ULTRAFILTER SEMIGROUPS Onesmus Shuungula Supervisors: Prof. Yevhen Zelenyuk Dr. Yuliya Zelenyuk A thesis submitted in ful llment of the requirements for the degree of Doctor of Philosophy in Mathematics School of Mathematics University of the Witwatersrand February 8, 2011 Declaration I declare that this thesis was composed by myself while registered as a stu- dent at The University of the Witwatersrand, and that I have acknowledged all sources of information contained herein. One of the main results of this thesis was presented at the annual congress of the South African Mathemat- ical Society in 2009 and another will be presented at the same congress in November 2010. Both results were also presented at the seminars of The School of Mathematics, University of the Witwatersrand. ................................ Onesmus Shuungula ii Contents Contents iii Acknowledgements v Abstract vi Introduction vii Historical background . . . . . . . . . . . . . . . . . . . . . . . . . vii Synopsis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiv 1 The Semigroup S 1 1.1 Ultra lters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.3 The structure of a completely simple semigroup . . . . . . . . 20 1.4 Right topological semigroups . . . . . . . . . . . . . . . . . . . 31 1.5 The Stone- Cech compacti cation of a discrete semigroup . . . 37 1.6 K( S) and its closure . . . . . . . . . . . . . . . . . . . . . . 49 2 Ultra lter semigroups and local homomorphisms 53 2.1 De nition and basic properties . . . . . . . . . . . . . . . . . . 53 2.2 Left invariant topologies de ned by idempotents . . . . . . . . 60 2.3 Local homomorphisms . . . . . . . . . . . . . . . . . . . . . . 72 iii 2.4 The ultra lter semigroup H . . . . . . . . . . . . . . . . . . . 86 2.5 Local isomorphisms of direct sums . . . . . . . . . . . . . . . . 93 2.6 The ultra lter semigroup 0+ . . . . . . . . . . . . . . . . . . . 99 3 The smallest ideal and its closure 103 3.1 Combinatorial characterizations of K(T) and its closure . . . . 103 3.2 Combinatorial characterizations of K(0+) and its closure . . . 113 3.3 Generalization of combinatorial characterizations . . . . . . . 117 3.4 c? K(T ) is not an ideal . . . . . . . . . . . . . . . . . . . . . . 121 4 Finite ultra lter semigroups 128 4.1 Almost maximal spaces and almost maximal left topological groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 4.2 Projectives in the category of nite semigroups . . . . . . . . . 141 4.3 Almost maximal topological groups . . . . . . . . . . . . . . . 145 Bibliography 153 iv Acknowledgements I would like to thank my mother, my whole family and my friends for their wonderful support during my research. I am hugely indebted to my super- visors Prof. Yevhen Zelenyuk and Dr. Yuliya Zelenyuk for their excellent guidance and support. I feel privileged to have been their student. I am very grateful to the School of Mathematics at The University of the Witwatersrand for o ering me excellent facilities and a wonderful research environment. Fi- nally, I would like to thank The University of Namibia for granting me a study leave and nancial support. v Abstract (1) Given a local left topological group X with a distinguished element 0, denote by Ult(X) the subsemigroup of X consisting of all nonprinci- pal ultra lters on X converging to 0. Any two countable nondiscrete zero-dimensional local left topological groups X and Y with count- able bases are locally isomorphic and, consequently, the subsemigroups Ult(X) X and Ult(Y ) Y are isomorphic. However, not every two homeomorphic zero-dimensional local left topological groups are locally isomorphic. In the rst result of this thesis it is shown that for any two homeomorphic direct sums X and Y , the semigroups Ult(X) and Ult(Y ) are isomorphic. (2) Let S be a discrete semigroup, let S be the Stone- Cech compacti- cation of S, and let T be a closed subsemigroup of S. The second and main result of this thesis consists of characterizing ultra lters from the smallest ideal K(T ) of T and from its closure c? K(T ), and show- ing that for a large class of closed subsemigroups T of S, c? K(T ) is not an ideal. This class includes the subsemigroups 0+ Rd and H ( L Z2). vi Introduction Historical background N, the Stone- Cech compacti cation of the discrete space N, can be described as follows. (a) The elements of N are the ultra lters on N, with elements of N being identi ed with principal ultra lters. (b) The base for the topology consists of subsets of the form A = fp 2 N : A 2 pg; where A N. Being the Stone- Cech compacti cation of N, every (continuous) function f : N ! Y into a compact Hausdor topological space Y factors uniquely through N, i.e. there exists a unique continuous function f : N! Y such that fjN = f . In particular, every compacti cation of N is a quotient of N. The semigroup operation + on N can be extended to N as follows. For any p; q 2 N, p+ q is de ned by A 2 p+ q :() fx 2 N : x+ A 2 qg 2 p; where for every x 2 N and every A N, x+A = fy 2 N : x+y 2 Ag. Then for each p 2 N, the translation p : N 3 q 7! q + p 2 N is continuous, and for each n 2 N, the translation n : N 3 q 7! n+ q 2 N is continuous. Although the same extension can be done for (N; ), unless otherwise stated, when we write N, we mean ( N;+). vii This extension works for any discrete semigroup S, making the Stone- Cech compacti cation S of S a right topological semigroup with S contained in its topological center. That means for each p 2 S, the right translation p : S 3 q 7! qp 2 S is continuous, and for each x 2 S, the left translation x : S 3 q 7! xq 2 S is continuous, respectively. This extension was explicitly established by Day [10] in 1957 using the methods of Arens [1]. For any semigroup S with a smallest ideal, we use K(S) to denote that smallest ideal. As any compact right topological semigroup, S has a small- est ideal K( S). In addition, since S is dense in S and S is a subset of the topological center of S (the set of all ultra lters p 2 S such that p : S ! S is continuous), the topological center of S is dense in S and, thus, the closure c? K( S) is also an ideal of S. The semigroup S has an algebraic structure of extraordinary complexity which makes S interesting for its own sake. The semigroup S is also in- teresting for its important applications to Ramsey Theory and topological dynamics. A typical problem in Ramsey Theory can be thought of as follows. One has a set X and a collection G of certain subsets of X. One then asks whether, whenever X is partitioned into nitely many pieces, one piece must contain a member of G. Viewed in this way one can see why we prefer the con- struction of N with ultra lters on N as points. If p is an ultra lter on N and N = Sn i=1Ai is a nite partition of N, then Ai 2 p for exactly one i 2 f1; ; ng. The Graham-Rothschild conjecture, now known as the Finite Sum Theorem, is one of the concepts that connect N and Ramsey Theory. This theorem says the following. Theorem 0.0.1. Let r 2 N and let N = Sr i=1Ai be a partition of N. Then there exists i 2 f1; 2; ; rg and a sequence (xn)n2N in N such that FS((xn)n2N) 2 Ai. For any nonempty subset B N, FS(B) = f F : F is a nonempty nite subset of Bg viii and FP (B) = f Y F : F is a nonempty nite subset of Bg: More generally, let S be a semigroup, let D be a linearly ordered set, and let (xs)s2D be a D-sequence in S. (a) For F 2 Pf (S), Q s2F xs is the product in the increasing order of indices. (b) FP ((xs)s2D) = f Q s2F xs : F 2 Pf (S)g. (c) The D-sequence (xs)s2D is said to satisfy the distinct nite products property if and only if whenever F;G 2 Pf (S) and Q s2F xs = Q s2G xs, one has F = G. (Given a set X, we use P(X) to denote the powerset of X, and Pf (X) to denote the collection of all nonempty nite subsets of X.) The Graham-Rothschild conjecture had been open for some decades, even though several mathematicians (including Hilbert) worked on it. Fred Galvin worked on this conjecture around the year 1970. At some point Galvin wanted to know whether there could exist p 2 N such that whenever A 2 p, fx 2 N : x+A 2 pg 2 p. Galvin called such ultra lters \almost downward translation invariant ultra lters". In current terminologies Galvin wanted to know if there could exist an idempotent in N, i.e. p 2 N such that p+ p = p. He had a construction showing that any element of such an ultra- lter contains FS(B) for some in nite subset B N. In fact, using current terminologies, what Galvin knew amounts to the following fact: The set = fp 2 N : for all A 2 p there exists an in nite B A with FS(B) Ag is the closure of the set fp 2 N : p+ p = pg in N. Around that time Steven Glazer was one of the few mathematicians who knew about the extension of the semigroup operation from a discrete semi- group S to S, and who were also aware of the following theorem of Ellis [15]: Every compact right topological semigroup contains an idempotent. These mathematicians also viewed the underlying set of S as the set of all ultra l- ters on S. So, one day in 1975 Galvin learned from Glazer that idempotents exist in N and together, using this information, they devised a simple and elegant proof of the Finite Sum Theorem. ix Going back a little, around 1972, while trying to answer the question of Galvin about the existence of almost downward translation invariant ultra- lters in N, Neil Hindman established the following result, where = fA N : there exists an in nite B A with FS(B) Ag: Theorem 0.0.2. The Graham-Rothschild conjecture holds if and only if there is an ultra lter p on N such that p . The ultra lter produced in Theorem 0.0.2 does not necessarily answer Galvin?s question even if the Graham-Rothschild conjecture is assumed to be true. It is in fact possible to produce, under this assumption, an ultra lter contained in which has an element A such that x+A =2 for every x in A. However Hindman established the following result. Theorem 0.0.3. Assume the continuum hypothesis. The Graham-Rothschild conjecture implies, and therefore is equivalent to, the existence of an almost downward translation invariant ultra lter on N. In fact, Galvin also asked whether what he called \almost upward transla- tion invariant ultra lters" exist. This question was answered in the negative by Hindman. Soon thereafter, Hindman showed directly, using a combinato- rial argument, that the Graham-Rothschild conjecture is true. However, the proof was of enormous complexity. Since the proof by Galvin and Glazer, there have been numerous applications of the algebraic structure of N to Ramsey Theory, various strong combina- torial results have been obtained using the algebraic structure of N, and the Finite Sum Theorem has been extended using the combination of addition and multiplication in N. Other famous combinatorial theorems that were initially established using combinatorial methods, such as the van der Waer- den?s Theorem, were reestablished elegantly using the algebraic structure of N. In the process more came to be known about the algebraic structure of S, for a general semigroup S. Among other interesting results and questions about the structure of N, we mention the following. (1) The center of N is N. Even more than that, N contains copies of the free group on 22 ! generators. (We use ! to denote the rst in nite cardinal, i.e. ! = f0; 1; 2; 3; g.) x (2) N has 22 ! minimal left ideals and 22 ! minimal right ideals. (3) It is not yet known whether N contains any element of nite order, other than the idempotents. (4) Only recently it was shown that there are no nite nontrivial groups in N. See [63] and [64]. (5) Let H = \ n2! c? N 2 nN = \ n2! 2nN ; T = \ n2N c? N nN = \ n2N nN: Then, (a) T and H are subsemigroups of N with T H. (b) All idempotents of N are in T. (c) If A is a left ideal, a right ideal, a minimal left ideal, a mini- mal right ideal, or the smallest ideal of N, then A \ T is the corresponding object in T. (c) There are functions f : N ! S, where S is a compact Hausdor right topological semigroup, such that f is not a homomorphism but f(p+q) = f(p)+ f(q) for every p 2 N and every q 2 H (where f : N ! S is the continuous extension on f). In particular the restriction of f to H is a homomorphism. We now look at the applications of S to topological dynamics. A topological dynamical system is a pair (X; (Ts)s2S) where (1) X is a compact Hausdor space, (2) S is a semigroup, (3) for each s 2 S, Ts is a continuous function from X to X, and (4) for each s; t 2 S, Ts Tt = Tst. For our introductory purpose we consider the case when S = (!;+). There- fore, unless otherwise stated, by a topological dynamical system we will mean a topological dynamical system of the form (X; (Tn)n2!) with T0 equal to the identity function of X. In this case, let T = T1. Then, for any n 1, Tn = T1+ +1 = T1 T1 = T n, i.e. Tn is the nth iterate of T . We de ne xi T 0 = T0, the identity function of X. We will write (X;T ) for (X; (Tn)n2!). Topological dynamics is concerned with the behavior of the iterates of T . To study the limiting behavior of the iterates of T for large n, one de nes for each ultra lter U 2 !, T U : X ! X ; T U(x) = U - lim n2N T n(x); (see Chapter 1, De nition 1.1.13). It is clear that if U is an ultra lter generated by m, then T U = Tm. We also have T U(T V(x)) = T U+V(x) for every x 2 X and for every U ;V 2 !. Regarded as a function of U 2 !, for a xed x 2 X, this is the continuous extension to ! of the function ! 3 n 7! T n(x) 2 X. It follows that fT U(x) : U 2 !g is the closure of the forward orbit fT n(x) : n 2 !g of the point x. More generally, a semigroup S of continuous functions acting on a compact Hausdor space X has a closure in XX called the enveloping semigroup of S, [15]. The enveloping semigroup is a compact right topological semigroup and, therefore, it is a quotient of S. In some important cases it is equal to S. The algebraic properties of S have implications for the dynamical behavior of the topological dynami- cal system. The topological space ! with the function P : ! 3 U 7! 1 + U 2 ! is a topological dynamical system. One observes that addition in ! is just the ultra lter iteration of P , i.e. for any U ;V 2 !, P U(V) = U + V . For any topological dynamical system (X;T ) and for any x 2 X, the function f : ! ! X; U 7! T U(x) is the unique continuous function from ! to X with f P = T f and f(0) = x. ! f // P X T ! f // X: Therefore ( !; P ) may be regarded as the free topological dynamical system on one generator. Let (X;T ) be a topological dynamical system and let x 2 X. Then, (1) x is called recurrent if for each neighborhood G of x there are in nitely many n 2 ! such that T n(x) 2 G, and xii (2) x is called uniformly recurrent if for each neighborhood G of x there exists M 2 ! such that for any n 2 ! there is k < M such that T n+k(x) 2 G. Theorem 0.0.4. Let (X;T ) be a topological dynamical system. (1) A point x 2 X is recurrent if and only if T U(x) = x for some nonprin- cipal ultra lter U on !. (2) A point x 2 X is uniformly recurrent if and only if for every ultra lter V on ! there is an ultra lter U on ! such that T U(T V(x)) = x. The interaction with topological dynamics works both ways. Several com- binatorial notions which originated in topological dynamics such as syndetic and piecewise syndetic sets, are important in describing the algebraic struc- ture of S. In a semigroup S a subset A S is said to be syndetic if there is a nite subset G S such that S = G 1A, where G 1A = fx 2 S : tx 2 A for some t 2 Gg: For example, one can easily verify that in a topological dynamical system (X;T ), a point x 2 X is uniformly recurrent if and only if for every neigh- borhood G of x, the set fn 2 ! : T n(x) 2 Gg is a syndetic subset of !. A is said to be piecewise syndetic if there is a nite subset G S such that for every nite subset F S, there exists x 2 S with Fx G 1A. We have the following very important theorem. Theorem 0.0.5. Let S be a semigroup, let K( S) be the smallest ideal of S and let p 2 S. The following statements hold. (1) p 2 K( S) if and only if for every A 2 p, fx 2 S : x 1A 2 pg is syndetic. (2) p 2 c? K( S) if and only if every A 2 p is piecewise syndetic. These combinatorial characterizations of members of K( S) and c? K( S) have been extended from S to certain important closed subsemigroup of S, namely, to H and 0+ Rd. Here Rd is the additive group of real num- bers reendowed with the discrete topology and 0+ = T n2N c? Rd(0; 1 n). As mentioned earlier, H contains most of the information about N and one can expect it to ?behave? in the same way as N. Surprisingly, as distinguished xiii from c? K( S), c? K(H) is not a left ideal of H [22]. The question of whether c? K(0+) is not a left ideal of 0+ remained open [25]. We have now answered this question and this answer is the main result in this thesis. In [56], J. Pym noticed that certain important subsemigroups of S de- pend very little on the operation of S and they can be de ned in terms of direct sums of sets with a distinguished element. We have obtained a result which says that if X and Y are homeomorphic direct sums, then the subsemi- groups of X and Y consisting of all nonprincipal ultra lters converging to the distinguished element are isomorphic. Synopsis In Chapter 1 we review the elementary concepts about ultra lters and semi- groups. One of the important results presented in this chapter is the structure of a completely simple semigroup. We then present the construction of the Stone- Cech compacti cation S of a discrete semigroup S and the exten- sion of the semigroup operation from S to S. The elements of K( S) and c? K( S) are given combinatorial characterizations as stated in Theorem 0.0.5. Chapter 2 deals with ultra lter semigroups. These are de ned as follows. Let G be a group and let T be a topology on G such that the left shifts a : G 3 x 7! ax 2 G, where a 2 G, are continuous. The nonprincipal ultra- lters on G that converge to the identity of G in this topology form a closed subsemigroup of the Stone- Cech compacti cation of the discrete semigroup G. We call this subsemigroup the ultra lter semigroup of (G; T ). Indeed, this can be de ned for any local left topological group. Ultra lter semigroups proved useful in studying the structure of the semigroups S. They are also a main tool in the proof of our main result. Homomorphisms of ultra lter semigroups are induced by what are called local homomorphisms. We also present two ways of constructing left invariant topologies on a group G, using idempotents from G n G. In Section 2.5 we prove our rst result, that the ultra lter semigroups of any two homeomorphic direct sums are isomorphic. Chapter 3 starts by presenting old results about the smallest ideals of T and 0+ and the closures of these smallest ideals. We extend the combinatorial characterizations of elements of K( S) and c? K( S) to an arbitrary closed xiv subsemigroup T of S. We then move on to our main result, showing that for a large class of closed subsemigroups T of S, c? K(T ) is not a left ideal of T . This class includes the semigroups 0+ and H . For every in nite cardinal , H is the ultra lter semigroup of the Boolean group L Z2 in the topology having as the basic neighborhood of the identity the subgroups of the form H = fx 2 M Z2 : x( ) = 0 for all < g; where < . Chapter 4 deals with nite ultra lter semigroups. The ultra lter semigroup of a left topological group G re ects quite good topological properties of G, especially when the ultra lter semigroup is nite. A topological space X is called almost maximal if it is without isolated points and for every element