e and every a < P. By the observation (*), (l{ W, n A n Xa : A E N and a < #} # 0 so by Corollary 12.12, there exists an infinite subset Xf of N such that X0* 9 WW n A n Xa for every A E 3e and every a < 0. We may suppose that
244
12 Ultrafilters Generated by Finite Sums
X# C W. We may also suppose that Xp E -3, because x,6 fl mN 0 for every m E N, and so X# contains a set in 4. It is clear that conditions (a) and (b) are satisfied with P in place of a.
We put X = XK. If a < K, X\Xa is finite and so there is some n E N such that FSn(X) C FS(Xa) and thus FS... (X) C FS(Xa) C Aa. Corollary 12.28. Let F be a family of subsets of N contained in an idempotent p for
which FS((n!)n° t) E p. Let L be any infinite subset of N and let co < K < c. If MA(K) and IFI < K, then there exists X E -3 with FS(X) C FS((n!)n° 1) such that FSOO(X) C A for every A E .F and L\ UxEFS(x) fsupp(x) is infinite.
Proof. We may suppose that F is closed under finite intersections. By Exercise 12.2.1, we may suppose that FS ((n !) n m) E F for every m E N.. _ By Lemma 12.27, there exists Y E -8 such that FSQO(Y) C A for every A E Y.
We claim that Y fl FS((n!)' m) # 0 for every m E N. To see this, let m E N. Since FS0O(Y) = nkENFSk(Y) C (FS(n!)n°_m), there exists k E N for which FSk(Y) C FS((n!)n_m). Thus we can inductively choose a sequence (xn)n°1 in Y fl FS((n!)n° 1) with the property that, for each n E N, max fsupp(xn) < in <min fsupp(xn+1) for some in E L. If X = {xn : n E N}, then X has the properties required.
Notice that, while the previous results only use MA(K) for a fixed K < c, the following theorem uses the full strength of Martin's Axiom. Theorem 12.29. Let F be a family of subsets of N contained in some idempotent q E N*
for which FS((n!)n° 1) E q. If IFS < c, Martin's Axiom implies that there is a special strongly summable ultrafilter p for which F C p. Proof. We assume Martin's Axiom. Let (Yu)a« be an enumeration of JP (N) and let (La)a« be an enumeration of the infinite subsets of N. We can choose Zo E {Yo, N\Yo} such that Zo E q. By Corollary 12.28 applied to F U {Zo}, we can choose X0 E i with FS(Xo) C FS((n!)n° 1) such that FSC,(Xo) C A fl Zo for every A E F and Lo\ UxCFS(xo)fsupp(x) is infinite. We then make the inductive assumption that 0 < P < c and that Xa E -3 with FS(Xa) C_ FS((n!)n 1) has been defined for every a < B so that the following conditions hold: (a) FSOO(Xa) C T,,, or FS,,, (X,,) C N\ Y,,;
(b) if 8 < a, then FSOO(Xa) C FS,,,(XS); and (c) La\ UxEFS(xa)fsupp(x) is infinite. By Lemma 5.11, for each a < ,B, FS,,.(Xa) is a compact subsemigroup of N* and hence so is na <,6 FS,, (Xa) which therefore contains an idempotent r. Since FS (Xo) C FS((n!)n° 1), FS((n!)n° 1) E r. We can choose Zf E {YS, N\ YO) satisfying ZS E r. By Corollary 12.28 (applied to {FSm(Xa) : a < 6 and m E N} U {ZS) in place of F) we
12.3 Strongly Summable Ultrafilters - Independence
245
can choose X, E -3 with FS(Xa) C FS((n!)O 1) such that Lp\ UXEFS(xs)fsupp(x) is infinite, FS00(Xp) c Zp-, and FS,,0, (Xp) C FSm(Xa) for every a < 0 and every m E N. Thus FS00(X,6) c FS.(Xa) for every a < P. So conditions (a)-(c) are satisfied with f in place of a. This shows that we can define Xa E -3 for every a < c so that conditions (a)-(c) are satisfied. We put p = (B C N : FSOO(Xa) C_ B for some a < c). It is clear that
p is a filter. For every Y C N, Y E p or N\Y E p, and so p is an ultrafilter. Since FSoo (Xo) c fAE.J1 A, we have .F C p.
Since FS(Xo) C FS((n!)' 1), we have by Exercise 12.2.1 that for each m E N, FS((n!)n_m) E p. Given A E p, A = Ya for some a and so FSOO(Xa) c A and thus for some m, FSm(Xa) C A. Condition (3) of the definition holds directly, and so we have that p is a special strongly summable ultrafilter.
Exercise 12.2.1. Show that if p is an idempotent in N* and 1) E p, then for all m E N, FS((n!)n_m) E p. (Hint: Use Lemma 6.6 and consider the proof of Lemma 12.18.)
12.3 Strongly Summable Ultrafilters - Independence In the last section we saw that Martin's Axiom implies that strongly summable ultrafilters exist. In this section we show that their existence cannot be deduced in ZFC.
Definition 12.30. U is a union ultrafilter if and only if 'U is an ultrafilter on Jr'f (N) and for every A E U there is a sequence (FF)n 1 of pairwise disjoint members of 9'f (N) such that FU((Fn)n° 1) E U and FU((Fn)n° 1) C A.
It would appear that union ultrafilters and strongly summable ultrafilters are essentially equivalent notions. This is indeed true, as we shall shortly see. The correspondence in one direction is in fact immediate.
Theorem 12.31. Let 2( be a union ultrafilter. Let
P={(ErEF2r-1:FE.A):AEU}. Then p is a strongly summable ultrafilter.
Proof. The function f : J'f(N) ->. 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.
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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 <
Xb+1
<2
Zk = EnEHk,eflHk,e+l Xn + EnEHk,e\Hk,e+l Xn - Ye = EnEFk Xn + EREHe,t\He,e+1 Xn - Ye = EnEFk X. + C.
Also zt + Yk =
EnEHI,k Xn SO
Yk = EnEHe,knHe+l,k Xn + EnEHe.k\Ht+l,k Xn - Ze
= EnEGk Xn + EnEHH.e\He+l.e Xn - Ze = EnEGk xn +s.
To complete the proof, it therefore suffices to show that s = -t. From the derivations above we have that Zk + Yk = EnEFk Xn + EnEGk xn + t + s and we also know that Zk + Yk = EnEHk,k Xn. Thus EnEHk.k Xn - EnEFk xn - EnEGk Xn = s + t. Let K1 = (Fk\Gk) U (Gk\Fk) and let K2 = Fk fl Gk. Then S + t = EnEHk k Xn - EnEK1 Xn - EnEK2 2xn
and min(Hk,k U K1 U K2) = min(Hk,k U Fk U Gk) ? b + 1 because Fk c Hk,e+l and Gk c He+1,k. Since Is + t l < XX , Lemma 12.40 tells us that s + t = 0 as required. II
As a consequence of the next theorem (and Theorem 12.29) it is consistent with ZFC
that there are maximal groups of (,BIN +) that are just copies of Z. We do not know whether the existence of such small maximal groups can be established in ZFC.
Theorem 12.42. Let p be a strongly summable ultrafilter. If q E N* and there exists
r E N* such that q + r = r + q = p, then q E Z + p. In particular, H(p) = Z + p. Proof, Let us notice first that since, by Theorem 4.23, Z is contained in the center
of (fl7L, +), Z + p is a subgroup of (fl7L, +). And, by Exercise 4.3.5, 7G + p c ON. Consequently Z + p S H(p). The "in particular" conclusion then follows from the
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252
first assertion immediately, because any member of H (p) is in N* and is invertible with respect to p.
So assume we have q and r in N* such that q + r = r + q = p and suppose that q 0 Z + p. Then likewise, p ¢ 7G + q. Since p 0 { 1 + q, q, -1 + q), pick A l E p such
that -1 + Al 0 q, Al 0 q, and 1 + Al ¢ q. By Lemma 12.20 we may presume (by passing to a subset) that Al = FS((xi,n)n°-1) where for each n, x1,n+1 > 4 EJ=1 x1, j,
and for each m, FS((xl,n)n_m) E p. Let B1 = {x E N : -x + Al E q}. Since
p =r+q, BiEr.PickyiEB1.Let C1 = (-yl + A1) n {x E N : -x + Al E r} n ni__1(N\(-i + A1)). Since yi E B1 and p = q +r, C1 E q. Pick zl > Yl with z1 E C1. Since yi +zl E Al, pick J1,1,1 E Rf(N) such that Y1 + zl = EnEJ1,1,1 xl,n- Let bi = max J1,1,1 and let ai = b1 + 1. Note that al is the smallest member of N with xl,a1 > EnEJ1,1,1 X1,n Assume now that k > 1 and we have chosen Ak_I = FS((xk_ l,n )n_1) E p, 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
EnEFk" X1,n - EnEL1 X1,n = EnEL2 X1,n + t - t'.
Likewise, from the expressions for yk we have EnEGk" X1,n - EnEMI X1,n = EnEM2 XI,n + t' - t.
Since min(Fk" U L 1) > at + 2 and min(Gk" U Ml) > at + 2, we have by Lemma 12.40 that
(1) EnEL2 X1,n + t - t' = 0 or I EnEL2 Xl.n + t - t'I > ZXl,ae+2 and (2) EnEM2 X1,n + t' - t = 0 Or I EnEM2 X1,1 + t' - tI > ?Xi,ae+2 Now 0 < EnEL2 X1,n < Enel XI,n < 4X1,ae+2 We have already seen that
12 Ultrafilters Generated by Finite Sums
254
Itl <
1
1
1
1
1
16Xl,ae+1 < -XI,at+2,
4-Xl,a,
t = (EnElt,t.e\Je,e.x XP,n) - YP < EnEJe,e.e XP,n = YP + Ze
t=(
< xl,at, and
nEJt.e,e\Je.t.k XP.n) - YP 2: -yP > -(ZP + yP) > -xl.at
Thus I EnEL2 XI,. + t - t'I < g4XI,at+2 < jXl.at+2 SO It'I < X1,ae < 16x1 SO EnELZ Xl,n + t - t' = 0. Likewise, EnEMZ xl,n + t' - t = 0. Thus EnEL2 Xl,n + EnEMZ xl,n = 0 so that L2 = M2 = 0 and hence t = t' as required. 'ae+2
We show in the remainder of this section that it is consistent with the usual axioms of set theory that there exist idempotents in (#N, +) which can only be written as a sum of elements of IBN in a trivial fashion.
Lemma 12.43. Let (xn)n°_1 be an increasing sequence in N with the property that for each n E N, xn divides xn+1. If a, b, m E N, a < xn xn,+l divides b, and a + b E FS((xn)n° 1), then a E FS((xn)n° 1) and b E FS((xn)' 1). P r o o f . Pick F E .Pf(N) such that a + b = EtEF x,. Let G = F fl {1, 2, ... , m} and
let H = F\{1, 2, ... , m}. Then a - E,EG Xt = EtEH xt - b (where EIEO X, = 0). Then xn,+l divides E,EH xt - b while Ia - EIEG XII < xtn+l, so a - EIEG Xt =
ErEHxt-b=0. Lemma 12.44. Let p be a special strongly summable ultrafilter, let q E fN, and let r E nc,'I 1 cf (Nn). If p = q + r, then q = r = p. Proof By Theorem 12.39 it suffices to show that r = p. So suppose not and pick A E p\r. Pick an increasing sequence (xn)n°_1 such that FS((xn)n° 1) e A, FS((xn)n_1) E p, and for each n E N, xn divides xn+1. Then FS((xn)n° I) E q + r so pick a E N such that -a + FS((xn )n° 1) E r. Pick m E N such that xn a. Pick b E (-a +FS((xn)n°O__I)) fl Nxn+1 fl (N\A).
Then a + b E FS((xn)n° 1), a < xn,, and x,n+l divides b so by Lemma 12.43, b E FS((xn )n° 1) c A, a contradiction.
Of course, any idempotent p in fiN can be written as (p - n) + (p + n) for any n E Z. We see now that if p is a special strongly summable ultrafilter, then this is the only way that p can be written as a sum.
Theorem 12.45. Let p be a special strongly summable ultrafilter and let q, r E fiN. If
p=q+r,then gandrarein7L+p. Proof. We claim that it suffices to show that r E 7L + nrt_1 c2(Nn). For assume we
have k E Z such that -k + r E nn ct(Nn). Then (k + q) + (-k + r) = p so by Lemma 12.44, k + q = -k + r ='p. I
12.4 Algebraic Properties
Thus we suppose that r o Z + f n° 1 ct(Nn), so that also q o z + I
255 no=1 cZ(Nn).
Define a : N -* X00110, 1,...,t}
by the equation x = E°_ 1 a (x) (t) t ! (It is easy to see that any positive integer has a unique such factorial expansion.) Let a be the continuous extension of a to #N.
Let M = it E N : a(q)(t) 0 t} and let L = It E N : a(q)(t) 0 0}. We claim that both M and L are infinite. To see that M is infinite, suppose instead we have some k E N such that f o r all t > k, a(q)(t) = t. Let z = (k + 1)! - Ei1 a(q)(t) t!. We claim that q + Z E fn_1 ct(Nn), a contradiction. To see this, let n E N be given with
n > k. We show that -z + N(n + 1)! E q. Now, by the continuity of a, X
foralltE{1,2,...,n},
a(x)(t) = a(q)(t)) E q. We claim that
N(n + 1)! + En=1 a(q)(t) t! C -z + N(n + 1)!. To see this, let w E N(n + 1)! + E°=1 a(q)(t) P. Then
so
as required.
Similarly, if L is finite so that for some k, one has a(q)(t) = 0 for all t > k, and
z = Ek,=t a(q)(t) fl, then q - z E fn__1 ct(Nn). Since p is a special strongly summable ultrafilter, pick a sequence (zn) 1 such that FS((zn)n_1) S FS((t!)°O1), FS((zn)O_1) E p and M\ U(fsupp(y) : y E FS((zn)n_1)} is infinite. Let K = U{fsupp(y) : y E FS((zn)001)}.
Now FS((zn)n_1) E q + r so pick a E N such that -a + FS((zn)n_1) E r. Pick m E N such that a < m! and pick f> s > m such that f E M\K and S E L. Now FS((t!)°Ot+1) E p so
{x E N : -x +FS((t!)i__e+1) E r} fl (N(f + 1)! + Ei_1 a(q)(t) t!) E q so pick b E N(f + 1)! + Ei=1 a(q)(t) t! such that -b + FS((t!)°Ot+1) E r. Pick x E (-b + FS((t!)°_e+1)) fl (-a + FS((Zn)n__1)). Then b + x = EIEG t! where min G > f + 1 and for some d E N,
b = d (2+1)!+EI_1a(q)(t)t!.
256
12 Ultrafilters Generated by Finite Sums
Since s E L, a(q)(s)
0 so
and
Ei_1 a(q)(t) t! - a < Ei_1 a(q)(t) t! < (s + 1)! so E1
a(q)(t) t! - a = Es1 =1 ct t! where each c1 E {0, 1, ... , t} and not all ct's
are 0.
Now b + x E FS((Zn)n° 1) + (b - a) so for some H E
ErEGt! = b+x EnEH Zn + (b - a)
= EnEH Zn +d (t + 1)! + EI=5+1 &(q)(t) t! + Es -1 a(q)(t) t! - a
= E JP f ( N) such that EnEH Zn = EtEJ t ! and let 11 = I fl { 1 , 2, ... , t} and 12 = I\{1,2,...,t}. Then EtEG t! - d (t + 1)! - EtE12 t! = EtE1, t! + 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)
El=zt!+t < (t + l)!
a contradiction.
Notes The presentation of Martin's Axiom is based on that by K. Kunen in [171]. See the book by D. Fremlim [94] for substantial information about Martin's Axiom. It was shown in 1972 in [ 117] that the (then unproved) Finite Sums Theorem together with the continuum hypothesis implied the existence of an ultrafilter p on N such that,
for every A E p, {x E N : -x + A E p} E p. And A. Taylor (unpublished) showed that the continuum hypothesis could be weakened to Martin's Axiom. Of course, we now know that such an ultrafilter is simply an idempotent in ($N, +). (See the notes to Chapter 5 for the history of the discovery of this fact.) With this knowledge it appeared that nothing interesting remained in [117]. Then, in 1985, E. van Douwen noted in
Notes
257
conversation that in fact the ultrafilter produced there had the stronger property that it was generated by sets of the form FS((x,,) 1), and wondered if the existence of such an ultrafilter could be proved in ZFC. This question led to the investigation of strongly summable ultrafilters, beginning with [131]. Union ultrafilters were introduced by A. Blass in [44]. Theorem 12.37, one of three
nonelementary results used in this book that we do not prove, is due to S. Shelah. Corollary 12.38 is due to P. Matet [181]. The proof of Corollary 12.38 presented here is from [47], a result of collaboration with A. Blass. Theorem 12.39 is from [46], a result of collaboration with A. Blass. Theorem 12.42 is from [136]. Some of the results of this chapter have been generalised in [151] (a result of collaboration with I. Protasov) to countably infinite abelian groups. If G is a group of this kind, the existence of strongly summable nonprincipal ultrafilters on G follows from Martin's Axiom, but cannot be demonstrated in ZFC. If G can be embedded in the unit circle, then any strongly summable ultrafilter p E G* has the property that the equation
p + x = p has the unique solution x = p in G*, and so does the equation x + p = p. It also follows from Martin's Axiom that there are nonprincipal strongly summable ultrafilters p E G* with the remarkable algebraic property that, for any x,_y E G*, x + y = p implies that x and y are both in G + p. If G is a countably infinite Boolean group, the existence of a nonprincipal strongly summable ultrafilter p on G has an interesting consequence. The ultrafilter p can be used to define a non-discrete extremally disconnected topology on G for which G is a topological group. It is not known whether the existence of a topological group of this kind can be demonstrated in ZFC.
Chapter 13
Multiple Structures in fS
We deal in this chapter with the relationships among different operations on the same set S, as for example the semigroups (N, +) and (N, ). We also include a section considering the relationships between the left continuous and right continuous extensions of an operation on S.
13.1 Sums Equal to Products in flZ The earliest applications of the theory of compact right topological semigroups to Ram-
sey Theory involved the semigroups (N, +) and (N, ). (See Chapters 5 and 17 for several examples.) In investigating these applications, the question arose whether the equation q + p = s r has any solutions in N*, or indeed in V. These questions remain open, but we present several partial results in this section. We begin with the only nontrivial instance of the distributive law known to hold in PZ. (For more information about the distributive law see Section 13.2.) Since we are dealing with two operations, the notations XP and pp are now ambiguous. We adopt the convention in Sections 13.1 and 13.2 that pp(q) = q + p and
Ap(q) = p + q and introduce the notations rp(q) = q p and 2p(q) = p q.
IfpEfZ, -p denotes Lemma 13.1. Let n E 7G and let p, q E PZ. Then n
(p +q)
(p + q) = n p + n q and
n.
Proof. Since, by Theorem 4.23, 7L is contained in the center of (fZ, ), the second
assertion follows from the first.
Now n (p + q) = (en o pq) (p) and n p + n . q = (pn.q o en) (p) so it suffices to show that the functions in o Pq and pn.q o in agree on Z. So let m E Z be given. Then
and
(Pn.goen)(m)=nm+nq=()nmofn)(q)
13.1 Sums Equal to Products in fiZ
Since the distributive law holds in Z, the functions t, o 1. and
259
o en agree on Z.
We have occasionally used the canonical mapping from Z to Z before. In this chapter it will be used in several proofs, so we fix some notation for it.
Definition 13.2. Let n e N. Then hn : Z -+ Z denotes the canonical mapping. Notice that the continuous extension W. : fiZ -+ Zn of hn is a homomorphism both from OZ, +) to (Zn, +) and from (f Z, -) to (Zn, -) (by Lemma 2.14).
Definition 133. Let b be a real number satisfying b > 1. We define 4b : N - co by putting 'b(n) = [logb(n)J. As usual, b : fiN --o- fim denotes the continuous extension of ¢b.
L e m m a 13.4. Let p E N* and let b > 1.
(i)-b(q+P) EOb(P)+{-1,0, 1}forevery q E fZ; (ii)
E b(q) + Ab(p) + (0, 1} for every q E #N.
Proof. (i) Choose E > 0 with the property that I logb(1 + t) I < 1 if t E (-E, e). If m E 7G, n E N, and Imp < En, we have logb(m +n) = logb n +logb(1 + n) and hence 4Pb(m + n) E Ob(n) + 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.
13 Multiple Structures in 8S
260
Lemma 13.6. Let p, r E 7G*, let s, t E ,8Z, and let m, n E Z. Ifs + p = m r and
t+p= n- r, then m=n. Proof. We may suppose that p E N*, since otherwise, by Lemma 13.1, we could replace
p, s, t, and r by -p, -s, -t, and -r respectively. This implies that s + p and t + p are in N*, by Exercise 4.3.5, and hence that in ¢ 0 and n 0 0. Therefore r, m, and n are all in fl N or all in -,6N. We may suppose that r, m, n E fiN, since otherwise we could replace r, m, and n by -r, -m, and -n respectively. If n # m, we may suppose without loss of generality that n > m. We can choose b E 1l satisfying b > 1 and mb4 < n. So ¢b(n) > cbb(m) + 4. By Lemma 13.4, ¢b (n r) = ob (m r) + k for some k > _3 in N. We also know that ob (s + p) and b (t + p) are both in,b (p) + {-1, 0, 1). So ¢b (t + p) = Ob (s + p) + f for some f E {-2, -1, 0, 1, 2). Since f ¢ k, this contradicts Lemma 6.28. This establishes that n = m. Lemma 13.7. Let p, r E 7G* and letm, n, m', n' E Z. Ifm+p = n r and m'+p = n'-r, then (m, n) = (m', n'). Proof. By Lemma 13.6, n = n'. It then follows from Lemma 6.28 that m = m'. As we remarked at the beginning of this section, we are primarily concerned with two questions, namely whether there exist solutions to the equation q + p = s - r in N* and whether there exist such solutions in V. Now, of course, an affirmative answer to the first question implies an affirmative answer to the second. Further, the reader is asked to show in Exercise 13.1.1 that if p, r, s E Z* and there exists q E Z* such that q + p = s r, then there exists q' E Z* such that q' + I pI = IsI Ir1, where I p I has its obvious meaning. Thus the questions are nearly the same. However, it is conceivable that one could have p, r, s E N* and q E -N* such that q + p = s - r but not have any q' E N* such that q' + p = s r. Accordingly, we address both questions.
Lemma 13.8. Let p, q, r E 7L* and let H = {m E 7Z : there exists t E i7G such that in + p = t r}.
There exists s E Z* such that q + p = s - r if and only if H E q.
Proof. Necessity. Picks E Z* such that q + p = s - r. We note that, by Lemma 13.6, there is at most one a E 7L for which a r E f 7L+ p. Thus, if S = in E Z : n r 0 ,67L+p},
we have S E S. Suppose that H 0 q and let B = Z\H E q. Since q + p E cf(B + p) ands r E c2 (S r), it follows from Theorem 3.40, that m + p E #Z r for some m E B, or else n r E f87G + p for some n E S. The first possibility can be ruled out because it implies that m E B fl H. The second can be ruled out by our choice of S.
Sufficiency. We note that there is at most one m E H for which m + p E Z - r (by Lemma 13.7). Let H' = {m E 7L : m + p c Z* - r). Then H' E q and hence
q + p E cPSz(H' + p) 9 cfpz(Z* r) = Z* - r.
13.1 Sums Equal to Products in PZ
261
Theorem 13.9. Let p, r E Z*, let G = {q E Z* : there exists s E Z*such that q + p = s r} and let
H = {m E 7L : there exists t E ,8Z such that m + p = t r}. Then the following statements are equivalent.
(a) G#0. (b) IHI = co.
(c) IGI = 2` Proof. (a) implies (b). Pick q E G. By Lemma 13.8, H E q so, since q E Z*, I H I = Co. (b) implies (c). By Theorem 3.58, I{q E 7Z* : H E q}I = 2` so by Lemma 13.8, IGI = 2c. That (c) implies (a) is trivial.
Theorem 13.10. Let p, r E N*, let G = {q E N* : there exists s E N* such that q + p = s r} and let
H = {m E N : there exists t E ON such that m + p = t r}. Then the following statements are equivalent.
(a)G00. (b) IHI = w. (c) IGI = 2`. Proof. (a) implies (b). Pick q E G. By Lemma 13.8, {m E 7L : there exists t E fl Z such that m + p = t r} E q. Since also N E q, one has
{m E N : there exists t E ,87Z such that m + p = t r} E q. Now, given m E N and t E OZ such that m + p = t r, one has m + p E N*, so one also must have t E ON. (Otherwise t- r E -N* U {0}.) Thus H E q and therefore I H I= co. (b) implies (c). By Theorem 3.58, 1{q E N* : H E q}I = 2`. Further,
H C {m E 7G : there exists t E PZ such that m + p = t r}.
So by Lemma 13.8, if q E N* and H E q, then there is some s E Z* such that q + p =s r. Since r and q + p are in N*, s EN*, sothatq E G. To put Theorems 13.9 and 13.10 in context, notice that we do not know whether the set H ever has two members, let alone infinitely many members. Notice also that if H,
p, and r are as in Theorem 13.10, then H + p SON r. Theorem 13.12 characterizes the existence of such H.
13 Multiple Structures in fiS
262
Definition 13.11. Let 0 0 H c N. Then
TH = (A c N : there exists F E [H]`w such that N = UnEF(-n + A)). Theorem 13.12. Let H C N with CHI > 1. Then there exist p, r e N* such that H + p S ON r if and only if there is a choice of xA E N for each A E TH such that (XA-' A : A E TH ) has the infinite finite intersection property.
Proof Sufficiency. Pick by Corollary 3.14 r E N* such that {xA-' A : A E TH } C ,Recall that C(r) = {A C N : for all x E N, 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.
13 Multiple Structures in fl S
264
Lemma 13.16. Let u # v in 7L\{0}. There do not exist p E I that
nN and q E N* such
uq + p = vq + p. Proof. Suppose we have such elements p and q and pick a prime r > lu I + I v I. For i E (1,2..... r - 11, let C1 = {r' (rk + i) : n, k E w}. Then C1 is the set of x E N whose rightmost nonzero digit in the base r expansion is i. Pick i such that C, E q. For each m E N, let f (m) denote the largest integer in co for which rf (m) is a factor
of m. Since rf (')+1 N E p, it follows from Theorem 4.15 that {um + s : m E C, and
s E r f(m)+1N} E uq + p and {vn + t : n E C, and t E
rf(n)+1 N}
E vq + p. Thus
there exist m,n E Ci,s E rf(m)+1 N, and t E rf(n)+1N such that um+s = vn + t. We observe that f (um + s) = f (m), because rf(m) is a factor of um + s, but rf(m)+1 is not. Similarly, f(vn + t) = f (n). So f (m) = f (n) = k, say.
We have m = irk (mod rk+1) and n = irk (mod rk+1) So um + s = uirk This implies that u = v (mod r), a (mod rk+1) and vn + t = virk (mod rk+1).
contradiction.
Theorem 13.17. Let u # v in 7L\{O}. There do not exist p, q E N* such that
uq + p = vq + p. Proof. Suppose we have such elements p and q. Pick r E K (MN, +). Then p + r E K (8N, +) so by Theorem 1.64 there is some maximal group G in K (f N, +) with p + r E G. Let e be the identity of G and let s be the (additive) inverse of p + r in
G. Then adding r + s on the right in the equation uq + p = vq + p, one obtains uq + e = vq + e. This contradicts Lemma 13.16, because hn (e) = 0 for every n E N, and so e E I InEN nN. Exercise 13.1.1. Given p E ,BZ, define
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')<(m,n) (n' + m'A)) and let
/
D = U(m,n)E(Nxw)\B 1(n + mA) \ U(m',n')<(m,n) (n' + m'A)).
13.2 The Distributive Laws in,6Z
267
Notice that c fl D = 0. Thus without loss of generality we have N\C E q. Pick
(m,n)EBsuch that and for every (n', m') -< (n, m), I(n' + m'A) fl (n + mA)I < w since A is doubly thin. Thus
(n' + m'A) E n +m p
(n +mA) \ so C E n+ m- p, a contradiction. We have one final preliminary.
Lemma 13.25. Let A be a doubly thin subset of N and let p E A*. Then N* N* + N* p, and N* (N* + p) have pairwise disjoint closures.
p,
Proof. By Lemma 13.24, it suffices to observe that
and
n>m}.
C c2in
The first observation is immediate. For the second, just notice that if q, r E N*, then nEN
For the last observation, let q, r E N*, let B E q (r + p), and pick m E N such
that m-1 B E r + p. Then {n E N : -n + m-1 B Ep} E r so pick n such that
-n+m-'BEp.Then
Theorem 13.26. The set (P E N* : N* p, N* + N* p, and N* (N* + p) have pairwise disjoint closures} has dense interior in N*.
Proof. Let B be an infinite subset of N. Pick by Lemma 13.23 a doubly thin subset A of B. Then apply Lemma 13.25.
Corollary 13.27. The set (P E N* : for all q, r E N*, (q + r) - p
r (q + p) # r q + r p} has dense interior in N*. Proof. Given p, q, r E N*,
(q +
E N* E
so Theorem 13.26 applies.
-
q p + r p and
13 Multiple Structures in PS
268
13.3 Ultrafilters on }I8 Near 0 We have dealt throughout this book with the Stone-tech compactification of a discrete space, especially a discrete semigroup. One may naturally wonder why we restrict our attention to discrete spaces, especially given that the Stone-tech compactification exists for a much larger class of spaces. We shall see in Theorem 21.47 that for many
nondiscrete semigroups S, including (R, +), the Stone-tech compactification of S cannot be made into a semigroup compactification of S.
Of course, if S is a topological semigroup, one can give the set S the discrete topology and proceed as we always have. It would seem that by doing that, one would lose any information about the original topological semigroup. We shall see however in Section 17.5 that we can get information about partitions of the real interval (0, oo), every member of which is measurable, or every member of which is a Baire set, by first giving (0, oo) the discrete topology and passing to its Stone-tech compactification. Because of these interesting applications, we shall investigate the algebra of (,BIIld, +) and (8Rd, ), where Rd is the set IIl with the discrete topology. We shall be primarily concerned with (,BIIld, +).
Definition 13.28. (a) Given a topological space X, Xd is the set X endowed with the discrete topology.
(b) Let a : ,JRd
[-oo, oc] be the continuous extension of the identity function.
(c) B(R) _ {p E find : a(p) 0 {-oc, oo}}. (c) Given X E IIl,
x+= {pEB(JR)a(p)=xand(x,00)Ep}and x- = {p E B(R) : a(p) = x and (-oo, x) E p}. (d) U = UXER x+ and D = UXEIIB X-The set B(R) is the set of "bounded" ultrafilters on R. That is, an ultrafilter p E 6 is in B(R) if and only if there is some n E N with [-n, n] E p. We collect some routine information about the notions defined above.
Lemma 13.29. (a) Let X E IIl and let p E 14II8d. Then p E x+ if and only if for every E > 0, (X, X + E) E p. Also, p E X if and only if for every E > 0, (x - E, X) E p.
(b) Let p, q E B(R), let x = a(p), and let y = a(q). If P E x+, then p + q E (x + y)+. If p E x-, then p + q E (x + y)-. (c) B(IR)\IR = U U D and U and D are disjoint right ideals of (B(IR), +). In particular, B(R) is not commutative. (d) The set 0+ is a compact subsemigroup of (B(R), +).
(e)Ifx E Rand p E,BIIBd, then x+p = p+x. (f) The set 0+ is a two sided ideal of the semigroup ($ (0, 1)d, ). Proof. The proof of (a) is a routine exercise, (e) is a consequence of Theorem 4.23, and (c) and (d) follow from (b). We establish (b) and (t).
To verify (b) assume first that p E x+. To see that p + q E (x + y)+, let E > 0 be given and let A = (x + y, x + y + E). To see that A E p + q, we show that
13.3 Ultrafilters on R Near 0
269
(X,x+E) c {z ER : -z+A E q}. Soletz E (x, x+e). LetS = min{z-x, x+e-z}. Since a (q) =y, we have (y - S, y + S) E q and (y - S, y + S) C -z+ A so -z+ A E q as required. The proof that p + q E (x + y)- if p E x- is nearly identical. To verify (f), let p E 0+ and let q E 0 (0, 1)d. To see that p q E 0+ and q p E 0+,
let E > 0 be given. Since (0 E) (0, 1) = (0, E), we have (0, E) C (x E (0, 1) : x-1(0, E) E q} (so that (0, E) E p q) and (0, 1) C {x E (0, 1) : x' 1(0, E) E p} (so that (0, E) E q p). We shall restrict our attention to 0+, and it is natural to ask why. On the one hand, it is a subsemigroup of (8Rd, +) and for any other x E R, x+ and x - are not semigroups. On the other hand we see in the following theorem that 0+ holds all of the algebraic structure of (B(R), +) not already revealed by R.
Theorem 13.30. (a) (0+, +) and (0-, +) are isomorphic. (b) The function rp : Rd x ({0} U 0+ U 0-) -+ B(R) defined by cp(x, p) = x + p is a continuous isomorphism onto (B(R), +). Proof. (a) Define r
: 0+ -+ 0- by r(p) = -p, where -p = {-A : A E p}.
It
is routine to verify that r takes 0+ one-to-one onto 0-. Let p, q E 0+. To see that r (p +q) = r (p) + r (q) it suffices, since r (p +q) and r (p) + r (q) are both ultrafilters,
to show that r(p+q) c r(p) + r(q). So let A E r(p + q). Then -A E p + q so
B={x ER:-x+-AEq}Ep and hence -B E r(p). Then -B C {x E R : -x + A E r(q)} so A E r(p) + r(q) as required. (b) To see that q is a homomorphism, let (x, p) and (y, q) be in Rd x ({0} U0+ U0-).
Since p + y =y+p,wehave rp(x, p)+'p(y,q) ='P(x+y, p + q). To see that cp is one-to-one, assume we have 'p (x, p) = V (y, q). By Lemma 13.29(b)
wehavex=a(x+p)=a(y+q)=y.Thenx+p=x+q sop=-x+x+q=q. To see that 'p is onto B(R), let q E B(R) and let x = a(q). Let p = -x + q. Then
q=x+p='P(x, p).
To see that rp is continuous, let (x, p) E Rd x ({0} U 0+ U 0-) and let A E x + p. Then -x + A E p so {x} x (cf#Rd (-x + A)) is a neighborhood of (x, p) contained in c-1 [cf $RdA].
As we have remarked, 0+ has an interesting and useful multiplicative structure. But much is known of this structure because, by Lemma 13.29(t), 0+ is a two sided
ideal of (f(0, 1)d, ), so K(f(0, 1)d, ) C 0+ and hence by Theorem 1.65, K(0+,) _ K(48(0, 1)d, ')On the other hand, 0+ is far from being an ideal of (B(S), +), so no general results apply to (0+, +) beyond those that apply to any compact right topological semigroup. We first give an easy characterization of idempotents in (0+, +), whose proof we leave as an exercise.
Theorem 13.31. There exists p = p + p in 0+ with A E p if and only if there is some 9 A. 1 in (0, 1) such that ER_I x converges and
sequence
270
13 Multiple Structures in SS
Proof. This is Exercise 13.3.1.
As a compact right topological semigroup, 0+ has a smallest two sided ideal by Theorem 2.8. We now turn our attention to characterizing the smallest ideal of 0+ and its closure. If (S, +) is a discrete semigroup we know from Theorem 4.39 that a point p E 16S
is in the smallest ideal of fS if and only if, for each A E p, (x E S : -x + A E p) is syndetic (and also if and only if for all q E fS, p E PS + q + p). We obtain in Theorem 13.33 a nearly identical characterization of K(0+, +).
Definition 13.32. A subset B of (0, 1) is syndetic near 0 if and only if for every e > 0
there exist some F E ?f((0, e)) and some 8 > 0 such that (0, 8) C U:EF(-t + B). Theorem 1333. Let p E 0+. The following statements are equivalent.
(a) p E K(0+,+). (b) For all A E p, {x E (0, 1) : -x + A E p) is syndetic near 0.
(c) For allrE0+, pE0++r+p. Proof (a) implies (b). Let A E p, let B = {x E (0, 1) : -x + A E p}, and suppose that B is not syndetic near 0. Pick e > 0 such that for all F E .Pf ((0, e)) and all 8 > 0,
(0, 8)\ UIEF(-t + B) 96 0. Let = ((0, 8)\ U,EF(-t + B) : F E R f ((0, e)) and 8 > 0}. Then g has the finite intersection property so pick r E P(0, 1)d with g e r. Since {(0, 8) : 8 > 015; r we have r E 0+. Pick a minimal left ideal L of 0+ with L C 0+ + r, by Corollary 2.6. Since K(0+) is the union of all of the minimal right ideals of 0+ by Theorem 2.8, pick a minimal right ideal R of 0+ with p E R. Then L n R is a group by Theorem 2.7, so let q be the identity
of L n R. Then R= q +O+, so p E q +O+ sop = q+ p so B Eq. Also q E 0++r so
pick wE0+such that q=w+r.Then (0,e)Ewand{tE(0,1):-t+BEr)Ew so pick t E (0, e) such that -t + B E r. But (0, 1)\(-t + B) E g C r, a contradiction. (b) implies (c). Let r E 0+. For each A E p, let B(A) = {x E (0, 1) : -x + A E p} and let C(A) = It E (0, 1) : -t + B(A) E r}. Observe that for any A1, A2 E p, one has B(A1 n A2) = B(A1) n B(A2) and C(A1 n A2) = C(Ai) n C(A2).
We claim that for every A E p and every e > 0, C(A) n (0, e) # 0. To see this, let A E p and e > 0 be given and pick F E Pj((0, E)) and 8 > 0 such that (0, 8) C U,EF(-t + B(A)). Since (0, 8) E r we have U,EF(-t + B(A)) E r and hence there is some t E F with -t + B(A) E r. Then t E C(A) n (0, e). Thus ((0, e) n C(A) : E > 0 and A E p} has the finite intersection property so pick q E f(0, 1)d with {(0, e) n C(A) : E > 0 and A E p} C q. Then q E 0+. We claim that p = q + r + p for which it suffices (since both are ultrafilters) to show that
p e q + r + p. Let A E p be given. Then {t E (0, 1) : -t + B(A) E r} =C(A)E q so B(A) E q + r, so A E q + r + p as required. (c) implies (a). Pick r E K(0+). -
13.3 Ulirafilters on R Near 0
271
If (S, +) is any discrete semigroup, we know from Corollary 4.41 that, given any discrete semigroup S and any p E 13 S, p is in the closure of the smallest ideal of PS if and
only if each A E p is piecewise syndetic. We obtain a similar result in 0+. Modifying the definition of "piecewise syndetic" to apply to 0+ is somewhat less straightforward than was the case with "syndetic". Definition 13.34. A subset A of (0, 1) is piecewise syndetic near 0 if and only if there exist sequences (F,,) n° t and (Sn)n° t such that
(1) for each n E N, Fn E Pf ((0, 1/n)) and S E (0, 1/n), and (2) for all G E Pf((0, 1)) and all µ. > 0 there is some x E (0, µ) such that for all n E N, (G n (0, s,)) +x C U:EFF(-t + A).
Theorem 13.35. Let A C (0, 1). Then K(0+) n C 0(0,lld A 0 0 if and only if A is piecewise syndetic near 0.
Proof. Necessity. Pick p E K(0+)nct$(o,1)d A and let B = {x E (0, 1) : -x+A E p}. By Theorem 13.33, B is syndetic near 0. Inductively for n E ICY pick Fn E Pf ((O, 1 /n))
and Sn E (0,11n) (with S < 6n_1 if n > 1) such that (0, Sn) c UIEF (-t + B). Let G E Pf ((0, 1)) be given. If G n (0, 31) = 0, the conclusion is trivial, so assume
Gn(0,Si)#0andletH=Gn(0,S1).Foreach yEH,let m(y)=max{n EN:y <Sn}. For each y E H and each n E {1, 2, ... , m(y)}, we have y E
B) so pick
t(y,n) E Fn such that y E -t(y,n)+B. Then given y E H andn E {1,2,...,m(y)}, one has -(t(y, n) + y) + A E P. Now let µ > 0 be given. Then (0, µ) E p so pick
x E (0, µ) n nyER nn (i)(-(t(y, n) + y) + A). Thengivenn E Nandy E Gn(O,Sn),onehasy E Handn < m(y)sot(y,n)+y+x E A so
y+x E -t(y,n)+A c Sufficiency. Pick (F,,)', and (Sn)n_i satisfying (1) and (2) of Definition 13.34. Given G E Pf((0, 1)) and µ > 0, let
C(G, µ) = {x E (0, IL) : for all n E N, (G n (0, Sn)) + x c U:EF, (-t + A)).
By assumption each C(G, µ) 96 0. Further, given Gt and G2 in Rf ((0, 1)) and µl,µ2 > 0, one has C(Gi U G2, min{µl, µ2}) c C(GI, µi) n C(G2, µ2) so {C(G, µ) : G E Pf((0, 1)) and p, > 0) has the finite intersection property so pick p E 6(0, 1)d with {C(G, µ) : G E Pf ((0, 1)) and 1L > 0) c p. Note that since each C(G, A) C (0, µ), one has p E 0+.
272
13 Multiple Structures in PS
Now we claim that for each n E N, 0+ + p S A)), so let n E N and let q E 0+. To show that UtEF (-t + A) E q + p, we show that (0, Sn) C (y E (0, 1) : -y + UtEF (-t + A) E p}.
So let y E (0, &). Then C({y}, Sn) E p and C({y}, SO c -y + UtEF, (-t + A).
Now pick r E (0+ + p) fl K(0+) (since 0+ + p is a left ideal of 0+). Given n E N, (-t + A) E r so pick t E Fn such that -tn + A E r. Now each to E Fn C (0, 1/n) so nlimntn = 0 so pick q E0+ f1 ctf(o,1)d{tn : n E N}. Then
q+rEK(0+)and {to:nENJ C(tE(0,1):-t+AEr)so AEq+r. Since (0+, +) is a compact right topological semigroup, the closure of any right ideal is again a right ideal. Consequently cfo+ K(0+) = c2B(o,I)d K(0+) is a right ideal of 0+. On the other hand, if S is any discrete semigroup, we know from Theorem 4.44 that the closure of K(PS) is a two sided ideal of OS. We do not know whether cto+ K(0+) is a left ideal of 0+, but would conjecture that it is not.
Exercise 13.3.1. Prove Theorem 13.31. (Hint: Consider Lemma 5.11 and either proof of Theorem 5.8.)
13.4 The Left and Right Continuous Extensions of One Operation Throughout this book, we have taken the operation on PS to be the operation making (flS, ) a right topological semigroup with S contained in its topological center. Of course, the extension can also be accomplished in the reverse order to that used in Section 4.1. That is, given x E S and P E PS we can define pox = p- lim y x (so yES
that if p E S, then pox = p x) and given p, q E PS,
Then (PS, o) is a compact left topological semigroup and for each x E S, the function rx : ,BS fS defined by rx (p) = pox is continuous. Further, Theorems 2.7 through 2.11 apply to (OS, o) with "left" and "right" interchanged. Notice that, given p, q E 46S and A C S,
A E p o q if and only if {x E S : 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
Thus
280
14 The Central Sets Theorem
By Theorem 2123, K(Y) = Xi_0K(,6N). We now claim that E n K(Y) # 0 so that by Theorem 1.65, K(E) = E n K(Y). In fact we show that
(*) if p E K(#N) and P = (p, p...., p), then p E E. To establish (*), let U be a neighborhood of p and fort E {0, 1, ..., t} pick At E p
such that X 0Ar c U. Then ne=0 At EP so nr-0 At 0 0 so pick a En't=O A,. Then (a,a,...,a) E Ufl E. Since K(E) = E fl K(Y) = E fl XOK(PN) and I is an ideal of E, we have En X,=0K(,BN) C 1. Now since A is piecewise syndetic, pick by Theorem 4.40 some p E A fl K(,BN) and let (p, p, ... , p). Then by (*) p E E fl X i=OK(fN) C I.
Let U = X ,=OA. Then U is a neighborhood of p so u n 1°
0 so pick a, d E N
such that (a,a+d,a+2d,...,a+td) E U. The above proof already illustrates the most startling fact about the Central Sets Theorem proof, namely how much one gets for how little. That is, one establishes the
statement (*) based on the trivial fact that E° contains the diagonal of X0N. Then one concludes immediately that p E I and hence all members of p contain interesting configurations. It is enough to make someone raised on the work ethic feel guilty. Note that if r E N and N = Ui=1 A1, then some Ai is piecewise syndetic. This can be seen in at least two ways. One can verify combinatorially that the union of two sets which are not piecewise syndetic is not piecewise syndetic. Somewhat easier from our point of view is to note that fiN = U;=1 Ti so some Ti fl K(fiN) 0 0 so Theorem 4.40 applies. Corollary 14.2 (van der Waerden [238]). Let r E N and let N = U;=1 A. For each
t E N there exist i E {1,2,...,r}anda,d E Nsuch that {a,a+d,...,a+td} C A;. Proof. Some Ai is piecewise syndetic so Theorem 14.1 applies.
In Exercise 5.1.1 the reader was asked to show that the above version of van der Waerden's Theorem implies the apparently stronger version wherein one i is chosen which "works" for all t. In fact the current method of proof yields the stronger version directly.
Corollary 14.3 (van der Waerden [238]). Let r E N and let N = U1=1 Ai. There e x i s t s i E 11, 2, ... , r} such that for each t E N there exists a, d E N such that
{a,a+d,...,a+td} C A;.
Proof. Some A; is piecewise syndetic so Theorem 14.1 applies. One may make the aesthetic objection to the proof of Theorem 14.1 that one requires a different space Y for each length of arithmetic progression. In fact, one may do the proof once for all values oft by starting with an infinite product. (See Exercise 14.1.2.) We shall use such an approach in our proof of the Central Sets Theorem.
14.2 The Hales-Jewett Theorem
281
As a consequence of van der Waerden's Theorem, we establish the existence of a combinatorially interesting ideal of (ON, +) and of (14N, ), namely the set of ultrafilters every member of which contains arbitrarily long arithmetic progressions.
Definition 14.4. AY = {p E ON : for every A E p and every f E N, there exist
a,d E Nsuch that (a,a+d,a+2d,...,a+id) c A}. Theorem 14.5. 4
is an ideal of (1BIN +) and of (ON, ).
Proof. By Theorems 4.40 and 14.1 K(fiN, +) C AR so AeP 0. To complete the proof, let p E A5 and let q E fiN. Given A E q + p and f E N, pick x E N such
that -x+A E P. Pick a, d E Nsuch that {a, a+d,a+2d,...,a+Id} c -x + A. Then {x + a,x + a + d,x + a + 2d,...,x + a + Id} C A. Given A E p + q and f E N, pick a, d E Nsuch that (a,a+d,a+2d,...,a+Id) C {x E N: -x + A E
q}. Then n1-o(-(a + td) + A) E q so pick x E nf.0(-(a + td) + A). Then
{x+a,x+a+d,x+a+2d,...,x+a+Id} C_ A.
Given A E qp and f E N, pick x E N such that x-1 A E p. Pick a, d E N such that
(a, a+d, a+2d,... , a+Id} C x-1 A. Then {xa, xa+xd, xa+2xd, ... , xa+Ixd} C_ A Given A E pq and f E N, pick a, d E N such that (a, a + d, a + 2d, ... , a + Id ) C
{x E N : X-1 A E q}. Then n1p(a + td)-'A E q so pick x E n1=0(a + td) Then {xa, xa + xd, xa + 2xd,... , xa + Ixd} C A.
A.
Exercise 14.1.1. Using compactness (see Section 5.5) and Corollary 14.2 prove that given any k, I E N there exists n E N such that whenever 11, 2, ... , n} is r-colored, there must be a length f monochrome arithmetic progression.
Exercise 14.1.2. Prove Theorem 14.1 as follows.
Let Y = X 000N. Let E'
{(a, a+d, a+2d.... ) : a E N and d E N U {0} } and let I° _ {(a, a+d, a+2d, ...) : a, d E N}. Let E = cIY E° and let I = cIY I Prove that E is a subsemigroup of Y and that I is an ideal of E. Then prove
(*) if pEK(fN)and p=(p,p,p.... ),then pEE. Pick P E K(ON) fl A, let p = (p, p, p, ...), and show that p E I. Given I E N let
U={q"EY:foralliE{0,1,...,I},q; EA}andshow that UflI'
0.
14.2 The Hales-Jewett Theorem In this section we prove the powerful generalization of van der Waerden's theorem which is due to A. Hales and R. Jewett [113]. In this case it is worth noting that even though we work with a very noncommutative semigroup, namely the free semigroup on a finite alphabet, the proof is nearly identical to the proof of van der Waerden's Theorem. (There is a much more significant contrast between the commutative and noncommutative versions of the Central Sets Theorem.)
14 The Central Sets Theorem
282
Definition 14.6. Let S be the free semigroup on the alphabet A and let u be a variable which is not a member of A. (a) A variable word w(v) is a member of the free semigroup on the alphabet A U (v) in which v occurs. (b) S(v) = {w(v) : w(v) is a variable word). (c) Given a variable word w(v) and a member a of A, w(a) is the word in which each occurrence of v is replaced by an occurrence of a. Recall that the free semigroup on the alphabet A was formally defined (Definition 1.3)
as the set of functions with domain (0, 1, ... , n) for some n E w and range contained in A. Given a member w of the free semigroup on an alphabet A and a "letter" a E A, when we say that a occurs in w we mean formally that a is in the range of (the function) w. Given w(v) E S(v), and a E A, one then has formally that w(a) is the function with the same domain as w(v) such that for i in this domain, if W(V); v w(a), = j w(v)i a if W(V); = U.
Informally, if A = (1,2,3,4,5) and w(v) = M0251 v4, then w(3) = 1333325134 and w(2) = 1322325124. Theorem 14.7. Let A be a finite nonempty alphabet, let S be the free semigroup over A, and let B be a piecewise syndetic subset of S. Then there is a variable word w(v) such that (w (a) : a E A) C B. Proof.
Let f _ IAI and write A = (al, a2, ... , at). Let Y = X f=148S. Denote the
operation of Y (as well as that of S) by juxtaposition. Then by Theorem 2.22, Y is a compact right topological semigroup and if x" E X,=1 S, then Xi is continuous. Thus Y is a semigroup compactification of X e-1 S. Let 1° = ((w(ai), W 0 2 ) ,--- w(at)) : w(v) E S(v)} and let
E
I'U{(w,w,...,w): w r= S}.
Let E = cfy E° and let I = cdy V. Observe that E° is a subsenugroup of XI S and that I° is an ideal of E`. Thus by Theorem 4.17 E is a subsemigroup of Y and I is an ideal of E. We now claim that EflK(Y) 00 By Theorem 2.23 we have K(Y) = so that by Theorem 1.65, K(E) = E fl K(Y). In fact we show that
(*) if p E K(PS) and p = (p, p, ..., p), then p E E. To establish (*), let U be a neighborhood of p and fort E 11, 2, ... , e} pick Bt E p
such that Xt_IBt c_ U. Then nr=1 Bt E p so nt=1 Bt # 0 so pick w E n,t_1 Bt. Then (w, w, ... , w) E U fl E. Since K(E) = E fl K(Y) = E n X f_1 K(48S) and I is an ideal of E, we have En X ,=i K(48S) C I. Now since B is piecewise syndetic, pick by Theorem 4.40 some
14.3 The Commutative Central Sets Theorem
283
p E B f1 K(fS) and let p = (p, p, ... , p). Then by (*) p E E fl Xi_1 K($S) _c I. Let U = X e-1 B. Then U is a neighborhood of p so u fl I` # 0 so pick w(v) E S(v) such that (w(al), w(a2), ... , w(at)) E U. Corollary 14.8 (Hales-Jewett [113]). Let A be a finite nonempty alphabet, let S be the free semigroup over A, let r E N and let S = U, =1 B; . Then there exist i E (1, 2, ... , r) and a variable word w(v) such that (w(a) : a E A) C B,. Proof. Some B; is piecewise syndetic so Theorem 14.7 applies.
Exercise 14.2.1. Using compactness (see Section 5.5) and Corollary 14.8 prove that given any finite nonempty alphabet A and any r E N there is some n E N such that whenever the length n words over A are r-colored there must be a variable word w(v) such that {w (a) : a E A) is monochrome. Exercise 14.2.2. Derive van der Waerden's Theorem (Corollary 14.2) from the HalesJewett Theorem (Corollary 14.8). (Hint: Consider for example the alphabet (0, 1, 2, 3, 4) and the following length 5 arithmetic progression written in base 5: 20100010134, 20111011134,20122012134,20133013134,20144014134.)
14.3 The Commutative Central Sets Theorem In this section we derive the Central Sets Theorem for commutative semigroups, as well as some of its immediate corollaries. (One is Corollary 14.13 and two more are exercises.) We first encode some of the important ideas that underlie the proof of the Central Sets Theorem, some of which we have illustrated in the proofs of van der Waerden's Theorem and the Hales-Jewett Theorem. The lemma is more general than needed here because we shall use it again in later chapters. (We remind the reader that we have been using the notation x for a function constantly equal to x.)
Lemma 14.9. Let J be a set, let (D, <) be a directed set, and let (S, ) be a semigroup. Let be a decreasing family of nonempty subsets of S such that for each i E D and each x E Ti there is some j E D such that x Tj C Ti. Let Q = niED cfss Ti. Then Q is a compact subsemigroup of PS. Let and (I,)iED be decreasing families of nonempty subsets of X r EJ S with the following properties:
(a) foreachi E D, I; C E; C X,EJT,, (b) for each i E D and each x' E I, there exists j E D such that x Ej C 1,, and (c) for each i E D and each x E E, \I, there exists j E D such that x . Ej C E, and
- .!j CI,. Let Y = X,EJJ S, let E= n1ED c y E,, and let I= f;ED cfY I,. Then E is a subsemigroup of X,Ej Q and I is an ideal of E. If in addition, either
14 The Central Sets Theorem
284
(d) for each i E D, Ti = S and {a E S : a 0 E; } is not piecewise syndetic, or (e) for each i E D and each a E Ti, a E E,,
then given anypEK(Q),onehaspEEnK(XtEj Q)=K(E)CI. Proof. It follows from Theorem 4.20 that Q is a subsemigroup of 0 S.
By condition (a) and the fact that (I,)iED is decreasing we have 0 0 1 c E C X tEJ Q. To complete the proof that E is a subsemigroup of X (EJ Q and I is an ideal of E, we let p, q" E E and show that p q E E and if either p' E I or q" E I, then p" q" E I. To this end, let U be an open neighborhood of p" .4 and let i E D be given. We show that U fl E; # 0 and if p E I or q E I, then U fl I; 96 0. Pick a neighborhood V of p" such that V q" c U and pick x E E; n V with x E 1; if p E I. If! E I; pick
j E D such that z Ej c I. If x' E E; \I,, pick j E D such that x EJ C_ E; and 2 . I j C Ii. Now 2- g E U so pick a neighborhood W of q such that x- W C U and
picky E W n Ej with y E Ij if q E I. Then x" y E U n E, and if either p E I or
4EI,then . 5 Ufll,.
To complete the proof, assume that (d) or (e) holds. It suffices to establish
(*) if p E K(Q), then p E E. Indeed, assume we have established (*). Then p E E f1 X t ,j K (Q) and X tE K (Q) _ K (X tEJ Q) by Theorem 2.23. Then by Theorem 1.65, K (E) = E n K (X tE.t Q) and,
since I is an ideal of E, K(E) C I. To establish (*), let p E K(Q) be given. To see that p E E, let i E D be given and let U be a neighborhood of T. Pick F E P" f(J) and for each t E F pick some At E p such that ntEF rri 1 [ceps At] C U. Assume now that (d) holds. Since P E K(88S) and {a E S : a 0 E,} is not piecewise syndetic, we have by Theorem 4.40 that {a E S : a 0 E, } 0 p and hence
{aES:REE,}Ep.Pick aE(n,EFAt)n{aES:dEE,}.Then ii EUnE,. Finally assume that (e) holds. Since p E ce T,, pick a E (n,1F At) fl Ti. Then
aEUfE,.
Definition 14.10. 4) is the set of all functions f : N n E N.
N for which f (n) < n for all
We are now ready to prove the Central Sets Theorem (for commutative semigroups).
Theorem 14.11 (Central Sets Theorem). Let S be a commutative semigroup, let A be a central subset of S, and for each e E N, let (yy,n )n_ 1 be a sequence in S. There exist a sequence (an) in S and a sequence (Hn )n° in .% f (N) such that max H,, < min Hn+1 1
1
for each n E N_and such that for each f E 4', FP((an ntEH Yf(n),t)a--1) c A. Proof. Let Y = X o_=, PS. By Theorem 2.22, Y is a compact right topological semigroup and X e01 S C A(Y). Given a E S and H E R f (ICY) let
2(a, H) = (a i 1tEH Yl,t , a M EH Y2,t , a - 111EH Y3,t ,
)
285
14.3 The Commutative Central Sets Theorem
(so that x!(a, H) E X e01 S C Y). For each i E N, let T, = S, let I; = {x2(a, H) : a E S, H E Yf (N), and min H > i },
and let E; = I; U {a : a E S}. We claim that Ti, E,, and I, satisfy hypotheses (a), (b), (c), and (d) of Lemma 14.9. Indeed, (a) and (d) hold trivially. To verify (b), let i E N and X E I; be given. Pick a E S and H E Yf (N) such that min H > i and X = !(a, H). Let j = max H + 1. Then, since S is commutative, one sees easily that Ej C Ii. To verify (c), let i E N and z r= E; \I, be given. Pick a E S such that x"
cl;.
x" =i.Then 1-E;
Since A is central in S, pick an idempotent p E K (fl S) such that A E p. By Lemma
14.9pEI=niENcayII. We now construct the sequences (an)n°O_1 and (H,,)n° inductively. The construction
_
1
is reminiscent of the proof of Theorem 5.8.
Let A* = A*(p) = {x E A : x-1A E p} and let U = {q E Y : ql E A*}. By Theorem 4.12 A* E p so U is a neighborhood of p. Since also p E I, pick z' E U n I,.
Pick al E S and H1 E Pf(N) such that z' = z(al, Hl). Since x(al, Hl) E U we have that al n > 1 and assume that we have chosen (am)m 11 and (Hm)m=1 so that, if m E { 1, 2, ... , n - 2}, then max Hm < min H,,,+ 1, and so that for any f E 4), FP((am nrEHm Yf(m),r)11) = c A*. Let k = max Hn_1 + 1. Let
L = UJE1 FP((am
n/EH, Yf(m).t)m-11).
By Lemma 4.14 we have for each b E L that b-1 A* E p. Let B = A* n nbEL b-1 A*.
Since L is finite (although (D isn't) we have that B E p. Let U = {q" E Y : for all e E { 1 , 2, ... , n}, qe E B). Then U is a neighborhood of p so pick i E U n Ik. Pick an E S and Hn E J" f(N) with min Hn > k such that z = x' (an, Hn). To complete the proof we need to show that for any f E 4), FP((am
ntEHm Yf(m),t)n1=1) C A*.
So let f E 4) and let 0 0 F C (1, 2, ... , n). If n V F, then nmEF(am ' ntEH,,, Yf(m),t) E FP((am . ntEHm Yf(m),t)m 11) C
A*
so assume that n E F. Now an
n:EH,, Yf(n),t E B since z(an, Hn) = i E U. If F = {n}, then we have IImEF(am ' nzEH,,, Yf(m),t) =an ' ntEH Yf(n),t E B C_ A*. Otherwise, let G = F\{n} and let b = fImEG(am ' T7 111EH,,, Yf(m),t) Then b E L so an' fItEH Yf(n),t E B C b-' A* SO f ImEF(am' n1EHH Yf (m),t) = b'an ntEH Yf(n),t E A*.
Of course we have the corresponding partition result.
Corollary 14.12. Let S be a commutative semigroup and for each f E N, let (Ye,n)n°
1
be a sequence in S. Let r E N and let S = U;=1 A. There exist i E 11, 2, ... , r}, a sequence (an )n° 1 in S and a sequence (H)1 in J'f (N) such that max Hn < min Hn+1 for each n E N and for each f E 4), FP((an I1rEH Yf (n),t)n' 1) 9 A;
14 The Central Sets Theorem
286
Proof. Some Al is central. A special case of the Central Sets Theorem is adequate for many applications. See Exercise 14.3.2.
It is not surprising, since the proof uses an idempotent, that one gets the Finite Products Theorem (Corollary 5.9) as a corollary to the Central Sets Theorem (in the case S is commutative). For a still simple, but more interesting, corollary consider the following extension of van der Waerden's Theorem, in which the increment can be chosen from the finite sums of any prespecified sequence. Corollary 14.13. Let (xn)n° be a sequence in N, let r E N, and let N = U;=1 A. T h e n t h e r e is some i E {1, 2, ... , r} such that for all f E N there exist a E N and d E FS((xn )n° 1) with {a, a + d, a + 2d, ..., a + 1d) C A1. 1
Proof. Pick i E (1, 2, ..., r) such that A; is central in (w, +). For each k, n E N, let 1 as guaranteed by Theorem Yk,n = (k - 1) xn. Pick sequences (an)n° 1 and 14.11. Given f E N, pick any m > f and let a = a. and let d = EtEH,,, xt. Now, given k E (0, 1, ... , P}, pick any f E (D such that f (m) = k + 1. Then
a + kd = an +
E FS((an . EtEHn yf(n).t)
1) S A1. 13
Exercise 14.3.1. Let (xn)no l be a sequence in N, let r E N, and let N = Ui=1 A. Prove that there is some i E { 1, 2, ... , r) such that for all e E N there exist a E N and d E FS((xn)n° 1) and c r= FS((xn2)° 1) with {a, a + d, a + 2d, ..., a + id) U {a +
c,a+2c,...,a+tc} c A;.
Exercise 14.3.2. Let S be a commutative semigroup, let A be central in S, let k e N and for each P E {1, 2, ... , k}, let (ye,n)n__1 be a sequence in S. Prove (as a corollary to Theorem 14.11) that there exist a sequence (a,,)n° 1 and for each f E {1, 2, ..., k} a product subsystem (ze,n)n__1 of (ye,n)n_1 such that for each f E (1, 2, ... , k}, FP((an ze.n)n° t) S A. Show in fact that the product subsystems can be chosen in a uniform fashion. That is, there is a sequence (Hn)n_1 in Pf(N) with max H,, < min Hn+1 for
each n such that for each f E (1, 2, ... , k) and each n E N, ze,n = n,,H ye.t
14.4 The Noncommutative Central Sets Theorem In this section we generalize the Central Sets Theorem (Theorem 14.11) to arbitrary semigroups. The statement of the Central Sets Theorem in this generality is considerably more complicated, because the "a" in the conclusion must be split up into many parts.
(In the event that the semigroup S is commutative, the conclusion of Theorem 14.15 reduces to that of Theorem 14.11.)-
14.4 The Noncommutative Central Sets Theorem
287
Definition 14.14. Let M E N. Then
1m = {(H1, H2, ... , Hm) E d'f(N)m : if m > l and
t E {1,2,...,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
z'(a, H) = (w(a, H, 1), w(a", H, 2), w(a, H, 3),...).
(Observe that if S is commutative, a = fl' a,, H = Um1 Hi, and s(a, H) is as defined in the proof of Theorem 14.11, then x' (a, H) = z(a, H).)