x 2 X, there are only nitely many ultra lters on X converging to x. We look at almost maximal spaces, almost maximal left topological semigroups, projectives in the category of nite semigroups and almost maximal topolog- ical groups. xv Chapter 1 The Semigroup S 1.1 Ultra lters De nition 1.1.1. Let X be a set and let F ;B P(X). F is called a lter on X if the following properties are satis ed: (a) ; =2 F and X 2 F , (b) for every A;B 2 F , A \B 2 F , and (c) for any A 2 F and any B X with A B, B 2 F . B is called a lter base on X if the following conditions are satis ed: (i) ; =2 B and ; 6= B, and (ii) for any A;B 2 B, there exists C 2 B such that C A \B. An ultra lter on X is a lter on X that is not properly contained in any other lter on X. When dealing with lters, the following condition will serve the same purpose as condition (ii) in the de nition of a lter base. For any A;B 2 B, A \B 2 B. 1 1.1 Ultra lters 2 For any lter base B on X, there is a smallest lter on X that contains B. It is de ned as follows. FB := fA X j B A for some B 2 Bg: We call FB the lter generated by B and we call B a lter base (or simply a base) for FB. Let X be a topological space. For each x 2 X, let Nx denotes the family of neighborhoods of x. It is easy to check, using the de nition of a neighborhood and the de nition of an open set, that the family fNx : x 2 Xg satis es the following conditions. (N1) For every x 2 X and U 2 Nx, x 2 U . (N2) For every x 2 X and U 2 Nx, fy 2 X : U 2 Nyg 2 Nx. (N3) For every x 2 X and U; V 2 Nx, U \ V 2 Nx. (N4) For every x 2 X and U 2 Nx, U V X implies V 2 Nx. Remark 1.1.2. (1) Condition (N2) is equivalent to the following condition. (N2)0 For every x 2 X and for every U 2 Nx, there exists V 2 Nx such that U 2 Ny for every y 2 V . (2) (N1), (N3) and (N4) imply that Nx is a lter on X. Nx is called the neighborhood lter at (or of ) x and fNx : x 2 Xg is called the neighborhood system for X. Proposition 1.1.3. Let X be a set and let fNx : x 2 Xg be a family of lters on X satisfying (N1) and (N2). Then there exists a topology T on X such that Nx is precisely the neighborhood lter at x for each x 2 X. Proof. Let T = fU X : for every x 2 U there is W 2 Nx with W Ug. Clearly ; 2 T . Since for any x 2 X, Nx 6= ;, X 2 T . Let U; V 2 T and let x 2 U \ V . Pick W1;W2 2 Nx such that W1 U and W2 V . Then W := W1 \W2 2 Nx and W U \ V . Therefore U \ V 2 T . Let (Ui)i2I be a family in T and let x 2 S i2I Ui. Pick j 2 I such that x 2 Uj and pick W 2 Nx with W Uj. Then W S i2I Ui and, hence, S i2I Ui 2 T . It 1.1 Ultra lters 3 follows that T is a topology on X. Let x 2 X and denote by Fx the neighborhood lter at x in T . Let U 2 Nx. Then x 2 U . By (N2), W := fy 2 X : U 2 Nyg 2 Nx. By (N1) W U . Therefore U 2 T . Hence U 2 Fx. It follows that Nx Fx. Now let U 2 Fx. There exists V 2 T such that x 2 V U . Pick W 2 Nx such that W V . Then W U . Since Nx is a lter, U 2 Nx. It follows that Fx Nx. Hence Fx = Nx. Recall that a family A of sets is said to have the nite intersection property if for every nonempty nite subset F of A, T F 6= ;, and that a topological space X is compact if and only if for every family A of closed subsets of X such that A has the nite intersection property, T A 6= ;. A subset A P(X) is said to be maximal with respect to the nite intersection property if A has the nite intersection property and A is not properly contained in any other subset of P(X) having the nite intersection property. Theorem 1.1.4. [[29], Theorem 3.6] Let X be a set and let p P(X). The following statements are equivalent. (a) p is an ultra lter on X. (b) p has the nite intersection property and for every A X, A 2 p if and only if A \B 6= ; for every B 2 p. (c) p is maximal with respect to the nite intersection property. (d) p is a lter on X and for every nite A P(X), if S A 2 p, then A \ p 6= ;. (e) p is a lter on X and for every B X, either B 2 p or X nB 2 p. Proof. (a)) (b): Let p be an ultra lter on X. Since for each A;B 2 p, A \ B 2 p and ; =2 p, A \ B 6= ;. Therefore p has the nite intersection property. Let A X. If A 2 p, then clearly A \ B 6= ; for every B 2 p. Conversely, suppose A \B 6= ; for every B 2 p. Put q = fC X : A \B C for some B 2 pg. It is clear that q is a lter on X with A 2 q and p q. Since p is an ultra lter, p = q. Thus A 2 p. (b)) (c): Suppose (b) holds. Let U P(X) such that p U and U has the nite intersection property. Let A 2 U and let B 2 p. Then A;B 2 U and, therefore, A\B 6= ;. By (b), A 2 p. Since A was an arbitrary element of U , 1.1 Ultra lters 4 U p. Therefore p = U . (c)) (d): Suppose (c) holds. It is clear that ; =2 p. Let A;B 2 p and let C X with A C. It is easy to see that the families p[ fXg, p[ fA\Bg and p [ fCg have the nite intersection property and contain p. Therefore these families are equal to p. Hence p is a lter on X. Let A;B X such that A[B 2 p and suppose A =2 p and B =2 p. Since p $ p[fAg and p $ p[fBg, p [ fAg and p [ fBg do not have the nite intersection property. Pick C1; ; Cn; D1; ; Dm 2 p, for some n;m 2 N, such that A\C1\ \Cn = ; = D1 \ \ Dm \ B. Then F := fC1; ; Cn; D1; ; Dm; A [ Bg p with T F = ;, a contradiction. Therefore A 2 p or B 2 p. (d))(e): Suppose (d) holds. Let B X. Take A = fB;X n Bg. Then S A = X 2 p. Therefore B 2 p or X nB 2 p. (e))(a): Suppose (e) holds. Let q be a lter on X with p q. Assume p 6= q. Pick A 2 q n p. Then X n A 2 p q and, therefore, ; = A \ (X n A) 2 q, a contradiction. It follows that p = q and, thus, p is an ultra lter on X. Theorem 1.1.5. [[29], Theorem 3.8] Let X be a set and let A P(X) such that A has the nite intersection property. Then there is an ultra lter p on X such that A p. Proof. Let = fB P(X) : A B and B has the nite intersection propertyg: Since A 2 , 6= ;. Let C be a nonempty chain in . Then A S C. Let ; 6= F S C be nite. Since C is a chain, F B for some B 2 C. Since B has the nite intersection property, T F 6= ;. Therefore, S C has the nite intersection property. It is clear that S C is an upper bound of C in . By Zorn?s Lemma, has a maximal element U . Let V P(X) such that U V and V has the nite intersection property. Then A V and, therefore, V 2 . Since U is a maximal element of , U = V . Hence U is maximal with respect to the nite intersection property. Thus U is an ultra lter on X, with A U . Since every lter has the nite intersection property, we obtain the following. Corollary 1.1.6. Let X be a set. Every lter on X is contained in an ultra lter. Example 1.1.7. Given x 2 X, the set fA X : x 2 Ag is an ultra lter on X. It is called the principal ultra lter generated by x. More generally, for every nonempty subset A of X, fB X : A Bg is a lter on X. However, 1.1 Ultra lters 5 if A is not a singleton, then fB X : A Bg is not an ultra lter since for any a 2 A, fB X : A Bg ( fB X : a 2 Bg. A nonprincipal ultra lter is also called a free ultra lter. Note that an ultra- lter p is free if and only if T p = ;. De nition 1.1.8. Let X be a topological space and let F be a lter on X. A point x 2 X is called a cluster point of F if x belongs to the closure of every element of F . A point x 2 X is called a limit of F if F contains the neighborhood lter of x. If x is a limit of F , we say that F converges to x, and we write F ! x. Remark 1.1.9. Let X be a topological space, let x 2 X and let F and G be lters on X. (a) Every cluster point of an ultra lter is a limit of that ultra lter. (b) If F ! x and G F , then x is a cluster point of G. In particular if every ultra lter is convergent, then every lter has a cluster point. (c) If we combine the remarks (a) and (b), we obtain the following. Every ultra lter is convergent () every lter has a cluster point. Proof. (a) It su ces to recall that for an ultra lter F and A X, A 2 F if and only if A \B 6= ; for all B 2 F . (b) Let U be a neighborhood of x and let B 2 G. Then U;B 2 F . Hence U \B 6= ;. Therefore, x 2 c? B. (c) This is clear from (a) and (b). Remark 1.1.10. Let X be a topological space and let x 2 X. Denote the neighborhood lter at x by Fx. It is easy to see that Fx = \ fF j F is a lter on X and F ! xg: In other words, Fx is the smallest lter on X converging to x. 1.1 Ultra lters 6 Theorem 1.1.11. Let X be a topological space. (1) X is Hausdor if and only if every ultra lter on X has at most one limit. (2) X is compact if and only if every ultra lter on X has at least one limit. Proof. (1) Suppose X is Hausdor and let p be an ultra lter on X such that p has a limit. Suppose x and y are distinct limits of p. Pick neighborhoods U and V of x and y respectively such that U \ V = ;. Then U; V 2 p and, therefore, ; = U \ V 2 p, a contradiction. Conversely, suppose every ultra lter on X has at most one limit. Let x and y be distinct points of X. Let Nx and Ny be neighborhood lters of x and y respectively. Assume for every U 2 Nx and every V 2 Ny, U \ V 6= ;. Then Nx [ Ny has the nite intersection property. Pick an ultra lter F on X such that Nx [ Ny F . Then F converges to x and F converges to y, a contradiction. It follows that U \V = ; for some U 2 Ny and some V 2 Ny. Hence X is Hausdor . (2) Suppose X is compact. Let p be an ultra lter on X. Then the intersec- tion of all closed sets in p is nonempty, since every nite subintersection is nonempty. Note that fA : A 2 p and A is closedg = fc? A : A 2 pg: Pick x 2 T fc? A : A 2 pg and let U be a neighborhood of x. Then U \A 6= ; for every A 2 p. Since p is an ultra lter, U 2 p. Therefore p contains the neighborhood lter at x and, hence, p converges to x. Conversely, suppose every ultra lter on X has a limit. Let A be a collection of closed subsets of X satisfying the nite intersection property. Pick an ultra lter p on X such that A p and let x be a limit of p. Suppose there exists A 2 A such that x =2 A. Then x 2 X nA and X nA is open. It follows that X n A 2 p, a contradiction. Therefore x 2 T A and, hence, T A 6= ;. Thus X is compact. Theorem 1.1.11 shows that, if (X; T ) is a compact Hausdor topological space, then the operation of taking limits of ultra lters de nes a function from the set of all ultra lters on X to X. Moreover, the proof of Theorem 1.1.11 shows that, for any ultra lter p on X, T fc? A : A 2 pg = f (p)g. Since X is regular, it follows that (p) 2 U for some open set U if and only if there is an open set V 2 p such that c? V U . Thus, 1(U) = fp : c? V U for some open V 2 pg: 1.1 Ultra lters 7 In Section 1.5 the set of all ultra lters on X will be endowed with a topology having sets of the form fp : p 2 V g, where V X, as basic open sets. It follows that is continuous with respect to that topology. Let X and Y be topological spaces, let p be an ultra lter on X and let f : X ! Y be a function. Then fA Y : f 1(A) 2 pg is an ultra lter on Y (we will prove this in Theorem 1.5.7), and we have the following lemma. Lemma 1.1.12. Let X and Y be compact Hausdor topological spaces. A function f : X ! Y is continuous if and only if for every ultra lter p on X, f( (p)) = (fA Y : f 1(A) 2 pg) (i.e. f preserves limits of ultra lters). We have the following, more general, concept. De nition 1.1.13. Let X be a topological space, let p be an ultra lter on X, let f : X ! Y be a function into a topological space Y and let y 2 Y . y is called a p -limit of f if for every neighborhood U of y, f 1(U) 2 p. If a p -limit y of f is unique we denote it by p - lim x2X f(x) or p - lim f(x) or p - lim f or lim x!p f(x): Theorem 1.1.14. Let f : X ! Y be a function from a topological space X into a topological space Y and let p be an ultra lter on X. (1) If Y is Hausdor and a p -limit of f exists, then it is unique. (2) If Y is compact, then f has a p -limit. Proof. (1) Suppose Y is Hausdor and suppose y1 and y2 are p - limits of f in Y . Suppose y1 6= y2. Pick two disjoint open subsets U; V Y such that y1 2 U and y2 2 V . Then ; = f 1(U) \ f 1(V ) 2 p, a contradiction. It follows y1 = y2. (2) Suppose Y is compact. Let Ap = fc? f(A) : A 2 pg. Let A;B 2 p. Then c? f(A)\c? f(B) c? f(A\B) f(A\B) 6= ;. Therefore Ap has the nite intersection property. Since Y is compact, T Ap 6= ;. Pick y 2 T Ap and let U be a neighborhood of y. Then, U \ f(A) 6= ; for every A 2 p. Therefore f 1(U) \ A 6= ; for every A 2 p. Since p is an ultra lter, f 1(U) 2 p. It follows that y is a p -limit of f . Theorem 1.1.15. Let f : X ! Y and g : Y ! Z be functions between topological spaces with Y and Z satisfying the Hausdor separation property. Let p be an ultra lter on X. If g is continuous and y is a p - limit of f , then g(y) is a p - limit of g f . 1.1 Ultra lters 8 Proof. Let W be a neighborhood of g(y). By continuity of g, g 1(W ) is a neighborhood of y. Therefore, f 1(g 1(W )) 2 p. Remark 1.1.16. Let X be a topological space, let p be an ultra lter on X and let y 2 X. Then y is a limit of p in X if and only if y is a p -limit of idX ; where idX is the identity function on X. A lter F on a topological space X is called open (respectively, closed) if F has a base consisting of open (respectively, closed) subsets of X. Theorem 1.1.17. Let X be a set and let Y be a nonempty subset of X. There is a bijective correspondence between lters on X containing Y and lters on Y . This correspondence restricts to a bijection between ultra lters on X that contain Y and ultra lters on Y . Proof. Consider the correspondence F 7 ! F \ Y := fA \ Y : A 2 Fg; where F ranges over lters on X containing Y ; G 7 ! G := fA X : U A for some U 2 Gg; where G ranges over lters on Y . It is easy to see that this correspondence satis es the required conditions. Note that the correspondence in the proof of Theorem 1.1.17 sends principal ultra lters to principal ultra lters. De nition 1.1.18. Let be a cardinal and let p be an ultra lter on . The norm of p, denoted kpk, is de ned by kpk = minfjAj : A 2 pg: It is clear that 1 kpk for any ultra lters p on . 1.1 Ultra lters 9 De nition 1.1.19. Let and be cardinals and let p be an ultra lter on . p is called -uniform if kpk . We denote the set of -uniform ultra lters on by U ( ) and simply use U( ) if = . The elements of U( ) are simply called uniform ultra lters, instead of -uniform ultra lters. Note that an ultra lter p on is nonprincipal if and only if p is !-uniform. It follows that n = U!( ). In particular, n = ; if and only if < !. De nition 1.1.20. Let and be cardinals and let F P( ). F is said to have the -uniform nite intersection property if T F 6= ; and j \ i n Aij ; whenever n < ! and Ai 2 F for i n. If = , we say F has the uniform nite intersection property, instead of saying F has the -uniform nite intersection property. De nition 1.1.21. Let and be cardinals. A lter F on is -complete if T G 2 F whenever G F with jGj < . It follows that every lter is !-complete. Remark 1.1.22. Let be a cardinal and let F P( ) such that T F 6= ;. Then, F has the nite intersection property if and only if F has the 1-uniform nite intersection property. De nition 1.1.23. Let and be cardinals. Then we de ne P ( ) = fA 2 P( ) : j n Aj < g: If ! , then P ( ) is a lter on , called the (generalised) Fr echet lter on . Lemma 1.1.24. Let ! . Then (a) an ultra lter p on is -uniform if and only if P ( ) p; (b) there is a -uniform ultra lter on ; and (c) each family of subsets of with the -uniform nite intersection prop- erty is contained in a -uniform ultra lter on . 1.2 Semigroups 10 Proof. (a) Suppose p is -uniform. Let A 2 P ( ). Then j n Aj < and hence n A =2 p. Thus A 2 p. Conversely, suppose P ( ) p. Let B 2 p and suppose jBj < . Then j n Bj and, therefore, n B 2 p. Then ; = B \ ( nB) 2 p, a contradiction. It follows that jBj . (b) Since ! , P ( ) is a lter on . Pick an ultra lter p on with P ( ) p. Then, by (a), p is -uniform. (c) Let F be a family of subsets of with the -uniform nite intersection property. Let m;n 2 N and let fAk : k ng F and fBi : i mg P ( ). Then j T k nAkj and j n T i mBij < . Therefore, ( T k nAk) \ ( T i mBi) 6= ;. It follows that F [P ( ) has the nite intersection property. Pick an ultra lter p on such that F [ P ( ) p. Then, by (a), p is -uniform. De nition 1.1.25. Let be a limit ordinal. The co nality of , denoted cf( ), is the least ordinal for which there is an order-preserving function f : ! such that f is unbounded, i.e. sup < f( ) = . An in nite cardinal is called regular if = cf( ). Otherwise, is called singular. 1.2 Semigroups De nition 1.2.1. A semigroup (S; ) is a nonempty set S equipped with a binary operation such that x (y z) = (x y) z for all x; y; z 2 S; i.e. is associative. We will often write xy instead of x y and S instead of (S; ), unless if we would like to emphasize the binary operation in question. For any A;B S we de ne AB := fab j a 2 A; b 2 Bg; with aB := fagB and Ab := Afbg for all a; b 2 S. We will often write A2 for AA. If A S and x 2 S, we denote by x 1A the set fy 2 S : xy 2 Ag. It follows from the associative law of the binary operation on S that for any x; y 2 S and any A S, (xy) 1A = y 1(x 1A). A semigroup S is said to be commutative if is commutative (i.e. xy = yx for all x; y 2 S.) De nition 1.2.2. Let S be a semigroup and let 0; e 2 S. 1.2 Semigroups 11 (a) e is called a left identity of S if ex = x for every x 2 S. e is called a right identity of S if xe = x for every x 2 S. e is called an identity of S if e is both a left and a right identity of S; i.e. ex = x = xe for every x 2 S. (b) 0 is called a left zero of S if 0x = 0 for every x 2 S. 0 is called a right zero of S if x0 = 0 for every x 2 S. 0 is called a zero of S if 0 is both a left and a right zero of S; i.e. 0x = 0 = x0 for every x 2 S. (c) Let x; y; z 2 S. x is called a left z-inverse of y (and y is called a right z-inverse of x) if xy = z. If e is an identity of S, then x is called an inverse of y if xy = e = yx. Proposition 1.2.3. Let S be a semigroup. If S has a left identity (left zero) and a right identity (right zero) then they are equal. In particular every semigroup has at most one identity (zero). Proof. Let eL be a left identity of S and eR be a right identity of S. Then, eL = eLeR = eR. Similarly, if 0L is a left zero and 0R is a right zero of S, then 0L = 0L0R = 0R. Uniqueness of an identity ( a zero) follows easily from these equations since an identity (a zero) is both a left identity (left zero) and a right identity (right zero). If a semigroup S does not have an identity element, we can adjoin an extra element e to S to obtain the set S [ feg and extend multiplication on S to S [ feg by xe = ex = x and ee = e; for every x 2 S. This makes S [ feg a semigroup and we denote this semi- group by S1. Similarly, if a semigroup S does not have a zero, we can adjoin an extra element 0 to S to obtain the semigroup S0 := S [f0g with multiplication on S extended to S0 by x0 = 0 = 0x and 00 = 0; for every x 2 S. If G is a group, then G0 is called a 0-group. Indeed G0 is not a group. Unless otherwise stated, we will not assume that a given semigroup has an identity or a zero. 1.2 Semigroups 12 De nition 1.2.4. Let S be a semigroup and let a 2 S. We say that a is (an) idempotent if a2 = a. We de ne E(S) := fx 2 S j x2 = xg. De nition 1.2.5. Let S be a semigroup. S is called a left zero semigroup if for every x; y 2 S, xy = x. S is called a right zero semigroup if for every x; y 2 S, xy = y. Example 1.2.6. (a) Let S be a nonempty set. De ne multiplication on S by xy := y. Then S with this multiplication is a right zero semigroup. (b) Let S be a semigroup with a left identity and let T = fx 2 S j xy = y for all y 2 Sg; i.e. T is the set of all left identities of S. Then T is a right zero semigroup. Similarly, in a semigroup S with a right identity, P = fy 2 S j xy = x for all x 2 Sg; i.e. the set of all right identities of S, is a left zero semigroup. Note that T and P are subsets of E(S). In particular, if every idempotent is a left (respectively, right) identity, then E(S) is a right (respectively, left) zero semigroup. Let A and B be nonempty sets. De ne a binary operation on the cartesian product A B by (a1; b1)(a2; b2) = (a1; b2): This operation is clearly associative. A B equipped with this operation is called a rectangular band. If S is a rectangular band, we sometimes refer to S as a rectangular semigroup. Note that if jAj = 1, then A B is a right zero semigroup, while if jBj = 1 then A B is a left zero semigroup. In particular, every left(right) zero semigroup can be regarded as a rectangular band (what we mean here will become clear after we have de ned semigroup isomorphisms, see De nition 1.2.19). De nition 1.2.7. A monoid is a semigroup which has an identity element. Let M be a monoid with identity e. It is clear that every element of M has at most one inverse. If x 2M has an inverse we call x invertible and we will denote the inverse of x by x 1. A group is a monoid in which every element is invertible. If S is a monoid, we will denote the set of its invertible elements by I(S). 1.2 Semigroups 13 Proposition 1.2.8. Let S be a semigroup.Then S is a group () aS = Sa = S for every a 2 S: Proof. ): Suppose S is a group. Let e be the identity of S. We need to show that S aS \ Sa for every a 2 S. Let a; x 2 S. x = ex = (aa 1)x = a(a 1x) 2 aS and x = xe = x(a 1a) = (xa 1)a 2 Sa. Hence S aS \ Sa. It follows that aS = S = Sa. (: Suppose aS = Sa = S for every a 2 S. Let a 2 S. There exists b; c 2 S such that ab = a = ca. We will show that b is a right identity of S and c is a left identity of S and hence we have an identity. Let z 2 S. There exists x; y 2 S such that ax = z = ya. Then, cz = c(ax) = (ca)x = ax = z and zb = (ya)b = y(ab) = ya = z. Thus we have an identity e := b(= c). Now we show that every element of S has an inverse. So, let t 2 S. Since e 2 S = tS = St, there exist g; h 2 S such that tg = e = ht. Then g = eg = (ht)g = h(tg) = he = h. Hence we have t 1 := h(= g). It follows that S is a group. Proposition 1.2.9. Let S be a semigroup. The following statements are equivalent. (a) S is a group (b) S has a left identity e and every element of S has a left e-inverse. (c) S has a right identity e and every element of S has a right e-inverse. Proof. The implications (a) ) (b) and (a) ) (c) are obvious. We prove (b) ) (a) and then (c) ) (a) follows by a left-right switch. So, suppose (b) holds. Let x 2 S and let y be a left e-inverse of x. Let z be a left e-inverse of y. Then xy = e(xy) = (zy)(xy) = z(y(xy)) = z((yx)y) = z(ey) = zy = e. Hence y is also a right e-inverse for x. In addition, we get: xe = x(yx) = (xy)x = ex = x. Hence e is also a right identity for S. It follows that S is a group. De nition 1.2.10. Let S be a semigroup and let T be a nonempty subset of S. T is called a subsemigroup of S if T 2 T , (i.e. T is closed under the binary operation). T is called a submonoid of S if T is a subsemigroup of S and T has an identity element. T is called a subgroup of S if T is a submonoid of S and every element of T has an inverse in T . Example 1.2.11. 1.2 Semigroups 14 (a) Let S be a semigroup. For every x 2 S, xS and Sx are subsemigroups of S. (b) Let S be a semigroup and let x 2 S. fxg is a subsemigroup (in fact a subgroup) of S if and only if x2 = x. In particular, a singleton consisting of a left(right) identity or a left (right) zero of S is a subgroup of S. (c) For any monoid M , I(M) is a subgroup of M (where I(M) is as de ned in De nition 1.2.7) . We will call a subsemigroup T of a semigroup S proper if T 6= S. As a corollary to Proposition 1.2.8 we have the following. Corollary 1.2.12. Every semigroup containing more than one element has a proper subsemigroup. Proof. Let S be a semigroup containing more than one element and suppose S has no proper subsemigroup. Then for every a 2 S, aS = S = Sa. Then, by Proposition 1.2.8, S is a group. Let e be the identity of S. Then feg is a proper subsemigroup of S. This is a contradiction. Therefore, S has a proper subsemigroup. A subsemigroup T of a semigroup S is said to be left saturated in S if (S n T )T \ T = ;: Remark 1.2.13. Let S be a semigroup. (a) Every left (right) zero and every left (right) identity of S is an idem- potent. (b) If S is commutative and E(S) 6= ;, then E(S) is a subsemigroup of S. (c) If S has a left identity e, then every idempotent which has a right e-inverse is a left identity. (d) If S has a right identity e, then every idempotent which has a left e-inverse is a right identity. (e) If S has an identity, then the only invertible idempotent of S is the identity. In particular, every group has exactly one idempotent. 1.2 Semigroups 15 Proof. (a) is obvious, (d) is similar to (c), and (e) follows from (c) and (d). We will prove (b) and (c). (b): Suppose S is commutative and E(S) 6= ;. Let x; y 2 E(S). x2 = x and y2 = y. Then (xy)(xy) = x(yx)y = x(xy)y = (xx)(yy) = xy. Hence xy 2 E(S).Thus E(S) is a subsemigroup of S. (c): Let x 2 E(S) and suppose x has a right e-inverse z. Let y 2 S. Then xy = xey = x(xz)y = (xx)zy = (xz)y = ey = y. Hence x is a left identity. A semigroup in which every element is idempotent is called a band. A rect- angular component of a band is a maximal rectangular subsemigroup. As suggested by the name, distinct rectangular components are disjoint. The rectangular components of a band are partially ordered by the following re- lation, [[43], Theorem 1]: P Q if and only if PQ P , equivalently QP P . For any two semigroups S and T , the direct product of S and T is the carte- sian product S T , equipped with the componentwise operation: (a1; b1)(a2; b2) = (a1a2; b1b2): At this point it is reasonable to ask the following question. Given a semigroup S and e 2 E(S), is there a subgroup of S in which e is the identity? There are two canonical answers to this question. We call a subgroup of S maximal if it is maximal with respect to the inclusion order among subgroups of S. De nition 1.2.14. Let S be a semigroup and let e 2 E(S). We de ne H(e) := [ fG j G is a subgroup of S and e 2 Gg: Equivalently, H(e) := S fG j G is a subgroup of S and e is the identity of Gg: Theorem 1.2.15. [[29], Theorem 1.18] Let S be a semigroup and let e 2 E(S). Then feg is the smallest subgroup of S with e as the identity and H(e) is the largest subgroup of S with e as the identity. In addition, H(e) is a maximal subgroup of S. Proof. It is obvious that feg is the smallest subgroup of S with e as the identity and that every subgroup of S with e as the identity is a subgroup of H(e). It is also obvious that any subgroup of S which contains H(e) contains 1.2 Semigroups 16 e and, hence, is contained in H(e). It remains only to show that H(e) is a subgroup of S. In this regard the only property that is not immediately clear is whether H(e) is closed under multiplication. Let a; b 2 H(e). a 2 G1 and b 2 G2 for some subgroups G1 and G2 of S with e as their identity. Let G = f nY i=1 xi j n 2 N and x1; ; xn 2 G1 [G2g: G is obviously closed under multiplication and contains e. Let x = Qn i=1 xi 2 G. Then, y := Qn i=1 x 1 n i+1 2 G and xy = e = yx. It follows that G is a subgroup of S with identity e and ab 2 G. Therefore, ab 2 H(e). Hence H(e) is a subgroup of S. We have just seen that H(e) is a maximal subgroup of S. In fact, as we will see shortly, all maximal subgroups of S are of this form. We will refer to them as maximal groups. Proposition 1.2.16. Let S be a semigroup. Every subgroup of S is contained in a group H(e) for some idempotent e of S. Proof. Let G be a subgroup of S and let e be the identity of G. Then e is an idempotent and G H(e). Corollary 1.2.17. Let S be a semigroup. The maximal subgroups of S are exactly the elements of the set fH(e) j e 2 E(S)g. At this point we would like to answer another question arising from what we have seen before. Let S be a semigroup with a left identity e such that every element of S has a right e-inverse. Is S necessarily a group? The answer is negative, but we will see that S has a special structure. Example 1.2.18. Let S be a right zero semigroup. The statement: xy = y for every x; y 2 S, implies that every element of S is a left identity and for every x; y 2 S, y is a right y-inverse of x. Now x e 2 S (as a left identity). Let x 2 S. If x 6= e, then x has no left inverse since for every y 2 S, yx = x 6= e. Hence, if jSj > 1, then S is not a group. This provides a counterexample to our question. It follows from this example that a right zero semigroup S has a unique left identity if and only if S is a singleton, if and only if S is a group. We now show that the above example is essentially the only instance of a semigroup with a left identity e such that every element has a right e-inverse. First we need the following de nition. 1.2 Semigroups 17 De nition 1.2.19. Let S and T be semigroups and let f : S ! T be a func- tion. f is called a semigroup homomorphism if f(xy) = f(x)f(y) for every x; y 2 S. A semigroup isomorphism is a bijective semigroup homomorphism. Two semigroups are isomorphic if there is a semigroup isomorphism between them. If S and T are isomorphic, we write S T . Theorem 1.2.20. [[29], Theorem 1.40] Let S be a semigroup with a left identity e such that every element of S has a right e-inverse. Then E(S) is a right zero semigroup, Se is a group and S = Se E(S) Se E(S). Proof. We know that if an idempotent has a right e-inverse, then it is a left identity (see Remark 1.2.13 (c)). Therefore the right zero property xy = y is satis ed in E(S). (In fact, xy = y for every x 2 E(S) and every y 2 S). Let x; y 2 E(S). Then, (xy)(xy) = yy = y = xy. Therefore E(S) is closed with respect to multiplication. It follows that E(S) is a right zero semi- group. Since e is an idempotent, e is a right identity for Se. Let x; y 2 S. (xe)(ye) = (xey)e = xye 2 Se. Therefore, Se is closed under multiplication. Let x 2 S. Let y be a right e-inverse of x. Then, (xe)(ye) = x(ey)e = xye = (xy)e = ee = e. It follows that Se is a group. Now de ne ? : Se E(S)! S by ?(g; y) = gy. Let (g1; y1); (g2; y2) 2 Se E(S). Then, ?(g1g2; y1y2) = g1g2y1y2 = g1g2y2 = g1(y1g2)y2 = (g1y1)(g2y2) = ?(g1; y1)?(g2; y2). There- fore, ? is a homomorphism. Let a 2 S and let b be the inverse of ae in Se. Then, baba = ba(eb)a = b(aeb)a = bea = ba. Therefore ba 2 E(S). We will show that ? 1(a) = f(ae; ba)g, thus, showing that ? is bijective. ?(ae; ba) = (ae)(ba) = (aeb)a = ea = a. Therefore, ? is surjective and, thus, S = Se E(S). Now let (g; y) 2 Se E(S) such that gy = a. Then, g = ge = gye = ae and y = ey = baey = bgyey = bgyy = bgy = ba. Therefore, ? is injective. It follows that ? is bijective. (Note that we have already seen that if every idempotent is a left identity, then E(S) is a right zero semigroup. Thus, the proof of Theorem 1.2.20 can be shortened.) Corollary 1.2.21. Any semigroup S with a unique left identity e such that every element of S has a right e-inverse is a group. Proof. In Theorem 1.2.20, if e is the only left identity of S, then E(S) = feg and, hence, S Se feg is a group. De nition 1.2.22. Let S be a semigroup and let a 2 S. We say that a is left (right) cancelable if whenever x; y 2 S are such that ax = ay (respectively, 1.2 Semigroups 18 xa = ya), then x = y. S is left (right) cancellative if every element of S is left (respectively, right) cancelable. S is cancellative if it is both left cancellative and right cancellative. Remark 1.2.23. Let S be a semigroup and let e 2 E(S). If S is left (right) cancellative, then e is a left (right) identity of S. If S is cancellative, then e is an identity of S. Proof. Suppose S is left cancellative. Let x 2 S. Since ex = eex, x = ex. Hence e is a left identity. Similarly, suppose S is right cancellative. Let x 2 S. Since xe = xee, x = xe. Hence e is a right identity. This su ces. Remark 1.2.24. Let S and T be semigroups and let N be a subsemi- group of S and M a subsemigroup of T . Let f : S ! T be a semigroup homomorphism. We have the following. (a) If f 1(M) 6= ;, then f 1(M) is a subsemigroup of S. (b) f(N) is a subsemigroup of T . In particular, the image of any idempo- tent under a semigroup homomorphism is an idempotent. (c) If N has an identity e, then f(e) is an identity for f(N). (d) If N has an identity, then f(x 1) = f(x) 1 for every x 2 I(N). We have the following observation. Proposition 1.2.25. A semigroup is a rectangular band if and only if it is a direct product of a left zero semigroup and a right zero semigroup. Remark 1.2.26. Let S be a semigroup with an idempotent a and let T be a semigroup. Then, there exists a semigroup homomorphism from S to T if and only if T has an idempotent. Proof. ): (This is actually part of Remark 1.2.24 (c).) Let f : S ! T be a semigroup homomorphism and let b := f(a). Then b2 = f(a)f(a) = f(aa) = f(a) = b. Therefore b is an idempotent. (: Suppose T has an idempotent b. De ne f : S ! T by f(x) := b for every x 2 S. Then for every x; y 2 S, f(x)f(y) = bb = b = f(xy). Therefore f is a semigroup homomorphism. 1.2 Semigroups 19 De nition 1.2.27. Let S be a semigroup and let I be a nonempty subset of S. I is called a left ideal of S if SI I. I is called a right ideal of S if IS I. I is called an ideal (or a two sided ideal) of S if I is a left ideal and a right ideal of S. Example 1.2.28. (a) For any a 2 S, Sa is a left ideal of S, aS is a right ideal of S and SaS is an ideal of S. More generally, we have the following. (b) Let L be a left ideal of S, let R be a right ideal of S and let a 2 S. Then, La is a left ideal of S, aR is a right ideal of S and LaR is an ideal of S. Remark 1.2.29. Let S be a semigroup, let L S and let a 2 L. (a) If L is a left ideal of S, then Sa L. (b) If L is a right ideal of S, then aS L. (c) If L is an ideal of S, then SaS L. Note that the relation \ T is a left ideal of L" among the subsemigroups of a semigroup S is not transitive. However, if L and T are left ideals of S such that T L, then T is a left ideal of L. These facts hold for the right ideals and ideals as well. Proposition 1.2.30. Let (J ) 2 be a nonempty family of ideals of a semi- group S. De ne J := T 2 J . If J 6= ;, then J is an ideal of S. Proof. The statement is obvious. Corollary 1.2.31. Let S be a semigroup and let ; 6= A S. Consider A := fI S j I is an ideal and A Ig and hAi := \ A: Then, hAi is the smallest ideal of S containing A. We call hAi the ideal generated by A. For any a 2 S, we call hfagi the principal ideal generated by a. In a similar way we obtain the left ideal generated by A and the right ideal generated by A. Note that for any a 2 S, the principal left ideal generated by a 1.3 The structure of a completely simple semigroup 20 is Sa [ fag, the principal right ideal generated by a is aS [ fag and the principal ideal generated by a is Sa [ aS [ SaS [ fag. If S is a monoid, then Sa [ aS [ SaS [ fag = SaS. If a 2 E(S), then a 2 Sa and a 2 aS, in particular, Sa [ fag = Sa and aS [ fag = aS. Let I be a left ideal of a semigroup S. I is said to be a proper left ideal of S if I 6= S. Similarly, we de ne a proper right ideal and a proper ideal of S. 1.3 The structure of a completely simple semi- group De nition 1.3.1. Let S be a semigroup, let L be a left ideal of S, let R be a right ideal of S, and let I be an ideal of S. (a) L is a minimal left ideal of S if whenever J L for some left ideal J of S, then J = L. R is a minimal right ideal of S if whenever J R for some right ideal J of S, then R = J . (b) S is left simple if S has no proper left ideal. S is right simple if S has no proper right ideal. S is simple if S has no proper ideal. If S has a zero 0, then S is called 0-simple if its only ideals are S itself and f0g, and S2 6= f0g. (c) I is a minimal ideal of S if whenever J I for some ideal J of S, then I = J . As we will see later in Theorem 1.3.21, a semigroup S can have at most one minimal ideal and, as mentioned in the introduction, it is denoted by K(S). Remark 1.3.2. Let S be a semigroup and let L and J be left ideals of S and let R be a right ideal of S. We have the following. (i) S(L\ J) L\ J . In particular, if L\ J 6= ;, then L\ J is a left ideal of S. (ii) LR is an ideal of S. (iii) RL L \R. In particular, L \R 6= ;. Note that every left, right or two sided ideal of S is a subsemigroup of S but the converse is false. 1.3 The structure of a completely simple semigroup 21 Remark 1.3.3. Let S be a semigroup and let e 2 E(S). (a) Se = fx 2 S j xe = xg (b) eS = fx 2 S j ex = xg (c) eSe = fx 2 S j xe = x = exg In particular, e is a left identity for eS, a right identity for Se and an identity for eSe. Note that in this case Se is a subsemigroup of S with a right identity, eS is a subsemigroup of S with a left identity and eSe is a monoid. If S is left simple then e is a right identity for S and, if S is right simple, then e is a left identity for S. Remark 1.3.4. Let S be a semigroup and let e 2 E(S). Se is the smallest left ideal of S containing e. Proof. Since e = ee, e 2 Se. Now let L be a left ideal of S with e 2 L. Then, Se SL L. De nition 1.3.5. Let S be a semigroup. We de ne the relations L; R and on E(S) as follows. (1) f L e if and only if f = fe. (2) f R e if and only if f = ef . (3) f e if and only if f L e and f R e. Proposition 1.3.6. Let S be a semigroup and let e; f 2 E(S). (a) f L e if and only if for any left ideal L with e 2 L, f 2 L. (b) f R e if and only if for any right ideal R with e 2 R, f 2 R. (c) f e if and only if for any ideal I with e 2 I, f 2 I. Proof. (a) Suppose f = fe and let L be a left ideal with e 2 L. Then f = fe 2 SL L. Conversely, suppose for any left ideal L with e 2 L, f 2 L. Take L = fx 2 S : x = xeg. Since ee = e, e 2 L. Thus L 6= ;. Let x 2 L and let a 2 S. Then ax = a(xe) = (ax)e. Therefore ax 2 L. It follows that L is a left ideal. Since e 2 L, f 2 L; i.e f = fe. Thus f L e. (b) This is similar to (a). 1.3 The structure of a completely simple semigroup 22 (c) This follows easily from (a) and (b), using the fact that an intersection of a left ideal and a right ideal is nonempty, and it contains an ideal (see Remark 1.3.2 (iii)). Remark 1.3.7. Let S be a semigroup, let L be a minimal left ideal of S and let e; f 2 E(S). If e; f 2 L, then Se = L = Sf . In particular, if e; f 2 L, then e L f and f L e. The converse is also true but obvious. Proof. Suppose e; f 2 L. Then Se SL L. Since Se is a left ideal and L is a minimal left ideal, L = Se. Similarly, Sf = L. Now let J be a left ideal with f 2 J . Then Se = L = Sf J . Since e = ee 2 Se, e 2 J . Hence e L f . Similarly, f L e. Remark 1.3.8. Let S be a semigroup. The relations L; R and on E(S) are re exive and transitive. In addition, is antisymmetric. De nition 1.3.9. Let S be a semigroup, let be a relation on E(S) and let e 2 E(S). We say that e is minimal with respect to if and only if for every f 2 E(S), if f e then e f . Proposition 1.3.10. [[29], Theorem 1.36] Let S be a semigroup and let e 2 E(S). Then e is minimal with respect to L if and only if e is minimal with respect to R if and only if e is minimal with respect to . Proof. Suppose e is minimal with respect to L and let f e. Then f L e and f R e. Since e is minimal with respect to L, e L f . We obtain: e = ef = e(ef) = (ee)f = ef = f . Hence f = e. Hence e is minimal with respect to . Now suppose e is minimal with respect to and let f L e, i.e. f = fe. Let g = ef . Then gg = efef = e(fe)f = eff = ef = g, i.e. g 2 E(S). Also g = ef = efe. Therefore ge = efee = efe = g = eefe = eg. So, g e. Hence g = e; i.e. ef = g = e. Thus e L f . Therefore that e is minimal with respect to L. It follows that e is minimal with respect to if and only if e is minimal with respect to L. By a left-right switch in the above arguments we obtain: e is minimal with respect to if and only if e is minimal with respect to R. We will say that an idempotent is minimal if it is minimal with respect to any of the three relations de ned above. As a corollary to Remark 1.3.3 and Proposition 1.3.6, we have the following. Corollary 1.3.11. Let S be a semigroup and let e 2 E(S). If e 2 L for some minimal left ideal L of S (equivalently, if Se is a minimal left ideal of S), then e is a minimal idempotent. 