For each i E N, let T, = S, let I; = {i(aa, H) : there exists m E N such that a E S'n}1, H E 1m, and min H1 > i } and let E; = I, U {a : a E S}. We claim that Ti, E,, and I, satisfy hypotheses (a), (b), (c), and (d) of Lemma 14.9. Indeed, (a) and (d) hold trivially.
To verify (b), let i E N ands E Ii be given. Pick M E N, a E 5m+1, and H E 1m i and x' = i(a, H). Let j = max Hm + 1. To see that x" Ei C I,, let E Ej. If = s f o r some s E S, let b ' = (a1, a2, ...,am, am+1 s). Then such that min Hl
= i(bb, if) E I,. So assume y E Ij, pick k E N, E E Sk+l, and G E 1k such that min G1 > j and y" = i(b, G). Let c = (a1, a2, ... , am, am+1 b1. b2, ... , bk+l ) and let K = (H1, H2,..., Hm, G1, G2,..., Gk). Then c' E Sk+m+t, K E 1m+k and x"
To verify (c), let i e N and l E E, \I, be given. Pick a E S such that s = a. Then as above one easily sees that x" E, c E, and x" I, C I,. Since A is central in S, pick an idempotent p E K (PS) such that A E p. By Lemma
14.915 EI=fiElyctyIi. Now we construct the sequences (m(n))n° t, (a""n)n° and (l~In)n_ inductively. Write A* = A*(p). Let U = {q" E Y q1 E A*). By Theorem 4.12 A* E P SO U is a neighborhood of T. Since also 15 E I, pick y" E U fl 11. Pick m(1) E N, a"1 E Sm(1)+l, and H1 E 1m(1) such that y' = z(a1, H1). Since i(a"1, H1) E U we have that w(a1, H1, 1) E A*. Inductively, let n > 1 and assume that we have chosen (m(j))l=', and so that, if j E { 1, 2, ... , n - 2}, then max Hj,m(j) < min Hj+1,1, and so that
for any f E 4',FP((w(aj, Hj, f(j)))j=1) c A.
14 The Central Sets Theorem
288
Let k =
Let E = UfE.v FP(.(w(dj, flj, By Lemma 4.14 we have for each b E E that b-IA* E p. Let B = A* nnb,E b- 'A*. Since E is finite we h a v e that B E p. Let U = { q ' E Y : f o r all f E {1, 2, ... , n}, qe E B}. Then U is a neighborhood of p so pick y' E U n Ik. Pick m(n) E N, 1.
do E Sm(n)+1 and Hn E .tm(n) with min Hn,I > k such that Y = Z(dn, Hn). To complete the proof we need to show that for any f E CD,
FP((w(aj, Hj, f(j)));_1) C A'. (1,2,
. . .
, nn}. ).
If n 0 F, then
njEF w(dj, Hj, f(j)) E FP((w(dj, Hj, f(j)))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 <
22k
and 22k ly, then x + y E A.
Consequently for f E 0 one has FS((an + EfEH Yf(n),t)°O_1) c A. We now proceed to derive a combinatorial characterization of central sets. This characterization involves the following generalization of the notion of a piecewise syndetic set.
Definition 14.19. Let (S, .) be a semigroup and let A c P(S). Then A is collectionwise piecewise syndetic if and only if there exist functions G : Pf(A) -* Pf(S) and x : Pf(A) x Pf(S) -a S such that for all F E Pf(S) and all .F and 3e in Pf(A) with
F c 3e one has F x(3e, F) c UZEG(.)t-' (n P) Note that a subset A is piecewise syndetic if and only if {A} is collectionwise piecewise syndetic. An alternate characterization is: A is collectionwise piecewise syndetic if and only if there exists a function G :.Pf(A) --> Pf(S) such that {y-1(G(P))-'(n F) : y E Sand' E Pf(A)}
has the finite intersection property.
(Here y-' (G(.F))-' ( n F) = UIEG(.F)Y I t-' ( n F), see Exercise 14.5.1.) In the event that the family and the semigroup are both countable, we have a considerably simpler characterization of collectionwise piecewise syndetic.
Lemma 14.20. Assume the semigroup S = (an : n E N} and {An : n E NJ C P(S). The family {An : n E N} is collectionwise piecewise syndetic if and only if there exist a sequence (g(n))n° 1 in N and a sequence (yn)n° 1 in S such that for all n, m E N with
m>n, {alym,a2Ym,...,amym} c U8(l a!-'(nn IAi). Proof For the necessity, pick functions G and x as guaranteed by Definition 14.19. Given n E N, let
g(n) = min{k E N : G({A1, A2, ... , An)) c {al, a2, ..., ak}} and y n = x({A1, A2, ... , An}, {al, a2, ..., an}). Then if m > n, letting
F _ {A1, A2, ... , An},
3e = {A1, A2, ... , Am},
and F = {al, a2, ... , am},
one has F C Re and
F x(Je, F) c UrEG(.y)t-' (n r)
{alym, a2Ym,... , amym} =
s(n)
c U,=1
ai_
1
(n,=1
A,).
For the sufficiency, let (g(n))n° and (yn)R I be as in the statement of the lemma. I
Given F E Pf((An : n E N}), let m = max{k E N : Ak E F} and let G(F) _
291
14.5 Combinatorial Characterization
{a1, a2,. - ., ag(m)). Given 3e E ,?'f((An : n E N)) and F E 3'f (S), let m = max(k E N : Ak E 3e or ak E F) and let x(3e, F) = ym. Then if .F, 3e E 3'f({An : n E N)),
FEpf(S), "cJe,m=max{kEN:AkE3eorakEF},andn=max{kEN: Ak E F), then n<mso F -x(3e, F) C {a1Ym,a2Ym,
- , amYm}
c Ui=t aj- (nf=t A;) 8(n)
n
1
c utEG(y)t_'(nF)
11
Theorem 14.21. Let (S, ) be an infinite semigroup and let A C .P (S). There exists p E K(f S) with A c p if and only if A is collectionwise piecewise syndetic.
Proof Necessity. Pick P E K ($ S) such that A c p. For each F E 3'f (.A), n F E p,
so by Theorem 4.39, B(F) = It E S : t-' (n .F) E p} is syndetic. Pick G(F) E UZEG(F)t-' B(.F). For each y E S and each F E ?f (A), pick ,Pf(S) such that S = F))-1 B(F) (so that (t (y, .F) -y)-' (n .F) E P). t(y, F) E G(F) such that y E (t(y, Given 3e E R f (A) and F E ,Pf (S), { (t (y, F) y)-' (n 3:') : y E F and 0 # F c 3e}
is a finite subset of p so pick
x(3e, F) E n{(t(Y, F) - Y)-1 (n F) : y E F and O :A F C 3e}.
To see that the functions G and x are as required, let F E ?(S) and let F, 3e E UZEG(.n)t-' (n F), let y E F. Then P f (,A) with .F c 3e. To see that F x (3e, F) c x(3e, F) E (t (y, .F) Y)-1(n .F), so
y . x(3e, F) E (t(y, F))-1(n F) c UtEG(T)t-`(n F). Sufficiency. Pick G : .i'1(A) -> Pf(S) andx : J'f(A) x pf(S) -> S as guaranteed by the definition. For each F E ?f (.A) and each y E S, let
D(F,y)={x(3e,F):3eEJ"f(A), FEJ"f(S), .FcJ?, andyEF). Then {D(F, y) :.F E ?1(A) and y E S) has the finite intersection property so pick u E PS such that {D(.F, y) :.F E 5'f (,A) and y E S) c u. We claim that (*) for each F E .P f(,A), S S. u C- U,EG(.x)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.
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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.
298
15 Partition Regularity of Matrices
Theorem 15.5. Let (S, +) be an infinite commutative semigroup with identity 0, let u, v E N, and let A beau x v matrix with entries from w which satisfies the first entries condition. Let C be central in S. If for every first entry c of A, cS is a central* set, then there exist sequences (xl,n)n° 1, (x2,n)n__l, ..., (x,,,n)n°_1 in S such that for every F E ?f (N), xF E (S\{0})v and AIF E Cu, where EnEF xl.n
XF=
EPEF x2.n
EnEF xv,n
Proof S\{0} is an ideal of S so by Corollary 4.18, 0 is not a minimal idempotent, and thus we may presume that 0 f C. We proceed by induction on v. Assume first v = 1. We can assume A has no repeated rows, so in this case we have A = (c) for some c E N such that cS is central*. Then C fl cS is a central set so pick by Theorem 14.11 (with the sequence yl,n = 0 for each n and the function f constantly equal to 1) some sequence (kn)n°_1 with FS((kn)n°-l) c C fl cS. (In fact here we could get by with an appeal to Theorem 5.8.) For each n E N pick some xl,n E S such that k,, = cxThe sequence (xl,,,)n° l is as required. (Given F E 9f(N), EnEF kn 0 0 so EnEF X1,n 0 Now let v E N and assume the theorem is true for v. Let A beau x (v + 1) matrix with entries from co which satisfies the first entries condition, and assume that whenever
c is a first entry of A, cS is a central* set. By rearranging the rows of A and adding additional rows to A if need be, we may assume that we h a v e some r E 11, 2, ... , u and some d E N such that -
_
0
1}
ifi E (1,2,...,r}
a`'l - d ifi E {r+1,r+2,...,u}. Let B be the r x v matrix with entries bij = aij+l. Pick sequences (zl,n)n=1, (z2,n)n_-l,
, (zv,n)n_l in S as guaranteed by the induction hypothesis for the matrix
B. For each i E {r+1,r+2,...,u} andeachn E N, let u+l
Yi,n = EJ=2 aij - 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
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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 <£2 <... <1msuchthat{11,12,..., Im}isapartitionof(1,2,...,v) and for each t E {1, 2, ... , m} and each j E I,, I(xj) = t,. For n E {2, 3, ... , m}, if any, let Jn = Ui=i We now show that
(1) EjEl, c = 6 and (2) given n E (2, 3, ... , m), E jE,, cj is a linear combination over Q of (c"j) jEjq, so that C satisfies the columns condition over Q.
To establish (1), let d = 11. Then for any j E 11 let ej = b(xj) and note that xj = a qd + ej qd+1. For j E (1, 2, ..., v)\Il we have f(xj) > d so xj = ej qd+l for some e j E N. Then
0 = Ej_l xj
c"j
EJEI, a . qd , Cj + EJ_l ej qd+l , cj.
Suppose that EjEl, c'j Then
0 and pick some i E 11, 2, ... , u) such that EJEl, ci, j 96 0.
so, since q Xa we have qI EjEI, ci, j, contradicting our choice of q.
Now consider (2). Let n E (2, 3, ... , m) be given and let d = E,,. For j E In, let ej = b(xj) and note that xj = a qd +ej . qd+l. For j E U7 n+l Ii, if any, pick ej E N
such that xj = ej qd+t Thus
6 =Ev Let
UT
EjEU.iint_(-ej) - cj. Then
-
qd+I
i = EjEJn xj
j + EjEln a qd
.j
Since q 0 E(C, Jn, 1n) we have that EjEln c'j is a linear combination of required. (c) implies (a). This follows from Theorem 15.16(c).
as
15.3 Kernel Partition Regularity Over N
307
To illustrate the use of Theorem 15.20 we establish the case t = 4 of van der Waerden's Theorem. That is, we show that whenever N is finitely colored we can get
some a, d E N with a, a + d, a + 2d, a + 3d monochrome. So one lets xt = a,
x2 =a+d,x3 =a+2d,andx4 = a+3d. Then x2-xi =d SO X3 =xi+2(x2-xi) and x4 = X1 + 3(x2 - x i ). That is we have the equations
xi -2x2+x3 = 0 2xi - 3x2 + X4 = 0 so we are asking to show that the matrix 1
-2
1
2 -3 0
0 1
is kernel partition regular. Indeed it satisfies the columns condition because the sum of its columns is 0.
But there is a problem! The assignment xi = x2 = X3 = X4 = 1 (or any other number) solves the equations. That is, these equations allow d = 0. So we strengthen the problem by asking that d also be the same color as the terms of the progression. Let xi = a, x2 = d, x3 = a +d, x4 = a + 2d, and x5 = a + 3d. Then we get the equations
X3 = xl +X2 X4 = xi + 2x2 X5 = xi + 3x2 so we need to show that the matrix 1
1
1
2
1
3
-1
0
0 0
-1
0 0
0
-1
is kernel partition regular. For this, let Ii = 11, 3, 4, 5}, 12 = 12}, 52,1 = 0, 82,3 = -1,
52,4 = -2, and 52,5 = -3. Exercise 15.3.1. Derive the finite version of the Finite Sums Theorem as a consequence
of Rado's Theorem. That is, show that for each n, r E N, whenever N = Ui=1 Aj, there exist i E 11, 2, ..., r) and (xt)" 1 such that FS((xt)" 1) c A. For example, the case n = 3 requires that one show that the matrix
is kernel partition regular.
1
1
0 -1
1
0
1
0
1
1
1
1
1
0 0 0
0
-1 0 0
0 0
-1
0 0 0
0
-1
15 Partition Regularity of Matrices
308
15.4 Image Partition Regularity Over N We concluded the last section with a verification of the length 4 version of van der Waerden's Theorem via Rado's Theorem. To do this we had to figure out a set of equations to solve and had to strengthen the problem. On the other hand, we have already seen that the original problem is naturally seen as one of determining that the matrix 1
0
1
1
1
2
1
3
is image partition regular. (And the strengthened problem clearly asks that the matrix 0
1
1
0
1
1
1
2
1
3
be image partition regular.) Since many problems in Ramsey Theory are naturally stated as deciding whether certain matrices are image partition regular, it is natural to ask for a determination of precisely which matrices are image partition regular. We do this in this section. Since we are now dealing with the semigroup (N, +), we extend Definition 15.1 to allow entries from Q.
Definition 15.21. Let u, v E N, and let A beau x v matrix with entries from Q. Then A is image partition regular over N if and only if whenever r E N and N = Uii=1 Ei, there exist i E { 1 , 2, ... , r} and x E N' such that Ax E E, u. We need a preliminary lemma which deals with a specialized notion from linear algebra.
Definition 15.22. Let u, v E N. let cl, c2, ... , c" be in Q", and let I c {1, 2, ... , v}. The I-restricted span of (c'1, c'2, ... , {E
1 ai ci
:
is
each a1 E 1R and if i E I, then ai > 0}.
Lemma 15.23. Let u, v E N, let c ' 1 , c ' 2 , ... , c " be in Qu, and let 1 c { 1, 2, ... , v}. Let
S be the I-restricted span of (cl, c2, ... , c'v). (a) S is closed in R".
(b) If y E S rl Q", then there exist 81, 32, ... , S in Q such that y = E Si > O for each i E I.
Si
c'i and
Proof. (a) We proceed by induction on I I I (for all v). If I = 0, this is simply the assertion
that any vector subspace of R' is closed. So we assume 10 0 and assume without loss
309
15.4 Image Partition Regularity Over N
of generality that 1 E I. Let T be the (I\(1))-restricted span of (c2, c3, ... ,
By the induction hypothesis, T is closed. To see that S is closed, let b E C1 S. We show b E S. If cj = 0, then b E cZ S = such that c2 T = T = S, so assume that cl 0 0. For each n E N, pick (ai (n)) ai (n) > 0 for each i E 1 and 11b - E I a; (n) c"; II < 1/n. Assume first that {aI (n) : n E N} is bounded. Pick a limit point S of the sequence
(ai(n))n° t and note that S > 0. We claim that b - 8 cl E T. To see this we show that b - S c"i E ct T. So let E > 0 be given and pick n > 2/c such that jai (n) - SI < E/(21Ict II). Then IIb - S c1 - Ei_2 ai (n) - ci 11
llb - Ei=t a; (n) iF; Il + llai (n)cl - Scj 11 < E.
Sinceb-
E T, we have Now assume that {al (n) : n E N) is unbounded. To see that S is closed, it suffices to
show that -cj E T. For then S is in fact the (I\(l))-restricted span of (cl, c'2, ... , which is closed by the induction hypothesis.
To see that -ci E T, we show that -c1 E ct T. Let E > 0 be given and pick n E N such that al (n) > (1 + IIbII)/E. For i E {2, 3, ... , v} let Si = a; (n)/c i (n) and note that for i E I \(1), S; > 0. Then
II-el - E2 Si
II
(b) Again we proceed by induction on III. The case I = 0 is immediate, being merely the assertion that a rational vector in the linear span of some other rational vectors is actually in their linear rational span (which is true because one is solving linear equations with rational coefficients).
So assume 1 # 0. Let X = {x E ]E8°
:
Ziv= I xi
c;
Thus X is an affine
subspace of Rv, and we are given (by the fact that y E S) that there is some x' E X such that xi > 0 for all i E 1. Also (by the case I = 0) there is some 'z E X such that z; E Q f o r all i E 11, 2, ... , v). If z; > 0 for all i E I, then we are done, so assume there is some i E I such that zi < 0. Given i E I, {t E [0, 11 : (1 - t)xi + tzi > 0} is [0, 1] if
zi > 0 and is [0, xi/(xi - zi)J if zi < 0. Let t be the largest member of [0, 1] such that the vector w = (I - t) x + t - z satisfies wi > 0 for all i E I. Then for some i E I we have that wi = 0. We assume without loss of generality that 1 E I and wI = 0. Then y = Ei_2 wi c ; so y ' is in the (I\{I})-restricted span of (c2, c3, ... , c) so by the induction hypothesis we may pick S2, 33, ..., S in Q such that y' = Ei=2 Si ci and Si > 0 for each i E I\[ 1). Letting S i = 0, we are done.
Theorem 15.24. Let u, v E N and let A be a u x v matrix with entries from Q. The following statements are equivalent.
(a) A is image partition regular over N.
15 Partition Regularity of Matrices
310
(b) Let c"t, cZ...... be the columns of A. There exist st, s2, ... , sv E {x E Q : x > 0} such that the matrix
-1
0
0
M=
1
... ...
0 0
Sv'Cv 0
0
... -1
is kernel partition regular over N.
(c) There exist m E N and a u x m matrix B with entries from Q which satisfies the first entries condition such that given any 3 E N' there is some x E Nv with
Al =By. (d) There exist m E N and a u x m matrix C with entries from Z which satisfies the first entries condition such that given any y' E N' there is some x" E N° with Ax" = Cy. (e) There exist m e N and a u x m matrix D with entries from w which satisfies the first entries condition such that given any y' E N' there is some x" E N" with
Ax"=Dy'. (f) There exist m E N, a u x m matrix E with entries from w, and c E N such that E satisfies the first entries condition, c is the only first entry of E, and given any y E N' there is some x' E N° with A. = Ey. Proof. (a) implies (b). Given any p E N\{1}, we let the start base p coloring of N be the function
op:N -. {1,2,...,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.
312
15 Partition Regularity of Matrices
Now let k E {2, 3, ... , m) and assume we have chosen E1, E2, ... , Ek_ I e {1, 2, ... , v}, Si E Q+ for i E U;=1 Ei, and Ii = Ei U (v + Di) as required for the columns condition. Let Lk = U Ei and let Mk = U;=, Di and enumerate Mk in order as q(1), q(2), ..., q(r). We claim that it suffices to show that EiEDk di is in the { 1 , 2, ... , v)-restricted span of (c1, c " 2 , ... , IF,,, dq(1), dq(2), ... , dq(r)), which
we will again denote by S. Indeed, assume we have done this and pick by Lemma 15.23(b) a1, a2, ... , a in {x E Q : x > 01 and 6. (1), 8 q ( 2 ) . . . . . Sq(r) in Q such that EiEDk di = Ei I ai ci + Ei=1 Sq(i) dq(i). Let Ek = (i E {1, 2, ... , v}\Lk : ai > 0) and for i E Ek, let si = ai. Let Ik = Ek U (v + Dk). Then + EiEDk -di EiEIk bi = EiEEk ai = EiEEk of 61 + Ei=1 -ai ' ,F; + Ei=1 -Sq(i) dq(i) Now, if i E (1, 2, ... , v) and ai # 0, then i E Lk U Ek so
EiEEk ai'ci+E° 1 -a, ci+Ei=1 -Sq(i)' dq(i) = EiELk
ci+Ei=1 -Sq(i)' dq(i)
and
EiELk -ai ' 4 + E1=1 -Sq(i) Let jai = -ai /si if i E Lk and let E.
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.
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16 EP, EP*, Central, and Central* Sets
(a) A is an JP* set. (b) A is a member of every idempotent of 14S.
(c) A fl B is an IP set for every IP set B of S.
Proof. (a) implies (b). Let p be an idempotent of CBS and suppose that A f p. Then in S with FP((xn )n° t) C S\A. S\A E p so by Theorem 5.8 there is a sequence (xn That is, S\A is an IP set which misses A, a contradiction. (b) implies (c). Let B be an IP set and pick by Lemma 5.11 an idempotent p of fS such that B E p. Then A fl B E p so A fl B is an IP set by Theorem 5.8. 1
That (c) implies (a) is trivial. As a trivial consequence of Lemma 15.4, Theorems 16.4 and 16.6, and the definition of central, one has that IP*
central* = central
IP.
One also sees immediately the following:
Remark 16.7. Let S be a semigroup and let A and B be subsets of S. (a) If A and B are IP* sets, then A fl B is an IP* set. (b) If A and B are central* sets, then A fl B is a central* set.
We see now that in most reasonable semigroups, it is easy to produce a specific central* set which is not an IP* set. Theorem 16.8. Let S be a semigroup and assume that (x n ) is a sequence in S such that FP((xn )n°_ 1) is not piecewise syndetic. Then S\ FP((xn) n 1) is a central* set which is not an IP* set.
Proof. Trivially S\FP((xn)n° 1) is not an IP* set. Since FP((xn)n° 1) is not piecewise syndetic, we have by Theorem 4.40 FP((xn)n_1) fl K($S) = 0 so for every minimal idempotent p, S\ FP((xn )n_1) E P. Of course, since the notions of central and IP are characterized by membership in an idempotent, they are partition regular notions. In trivial situations (see Exercises 16.1.2 and 16.1.3) the notions of central* and IP* may also be partition regular.
Lemma 16.9. Let S be a semigroup. (a) The notion of IP* is partition regular in S if and only if fS has a unique idempotent. (b) The notion of central* is partition regular in S if and only if K(fS) has a unique idempotent.
Proof We establish (a) only, the other proof being very similar. Necessity. Suppose that fS has two idempotents p and q and pick A E p\q. Then A U (S\A) is an IP* set while neither A nor S\A is an IP* set.
16.1 Sets in Arbitrary Semigroups
323
Sufficiency. Let p be the unique idempotent of fS. Then a subset A of S is an IP* set if and only if A E p. As a consequence of Lemma 16.9 we have the following.
Remark 16.10. Let S be a semigroup. If the notion of IP* is partition regular in S, then so is the notion of central*.
Exercise 16.1.3 shows that one may have the notion of central* partition regular when the notion of IP* is not. We see now that in more civilized semigroups, the notions of EP* and central* are not partition regular. (Theorem 16.11 is in fact a corollary to Corollary 6.43 and Lemma 16.9, but it has a simple self contained proof, so we present it.) Theorem 16.11. Let S be an infinite weakly left cancellative semigroup. There exist disjoint central subsets of S. Consequently, neither the notions of central* nor IP* are partition regular in S.
Proof Let K = IS1 and enumerate Pf(S) as (Fa)a
(Uo
xa E Ssuch that
0
andchoose ya ESsuch that By Theorem 16.1 we have that ( Jo
ya) are disjoint
In any semigroup, we see that IP* sets satisfy a significantly stronger combinatorial conclusion than that given by their definition.
Theorem 16.12. Let S be a semigroup, let A be an IP* set in S, and let (xn)n° 1 be a sequence in S. There is a product subsystem (yn)n°1 of (xn)n 1 such that FP((Yn)n°_1) C A.
Proof. Pick by Lemma 5.11 an idempotent p E fS such that FP((xn)n°_m) E p for every m E N. Then A E p by Theorem 16.6, so Theorem 5.14 applies.
Exercise 16.1.1. Prove Corollary 16.2. Exercise 16.1.2. Let S be a set and let a E S. Define xy = a for all x and y in S. Show that a subset A of S is an IP set if and only if a E A and consequently that the notions of IP* and central* are partition regular for this semigroup.
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16 IP, IP*, Central, and Central* Sets
Exercise 16.1.3. Recall the semigroup (N, A) where n Am = min(n, m). (See Exercise 4.1.11.) Verify each of the following assertions. (a) In (flN, A) every element is idempotent and K(f N, A) _ {1}. (b) The IP sets in (N, A) are the nonempty subsets of N and the central sets in (N, A) are the subsets A of N with 1 E A. (c) The only IP* set in (N, A) is N, while the central* sets in (N, A) are the same as the central sets. Consequently, the notion of IP* is not partition regular in (N, A) while the notion of central* is partition regular in (N, A).
Exercise 16.1.4. Recall the semigroup (N, v) where nvm = max{n, m}. (See Exercise 4.1.11.) Verify each of the following assertions. (a) In (fN, v) every element is idempotent and K(fN, v) = N*. (b) The IP sets in (N, v) are the nonempty subsets of N and the central sets in (N, v) are the infinite subsets of N. (c) The only IP* set in (N, v) is N, while the central* sets in (N, v) are the cofinite subsets of N. Consequently, neither the notions of IP* nor the notions of central* are partition regular in (N, v).
16.2 IP* and Central Sets in N In this section we compare the additive and multiplicative structures of N. We begin with a trivial observation.
Lemma 16.13. Let n E N. Then Nn is an IP* set in (N, +). Proof. This is an immediate consequence of Theorem 16.6 and Lemma 6.6.
0
Now we establish that central sets in (N, +) have a richer structure than that guaranteed to an arbitrary commutative semigroup (and in fact richer than that possessed by central sets in (N, ) ).
Theorem 16.14. Let B be central in (N, +), let A be a u x v matrix with entries from Q. (a) If A is image partition regular, then there exists y E N' such that all entries of Ay are in B. (b) If A is kernel partition regular, then there exists y" E B° such that Ay' = 0. Proof. (a) Pick by Theorem 15.24 some m E N and a u x m matrix D satisfying the first
entries condition with entries from w such that for any i E N' there is some y E N" with Ayy = Di. By Lemma 16.13, for each n E N, Nn is an IP* set and hence a central* set. Consequently, by Theorem 15.5, pick 'z E N' such that all entries of Dz are in B and pick y' E N' such that Ay" = Dzz. (b) Let d E N be a common multiple of all of the denominators of entries of A. Then dA is a matrix with entries from 7G which is kernel partition regular. (If N = u =1 C,,
325
16.2 IP* and Central Sets in N
pick i E (1, 2, ... , r) and x' E C`;' such that Ax' = 0. Then (dA)x'. = 0.) Thus by Rado's Theorem (Theorem 15.20) dA satisfies the columns condition over Q so, by Lemma 15.15, pick m E N and a v x m matrix C with entries from w which satisfies the first entries condition such that d AC = 0. As in part (a), pick x E NM such that all entries of Cxi are in B and let j = Cx'. Then dAy = dAC! = 01 = 0 so Ay" = 0. In fact, central sets in (N, +) not only contain images of all image partition regular matrices, but all finite sums choosing at most one term from each such image as well. be a sequence of subsets Definition 16.15. Let (S, +) be a semigroup and let of S. Then FS((Y,,)n° 1) = {EfEF an : F E .Pf(N) and for all n E F, an E 1
We define FS((Y,,)n 1) and
analogously.
Theorem 16.16. Let B be central in (N, +). Let (A(n))n° enumerate the image partition regular matrices with entries from Q and for each n, let m(n) be the number of columns of A(n). There exists for each n r= N a choice of z(n) E N,(n) such that, if Yn is the set of entries of A(n)x(n), then FS((Yn)' 1) c B. 1
Proof. Pick a minimal idempotent in ($N, +) such that B E p. Pick by Theorem 16.14 some 7C(I) E N'(1) such that all entries of A(I)i(l) are in B* = {a E B : -a + B E P). Let Y1 be the set of entries of A(1)x"(1).
Inductively, let n E N and assume that we have chosen x(k) E N, (k) for each k E {1, 2, ... , n} so that, with Yk as the set of entries of A(k)x'(k), one has FS((Yk)k=1) c
B*. By Lemma 4.14, for each a E B*, -a + B* E p so
B* fl n{-a + B* : a E FS((Yk)k_1)} E p so pick x"(n + 1) E Nm(n+1) such that all entries of A(n + 1)x"(n + 1) are in
B* fl n{-a + B* : a E FS((Yk)k=1)}. Letting Yn+l be the set of entries of A(n + 1). (n + 1) one has FS((Yk)k±i) C B. A similar result applies to kernel partition regular matrices.
Theorem 16.17. Let B be central in (N, +). Let (A(n))n° 1 enumerate the kernel partition regular matrices with entries from Q and for each n, let m(n) be the number of columns of A(n). There exists for each n E N a choice of x' (n) E N'(') such that A(n)z(n) = 0 and, if Yn is the set of entries of. (n), then FS((Yn)' 1) C B. Proof. This can be copied nearly verbatim from the proof of Theorem 16.16.
The contrast with Theorems 16.14, 16.16, and 16.17 in the case of sets central, in fact central* in (N, ) is striking. Recall from Section 15.3 the notation za = 1 which represents the multiplicative analogue of the equation Az = 0.
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16 IP, IP*, Central, and Central* Sets
Theorem 16.18. Let C = N\(x2 : x E N), let A = (2) and let B = (2 -2 1). Then C is central* in (N, ), A is image partition regular in (N, ), and B is kernel partition regular in (N, ). However, for no x E N is X2 E C and for no z E C3 is
xe=1. Proof The reader was asked to show in Exercise 15.1.2 that (x2 : x E N) is not central in (N, ). Consequently, its complement must be central*. Trivially A is image partition regular in (N, ) and B satisfies the columns condition over Q, so is kernel partition regular in (N, ) by Theorem 15.20. The following lemma establishes that multiplication preserves IP* sets in (N, +).
Lemma 16.19. Let A C N and let n E N. The following statements are equivalent.
(a) A is an IP* set in (N, +). (b) n-1A is an IP* set in (N, +). (c) nA is an IP* set in (N, +). Proof. (a) implies (b). Assume that A is an lP* set, and let B be an IP set. Pick a sequence
(xt)°_, such that FS((xt)t°O1) c B. Then FS((nxt)°1) fl A 96 0 so n-1A n B # 0. (b) implies (a). Assume that n-1A is an IP* set, and let B be an EP set. Pick
a sequence (x,)', such that FS((xt)°°,) C B. By Lemma 16.13, Nn is an IP* set in (N, +) so pick by Theorem 16.12 a sum subsystem (yt)°°, of (x,)°_1 such that FS((yt)°_°,) C Nn. In particular n divides yt for each t. Consequently
FS((A )°_-1) fl n-1A # 0 so that FS((yt)°O,) fl A 54 0. Since FS((yt)°_,) S; FS((xt)°__i) c B, we are done. Since n-1(nA) = A, the equivalence of (a) with (c) follows from the equivalence of (a) with (b). We see now that IP* sets in (N, +) are guaranteed to have substantial multiplicative structure. Theorem 16.20. Let -3 be the set of finite sequences in N (including the empty sequence)
and let f : -8 -s N. Let (yn )n° , be a sequence in N and let A be an IP* set in (N, +). Then there is a sum subsystem (xn)n° , of (yn)n°_, such that whenever F E J'f(N), Z = f ((xk)k 1 F-1), and t E (1, 2, ... , 8), one has t EnEF Xn E A. Proof. Pick by Lemma 5.11 some p = p + p in fiN such that FS((yn)n_m) E p for each m E N. Then by Lemma 16.19, we have for each t E N that t-1 A is an lP* set in (N, +) and hence is in p. Let B1 = FS((yn)n° 1) fl n` 10) tA and note that B1 E P. Pick X1 E B1* and pick H1 E .% f(N) such that x1 = EtEH, Yt Inductively, let n E N and assume that we have chosen (x i) l , , (H,)=,, and (B,)"_, such that for each i E { 1, 2, ... , n}:
16.2 IP* and Central Sets in N
327
(1) Xi = ErEH; yr, (2) if i > 1, then min Hi > max H1 _ 1,
(3) Bi E p, (4) if 0 F c {1, 2, ... , i} and m = min F, then EjEF xj E Bm*, and
(5) if i > 1, then Bi C n;((Xj)'') CIA. Only hypotheses (1), (3), and (4) apply at n = I and they hold trivially. Let k = max Hn + 1. By assumption FS((yt)°__k) E p. Again we have by Lemma 16.19 that for each t E N, 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 <
E A.