1.3 The structure of a completely simple semigroup 23 Proof. Suppose Se is a minimal left ideal of S. Let f 2 E(S) such that f L e. Then f 2 Se. Since Se is minimal, Sf = Se. Then, ef = e and, thus, e L f . It follows that e is minimal. De nition 1.3.12. A semigroup S is called completely simple if S is simple and contains a minimal idempotent. A semigroup S with a zero 0 is called completely 0-simple if S is 0-simple and has a minimal idempotent di erent from 0. Theorem 1.3.13. Let S be a semigroup and let L S be a left ideal. L is a minimal left ideal of S if and only if Sa = L for every a 2 L. Proof. Suppose L is a minimal left ideal. Let a 2 L. Since Sa is a left ideal and Sa L, Sa = L. Conversely, suppose Sa = L for every a 2 L. Let J S be a left ideal with J L. Let a 2 J . Then L = Sa J . Hence J = L. It follows that L is a minimal left ideal. Corollary 1.3.14. Let S be a semigroup. Minimal left ideals of S are exactly left ideals of S of the form Sa, for some a 2 S, satisfying the property: for every b 2 S, Sba = Sa. Proof. Let L be a minimal left ideal of S and let a 2 L. By Theorem 1.3.13, L = Sa. Let b 2 S. Then ba 2 L. Therefore, Sba SL L. Since Sba is a left ideal of S, Sba = L. Conversely, let a 2 S such that Sba = Sa for every b 2 S. Let x 2 Sa. x = ba for some b 2 S. Then Sx = Sba = Sa. By Theorem 1.3.13, Sa is a minimal left ideal of S. Note that in a right zero semigroup every left ideal of the form Sa is minimal and it is a singleton. Since left ideals, right ideals and ideals are semigroups, we can apply to them what we already know about semigroups. Theorem 1.3.15. [[29], Theorem 1.42] Let S be a semigroup and assume that there is a minimal left ideal L of S which has an idempotent e. Then X := E(L) is a left zero semigroup, G := eL = eSe is a group and L = XG = X G. Proof. Since Le = L, e is a right identity for L. To see that eL = eSe, let x 2 L. Then ex = e(xe) 2 eSe. Therefore, eL eSe. Conversely, let x 2 S. Then xe 2 L and, therefore, exe 2 eL. Thus eSe eL. Let x 2 L. Since Lx = L, there exists y 2 L such that yx = e. Thus, L is a semigroup with a right identity e and every element of L has a left e-inverse. Therefore, by the right-left switch, Theorem 1.2.20 is applicable. 1.3 The structure of a completely simple semigroup 24 It is routine to show that maximal subgroups of X G are of the form fxg G, where x 2 X. Proposition 1.3.16. Let S be a semigroup and let L be a left ideal of S. L is a minimal left ideal of S if and only if L is left simple. Proof. ): Suppose L is a minimal left ideal of S and let T be a left ideal of L. Let t 2 T . Then Lt is a left ideal of S, Lt T L. Therefore, L = Lt T . It follows that L = T . Hence L is left simple. (: Suppose L is left simple. Let T be a left ideal of S with T L. Then T is a left ideal of L. Hence L = T . It follows that L is a minimal left ideal of S. Recall that we remarked that in a given semigroup S, a left ideal of a small subsemigroup is not necessarily a left ideal of a bigger subsemigroup. For example, a subsemigroup of S is a left ideal of itself but not necessarily a left ideal of S. However the following proposition gives a su cient condition for a left ideal of a small subsemigroup to be a left ideal of a bigger subsemigroup. Proposition 1.3.17. Let S be a semigroup, let L be a left ideal of S and let T be a minimal left ideal of L. Then T is a left ideal of S. Proof. Let t 2 T . Then Lt is a left ideal of L and Lt T . Since T is minimal, Lt = T . Then, ST = S(Lt) = (SL)t Lt T . Hence T is a left ideal of S. Lemma 1.3.18. Let S be a semigroup and let L be a minimal left ideal of S. Then, L is contained in every ideal of S. Proof. Let I be an ideal of S. Then IL is a left ideal of S and IL I \ L. Since L is minimal, IL = L. Hence, L = IL I. Theorem 1.3.19. Let S be a semigroup and let L be a minimal left ideal of S. Then the minimal left ideals of S are exactly members of the family fLa j a 2 Sg: Proof. Let a 2 S. We already know that La is a left ideal of S. Now, let T be a left ideal of S with T La. Consider A = fs 2 L j sa 2 Tg. Since T La, A 6= ;. If sa 2 T , then tsa 2 T for any t 2 S. Hence A is a left ideal of S. Since A L and L is minimal, A = L. Thus La T . It follows that La is a minimal left ideal of S. Conversely, let T be a minimal left ideal of S. Let a 2 T . Then, La is a left ideal of S and La T . Therefore, T = La. 1.3 The structure of a completely simple semigroup 25 Corollary 1.3.20. Let S be a semigroup. If S has a minimal left ideal, then every left ideal of S contains a minimal left ideal. Proof. Let L be a minimal left ideal of S and let J be a left ideal of S. Pick a 2 J . Then La is a minimal left ideal of S and La J . Theorem 1.3.21. [[29], Theorem 1.51] Let S be a semigroup. If S has a minimal left ideal, then K(S) exists and K(S) = S fL : L is a minimal left ideal of Sg. Proof. Let I = S fL : L is a minimal left ideal of Sg. Then, by Lemma 1.3.18, I J for every ideal J of S. Therefore it su ces to show that I is an ideal of S. By assumption I 6= ;. Let x 2 I and let a 2 S. Pick a minimal left ideal L of S such that x 2 L. Then ax 2 L I. Also, by Theorem 1.3.18, La is a minimal left ideal of S. Therefore, La I. It follows that I is an ideal of S. Similarly, one can show that if K(S) exists, then K(S) = [ fR : R is a minimal right ideal of Sg: Some examples of semigroups which do not have a smallest ideal are (N;+) and (N; :). We note that any two distinct minimal left (right) ideals are disjoint, see Remark 1.3.2 (i). We also note that the semigroups of the form K(S) are exactly the completely simple semigroups. Therefore we have the following. Corollary 1.3.22. Every completely simple semigroup is a disjoint union of its minimal left ideals and a disjoint union of its minimal right ideals. Lemma 1.3.23. Let S be a semigroup, let L be a left ideal of S, and let I be an ideal of S. (1) L is minimal if and only if Lx = L for every x 2 L. (2) I is the smallest ideal of S if and only if IxI = I for every x 2 I. Proof. (1) Suppose L is minimal. Let x 2 L. Then, by Theorem 1.3.19, Lx is a minimal left ideal of S. Since L is a left ideal, Lx L. It follows that 1.3 The structure of a completely simple semigroup 26 L = Lx. Conversely assume that Lx = L for every x 2 L. Let J be a left ideal of S with J L. Let y 2 J . Then L = Ly LJ J . Thus L = J . It follows that L is minimal. (2) Suppose I is the smallest ideal of S. Let x 2 I. Since I is a left ideal of S, IxI I. Since IxI is an ideal of S, it follows that IxI = I. Conversely suppose IxI = I for every x 2 I. Let J be an ideal of S, with J I. Let y 2 J . Then I = IyI IJI J . Thus I = J . It follows that I is the smallest ideal of S. Theorem 1.3.24. [[29], Theorem 1.53] Let S be a semigroup. If L is a minimal left ideal of S and R is a minimal right ideal of S, then K(S) = LR. Proof. It is clear that LR is an ideal of S. Let x 2 LR. By Lemma 1.3.23, it su ces to show that LRxLR = LR. Now LRxL is a left ideal of S which is contained in L. So LRxL = L and, thus, LRxLR = LR. Theorem 1.3.25. [[29], Theorem 1.54] Let S be a semigroup and assume that K(S) exists and e 2 E(S). The following statements are equivalent. (a) e 2 K(S). (b) K(S) = SeS. Proof. (a) ) (b): Since SeS is an ideal, then K(S) SeS. Since, by assumption, e 2 K(S), then SeS K(S). (b) ) (a): By assumption, e = eee 2 SeS = K(S). Thus, e 2 K(S) as required. Theorem 1.3.26. [[29], Theorem 1.56] Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Then every minimal left ideal of S has an idempotent. Proof. Let L be a minimal left ideal with an idempotent e and let J be a minimal left ideal. J = La for some a 2 S. Then, by Theorem 1.3.15, eL = eSe is a group. Let b 2 L such that b = ebe is the inverse of eae in eSe, i.e. (eae)(ebe) = e = (ebe)(eae). Then (ba)(ba) = (be)a(eb)a = b(eae)ba = ba. Hence ba is an idempotent of J . Let S be a semigroup and let e 2 E(S). It can be shown that Se is left simple if and only if eS is right simple, see for example [[29], Theorem 1.48]. 1.3 The structure of a completely simple semigroup 27 Lemma 1.3.27. [[29], Theorem 1.57] Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Then there is a minimal right ideal of S which has an idempotent. Proof. Let L be a minimal left ideal of S with an idempotent e. Then Se = L is a minimal left ideal of S and, thus, eS is a minimal right ideal of S and e is an idempotent of eS. Theorem 1.3.28 ([29], Theorem 1.58). Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Let T S. (1) T is a minimal left ideal of S if and only if there is some e 2 E(K(S)) such that T = Se. (2) T is a minimal right ideal of S if and only if there is some e 2 E(K(S)) such that T = eS. Proof. Let L be a minimal left ideal with an idempotent f 2 L. (1) Necessity. Since Sf is a left ideal contained in L, Sf = L. Therefore, by Theorem 1.3.15 fSf is a group with identity f . Pick any a 2 T . Then faf 2 fSf . Pick x 2 fSf such that x(faf) = f . Then xaxa = (xf)a(fx)a = (xfaf)xa = fxa = xa. Hence xa is idempotent. Since xa 2 T K(S), xa 2 E(K(S)). Since Sxa is a left ideal with Sxa T and T is a minimal left ideal, Sxa = T . Su ciency. Since e 2 K(S), there is a minimal left ideal I of S with e 2 I. Then T = Se = I. Therefore T is a minimal left ideal of S. (2) By Lemma 1.3.27, there is a minimal right ideal of S which has an idempotent. The result then follows from (a) by left-right switch. The following theorem summarizes some of the results we have so far in this section. Theorem 1.3.29. [[29], Theorem 1.59] Let S be a semigroup, assume that there is a minimal left ideal of S which has an idempotent, and let e 2 E(S). The following statements are equivalent. (1) Se is a minimal left ideal. (2) Se is left simple. (3) eSe is a group. 1.3 The structure of a completely simple semigroup 28 (4) eSe = H(e). (5) eS is a minimal right ideal. (6) eS is right simple (7) e is a minimal idempotent. (8) e 2 K(S). (9) K(S) = SeS. Proof. See [[29], Theorem 1.59] Theorem 1.3.30. [[29], Theorem 1.60] Let S be a semigroup, assume that there is a minimal left ideal of S which has an idempotent, and let e be an idempotent in S. There is a minimal idempotent f of S such that f e. Proof. Se is a left ideal of S. Pick a minimal left ideal L of S and an idempotent g such that g 2 L Se. Since e is a right identity for Se, g = ge. Let f = eg. Then ff = egeg = egg = eg = f . Thus f is an idempotent. In addition, ef = eeg = eg = f and fe = ege = eg = f . Hence f e. Theorem 1.3.31. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Given any minimal left ideal L of S and any minimal right ideal R of S, there is an idempotent e 2 R \ L such that R \ L = RL = eSe and eSe is a group. Proof. See [[29], Theorem 1.61] Lemma 1.3.32. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Then all minimal left ideals of S are isomorphic. Proof. See [[29], Lemma 1.62] De nition 1.3.33. Let G be a group. Let I and be non-empty sets and let P be a matrix with entries in the 0-group G0 which is ?regular? in the sense that each row and each column contains at least one non-zero entry. The I Rees matrix semigroup over G0 with sandwich matrix P, which 1.3 The structure of a completely simple semigroup 29 we denote by M0(G; I; ; P ), is the set I G [ f0g, together with the binary operation given by: 0(i; g; ) = 0 = (i; g; )0 and (i; g; )(j; h; ) = 8 < : (i; gp jh; ) if p j 6= 0 0 if p j = 0; for every (i; g; ); (j; h; ) 2 I G . We de ne a Rees matrix semigroup without 0 simply by deleting all reference to 0 in the above de nition, dropping the insistence on P being regular and de ning multiplication simply by (i; g; )(j; h; ) = (i; gp jh; ). Theorem 1.3.34. Let S be a Rees matrix semigroup over a 0-group with a regular sandwich matrix. Then S is completely 0-simple. Conversely, every completely 0-simple semigroup is isomorphic to such a Rees matrix semi- group. Proof. See [33]. For clarity, we state the completely simple version of this theorem separately. Theorem 1.3.35. Let S be a Rees matrix semigroup without 0. Then S is completely simple. Conversely, every completely simple semigroup is isomor- phic to such a Rees matrix semigroup. Proof. See [33]. Theorem 1.3.36. Let X be a left zero semigroup, let Y be a right zero semigroup, and let G be a group. Let e be the identity of G, x u 2 X and v 2 Y and let [ ; ] : Y X ! G be a function such that [y; u] = [v; x] = e for all y 2 Y and all x 2 X. Let S = X G Y and de ne an operation on S by (x; g; y) (x0; g0; y0) = (x; g[y; x0]g0; y0). Then S is a simple semigroup (so that K(S) = S = X G Y ) and each of the following statements holds. (a) Idempotents of S are exactly elements of the form (x; [y; x] 1; y), where (x; y) 2 X Y . In particular, the idempotents in X G fvg are of the form (x; e; v) and the idempotents in fug G Y are of the form (u; e; y). 1.3 The structure of a completely simple semigroup 30 (b) For every y 2 Y , X G fyg is a minimal left ideal of S and all minimal left ideals of S are of this form. (c) For every x 2 X, fxg G Y is a minimal right ideal of S and all minimal right ideals of S are of this form. (d) For every (x; y) 2 X Y , fxg G fyg is a maximal group in S and all maximal groups in S are of this form. (e) The minimal left ideal X G fvg is the direct product of X, G and fvg and the minimal right ideal fug G Y is the direct product of fug; G and Y . (f) All maximal groups in S are isomorphic to G. (g) All minimal left ideals of S are isomorphic to X G and all minimal right ideals of S are isomorphic to G Y . Proof. See [[29], Theorem 1.63] Therefore, we have the following result. Theorem 1.3.37. (The Structure Theorem) Let S be a semigroup and as- sume that there is a minimal left ideal of S which has an idempotent. Let R be a minimal right ideal of S, let L be a minimal left ideal of S, let X = E(L), let Y = E(R), and let G = RL. De ne an operation on X G Y by (x; g; y) (x0; g0; y0) = (x; gyx0g0; y0). Then X G Y satis es the conclusions of Theorem 1.3.36 (where [y; x] = yx) and K(S) X G Y . In particular, (1) The minimal right ideals of S partition K(S) and the minimal left ideals of S partition K(S). (2) The maximal groups in K(S) partition K(S). (3) All minimal right ideals of S are isomorphic and all minimal left ideals of S are isomorphic. (4) All maximal groups in K(S) are isomorphic. Proof. See [[29], Theorem 1.64] Theorem 1.3.38. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Let T be a subsemigroup of S and assume also that T has a minimal left ideal with an idempotent. If K(S)\T 6= ;, then K(T ) = K(S) \ T . 1.4 Right topological semigroups 31 Proof. See [[29], Theorem 1.65] Theorem 1.3.39. Let S be a semigroup and assume that there is a minimal left ideal of S which has an idempotent. Let e; f 2 E(K(S)). If g is the inverse of efe in eSe, then the function : eSe ! fSf de ned by (x) = fxgf is an isomorphism. Proof. See [[29], Theorem 1.66] 1.4 Right topological semigroups De nition 1.4.1. Let S be a semigroup. Each a 2 S gives two functions a; a : S ! S de ned as follows. a(x) := ax ; a(x) := xa for every x 2 S: a is called the left shift (by a) and a is called the right shift (by a). Shifts are also called translations. De nition 1.4.2. Let S be a semigroup and let T be a topology on S. (S; T ) is called a right topological semigroup if for every a 2 S, the right shift a is continuous. (S; T ) is called a left topological semigroup if for every a 2 S, the left shift a is continuous. We often write S instead of (S; T ) if it is understood which topology we are referring to. Remark 1.4.3. Let S be a semigroup. (a) If S is commutative, then a = a for every a 2 S. (b) For any a 2 S and U S, we will often write a 1U for 1a (U). Thus a 1U = fx 2 S j ax 2 Ug: One should note that in this case a 1 is simply a notation, it is not necessarily the inverse of a. In fact the inverse of a need not exist. Similarly, we will often write Ua 1 for 1a (U). (c) Let a 2 S and let U S. If a is invertible, then we have the following. 1a (U) = a 1U = fx 2 S j ax 2 Ug = fa 1u j u 2 Ug = a 1U = a 1(U) 1.4 Right topological semigroups 32 where a 1 is the last three expressions stands for the inverse of a. Thus 1a = a 1 and our notation in (b) is consistent with the invertible case. Similarly, if a is invertible, then 1a = a 1 . (d) Let S be a left (respectively, a right) topological semigroup. If S is a group, then we also call S a left (respectively, a right) topological group. De nition 1.4.4. Let T be a right topological semigroup. The topological center of T , denoted (T ), is de ned as (T ) := fa 2 T j a is continuousg: Similarly, one de nes the topological center of a left topological semigroup. Remark 1.4.5. (a) Since for an invertible a, 1a = a 1 , a is continuous if and only if a 1 is open. Similarly, a is continuous if and only if a 1 is open. Thus we have the following. (b) A group G (equipped with a topology) is right (respectively, left) topo- logical if and only if for each a 2 G, a (respectively, a) is open. (c) Let T be a right topological semigroup. If (T ) 6= ;, then (T ) is a subsemigroup of T . Of particular importance in our study are compact right topological semi- groups. Theorem 1.4.6. (Ellis?s Theorem)[[29], Theorem 2.5] Every compact Haus- dor right topological semigroup has an idempotent. Proof. Let T be a compact Hausdor right topological semigroup. Let A = fA T : AA A; A 6= ;; and A is compactg; i.e. A is the family of all compact subsemigroups on T . Since T 2 A, A 6= ;. Let C be a chain in A. Let x; y 2 T C. Then xy 2 C for every C 2 C. Therefore, xy 2 T C. Thus ( T C) ( T C) T C. Since C is a chain of nonempty sets, C has the nite intersection property. Since T is compact, T C 6= ;. Since T C is closed and T is compact, T C is compact. It follows that T C is a lower bound for C in A. By Zorn?s Lemma, A has a minimal element A. Let x 2 A and let B = Ax. Then B 6= ; and B = x(A). Since 1.4 Right topological semigroups 33 x : T ! T is continuous and A is compact, B is compact. In addition, BB = AxAx AAAx AAx Ax = B. Therefore B 2 A. Since B = Ax AA A and A is minimal in A, B = A. Hence A = Ax for every x 2 A. Let D = fy 2 A : yx = xg. Since x 2 A and Ax = A, x = yx for some y 2 A. In particular, D 6= ;. Since D = A \ 1x (fxg), D is compact. For any y; z 2 D, (yz)x = y(zx) = yx = x. Therefore, DD D. It follows that D 2 A. Since D A, D = A. Thus Ax = fxg. In particular, xx = x. Hence x is an idempotent of T . Corollary 1.4.7. Let T be a compact Hausdor right topological semigroup. Every left ideal of T contains a minimal left ideal. Minimal left ideals are closed, and each minimal left ideal has an idempotent. Theorem 1.4.8. [[29], Theorem 2.7] Every compact Hausdor right topo- logical semigroup has a smallest ideal. Proof. This follows from Corollary 1.4.7 and Theorem 1.3.21. Theorem 1.4.9. [[29], Theorem 2.9] Let T be a compact right topological semigroup and let e be an idempotent in T . Then e is minimal with respect to any (and hence to all) of the orders R; L or if and only if e belongs to the smallest ideal of T . Proof. We know that K(T ) = S fL : L is a minimal left ideal of Tg. Necessity. Since Te is a left ideal of T , there exists a minimal left ideal L of T with L Te. By Corollary 1.4.7, L is closed. Therefore there exists an idempotent f 2 L. Since f 2 Te, f = ge for some g 2 T . Then fe = gee = ge = f . Thus, f L e and, hence, f = e. It follows hat e 2 K(T ). Su ciency. This follows from Corollary 1.3.11. Theorem 1.4.10. Let T be a compact right topological semigroup and let e be an idempotent in T . There is a minimal idempotent f such that f e. Proof. As in the proof of \necessity" of Theorem 1.4.9, pick a minimal left ideal L of T with L Te and pick an idempotent f 2 L. Then f = fe. Take r = ef . Then r 2 L and we have rr = efef = eff = ef = r. Therefore r is a minimal idempotent. In addition, er = eef = ef = r and re = efe = ef = r, so r e. Theorem 1.4.11. Let T be a right topological semigroup and let R be a right ideal of T . Then c?