Theorem 16.20 and Corollary 16.21 establish that an IP* set in (N, +) must have substantial multiplicative structure. One may naturally ask whether similar results apply
to central* or central sets. On the one hand, we shall see in Corollary 16.26 that sets
328
16 IF, IP*, Central, and Central* Sets
which are central* in (N, +) have significant multiplicativestructure, in fact are central in (N, ). On the other hand, central sets in (N, +) need have no multiplicative structure at all (Theorem 16.27) while central sets in (N, ) must have a significant amount of additive structure (Theorem 16.28).
Definition 16.22. M = {p E ON: for each A E p, A is central in (N, +)}. As an immediate consequence of the definition of central we have the following.
Remark 16.23. M = cf E(K(6N, +)). Theorem 16.24. M[ is a left ideal of (13N, ). Proof. Let P E M[ and let q E flN. To see that qp E MI, let B E qp and pick a E ICY such that a-1 BE p. Then a-1 B is central in (N, +) so pick, by Theorem 14.25 a decreasing
sequence (C,)', such that (i) for each n E N and each x E Cn, there is some m E N such that Cm g -x + Cn and (ii) {Cn : n E NJ is collectionwise piecewise syndetic.
Then (aCn)n 1 is a decreasing sequence of subsets of B and immediately one has that for each n E ICY and each x E a Cn, there is some m E N such that aCm c -x +a Cn .
It thus suffices by Theorem 14.25 to show that {aCn : n E N} is collectionwise piecewise syndetic. That is, using Lemma 14.20, that there exist sequences (g(n))n° and (xn)' 1 in N such that for all n, m E N with m > n, 1
{Xm+1,xm+2,...,Xm+m} c US(n1(-.1+ni=l aCi) If a = 1, the conclusion is immediate from the fact that {Cn : n E N } is collectionwise piecewise syndetic so assume that a > 1. Since {Cn : n E N} is collectionwise piecewise syndetic, choose sequences (h (n))' n=1 and (yn )n° 1 in N such that for all n, m E N with m > n,
For each n E N, let xn = ayn and let g(n) = ah(n) + a. Given n, m E N with m > n and given k E [1, 2, ... , m }, pick r E {0, 1, ..., a - 11 such that k + r = al for some
I E {1,2,...,m}. Then for some j E {1,2,...,h(n)}wehaveym+I 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.
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Proof Let xt = 1 and for n E N pick some xn+1 > nxn. Let
A={kxn:n,kENandk
system of (yn )'1 can be chosen. Pick m, n, and r in N such that n < r, Ym E {xn, 2xn, 3xn, ... , nxn}, and ym+t E {xr, 2Xr, 3Xr, ... , rxr}. Pick k E (1, 2, ... , r} such that ym+t = kxr. Then
kxr < ym + ym+t <- kxr + nxn < kxr + xr = (k + 1)Xr
sOym+Ym+i 0A. Given any central set A in (N, +), one has by definition that there is a minimal idempotent p in (fiN, +) such that A E p. Since p = p + p, one has that
{xEN:-x+AEp}Ep and so, in particular, (x E N : -x + A is central) is central. That is, every central set often translates down to a central set.
Recall that for any p E ON, -p = (-1) p E fiZ. Since by Exercise 4.3.5, -N* is a left ideal of (f Z, +), one cannot have p = p + (-p) for any p E N*. If one had p = (-p) + p for some minimal idempotent pin (fiN, +) one would have as above that for any A E p, {x E N : x + A E p) E p and hence {x E N :_x +A is central} is central. However, according to Corollary 13.19 such an equation cannot hold. Nonetheless, we are able to establish that for any central set A in (N, +), {x E N : x + A is central) is central.
Theorem 1630. Let P E K(fiN, +) and let L be a minimal left ideal of (ON, +). There is an idempotent q E L such that p = (-q) + p.
Proof. Let T = {q E L : (-q)+p = p}. Givenq, r E ON, -(q+r) = (-q)+(-r) by Lemma 13.1, soifq,r E T,thenq+r E T. Define : f7G -+ fiZbye(p) = -p. (That is,inthesemigroup(f7G, ),2 =.1._1.) ThentiscontinuousandT = Lfl(ppot)-1[{p}] so T is compact. Therefore, it suffices to show that T # 0. For then, T is a compact subsemigroup of ON, hence contains an idempotent by Theorem 2.5.
Since P E K(fN, +), pick a minimal left ideal L' of ON such that p E L'. Now N* is a left ideal of f7L by Exercise 4.3.5 so by Lemma 1.43(c), L and L' are left ideals
of fZ. In particular -L + p S L. We claim that -L + p is a left ideal of ON so that
-L+p = L'. Toseethis,letq E -Landletr E ON. Weshowthatr+q+p E -L+p. Now -q E L so -r + -q E L. Since -r + -q = -(r + q) by Lemma 13.1 we have
r + q E -L so r + q + p E -L + p. Since -L+p=L'andpEL'wehave T#0 as required.
16.2 IP* and Central Sets in N
331
Corollary 16.31. Let A C N be central in (N, +). Then (x E N : x + A is central) is central:
Proof. Pick a minimal idempotent p in (fN, +) such that A E p and pick, by The-
orem 16.30, a minimal idempotent q in (,ON, +) such that p = (-q) + p. Then
{xEN:x+AEp}Egand Ix EN:x+AEp}C{xEN:x+A is central}. We do not know whether every central* set often translates either up or down to another central* set. However we do have the following strong contrast with Corollary 16.31 for IP* sets. Theorem 16.32. There is an JP* set A of (N, +) such that for all n E N, neither n + A nor -n + A is an JP* set.
Proof. Let (Dn)nEZ be a sequence of pairwise disjoint infinite sets of positive even integers such that for each n, min D,, > 12n 1. For each n E Z enumerate D. in 2a(n,2k-1) increasing order as (a(n, k))k_1 and for each k E N, let Yn,k = For each n E 76\{0}, let Bn = FS((yn,k)k° 1). Let C = {n +z : n E 76\{0} and z E Bn}.
Notice that if n E Z\{0} and Z E Bn, then Ini < z so C c N. Let A = N\C. Then given any n E 7G\(0), (A + n) fl B_n = 0, so A + n is not an IP* set. We now claim that A is an EP* set, so suppose instead we have a sequence (xn)n_1 with FS((xn)n° 1) fl A = 0. That is, FS((xn)n° 1) c C. By passing to a suitable sum subsystem we may presume that the sequence (xn)n° 1 is increasing. We first observe that if n E Z\{O}, U E n + Bn, and u = EtEF 2', then max F E D.
Indeed u = n + z where z E Bn and z = EtEH 2' where H C Dn and H has at least two members (because each yn.k has two binary digits). Thus, if n < 0, borrowing will not reach max H. Since then min H > 12n I one has max F = max H. We now show that one cannot haven E 7.\{0} and i < j with {xi, x!) C n + Bn, so
suppose instead that we do. Now xj = E,EF 2' where max F E D. Let k = max F. Since xi < xj we have that xj + xi = E,EH 2` where either max H = k or max H = k + 1. But as we have just seen, given any member of C, the largest element of its binary support is even, so the latter case is impossible. The former case tells us that xj + xi E n + B. But now {xi - n, xj - n, xi + xj - n) C Bn, so if P = min Dn we have that 2e divides each of xi - n, xj - n, and xi + xj - n, and hence 21 divides n, a contradiction.
Consequently, we may choose for each i E N. some n(i) E Z\{0} such that xi E
n(i) + Bn(,) and n(i) 96 n(j) for i # j. Choose i such that Jn(i)i > x1. Then xi = E,EF 2t and xi + xl = E,EG 21 where max F = max G and consequently, Xi + X1 E n (i) + Bn(i ). But now x1 = (xi + x1) - xi is a difference of two members of B,,(i) and is hence divisible by 2e, where f = min D(1). This is a contradiction because
f > 12n(i)I > xi. Exercise 16.2.1. In the proof of Lemma 16.19 we used the fact that in the semigroup (N, +), one must have 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.
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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)}.
334
16 EP, IP*, Central, and Central* Sets
Only hypotheses (1), (3), and (4) apply at n = 1 and they hold trivially. Let k = max Hn + 1. By assumption FS((yt)°_k) E p. Again we have by Lemma 16.35 that for each a E S, 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
338
16 IP, EP*, Central, and Central* Sets
one to one, ga,y[A] fl ga,y[B] = 0 while by Theorem 16.42(d), ga,y[B] is an IP set, a contradiction. (c) The necessity is Theorem 16.42(c). Assume that ga,y[A] is a central set and suppose that A is not a central set. Then N\A is a central* set so by Theorem 16.42(b), ga,y[N\A] is a central* set. But this is a contradiction because ga,y[A] fl ga,y[N\A] _ 0. Because for a > 1 the output of ga,y is the same kind of set as the input, one may iterate the functions at will. Thus, for example, if A is a set which is central* but not IP* (such as one given by Theorem 16.8), then {L[n
+.2Jrr+ 1i:nEA} n
is a central* set which is not IP*. Notice also that the description of a set ga, y [N] is effective. That is, one only needs a sufficiently precise decimal approximation to a and y in order to determine whether a specified number is a member of ga, y [N].
Notes The notions of IP, IP*, central, and central* are due to H. Furstenberg in [98] where central sets were defined in terms of notions from topological dynamics. See Chapter 19 for a proof of the equivalence of the notions of central. Theorem 16.20 is from [ 120] where it had a purely combinatorial proof. Most of the remaining results of Section 16.2 are from [31 ], obtained in collaboration with V. Bergel-
son, except for Theorem 16.16 (which is from [78] and was obtained in collaboration with W. Deuber) and Theorem 16.30 which, while previously unpublished, is from an early draft of [33], obtained in collaboration with V. Bergelson and B. Kra.
The results of Section 16.3 are due to E. Terry in [234]. The left weak ring (hom(S, S), ®, o) is due to J. Clay in (65] where it is given as an example of a left nearring which is not a right nearring. (The notion of nearring is a stronger notion than that of a weak ring.) The results of Section 16.4 are from [33], results of collaboration with V. Bergelson and B. Kra. H. Furstenberg [98, Proposition 9.4] gives an explicit description of certain IP* sets in terms of topological dynamics, and other examples can be deduced from his paper [100] with B. Weiss. Spectra of the form ([na + y j : n E N) were introduced by T. Skolem [224] and given the name the 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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340
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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344
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(-r)+2a(x,)-k+l
XrXn < 2a(x,)+a(xn) + 2a(x,)+a(xn)-k+2 + 2a(xr)+a(xn)-2k+2 = 2a(x,xn) + 2a(xrxn)-k+2 + 2a(x,xn)-2k+2 < 2a(xrxn) + 2a(xrxn)-k+3
SOb(xrXn) 5 a(xrx,)-k+2. Thusa(xrxn)-b(xrxn) E {k-1,k-2}so a(xrxn)b(xrxn) # a(Xr +xn) - b(Xr +xn) (mod 3), a contradiction. Finally assume that {xn : n E N} c A3 and {a(xn) - c(xn) : n E N) is bounded. We may presume that we have some k E N such that a(xn) - c(xn) = k for all n E N. By the pigeon hole principle, pick n < r in N such that either (a) x, < 2a(xn)+1(1
- 2-k) 2 and xr < 2a(xr)+t (1 - 2-k)2 or
(b) x, > 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) <
a(XrXn) - k. By Lemma 17.10 a(xrXn) - C(XrXn) < k so a(xrxn) - c(xrxn) = k.
Notice that (1 - 2-k-1 - 2-2k-2) < (1 - 2-k)2, a fact that may be verified by squaring both sides. Since xn < 2a(xn)+1 - 2 a(x.)-k and xr < 2a(xr)+l _ 2a(xr)-k we have
Xrxn < 2a(xrxn)+1 _2a(x,xn)-k+1
+2a(x,xn)-2k-1
-
< 2a(xrxn)+1 - 2a(x,xn)-k 2-k-1 2-2k-2) = 2a(xrxn)+1(1
-
< 2a(xrxn)+I (1 - 2-k) 2 = 2a(xrxn)+1 (I - 2r(xrxn)-a(xrx0 )2
2a(x,xn)-2k-1
346
17 Sums and Products
Thus since XrX,, E Bi we have that k = a(xrxn) - c(xrxn) - i =- k - 1 (mod 3), a contradiction.
Recall from Theorem 13.14 that if p E ON and Nn E p for infinitely many n, then there do not exist q, r, and s in N* such that q p = r + s. The following corollary has a much weaker conclusion, but applies to any p e N*.
Corollary 17.17. Let p E N*. Then p + p 0 p p. Proof. Suppose instead that p + p = p p and pick some A E R such that A E p p. Let B = {x E N : -x + A E p} f1 {x E N : x-t A E p} and pick xl E B. Inductively, let n E N and assume we have chosen
Pick
Xn+1 E (B l (l; 1(-x, + A) fl (x,-1 A))\{xt, x2, ... , X,,)Then PS((xn)n_1) U PP((xn)' 1) C A, contradicting Theorem 17.16. Theorem 17.16 establishes that one cannot expect any sort of combined additive and multiplicative results from an infinite sequence in an arbitrary finite partition of N. On the other hand, the following question remains wide open.
Question 17.18. Let r, n E N. If N = U;=1 A;, must there exist i E {1, 2, ... , r} and 1 such that FS ((x, )1 1) U FP((xl) l 1) c A; ?
a one-to-one sequence (x1)
We would conjecture strongly that the answer to Question 17.18 is "yes". However, the only nontrivial case for which it is known to be true is n = r = 2. (See the notes to this chapter).
17.3 Sums of Products Shortly after the original (combinatorial) proof of the Finite Sums Theorem and its corollary, the Finite Products Theorem, Erdo"s asked [89] whether, given any finite partition of N, there must exist one cell containing a sequence and all of its "multilinear combinations". We know, of course, that whatever the precise meaning of "multilinear combinations", the answer is "no". (Theorem 17.16.) We see in this section, however, that a certain regularity can be imposed on sums of products of a sequence.
Recall that, if n E N and p E ON, then n p is the product of n and p in the semigroup (ON, ) which need not be the same as the sum of p with itself n times. (We
already know from Theorem 13.18 that if p E N*, then p + p # 2 p.) Consequently we introduce some notation for the sum of p with itself n times.
Definition 17.19. Let P E ON. Then al (p) = p and, given n E N, an+1 (p) _ an(P) + p.
17.3 Sums of Products
347
Of course, if p E N, then for all n, n p = an (p). The question naturally arises as to whether it is possible to have n p = Q, (p) for some n E N\{ 1) and p E N*. We shall see in Corollary 17.22 that it is not possible.
Lemma 17.20. Let n E N, let p E N*, let A E p and let B E on(p). There is a one-to-one sequence (x,)001 in A such that, for each F E [N]", E:EF Xt E B. Proof. This is Exercise 17.3.1.
Theorem 17.21. Let n E N\ { 11. There is a finite partition ,R of N such that there do not exist A E JR and a one-to-one sequence (x`)001 such that
(1) for each t E N, n x, E A, and (2) whenever F E [N]", FIEF Xt E A. Proof For i E (0, 1, 2, 31, let A = Uk=O °O {x E N : n2k+i/2 < x < n 2k+(i+l)/2 JI
and let R = {A0, Al, A2, A3). Suppose that one has a one-to-one sequence (x,)0_1 and i E 10, 1, 2, 3) such that
(1) foreacht E N,n - x, E A,,and (2) whenever F E [N]", EiEF Xt E A. By the pigeon hole principle, we may presume that we have some j E (0, 1, 2,3) such
that (x,:tEN)cAj. Now if n2k+j/2 < x, < n2k+(j+1)12 then n2k+(j+2)/2 < n . x, < n2k+(j+3)/2 so that i = j + 2 (mod 4). On the other hand, for sufficiently large t, if n2k+j/2 < X, < n2k+(j+l)/2 then n2k+j/2 < X1 + X2 +
+Xn_l +xt < n2k+(j+2)/2
so that i # j + 2 (mod 4), a contradiction. Corollary 17.22. Let P E N* and let n E N\( 11. Then n p # a" (p). Proof. Suppose that n p = an (p) and pick A E ,R such that A E n p. Then n-' A E p so, by Lemma 17.20, choose a sequence (xt)001 in n-1 A such that for each F E [N]", E,EF Xt E A. This contradicts Theorem 17.21.
Recall that, given F, G E Pf (N) we write F < G if and only if max F < min G. Definition 17.23. Let (x,)001 be a sequence in N and let m E N. Then I
F m E 2f(N) and Fl < F2 <.
< Fm}.
17 Sums and Products
348
Theorem 17.24. Let p p = p E flN, let m E N, and let-A E am (p). Then there is a sequence (xt) °O 1 such that SPm ((xt) t0O 1) C A.
Proof If m = 1, this is just the finite products theorem (Theorem 5.8). In the proof, we are dealing with two semigroups, so the notation B* is ambiguous. We shall use it to
refer to the semigroup (fN, ), so that B* = {x E B : x- B E p}. We do the m = 2 case separately because it lacks some of the complexity of the
general theorem, so assume A E p + p. Let B1 = {x E N : -x + A E p). Then B1 E p so B1* E p. Pick xl E Bl*. Inductively, let n E N and assume that we have chosen (xt) in N and B1 ? B2 ? D Bn in p such that, if 0 0 F c {1, 2, ... , n}, 1
k = min F, andf =maxF,then (1) ITtEFXt E Bk* and
(2) if f < n, then Be+1 C (- IItEF Xt + A).
Fork E{1,2,...,n},letEk={IItEFxt::F(z {1,2,...,n}and k=min F}.Let Bn+1 = Bn nn Ik=1 laEEk ((-a + A) fl a-l Bk*).
Now, given k E {1, 2, ... , n} and a E Ek, a E Bk* c_ B1 so -a + A E p and a-1 Bk* E p so Bn+1 E p. Choose Xn+l E Bn+1* T o verify the induction hypotheses, let 0 F C { 1 , 2, ... , n + 1), k = min F, and
f = max F. If e < n, then both hypotheses hold by assumption. Assume that f = n. Then (1) holds by assumption and I1tEF xt E Ek so (2) holds. Finally assume that f = n + 1. Then (2) is vacuous. Ilk = n + 1, then IItEF X. = Xn+1 E Bn+l*. If k < n, then let G = F\{k}. Then fIfEG xt E Ek SO Xn+l E Bn+l C A EG X0 -' Bk* and thus (1) holds.
The construction being complete, let F1, F2 E . f (N) with F1 < F2. Let k = min F2 and let e = max F1. Then
11tEF2 Xt E Bk* C Be+1 C (- ftEF, xt + A). This completes the proof in the case m = 2.
Now assume that m > 3. Let B1 = {x E N : -x + A E a,,, -I (p)} and choose x1 E B1*. Inductively, let n E N and assume that we have chosen (Xk)k=1 in N and (Bk)k=1 in
p so that for each r E {1, 2, ... , n} each of the following statements holds.
(I) If 0 F C (1, 2, ... , r} and k = min F, then IIrEF Xt E Bk*. (II) If r < n, then Br+1 C Br
(III) Iff E {1,2,...,m-1},F1,F2,...,Fe E JPf({1,2,...,r}), and F1 < F2 < < Ft, then - El _1'1teF Xt + A E am-1(p).
(IV) If F1,F2,...,Fm_1 EPf({l,2,...,r}), F1 < F2 < ...
17.3 Sums of Products
349
(V) If f E (1,2. , m - 2), F1, F2, ... , Ft E .Pf({12. , r}) with F1 < F2 <
At n = 1, hypothesis (I) says that x1 E Bl*. Hypotheses (II), (IV), and (V) are vacuous, and hypothesis (III) says that -x1 + A E am-1(p).
Fort E{1,2,...,m-1},let Y = {(Fl, F2, ... , Ft) : F1, F2, ... , Ft E ,Pf({1, 2, ... , n})
and F1
F C {1,2,...,n} and minF = k}.
Ek = {Il EFX, : 0
Given a E Ek, we have that a E Bk* by hypothesis (I) and so a-1Bk* E p by Lemma 4.14. If (F1, F2, ... , Fm_1) E Fm -1, then by (III) we have
Ei=, 1 REF- Xt + A E P. If f E ( 1 , 2, ... , m - 2} and (F1, F2, ... , Ft) E Fe we have by hypothesis (III) that
{X E N : -x + (- Eti=1 REF, Xt + A) E am-a-1(P)} E P. So we let
Bn+1 = B. fl
n
Ik=1n
'
I
nn(FI,F2,,
IaEEk a-1 Bk*
. ,FF_,)EF,_,
-1
Em=1
(-
n I Ie=12I I(F,, F2, ...Fe)EFe {X E ITV
fltEF xt + A)
: -x +
fltEF Xt + A) E am-a-1(P)}
and notice that Bn+1 E p. For simplicity of notation we are taking n 0 = N in the above. Thus, for example,
ifn =m-3,then .Fm-2=pm-1 = 0 and so n
Bn+1 = B,n nn IInk=1 nIaEE,p a
*
1
Bk
nI 1,=13l(F1,F2,.. ,Ft)E.F, {X E N : -x + (- Ef=1 IEEE xt + A) E am-e-1(P)}.
Choose xn+1 E Bn+I*. T o verify hypothesis (I), let 0 # F C 11, 2, ... , n + 1} and let
k = min F. If n + 1 0 F, then (I) holds by assumption, so assume that n + I E F. If k = n + 1, we have xn+1 E Bn+1* directly, so assume that k < n + 1 and let G = F\{n + 11. Then rltEG xt E Ek so xn+1 E xt)-1 Bk* Hypothesis (II) holds trivially and hypotheses (IV) and (V) hold directly.
T o verify hypothesis (III), let f E ( 1 , 2, ... , m - 1 ) and let F1, F2, ... , Fe E
Yf({1, 2, ... , n + 1}) with F1 < F2 < ... < Fe. If f = 1, then by hypothesis
17 Sums and Products
350
(I) and (U), fIrEF1 X, E Bt' C B1 SO - f1EEF, Xt +A E am-1(P). So assume that
I > 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.
17.3 Sums of Products
351
Proof. For X E N, define a(x) and d(x) in w by ma(x) < x < ma(x)+1 and d(x) = max{t E w : m'jx}. Thus when x is written in base in, a(x) is the position of the leftmost nonzero digit and d(x) is the position of the rightmost nonzero digit. Let
Ao =N\Nm={xEN:d(x)=0} A1= {xEN:d(x)=1}
A2={mt:rENandt>1)
A3 = {x E N : a(x) is even, d(x) > 1 and ma(x) < x < maw + ma(x)-d(x)} ma(x)-d(x) < x} A4 = {x E N : a(x) is even, d(x) > 1 and ma(x) + A5 = {x E N : a(x) is odd, d(x) > 1 and ma(x)+1 - ma(x)-d(x) < x} A6 = {x E N : a(x) is odd, d(x) > 1 and maW < x < ma(x)+1 _ ma(x)-d(x)} Then {Ao, Al, A2, A3, A4, A5, A6} is a partition of N. Trivially, neither Ao nor A2 and Al does not contain SPi((xt)°_1) = FP((xt)' 1) for any contains sequence (xt) °O 1 in N.
We now claim that A3 does not contain any FP((xt)°O_1), so suppose instead we have some 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')
17 Sums and Products
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(x,)+2
and consequently a(r1k=m xt) = Ek m a(xt) + 1. Since, for each t, d(xt+1) > a(xt), we have that for each s E {m, m + 1, ..., k}, a (x1 + X2 + + xm _ l + xs) = a (xs) and, since +X2+...+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 <
ma(xm-I)+1
< ma(xi)-d(xl)-1 so
y < ma(xt) < ma(xi)
+ma(x,)-d(Xi)-1
+ma(xt)-d(xl)-1
+ma(xt)-d(xl)
= Ma(y) + ma(y)-d(y)
contradicting the fact that y E A4. Now we claim that A5 does not contain any FP((xt)°Ol). So suppose instead that
we have a one-to-one sequence (xt)°°1 such that FP((xt)°Oi) c A5. As before, we may presume that for each t, d(xt+1) > a(xt). Now x1 < ma(xi)+l - md(x,) as can be seen by considering the base m expansion of x1. Also, X2 < ma(x2)+l so that X1X2 < ma(xl)+a(x2)+2 - ma(x2)+d(x,)+1
Since also XJX2 > ma(xi)+a(x2) and a(x1),
a(x2), and a(xlx2) are all odd, we have a(xlx2) = a(xl) + a(X2) + 1. Also
a(x2) +d(x1) + 1 = a(X1x2) -a(XI)+d(xl) > a(xlx2) - d(xIX2) because d(x1x2) > d(x1) + d(X2) > a(x1) - d(xl). Thus
XIX2 < ma(XIX2)+1 - ma(xIx2)-d(xlx2)
173 Sums of Products
353
contradicting the fact that xlx2 E A5. Finally we show that A6 does not contain any SPm ((xt) ° t ), so suppose instead that we have a 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.
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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.
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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
-Vj_laiErEF;xr+AEae+1P+ae+2P++amp.
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357
(IV) If F1, F2, ... , Fm-1 E Pf({1, 2, ... , r}), F1 < F2 < : < Fm_1, and r < n, then
Br+l Cam
1
(-
Emi=-1Iai
EtEF xt + A).
(V) IftE{1,2,...,m-2},F1,F2,...,FtE?f({1,2,...,r}),FI Fe, and r
Br+1C {xEN: -ae ai EtEF Xt+A) E ae+2P+ae+3P+" '+amP}. a+Ix+(-Ei_I e
(VI) Xr = EtEH, Yr.
At n = 1, hypotheses (I) and (VI) hold directly, hypotheses (II), (IV), and (V) are vacuous, and hypothesis (III) says that -a 1x1 + A E a2 P + a3P + +amp which is true because x1 E B1.
Fore E {l,2,...,m- 11, let Fe = {(F1, F2,..., Fe) : F1, F2,..., Fe E Pf({1,2,...,n}) and F1 < F2 < . < Fe} and for k E {1, 2, ... , n}, let
Ek={EtEFXt:0#FC {l,2,...,n}and minF=k}. Given b E Ek, we have b E Bk* by hypothesis (I), and so -b + Bk* E p. If (FI, F2, ... , Fm_l) E Fm-j, then by (III) we have (- Em1l ai EtEF xt + A) E amp so that am-1(- Em11 ai EtEF xt + A) E p. If e E {1, 2, ... , m - 2} and (F1, F2, ..., Fe) E Fe we have by (III) that Ej=1 ai ErEF xr + A E ae+1 P + ae+2P +
+ am p
so that {
e xEN:-ae+lx + (- Ei=1 ai ErEF xt + A) E ae+2P +ae+3P + ....{_ amp} E p.
Let s = max Hn + 1. Then we have that Bn+1 E p, where
Bn+1 = B,, n FS((yt) i) n I
n I'(F,,F2,
Ik=l I
m-i)E m_I am
IbEEk(-b + Bk*) n I(- Em-1Iai EtEF, Xr + A) n
n f_1 n(F,,F2,...,Ft)EY'e{X E ICY : -ae+lx + (- Eje=1 ai ErEF,, xt + A)
E ae+2P+ae+3P+---+amp}. Here again we use the convention that n 0 = N so, for example, if m = 2, then
Bn+l = Bn n FS((yt)OQ,) n nk=1 nbEEk(-b + Bk*) n nO#Fcj1,2,....n) a2-1(-a1 EIEF Xt + A).
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and pick E a (s, s + 1, s + 2, ...) such that x,,+1 = Yt Hypothesis (I) can be verified as in the second proof of Theorem 5.8 and hypothesis
Choose
(II) holds trivially. Hypotheses (IV), (V) and (VI) hold directly. T o v e r i f y hypothesis (III), let t E { 1, 2, ... , m - 1) and let Ft < F2 < ... < Fe in
d'f((1, 2, ... , n + 1)). If f = 1, we have by hypotheses (I) and (II) that EtEF, xt E Bt so that -al EtE F, xt + A E a2 P + a3 P + amp as required. So assume that f > 1 andletk=min Fe and j =maxFe_1. Then EtEF, Xt E Bk* a Bj+1
a {x E N : -aex + (- Ef_j a; EtEFF xt + A)
E ae+lp+at+2P+...+amp} (by hypothesis (V) at r = j) so
-(E;_1a; EteF,Xr)+AEae+1P+ae+2P +
a, p
as required.
The induction being complete, let Fl < F2 < ... < F. in Pf(N) and let k = min Fm and j = max F,,,- 1. Then EtEF,R Xt E Bk* C Bj+1 C am-1(-
Emil a; EtEF;
xt + A)
So Em1 a; EtEF; Xt E A.
Just as Theorem 17.26 was a partial converse to Theorem 17.24, we now obtain a partial converse to Theorem 17.31. Notice that p is not required to be an idempotent.
Theorem 17.32. Let a = (a,, a2, ... , am) E A, let (x,)' 1 be a sequence in N, and let
PEn
1
FS((xr)°-°_k). Then MT(a, (Xr)°__1) E al p + a2P { ...
amp,
P r o o f . W e show, by downward induction on f E ( 1 , 2, ... , m) that for each k E N,
{Emea;EtEF,xt:Fe,Fe+1,...,FmERf({k,k+1.... })and Fe
E aeP+ae+tP+...+amp. For E = m, we have FS((xt)°Ok) E p so that FS((amxt)°°k) E amp. So let f E { 1, 2, ... , m - 1) and assume that the assertion is true fore + 1. Let k E N and let
A = (Eat EtEF; x, : Fe, Fe+t, ... , Fm E d'f({k, k + 1, ...}) and Fe
FS((atxt)°_k) 9 {x E N : -x + A E ae+lp +ae+2p +
+ amp).
17.4 Linear Combinations of Sums
359
So let b E FS((aext)O°k) and pick Fe E 1f ({k, k + 1, ...}) such that b = at EtEFL xt-
Let r = max Ft + 1. Then
{E;"a+t ai EtEF; xt : Fe+1, Fe+2.... , F. E .Pf({r, r + 1, ...})
andFt+i
+ amp as required.
13
We see now that Milliken-Taylor systems are themselves partition regular.
Corollary 17.33. Let a E A, let B C N, let (x1)0°_1 be a sequence in N, and assume that MT(d, (xt)O° 1 ) C B. I f r E N and B.= B;, then t h e r e e x i s t i E {1, 2, ... , r} and a sum subsystem (y,)1 of (x1) 001 such that MT (a, (y)1) a Bi.
Proof. Pick by Lemma 5.11 some p + p = p E nk FS((xt)O°k). By Theorem 1
17.32, MT(a, (x,)-,) E al p + a2 P + + am p, where as = (a1, a2, ..., am ). Pick i E (1, 2, ... , r) such that B, E al p + a2 P + + am p, and apply Theorem 17.31. We observe that certain different members of A always have Milliken-Taylor systems in the same cell of a partition.
Theorem 17.34. Let a, b E A, and assume that there is a positive rational s such that a = sb. T h e n whenever r E N and N = U;=1 Ai, t h e r e e x i st i E (1, 2, ... , r) and sequences (xt)O°1 and (yt)OD1 such that MT(a, (x,)0°1) C A; and MT(b, (yt)O°1) e Ai.