(R) is a right ideal of T . 1.4 Right topological semigroups 34 Proof. Let x 2 c?(R), let a 2 T and let U be an open neighborhood of xa. Since a is continuous, there exists an open neighborhood V of x such that V a U . Since x 2 c?(R), V \ R 6= ;. Pick y 2 V \ R. Then, ya 2 V a \ Ra U \ R. Therefore U \ R 6= ;. Thus xa 2 c?(R). It follows that c?(R) is a right ideal of T . Remark 1.4.12. Let T be a right topological semigroup and let L be a left ideal of T . The closure of L need not be a left ideal, as illustrated by the following example. Example [[4], Example V.1.1]: Let X be a compact Hausdor space with a non-closed subset D X such that c?D 6= X. De ne a binary operation on X as follows. xy = y if y 2 D; x if y =2 D: Let x; y; z 2 X. If z =2 D, then (xy)z = xy = x(yz). If z 2 D, then (xy)z = z = xz = x(yz). Therefore, X with this binary operation is a semigroup. Let x 2 X and let U be an open subset of X. If x =2 D, then 1x (U) = U . If x 2 U \ D, then 1 x (U) = X. If x 2 D \ (X n U), then 1x (U) = ;. Therefore x : X ! X is continuous. Thus X with this binary operation is a right topological semigroup. In addition (T ) = ;. Since XD = D, D is a left ideal. Let x 2 D and let y 2 X. If y 2 D, then xy = y 2 D. If y =2 D, then xy = x 2 D. Therefore D is also a right ideal. Hence D is an ideal. Given any ideal J of X, D = JD J . It follows that D is the smallest ideal of X. Let I be a proper ideal of X and suppose I 6= D. Pick x 2 I nD and pick y 2 X n I. Then y = yx 2 I, a contradiction. It follows that I = D. Hence D is the only proper ideal of X. Since D 6= c?(D) 6= X, then c?(D) is not an ideal. By Theorem 1.4.11, c?(D) is a right ideal. Therefore, c?(D) is not a left ideal. However, we have the following theorem. Theorem 1.4.13. Let T be a right topological semigroup such that the topo- logical center (T ) is dense in T . Then, the closure of every left ideal of T is a left ideal of T . Proof. Let L be a left ideal of T , let x 2 c?(L), let a 2 T and let U be an open neighborhood of ax. Then V := Ux 1 is an open neighborhood of a with V x U . Since (T ) is dense in T , there exists b 2 V \ (T ). Then bx 2 U . Since b is continuous, W := b 1U is a neighborhood of x with 1.4 Right topological semigroups 35 bW U . Then W \ L 6= ;. Pick c 2 W \ L. Then, bc 2 bW U and bc 2 L. Therefore, U \ L 6= ;. Hence ax 2 c?(L). It follows that c?(L) is a left ideal of T . De nition 1.4.14. A topological semigroup is a semigroup S equipped with a topology T such that multiplication S S ! S ; (x; y) 7! xy is continuous. De nition 1.4.15. A topological group is a group G equipped with a topol- ogy T such that multiplication and inversion are continuous functions, i.e. the functions G G! G ; (x; y) 7! xy and G! G ; x 7! x 1 are continuous. In particular every topological group is a topological semi- group. If (G; T ) is a topological group, we say that T is a group topology. Note that continuity of the group multiplication implies continuity in each variable separately. Therefore, every topological group is a left(right) topo- logical semigroup. We recall that a topological space X is homogeneous if for every x; y 2 X, there is a homeomorphism f : X ! X such that f(x) = y. Proposition 1.4.16. Every topological group is homogeneous. Proof. Let G be a topological group and let a; b 2 G. Consider the homeo- morphism ab 1 : G! G. Then, ab 1(b) = (ab 1)b = a(b 1b) = a. In the next proposition and in the next theorem the topological groups are assumed to be Hausdor . Proposition 1.4.17. Let G be a topological group and let F be the neigh- borhood lter of the identity e 2 G. For every x 2 G, xF := fxU : U 2 Fg is the neighborhood lter of x. Thus, the topology is completely determined by the neighborhood lter of the identity. 1.4 Right topological semigroups 36 Proof. Let x 2 G. Let U 2 F . There exists an open set V such that e 2 V U . Then xV is open, x = xe 2 xV and xV xU . Hence xU is a neighborhood of x. Conversely, let W be a neighborhood of x. Then x 1W 2 F and W = x(x 1W ). Thus, the rst statement holds. Since a topology is determined by the neighborhood lters at all the points of the space, the second statement holds. The next theorem characterizes the neighborhood lter of the identity of a topological group. Theorem 1.4.18. Let G be a group and let e be the identity of G. Then for every group topology T on G, the neighborhood lter F of e in T satis es the following conditions. (1) T F = feg, (2) for very U 2 F , there exists V 2 F such that V 2 U , (3) for every U 2 F , there exists V 2 F such that V 1 U , (4) for every U 2 F and for every x 2 G, there exists V 2 F such that xV x 1 U . Conversely, for every lter F on G satisfying conditions (1)-(4), there is a unique group topology T on G in which F is the neighborhood lter at e. Proof. Necessity: (1) follows by the Hausdor property. (2) follows from the fact that for every U 2 F , ee 2 U . (3) follows from the fact that for every U 2 F , e 1 2 U . (4) follows from the fact that for every x 2 G, x x 1 is continuous and x x 1(e) = e. Su ciency: Consider the system of lters Fx := xF , x 2 G. We have (i) T Fx = fxg, (ii) for every U 2 Fx, there exists V 2 Fx such that U 2 Fy for all y 2 V . Therefore, (Fx)x2G determines a topology on G with Fx being the neighbor- hood lter at x. 1.5 The Stone- Cech compacti cation of a discrete semigroup 37 1.5 The Stone- Cech compacti cation of a dis- crete semigroup De nition 1.5.1. Let D be a discrete topological space. We de ne the following. (a) D := fp P(D) : p is an ultra lter on Dg. (b) Given A D, A := fp 2 D : A 2 p g. We have a function e : D ! D de ned by e(x) = fA D : x 2 Ag, i.e. e(x) is the principal ultra lter generated by x. It is clear that e is injective. Lemma 1.5.2. [[29], Lemma 3.17] Let D be a discrete topological space. For any A;B D the following statements hold. (1) A \B = A \B. (2) A [B = A [B. (3) D n A = D n A. (4) A = ; if and only if A = ;. (5) A = D if and only if A = D. (6) A = B if and only if A = B. Proof. The statements follow easily from the de nition of A and properties of ultra lters. As a consequence of Lemma 1.5.2 (1), the set B := f A : A D g forms a basis for a topology T on D. From now on, unless otherwise stated, we will always assume that D is equipped with this topology. In addition, for each x 2 D, we will identify e(x) with x. Theorem 1.5.3. [[29], Theorem 3.18] Let D be a discrete topological space. (1) D is a compact Hausdor space. (2) For every A D, A is a clopen subset of D. 1.5 The Stone- Cech compacti cation of a discrete semigroup 38 (3) For every A D, A = c? D e(A) or, which is the same, for any A D and any p 2 D, p 2 c? D e(A) if and only if A 2 p. (4) e(D) is a dense subset of D whose points are precisely the isolated points of D. (5) If U is an open subset of D, then c? D U is also open. Proof. Veri cation of these statements is an easy common routine. See for example [[29], Theorem 3.18.] In particular, it follows that e : D ! D is an embedding. Since we identify e(x) by x, we will often write c? D A instead of c? D e(A), where A D. Sometimes, to avoid confusion, we use eD instead of e. Lemma 1.5.4. Let D be a discrete topological space. For every p 2 D, lim x!p e(x) = p. Proof. The statement to be proven is equivalent to the statement: For any p 2 D, and for any U D, if U is a neighborhood of p then e 1(U) 2 p. So, let U be a neighborhood of p. There exists A D such that p 2 A U , i.e. A 2 p and for any q 2 D if A 2 q then q 2 U . Therefore e(x) 2 U for all x 2 A and, thus, A e 1(U). Since p is a lter, e 1(U) 2 p. Lemma 1.5.5. Let D be a discrete topological space and let f : D ! Y be a function into a compact Hausdor topological space. For every p 2 D, let Ap = fc?Y f(A) : A 2 pg. Then, \ Ap = f lim x!p f(x) g: In particular for every t 2 D, lim x!e(t) f(x) = f(t): Proof. Let p 2 D and let y 2 T Ap. Let W be an open neighborhood of y. Since W \ f(A) 6= ; for all A 2 p, f 1(W ) \ A 6= ; for all A 2 p. Since p is an ultra lter, f 1(W ) 2 p. It follows that y = limx!p f(x). To show thatT Ap 6= ;, it su ces to show that Ap has the nite intersection property. Let A1; ; An 2 p for some n 2 N. We have c?Y f(A1) \ \ c?Y f(An) c?Y (f(A1) \ \ f(An)) c?Y (f(A1 \ \ An)) f(A1 \ \ An) 6= ;. Therefore Ap has the nite intersection property. Since Y is compact, 1.5 The Stone- Cech compacti cation of a discrete semigroup 39 T Ap 6= ;. Hence the equality follows (see also Theorem 1.1.14 (2)). For any t 2 D we have T Ae(t) c?Y f(ftg) = c?Y ff(t)g = ff(t)g. It follows that T Ae(t) = ff(t)g. Theorem 1.5.6. Let D be a discrete topological space. D is the Stone- Cech compacti cation of D. Proof. It remains to show that for any function f from D into a compact Hausdor topological space Y there exists a unique continuous function f : D ! Y such that the diagram D e // f !! C C C C C C C C D f Y commutes, i.e. f e = f . Let Y be a compact Hausdor space and consider a function f : D ! Y . De ne f : D ! Y by f(p) = lim x!p f(x): If follows from Lemma 1.5.5 that f(e(t)) = f(t) for all t 2 D. For continuity of f , let p 2 D and let U be a neighborhood of f(p). Since Y is regular, pick a neighborhood V of f(p) such that c?Y V U and let A = f 1(V ). Suppose D n A 2 p. Then f(p) 2 c?Y f(D n A). Then V \ f(D n A) 6= ;, contradicting the fact that A = f 1(V ). Thus A is a neighborhood of p. Let q 2 A and suppose that f(q) =2 U . Then Y n c?Y V is a neighborhood of f(q) and f(q) 2 c?Y f(A). Then, (Y n c?Y V )\ f(A) 6= ;, again contradicting the fact that A = f 1(V ). Hence f is continuous. The uniqueness of f follows from its de nition. As mentioned earlier, we will identify principal ultra lters on D with points of D, and D := D nD, i.e. the elements of D are the free ultra lters on D. Lemma 1.5.7. Let D and E be discrete topological spaces and let f : D ! E be a function. For every p 2 D, fA E : f 1(A) 2 pg 2 E. If, in addition, f is injective, then for every p 2 D , fA E : f 1(A) 2 pg 2 E . Proof. Let p 2 D and let q = fA E : f 1(A) 2 pg. Since f 1(;) = ; =2 p, ; =2 q. Since f 1(E) = D 2 p, E 2 q. Let A 2 q and let B E with A B. 1.5 The Stone- Cech compacti cation of a discrete semigroup 40 Since f 1(B) f 1(A) 2 p, f 1(B) 2 p. Therefore B 2 q. Let A;B 2 q Since f 1(A \ B) = f 1(A) \ f 1(B) 2 p, A \ B 2 q. It follows that q is a lter on E. Let G be a lter on E such that q G and assume q 6= G. Pick B 2 G n q. Then f 1(B) =2 p. Pick U 2 p such that f 1(B) \ U = ;. Since U f 1(f(U)), f 1(f(U)) 2 p and, therefore, f(U) 2 q. On the other hand f(U)\B = ;. This is not possible since f(U); B 2 G. Therefore q = G. It follows that q 2 E. Now suppose f is injective. Let p 2 D and assume fA E : f 1(A) 2 pg 2 E. Then f 1(B) 2 p for some nite subset B of E. Since f is injective, jf 1(B)j jBj and, therefore, f 1(B) is nite. This contradicts the fact that f 1(B) 2 p 2 D . Therefore, fA E : f 1(A) 2 pg 2 E . Proposition 1.5.8. Let D and E be discrete topological spaces and let f : D ! E E be a function. The continuous extension f : D ! E of eE f : D ! E is given by f(p) = fA E : f 1(A) 2 pg for every p 2 D. Proof. Let g : D ! E, g(p) := fA E : f 1(A) 2 pg. It remains to show that g is a continuous extension of eE f . Let x 2 D. g(eE(x)) = fA E : f 1(A) 2 eE(x)g = fA E : x 2 f 1(A)g = eE(f(x)). Therefore, gjD = eE f . Now to show that g is continuous. Let A E and consider the basic open subset A E. g 1(A) = fp 2 D : g(p) 2 Ag = fp 2 D : A 2 g(p)g = fp 2 D : f 1(A) 2 pg = f 1(A) is open in D. It follows that g is the continuous extension of eE f and, hence, g = f . Note that in Proposition 1.5.8 we used f but, formally, the correct notation is eE f , as one can see from the following diagram. D eD // f D eE f E eE // E Furthermore, since is a functor, one usually use the notation f instead of eE f . Proposition 1.5.9. Let D and E be discrete topological spaces and let f : D ! E E be a function. The continuous extension f : D ! E is injective if f is injective, surjective if f is surjective and a homeomorphism if f is bijective. 1.5 The Stone- Cech compacti cation of a discrete semigroup 41 Proof. Suppose f is injective. Let p; q 2 D with p 6= q. There exists A 2 p n q. Since A = f 1(f(A)) 2 p n q, f(A) 2 f(p) n f(q). Therefore, f(p) 6= f(q). Hence, f is injective. Suppose f is surjective. Let r 2 E and consider p := fA D : f 1(X) A for some X 2 rg: It is easy to see that p is a lter on D and, therefore, p is contained in some ultra lter q on D. Let X 2 r. Then, f 1(X) 2 p q. Therefore, r f(q). Conversely, let X 2 f(q) and let Y 2 r. Then, f 1(X \ Y ) = f 1(X) \ f 1(Y ) 6= ;. Therefore, X \ Y 6= ;. It follows that X 2 r. Thus f(q) r. It follows that f(q) = r. Hence, f is surjective. Now, suppose f is bijective. It remains to show that f 1 is continuous or, equivalently, that f is open. Let A D . It is easy to see that f(A) = f(A), where f(A) f(A), follows from injectivity while f(A) f(A) follows from surjectivity. Therefore, f is open. De nition 1.5.10. Let D be a discrete topological space and let F be a lter on D. We de ne F := fp 2 D : F pg; equivalently F := \ fA : A 2 Fg: F is called the subspace (or subset) induced by F . Remark 1.5.11. Note that \ p2 S p = fSg and, hence, S = fSg: Theorem 1.5.12. Let D be a discrete topological space. Given a lter F on D, F is a nonempty closed subset of D. Conversely, any nonempty closed subset of D can be uniquely represented as F for some lter F on D. In particular, for any two lters F and G on D, G F if and only if F G. Proof. Let F be a lter on D. Since every lter is contained in an ultra lter, F 6= ;. Let q 2 D n F . There exists A 2 F such that A =2 q. Then q 2 D n A and D n A D n F . Thus D n F is open and F is closed. (One can also see that F is closed from F = T fA jA 2 Fg, since X is closed for any X D.) Conversely, let ; 6= F D. Put F = T F. Since F is an intersection of lters, F is a lter. It then follows, from the rst part, that 1.5 The Stone- Cech compacti cation of a discrete semigroup 42 F is closed and, hence, F cl DF. Now, let q 2 F and let A 2 q. Then D n A =2 q. There exists p 2 F such that D n A =2 p. Then A 2 p. Thus p 2 A \ F. Therefore q 2 cl DF. Hence F cl DF. It now follows that F = cl DF. In particular, if F is closed then F = F , i.e. F = T F. For uniqueness let F and G be lters on D with F = G and assume F * G. Pick A 2 F n G. Then G [ fD n Ag U for some U 2 D. But then A 2 F U , a contradiction. It follows that F G. Similarly, G F . Hence F = G. To prove the last part, suppose G F . Then F = T F T G = G. Conversely, suppose F G. Then it is clear that G F . Now we would like to extend a binary operation on a discrete topological space D to D. Theorem 1.5.13. [[29], Theorem 4.1] Let D be a discrete topological space and let be a binary operation on D. There is a binary operation on D satisfying the following conditions. (a) extends ; i.e. s t = s t for all s; t 2 D. (b) For each q 2 D, the function from D into D, p 7! p q is contin- uous. (c) For each s 2 D, the function from D into D, p 7! s p is continuous. Proof. Given s 2 D, s de nes the function s : D ! D D, s(t) = st. By the universal property of D, there exists a unique continuous function ls : D ! D such that ls e = s, i.e. such that the diagram D e // s !!B B B B B B B B D ls D commutes. For s 2 D and p 2 D, we de ne s p := ls(p) = lim t!p st = fA D : s 1A 2 pg: Then for every B D we have (s ( )) 1(B) = l 1s (B) = s 1B: 1.5 The Stone- Cech compacti cation of a discrete semigroup 43 In particular condition (c) is satis ed. Since ls e = s, s t = ls(t) = s(t) = st for every t 2 S. Therefore, condition (a) is satis ed. Since D is dense in D and ls is a continuous extension of s, ls is unique. Now, we would like to extend to D D. Given q 2 D, de ne Rq : D ! D by Rq(s) = s q = ls(q). Then, there is a unique continuous extension rq : D ! D, such that rq e = Rq. D e // Rq !!B B B B B B B B D rq D For p 2 D n D we de ne p q := rq(p). Note that if t 2 D, then rq(t) = Rq(t) = t q. Thus, for all p 2 D, rq(p) = p q. Hence condition (b) is satis ed. By the uniqueness of a continuous extension of a function de ned on a dense subset, rq is unique. Hence we have the required binary operation . For any binary operation on a discrete topological space D we adopt the custom of denoting its extension to D by the same symbol. Remark 1.5.14. Let be a binary operation on a discrete topological space D. (a) For every s 2 D and every q 2 D, sq = lim t!q st. (b) For every p; q 2 D, pq = lim s!p lim t!q st. Proof. For s 2 D and for q 2 D let the functions s, ls Rq and rq be as in the proof of Theorem 1.5.13 (a) Let s 2 D and let q 2 D. D e // s D ls D e // D Let U be an open neighborhood of sq. Since ls is continuous, l 1s (U) is an open neighborhood of q. Therefore, there exists A S such that 1.5 The Stone- Cech compacti cation of a discrete semigroup 44 q 2 A l 1s (U). Then, A 2 q and for any t 2 A, ls(e(t)) 2 U . Since ls e = s e, for any t 2 A, ( s e)(t) 2 U , i.e. st 2 U . Hence A ft 2 D : st 2 Ug. Since A 2 q and q is a lter, ft 2 D : st 2 Ug 2 q. (b) Let p; q 2 D and let U be an open neighborhood of pq. D e // Rq !!B B B B B B B B D rq D Recall that Rq(s) = ls(q). By (a), lim t!q st = sq. Therefore, we would like to show that pq = lim s!p sq. So, let U be an open neighborhood of pq(= rq(p)). Since rq is continuous, there exists A D such that p 2 A rq 1(U). Then A 2 p and for every s 2 A, sq = rq(s) 2 U . Therefore, A fs 2 D : sq 2 Ug. Since p is a lter, fs 2 D : sq 2 Ug 2 p. Statement (b) in Remark 1.5.14 can be generalized as follows. Remark 1.5.15. Let D be a discrete topological space and let p; q 2 D. Let P 2 p and let Q 2 q. The following statement holds. pq = p - lim s2P ( q - lim t2Q st): In particular, for any compact Hausdor topological space Y and for any function f : D ! Y , lim s!p f(y) = p - lim s2D f(s): Proof. Let U be an open neighborhood of pq. We would like to show that fs 2 P : ft 2 Q : st 2 Ug 2 qg 2 p. From the previous remark we obtain the following: ft 2 Q : st 2 Ug = Q \ ft 2 D : st 2 Ug 2 q and fs 2 P : ft 2 Q : st 2 Ug 2 qg = P \ fs 2 D : ft 2 Q : st 2 Ug 2 qg 2 p: Thus the rst statement holds. Now let f : D ! Y be a function into a compact Hausdor space Y , let z = lims!p f(s) and let U be a neighborhood of z. Then f 1(U) 2 p and, therefore, fs 2 B : f(s) 2 Ug = B\f 1(U) 2 p. Thus the second statement follows. 