Proof. Assume that s = n where m, n E N and let c = mb. Pick by Corollary 17.33
some i E (1,2,...,r) and a sequence (zt)O°1 such that MT(F, (zt)O1) e Ai. For each t E N, let x, = nzt and yt = mz,. Then MT(a", (x1)001) = MT (b, (y,)°01) _ MT(F, (zt)O°1) We show now, as we promised earlier, that we can separate certain Milliken-Taylor systems.
Theorem 17.35. There is a set A C N such that for any sequence (x1)O° 1 in N, MT((1,2), (x,)001) A and MT((l, 4), (x1)0°1) N\A. Proof. For X E N, let supp(x) be the binary support of x so that x = E,ESUPP(X) 21. For each x E N, let a(x) = max supp(x) and let d (x) = min supp(x). We also define a function z so that z(x) counts the number of blocks of zeros of odd length interior to the binary expansion of x, that is between a(x) and d(x). Thus, for example, if in
binary, x = 1011001010001100000, then z(x) = 3. (The blocks of length 1 and 3 were counted, but the block of length 5 is not counted because it is not interior to the expansion of x.)
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Let A = {x E N : z(x) = 0 (mod 4) pr z(x) - 3 (mod .4)}. Suppose that we have a sequence (X: ) °O I such that either
MT((1, 2), (x1)001) C A
or
MT((1, 4), (xt)O01) C N\A.
Pick a sum subsystem (yt)°_I of (xt)I01 such that for each t, d(yt+i) > a(yt). (See Exercise 17.4.1.) For i E (0, 1, 2, 3) let
Bi={xEN:z(x)-i (mod4)}. By Corollary 17.33 we may presume that if
MT((1, 2), (xt)101) c A, then
MT((1, 2), (Yt)`O1) c Bp
or
MT((1, 2), (yt)°O1) S B3
and, if
MT((1,4), (x,)°01) c N\A, then
MT((1, 4), (yt)JO1) S B1
or
MT((1, 4), (Yt)°O1) c B2.
By the pigeon hole principle we may presume that for t, k E N,
a(yt) = a(Yk) (mod 2),
d(Yt) = d(Yk) (mod 2), and z(yt) = z(Yk) (mod 4).
Assume first that fort E N, a(yt) - d(yt) (mod 2). Now if MT((1, 2), (yt)to 1) c A, we have y1 + Y2 + 2y3 and Y2 + 2y3 are in the same B; so
z(YI + Y2 + 2Y3) = z(Y2 + 2y3) (mod 4).
Since a(yi) = d(y2) (mod 2) there is an odd block of zeros between yt and Y2 in YI+Y2+2Y3,soz(Y1+Y2+2Y3) = z(yI)+1+z(y2+2Y3) and consequentlyz(yt) = 3 (mod 4). Butd(2Y3) = d(y3)+ 1 #a (y2) (mod 2) so z(y2+2y3) = z(y2)+z(y3) 3 + 3 = 2 (mod 4), a contradiction to the fact that MT((1, 2), (yt)°O 1) S A. If MT((1, 4), (yt)°O 1) c N\A, we have as above that
z(yt + y2 + 4y3) = z(y2 + 4Y3) (mod 4)
and z(yt + y2 + 4y3) = z(yt) + 1 + z(y2 + 4y3) so that z(yt) = 3 (mod 4). Since d(4y3) = d(y3) +2 = a(y2) (mod 2) we have z(y2 +4y3) = z(y2) + I + z(y3) = 3 (mod 4), a contradiction to the fact that MT ((1, 4), (yt)O° 1) c N\ A.
Now assume that fort E N, a(yt) 0- d(yt) (mod 2). If MT((1, 2), (yt)°O1) c A we have z(yt + y2 + 2Y3) = z(y2 + 2Y3) (mod 4) and, since d (y2) # a(yt) (mod 2), z(yt + y2 + 2y3) = z(yt) + z(y2 + 2y3) so that z(y1) = 0 (mod 4). Since d(2Y3) = d(Y3) + 1 = a(y2) (mod 2)
17.5 Sums and Products in (0, 1)
361
we have that z(y2 + 2y3) = z(y2) + 1 + z(y3) = 1 (mod 4), a contradiction. If MT((1, 4), (y,) f ° t) C N\A, we have as above that z(yt) = 0 (mod 4) and since d(4y3) = d(y3) + 2 0 a(y2) (mod 2), z(y2 + 4Y3) = z(Y2) + z(Y3) = 0 (mod 4), a contradiction. As a consequence of Theorem 17.35, we have the following contrast between infinite and finite image partition regular matrices.
Corollary 17.36. There exist infinite image partition regular matrices B and C and a two cell partition of N, neither cell of which contains all of the entries of Bs and of C y for any 1, y E N°'. Proof. The matrices for the Milliken-Taylor systems MT((1, 2), (xt)°O1) and MT((1, 4), (yt) °O 1) are image partition regular.
In Section 15.4, we found several characterizations of finite image partition regular matrices.
Question 17.37. Can one find a reasonable combinatorial characterization of infinite image partition regular matrices? Exercise 17.4.1. In Exercise 5.2.2, the reader was asked to show combinatorially that given any sequence (xt)1°1 there is a sum subsystem (yt)°_1 of such that whenever 2k < yt, one has 2Jc+1IYt+1. Use the fact that, by Lemma 6.8, given any
p + p = p E I lkk,--1 FS((xt)O0k), one has p E IHII, to show that there is a sum subsystem (yt)°° 1 of (xt)°O 1 such that for each t E N, a(yt) + 2 < d(yt+1).
Exercise 17.4.2. Prove that if H is a finite subset of A, r E N, and N = A then there exist a function f : H --> {1, 2, ... , r} and a sequence (xt)°° 1 such that for each a E H, MT(a, (xt)O01) C Af(a). (Hint: Use Corollary 17.33.)
17.5 Sums and Products in (0,1) Measurable and Baire Partitions We show in this section that if one has a finite partition of the real interval (0, 1) each member of which is Lebesgue measurable, or each member of which is a Baire set, then one cell will contain 1) for some sequence in 1) U (0, 1). (The terminology "Baire set" has different meanings in the literature. We take it to mean members of the smallest a-algebra containing the open sets and the nowhere dense sets.) In fact, we show that certain sums of products and products of sums can also be found in the specified cell. We use the algebraic structure of the subset 0+ of 0(0, 1)d which we discussed in Section 13.3. The results that we obtain are topological in nature, so it is natural to 1
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362
wonder how one can obtain them starting with the discrete topology on (0, 1). The answer is that we use certain topologically characterized subsets of 0+. Recall that a subset A of R is meager (or first category) if and only if A is the countable union of nowhere dense sets.
Definition 17.38. (a) A set A e (0, 1) is Baire large if and only if for every E > 0, A n (0, e) is not meager.
(b) B = (p E 0+ : for all A E p, A is Baire large).
Lemma 17.39. B is a left ideal of (0+, ). Proof Since the union of finitely many meager sets is meager, one sees easily that whenever (0, 1) is partitioned into finitely many sets, one of them is Baire large.
Consequently, by theorem 5.7, it follows that {p E fl(0, 1)d : for all A E p, A is Baire large) # 0. On the other hand, if p E P (O, 1)d\0+ one has some e > 0 such that (e, 1) E p and (e, 1) is not Baire large. Thus B # 0 . To see that B is a left ideal of
0+, let p E 2 and q E 0+ and let A E q- p. Pick x E (0, 1) such that x-tA E p. Given e > 0, x-tA n (0, e) is not meager so x(xA n (0, e)) is not meager and x(x-'An ( O, E)) c A n (0, E). The following result, and its measurable analogue Theorem 17.46, are the ones that allow us to obtain the combined additive and multiplicative results. Recall that a subset of R is a Baire set if and only if it can be written as the symmetric difference of an open set and a meager set. (Or do Exercise 17.5.1.)
Theorem 17.40. Let p be a multiplicative idempotent in B and let A be a Baire set which is a member of p. Then {x E A : x-' A E p and -x + A E p} E p.
Proof. Let B = {x E A: x-1A E p}. Then since p = p- p, B E p. Also A is a Baire set so pick an open set U and a meager set M such that A = UOM. Now M\U is meager so M\U 0 p so U\M E p. We claim that
(U\M)nBc{xEA:x-'AEpand -x+AEp}. So let x E (U\M) n B and pick E > 0 such that (x, x + e) e_ U. To see that -x + A E p, we observe that (0, 1)\(-x + A) is not Baire large. Indeed one has ((0, 1)\(-x + A)) n (0, E) c -x + M, a meager set. (Given y E (0, E), x + y E U so, if x+ y 0 A, then x+ y E M.) Now we turn our attention to deriving the measurable analogue of Theorem 17.40. We denote by µ(A) the Lebesgue measure of the measurable set A, and by µ*(B) the outer Lebesgue measure of an arbitrary set B. We assume familiarity with the basic facts of Lebesgue measure as encountered in any standard introductory analysis course. The notion corresponding to "Baire large" is that of having positive upper density near 0. We use the same notation for upper density near 0 as we used for upper density of sets of integers.
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363
Definition 17A1. Let A e (0, 1). (a) The upper density near 0 of A, d (A) is defined by
a (A) = lim sup
µ* (A n (0, h)) h
hlo
(b) A point x E (0, 1) is a density point of A if and only if Urn
µ* (A n (x - h, x + h)) - 1 2h
hyo
The following result, while quite well known, is not a part of standard introductory analysis courses so we include a proof.
Theorem 17.42 (Lebesgue Density Theorem). Let A be a measurable subset of (0, 1). Then Et, ({x E A : x is not a density point of A}) = 0.
Proof. If x is not a density point of A, then there is some e > 0 such that lim
µ(A n (x - h, x + h)) ao
2h
<
Since the union of countably many sets of measure 0 is again of measure 0, it suffices to let e > 0 be given, let
B={xEA:limihnfo
µ(A n (x - h, x + h)) 2h
<1-E}
and show that µ(B) = 0. Suppose instead that IL*(B) > 0. Pick an open set U such that B C U and µ(U) < ui tee We first claim that
(t) if (If)nEF is a (finite or countably infinite) indexed family of pairwise disjoint
intervals contained in U and for each n E F, p(A n In) < (1 - e)p(In), then µ*(B\ UnEF 1n) > 0. To see this, since in general µ*(C U D) < µ*(C) +µ*(D), it suffices to show that µ*(B n UnEF In) < µ*(B). And indeed
µ.*(B n UnEF 1n) .5 µ(A n UfEF In) = EnEF P(A n in) < FinEF(1 - E)Fl(In) < (1 - 011 (U)
< A*(B). Now choose XI E B and h 1> 0 such that [x 1- h 1, x, + hl ] e U and
12(An(xi -h1,x1 + hi)) < (1 -e)2h1.
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Inductively, let n E N and assume we have chosen x1, x2, ... , xn in B and positive h 1, h2..... hn such that each [xi - h1, xi + hi ] c U, each u, (A n (xi - h i , xi + hi )) <
(1-E)2hi,and[xi-hi,xi+hi]n[xj-hj,xj+hj]=0fori 96 j. Let
do = sup{h : there exist x E B such that [x - h, x + h] c U,
[x-h,x+h]nU"=1[xi-hi,xi+hi]=0,andµ(An(x-h,x+h)) <(1-E)2h}. Notice that do > 0. Indeed, by (t) one may pick x E B\ U"=1 [xi - hi, xi + hi] and then pick 6 > 0 such that (x - d, x + S) n Un 1 [xi - hi, xi + hi ] = 0 and (x - 6, x + 3) c U. Then, since x E B, one may pick positive h < 6 such that µ(A n (x - h, x + h)) < (1 - E)2h. Pick xn+1 E B and hn+1 >
such that [xn+i - hn+1, xn+1 + hn+1]
2 [xn+l - hn+t , xn+1 + hn+1 ] n U"=1 [xi - hi, xi + hi] = 0,
U,
and µ(A n (xn+l - hn+1, xn+1 + hn+1)) < (1 - E)2hn+1 The inductive construction being complete, let C = B\ Un° [xn - hn, x, + hn]. Then by (]') µ*(C) > 0. Also, since ([xn - hn, xn + hn})n°1 is a pairwise disjoint 1
collection contained in (0, 1), En° 1 hn converges, so pick m E N such that
zoo Then µ(Un=m+1 [xn 3hn, xn + 3hn]).
m+l
hn < g*(C)/6.
- 3h, xn + 3hn]) < jA,* (C) so pick x E C\(Un
m+1 [xn -
Then X E B\ Un t [xn - hn, xn + hn] so pick h > 0 such that [x - h, x + h] c U,
[x-h,x+h]nUn
1[xn-hn,xn+hn]=0,andis(An(x-h,x+h)) <(1-E)2h.
Now there is some k-such that [x - h, x + h] n Un=t [xn - hn, xn + hn] # 0 (since otherwise each dk > h and consequently En_ 1 hn diverges) so pick the first k such that
[x - h, x + h] n [xk - hk, xk + hk] # 0 and notice that k > m. Then hk ? dk2' > 2
so that Ix - xk I < h + hk 3hk and hence x E [xk - 3hk, xk + 3hk], contradicting the fact that x ¢ Un°_m+t [xn - 3hn, xn + 3hn]. We also need another basic result about measurable sets.
Lemma 17.43. Let A be a measurable subset of (0, 1) such that d (A) > 0. There exists
B c A such that B U (0) is compact and d(A\B) = 0.
Proof. Foreachn E N, let An = An(1/2n, 1/2n-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 <_ x < 1/2n-1 and note that n > m. Then
1
A((A\B) n (0, x)) < Ek_n s(An\Bn) < Eko n 4k+t = 1 3 .4n
17.5 Sums and Products in (0, 1)
and x?
-
365
so
p ((A\B) n (0, x))/x <
2n < 3 2m'
a contradiction.
Definition 17.44. £ = (p E 0+ : for all A E p, d(A) > 0}. Lemma 17.45. £ is a left ideal of (0+, ). Proof. It is an easy exercise to show that if d(A U B) > 0 then either d(A) > 0 or
d(B) > 0. Consequently, by Theorem 3.11, it follows that {p E 18(0, 1)d : for all A E p, d(A) > 0) 0 0. On the other hand, if p E $(0, 1)d\0+ one has some e > 0 such that (e, 1) E p and d((e, 1)) = 0. Thus £ 0. Let p E £ and let q E 0+. To see that q p E £, let A E q p and pick x such that x-1 A E p. Another easy exercise establishes thatd(A) = d(x-1 A) > 0. Theorem 17.46. Let p be a multiplicative idempotent in £ and let A be a measurable member of p. Then {x E A : x-1 A E p and -x + A E P) E P.
{yEA:y is not a density point of Al. By Theorem 17.42, A(C) = 0. Consequently since p E X,
C gE p so B\C E P. We claim that B\C C {x E A x-1 A E p and -x + A E p}. Indeed, given x IE C one has 0 is a density point of -x + A so by an easy computation,
a ((O, 1)\(-x + A)) = 0 so (0, 1)\(-x + A) 0 p so -x + A E P. Now let us define the kind of combined additive and multiplicative structures we obtain. Definition 17.47. Let (xn)n° be a sequence in (0, 1). We define FSP((xn)n° ) and a : FSP((xn )n° 1) - * Y (5f (N)) inductively to consist of only those objects obtainable by iteration of the following: 1
(1) If m E N, then xm E FSP((xn)n° 1) and (m) E a(xm).
(2) Ifx E FSP((xn)n° 1),m E N, F E a(x),andminF > m, then{xm x,xm+x} C FSP((xn)n° 1), F U {m} E a(xm x), and F U {m} E a(xm +x). For example, if z = X3 + X5 X7 (x8 +x10 x11), then z E FSP((xrz)n 1) and 13, 5, 7, 8, 10, 111 E a (z). (Of course, it is also possible that z = x4 +x12 x 13, in which case also 14, 12, 13) E a(z).) Note also that (x3 +x5) x7 is not, on its face, a member 1) UFP((xn)n° 1) C FSP((xn)n 1). of FSP((xn)n°1). Notice that trivially The proof of the following theorem is reminiscent of the first proof of Theorem 5.8.
Theorem 17.48. Let p be a multiplicative idempotent in 2 and let A be a Baire set which is a member of p. Then there is a sequence (xn)n' 1 in (0, 1) such that FSP((xn)n° 1) C A
17 Sums and Products
366
Proof. Let Al = A. By Theorem 17.40, {x E Al : x-1A1 E. p and -x +At E p) E p so pick x, E Al such that x1 -' A 1 E p and -x, + A, E p. Let
A2 = A, nx,-'A, fl (-x, +A,). Then A2 E p and, since multiplication by x,
and addition of -x, are homeomor-
phisms, A2 is a Baire set. Inductively, let n E N\{1} and assume that we have chosen (x,)°_1 and (At)' 1 so
that An is a Baire set which is a member of p. Again invoking Theorem 17.40, one
has that {x E An : x-' An E p and -x + An E p} E p so pick xn E An such that xn-' An E p and -xn + A. E p. Let An+1 =An n xn-'Ann (-xn + An). Then An+1 is a Baire set which is a member of p. The induction being complete, we show that FSP((xn )n_1) c A by establishing the following stronger assertion: If
z E FSP((xn)n 1), F E o(z), and m = min F, then Z E Am. Suppose instead that this conclusion fails, and choose
Z E FSP((xn)n° 1), F E o(z), and m= min F such that z 0 Am and IF I is as small as possible among all such counterexamples. Now
if F = {m}, then z = X. E Am, so we must have BFI > 1. Let G = F\{m} and pick y E FSP((xn)n° 1) such that G E a(y) and either z = xm + y or z = xm y. Let r = min G. Since IGI < BFI, we have that
y E Ar C Am+1 C 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.
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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 <... <mt
m, < m2 <
< mt < n + 1, let such t and ml, m2, ... , m, be given. If m, < n, then the induction hypothesis applies, so assume that mt = n + 1. If t = 1, the conclusion holds because xn+, E B, (0, i). If t > 1, the conclusion holds because Xn+1 E Bt ({Xm1, Xm2, ... , Xmj_, }, 0-
The sequence (xn)n° having been chosen, let ml < m2 < 1
< Mk- Then
Xmk E Bk ({Xm, , Xm2 , ... , Xmk_ i }, I) SO {X.,, X.,_., Xmk) E Ai.
In order to establish some generalizations of Ramsey's Theorem, we introduce some special notation.
Definition 18.3. Let S be a set, let k E N, and let tp : S -+ N. (a) If (Dn)n° 1 is a sequence of subsets of S, then
[(D.)-1,tp]k
(xi,x2,...,xk):xi 0xjfori
jin{1,2,...,k}
and there exist m, < m2 <
< ink in N such that f o r each j E {1, 2, ... , k}, Xj E Dmi and for each
j E {2, 3, ...,k}, mj > tp(xj_l) (b) If t E N and (D.)n [(Dn)n_1, tp]<
is a sequence of subsets of S, then
(xl, x2, ... , xk} : xi # xj for i # jin {1, 2, ..., k} and there exist m, < m2 < < ink in (1, 2, ..., t} such that for each j E {1,2..... k}, xj E Dmi and f o r each j E 12, 3, ... , k}, mj > tp(xj_1)}.
For example, if S = N, tp(n) = n, and Dn = (m E N : m > n) for each n E N, then
[(Dn)n°=1,tp]<={{x,,x2,...,xk}9N:x,
<x2<<xk}=[N]k.
18.1 Ramsey's Theorem and Generalizations
371
Theorem 18.4. Let S be an infinite set, let p E S*, let k, r E N, and assume that [S]k = Ut=1 A;. Let rp : S -> N and assume that there is some C E p on which rp is 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<,
_
Dm C Bt+1(E,i) Notice that both hypotheses are vacuous when n = 1.
W e now claim that if t E (1, 2, ... , k - 1) and E E [(Dj n Hn)j=1,
then
Bt+1(E, i) E p. So let such t and E be given. Then there exist x,, x2, ... , xt in S and m, < m2 < . < mt in {1, 2, ... , n} such that E = {xi, x2, ... , xt}, Xj E Dmj n Hn for each j E {1, 2, ... , t}, and for each j E {2, 3, ..., t}, mj > rp(xj_,). If t = 1, we have x1 E Dm, c D, S; B1(0, i), so B2({x11, i) E p. So now assume
thatt > landletF = {x,,x2, ...,xt_,}. WeclaimthatF E
1.
Indeed, one only needs to verify that f o r each j E { 1, 2, ... , t - 1), x j E Hm,_, . To see this, notice that rp(xj) < mj+1 < mt. Since F E [(Dj n V]'- 1, we have by hypothesis (2) that Xt E Dm, S B,(F, i)
so that Bt+1(E, i) E pas required. Now f o r each t E 11, 2, ... , k - 1}, [(Dj n
rp] is finite (because C n H,,
is finite), so we may choose Dn+i E p such that D,+1 c D. and for each t E (1, 2, ..., k - 1) and for each E E [ (D j n Hn
rp] <, Dn+1 C Bt+1(E, i ). The construction being complete, we now only need to show that [(Dn)n° 1, (p]< c A. So let E E [(Dn)n° 1, rp]k and pick xl, x2, ... , xk, and m 1 < m2 < < Ink such that for each j E ( 1, 2, ..., k}, X j E Dmj , and for each j E {2, 3, ..., k}, mj > rp (x j _, ),
and E = {x,, x2, ... , xk}. Let F = (x,, x2, ... , xk_, }. Then for each j E (1, 2, ... , k - 1), cp (x j) < m j+1 <
Mk so F E [(DinH,nk_i)k1',rp]<' SO Xk E Dmk C Bk(F, i), so E E A, as required. 0
18 Multidimensional Ramsey Theory
372
As the first application of Theorem 18.4, consider the following theorem which is a common generalization of Ramsey's Theorem and van der Waerden's Theorem (Corollary 14.2). To prove this generalization, we use an ultrafilter which has the property that every member contains an arbitrarily long arithmetic progression.
Theorem 18.5. Let k, r E N and let [N]k = Ui_1 At. Then there exist sequences (an)n° 1 and (dn)n in N such that whenever nl < n2 < < nk and for each 1
j E{1,2,.. ,k},tj E{0,1,.... nj }, one has {ant +tido,,ant+t2dn2,...,ank+tkdnk} E A;. Proof. Let cy : N -) N be the identity map and pick p E A.P (which is nonempty by
Theorem 14.5). Pick i E (1, 2, ... , r} and a sequence (D.)1 °11 of members of p as guaranteed by Theorem 18.4. Pick ai and dl such that (a1, ai+di} C DI andlete(1) = 1. Inductively assume that (a.)m_1, (dm)m, and (2(m))m_1 have been chosen so that for each m E (1, 2, ... , n),
{am,am+dm,am+2dm,...,am+mdm} C_ Dt(m). Lete(n+1) =max{e(n)+l,an+ ndn + 1) and pick an+i and do+i such that {an+1, an+1 + do+1, an+l + 2dn+1,
Assume now that n j 10, 1, ..., nj}. Then
< n2 <
, an+1 + (n + 1)dn+i } C Dt(n+i )
< nk and for each j E (1, 2, ... , k) tj
{any + tidn,, ant + t2dn2, .... ank +tkdnk} E [(Dn)
1, (p]< C A1.
E
o
The case k = 2 of Theorem 18.5 is illustrated in Figure 18.1, where pairs of integers are identified with points below the diagonal. In this figure arithmetic progressions of length 2, 3, and 4 are indicated and the points which are guaranteed to be monochrome are circled. It is the content of Exercise 18.1.1 to show that Theorem 18.5 cannot be extended in the case k = 2 to require that pairs from the same arithmetic progression be included. A significant generalization of Ramsey's Theorem, which has often been applied to obtain other results, is the Milliken-Taylor Theorem. To state it, we introduce some notation similar to that introduced in Definition 18.3. We state it in three forms, depending on whether the operation of the semigroup is written as "+", " ", or "U".
Definition 18.6. (a) Let (S, +) be a semigroup, let (xn)n° 1 be a sequence in S. and let k E N. Then [FS((Xn)nn i)]<
{EtEHI Xt, `tEH2 Xt, ... , EtEHk xt } f o r each j E { 1 , 2, ... , k}, Ht E P'f(N)
and if j < k, then max Hj < min Hj+i }. (b) Let (S, ) be a semigroup, let (x)1 be a sequence in S, and let k E N. Then 7
[FP((Xn)oo 1)]< = { {11tEH1 Xt, iltEH2 Xt, ... , 171tEHk Xt}
for each j e {1,2,...,k}, Ht E.l f(N) and if j < k, then max Hj < min Hj+i }.
18.1 Ramsey's Theorem and Generalizations
373
Figure 18.1: Monochrome Products of Arithmetic Progressions
(c) Let (Fn)n-1 be a sequence in the semigroup (PPf(N), U) and let k E N. Then
[FU((Fn)n 1)]<
{UtEH, Ft, UtEH2 Ft, ... , UtEHk Ft} : for each j E (1, 2, ... , k), Ht E J" f(N) and if j < k, then max H1 < min Hj+1 }.
The Milliken-Taylor Theorem is proved by replacing the arbitrary nonprincipal ultrafilter used in the proof of Theorem 18.2 by an idempotent. Theorem 18.7 (Milliken-Taylor Theorem - Version 1). Let k, r E N, let (xn) n° be a sequence in N, and assume that [N]k = U =1 A. Then there exist i E (1, 2, ... , r) and a sum subsystem (yn)n°1 of (xn)n° such that 1
1
[FS((Yn)n 1)]< C A;. Proof. By passing to a suitable sum subsystem if necessary, we may assume that (xn) n° 1 satisfies uniqueness of finite sums. (That is, if F, G E Rf(N) and EnEF Xn = EnEG Xn,
then F = G.) Define tp : N -> N so that for each F E J"f(N), cp(EnEF Xn) = max F, defining tp arbitrarily on N\ FS ((xn )n 1). Notice that (p is finite-to-one on FS( (xn )n° 1).
Pick by Lemma 5.11 an idempotent p EON such that FS((X-)°,°=n) E p for every
n E N. Pick by Theorem 18.4 some i E (1, 2, ... , r) and a sequence (Dn)n° 1 of members of p such that Dn+1 c Dn for each n and [(Dn)n° 1, tp]' c A.
18 Multidimensional Ramsey Theory
374
Let B1 = D1 f1FS((xm)m 1) and pick y1 E Bj*. Inductively, let n E N and assume that we have chosen (yt)r in N and (B,)'=, in p. Let 1
B.+1 = Bn n (-Yn + Bn) n
fl FS((xm)m (p(Y,)+1)
and pick y.+1 E B.+1*.
Since, for each n E N, yn+l E FS((xm)m we have immediately that (yn)n° is a sum subsystem of (x,,)' 1. One can verify as in the first proof of Theorem 5.8 that whenever F E ,Pf(N) and in = min F, one has EnEF yn E Bm. To complete the proof, it suffices to show that 1
[FS((Yn)- 1)]< c [(Dn)- 1,V]< So, let a E [FS((yn)n°_1)]< and pick F1, F2, ... , Fk E ,f f(N) such that
maxFj <minFj+1 foreach j E {1,2,...,k- 1} and
a = {EtEF, Yt, EIEF2 Y1,
,
EtEFk Yd-
For each j E (1, 2, ..., k), let tj = min Fj, let ej = max Fj, and let mj = tp(Ytj-1) + 1. Then EZEF; Yt E Btj C Dmj. Further, if j E 12, 3, ... , k), then tp(E,EFj_I Yr) = tp(Ytj_I)
tp(Ytj-1) < mj
Corollary 18.8 (Milliken-Taylor Theorem - Version 2). Let k, r E N and let [Pf(N)]k = Ui=1 A,. Then there exist i E {1, 2, ... , r} and a sequence (Fn)n_1 in ?f(N) such that max Fn < min Fn+1 forall n and [FU((Fn)n° 1)]< C A,. P r o o f For each i E 11, 2, ... , r), let
Bi = {(EIEHI 21-t, E,EH2
2t-1, . , , EtEHk
2r-1) : {H1, H2,..., Hk) E A, ).
Then [N]k = Ui=1 Bi so pick by Theorem 18.7, i E (1, 2, ... , r) and a sum subsystem (yn)n of(2n'1)n° 1 suchthat[FS((yn)n° 1)]< C B. Since(y,,)n°-1 isasumsubsystem 1
of (2n-1)n° 1, pick for each n, some F,, E J f(N) such that y,, = E1EF 21'1 and maxFn < min F,,+,. To see that [FU((Fn)°1)]< C Ai, let H1, H2,..., Hk E ?f(N) with max Hj < min Hj+1 f o r each j E { 1, 2, ... , k - 1). Then (EtEHI Yt, EIEH2 y , . . . . . EIEHk YI) E [FS((Yn)n°_1)]< C Bi
so there exist (K1, K2, ... , Kk) E Ai such that {E/EHI Yt, EIEH2 Yt,
,
EtEHk YI} _ {E1EKI
2t-1
EIEK2 2
t-1
.. , EIEKk
2`-1}.
18.1 Ramsey's Theorem and Generalizations
375
Now, given j E (1, 2,.-., k}, let Gj = UnEHJ F. Then EIEHj Yt = EIEG; 2t-1 and if j < k, then max Gj < min G1+1. Thus (EtEK, 2f-1, EtEK22`-1,
, EtEKk 2t-1}
= {E1EG, 2`1, E,EG221-1, .., EIEGk 2r-1}
so (G 1 , G2,..., Gk) = (K1, K2, ... , Kk ) (although, we don't know for example that G1 = Kt-) so that {G1, G2, ... , Gk} E Al as required. It is interesting to note that we obtain the following version of the Milliken-Taylor Theorem (which trivially implies the first two versions) as a corollary to Corollary 18.8, which was in turn a corollary to Theorem 18.7, but cannot prove it using the straight
forward modification of the proof of Theorem 18.7 which simply replaces sums by products. The reason is that the function tp may not be definable if the sequence (x n )n_t
in the semigroup (S, ) does not have a product subsystem which satisfies uniqueness of finite products.
Corollary 18.9 (Milliken-Taylor Theorem - Version 3). Let k, r E N, let (S, ) be a semigroup, let 1 be a sequence in S, and assume that [S]k = U;=1 A. Then there exist i E (1, 2, ..., r) and a product subsystem (yn)R° of (xn)°O 1 such that 1
[FP((Yn)n° 1)]< C Ai. P r o o f . F o r each i E [1, 2, ... , r}, let
Bi = {(Fl, F2, ... , Fk} E [pf (N)]k : (REF, Xt, 11lEF2 Xt, ... , 11tEFk Xt } E Ai }.
Then [ 7'f (N)]k = Ui=1 Bi so pick by Corollary 18.8 i E 11, 2, ... , r) and a sequence (Fn)n°-1 in .9f (N) such that max Fn < min Fn+1 for all n and [FU((FF,)n 1)]< C Bi. For each n E N, let yn = IItEF x,. Then [FP((yn)n° 1)]< C A. (The verification of this assertion is Exercise 18.1.2.)
Exercise 18.1.1. Let
Al={ (X, y)E[N]2:x
A2 = {{x, y} E [N]2:y>2x}. Then [N]2 = Al U A2. Prove that there do not exist i E 11, 2) and sequences (an)n°1 and (dn)n° 1 in N such that (1) whenever n t < n2 and f o r each j E 11 , 2) and t j E {1, 2, ..., nj) one has {ant +tldn,,an2 +t2dn2} E Ai and (2) whenever n E N and 1 < t < s < n one has {an + tdn , an + sdn } E A; . Exercise 18.1.2. Complete the proof of Corollary 18.9 by verifying that
[FP((Y.)n 1)]< C A1.
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18 Multidimensional Ramsey Theory
Exercise 18.1.3. Apply Theorem 18.4 using a minimal idempotent in a countable semigroup S. See what kind of statements you can come up with. Do the same thing using
a combinatorially rich ultrafilter in fiN. (See Section 17.1.) (This is an open ended exercise.)
18.2 IP* Sets in Product Spaces We show in this section that IP* sets in any finite product of commutative semigroups contain products of finite subsystems of sequences in the coordinates. We first extend the notion of product subsystems to finite sequences.
Definition 18.10. Let (yn)' 1 be a sequence in a semigroup S and let k, m E N. Then (xn)n I is a product subsystem of FP((yn)n°-k) if and only if there exists a sequence (Hn)n I in .I f(N) such that
(a) min HI > k, (b) max Hn < min Hn+1 for each n E { 1, 2, ... , m - 1), and
(c) Xn = n,,H Yr for each n E (1,2..... m}. The proof of the following theorem is analogous to the proof of Theorem 18.4.
Theorem 18.11. Let f E N\{1} and for each i E {1, 2, ... , e} let Si be a semigroup and let (yi,n)no 1 be a sequence in Si. Let m, r E N and let X; _1S; = U;=1 CJ. There e x i s t s j E ( 1 , 2, ... , r) and f o r each i E (1, 2, ... , f - 1), there exists a product subsystem (xi,n )' of FP((y,,, )' I) and there exists a product subsystem (xe,n )R° 1 of FP((ye,n) 1) such that (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},
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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
=0.
For i E {1, 2,_. , e} and n E {1, 2, ..., m}, let Li,n = Hi,n U Ki,n. Then i itELi,n yi,t = ftEH,,,, yi,f . flfEK,,,, yi,[ = Xi,f ' zi,t
and if 1 < n < s < m, then Li,n fl Li,s = 0. Thus for each i E (1, 2, ... , e}, (xi,t zi,t) 1 is a weak product subsystem of FP((yi,n)n' k).
Finally we claim that X E_1 FP((xi,n zi,n)n 1) S A. To this end let c E X;_1 FP((xi,n zi,n)n 1) be given. For each i E (1, 2, ... , e}, pick F, c 11,2,..., m} such that ci = IIIEFi (Xi,t Zia) and let ai = fItEF Xi,t and bi = f1:EFj zi,t Then b E XE_1 FP((zi,n)n 1) so b E B. Since a E Xe_1 FP((Xi,n)"'= 1) one has that b E as -' A so that a b E A. Since each Si is commutative, we have for each i E (1, 2, ..., f) that ntEFi (Xi.t zi,t) = (fl,EF xi,t) . (f1,EF zi,t) so that c" = a b as required.
13
379
18.2 IP* Sets in Product Spaces
Theorem 18.14. Let I E N and for each i E ( 1, 2, ..., e), let Si be a commutative e semigroup and let (y;,,,)' i be a sequence in Si. Let C be an IP* set in Xi_1S, and let m E N. Then for each i E (1, 2, ..., t) there is a weak product subsystem (xi,,,)n 1 of FP((Yi,n)n°_I) such that Xc_1 FP((xi,n)n 1) C C. Proof. Let X be as in Lemma 18.13. Then X is a compact subsemigroup of X i=1 Si so by Theorem 2.5 there is an idempotent p E X. By Theorem 16.6, C E p. Thus by the definition of X for each i E {1, 2, ..., B} there is a weak product subsystem (xi,n)n1
of FP((yi,n)n_1) such that XFP((xi,n)n 1) C C. Three natural questions are raised by Theorem 18.14. (1) Can one obtain infinite weak product subsystems (defined in the obvious fashion) L such that X i_ 1 FP((xi n ) 1) _ C? (2) Can one replace "weak product subsystems" with "product subsystems"? (3) Can one omit the requirement that the semigroups Si be commutative? We answer all three of these questions in the negative. The first two questions are answered in Theorem 18.15 using the semigroup (N x N, +). Since the operation is addition we refer to "sum subsystems" rather than "product subsystems".
Theorem 18.15. There is an IP* set C in N x N such that.
(a) There do not exist z E N and a sequence (x)1 in N such that either {z} x FS((xn)n_1) C C or FS((xn)n° 1) x {z} C C. (In particular there do not exist infinite weak sum subsystems (xl,n)n° 1 and (X2,n)n°_1 of FS((2n)n° 1) with FS((xl,n)' 1) x FS((x2,n)n° 1) S C.) (b) There do not exist sum subsystems (xn)n=1 and (yn)rs_1 of FS((2n)n° 1) with {xl, x2} x {yl, y2} c C. Proof. Let
C = (N X N)\
2n, EnEG 2n) : F, G E ?(co) and max F < min G or max G < min F}.
To see that C is an IP* set in N x N, suppose instead that we have a sequence ((xn , yn)) n_
in N x N with
FS(((xn, Yn))' 1) C {(EnEF 2n, EnEG 2n) : F, G E ?(w) and max F < min G or max G < min F}. Pick Fl and G I in R(w) such thatxl = E,EF, 2' and yl =
2'. Letk = max(F1 U
G 1). Choose, by Lemma 14.16, H E R f (N) such that min H > 1, 2k+1 I EnE H xn, and
2k+1I EnEH Yn Pick F', G' E P(w) such that
xn = EnEF 2' and EnEH Yn =
380
18 Multidimensional Ramsey Theory
EnEG' 2". Then k + 1 < min(F' U G'), so xl + EnEH xn = EnEF, UF' 2" and Yl + EnEH Yn = EnEG,UG' 2". Also EnEl1)UH(xn, Yn) = (xl + EnEH Xn, Y1 + EnEH Yn) E {(EnEF 2", EnEG 2") : F, G E
`J'and (co)
max F < min G or max G < min F}.
Thus k + 1 < max(F1 U F') < min(G1 U G') < k or k + 1 < max(G1 U G') < min(F1 U F') < k, a contradiction. To establish (a), suppose that one has z E N and a sequence (x n )n° in N such that either (z) x FS((xn)n° 1) c C or FS((xn)n° 1) x {z} c C and assume without loss of
generality that {z} x FS((xn)n° 1) c C. Pick F E P(w) such that z = EtEF 2' and let k = max F. Pick H E pf(N) such that 2k+1 I EnEH xn. Pick G E J.P(w) such that EnEH xn = E,EG 2'. Then max F < min G so (z, EnEH xn) 0 C. To establish (b) suppose that one has sum subsystems (xn)2of n - and (yn )2= n FS((2n)n° 1) with {xl, x2} x {y1, y2} c C. Pick F1, G1, F2, G2 E .P(w) such that 2", yl = EnEG, 2", y2 = EnEG2 2", max Fl < min F2, xl = EnEF, 2", X2 = and max G l < min G2. Without loss of generality, max F1 > max G 1. But then we have max GI < max Fl < min F2 so (x2, yt) 0 C, a contradiction. 1
A striking contrast is provided by Theorem 18.15(b) and the case t = 2 of Theorem 18.11. That is IP* sets are not in general partition regular. It is easy to divide most semigroups into two classes, neither of which is an IP* set. Consequently it is not too surprising when one finds a property that must be satisfied by an IP* set in a semigroup S (such as containing a sequence with all of its sums and products when S = N) which
need not be satisfied by any cell of a partition of S. In this case we have a property, namely containing FS((x1,n)n 1) x FS((x2,n)n° 1) for some sum subsystems of any given sequences, which must be satisfied by some cell of a partition of N x N, but need not be satisfied by IP* sets in N x N. The third question raised by Theorem 18.14 is answered with an example very similar to that used in the proof of Theorem 18.15.
Theorem 18.16. Let S be the free semigroup on the alphabet {yl, Y2, Y3, ...}. There is an 1P* set C in S x S such that there do not exist weak product subsystems (x")n-1 and (wn)n1 of FP((yn)n° 1) such that {x1, x2} x {wl, w2} c C.
Proof Let
C=(SXS)\{(nnEFYn,nnEGYn):F,GCE ,%f(N) and max F < min G or max G < min F). To see that C is an IP* set, suppose one has a sequence ((xn, wn))n°l with FP(((xn, wn))
1) C {(nnEF Yn, nnEG Yn) : F, G E Jnf(N) and max F < min G or max G < min F}.
18.3 Spaces of Variable Words
381
Givenanyi < j inN,pick Fi, Fj, Gi, Gj, Fi,j, Gi, j E J'f(N) such thatxi = fIfEFF yn, Xi = nnEFj Yn, Wi = nnnEG, Yn, Wj = nnEGj yn, xixj = nnEFi,j yn, and wi wj = flnEGi,j Yn. Sincex1xj = nnEFi,j yn, we have thatmax F1 < min Fj andFF,j = FFUFJ and similarly max Gi < min Gj and Gi, j = Gi U Gj. Thus we may pick j E N such that max(F1 U G1) < min(Fj U Gj). Then
(Xixj, wiwj) 0 {(fIfEF Yn, nnEG Yn) : F, G E .! f(ly)
and maxF <minGor maxG <minF}, a contradiction.
Now suppose we have weak product subsystems (xn)n-1 and (wn)n=1 of FP((yn)'1) such that {xl, x2} x (wl, w2) C. Pick F1, F2, G1, G2 E J" f(N) such that xl = I1nEF, Yn, X2 = nnEF2 Yn, WI = nfEG, Yn, and w2 = nnEG2 Yn Since x1x2 E FP((yn )n° 1), we must have max F1 < min F2 and similarly max G 1 < min G2. Without loss of generality, max F1 > max G 1. But then we have max G 1 < max F1 < min F2 so (x2, WO 0 C, a contradiction. 0
18.3 Spaces of Variable Words In this section we obtain some infinite dimensional (or infinite parameter) extensions of the Hales-Jewett Theorem (Corollary 14.8) including the "main lemma" to the proof of Carlson's Theorem. (We shall prove Carlson's Theorem itself in Section 18.4.) These extensions involve sequences of variable words. Recall that we have defined a variable word over an alphabet %P to be a word over %P U {v} in which v actually occurs (where v is a "variable" not in 'Y). We introduce new notation (differing from that previously used) for variable words and nonvariable words, since we shall be dealing simultaneously with both.
Definition 18.17. Let W be a (possibly empty) set (alphabet). (a) W (T; v) is the set of variable words over 'P. (b) W (%P) is the set of (nonvariable) words over %P.
Notice that W(4'; v) U W('Y) = W('Y U {v}). Definition 18.18. Let %P be a set and let (sn (v)) n° 1 be a sequence in W('P; v). (a) The sequence (t, (v))n° 1 is a variable reduction of (sn (v))n° 1 if and only if there exist an increasing sequence of integers (ni )°O 1 with n j = 1 and a sequence (a1)O° in 'P U {v} such that for each i, 1
v E {ani , ani+1 , ... , ani+1- t } and
ti (v) = Sni (ani)^Sni+l (ani+l)- ... ^Sni+I -1 (ani+i-1).
18 Multidimensional Ramsey Theory
382
(b) The sequence (tn(v))n 1 is a variable extraction of (sn(v))n°1 if and only if (tn(v))n°1 is a variable reduction of some subsequence of (sn(v))n° 1. Thus, a variable reduction of (sn(v))n_1 is any infinite sequence of variable words
obtained from (sn(v))n° 1 by replacing each s,(v) by one of its instances, dividing the resulting sequence (of words over W U {v}) into (infinitely many) finite blocks of consecutive words, and concatenating the members of each block, with the additional requirement that each of the resulting words has an occurrence of v. For example, if (sn(v))n were the sequence (avbv, v, bav, abvb, vavv, ...) then (tn(v))n° 1 might 1
be (aabavbab, abvbbabb, ... Definition 18.19. Let W be a set and let (sn(v))n° 1 be a sequence in W(W; v). (a) The word w is a reduced word of (sn 1 if and only if there exist k E N and
a1, a2, ... , ak in W U (v) such that w = s, (al)'s2(a2)- `'sk(ak) (b) A variable reduced word of (sn(v)) ° 1 is a variable word which is a reduced word of (sn(v))°O 1.
(c) A constant reduced word of (sn(u))n° 1 is a reduced word of (sn(v))n° 1 which is not a variable reduced word.
Reduced words, variable reduced words and constant reduced words of a finite sequence (sn (v))n=1 are defined analogously. Notice that a word is a variable reduced word of (sn (v))n° 1 if and only if it occurs as the first term in some variable reduction of (sn (v)) n° 1.
Definition 18.20. Let q, be a set and let (sn(v))n° 1 be a sequence in W(W; v). (a) The word w is an extracted word of (sn (v))n°1 if and only if w is a reduced word of some subsequence of (sn (v))n' 1. (b) A variable extracted word of (sn (v))n° 1 is a variable word which is an extracted word of (sn(v))n' 1. (c) A constant extracted word of (sn (v))R° 1 is an extracted word of (s n (v))n° which is not a variable extracted word. 1
Notice that w is an extracted word of (sn (v))n°1 if and only if there exist k E N, < nk in N, and al, a2, ..., a/, E W U {v) such that
n1 < n2 <
w = Sni (a1)-sn2(a2)' ...^Snk(ak) Extracted words are defined in the analogous way for a finite sequence (sn (v))n=1 We now introduce some special notation which will be used in the proof of the main theorem of this section. The notions defined depend on the choice of the alphabet W, but the notation does not reflect this dependence.
Definition 18.21. Let W be a set, let k E N U loo) and let (sn(v))n_1 be a sequence in W (W ; v). Let S be the semigroup (W (W U {v}), "). (a) Foreach M E N, A((sn(V)),k,=,n) is the set of extracted words of (sn(v))n=m
(b) For each m E N, V((sn(v))R=m) is the set of variable extracted words of (Sn(v))n=m
18.3 Spaces of Variable Words
383
(c) For each m E N, C((sn(v))k=m) is the set of constant extracted words of (sn (v))n=m A((sn(v))n_m).
(d) A((sn(v))n_1) = nm=1 ctps (e) V((sn(v))' 1) = nm=1 cE,6s V((sn(v))n°=m). (0 C((sn(v))n= 1) = nm=1 CIfS C((sn(v))n m).
Notice that 4((sn(v))n° 1) = 'V((sn(v))' 1) U G((sn(v))R° 1). Notice also that if tY = 0, then each C((sn(v))k=m) = 0. Lemma 18.22. Let %P be a set, let (sn(v))rs_1 be a sequence in W(41; v), and let S be
the semigroup (W(W U {v}), "). Then A((sn(v))' 1) is a compact subsemigroup of 18S, and V((sn(v))n° 1) is an ideal of A((sn(v))n° 1). If 9l 0, then C((sn(v))°O 1) is a compact subsemigroup of PS.
Proof. We use Lemma 14.9 with J = (11, D = N, and Tm = S for each m E N. To see that ,it((sn (v))n° 1) is a subsemigroup of PS and V((sn (v))n°1) is an ideal of ,4((sn(v))n°_1), for each m E N let Em = A((sn(v))n°_m) and let I. = V ((sn(v))n°=m). In order to invoke Lemma 14.9 we need to show that
(1) for each m E N, Im c Em c Tm, (2) for each m E N and each W E Im, there is some k E N such that w"Ek C Im, and (3) for each m E N and each w E Em \ Im , there is some k E N such that w- Ek S Em
and w"'Ik c Im. Statement (1) is trivial. To verify statement (2), let m E N and W E Im be given. Nwith nl>m,and al,a2,...,a,E'IU(v)(with Pick at least one a; = v) such that W = sn, (a1)^snz(a2)^ ... -sn,(at).
Let k = nt + 1. Then w"Ek c Im as required. The verification of statement (3) is identical except that a1, a2, ... , at E 4'. Now assume that 41 0 0. To show that C ((sn (v))n°1) is a subsemigroup it suffices by Theorem 4.20 to show that for each m E N and each w E C((sn(V))°n°__m) there is some k E N such that w"'C((sn(v))n° k) c C((sn(v))°_m). The verification of this assertion is nearly identical to the verification of statement (2) above. o By now we are accustomed to using idempotents to obtain Ramsey-theoretic results. In the following theorem, we utilize two related idempotents in the case 9! 0.
Theorem 18.23. Let 41 be a finite set and let (sn (v))n° 1 be a sequence in W(4'; v). Let
Y and g be finite partitions of W(W; v) and W(4') respectively. There exist V E ,F and C E $, and a variable extraction (wn(v))n_1 of (sn(v))' such that all variable 1
extracted words of (wn(v))n° 1 are in V and all constant extracted words of (wn(v))n° are in C.
18 Multidimensional Ramsey Theory
384
Proof Assume first that 'Y = 0. Then W(i/r; v) = {vk : k E N} and W(AY) = {0}. Given n E N, pick k(n) E N such that sn(v) = vkini. For each V E F let B(V) = {k E N : vk E V}. Pick by Corollary 5.15 V E F and a sum subsystem (f(n))n° of (k(n))n° t such that FS((2(n))n° r) C B(V). For each n E N let wn(v) = Onl. Then (wn(v))n is a variable extraction of (sn(v))n° and the variable extracted words of 1
1
1
(wn(v))n° 1 correspond to the elements of FS((e(n))n° 1) so that all variable extracted
words of (wn(v))' 1 are in V. The conclusion about constant extracted words of (wn(v))n° 1 holds vacuously. Now assume that 'I 0. By Lemma 18.22, C ((sn(v))n° 1) is a compact subsemi-
group of ,BS, where S is the semigroup (W ('Y U {v}), " ). Pick by Theorem 1.60 an idempotent p which is minimal in C((sn(v))n° 1). Then p E A((sn(v))n° 1) so pick, again by Theorem 1.60, an idempotent q which is minimal in As n (v))n° 1) such that q < p. Since, by Lemma 18.22, V((sn(V)),° 1) is an ideal of A((sn(v))n° 1), we have that q E V ((sn(v))n-1). Now, for each a E 'Y, define the evaluation map Ea : W(' U {v}) W(4<) as follows. Given W E W (tP U {v}), Ca (w) is w with all occurrences of v (if any) replaced by a. Thus, if w = w(v) E W ('Y; v), then Ea (w (v)) = w(a), while if w E W (%P), then Ea(W) = w. Notice that Ca is a homomorphism from W (4Y U {v}) to W (%P) and hence,
by Corollary 4.22, Ea is a homomorphism from,(W(q, U {v})) to,n(W(q')). We claim that for each a E 'If, Ea(q) = p. Since EQ is a homomorphism, we have that Z. (q) < EQ (p). Since ca is equal to the identity on W (*), we have EQ (p) = p and consequently EQ (q) < p. Since p is minimal in C ((sn (v))00 1), to verify that Ea (q) = p
we need only show that a(q) E e((sn(u))n° 1). To see that Ea(q) E C((sn(v))n 1), let m E N be given. Now V((sn(V));_,,,) E q SO Ca IV ((Sn (V))n° m)] E Ea (q) by Lemma 3.30. It is routine to verify that
Ea[V((Sn(V))nm)] C C((Sn(V))nm) m) E Q(q) as required. We shall, according to our usual custom, use " for the binary operation on PS which extends the operation " on S. SO C((sn(V))
Let V E F n q and let C E gn p. We put V a= {y E V : Y- P E V and y"q E V) and CO = {x E C : x"p E C and x"q E V}. Since q"'q = p"q = q"p = q and p- p = p, it follows that V G E q and Co E P.
We claim that for each y E Va and each x E Ca, y-'(Vt) E q, y-'(Vt) E p, x-' (Vtt) E q, and X-' (CO) E p. To see this, let y E Vt and X E Ct be given. To see that Y-1 (0) E q, we observe that ky' pP' IV] is a neighborhood of q in PS, and so is x -'p-1 Y
4
[V]. Also y-'V Eq. Since y-1 (0) = Y-'V n,l Y 'p-'{V] n xY p-` {V], P
it it follows that y-1(Vt) E q. Likewise, since p E Ay'pp'[V] n xyIp4'[V], and
y-' V E p, y-1(Vt) E P. Similarly x-1(Vt) = x-'VnAX' pP I[V]nAx I py l[V] E q andx-1(Ct)=x-1CnAXlpp'[C]nAX1pq'[V]
E P.
If x, y E A((sn(u))n° 1), we shall write x < < y if there exists m E N for which x E A((sn(v))n 1) and y E A((sn(v))n°m+i). We observe that, for any given x E A((Sn(V))n°-1), {Y E A((s(v))n° 1) : x << y} E q.
18.3 Spaces of Variable Words
385
We now inductively construct a variable extraction (wn(v))n° 1 of (sn(v))n° 1 for which V((wn(v))nm 1) c Vt and C((wn(v))n° 1) c C. We first choose w1 (v) E Va ifl naEV1 e;' [CU]. We note that this is possible because
ea1[Ct] E q for every a E'Y. W e then let n E N and assume that we have chosen w; (v) f o r each i E { 1, 2, ... , n) so that wi(v) < < w;+1(v) for every i E (1,2,...,n - 11, V((w;(v))" 1) e Va and
C((w; (v))i 1) C C. We can then choose wn+1 (v) satisfying the following conditions.
(1) wn+1(v) E Va n naEgv ea-1[Ct], (ii) wn (v) < < wn+1(v), (iii) wn+1(v) E y-1(Va) for every y E V((wi(v))n_1), (iv) wn+1(v) E ea-1[y-1(Vt)] for every y E and every a E'Y, (v) wn+1(v) E x-1(Vt) for every x E C((w; (v))" 1), and (vi) wn+l (v) E ea-1 [x-1(0)] for every x E C((wi (v))"=1) and every a E %P. This choice is possible, because each of these conditions for wn+l (v) is satisfied by all the elements of some member of q.
To see that V((wi(v))n+i) C VU, let u E and pick k and ml < < Mk in N and al, a2, ..., ak E 'Y U (v) such that some ai = v and
m2 <
u = wm i (a l) ^ wm2 (a2)^ ... ^ wmk (ak)
If mk < n + 1, then u E Vt by the induction hypothesis so assume that mk = n + 1. If k = 1, then u E V t by condition (i), so assume that k > 1. If ak E W, then u E V O by condition (iv). If ak = v and {a1, a2, ... , ak_ 1 } c 'P, then u E V t by condition (v). If ak = v and v E jai, a2, ... , ak _ 1), then u E V1 by condition (iii). That C((w; (v))n+11) c Ct follows from the induction hypotheses and conditions (i) and (vi). Finally, notice that (wn(v))n 1 is a variable extraction of (sn(v))n° 1 by conditions (i) and (ii).
The following corollary includes the "main lemma" to the proof of Carlson's Theorem. Corollary 18.24. Let kY be a finite set and let (sn (v)) ' 1 be a sequence in W (%P; v). Let
F and g be finite partitions of W (4Y; v) and W ('Y) respectively. There exist V E F and C E 9 and a variable reduction (tn(v))n° 1 of (sn(v))n° 1 such that all variable reduced words of (tn (v))n° 1 are in V and all constant reduced words of (tn (v))' 1 are in C. Proof. Define cp : W(W U {v}) --> W
U {v}) by
rp(a1a2 ... ak) = sI (a1)^s2(a2)^ ...
sk(ak)
where a 1,a2,...,akare in xPU{v}. Note that cp[W(%Y; v)] c W(W; v)andrp[W('P)] C
W(W). Let 3f = IV-' [B] : B E F) and 3C = {rp-1[B]
B E }. Then 3f and X
386
18 Multidimensional Ramsey Theory
are finite partitions of W(%P; v) and W(%P) respectively. Pick V' E X, C' E X, and a variable extraction (wn(v))'1 of the sequence (v, v, v, ...) as guaranteed by Theorem 18.23. (That is, (wn (v)) °_ 1 is simply a sequence of variable words, all of whose variable
extracted words are in V' and all of whose constant extracted words are in C'.) Pick V E F and C E a such that V' = q- [V ] and C' = (p-1 [C]. For each n E N pick kn e N and an,1, an,2, ... , an,kn E'P U {v) such that wn = an, I an,2 ... an,kn .
Let 81 = 0 and for each n > 1, let In = E7. k; . For each n E N let to (v) = St.+I
(an.2)-
. -st.+kn (an.kn )
and notice that (tn(v))°_1 is a variable reduction of (sn(v))n° 1. Notice further that if
b1,b2,...,bn E WU{v},then op(wl (bt)_w2(b2)_ ... -wn (bn)) = t1(b1)-t2(b2)- ... --tn(bn) so if b1b2 ... bn E W (%P; v), then t1(b1)"t2(b2)"
-tn (bn) E V and if b1b2 ... bn E
W M, then t100-t202)_ ... -tn (bn) E C. Theorem 18.23 produces a sequence (wn (v))n_ 1 with a stronger homogeneity prop-
erty than that in Corollary 18.24, but to obtain such a (wn(v))n° 1 we must take an extraction of (sn(v))°_j; a reduction won't do. (To see why it won't do, consider (sn(v))n'1 = (av, by, by, by, ...)and partition the variable words according to whether the first letter is a or not.) Exercise 183.1. Derive the Finite Sums Theorem (Corollary 5.10) as a consequence of Corollary 18.24. (Hint: Consider the function f : W(W; v) -* N defined by the condition that f (w) is the number of occurrences of v in w.)
18.4 Carlson's Theorem In 1988 T. Carlson published [58] a theorem which has as corollaries many earlier Ramsey-theoretic results. We prove this theorem in this section. Definition 18.25. Let %P be a (possibly empty) finite set and let 3 be the set of all infinite sequences in W (W ; v). Denote the sequence (sn )' 1 by i. For s" E 8, let
B0(s) = (7 E 8 : t is a variable reduction of s), and, for each n E N, let Bn ( s ' ) = (d E B o ( s ' ) : for each i E { 1, 2, ... , n}, t; = si }.
18.4 Carlson's Theorem
387
Remark 18.26. For every ss", t E s and m, n E w, if tt E Bm(i) and if m < n, then Bn (F) C B. (i)
-
The following lemma is an easy consequence of Remark 18.26. Lemma 18.27. A topology can be defined on s by choosing {Bn(ss) as a basis for the open sets.
E sand n E w}
Proof. Let s, t E s, let m, n E w, and let i E Bn (s) n Bn (t'). Let k = maxim, n}. Then by Remark 18.26 Bk (i) C Bn (s) n Bn (t). Definition 18.28. The Ellentuck topology on ,& is the topology with basis {Bn(i) : i E s and n E W). For the remainder of this section, we shall assume that ,& has the Ellentuck topology.
Definition 18.29. A set X c s is said to be completely Ramsey if and only if for every n E w and every s" E s, there is some i E B. (s) such that either Bn (t') c X or
Bn(t')nx=0. Recall that we take the Baire sets in a topological space to be the a-algebra generated by the open sets and the nowhere dense sets. (Other meanings for the term Baire sets also exist in the literature.) Recall also (see Exercise 17.5.1) that a set is a Baire set if and only if it can be written as the symmetric difference of an open set and a meager set, where the meager (or first category) sets are those sets that are the countable union of nowhere dense sets. Carlson's Theorem (Theorem 18.44) is the assertion that a subset of s is completely Ramsey if and only if it is a Baire set (with respect to the Ellentuck topology). The proof in one direction is simple.
Theorem 18.30. If X is a completely Ramsey subset of s, then X is a Baire set. In fact X is the union of an open set and a nowhere dense set. Proof Let X° denote the interior of X. We shall show that X is a Baire set by showing that X \X' is nowhere dense so that X is the union of an open set and a nowhere dense set.
If we assume the contrary, there exists i E s and n E co for which Bn (s') c cfs (X \X °). Since X is completely Ramsey, there exists t' E B. (s') such that Bn (t) c X
or Bn(t) c &\X. However, if Bn(t) X, then Bn(t) c X° and hence B,,(F) n cfs(X\X°) = 0. This is a contradiction, because Bn (t) c Bn (i ). If B,, (t) c -3 \X, then Bn (t) n ces(X) = 0. This is again a contradiction, because
Bn(t) c B. (g) c cfs(X). Lemma 18.31. The completely Ramsey subsets of s form a subalgebra of .l (s).
18 Multidimensional Ramsey Theory
388
Proof It is immediate from the definition of a completely Ramsey set that a subset of -3 is completely Ramsey if and only if its complement is. So it is sufficient to show that the completely Ramsey subsets of 3 are closed under finite unions. To this end let X and Y be completely Ramsey subsets of -3. To see that X U Y is completely Ramsey, lets E -8 and n E w be given. Pick t E Bn (s") such that B (t') C X
or B.(1) c .8\X. If B.(!) C X, then Bn(i) c X U Y and we are done, so assume that B1(!) C -3 \X. Since Y is completely Ramsey, pick u" E B, (7) such that Bn(ua) C Y or Bn(u) C 45\Y. Now Bn(u') C B,, (F) and so if Bn(u') C Y, then Bn(ua) C X U Y and if
B. (a) C 3\Y, then Bn(i) C 3\(X U Y). Remark 18.32. Let (7m )m l be a sequence in -3 with the property that t"'+1
E
Bm+l(?n) for every m E N, and let t = (t11, t22, t33, ...). Then t E Bm(t'") for every m E Co.
We now need to introduce some more notation.
Definition 18.33. Let n E w. (a) Ifs E 45 and n > 0, then sin = ( 5 1 , 5 2 . . . . . sn) and s'1o = 0. (b) If s' E 8, then tails (s) = (sn+1, sn+2, ) (c) If s' E W (%P; v)", then Is ] = n. (d) Ifs is a finite sequence and t is a finite or infinite sequence, then (s, t) will denote the sequence in which s' is followed by F. (Thus, if s = (sl , s2, ... , sn,) and
t = (ti, t2, ...), then (s", t) = (s1, s2, ... , sm, ti, t2....).) (e) A subset X of -8 is said to be almost dense in Bn (s") if X n Bn (t') t E Bn (s').
0 whenever
Notice that the notation iijn differs slightly from our formal viewpoint since formally n = {0, 1, ... , n - 1) and {0, 1, ... , n - 1) is not a subset of the domain of the function s. Definition 18.34. We define a condition (*) on a sequence (T")nE,,, of subsets of 43 as follows: (*) For every n E w and every s' E -3
(a) B1(s) n T" 0 and (b) if s' E T", then B1(s) C T". We make some simple observations which will be useful in the proof of Carlson's Theorem.
Remark 18.35. Lets E 45. (a) If X is almost dense in Bn (i), then X is almost dense in Bn (I)for every t' E Bn (s).
(b) If X n Bn (s) = 0, then X n Bn (t) = O for every t' E Bn (s). (c) For any t r= 43, t E Bn (s') if and only if t1n = Sin and tails (t) E Bo(tailn (s )).
18.4 Carlson's Theorem
389
Lemma 18.36. Let (T")nE(o be a sequence of subsets of .8 which satisfies (*) and let s' E -3. Then for each k E to, there exists t E Bk (s') such that, for every n > k in N and every variable reduction r' of tin, we have (F, tailn(t)) E TM. Proof. We first show that
(t) for every t E & and every n E N, there exists a E Bn(t) such that (r, tailn(a)) E T1'1 for every variable reduction r" of in.
We enumerate the finite set of variable reductions of in as {r' 1, r' 2, We put r" ° = tails (t) and then inductively choose i 1, i 2, E .8 so that the following properties hold f o r each i E { 1 , 2, ...
(i) i' E Bo(?i-') and (ii) (F', V) E TI't1. Suppose that i E (0, 1, 2, ... , m - 1} and that we have chosen V. Let I = Ir'i+1I. By condition (a) of (*), we can choose u E B1((i'+1, i')) n TI. Let V+' = taill(u'). r+1) = u' Then V+' E Bo(i `) and (i'+, V+')
E TI' i+' I.
Having chosen z "', we put o = (tin, T "`). Since a""` E Bo(tailn(t)), we have E Bn(t). Now let r be a variable reduction of tin. Then r" = r"' for some i E {1, 2, ... , m}.
We note that s"m E Bo(i'), and hence that (r', ?m) E BI;i ((r"', i')). So (r`, i") _ (r"', tailn(a)) E TI'`1, by condition (b) of (*). Thus (t) is established. Now let k E co be given. By (f) we can inductively define elements TO, 71, t'2, of -8 with the following properties:
. .
(i)t°=t'=.. =tk=S, (ii) t " E Bn (t'"-') for every n > k, and (iii) for every n > k and every variable reduction r of t n-1 In, (r', tail,, (in )) E T1'I.