1.5 The Stone- Cech compacti cation of a discrete semigroup 45 As a result of the last two remarks above, we have the following corollary. Corollary 1.5.16. Let D be a discrete topological space endowed with a binary operation and let p; q 2 D. The following statements hold. (i) For every U D, U 2 pq () fs 2 D : ft 2 D : st 2 Ug 2 qg 2 p: (ii) pq has a base consisting of subsets of D of the form [ fxBx : x 2 Ag; where A 2 p and Bx 2 q. Proof. (i) Let U D. Since lim t!q st = sq and, hence, pq = lim s!p lim t!q st = lim s!p sq, sq 2 U () ft 2 D : st 2 Ug 2 q: Therefore, U 2 pq () fs 2 D : sq 2 Ug 2 p () fs 2 D : ft 2 D : st 2 Ug 2 qg 2 p: (ii) Let p; q 2 S, let A 2 p and, for each x 2 A, let Bx 2 q. For each s 2 S de ne Vs by Vs = ft 2 S : st 2 [ x2A xBxg; and let W = fs 2 S : Vs 2 qg: We would like to show that W 2 p. Let s 2 S. Then Bs Vs. Since Bs 2 q, Vs 2 q. Therefore, A W . Since A 2 p, W 2 p. It follows that S x2A xBx 2 pq. Now let U 2 pq and let B 2 q. Then V := fx 2 A : fy 2 B : xy 2 Ug 2 qg 2 p. In particular, V 6= ;. Pick x 2 V and let Bx = fy 2 B : xy 2 Ug. Then, Bx 2 q and xBx U . Hence S x2V xBx U . It follows that the family [ x2A xBx ! A2p form a base for pq. 1.5 The Stone- Cech compacti cation of a discrete semigroup 46 Remark 1.5.17. In Corollary 1.5.16 (i) if we take p to be a principal ultra- lter x, then the statement becomes U 2 xq () ft 2 D : xt 2 Ug 2 q: On the other hand, if we take q to be a principal ultra lter y, then the statement becomes U 2 py () fs 2 D : sy 2 Ug 2 p: Now we would like to look at the extended binary operation on S where S is a semigroup, regarded as a discrete space (we will call such a semigroup a discrete semigroup). So, let S be a discrete semigroup. As we have seen above, the binary operation on S can be extended to S as follows. For p; q 2 S pq = lim s!p lim t!q st; where s; t 2 S. Lemma 1.5.18. [[29], Theorem 4.5] Let S be a discrete semigroup, let f : S ! X be a function into a Hausdor space, and let p; q 2 S. If all limits involved exist, then lim v!pq f(v) = lim s!p lim t!q f(st). Proof. Suppose all limits concerned exist. Let x0 = lim s!p lim t!q f(st) and let U be a neighborhood of x0. Then, fs 2 S : lim t!q f(st) 2 Ug 2 p. On the other hand, lim t!q f(st) 2 U ) ft 2 S : f(st) 2 Ug 2 q , ft 2 S : st 2 f 1(U)g 2 q , ft 2 S : t 2 s 1(f 1(U))g 2 q , s 1(f 1(U)) 2 q: Therefore, fs 2 S : lim t!q f(st) 2 Ug 2 p , fs 2 S : s 1(f 1(U)) 2 qg 2 p , f 1(U) 2 pq: It follows that x0 = lim v!pq f(v). 1.5 The Stone- Cech compacti cation of a discrete semigroup 47 Theorem 1.5.19. Let S be a discrete semigroup. Then, with the extended binary operation, S is a semigroup. Proof. Let p; q; r 2 S. (pq)r = lim x!pq lim y!r xy = lim s!p lim t!q lim y!r (st)y = lim s!p lim t!q lim y!r s(ty) = lim s!p lim v!qr sv = p(qr) For any semigroup S, unless otherwise stated, when we refer to the semigroup S, we are considering the extension of the binary operation of S. Theorem 1.5.20. Let S be a discrete semigroup. Then, S is a compact right topological semigroup, with S contained in its topological center. Proof. It remains to show that S is right topological and that S ( S), but these follow from Theorem 1.5.13 (b) and (c) respectively. Remark 1.5.21. Let S be a discrete semigroup. Let p; q 2 S and let A S. (a) Recall that x 1A = fy 2 S : xy 2 Ag. Therefore, using these notations, we have the following. A 2 pq () fx 2 S : x 1A 2 qg 2 p: In particular for any p 2 S and for any x 2 S, A 2 xp () x 1A 2 p and A 2 px () Ax 1 2 p; where Ax 1 = fy 2 S : yx 2 Ag. 1.5 The Stone- Cech compacti cation of a discrete semigroup 48 (b) A 2 xq for every x 2 S if and only if A 2 rq for every r 2 S. To see this, suppose A 2 xq for every x 2 S. Then x 1A 2 q for every x 2 S. Let r 2 S. Since fx 2 S : x 1A 2 qg = S 2 r, A 2 rq. The converse is obvious. (c) We will use A (p) to denote the set fx 2 A : x 1A 2 pg. Thus, if A (q) 2 p then A 2 pq. In particular, for A 2 p, A (q) 2 p if and only if A 2 pq. (d) It follows that pp = p if and only if fA (p) : A 2 pg p. Lemma 1.5.22. [[29], Lemma 4.14] Let S be a semigroup, let p 2 S such that pp = p, and let A S. For each x 2 A (p), x 1(A (p)) 2 p. Proof. Let x 2 A (p). Then B := x 1A 2 p. Since pp = p, B (p) 2 p. We show that B (p) x 1(A (p)). So let y 2 B (p). Then y 2 B, i.e. xy 2 A, and (xy) 1 = y 1(x 1A) = y 1B 2 p. Therefore, xy 2 A (p) and, hence, y 2 x 1(A (p)). It follows that B (p) x 1(A (p)). Since B (p) 2 p, x 1(A (p)) 2 p. Theorem 1.5.23. Let S be a discrete cancellative semigroup. Then, S is an ideal of S. Proof. Let p 2 S and let q 2 S . Suppose pq 2 S, say pq = fA S : a 2 Ag for some a 2 S. Since fag 2 pq, fs 2 S : s 1a 2 qg 2 p. In particular, fs 2 S : s 1a 2 qg 6= ;. Pick s 2 S such that s 1a 2 q. Then, the equation sb = a has in nitely many solutions. This is a contradiction, since S is cancellative. Therefore pq 2 S . It follows that S is a left ideal of S. Similarly, S is a right ideal of S. It follows that S is an ideal of S. Theorem 1.5.24. [[29], Theorem 4.20] Let S be a discrete semigroup and let A P(S) such that A has the nite intersection property. If for each A 2 A and each x 2 A, there exists Bx 2 A such that xBx A, then \ A2A A is a subsemigroup of S. Proof. Let T = T A2A A. Since A has the nite intersection property, there is p 2 S such that A p. Then p 2 T and, therefore, T 6= ;. Let p; q 2 T 1.6 K( S) and its closure 49 and let A 2 A. By assumption, for each x 2 A there exists Bx 2 A such that xBx A. Since S fxBx : x 2 Ag belongs to a base for pq, A 2 pq. It follows that pq 2 T . Therefore, T is a subsemigroup of S. Example 1.5.25. Let S be a discrete semigroup and let (xi)i2N be a sequence in S satisfying the distinct nite products property. For each m 2 N, let Am = FP ((xi)i m): Since An+1 An and An 6= ; for all n 2 N, the family (An)n2N has the nite intersection property (see page 2). Let m 2 N, let F N be nite with min(F ) m, and let x = i2F xi: Let n 2 N such that n > max (F ). Let E N with min(E) n and let G = F [ E. Then, G is a nite subset of N, min(G) m and x i2E xi = i2F xi i2E xi = i2G xi. Therefore, x An Am. It follows that T := \ m2N c? S FP ((xi)i m) is a subsemigroup of S. 1.6 K( S) and its closure It follows from Theorem 1.4.8 that for any semigroup S, S has a smallest ideal K( S), which is a completely simple semigroup, and that c? K( S) is also an ideal of S. We would like to present the combinatorial characteri- zations of these ideals. De nition 1.6.1. Let S be a semigroup and let A S. (a) A is called syndetic if there exists a nite subset G of S such that for every x 2 S, tx 2 A for some t 2 G, i.e. G 1A = [ t2G t 1A = S: (b) A is called piecewise syndetic if there exists a nite subset G of S such that the family y 1 [ t2G t 1A !! y2S has the nite intersection property. 1.6 K( S) and its closure 50 Remark 1.6.2. Let S be a semigroup and let A S. (1) It is clear that every syndetic subset of S is piecewise syndetic. (2) Let A;G S and let x; y 2 S. x 2 y 1 [ t2G t 1A ! () yx 2 [ t2G t 1A () tyx 2 A for some t 2 G: Therefore, y 1 S t2G t 1A y2S has the nite intersection property if and only if for any nite subset fy1; ; yng of S there exists x 2 S such that x 2 y 11 S t2G t 1A \ \ y 1n S t2G t 1A , if and only if there exists a nite subset ft1; ; tng of G such that t1y1x; ; tnynx 2 A . (3) It follows easily from (2) that the following statement holds. A subset A of a semigroup S is piecewise syndetic if and only if there is a nite G S such that for every nite F S there is some x 2 S with Fx [ t2G t 1A: Elements of K( (S)) are characterized by the following theorem. Theorem 1.6.3. [[29], Theorem 4.39] Let S be a semigroup and let p 2 S. The following statements are equivalent. (a) p 2 K( S). (b) For every A 2 p, fx 2 S : x 1A 2 p g is syndetic. (c) For every q 2 S, p 2 Sqp. Proof. (a) ) (b): Let p 2 K( S) and let A 2 p. Put B = fx 2 S : x 1A 2 pg. Let L be the minimal left ideal of S with p 2 L. For every q 2 L, we have p 2 Sq = c? S(Sq). Since A is a neighborhood of p in S, tq 2 A for some t 2 S and so q 2 t 1A. Thus the sets of the form t 1A cover the compact set L and hence L S t2G t 1A for some nite subset G of S. We show that S S t2G t 1B. Let a 2 S. Then ap 2 L. Pick t 2 G such that ap 2 t 1A. Then t 1A 2 ap so that (ta) 1A = a 1(t 1A) 2 p and so ta 2 B and thus a 2 t 1B. (b) ) (c): Let q 2 S and suppose that p =2 Sqp. Pick A 2 p such that A \ Sqp = ;. Let B = fx 2 S : x 1A 2 pg and pick G 2 Pf (S) such that 1.6 K( S) and its closure 51 S = S t2G t 1B. Pick t 2 G such that t 1B 2 q. Then B 2 tq. That is, fx 2 S : x 1A 2 pg 2 tq. So A 2 tqp, a contradiction. (c)) (a): Take q 2 K( S). Then Sqp K( S). Hence p 2 K( S). Theorem 1.6.4. [[29], Theorem 4.40] Let S be a semigroup and let A S. Then, A is piecewise syndetic () A \K( S) 6= ;: Proof. (: Let p 2 A \ K( S) and let B = fx 2 S : x 1A 2 pg. Then B is syndetic. Pick a nite subset G of S such that S = S t2G t 1B. For each a 2 S pick t 2 G such that a 2 t 1B, i.e. ta 2 B. Then a 1(t 1A) = (ta) 1A 2 p. Then a 1( S t2G t 1A) = S t2G a 1(t 1A) 2 p. It follows that fa 1( S t2G t 1A) : a 2 Sg has the nite intersection property. Hence A is piecewise syndetic. ): Suppose A is piecewise syndetic. Pick a nite subset G S such that fa 1( S t2G t 1A) : a 2 Sg has the nite intersection property. Pick q 2 S such that fa 1( S t2G t 1A) : a 2 Sg q. Then S t2G t 1A 2 aq for every a 2 S and, hence, S t2G t 1A 2 pq for every p 2 S, i.e. ( S)q S t2G t 1A. Pick y 2 K( S) \ ( S q). Then, y 2 t 1A for some t 2 G and so ty 2 A \K( S). Hence A \K( S) 6= ;. Corollary 1.6.5. Let S be a semigroup and let A S. Then, p 2 c? K( S) () every A 2 p is piecewise syndetic: De nition 1.6.6. Let S be a semigroup and let A S. A is said to be central in S if there is some idempotent p 2 K( S) such that A 2 p. Theorem 1.6.7. [[29], Theorem 4.43] Let S be a semigroup and let A S. The following statements are equivalent. (1) A is piecewise syndetic. (2) The set fx 2 S : x 1A is centralg is syndetic. (3) There is some x 2 S such that x 1A is central. Proof. (a))(b): Since A is piecewise syndetic, there exists p 2 K( S) such that A 2 p. Pick a minimal left ideal L of S with p 2 L and pick an idempotent e 2 L. Then A 2 p = pe. So fy 2 S : y 1A 2 eg 2 p. In particular fy 2 S : y 1A 2 eg 6= ;. Pick y 2 S such that y 1A 2 e. Since e 2 K( S), B := fz 2 S : z 1(y 1A) 2 eg is syndetic. Pick a nite G S such 1.6 K( S) and its closure 52 that S = S t2G t 1B. Let D = fx 2 S : x 1A is centralg. We claim that S = T t2yG t 1D. Let x 2 S. Pick t 2 G such that tx 2 B. Then (tx) 1(y 1A) 2 e. Thus (tx) 1(y 1A) is central. But (tx) 1(y 1A) = (ytx) 1A. Thus ytx 2 D. Hence x 2 (yt) 1D. It follows that S = T t2yG t 1D. (b))(c): Since ; is not syndetic, fx 2 S : x 1A is centralg 6= ;. (c))(a): Pick x 2 S such that x 1A is central and pick an idempotent p 2 K( S) such that x 1A 2 p. Then A 2 xp and xp 2 K( S). Thus A \ c?K( S) 6= ;. Hence A is piecewise syndetic. Chapter 2 Ultra lter semigroups and local homomorphisms Let G be an in nite group with identity e and let T be a left invariant topology on G. We obtain a very important subsemigroup Ult(T ) of G consisting of all nonprincipal ultra lters on G converging to e in T . In par- ticular, Ult(T ) carries a lot of information about T . In Section 2.1 we present the basic information about Ult(T ). In Section 2.2 we look at left invariant topologies on a group G determined by idempotents in G. In Section 2.3 we look at local homomorphisms. These give rise to homomorphisms of ul- tra lter semigroups. In Section 2.5 we prove that for any two homeomorphic direct sums X and Y , Ult(X) and Ult(Y ) are isomorphic. In Section 2.4 and in Section 2.6 we present two very important ultra lter semigroups. 2.1 De nition and basic properties Although some concepts in the following de nition have been de ned before, we state them again here for easy reference. De nition 2.1.1. Let G be a group and let T be a topology on G. T is said to be left invariant (respectively, right invariant) if for every a 2 G, and every U 2 T , aU 2 T (respectively, Ua 2 T ). T is said to be inverse invariant if for every U 2 T , U 1 2 T . T is said to be invariant if it is left invariant, right invariant and inverse invariant. A group endowed with a left invariant topology is called a left topological group. 53 2.1 De nition and basic properties 54 We have the following easy observation. Remark 2.1.2. Let G be a group and let T be a topology on G. The following are equivalent. (a) T is left invariant. (b) a is open for every a 2 G. (c) a is continuous for every a 2 G. (d) G is a left topological (semi)group. (e) a is a homeomorphism for every a 2 G. Proof. The implications follow directly from the de nitions and the fact that for any a 2 G, 1a = a 1 Remark 2.1.3. Let G be a group. (a) Since G is a cancellative semigroup, G is an ideal of G (see Theorem 1.5.23). (b) For any A G and for any x 2 G, x 1A = fy 2 G : xy 2 Ag = fx 1a : a 2 Ag; where x 1 in x 1a stands for the inverse of x in G. Note that a left topological group is homogeneous. Also note that a homo- geneous space X is regular if and only if there exists a point x 2 X such that the neighborhood lter at x is closed. In particular, a left topological group is regular if and only if the neighborhood lter at the identity is closed. Unless otherwise stated, we will be using e to denote the identity of a group. In addition, unless otherwise stated, all topologies are assumed to satisfy the T1 separation axiom. Proposition 2.1.4. Let G be a group, let T be a left invariant topology on G and let F be the neighborhood lter of e in T . F satis es the following conditions: (1) T F = feg, and 2.1 De nition and basic properties 55 (2) for every U 2 F , there exists V 2 F such that for every x 2 V , x 1U 2 F . Conversely, suppose there is a lter F on G satisfying conditions (1) and (2). Then, there is a left invariant topology T on G such that F is the neighborhood of e in T . Proof. Condition (1) follows from the T1 property. We show (2). Note that for every y 2 G, yF is the neighborhood lter of y in T . Let U 2 F . There exists V 2 F such that U 2 xF for every x 2 V . One then notices that U 2 xF if and only if x 1U 2 F . Conversely, let F be a lter on G satisfying conditions (1) and (2). Then, for every x 2 G, xF is a lter on G with T xF = fxg. In particular, x 2 U for every U 2 xF . Let x 2 G and let U 2 xF . Then, x 1U 2 F . Therefore, there exists V 2 F such that z 1U 2 F for every z 2 V . Then, xV 2 xF . Let y 2 xV . Then x 1y 2 V . Then y 1U = y 1xx 1U = (x 1y) 1(x 1U) 2 F . Hence U 2 yF . Therefore U 2 yF for every y 2 xV . It follows that there is a topology T on G such that xF is precisely the neighborhood lter at x for each x 2 G. To see that T is left invariant, let a; x 2 G and let U be a neighborhood of ax. U can be written as U = axU0 for some U0 2 F . Then xU0 2 xF and a(xU0) = U . It follows that the left shift a is continuous. Remark 2.1.5. Condition (2) of Proposition 2.1.4 can be restated in the following stronger form: (2)0 For every U 2 F , there is a V 2 F such that V U and for all x 2 V , x 1V 2 F . Proof. Assume (2) holds. For every U 2 F , let U0 = fx 2 U : x 1U 2 Fg: Then U0 U . By (2), there exists A 2 F such that x 1U 2 F for all x 2 A. Then A \ U 2 F . Since A \ U U0, U0 2 F . We claim that for every x 2 U0, x 1U0 2 F , i.e. (U0)0 U0 and, hence, (U0)0 = U0. Let x 2 U0. Then x 1U 2 F . Let V = x 1U . Then V 2 F and xV U . By (2), V 0 2 F . So, it su ces to show that xV 0 U0. Let y 2 V 0 and let W = y 1V . Then W 2 F and yW V . Consequently, xyW xV U , which implies that W (xy) 1U and, thus, (xy) 1U 2 F . Hence, xy 2 U0. It follows that xV 0 U0. Conversely, it is easy to see that (2)0 implies (2). 2.1 De nition and basic properties 56 De nition 2.1.6. Let G be a group, let T be a left invariant topology on G and let e be the identity of G. The ultra lter semigroup of T , denoted Ult(T ), is de ned as follows. Ult(T ) := fp 2 G j p converges to e in T g: Remark 2.1.7. Let G be a group, let T be a left invariant topology on G and let F be the neighborhood lter of the identity in T . It is clear that Ult(T ) = F n feg: Since F is a closed subset of G and e is an isolated point of G, F n feg is a closed subset of G . Sometimes we will also write Ult(G) for Ult(T ) if it is clear which topology we are referring to. Remark 2.1.8. Let G be a group. If T and T 0 are distinct left invariant topologies on G, the ultra lter semigroups Ult(T ) and Ult(T 0) are distinct. Proof. Let F and F 0 be the neighborhood lters of e in T and T 0 respectively. Since F 6= F , F 6= F 0 (see the proof of Theorem 1.5.12). Therefore, Ult(T ) = F n feg 6= F 0 n feg = Ult(T 0). Proposition 2.1.9. [[62], Proposition 2.5] Let G be a group, let T be a left invariant topology on G and let S = Ult(T ). Then, T is Hausdor if and only if for every a 2 G n feg, aS \ S = ;. Proof. Let F be the neighborhood lter of e, i.e. S1 = F , where S1 denotes Ult(T ) [ feg. Note that the topology T is Hausdor if and only if for every a 2 G n feg, there exists U 2 F such that aU \ U = ;. Suppose T is Hausdor . Let a 2 Gnfeg and let U 2 F such that aU\U = ;. Since aU = aU and G is extremally disconnected, aU \ U = ;. Hence, aS1 \ S1 = ; and so aS \ S = ;. Conversely, suppose aS \S = ; for every a 2 Gnfeg. Let a 2 Gnfeg. Then aS1\S1 = ;. Hence, aU \U = ; for some U 2 F , and so aU \U = ;. Thus, T is Hausdor . Lemma 2.1.10. [[62], Lemma 2.9] Let G be a group, let T be a left invariant topology on G and let U G. Then U is open if and only if U Ult(T ) U . Proof. Suppose U is open. Let p 2 U and let q 2 Ult(T ). It su ces to show that U contains a subset of the form S fxBx : x 2 Ag for some A 2 p such that Bx 2 q for each x 2 A. Take A = U . Since the left shifts 2.1 De nition and basic properties 57 G 3 y 7! xy 2 G are continuous and x = xe 2 U , for each x 2 U choose an open neighborhood Bx of e such that xBx U . Then S fxBx : x 2 Ug U and Bx 2 q for each x 2 U , as q converges to e. Hence U 2 pq. It follows that pq 2 U . Therefore U Ult(T ) U . Conversely, suppose U Ult(T ) U and let x 2 U . By continuity of the left shift G 3 q 7! xq 2 G, for every q 2 Ult(T ) pick Bq 2 q such that xBq U . Put V = S q2S Bq [ feg. Then V is a neighborhood of e and xV U . Therefore U is a neighborhood of x. It follows that U is open. Corollary 2.1.11. Let G be a left topological group with the neighborhood lter F of e and let R be a lter on G with a base of open sets such that F R. Then R F R. Proof. Let B be a base of R. Then R = \ U2B U: If B is open, then R F \ U2B U = R: Proposition 2.1.12. [[64], Lemma 2.4] Let G be a group and let T be a left invariant topology on G. Then Ult(T ) is a closed subsemigroup of G . Proof. It remains to show that Ult(T ) is a subsemigroup of G . Let p; q 2 Ult(T ). Since G is an ideal of G, pq 2 G . Let U be an open neighborhood of e. Then, Ult(T ) Ult(T ) U Ult(T ) U . Therefore, U 2 pq. Since pq is a lter, pq contains all neighborhoods of e. It follows that pq 2 Ult(T ). Thus Ult(T ) is a subsemigroup of G . Proposition 2.1.13. [[62], Proposition 2.4] Let G be a group and let S be a nite subsemigroup of G . Then there is a left invariant topology T on G with S = Ult(T ). Proof. Let F = T S1, so that S1 = F . By proposition 2.1.4 it su ces to show that F satis es the following conditions (noting that condition (2) implies condition (2) of Proposition 2.1.4): (1) T F = feg, and 2.1 De nition and basic properties 58 (2) for every V 2 F there exists W 2 F and W 3 x 7! Wx 2 F such that xWx V . Since F e, e 2 T F . Conversely, suppose x 2 T F for some x 2 G. Then F x. Therefore x 2 F = S1 = S [ feg. Since S G , x = e. It follows that feg = T F . Let V 2 F . For every p; q 2 S1, V 2 pq, i.e. pq 2 V . Since q : G! G is continuous, there exists Ap;q 2 p such that Ap;qq V . Let Ap = \ q2S1 Ap;q: Then Ap 2 p (since S1 is nite) and ApS1 V . Put W = [ p2S1 Ap: Then W 2 F and WS1 V . Next, for every x 2 W and q 2 S1, there exists Bx;q 2 q such that xBx;q V , so xBx;q V . Put Wx = [ q 2 S1Bx;q: Then Wx 2 F and xWx V . Lemma 2.1.14. [[64], Lemma 2.6]Let (G; T ) be a left topological group. If Ult(T ) has only one minimal right ideal, then T is extremally disconnected. Proof. Assume Ult(T ) has only one minimal right ideal and that T is not extremally disconnected. Then there are two disjoint open subsets U and V such that e 2 c?(U) \ c?(V ): But then, by Lemma 2.1.10, U \ Ult(T ) and V \ Ult(T ) are two disjoint right ideals of Ult(T ), which gives a contradiction. Lemma 2.1.15. [[64], Lemma 2.8] Let (G; T ) be a left topological group and let S = Ult(T ). If T is regular, then S1 is left saturated in G. Proof. It is clear that S1 is a subsemigroup of G. Let p 2 G n S1. There is a neighborhood U of e such that U =2 p. Since T is regular there exists a closed neighborhood V of e with V U . Then V =2 p. Let C = G n V . Then C is open and C 2 p. Since every element of S1 contains V , none of the elements of S1 contains C, i.e. C \ S1 = ;. Since C is open, we have C S1 = (C S) [ (C feg) C [ C = C. Since, pS1 C S1 C, pS1 \ S1 = ;. Hence S1 is left saturated in G. 2.1 De nition and basic properties 59 Since S S1, for any p 2 G we have pS \S pS1 \S1. Therefore, if S1 is left saturated in G then S is left saturated in G . In particular, by Lemma 2.1.15, if T is regular, then S is left saturated in G . Proposition 2.1.16. [[64], Proposition 2.10] Let G be an in nite group and let S be a closed subsemigroup of G . Suppose that S1 is left saturated in G and that S has a nite left ideal. Then there is a regular left invariant topology T on G with S = Ult(T ). Proof. Since feg = feg is closed, S1 = S [ feg is a union of two closed sets and, hence, closed. Therefore S1 = F for some lter F on G (speci cally, F = T S1). We will show that F satis es the conditions for generating (de ning) a left invariant topology on G (in which F will be a system of neighborhoods of e). It will then follow automatically that S = Ult(T ), and we will have to show that T is regular. (i) Since F e, e 2 T F . Conversely, suppose x 2 T F for some x 2 G. Then F x. Therefore x 2 F = S1 = S [ feg. Since S G , x = e. Therefore feg = T F . (ii) Now we show that for every U 2 F , there exists V 2 F such that for all x 2 V , x 1U 2 F . Let U 2 F and let L be a nite left ideal of S. Then G n U G nS1. Since S1 is left saturated in G, G n U S1 \ S1 = ;. For every q 2 L, choose Wq 2 F such that G n U q \Wq = ;, and put W = \ q2L Wq: Then W 2 F , as L is nite, and (G n U L) \W = ;. Next, since L is a left ideal of S, it follows that S1 L L. For every q 2 L, choose Vq 2 F such that Vq q W (This is possible since q is continuous), and put V = \ q2L Vq: Then V 2 F and V L W . We claim that for all x 2 V , x 1U 2 F . Indeed, otherwise xp 2 G n U for some x 2 V and p 2 S. Take any q 2 L. Then, on one hand xpq = xp q 2 G n U L G nW; 2.2 Left invariant topologies de ned by idempotents 60 and on the other hand, xpq = x pq 2 V L W; which is a contradiction. It follows from (i)-(ii) that there is a left invariant topology T on G in which F is the neighborhood lter of e, and so Ult(T ) = S. We now show that T is regular. Assume the contrary. Then there is a neighborhood, U , of e such that for every neighborhood V of e, c?(V ) n U 6= ;. For every open neighborhood V of e, choose xV 2 c?(V ) n U . Since xV 2 c?(V ), there is an ultra lter on G containing V and converging to xV . Consequently, there is a pV 2 S such that V 2 xV pV . Take q 2 L. By Lemma 2.1.10, V 2 xV pV q, and pV q 2 L, as L is a left ideal. Since L is nite, it follows that there exists q 2 L and p 2 G n U such that for every neighborhood V of e, V 2 pq, and so pq 2 S. We have obtained that S1 is not left saturated in G, which is a contradiction. 2.2 Left invariant topologies de ned by idem- potents In this section we will present two ways of generating left invariant topologies on an in nite group G using idempotents of G . To give a complete picture, in [40], A.Markov asked whether every in nite group admits a non-discrete Hausdor group topology. For Abelian groups and many other groups (see [6]) the answer is positive, but the problem for general groups was eventually solved in the negative in [58], [44]. On the other hand, for every in nite group G and for every idempotent p 2 G, the largest topology on G in which p converges to the identity e 2 G is zero-dimensional and Hausdor . In [71], it was shown that in fact every in nite group admits a non-discrete zero- dimensional Hausdor topology with continuous left and right translations and inversion. De nition 2.2.1. [51] Let G be a group with identity e, let ? be a lter on G, and let A G. The subset c?(A;?) = fx 2 G : xF \ A 6= ; for every F 2 ?g is called the closure of A in the direction of ? (or the closure of A by ?). 2.2 Left invariant topologies de ned by idempotents 61 The subset int(A;?) = fx 2 G : xF A for some F 2 ?g is called the interior of A with respect to ?. Note that int(A;?) c?(A;?) for every A G. Remark 2.2.2. Let G be a left topological group with the neighborhood lter of the identity. Then c?(A; ) = c?(A) and int(A; ) = int(A). Lemma 2.2.3. [[51], Lemma 1.1] Let G be a group, let A G and let p 2 G. Then, c?(A; p) = fx 2 G : A 2 xpg = int(A; p): Proof. This is obvious from the de nitions, recalling that A 2 xp if and only if x 1A 2 p. Lemma 2.2.4. [55] Let G be a left topological group with identity e, let A G and let x 2 G. Then x 2 c?(A) if and only if there exists p 2 Ult(G)[feg such that xF A for some F 2 p. Thus c?(A) = [ fc?(A; p) : p 2 Ult(G) [ feg g: Proof. Necessity. Let x 2 c?(A). Then for every neighborhood U of x, U \A 6= ;. Since neighborhoods of x are exactly sets of the form xW where W is a neighborhood of e, there exists an ultra lter q on G such that fxW : W is a neighborhood of eg [ fAg q: Then x 1q = fx 1B : B 2 qg is an ultra lter on G converging to e. In addition, x 1A 2 x 1q and x(x 1A) = A. Therefore we take p = x 1q and F = x 1A. Su ciency. Let U be a neighborhood of x. Since F 2 p and p 2 Ult(G)[feg, F \ V 6= ; for every neighborhood V of e. Therefore, xF \W 6= ; for every neighborhood W of x. In particular xF \ U 6= ;. Since xF A, A \ U 6= ;. Therefore, x 2 c?(A). Lemma 2.2.5. [[55], Lemma 2.1] Let G be a group, let A G and let ? be a lter on G. Then c?(A;?) = \ fAF 1 : F 2 ?g: 2.2 Left invariant topologies de ned by idempotents 62 Proof. Let x 2 G. x 2 c?(A;?) () xF \ A 6= ; for all F 2 ? () x 2 AF 1 for every F 2 ?: Lemma 2.2.6. [[55], Lemma 2.2] Let G be a group, let A G and let p 2 G. Then c?(A; p) AA 1x for every x 2 c?(A; p). Proof. Let x 2 c?(A; p). Pick a subset F1 2 p such that xF1 A. Let y 2 c?(A; p). Let F2 2 p such that yF2 A. Take F = F1 \F2. Then F 2 p, F F1 and yF yF2 A. We have y 2 AF 1 AF 1 1 AA 1x. Lemma 2.2.7. [[51], Lemma 1.1] Let G be a group and let ? be a lter on G. The following statements hold. (i) c?(A [B;?) = c?(A;?) [ c?(B;?) for all subsets A;B G. (ii) c?(;; ?) = ; and int(;; ?) = ;. (iii) c?(A;?) = S fc?(A; p) : p 2 G with ? pg for every A G. (iv) int(A;?) = T fint(A; p) : p 2 G with ? pg for every A G. Proof. (i) : Let x 2 c?(A;?) [ c?(B;?) and let F 2 ?. Then xF \ A 6= ; or xF \B 6= ;. Hence xF \ (A [B) = (xF \A) [ (xF \B) 6= ;. It follows that c?(A [B;?) c?(A;?) [ c?(B;?). : Let x 2 c?(A [ B;?) and let F 2 ?. Then xF \ (A [ B) 6= ;. Therefore xF \A 6= ; or xF \B 6= ;. We claim that if xF \B = ; then xE\A 6= ; for any other E 2 ?. This will imply that x 2 c?(A;?). So suppose xF \B = ; and let E 2 ?. Since x 2 c?(A[B;?) and E\F 2 ?, x(E\F )\(A[B) 6= ;. Therefore ; 6= x(E \F )\ (A[B) = (x(E \F )\A)[ (x(E \F )\B). Since x(E \F )\B xF \B = ;, x(E \F )\B = ;. Therefore x(E \F )\A 6= ;. Since x(E \ F ) \ A xE \ A, xE \ A 6= ;. It follows that if x =2 c?(B;?), then x 2 c?(A;?). Hence c?(A [B;?) c?(A;?) [ c?(B;?) (ii) Since xF \ ; = ; for any F 2 ? and for any x 2 G, c?(;; ?) = ; and int(;; ?) = ;. 2.2 Left invariant topologies de ned by idempotents 63 (iii) : Let x 2 c?(A;?). For every F 2 ?, xF \ A 6= ;. It follows that the family fxF : F 2 ?g [ fAg has the nite intersection property. Pick q 2 G such that fxF : F 2 ?g[fAg q. Then x 1q 2 G and ?[fx 1Ag x 1q. For any F 2 ?, x(F \ x 1A) = xF \ A A and F \ x 1A 2 x 1q. Hence x 2 int(A; x 1q) = c?(A; x 1q). : Let p 2 G with ? p and let x 2 c?(A; p). Let F 2 ?. Since ? p, F 2 p. Therefore xF \ A 6= ;. Thus x 2 c?(A;?). (iv) : Let x 2 int(A;?) and let p 2 G with ? p. There exists F 2 ? such that xF A. Since ? p, F 2 p and hence x 2 int(A; p). : Let x 2 G such that for every p 2 G with ? p, xF \ A 6= ; for any F 2 p. Suppose x =2 int(A;?). Then for every E 2 ?, xE \ (G n A) 6= ;. Pick q 2 G with ? [ fx 1(G n A)g q. Then x(x 1(G n A)) \ A 6= ;. We obtain ; 6= x(x 1(G n A)) \ A = (G n A)) \ A = ;, a contradiction. De nition 2.2.8. [51] Let G be group with identity e and let ? be a lter on G. ? is called pointed if e 2 F for every F 2 ?. Lemma 2.2.9. [[51], Lemma 1.1] Let G be a group, let A G and let ? be a pointed lter on G. Then A c?(A;?) and int(A;?) A. Proof. If A = ;, then c?(A;?) = ; = A and int(A;?) = ; = A. So, we may assume that A 6= ;. Let a 2 A and let F 2 ?. Since ? is pointed, a 2 aF . Therefore, since A 6= ;, aF \ A 6= ;. Hence A c?(A;?). Now let x 2 int(A;?). Pick F 2 ? such that xF A. Then x = xe 2 xF A. Hence int(A;?) A. Lemma 2.2.10. [[55], Lemma 2.3] Let G be a left topological group, let A G and let p 2 G. Then c?(A; p) is a neighborhood of the identity of G if and only if A 2 qp for every q 2 Ult(G) [ feg. Proof. Suppose c?(A; p) is a neighborhood of the identity and suppose A =2 qp for some q 2 Ult(G) [ feg. Then G n A 2 qp. Choose a subset B 2 q and for each b 2 B choose Cb 2 p such that S fbCb : b 2 Bg G n A. Then bCb \ A = ; for every b 2 B. Since q 2 Ult(G), c?(A; p) 2 q and, therefore, c?(A; p) \ B 6= ;. Pick b 2 c?(A; p) \ B. Since b 2 c?(A; p), bCb \ A 6= ;, a contradiction. Conversely, suppose A 2 pq for every q 2 Ult(G) [ feg and suppose that c?(A; p) =2 Ult(G) [ feg. Then in each neighborhood of the identity we may choose an element x such that xF \ A = ; for some F 2 p. Consequently, there is an ultra lter q 2 Ult(G) [ feg such that G n A 2 pq. Then ; = (G n A) \ A 2 pq, a contradiction. 2.2 Left invariant topologies de ned by idempotents 64 De nition 2.2.11. [51] Let G be a group and let ? and be lters on G. We de ne the product ? as follows. For every A G, A 2 ? () int(A; ) 2 ?. The following lemma shows that this product generalizes the product on G. Lemma 2.2.12. Let G be a group. For every p; q 2 G, p q = pq. Proof. Let A 2 pq. It su ces to show that A 2 p q. Put C = int(A; q) = fx 2 G : xF A for some F 2 qg and put B = fx 2 G : x 1A 2 qg. Then B 2 p. Let x 2 B. Then x 1A 2 q. Put F = x 1A. Then xF = x(x 1A) = A. It follows that x 2 C. Therefore, B C. Since B 2 p, C 2 p. Hence A 2 p q. Since A was an arbitrary element of pq, pq p q and, thus pq = p q. Remark 2.2.13. [51] The family f [ x2F xHx : F 2 ?;Hx 2 for every x 2 Fg forms a base for the lter ? . Proof. Let A 2 ? . Then int(A; ) 2 ?. Take F = int(A; ). For each x 2 F , there exists Hx 2 such that xHx A. Therefore S x2F xHx A. Lemma 2.2.14. [[51], Lemma 1.2] Let G be a group, let and ? be lters on G and let A G. The following statements hold. (i) int(int(A; ); ?) = int(A;? ). (ii) c?(c?(A; ); ?) = c?(A;? ). Proof. (i) Let y 2 int(int(A; ); ?). Take F 2 ? such that yF int(A; ). For every x 2 F pick Hx 2 such that yxHx A. Put P = S x2F xHx. Then P 2 ? and yP A. Therefore y 2 int(A;? ). It follows that int(int(A; ); ?) int(A;? ). Conversely, let y 2 int(A;? ). Choose F 2 ? and for each x 2 F choose Hx 2 such that yxHx A. Then yF int(A; ) and so y 2 int(int(A; ); ?). It follows that int(int(A; ); ?) int(A;? ). (ii) : Let x 2 c?(A;? ) and assume that x =2 c?(c?(A; ); ?). There exists 2.2 Left invariant topologies de ned by idempotents 65 F 2 ? such that xF \ c?(A; ) = ;. For every y 2 F take Hy 2 such that xyHy \ A = ;. Put B = S y2F yHy. Then B 2 ? and xB \ A = ;. This implies that x =2 c?(A;? ), a contradiction. : Let x 2 c?(c?(A; ); ?). Let F 2 ?. Then xF \ c?(A; ) 6= ;. Therefore, there exists y 2 F such that for every E 2 , xyE \ A 6= ;. For each z 2 F pick Hz 2 . Then x( S z2F zHz) \A 6= ;. Since F was an arbitrary element of , xB \A 6= ; for every basic element B of ? in the base described in Remark 2.2.13. Hence xC \ A 6= ; for every C 2 ? . Corollary 2.2.15. Let G be a group. For any lters ; ? and on G, (? ) = ( ?) : Proof. Using Lemma 2.2.14 (a), we have the following. A 2 (? ) () int(A;? ) 2 () int(int(A; ); ?) 2 () int(A; ) 2 ? () A 2 ( ?) : De nition 2.2.16. Let G be a group and let be a lter on G. is called left topological if is the lter of neighborhoods of the identity of G for some left invariant topology on G. Lemma 2.2.17. Let G be a group with identity e and let ? be a lter on G. Then fint(A;?) : A 2 ?g is a lter base on G. Proof. Let A 2 ?. Then e 2 int(A;?). Therefore int(A;?) 6= ;. It is easy to see that for any A;B G, int(A;?) \ int(B;?) = int(A \B;?). De nition 2.2.18. Let G be a group. Given any lter ? on G, (i) we denote by int(?) the lter with the base fint(A;?) : A 2 ?g, and (ii) we denote by C?? the mapping determined by the rule C??(A) = c?(A;?); A G: Remark 2.2.19. Let G be a group and let ? be a lter on G. The following statements hold. 2.2 Left invariant topologies de ned by idempotents 66 (i) int(?) is pointed. (ii) If ? is pointed, then ? int(?). Proof. (i) Let A 2 ?. Then e 2 int(A;?). It follows that e 2 F for every F 2 int(?). (ii) Suppose ? is pointed. Then, by Lemma 2.2.9, int(A;?) A for every A G. In particular, ? int(?). Theorem 2.2.20. [[51], Theorem 1.3] Let G be a group and let ? be a lter on G. The following statements are equivalent. (i) ? is left topological, (ii) ? is pointed and ? ? = ?, (iii) int(?) = ?, (iv) C?? is a closure operator. Proof. (i) ) (ii): Suppose ? is left topological. Then clearly ? is pointed. Let A 2 ? ?. Then int(A;?) 2 ?. Since ? is pointed, by Lemma 2.2.9, int(A;?) ?. Therefore, A 2 ? and, hence, ? ? ?. Now let A 2 ?. Choose an open neighborhood of the identity in (G;?) such that U A. For each x 2 U , pick Vx 2 ? with xVx U . Put V = S x2U xVx. Then V 2 ? ? and V A. Therefore A 2 ? ? and, hence, ? ? ?. It follows that ? = ? ?. (ii) ) (iii): Suppose ? is pointed and ? ? = ?. Then for every A G, int(A;?) A. Hence ? int(?). Now let B 2 int(?). Pick A 2 ? such that int(A;?) B. Since ? ? = ? and A 2 ?, int(A;?) 2 ?. Therefore B 2 ? and, hence, int(?) ?. It follows that int(?) = ?. (iii) ) (ii): Suppose int(?) = ?. Since int(?) is pointed, ? is pointed. Let A 2 ?. Then int(A;?) 2 int(?) = ?. Therefore A 2 ? ?. Thus ? ? ?. Conversely, let A 2 ? ?. Then int(A;?) 2 ?. Since ? is pointed, int(A;?) A. Therefore A 2 ?. Thus ? ? ?. It follows that ? ? = ?. 2.2 Left invariant topologies de ned by idempotents 67 (ii) ) (iv): Suppose ? is pointed and ? ? = ?. By Lemma 2.2.9, A c?(A;?) = c??(A) for every A G. It follows from the de nition that c??(A) = c?(A;?) c?(B;?) = c??(B) for every A;B G with A B. By Lemma 2.2.14 (ii), since ? ? = ?, we have c??(c??(A)) = c?(c??(A); ?) = c?(c?(A;?); ?) = c?(A;? ?) = c?(A;?) = c??(A). It follows that c?? is a closure operator. (iv) ) (i): Let T be the topology on G determined by the closure oper- ator c??. Let F G with c??(F ) = F and let x 2 G. We show that c??(xF ) = xF . So let y 2 c??(xF ) and let E 2 ?. Then yE \ xF 6= ;. Then x 1yE \ F 6= ;. Since E was an arbitrary element of ?, we conclude that x 1y 2 F . Therefore y 2 xF . Hence c??(xF ) xF . Since c?? is a closure operator, c??(xF ) xF . Thus c??(xF ) = xF . It follows that T is left invariant. We now show that ? is the neighborhood lter of the identity e 2 G in T . Let F G such that c??(F ) = F and e =2 F . Put U = G n F . Since e 2 U and int(U;?) = U , we get U 2 ?. Therefore, every open neighborhood of e is an element of ?. Hence, every neighborhood of e is an element of ?. Let V 2 ?. Since eV \ (G n V ) = ;, we have e =2 c?(G n V; ?). Hence, V is a neighborhood of e in T . It follows that ? is the neighborhood lter of e in the topology T . Given a group G and a lter ? on G, one can construct two left topological lters on G using ?. We now look at these two left topological lters. Lemma 2.2.21. [[51], Lemma 1.4] Let G be a group and let ? be a lter on G. Then int(?) = fp 2 G : ? p ?g: In particular, if p 2 G, then int(p) = fq 2 G : p = qpg. Proof. Let p 2 G such that int(?) p. Since fint(A;?) : A 2 ?g int(?), fint(A;?) : A 2 ?g p. It follows that ? p ?. Conversely, let p 2 G with ? p ?. Let A 2 ?. Then A 2 p ? and, therefore, int(A;?) 2 p. Hence int(?) p. Theorem 2.2.22. [[51], Theorem 1.5] Let G be a group. For every lter ? on G, int(?) is left topological. Proof. We will show that int(int(?)) = int(?). Since int(?) is pointed, int(?) int(int(?)). For the other inclusion, let B 2 int(?) and choose 2.2 Left invariant topologies de ned by idempotents 68 A 2 ? such that int(A;?) B. We have the following int(int(A;?); int(?)) = \ fint(int(A;?); p) : p 2 G with int(?) pg = \ fint(int(A;?); p) : p 2 G with ? ? pg = \ f(int(A; p ?) : p 2 G with ? ? pg By Lemma 2.2.21, int(A;?) int(A; p ?) for every p 2 G with ? p. Hence, int(A;?) int(int(A;?); int(?)) int(B; int(?)) and int(B; int(?)) 2 int(?). Theorem 2.2.23. For every ultra lter p on a group G, the left topologi- cal group (G; int(p)) is Hausdor and zero-dimensional. If pp = p, then (G; int(p)) is extremally disconnected. Proof. [[29], Theorem 9.13] Let G be a group and let p 2 G be an idempotent. In addition to Theorem 2.2.23, the topology of (G; int(p)) is the same as the topology onG induced by the restriction of the mapping p : G ! G G to G, see [[29], Lemma 7.6 and Theorem 9.13]. These functions can be seen in the commutative diagram G p // G G ? e OO ( p)jG // Gp; ? i OO where i is the set inclusion function. Let G be a group with identity e and let ? be a lter on G. Let be the family of all left invariant (respectively, invariant) topologies on G in which ? converges to e. Since f;; Gg 2 , 6= ;. Let G = S and let T be the topology generated by G on G. Then, clearly, T is left invariant (respectively, invariant) and ? converges to e in T . In addition, T is the largest left invariant (respectively, invariant) topology on G in which ? converges to e. In particular, the following de nition is justi ed. 2.2 Left invariant topologies de ned by idempotents 69 De nition 2.2.24. Let G be an in nite group with identity e and let ? be a lter on G. We denote by Tl(?) (respectively, T (?)) the largest left invariant (respectively, invariant) topology on G in which ? converges to e. De nition 2.2.25. Let G be a group and let ? be a lter on G. We denote by hull(?) the nest left topological lter on G contained in ?. It is clear that hull(?) is the neighborhood lter of e in Tl(?). De nition 2.2.26. [[71], De nition 3] Let G be a group. For every map M : G! P(G) and every x0 2 G, de ne [M ]x0 2 P(G) by [M ]x0 = fx0x1 xn : xi+1 2M(x0 xi); i < n < !g: The following lemma gives an explicit description of hull(?). Lemma 2.2.27. [[71], Lemma 1] Let G be a group with identity e and let ? be a lter on G. The subsets of the form [M ]x0, where M : G ! ?, are precisely the open neighborhoods of x0 in Tl(?). Proof. Let U be a neighborhood of x0 2 G in Tl(?). De ne M : G! ? by M(x) = x 1U if x 2 U; G otherwise. Then clearly [M ]x0 = U . Conversely, the subsets [M ]x0 form a base of open neighborhoods of x0 for a topology T on G since (1) if y 2 [M ]x0 then [M ]y [M ]x0 , and (2) [M ]x0 \ [N ]x0 [K]x0 , where K : G! ? is de ned by K(x) = M(x) \N(x): The topology T is left invariant as (3) for every a 2 G, a[M ]x0 = [N ]ax0 , where N : G ! ? is de ned by N(x) = M(a 1x). The lter ? converges to e in T since M(e) 2 ? and M(e) [M ]e. It follows that T Tl(?), and so T = Tl(?). 