Let t = (t11, t22, t3, ...). Then t E Bk(s'). Let n > k be given and let r' be a variable reduction of tin. Now t1n =
t"-1
,, so r" is a variable reduction of t n-1In and
thus (r', tailn(i")) E TI'I. Also (r, tailn(t)) E Blr, ((r, tailn(t"))) and so by condition (b) of (*), (r', tails (t)) E T171.
Corollary 18.37. Let s' E -3, k E w, and X c 3. Then there exists t e Bk(s') such that, for every n > k and every variable reduction r' of in, either X is almost dense in Bl;, ((r', tailn(t))) or else X fl Bl;l ((r", tailn(t))) = 0. Proof. For each n E w we define T" C 8 by stating that i E T" if and only if either X is almost dense in Bn(r') or X fl Bn(?) = 0. We shall show that (T"),E,,, satisfies (*), and our claim will then follow from Lemma 18.36. Let n E co and i E -3. If X is almost dense in Bn (r'), then i E Bn (r") fl T" . Otherwise there exists i E Bn (i) for which Bn (u') fl X = 0 and soil E Bn (r') fl T". Thus (T" )nE,,, satisfies condition (a) of (*).
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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
394
18 Multidimensional Ramsey Theory
many pieces, there exists an infinite dimensional subspace of W('Y) all of whose ddimensional subspaces lie in the same cell of the partition. This result can be established by methods similar to those of Section 18.3. (See [20].)
Part IV Connections With Other Structures
Chapter 19
Relations With Topological Dynamics
We have already seen that the notions of syndetic and piecewise syndetic, which have their origins in topological dynamics, are important in the theory of OS for a discrete semigroup S. We have also remarked that the notion of central, which is very important in the theory of f3S and has a very simple algebraic definition, originated in topological dynamics. In this chapter we investigate additional relations between these theories. In particular, we establish the equivalence of the algebraic and dynamical definitions of "central".
19.1 Minimal Dynamical Systems The most fundamental notion in the study of topological dynamics is that of dynamical system. We remind the reader that we take all hypothesized topological spaces to be Hausdorff.
Definition 19.1. A dynamical system is a pair (X, (TS)SEs) such that
(1) X is a compact topological space (called the phase space of the system); (2) S is a semigroup; (3) for each s E S, T5 is a continuous function from X to X; and (4) for all s, t E S, TS o Tr = TSr If (X, (TS)SEs) is a dynamical system, one says that the semigroup S acts on X via (TT)SES.
We observe that we are indeed familiar with certain dynamical systems.
Remark 19.2. Let S be a discrete semigroup. Then (,OS, (X )IEs) is a dynamical system.
Definition 19.3. Let (X, (TS)SEs) be a dynamical system. A subset Y of X is invariant if and only if for every s E S, TS [Y] c Y.
19 Relations With Topological Dynamics
398
Lemma 19.4. Let S be a semigroup. The closed invariant subsets of the dynamical system (VS, (A5)SEs) are precisely the closed left ideals of $S.
Proof Given a left ideal L of,8S, p E L, ands E S, one has XS(P) = sp E L so left ideals are invariant.
Given a closed invariant subset Y of fS and p E Y one has that (PS) p = ct(Sp) S
cl'Y=Y.
t]
Definition 19.5. The dynamical system (X, (TS)SEs) is minimal if and only if there are no nonempty closed proper invariant subsets of X.
A simple application of Zom's Lemma shows that, given any dynamical system (X, (TS)SEs), X contains a minimal nonempty closed invariant subset Y, and consequently (Y, (T5)sE5) is a minimal dynamical system, where T,s is the restriction of TS to Y.
Lemma 19.6. Let S be a semigroup. The minimal closed invariant subsets of the dynamical system (,BS, (AS)SES) are precisely the minimal left ideals of PS.
Proof. This is an immediate consequence of Lemma 19.4 and the fact that minimal left ideals of )SS are closed (Corollary 2.6). o
Of course, a given semigroup can act on many different topological spaces. For example, if X is any topological space, f is any continuous function from X to X, and is a dynamical system. And, since any for each n E N, T = f", then (X, dynamical system contains a minimal dynamical system, a given semigroup may act minimally on many different spaces. Definition 19.7. Let S be a semigroup. Then (X, (Ts)sEs) is a universal minimal dynamical system for S if and only if (X, (Ts)ses) is a minimal dynamical system and, whenever (Y, (Rs )SEs) is a minimal dynamical system, there is a continuous function (p
from X onto Y such that for each s E S one has that RS o v = w o TS. (That is, given any s E S, the diagram TS X (P
RS
Y
commutes.)
Theorem 19.8. Let S be a semigroup, let L be a minimal left ideal of,BS, and for each s E S, let )ls be the restriction of As to L. Then (L, (As)SEs) is a universal minimal dynamical system for S.
19.1 Minimal Dynamical Systems
399
Proof Let (Y, (Rs )sEs) be a minimal dynamical system and fix y E Y. Define g : S -> Y by g(s) = RA(y) and let V = gL. Then immediately we have that (p is a continuous function from L to Y. To complete the proof it suffices to show that for all s E S and all p E L, ((,ls(p)) = Rs ((p(p)). (For then R.[(p[L]] C W[L] so (p[L] is invariant and thus by minimality (p[L] = Y.) To see that tp(X's(p)) = Rs(rp(p)) for all s E S and all p E L, it suffices to show
that for each s E S, g o A = Rs o g on PS for which it in turn suffices to show that g o As = Rs o g on S. So let t E S be given. Then g(As(t)) = g(st) = Rsr(y) = Rs(Rt(y)) = Rs(g(t)). We now wish to show that (L, (As)SEs) is the unique universal minimal dynamical
system for S. For this we need the following lemma which is interesting in its own right.
Lemma 19.9. Let S be a semigroup and let L and L' be minimal left ideals of fS. If tp is a continuous function from L to L' such that ,ls o rp = cp o As for all s E S (where),' is the restriction ofAs to L), then there is some p E L' such that tp is the restriction of pp to L.
Proof. Pick by Corollary 2.6 some idempotent q E L and let p = ((q). Then pp and tP o Pq are continuous functions from fiS to L'. Further, given any s E S,
pp(s)=sp=stp(q)=(),sotp)(q)=(tpoA')(q)=W(sq)=(tpopq)(s) and thus pp and tp o pq are continuous functions agreeing on S and are therefore equal. In particular, ppjL = (tp o pq)IL. Now, by Lemma 1.30 and Theorem 1.59, q is a right identity for L, so (tp o pq)IL = tp and thus tp = PpIL as required. The following theorem is in many respects similar to Remark 3.26 which asserted the uniqueness of the 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 _
1
1
}
Notice that (since a2 +2 < a3), C n D = 0. We claim that (p)(x) E C and (q)(x) E D. To see this it suffices to show, since C and D are closed, that for each n E N, if n E A n B, then TT(x) E C, and if n f A n B, then Tn(x) E D. So let n E N be given. Then
Tn (x) = Z + E! ° 1 a ja"-J = Z + Ej n+l ajan-j since aja"-J E Z if j < n. Now assume that n E A n B. Then an+1 = 1 and (because aja"-j = a-1 + E n+3 aja"-3 So
n + 10 B) an+2 = 0. Thus E n+1
U
< Ej=n+1
and hence Tn (x) E C as claimed.
ajan-j
<
a
+
a3
404
19 Relations With Topological Dynamics
Next assume that n 0 A fl B. Then an+, = 0 and either an+2 = 0 or an+3 = 0. Consequently 0 < Ej n+la%an-J
T22 + a4
and hence T,, (x) E D as claimed.
If one views the circle group as {z E C : Izi = 11, then Theorem 19.19 says that the enveloping semigroup of the system (T, (f°)fEN) is isomorphic to $N, where
f(z)=z°.
19.3 Dynamically Central Sets The term "central set" was originally defined by Furstenberg [98, Definition 8.3] for subsets of N. Sets defined algebraically as elements of minimal idempotents were called "central" because many of the same theorems could be proved about them. The original definition (restricted to the semigroup N and applied only to metric dynamical systems) is as follows. We add the modifier "dynamically" because it is by no means obvious that the concepts are the same, even for subsets of N. Definition 19.20. Let S be a semigroup. A set C c S is dynamically central if and only if there exist a dynamical system (X, (TS)SES), points x and y in X, and a neighborhood U of y such that
(1) y is a uniformly recurrent point of X, (2) x and y are proximal, and (3) C= Is E S: T,.(x) E U}. The terms "uniformly recurrent" and "proximal" are notions from topological dynamics, defined as follows. Definition 19.21. Let (X, (TS)SES) be a dynamical system. (a) A point y E X is uniformly recurrent if and only if for every neighborhood U of y, IS E S : TS(y) E U} is syndetic. (b) Points x and y of X are proximal if and only if there is a net (s,)1Et in S such that the nets (TS, (x)),EI and (TS, (y)),EI converge to the same point of X.
The definition of "proximal" as standardly given for a dynamical system with a metric phase space (X, d) says that there is some sequence (s n )n° , such that lim d (T,,, (x), TS (y)) = 0. It is easy to see that this definition is equivalent to the n
00
one given above. The characterization of proximality provided by the following lemma will be very convenient for us.
19.3 Dynamically Central Sets
405
Lemma 19.22. Let (X, (TS)SEs) be a dynamical system and let x, y E S. Then x and y are proximal if and only if there exists a point p E CBS such that Tp(x) = Tp(y).
Proof Notice that if p E V, (S,61 is a net in S which converges to p, and z E X, then Tp(z) = limLEJ TS,(z). Thus if p E 6S and Tp(x) = Tp(y), then picking any net (s,),EJ in S which converges to p, one has limLE! Ts, (x) = lim,EI Ts, (y).
Conversely, assume that one has a net (s,),EJ in S such that limLEJTs,(x) _ lim,EJ Ts, (y). Let p be any limit point of (s,),E, in fS. By passing to a subnet we may presume that (S,),EJ converges to p. Then Tp(x) = lim,Ej Ts, (x) = lim,EI Ts, (y) _ Tp (Y).
In order to establish the equivalence of the notions of "central" we investigate relationships between some algebraic and dynamical notions. Theorem 19.23. Let (X, (Ts)SES) be a dynamical system, let L be a minimal left ideal of ,6S, and let x E X. The following statements are equivalent. (a) The point x is a uniformly recurrent point of (X, (Ts)SES) (b) There exists u E L such that Tu(x) = x. (c) There exists an idempotent e E L such that Te(x) = x. (d) There exists y E X and an idempotent e E L such that Te(y) = x. Proof (a) implies (b). Choose any v E L. Let .A be the set of neighborhoods of x in X. For each U E dV let BU = IS E S : Ts(x) E U). Since x is uniformly recurrent, each
BU is syndetic and so there exists a finite set FU c S such that S = U,EFU(t-'Bu). Pick to E FU such that tU - t BU E v so that tU v E BU. Let u be a limit point of the net (tUV)UE.N in , S and notice that u E L. We claim that T. (x) = x. By Remark 19.13, it suffices to show that B v E u for each V E X. So let V E X and suppose that By 0 u.
Pick U C V such that tuv E S\By. Then tUV E BU C By, a contradiction. (b) implies (c). Since u belongs to a group contained in L (by Theorem 2.8), there is an idempotent e E L for which eu = u. So (using Remark 19.13) Te (x) = Te (Tu (x)) _ Teu(x) = Tu(x) = X. That (c) implies (d) is trivial.
(d) implies (a). We note that Te(x) = Te(Te(y)) = Tee(y) = Te(y) = x. Let U be a neighborhood of x. We need to show that IS E S : Ts(x) E U} is syndetic. Pick a neighborhood V of x such that ce V C_ U and let A = Is E S : Ts(x) E V}. Since
x = e-lim Ts(x) by Remark 19.13, A E e. Let B = Is E S : se E A}. Then by sEs Theorem 4.39, B is syndetic. We claim that B C Is E S : Ts(x) E U}. Indeed, if
SEB,then T5(x)=Ts(Te(x))=TSe(x)E(T,(x):tEA)CVCU. Theorem 19.24. Let (X, (Ts)SEs) be a dynamical system and let x E X. Then there is a uniformly recurrent point y E ce{Ts (x) : s E S} such that x and y are proximal.
Proof. Let L be any minimal left ideal of CBS and pick an idempotent u E L. Let y = Tu(x). Then trivially y E ce{Ts(x) : s E S}. By Theorem 19.23 y is a uniformly
406
19 Relations With Topological Dynamics
recurrent point of (X, (Ts)SEs). By Remark 19.13 we have that T. (y) = Tu(Tu(x)) = y are proximal. Tuu (x) = T. (x) so by Lemma 19.22
Theorem 19.25. Let (X, (Ts)sEs) be a dynamical system and let x, y E X. If x and y are proximal, then there is a minimal left ideal L of $S such that Tu (x) = Tu (y) for all
uEL. Proof. By Lemma 19.22, {p E fS : Tp(x) = Tp(y)} # 0. It is a left ideal in $S, because, for every p, q E AS, Tp(x) = Tp(y) implies that Tqp(x) = Tq(Tp(x)) = Tq (Tp(Y)) = Tgp(Y)
Theorem 19.26. Let (X, (Ts)5Es) be a dynamical system and let x, y E X. There is a minimal idempotent u in fS such that Tu(x) = y if and only if x and y are proximal and y is uniformly recurrent. Proof. Necessity. Since u is minimal, there is a minimal left ideal L of $S such
that u E L. Thus by Theorem 19.23 y is uniformly recurrent. By Remark 19.13 Tu(y) = Tu(Tu(x)) = Tuu(x) = Tu(x) so by Lemma 19.22 x and y are proximal. Sufficiency. Pick by Theorem 19.25 a minimal left ideal L of 1S such that Tu (x) _ Tu (y) for all u E L. Pick by Theorem 19.23 an idempotent u E L such that Tu (y) = y. We are now prepared to establish the equivalence of the notions of central.
Theorem 19.27. Let S be a semigroup and let B C S. Then B is central if and only if B is dynamically central.
Proof Necessity. Let Q = S U {e} where e is a new identity adjoined to S (even if S already has an identity). Let S2 = Q{0, 1} and for s E S define Ts : 12 -+ f2 by TS(f) = fops. Then by Lemma 19.14, (f2, (T5)SEs) is a dynamical system. Let x = X B, the characteristic function of B g Q. Pick a minimal idempotent u in $ S such that B E u and let y = Tu (x). Then by Theorem 19.26 y is uniformly recurrent and x and y are proximal. Now let U = {z E f2 : z(e) = y(e)}. Then U is a neighborhood of y in Q. We note
that y(e) = 1. Indeed, y = Tu(x) so {s E S : Ts(x) E U} E u so choose some s E B such that T,(X) E U. Then y(e) = TS(x)(e) = x(es) = 1. Thus given any s E S,
5EB
.
x(s)=I Ts(x)(e) = 1
q TS(x)EU. Sufficiency. Choose a dynamical system (X, (Ts)SEs), points x, y E X, and a neighborhood U of y such that x and y are proximal, y is uniformly recurrent, and B = is E S : Ts(x) E U). Choose by Theorem 19.26 a minimal idempotent u in CBS such that Tu(x) = y. Then B E U. It is not obvious from the definition that the notion of "dynamically central" is closed
under passage to supersets. (If B = (s E S : TS (x) E U) and B C C, one can let V = U U {TS(x) : s E C). But there is no reason to believe that Is E S : Ts(x) E V) c C.)
407
19.4 Dynamically Generated IP* Sets
Corollary 19.28. Let S be a semigroup and let B C_ C c S. If B is dynamically central, then C is dynamically central. Proof. This follows from Theorem 19.27 and the fact that supersets of central sets are central. 0
Exercise 19.3.1. Let (X, (Ts)sEs) be a dynamical system and let x E X. The point x E X is periodic if and only if there is some s E S which is not an identity such that TS (x) = x. The point x E X is recurrent if and only if every neighborhood of x contains a point of the form T,s (x) for some s e S other than an identity.
Assume that x is not a periodic point of X and that S* is a subsemigroup of fS. (See Theorem 4.28.) Prove that the following statements are equivalent:
(i) x is recurrent. (ii) Tp (x) = x for some p E S*. (iii) Te(x) = x for some idempotent e E S. (iv) Given a neighborhood U of x, there exists a sequence (s )n t). of S such that T5(x) E U for every s E
of distinct elements
19.4 Dynamically Generated IP* Sets In this section we consider certain dynamically defined sets that are always IP* sets. In contrast with the results of Section 19.3, we see that the notion of a "dynamical IP* set" is not equivalent to that of "IP* set", but is rather significantly stronger. The dynamical notion with which we shall be concerned in this section is that of a measure preserving system.
Definition 19.29. (a) A measure space is a triple (X, B, A), where X is a set, 2 is a a-algebra of subsets of X, and /.t is a countably additive measure on 2 with µ(X) finite. (b) Given a measure space (X, 2, p,), a function T : X -+ X is a measure preserving
transformation if and only if for all B E 2, T-I [B] E 2 and µ(T-t [B]) = lt(B). (c) Given a semigroup S and a measure space (X, 2, A), a measure preserving action of S on X is an indexed family (TS)SEs such that each TS is a measure preserving
transformation of X and TS o T, = TS, for all s, t E S. It is also required that if S has an identity e, then Te is the identity function on X.
(d) A measure preserving system is a quadruple (X, 2, s, (Ts)sEs) such that (X, 2, µ) is a measure space and (TS)SEs is a measure preserving action of S on X.
We shall need a preliminary result about the semigroup (pf(N), U).
Definition 1930. Let (X, 2, p,) be a measure space. A monotone action of R f (N) on X is an indexed family (TF)FE.,f(N) such that
19 Relations With Topological Dynamics
408
(a) each TF is a measure preserving transformation of X and (b) if F, K E R f (N) and max F < min K, then TF o TK = TFUK
Notice that, because of the "max F < min K" requirement, a monotone action of Pf(N) on X need not be a measure preserving action of .l f(N) on X. Lemma 19.31. Let (TF)FE.Pf(N) be a monotone action of .9'f (N) on the measure space (X, B, g) and let A E B with g(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 <
I{y+1,y+2,...,y+m}fl(UtEG -t+A)I u
= E;=o I{y+ik+1,y+ik+2,...,y+(i+1)k}fl(UtEG -t+A)I
+I{y+vk+l,y+vk+2,...,y+m}fl(LJtEG -t+A)I < v(k-1)+k
so, since 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<
I(-yi+A)n{1,2,...,Xn}1 <2a. Xn
It suffices to show that for some pair (i, j) with 1 < i < j < k, one has
d((-yi + A) n (-yj + A)) > a2 - E since d((-yi + A) n (-yj + A)) = d(A n (-(yj - yi) + A)). Suppose instead that this conclusion fails. Then in particular for each pair (i, j) with 1 < i < j < k, there is some vi, j E N such that for all n > vi,1 I(-yi+A)n(-yj+A)n{1,2,...,x,}I xn
Let n = max({e} U (vi, j : 1 < i < j < k}) and let in = x,. For each i E 11, 2, ..., k},
letAi = (-yi+A)n{1,2,...,m} and foranyC c {1,2,...,m},let t(C) = ICI m Then foreachi E {l,2,...,k},b
x`(t)
A(C) = Em'M We claim that it suffices to show that
k2b2 < Ek_. 1i(Ai) + 2
Ei=k- I
=i+i t.(Ai n Aj ).
1
Indeed, once we have done so, we have k2b2 < 2ka + k(k - 1)(a2 - E) so that
2a > k(E - (a2 - b2)) + a2 - E > k(E - (a2 - b2)) = kz > 2a a contradiction.
20.3 A Density Version of the Finite Sums Theorem
421
To establish the desired inequality, observe first that zk=1 Em 1 XA. (t) m
k
kb < Ei=1 N,(Ai) = so that m
k
k2b2 < (Et=1 E,=1 XAi (t))
2
(1)
m2
Next one establishes by a routine induction on m that for any function f
{1,2,.. ,m}-* R, m Et"_, f2(t) - (Et _1 f(t))2 = Em2 Es=1 (f (t) -
f(s))2
and thus in Em f2(t) > (Em 1 f (t))2. Then in particular EmI(Ek_1 XA
(Eml Ek=1 XA (t))2 < m
(t))2
and thus by (1) we have Em 1(Ek=, XA` (t))2
k2b2 <
(2)
m
Now, for any t, (Ek=1 XA.
(t))2
= Ek=1 XA; 2(t) + 2 Ek=1 Ej=i+1 XA; (t)XAi (t) Ek=1 XA; (t) + 2 Ek=1 Ej=i+1 XA;nAj (t)
so that
E' Ek 1 XAi (t))2 = Ek=1 Em1 XA; (t) + m
2 Ek=1 Ej=i+1 E'17=1 XA;nAj (t)
m
m
Ek=l a(Ai)+2 Ek=1 Ej=i+1 AM f Aj). Now, using (2) the required inequality is established. Notice that in Lemma 20.15, one cannot do better than d(A)2. Indeed, choose a set A C N by randomly assigning (with probability each n E N to A or its complement. 2)
Then for any t E N, d(A fl (-t + A)) = a = d(A)2.
Lemma 20.16. Let P E E such that p + p = p, let A E p, and let e > 0. Then
{x E A : (-x + A) E p and d (A fl (-x + A)) > d(A)2 - e} E p.
422
20 Density - Connections with Ergodic Theory
Proof Let B = {x E N : d(A fl (-x + A)) > d(A)2 - 6/2). Since by Theorem 4.12, A* = (x E A : (-x + A) E p} E p, it suffices to show that B E P. Suppose instead that N\B E p and pick by Theorem 5.8 a sequence in N such that 1
FS((xn)n_t) c N\B. Let C = {E 1 xt : n E N} and pick by Lemma 20.15 some n < m such that E', xt - E% j xt E B. Since Em 1 xt - E' 1 xt = Emn+1 xt E FS((xt)°O1), this is a contradiction.
The proof of the following theorem is reminiscent of the proof of Lemma 14.24.
Theorem 20.17. Let P E A, let A E p and let 0 < S < J (A). There is a tree T in A such that (1) for each path g of T (a) FS((g(t))°O1) C A and
(b) for each F E 'f(w), d(A fl n{-y + A : y E FS((g(t))tEF)}) > 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.
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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
W=(oo)U({t":u+1
Next let 0 < t < K and let U be a neighborhood of (t, a) (t', s') = (t' + t, a). Assume first that t = to+l and pick F E J"f (S) with a E F such that {t'+to) x (S\ F) S
21.2 Right Topological Compactifications
431
U. Let G = IS E S : ss' E F} and notice that, since S is weakly right cancellative, G is finite. Let
W = ((t, a))U((to) x (S\(F U G))). Then ptt',3'l[W] C U. (If to = 0 and s E S\G, then ss' 0 F.) Now assume that t is a limit ordinal and pick to < t and F E .Pf(S) With a E F such that
{(t' + t, a)) U ((t" : t' + to < t" < t' + t) x S) U (ft' + to} x (S\F)) C U and let
W = {(t, a)) U ({t" : to < t" < t} x S) U ((to) x (S\F)). Then p(i's')[W] C U.
Thus T is a compact right topological semigroup. Define r : S --* T by r (s) _ (0, s). Trivially r is an injective homomorphism. Now let H be a closed subsemigroup of T containing r [S]. We claim that K x (a) C H. Suppose not, and pick the least t < K such that (t, a) f H. Notice that t > 0. Pick
a neighborhood U of (t, a) such that u fl H = 0. Then there exist some t' < t and some F E 9'f (S) such that {t'} x (S\F) C U. Then (t', a) E H so
Since S is weakly right cancellative, Sa is infinite, so ({t'} x Sa) fl u # 0, a contradiction.
The following corollary says in the strongest terms that without the requirement that r[S] C A(T) (or some other requirement), there could not be a maximal right topological compactification of S. Notice that the corollary does not even demand any relationship between the topology on X and the semigroup operation, nor are any requirements, either algebraic or topological, placed on the function V. (Although, presumably, the kind of result one would be looking for would have X as a compact right topological semigroup and would have (p as a homomorphism with V[S] dense in X.)
Corollary 21.8. Let S be an infinite discrete weakly right cancellative semigroup, let X be a semigroup with a compact topology, and let (p : S --o, X. There exist a compact right topological semigroup T and an injective homomorphism r : S -a T such that there is no continuous homomorphism rl : X -+ T with r) o rp = r. Proof. Pick a cardinal K > 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
Since (iSd, ) is a right topological semigroup, (,BSd, o) is a left topological semigroup,
and x b=xob for b E S, A, AMEx and Be Y. Choose al E Al and b1 E B1. Inductively, let r E N and assume that we have
chosen {a1,a2,...,ar}and{b1,b2,...,br} so that forallm,n E {1,2,...,r} (i) am E Am and bn E Bn, (ii) if m > n, then I f (an bn) - 2I <
(iii) if n > m, then If (am bn) - kI < m (iv) if m > n, then If (am bn) - 7(x bn) I <
(v) ifn>m,then
m, and
<,1-,
21.3 Periodic Compactifications as Quotients
437
Now for each n E { 1, 2, ... , r}, bn E Bn so {a E S : I f (a bn) -1I < 11 Ex. Also,
foreachn E {1,2,...,r},{a E S: I thatforeachn E {1,2..... r}, If (a,-+ I -bn)-f I< and If Similarly, choose
+t) Exsopickar+l such (x'bn)I < +1
n
rn}tl{b E S :
br+1 E nm=1({b E S :
_L })
Now, given n E N, one has lim f (am bn) = 7(x b,) and given m E N one has m-*oo
lim lim f(am bn) = k and lim lim f(am lim f (am b,) = f (am y). Alsom-goo n->"O n-.oo n-oo m-+oo bn) = f. So f = k. Theorem 21.19. Let S be a commutative semitopological semigroup. Then WAJ"(S) is also commutative. Furthermore, (112,'WAJ"(S)) is the maximal commutative semigroup compactification of S in the following sense: If (v, T) is any commutative semigroup compactification of S, then there is a continuous homomorphism t : 'WAR(S) -±
Tforwhich v=p.o112 Proof. By Theorems 21.5 and 21.14, we may assume that 'WAY(S) = j8 Sd/ _- and 772 = 7r1s. Now given (f, C) such that 02(f, C), one has by Theorem 13.37 that for all x, y E l4Sd, T (X y) = 7(y x) and thus 7r (x) x(y) = 7r (y) 7r (x).
Now, if (v, T) is a commutative semigroup compactification of S, then T is a semitopological semigroup so there exists a continuous homomorphism A : 'W A.l" (S) - . T
forwhich v=tto112. Theorem 21.20. Let S be an infinite commutative semitopological semigroup. Then K(1NA.P(S)) can be identified with 3A,P(S). Proof. Let K = K (WAJP(S)). By Theorem 21.19 and Corollary 2.40, K is a compact
topological group. Let e denote the unique minimal idempotent of WA.P(S). We define a continuous homomorphism r : WAIF(S) - K by r(x) = xe and define a continuous homomorphism It : S -+ K by p = r o r12. Let T be a compact topological group and let v : S -+ T be a continuous homomorphism. By the defining property of 'WAIF(S), there is a continuous homomorphism 0 : WAY(S) --* T for which 0 o ?72 = v. We note that 0(e) is the identity of T and so c1K o a = v. This shows that we can identify -3A,"(S) with K (by Theorem 21.5). We now discuss some simple properties of 'W AJ" (N). We use Theorems 21.5 and 21.14 to identify WAR(N) with PN/- and 172 with rr,N. If S = N U {oo}, the one point compactification of N, and oo + x = x + oo = oo for all x E S, then S is a semitopological semigroup. Thus, by Exercise 21.1.1, 112 : N -+ WAY(N) is an embedding. Consequently, we shall regard N as being a subspace of
WA(N). Lemma 21.21. Let x be in the interior in fiN of.8N\(N* + N*). Then I r(x)I = 1.
438
21 Other Semigroup Compactifications
Proof. Let Y E #N\{x}. We can choose a subset A of N for which x E A, Y
An(N*+N*)=0.
A, and
.7(q
Let f = X A : N --> (0, 11 so that f = X A . We claim that f (p + q) = + p) for all p, q E N. Indeed, if p E N or q E N, then p + q = q + p, while if p, q E N*, then
f (p + q) = 0 = f (q + p). Thus by Corollary 21.16, there is a continuous function g : MAP - (0, 1) such that f = g o 7rigj. Then f and g o 7r are continuous functions from fiN to (0, 1) agreeing on N and so f = g o jr. Since T (x) = 1 and .7(y) = 0, it(x) i4 7r (y). That is, y ¢ 7r (x). Theorem 21.22. There is a dense open subset U of N* with the property that I;r(x) I = 1 for every x E U.
x -+ oo and Proof By Exercise 4.1.7, if (x,,)n°_1 is a sequence in N for which if A = {xa : n E N}, then (ctfly A) n (N* + N*) = 0. Now any infinite sequence in N contains a subsequence with this property, and so there is a dense open subset U of N* for which U n (N* + N*) = 0. Theorem 21.23. If x is a P-point in N*, then tr(x) is P -point in 'WA?(N)\N.
Proof. Let U denote the interior of fN\(N* + N*). We shall show that x E U. Our claim will then follow from Lemma 21.21 and the observation that n[U] is open in 'WA,?(N) because n-1 [ir[U]] = U. For each k E N, we can choose a set Ak E x with the property that la - a'I > k whenever a and a' are distinct integers in Ak. (This can be seen from the fact that there exists i E 10, 1, 2, ... , k - 1) for which Nk + i E X.) Since x is a P-point in N*, there is a set A E x for which A\Ak is finite for every k. It follows from Exercise 4.1.7 that
An(N*+N*)=0. We now turn our attention to representing ALP(S) as a quotient of fSd.
Definition 21.24. Let S be a semitopological semigroup. 03 (f, C) is the statement "C is a compact semitopological semigroup, f is a continuous function from S to C, and there is a continuous function f * : fSd x fSd - C such that for all s, t E S, PS t) = f *(s, t)". 03(g, C) is the statement "there exists (f, C) for which 03(f, C)
holds and g = f : fSd -> 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.
440
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 <
e whenever (x, y), (u, v) E U X V. We can choose a finite family F of sets of this form which covers 0 Sd x 0 Sd. Let
'W={U: U X V E.Fforsome V}U{V: U x V EFforsome U}. Define an equivalence relation -- on S by s - t if and only if for all W E W, either {s, t} C W or is, t} fl w = 0. Then ~ is an equivalence relation with finitely many equivalence classes. Now assume that s -- s' and t t'. Pick U and V such that U x V E F and (s, t) E U x V. Then since s s' and t -- t', (s', t') E U x V and so I f * (s, t) - f * (s', t') I < E. <E. That
(3) implies (2). Define f *(x, y) = 7(x y) for all x, y E ,3Sd. Then trivially f * (s, t) = f (s t) for all s, t E S. To see that f* is continuous, let x, y E 13Sd and let e > 0 be given. Pick an equivalence relation ti on S with finitely many equivalence classes such that If (s
t) -
f (s' t') I < 3 whenever s -- s' and t N t'. Let ,R be the set of --equivalence classes and pick A, B E .R such that x E A and y E B. We claim that for all (u, v) E A x B, one has I f*(x, y) - f*(u, v) I < E. Since f(x y) = x- limy- lim f(s t), we can sES
(ES
choose (s, t) E A x B such that I f (x y) - f (s t) I < 3. Similarly, we can choose (s', t') E A x B such that I f (u v) - f (s' t') I < 3. Since s s' and t t', one has
3 and solf(x.y)-f(u.v)I <E.
21.4 Spaces of Filters
441
Exercise 21.3.1. Show that WA,P(N)\N contains a dense open set V disjoint from
('W,A2(N)\N) + ('WAR(N)\N). Exercise 21.3.2. Show that'W,A.P(N)\N contains weak P-points.
Exercise 21.3.3. Show that Wd P(N) is not an F-space. (Hint: An F-space cannot contain an infinite compact right topological group. See [182, Corollary 3.4.2].)
Exercise 21.3.4. We have seen that K (W A.l (N)) can be identified with -SAP(N) (by Theorem 21.20). Show that the same statement holds for K(,4.P(N)). Exercise 21.3.5. Let S be a semitopological semigroup and suppose that f E C(S) is almost periodic. Show that every sequence in S contains a subsequence (s,,,) for which the sequence of functions t r-* f (s,,,t) converges uniformly on S. (Hint: Use Theorem 21.28.)
21.4 Semigroup Compactifications as Spaces of Filters We have seen that if S is a discrete semigroup, then 8S is the aCA(C-compactification of S. We have also found it useful throughout this book to have a specific representation for the points of 14S, namely the ultrafilters on S.
As we noted in Section 21.3, the WAY-, ,4.P-, and i 4 P-compactifications are all quotients of 4S. The points of these compactifications correspond to closed subsets of fl S. Since the closed subsets of ,8S correspond to filters on S by Theorem 3.20, each of these compactifications - indeed any semigroup compactification - can be viewed as a set of filters. In this section we characterize those sets of filters that are, in a natural way, semigroup compactifications of S.
Remark 21.29. Let S be a discrete space, let X be a compact space, and let g be a continuous function from $S onto X. Then given any x E X and any neighborhood U
ofx, g-1[U] fl S E fg-1[{x}]. The first step in the characterization is to define an appropriate topology on a set of filters.
Definition 21.30. Let S be a discrete space, let 91 be a set of filters on S, and let A C S.
(a)At={AE9t:AEd}.
(b) The quotient topology on 91 is the topology with basis {A5 : A C S}.
Observe that if A, B C S, then A= fl B= = (A fl B)l so that {Att : A C S} does form a basis for a topology on 91. That the term "quotient topology" is appropriate is part of the content of the following theorem. Recall that, given a set 91 of sets, a choice function for 91 is a function f : 91 - U N such that for all A E 91, f (A) E A.
21 Other Semigroup Compactifications
442
Theorem 21.31. Let S be a discrete space, let X be a compact space, and let g be a in g-1 [{x}] : x E x). Then 9R is a set
continuous function from /3S onto X. Let of filters on S satisfying
(a) given any choice function f for 99 there exists I E 'f(91) such that S = UAEa f (A) and (b) given distinct A and B in
eE
there exist A E A and B E 2 such that whenever
either S\A E C or S\B E e.
Further, with the quotient topology on n g-1 [{x}] is a homeomorphism.
the function h : X -* 9R defined by h(x) _
Proof. Given X E X, g-1[(x)] is a set of ultrafilters whose intersection is therefore a filter.
Let f be a choice function for 91 and suppose that for each 3 E J"f (91),
S\ UA f (A) 0 0. Then (S\ f (A) : A a that [S\ f (A) A E
has the finite intersection property so pick p E fS such
c p. Let x = g(p) and let A= n g-1 [{x}]. Then
f (A) E A c p and S\ f (A) E p, a contradiction. Now let A and 2 be distinct members of 91 and pick x and y in X such that A= n g-1 [{x}] and B= n g-1 [{y}]. Pick neighborhoods U of x and V of y such
that ctUf1ceV = 0. LetA=g-1[U]f1SandletB=g-1[V]f1S. Then by Remark 21.29, A E A and B E S. Now let e E 91 and pick z E X such that C= n g-1 [{z}]. Then without loss of generality z 0 ct U so if C = g-1 [X\ cE U] fl S we have by Remark 21.29 that C E C. Then C fl A = 0 so, since C is a filter, S\A E C. Finally define h : X -> 91 by h(x) = ng-1[{x}]. Then trivially h is one-to-one and onto 91. Since X is compact, in order to show that h is a homeomorphism it suffices to show that 9% is Hausdorff and h is continuous. To see that 9% is Hausdorff, let A and 9 be distinct members of 9% and pick A E A and B E 2 as guaranteed by (b). Then At and Bt are disjoint neighborhoods of A and
2 respectively. To see that h is continuous, let x E X and let A C S such that Al is a basic open neighborhood of h(x). Let U = X\g[S\A]. Now g[S\A] is the continuous image of a compact set so is compact. Thus U is an open subset of X. We claim that x E U and h[U] C Aft. For the first assertion suppose instead that we have some p E S\A such
that g(p) = x. Then h(x) C p so A E p, a contradiction. Now let y E U. Then by Remark 21.29, we have g-1 [U] fl S E h(y). Since g-1 [U] fl S C Awe have A E h(y) so h(y) E Al as required. 13 The following theorem tells us that the description of quotients provided by Theorem 21.31 is enough to characterize them. Theorem 21.32. Let S be a discrete space and let 91 be a set of filters on S satisfying
21.4 Spaces of Filters
443
(a) given any choice function f for 91 there exists a E .Pf(91) such that S = UAE3 f (A) and (b) given distinct A and B in 91, there exist A E A and B E 2 such that whenever C'' E'91, either S\A E C or S\B E C. Then, with the quotient topology, 91 is a compact Hausdorf space. Further, for each
p E l S there is a unique A E 91 such that A c p and the function g
9 S --> 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),
444
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 <
e whenever (s, s') E U. By condition (ii) of Theorem 21.43, there exists V E U such that (S t, s't) E U for every (s, s') E V and every t E X. If x' E yuX, then I f (sx') - f (s'x') I = lim I f (st) - f (s't) I < e whenever (s, s') E V. Thus, ifs' E V(s) t ax'
and x' E ?.'[{y E yuX : I f (sx) - f (y) I < e}J, we have I f (sx) - f (sx') I < e and I f (sx') - f (s'x') I < e. So If (sx) - f (s'x') I < 2e. Theorem 21.45. Suppose that (X, 2() satisfies the conditions of Theorem 21.43. Suppose that Y is a compact right topological semigroup and that h : X -+ Y is a uniformly continuous homomorphism with h [X] C A (Y). Then h can be extended to a continuous homomorphism h : yuX -a Y.
Proof. By Exercise 21.5.4, h can be extended to a continuous function h : yuX -. Y. For every u, v E yuX, we have
h(uv) = lim lim h(st) = lim lim h(s)h(t) = lim h(s)h(v) = h(u)h(v), S->U 1_V
S_ U 1-V
S-!U
where s and t denote elements of X. So h is a homomorphism. We now show that, if X is a topological group, yu X is maximal among the semigroup compactifications of X which have a joint continuity property.
Theorem 21.46. Let X be a topological group and let (0, Y) be a semigroup compactification of X. Suppose that the mapping (x, y) H 9(x)y from X x Y to Y is continuous. Then there is a continuous homomorphism 9 : yuX -> Y such that 0 = Box.
450
21 Other Semigroup Compactifications
Proof We shall show that 8 is uniformly continuous. Let e denote the identity of X and let ,NQ be the set of neighborhoods of e. Suppose that 8 is not uniformly continuous. Then there exists an open neighborhood U of the diagonal in Y x Y such that, for every V E .14, there are points sv, tV E X satisfying svtV t E V and (8(sv), 8(tv))o U. The net ((8(s,), 8(tv)))vE,N has a limit point
(y, z) E (Y x Y)\U. Now sv = evty, where ev E V, and so 8(sv) = 8(ev)8(tv). Since ev -> e, 8(ev) -> 8(e), and 8(e) is easily seen to be an identity for Y. So our continuity assumption implies that y = z, a contradiction. Thus 8 is uniformly continuous and therefore can be extended to a continuous homomorphism 8 : Y (by Theorem 21.45). Theorem 21.45 shows that y. X can be identified with /3X if X is a discrete semigroup
with the trivial uniformity. However, there are many familiar examples in which y,,X and 3X are different. Suppose, for example, that X = (R, +). We shall see in Theorem 21.47 that PR cannot be made into a semigroup compactification of (R, +), It is natural, given a nondiscrete topological semigroup such as (R, +), to attempt to proceed as we did with a discrete semigroup in Theorem 4.1. And, indeed, one can do the first part of the extension the same way. That is, given any s E S, one can define es
: S -+ S c fl S by 4(t) = st. Then by Theorem 21.40, there is a continuous
function Ls : f S -> /3S such that Ls I S = is so one can define for s E S and q E P S, sq = Ls(q). As before one still has the function rq : S -> f S defined by rq(s) = sq. However, in order to invoke Theorem 21.40 to extend rq, one would need rq to be continuous. We now see that this condition does not hold in many familiar semigroups, including (R, +). Note that in the following theorem we do not assume that the operation on fS is associative.
Theorem 21.47. Let S be a semigroup which is also a metric space with a metric S
for which 8(st, st') = 8(t, t') and S(ts, t's) = 8(t, t') for every s, t, t' E S. Suppose that S contains a sequence (yn)n°_1 such that S(y,n, yn) > 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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[245] T. West, Weakly monothetic semigroups of operators in Banach spaces, Proc. Roy. Irish Acad. Sect. A 67 (1968), 27-37.
[246] E. Zelenuk, Finite groups in ON are trivial (Russian), Ukranian National Academy of Sciences Institute of Mathematics, Preprint 96.3 (1996). [2471 E. Zelenuk, Topological groups with finite semigroups of ultrafilters (Russian), Matematychni Studii 6 (1996), 41-52.
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
Wdb'(S) 428 smallest ideal, 437
P^RKg164
'WAR(N) 437
p-lim 63, 64, 65 existence, 63
x+ 268
-p 85
x- 268
Pf(A) 2
xn 11
(Pf(N), U) 95
Xd 268 Xci 207 XX 4, 8, 32, 42
r-coloring 101 R(S) 7
a topological semigroup, 32 right topological semigroup, 32
rp 258 rx 272
+2
Ya, 207
Rd 268 (R, A) 3
Za 207 subsemigroup of (,6N, ), 209
SPm ((x,) °O 1) 347 S*
(Z, +)
as ideal of 14S, 84 as left ideal of P S, 83 as right ideal of P S, 84 as right ideal of CBS, 83 as subsemigroup of P S, 82 minimal left ideals, 84 minimal right ideals, 84 smallest ideal, 84
center, 107 left ideals, 85
copies of Z* in Z*, 218
,9D53,56,57
-
size of, 66, 67 topological properties, 53 IS
as commutative semigroup, 81 as noncommutative semigroup, 81 as semigroup, 73 as universal semigroup compactification, 75 binary operation, 72
s-1A 76 Sd 432 supp 107, 111
-s + A 76 -8242
JAR compactification, 428
,SAP(S) 428 f 242 T 150, 206, 403
T 85 Ua 207 U,K 52
cardinality, 68 size of, 67 W(tP) 381
J9Sd 432
,BT
copies of ST in S*2217 identified with T, 57 additive, 107 topological copy, 67 additive, 107
List of Symbols
474
r 96, 240 and finite sums, 96
nn
1 c15rq(nN)
a copy of III, 149
left ideal of (fN, ), 96
'1InEN ceorq(3nN)
not semigroup under +, 112 not semigroup, 242
nfEN cepz(3'Z)
I
a copy of ]H[, 152 a copy of 1HI, 152
Yu 447
8 338 A 134, 413 finite sums and upper density, 422 A* 413 ideal, 414
A*(N, +) left ideal of (fN, ), 415 A *(S, ) as left ideal, 414 A*(S, as right ideal, 414 (n;, yjS) 427 K-uniform finite intersection property 68 K-uniform ultrafilter 52 A (S) 33, 38, 78
nnEZ ctpz(2'Z) a copy of iHi, 152 ® 5, 111, 152
X5 o 272 < 15
in K(fN), 184
Xx 7
nsEFX, 115
µ(A) 362 k*(A) 362
® 223 232
Px 7
IIPII 51
a-compact sets 61
an(p) 346 4 284 4b 259 %Pi 425
w2
II`uII 51
*-tree 292
0+269,361 0- 269
Index
action measure preserving, 407 monotone, 408 almost disjoint family 238 almost left invariant 121 almost periodic function 436, 439 alphabet 281 Anthony, P. 275 anti-homomorphism 4 anti-isomorphic 4 anti-isomorphism 4 Arens, R. 88 arithmetic progression 280 associative 3
associativity of (fS, ) 73 Auslander, J. 47, 411 Axiom of Choice 69
Baire set 361, 387 Baker, J. viii, 88, 135, 204, 221, 453 Balcar, B. 275, 411 Banach density 412 base filter, 48, 52 Baumgartner, J. 102 Bergelson, V. viii, 103, 221, 295, 319, 338, 367, 368, 393, 411, 424 Berglund, J. viii, 47, 135, 453 bicyclic 22 binary expansion 107 Blass, A. viii, 103, 185, 234, 257, 393 Boolean algebra 52, 71 homomorphism, 52 Budak, T. 135 Butcher, R. 88 c.c.c. 237
cancelability
and K(S), 30 and K(fS), 169 in PZ, 176 cancelable elements in CBS 175
cancelable elements in PZ 175 cancelable elements of S cancelable in PS, 158 cancelable elements 115 cancelable right, 161 cancellative 7 PS not, 158 weakly left, 122, 323 weakly right, 122 cardinality of U,, 67 cardinality of OD 67 cardinality of closed sets in fiD 67 cardinal 2 Carlson's Theorem 381, 386 Carlson, T. 386, 393
tech, E. 69 center 7, 38, 80
(Z, +), 107 of p(,6S)p, 129, 130 of p +,BN + p, 130 of U,,(S) empty, 124 of CBS equal to that of S, 125
Central Sets Theorem 279, 294 commutative, 284 noncommutative, 287 satisfied by noncentral set, 289 central* sets 297 examples, 337 centralizer nowhere dense in U (S), 124 central 86, 279, 284, 320
476
Index
combinatorial characterization, 293 dynamically, 404 equivalent to dynamically central, 406 examples, 320 not changed by adding 0, 301 often translates to central, 331 chain of ideals increasing, 133 characteristic function 43 Chou, C. 135 Christensen, J. 47 circle group 150, 206, 403 Civin, P. 88 Clay, J. 338 clopen 55 closed 3 closure and p-lim, 65 of subsemigroup, 40, 167 collectionwise piecewise syndetic 290, 291, 293, 294, 328 coloring 101 columns condition 302, 303, 305, 325 over Q, 303 over 7G, 303
over field, 316 combinatorially rich ultrafilter 339 commutative semigroup S contained in center of PS, 80 commutative 7 compactification 55
AY, 428 £,MC, 428 SAY, 428 'WAY, 428 and algebra of functions, 446 semigroup, 75, 77, 431
Stone-tech, existence, 447 Stone-tech, 56, 57, 69 topological, 447 compactness 101, 281, 283, 300, 318
compact
if p-lim exists, 65 compatible 237 completely regular 446 concatenation 4 continuity 66 and p-lim, 65 convergent ultrafilter 197 correspondence principle 417 Davenport, D. viii, 88, 135, 424 Day, M. 88 dense 236 density 134 Banach, 413 Deuber, W. viii, 318, 319, 338, 367 direct product 5 direct sum 5, 152 of countable groups, 152 distinct finite products 115 semigroup isomorphic to H, 115 of K sequence, 117 distinct finite sums 115 distributive laws in N*, 267 in OZ, 258 in 476, 264, 265 in El, 265 doubly thin 266 Douwen, E. van 135, 275 dynamical IP* set 408 contains FP((xn)n° 1) x FP((y,,)n° 1), 409 dynamical system 397 minimal, 398 dynamically central 404 equivalent to central, 406
El-Mabhouh, viii, A. 185, 275 Ellentuck topology 387 Ellentuck, E. 392, 393 Ellis' Theorem 46 Ellis, R. 32, 46, 47, 88, 411 embedding 55
477
Index
enveloping semigroup 42, 400 as image of flS, 400 19Sas,401 of actions on the circle group, 403 of closed left ideal of fS, 402 Erd6s, P. 346 extracted word 382 extraction variable, 382 extremally disconnected 61,62,71, 155, 194, 198 D* not, 61 PD, 61 F-space 71, 441 filter base 48, 52 filter 48, 49, 54, 236 in partially ordered set, 236 finite intersection property 49, 50, 66 K-uniform, 68 infinite, 52, 262 finite products theorem 92, 93, 286 finite version, 101 finite products 90, 115 and idempotents, 93 finite semigroups images of H, 108 finite sums and products in same cell, finite version, 101 in same cell, 97 Finite Sums Theorem 93, 95, 96, 369,
386,419 Galvin-Glazer proof, 92 finite sums 90, 115 finite unions 95, 97 finite-to-one 83 first category 362 first entries condition 297, 298, 299, 300, 302, 303, 315 over F, 315 fixed point 59 non-existence, 59 FP-tree 292 Franek, F. 411
free group 10, 151 mapped to finite group, 10 on 22 generators, 153, 155 on 2` generators in Cp, 181 on 2` generators in PN, 155 on 2` generators in 8S, 155 subgroup of compact group, 40 universal property, 10 free semigroup 4, 7, 130, 281, 282 defined by a right cancelable element, 167 in N*, 153 on 21 generators, 153 with identity, 4 Frollc, Z. 70, 235 functions equal at p not equal at any x, 160 Furstenberg, H. viii, 102, 294, 338, 393, 411, 412, 424 Galvin, F. 70, 102 Garcia-Ferreira, S. viii, 234 Glazer, S. 102 Graham, R. 102, 338, 367 Grainger, A. viii Grothendieck, A. 453 group 6, 17, 18 characterizations, 17 closure, 39 equal to eSe, 20, 24 in smallest ideal, 34, 35 locally compact, 32 maximal groups in minimal left ideal, 19 maximal, closed, 39, 42 maximal, 9, 35, 39 of quotients, 129 Gunderson, D. viii Hales, A. 281
Hales-Jewett Theorem 281, 282, 283, 295, 381 consequence of Central Sets Theorem, 288
Index
478
Hausdorff metric 43 Hausdorff all hypothesized spaces, vii, 31 Hilbert's Theorem 90, 296 Hilbert, D. 90 Hofmann, K. vii homeomorphism classes of maximal groups, 156 homeomorphism types in flD at least 21, 119 homeomorphism type 119 homomorphism 4, 75, 205 from T* to S*, 216, 217 from fl S to G*, 210 from SST to S*, 211 from fl T to fi S, 217 from ON to N*, 213
image of K(S), 30 of topological center, 37 on H, 108 on intersection, 78 to compact right topological semigroup, 78 Hong-ting, S. 411 Hong-wei, Y. 411
I-restricted span 308 ideal 12 equal to SxS, 14 smallest, 21, 29 idempotents 8, 321 2° right maximal, 187 21 strongly right maximal, 193 > minimal idempotents, 24
in K(fN), 184 wI chains, 201 and finite products, 93 at least 2', 118 chains in Cp, 181 closure, 96 commuting, 114
conditions equivalent to being minimal, 35 defining maximal topologies, 196 existence, 33 immediately below a given idempotent, 199
in K(flN), 184
in A,iff,94 in finite semigroup, 12 in III[, 109
infinite chains in H, 110 infinite chains, 118 left maximal, 186 maximal, 186 minimal for partial orders, 15 minimal, at least 22% 121 minimal, in intersection of left and right ideals, 34 minimal, in right ideal, 34 multiplicative idempotents in r, 97 nonminimal, 118 not closed, 169, 242 number in minimal left and right ideals, 121 right maximal, existence, 36 right maximal, qp not in smallest ideal, 191 right maximal, finite fibres, 189 right maximal, in GS set, 189 right maximal, 186, 188 strongly right maximal, existence, 192
strongly right maximal, 191 sum not idempotent, 112 topologies defined by strongly right maximal idempotents, 196 topologies defined by, 194 identity 5 adjoined, 8 in PS, 203 image partition regular matrices sums of terms, 325, 340 image partition regular 297, 298, 299, 300, 308, 309, 315.318, 324
479
Index
infinite, 354
over (N, +) and (N, ), 314 strongly, 300 vector spaces over a field, 316 weakly, 300 incompatible 237 infinite finite intersection property 52, 262 invariant 397 inverse 5 irrational with respect to S, 207 isomorphic 4, 35, 39 isomorphism 4 of principal left ideals, 219 of principal right ideals, 220 Jewett, R. 281 Junghenn, H. 453 Kal'as"ek, P. 275
Katetov, M. 70 Katznelson, Y. viii, 295, 393 Keisler, H. 235 kernel partition regular matrices sums of terms, 325, 340 kernel partition regular 301, 302, 303, 305, 318, 324
over (N, +) and (N, ), 304 over a field, 316 over a finite field, 317 Kra, B. 221, 338 Kunen, K. 70, 256
Lau, A. 204 Lawson, J. viii, 47, 411 Leader, I. viii, 275, 319, 367 Lebesgue measurable 361 Lebesgue density theorem, 368 Lefmann, H. viii, 367 left cancelability in fiN 176 does not imply right cancelability, 176
left cancelable elements in fiN left cancelable in fiZ, 173 left cancelable elements of fi S contain dense open subset of S*,
174
left cancelable 7, 174 left cancellative 7, 82, 83, 224 weakly, 323 left ideal and right ideal intersection, 13 left ideal 12, 19, 133
22',
120 S* a finite union of disjoint closed left ideals, 121 UK (S) a left ideal of fl S, 125 X., and Ya as left ideals, 208 closed left ideals defined by density, 134 closure, 38, 77 contains minimal left ideal, 20 decomposition of U,K (S) into, 125 equal to Sx, 14 infinite chains, 129 intersecting, 112 intersection, 13 isomorphic principal left ideals, 219 maximal principal left ideals, 127 maximal semiprincipal left ideals in N*, 128 minimal, all homeomorphic, 35 minimal, contains idempotent, 34 minimal, homeomorphic to those in fiww, 152
minimal, 19, 34, 35 principal, 218 semiprincipal, 112, 113 left identities 5 22', 203
for fS, not in S*, 203 left invariant topology 136, 141 Hausdorff, 142 zero dimensional, 143 left inverse 5 left simple 13, 20, 24, 35
Index
480
left topological groups partitions, 197, 199 left topological semigroup 31 left zero semigroup 26, 81 left zero 4, 5 length 4 limit 63, 66 Lin, C. 338 Lin, S. 338 Lisan, A. viii, 135, 157, 295, 411 local homomorphism 146 local isomorphism 146 local left group 146 locally isomorphic, 147 Maleki, A. viii, 135, 221, 275, 295 Martin's Axiom 237 MA(c) false, 239 and P-points, 239 maximal almost disjoint families have cardinality c, 239 non-empty GK subsets of N* have non-empty interior, 239 proper FK subsets of N* not dense in N*, 239 Matet, P. 257 matrices image partition regular, 309 maximal groups 9 in K(S) isomorphic, 35 in PG, 155
in8N, 155 in smallest ideal isomorphic, 29 isomorphic to 7G, 251
not closed, 157 number of homeomorphism classes, 156 maximal topologies 196 number, 197 meager 362 measurable 361 measure preserving action 407 measure preserving system 407 measure preserving transformation 407
measure space 407 measure 49 Milliken, K. 355, 393 Milliken-Taylor system 355 partition regular, 359 Milliken-Taylor Theorem 369, 373, 374, 375, 392
Milnes, P. viii, 88, 135, 221, 453 minimal (w.r.t. partial order) 15 minimal dynamical system 398 universal, 398 minimal idempotent 16, 20, 22, 24 and minimal left ideals, 16 at least 2`, 109 closure has elements outside S* S*, 169
existence, 23 minimal left and right ideals in Cp 180 minimal left and right ideal defined by idempotents, 24 intersection, 25 minimal left ideal 12, 14, 19, 20, 24 a group if commutative, 23 all homeomorphic, 35 and minimal idempotents, 16 as minimal closed invariant set, 398 at least 2`, 109, 121 infinite, 120 isomorphic, 25 same in S* and CBS, 84
without idempotents, 23 minimal right ideal 12, 20, 24 at least 2`, 109, 120, 121 contains idempotent, 23 number, 122 same in S* and PS, 84 monochrome 91, 101 monotone action 408 multilinear combinations 346
neighborhood 48 Neuman, B. 89 nonprincipal ultrafilter 50
481
Index
norm 51 Numakura, K. 47 oid 111 operator uniform, 64 order 11 ordinal 2 Owings, J. 367
P-point 60, 70, 175, 239 cancelable in $7L, 175 cancelable, 177 defined by union ultrafilter, 247 existence, 60 independence, 248 pairwise sums and products 344 Papazyan, T. 135, 204 Parsons, D. 135 partial order 236 partially ordered set 236 partition regularity and ultrafilters 51 partition regular 51, 52, 296, 322
image, 297, 298, 299, 300, 308, 315, 318 kernel, 301, 302, 303, 305, 318 strongly image, 300 weakly image, 300 path 416 periodic 408 piecewise syndetic 85, 86,87,279,282, 290, 301, 322 and central sets theorem, 289 collectionwise, 290, 291, 293, 294 in N, 87 left, 274 near 0, 271 partition has left and right, 275 partition regular, 280 right piecewise syndetic versus left piecewise syndetic, 274 right, 274
if and only if the closure meets K(PS), 86
pigeon hole principle 91 Plewik, S. 368 pointwise convergence 32 Polya, G. 424 positive upper density 340 principal closed ideal 131 increasing chain, 133 principal ideal 14 principal ultrafilter 50, 55 product in fJS as ultrafilter, 76, 77 product of a finite number of ultrafilters 79
product subsystem 94, 323, 375, 376 weak, 377 product topology 32
products nowhere dense in S*, 118 product of right topological semigroups, 40 proper ideal 12 Protasov, I. vii, 89, 157, 185, 192, 204, 257 - proximal 405 and minimal left ideals of ,6S, 406 Promel, H. 368 Pym, J. viii, 30, 135, 157, 185, 204, 275
quotient spaces 432 quotient topology 441 quotients 8 group of, 129
Rado's Theorem 91, 304, 305 Rado, R. 91, 318 Ramsey Theory 90, 296 Ramsey ultrafilter 235 Ramsey's Theorem 91, 369, 370 finite version, 102 Ramsey, F. 91, 370 rapidly increasing sequences 79 recurrent 407 uniformly, 404 reduced word 382
482
Index
reduction variable, 381 Rees, D. 30 right cancelability in PS and the Rudin-Keisler order, 165 characterizations, 165, 166 on dense open subset of flS, 166 relations with discreteness, 162 right cancelability in fiN characterization, 170 does not imply left cancelability, 173
right cancelability in ,8Z characterization, 170 stronger than in fN, 170 right cancelable elements of PS contain a dense open subset of U,K (S), 162
number, 162 right cancelable elements intrinsic characterization, 161 in closure of minimal idempotents, 169
right cancelable 7, 126, 161 right cancellation characterizations, 163, 234 right cancellative 7, 82, 83 right continuous 31 right ideals in ON closed, 112 right ideal 12, 14 2' disjoint in N*, 210 closed disjoint, 112 closure as ideal, 88 closure, 38, 39, 77
in PS Ua and Da as right ideals, 208
isomorphic principal right ideals, 220 minimal, 34, 35 minimal closed, 39 of right ideal, 19 principal, 218 right identities 5
22'
, 202
for S*, at least 21, 204 for S*, 203 unique, 203 right inverse 5 right maximal strongly, 248 right simple 13, 20, 24, 35 right topological semigroup 31, 32 compact, 33 right zero semigroup 26, 82 finite, 187 right zero 4, 5 Rothschild, B. 102 Rudin, M. vii, 135, 235 Rudin, W. 70 Rudin-Frolilc order 232
p c q implies p
P
Schur's Theorem 91, 296 Schur, I. 91 selective ultrafilter 235 semigroup compactification 75,77,431 and spaces of filters, 444
483
Index
maximal, 75 universal, existence, 427 universal, uniqueness, 428 semigroup 3 closure not semigroup, 167 free, 281, 282 simple, 26 semiprincipal left ideal 112 at least 21 maximal, 128 characterization of members, 114 infinite chains, 129 intersecting, 113 reverse well ordered, 129 semitopological group not topological semigroup, 199 semitopological semigroup 31 compact, 41 dense center, 41 separable 62 Shelah, S. vii, 70, 135, 257 Simon, P. viii, 70, 135 simple 13, 22, 26 Simpson, S. 393 Skolem, T. 338 smallest ideal of ON closure contains 21 nonminimal idempotents, 184
closure contains chains of idempotents, 184 smallest ideal 21, 182 equal to LR, 22 in product, 40 of (ON, ) and (ON, +), 263 of subsemigroup, 29 structure theorem, 28 Smith, G. 367 span I-restricted, 308 special strongly summable ultrafilter 242
p = q +r implies that p = q = r, 254
p = q + r implies that q, r E Z +
p,254 existence, 244
Stone space 71 Stone, M. 69
Stone-tech compactification 56,57,69 existence, 447 not a semigroup compactification, 450 of completely regular space, 447 strongly discrete 62, 162 subset, 62
strongly image partition regular 300 strongly right maximal idempotents 191, 248 2`, 193
defined by strongly summable ultrafilters, 248 defining a topology, 196 existence, 192 maximal topologies, 196 strongly summable ultrafilter 240 and union ultrafilter, 247 as idempotent, 241 as strongly right maximal idempotent, 248 independence, 248
maximal group isomorphic to 7G, 251
not in c8 K(ON, +), 241
special, p = q + r implies that
p=q= r, 254 special, p = q + r implies that
q,rEZ+p,254
special, existence, 244 special, 242 subgroup 8 subsemigroup 8 as intersection, 78 closure in fS, 77 closure not a semigroup, 41 closure, 41, 77 subsystem product, 376 sum, 326 sum subsystem 94, 326, 373
484
summable strongly, 240 weakly, 240 sums equal to products 258, 263 nowhere dense sets, 262 Suschkewitsch, A. 30 symmetric difference 43 syndetic subgroup 87 syndetic 85, 86 near 0, 270 piecewise, 279 Szemeredi's Theorem vii, 419 Szemeredi, E. 412 Tang, D. viii Taylor, A. 256, 355, 393 tensor product 223 as double limit, 223 Terry, E. viii, 338 topological center 33, 40, 80 equal to algebraic center, 80 of N* empty, 170 topological compactification 447 topological group 31 topological semigroup 31 topologically and algebraically isomorphic 35 topologies defined by idempotents 194 topology left invariant, 136 transformation measure preserving, 407 tree 292, 415 *, 292 FP, 292, 415 in A, 292
ultrafilter 48, 49, 50 x-uniform, 52 as measure, 49, 52 combinatorially rich, 339 nonprincipal, 50 principal, 50, 55 uniform, 52, 67
Index
Umoh, H. 185, 424 uniform compactification of I[8 and flZ, 451 ideals, 451 idempotents, 452 uniform compactification universal property, 449 as semigroup, 448 contains 6N, 447 existence, 447 joint continuity, 449 maximal, 449 uniform finite intersection property 68 uniform operator 64 uniform structure 445 uniform ultrafilter 52, 67 uniformity 445 uniformly continuous 445 uniformly recurrent 404 union subsystem 95 union ultrafilter 245 and strongly summable ultrafilter, 247 defines P-point, 247 universal minimal dynamical systems uniqueness, 400 universal minimal dynamical system 398 and minimal left ideals in PS, 398 upper density positive, 340
van der Waerden's Theorem 91, 279, 280, 295, 296, 369 consequence of Hales-Jewett, 283 increment from finite sums, 286 van der Waerden, B. 91, 280 van Douwen, E. 135, 204, 221, 256 van Mill, J. viii,135 variable extracted word 382 variable extraction 382 variable reduced word 382 variable reduction 381 variable word 282, 381 Voigt, B. 368
485
Index
Wallace, A. 47 Wallman, H. 69 weak P-point right cancelable, 164 weak product subsystem 377 weakly almost periodic function 436 weakly image partition regular 300 weakly left cancellative semigroup 116
weakly left cancellative 82, 122, 323 weakly right cancellative 82, 122 weakly summable ultrafilter 240 Weiss, B. viii, 102, 338, 368, 411
Woan, W. viii, 318 word 4 extracted, 382 reduced, 382 variable, 282, 381 Yood, B. 88
Zelenuk's Theorem 144 Zelenuk, E. 157, 191 zero dimensional 137 zero set 62