2.2 Left invariant topologies de ned by idempotents 70 We recall that a topological space X is said to be strongly extremally discon- nected if for every open nonclosed subset U X, there exists x 2 X nU such that U [ fxg is open. Notice that every strongly extremally disconnected space is extremally disconnected. Theorem 2.2.28. For every in nite group G and every ultra lter p on G, the left topological group (G; hull(p)) is strongly extremally disconnected. Proof. [[51], Theorem 4.12] Theorem 2.2.29. [[29], Theorem 9.14] Let G be an in nite discrete group with identity e and let p 2 G . Let N = fA [ feg : A 2 pg and T = fU G : for all a 2 U; a 1U 2 pg: The following statements hold. (i) T is a left invariant topology on G. (ii) N contains the neighborhood lter of e in T . (iii) T is the nest topology on G satisfying (i) and (ii). (iv) The neighborhood lter of e in T is equal to N if and only if p is idempotent. In this case T is Hausdor . Proof. (i) It is clear that ;; G 2 T . Let U; V 2 T . Since for any a 2 G, a 1(U \ V ) = a 1U \ a 1V , U \ V 2 T . Let (Ui)i2I be a family of elements of T . Since for any a 2 G, a 1( S i2I Ui) = S i2I a 1Ui, S i2I Ui 2 T . It follows that T is a topology on G. Now let U 2 T and let x 2 G. Let a 2 x 1U . Then xa 2 U . Therefore a 1(x 1U) = (xa) 1U 2 p. It follows that x 1U 2 T . Hence T is left invariant. (ii) Let W be a neighborhood of e in T . There exists U 2 T such that e 2 U W . Then U = e 1U 2 p and thus W 2 p. Since e 2 W , W = W [ feg 2 N . (iii) Let T 0 be a topology on G satisfying (i) and (ii). Let U 2 T 0 and let a 2 U . Then e 2 a 1U . Since T 0 is left invariant, a 1U 2 T 0. Therefore, a 1U is a neighborhood of e in T 0. It follows that a 1U = A [ feg for some A 2 p. Therefore, a 1U 2 p. Hence U 2 T and thus T 0 T . (iv) LetM be the neighborhood lter at e in T . SupposeM = N . Let A 2 p. Pick U 2 T such that e 2 U A [ feg. Since a 1U 2 p for any a 2 U , U = e 1U 2 p. Now let a 2 U . Since a 1U a 1(A [ feg) = a 1A [ fa 1g 2.2 Left invariant topologies de ned by idempotents 71 and a 1U 2 p, a 1A [ a 1feg 2 p. Since p 2 G , a 1A 2 p. Therefore U fa 2 G : a 1A 2 pg and, hence, fa 2 G : a 1A 2 pg 2 p. It follows that pp = p. Conversely, assume pp = p. Let V 2 N . Then V 2 p. Let B = fx 2 V : x 1V 2 pg = V \ fx 2 G : x 1V 2 pg. Then e 2 B. Since pp = p, B 2 p. It follows (see Lemma 1.5.21) that x 1B 2 p for every x 2 B. Therefore e 2 B [ feg = B 2 T . Since e 2 B V , V 2M. Hence N =M. Finally, assume pp = p and let a 6= e. Then, by [[29], Lemma 6.28], ap 6= ep. Pick B 2 p n ap. Let A = B n fa 1B [fa; a 1gg. Then aA[fag and A[feg are disjoint neighborhoods of a and e respectively. Corollary 2.2.30. Let G be an in nite discrete group with identity e and let p 2 G such that p is idempotent. Put ? = fA [ feg : A 2 pg. Then ? = hull(p). Proof. This follows directly from Theorem 2.2.29. Corollary 2.2.31. Let G be an in nite discrete group with identity e and let p 2 G such that p is idempotent. Put ? = fA [ feg : A 2 pg. Then ? satis es the following conditions: (i) T ? = feg, and (ii) for every A 2 p, there exists B 2 p such that for every x 2 B, x 1(A [ feg) 2 ?: Proof. This follows directly from Proposition 2.1.4 and Theorem 2.2.29. Given an in nite group G and an idempotent p 2 G , we denote (G; int(p)) by G[p] and we denote (G; hull(p)) by G(p). Lemma 2.2.32. Let G be an in nite discrete group with identity e and let p 2 G be an idempotent. Then Ult(G(p)) = fpg. Proof. It is clear that p converges to e. Let q 2 Ult(G(p)). Since fA [ feg : A 2 pg q and q 2 G , A \ B 6= ; for every A 2 p and every B 2 q. It follows that q = p. 2.3 Local homomorphisms 72 2.3 Local homomorphisms De nition 2.3.1. A topological space X with a partial binary operation and a distinguished element e is called a local left topological group if for every x 2 X there is an open neighborhood Ux of e such that (1) x y is de ned for all y 2 Ux, x e = x, xUx is an open neighborhood of x, and the mapping Ux 3 y 7! x y 2 xUx is a homeomorphism, and (2) (x y) z = x (y z) whenever y 2 Ux, z 2 Ux y \ Uy and y z 2 Ux. For a local left topological group, from this point on, when we write x y we assume that y 2 Ux and when we write xU , where U is a neighborhood of e, we assume that U Ux. We will write xy for x y. The de nition of a local left topological group includes the following important basic example. Example 2.3.2. LetG be a left topological group. Every open neighborhood X of e in G is a local left topological group. To see this, for every x 2 X take an open subset W of G such that x 2 W X and take Ux = x 1W \X. Given a local left topological group X with a distinguished element e, as for left topological groups, we let Ult(X) be the set of all nonprincipal ultra lters on X converging to e. We recall that a partial semigroup is a nonempty set S equipped with an associative partial binary operation (i.e. xy is not necessarily de ned for all x; y 2 S, but (xy)z = x(yz) whenever the products that are involved are all de ned). De nition 2.3.3. A mapping f : X ! S from a local left topological group X into a partial semigroup S is called a local homomorphism if for every x 2 X n feg, there is a neighborhood Ux of e such that f(xy) = f(x)f(y) for all y 2 Ux n feg. In case when S is also a local left topological group, we in addition assume that f(eX) = eS. A bijection f : X ! Y such that both f and f 1 are local homomorphisms is called a local isomorphism. By a topological local homomorphism (isomorphism) between local left topological groups we will mean a local homomorphism (isomorphism) f : X ! Y such that f is continuous (a homeomorphism). Example 2.3.4. 1. Let B = L ! Z2 be the countable Boolean group endowed with the topology induced by the product topology on !Z2, i.e. the topology 2.3 Local homomorphisms 73 on B having the neighborhood base at zero consisting of the subgroups H = fx 2 B : supp(x) \ = ;g; where < !. As usual, for any b = (an) 2 B, supp(b) = fn < ! : an 6= 0g. Observe that each nonzero element b 2 B has a unique canonical representation in the form b = b0 + + bk, where (a) for every l k, supp(bl) is a nonempty interval in !. (b) for every l k 1, max supp(bl) + 2 min supp(bl+1). If supp(b) is a nonempty interval in !, we say that b is basic. It follows that a mapping f0 from the set of basic elements of B into a semigroup S can be extended to the mapping f : B! S by f(b) = f(b0) f(bk); where b = b0 + + bk is the canonical representation of b. Let x; y 2 B such that max supp(x) + 1 < min supp(y) and let x = x0 + + xk and y = y0 + + ym be the canonical representations of x and y respectively. Then, x+ y = x0 + + xk + y0 + + ym is the canonical representation of x+ y. Therefore, f(x+ y) = f(x0 + + xk + y0 + + ym) = f(x0) f(xk)f(y0) f(ym) = (f(x0) f(xk))(f(y0) f(ym)) = f(x)f(y): It follows that f is a local homomorphism. 2. Let be an in nite cardinal. For every ordinal < , let G be a nontrivial group and let G = M < G : Let T0 denote the group topology on G with the neighborhood base of e consisting of the subgroups H = fx 2 G : supp(x) \ = ;g; 2.3 Local homomorphisms 74 where < . Let D = fx 2 G : jsupp(x)j = 1g: It is clear that every nonzero x 2 G has a canonical representation x = x1 xn where x1; ; xn 2 D are such that if supp(x1) = f 1g; ; supp(xn) = n, then 1 < < n. It is then clear that if x = x1 xn and y = y1 ym are such representations with max supp(x) < min supp(y), then x + y has the representation x + y = x1 xny1 ym. Therefore, every mapping f0 : D ! S into a semigroup S can be extended to the local homomorphism f : G! S de ned by f(x1 xn) = f(x1) f(xn): Remark 2.3.5. (1) Let X be a local left topological group. The partial operation on X can be extended naturally to X by pq = lim x!p lim y!q xy; where x; y 2 X, making X a compact right topological partial semi- group. The product pq is de ned if and only if fx 2 X : fy 2 X : xy is de nedg 2 qg 2 p: To see the necessity, suppose pq is de ned. Since pq is a lter on X, X 2 pq. By the de nition of pq, X 2 pq () fx 2 X : x 1X 2 qg 2 p () fx 2 X : fy 2 X : xy is de ned g 2 qg 2 p Conversely, suppose fx 2 X : fy 2 X : xy is de ned g 2 qg 2 p. Let F = fA X : fx 2 X : x 1A 2 qg 2 pg: It is clear that ; =2 F . The equality x 1X = fy 2 X : xy is de nedg together with the hypothesis implies that X 2 F . Let A;B 2 F . Since for any x 2 X, x 1(A\B) = x 1A\x 1B, A\B 2 F . For any A 2 F and any B X with A B, x 1A x 1B. Since p is a lter, B 2 F . It follows that F is a lter. Hence, pq is de ned. In particular, let p 2 X and let q 2 Ult(X). Then for every x 2 X, Ux 2 q. Since xy is de ned for every y 2 Ux, Ux fy 2 X : xy is de nedg: 2.3 Local homomorphisms 75 Therefore fy 2 X : xy is de nedg 2 q. It follows that fx 2 X : fy 2 X : xy is de nedg 2 qg = X 2 p: Hence pq is de ned. In particular, Ult(X) is a (closed) subsemigroup of X . (2) Recall that for any topological space X and any nonempty subset Y of X, there is a bijective correspondence between lters on X containing Y and lters on Y (see Theorem 1.1.17). This correspondence sends ultra lters to ultra lters and principal ultra lters to principal ultra- lters. Now, let G be a left topological group and let X be an open neighborhood of e in G. Let p 2 Ult(G). Then X 2 p. Therefore, for every neighborhood U of e in G, X \ U 2 p. The sets X \ U , where U is a neighborhood of e in G, are exactly the neighborhoods of e in X. It follows that p \X := fA \X : A 2 pg 2 Ult(X). Thus Ult(X) can be identi ed with Ult(G). In particular, if a local left topological group X is an open neighborhood of the identity in a left topological group G, then we will identify Ult(X) with Ult(G). Theorem 2.3.6. [[29], Theorem 4.21, [64], Lemma 3.2] Let S be a discrete semigroup and let A P(S) have the nite intersection property. Let T be a compact right topological semigroup and let ? : S ! T be a function such that ?(S) (T ). Assume that there is some A0 2 A such that for each x 2 A0, there is Bx 2 A such that ?(xy) = ?(x)?(y) for each y 2 Bx. Then, for every p 2 S and every q 2 T A2AA, ?(pq) = ?(p)?(q). In particular, ? := ?jTA2A A : \ A2A A! T is a homomorphism. Furthermore, if for every A 2 A ?(A) is dense in T , then ? is surjective. Proof. Let p 2 S and let q 2 T A2AA. Then for each x 2 A0, Bx 2 q. Therefore, lim y!q ?(xy) = q - lim y2Bx ?(xy) = q - lim y2Bx ?(x)?(y) = lim y!q ?(x)?(y): 2.3 Local homomorphisms 76 Therefore, we obtain the following. ?(pq) = ?(lim x!p lim y!q xy) = lim x!p lim y!q ?(xy) = lim x!p lim y!q ?(x)?(y) = lim x!p ?(x)?(q) because ?(x) 2 (T ) = ?(p)?(q): Now suppose for every A 2 A, ?(A) is dense in T . Let t 2 T . Then, for each neighborhood V of t and for each A 2 A, V \ ?(A) 6= ; and, therefore, ? 1(V ) \ A 6= ;. This implies that the family B := f? 1(V ) \ A : V is a neighborhood of t and A 2 Ag has the nite intersection property. Therefore, B p for some ultra lter p on S. It is clear that A B. Therefore A p, i.e. p 2 T A2AA. In addition since for every neighborhood V of t and every A 2 A, ? 1(V ) ? 1(V )\A, ? 1(V ) 2 p. It follows that ?(p) = t. Hence, ? is onto. Corollary 2.3.7. Let S be a discrete semigroup and let ? : S ! T be a homomorphism into a compact right topological semigroup T such that ?(S) (T ). Then, the continuous extension ? : S ! T is a homo- morphism. Proof. Let A = fSg. Then A = fSg = S and the result follows easily from Theorem 2.3.6. Corollary 2.3.8. Let f : X ! T be a local homomorphism from a local left topological group into a compact right topological semigroup such that f(X) (T ) and let f : X ! T be the continuous extension of f . Then f = f jUlt(X) : Ult(X) ! T is a homomorphism. Furthermore, if for every neighborhood U of e f(U n feg) is dense in T , then f is surjective. Proof. The rst part is obtained by taking A = fU n feg : U is a neighborhood of eg in Theorem 2.3.6 and noting that in that theorem it su ces if xy is de ned for x 2 A0 and y 2 Bx. The second statement follows easily from the last part of Theorem 2.3.6. 2.3 Local homomorphisms 77 Example 2.3.9. Consider the topological group G and D G in Example 2.3.4 (2). Let be a cardinal such that jG j for all < and let = maxf ; g. Let H be the ultra lter semigroup of T0, i.e. H = \ < c? Gfx 2 G n feg : min supp(x) g: Let T be a compact right topological semigroup containing a dense subset A such that jAj and A (T ). Choose a partition D = fD : < g of D such that D \ H 6= ; and choose any surjection g : ! A. De ne f0 : D ! A by f0(x) = g( ) if x 2 D ; and let f : G! T be a local homomorphism extending f0. Then f : H! T is a surjective homomorphism. Corollary 2.3.10. Let f : X ! T be a local homomorphism from a local left topological group into a nite discrete semigroup and let f : X ! T be the continuous extension of f . Then f = f jUlt(X) : Ult(X) ! T is a homomorphism. Proof. A nite discrete semigroup is a compact right topological semigroup which is equal to its topological center. Corollary 2.3.11. For every local homomorphism f : X ! S from a local left topological group into a partial semigroup, the continuous extension f : X ! S induces a homomorphism f = f jUlt(X) : Ult(X)! S. Proof. The composition \f followed by the embedding of S into S " is a local homomorphism into the compact right topological semigroup S, and the image of X is contained in the topological center. Theorem 2.3.12. Let X and Y be local left topological groups and let f : X ! Y be a topological local isomorphism. Then, f(Ult(X)) = Ult(Y ) and ef := f jUlt(X) : Ult(X)! Ult(Y ) is a topological isomorphism. Proof. It remains to show that f(Ult(X)) = Ult(Y ) and that ef is a home- omorphism. Let p 2 Ult(X) and let V be a neighborhood of eY . Since f is continuous and f(eX) = eY , f 1(V ) is a neighborhood of eX and, therefore, f 1(V ) 2 p . Hence V 2 f(p). It follows that f(p) 2 Ult(Y ). Therefore, 2.3 Local homomorphisms 78 f(Ult(X)) Ult(Y ). Conversely, let q 2 Ult(Y ). Since f is surjective, q = f(p) for some p 2 X. Let U be a neighborhood of eX . Since f(U) is a neighborhood of eY , f(U) 2 q. Then, U = f 1(f(U)) 2 p. Hence, p 2 Ult(X). Therefore, Ult(Y ) f(Ult(X)). Thus f(Ult(X)) = Ult(Y ). Since f is bijective, ef is a bijective. Let A X and B Y be open. Then, ef(Ult(X)\A) = f(Ult(X)\A) = f(Ult(X))\f(A) = Ult(Y )\f(A) is an open subset of Ult(Y ) and ef 1(Ult(Y )\B) = f 1 (Ult(Y ))\ f 1 (B) = Ult(X) \ f 1 (B) is an open subset of Ult(X). Therefore, ef is a homeomor- phism. The induced homomorphisms f : Ult(X) ! T and ef : Ult(X) ! Ult(Y ) in Corollary 2.3.10 and Theorem 2.3.12, respectively, are called proper. In particular, in Theorem 2.3.12, we say that Ult(X) is proper isomorphic to Ult(Y ). Theorem 2.3.13. [[62], Theorem 3.1] Let X be a countable nondiscrete regular left topological group, let S be a partial semigroup, and let f : X ! S be a local homomorphism. Then there is a continuous local isomorphism g : X ! B such that g(xy) = g(x)g(y) and f(xy) = f(x)f(y) whenever max supp(g(x)) + 2 min supp(g(y)). If X is rst countable, g can be chosen to be a homeomorphism. Proof. Let F be the set of words on the letters 0 and 1 with empty word ;. For any w = a0 an and v = b0 bm, de ne w + v = c0 ck by k = maxfn;mg and cl = al; if l n; bl; otherwise. Notice that each nonempty w 2 F has a unique canonical representation in the form = w0 + + wk, where (a) for every l k, wl = 0il1jl with il; jl < ! and il + jl > 0, (b) for every l k 1, jl > 0, (c) for every l k 1; il + jl < il+1. Words of the form 0i1j , where i; j < ! and j > 0, are called basic. Words of the form 0i , where i < !, are called zero. From now on when we write w = w0 + + wk, we mean that this is the canonical representation. For any w 2 F , jwj denotes the length of w. Since X is countable, enumerate X as fe; x1; x2; g. We shall assign to each w 2 F a point x(w) 2 X and a clopen neighborhood X(w) of x(w) such that 2.3 Local homomorphisms 79 (1) x(0n) = e and X(;) = X, (2) X(wa0) \X(wa1) = ; and X(wa0) [X(wa1) = X(w), (3) x(w) = x(w0) x(wk) and X(w) = x(w0) x(wk 1)X(wk), where w = w0 + + wk , (4) f(x(w)y) = f(x(w))f(y) for all y 2 X(0jwj+1) n feg (for nonzero w), (5) xn 2 fx(w) : jwj = ng. We take as X(0) a clopen neighborhood of e such that x1 =2 X(0). Put X(1) = X n X(0), x(0) = e and x(1) = x1. Fix n > 1 and suppose that X(w) and x(w) have been constructed for all w with jwj < n such that conditions (1)-(5) hold. Notice that the subsets X(w), jwj = n 1, form a partition of X. So, one of them, say X(u), contains xn. Let u = u0 + +ur . Then X(u) = x(u0) x(ur 1)(X(ur)) and xn = x(u0) x(ur 1)yn for some yn 2 X(ur). If yn = x(ur), we take as X(0n) a clopen neighborhood of e such that for each basic w with jwj = n 1, X(w) n x(w)X(0n) 6= ; and f(x(w)z) = f(x(w))f(z) for all z 2 X(0n) n feg. For each such w, put x(wa0) = x(w), X(wa0) = x(w)X(0n) and X(wa1) = X(w) nX(wa0), and take as x(wa1) any element of X(wa1). If yn 6= x(ur), we take X(0n) in addition such that yn =2 x(ur)X(0n), and put x(uar 1) = yn. For all nonba- sic nonzero w with jwj = n, we de ne X(w) and x(w) by condition (3). Then x(w) = x(w0) x(wk) 2 x(w0) x(wk 1)X(wk) = X(w), for all y 2 X(0n+1) n feg, f(x(w)y) = f(x(w0) x(wk)y) = f(x(w0) x(wk 1))f(x(wk)y) = f(x(w0)) f(x(wk 1)f(x(wk))f(y) = f(x(w))f(y); and if xn =2 fx(w) : jwj = n 1g, xn = x(u0) x(ur 1)x(uar 1) = x(u a1) 2 fx(w) : jwj = ng. 2.3 Local homomorphisms 80 To check (2), let jwj = n 1. Then x(wa0) = x(w0 + + wk + 0 n) = x(w0) x(wk)x(0 n) = x(w0) x(wk) = x(w); X(wa0) = x(w)X(0n) = x(w0) x(wk)X(0 n) = x(w0) x(wk 1)X(w a k 0); X(wa1) = x(w0) x(wk 1)X(w a k 1); so, X(wa0) [X(wa1) = x(w0) x(wk 1)[X(w a k 0) [X(w a k 1)] = x(w0) x(wk 1)X(wk) = X(w); X(wa0) \X(wa1) = ;: Now for every x 2 X, there is w 2 F with nonzero last letter such that x = x(w), so fv 2 F : x = x(v)g = fwa0n : n < !g. It follows that we can de ne g : X ! B by putting for every w = a0 an 2 F , g(x(w)) = w = (a0; ; an; 0; 0; ): It is clear that g is a bijection with g(e) = 0. Since for every w = a0 an 2 F , X(w) consists of all elements x 2 X such that g(x) = (a0; ; an; ), g is continuous. To check that g(xy) = g(x)g(y) and f(xy) = f(x)f(y) whenever max supp(g(x)) + 2 min supp(g(y)), let w and v be words with nonzero last letters such that x = x(w) and y = x(v), and let w = w0 + +wk and v = v0+ +vt. We have y 2 X(0jwj+1), so w+v = w0+ +wk+v0+ +vt and xy = x(w0) x(wk)x(v0) x(vt) = x(w + v). Hence, g(xy) = g(x(w + v)) = w + v = w + v = g(x(w)) + g(x(v)) = g(x) + g(y) and 2.3 Local homomorphisms 81 f(xy) = f(x(w + v)) = f(x(w0) x(wk)x(v0) x(vt)) = f(x(w0)) f(x(wk))f(x(v0)) f(x(vt)) = f(x(w))f(x(v)) = f(x)f(y): If X is rst-countable, we can choose fX(0n) : n < !g to be a neighborhood base of e, and then g will be a homeomorphism. Corollary 2.3.14. For every countable nondiscrete regular left topological group X, there is a continuous local isomorphism g : X ! B. If X is rst countable, then g can be chosen to be a homeomorphism. Proof. Take S = X and f = idX in Theorem 2.3.13 Corollary 2.3.15. Let X be a countable nondiscrete regular local left topo- logical group, let T and Q be nite semigroups, let g : T ! Q be a surjective homomorphism, and let f : X ! Q be a local homomorphism. Then there is a local homomorphism h : X ! T such that f = g h. Proof. Let k : X ! B be a local isomorphism guaranteed by Theorem 2.3.13. For every basic b 2 B, choose l(b) 2 g 1 f k 1(b) T . Extend the mapping b 7! l(b) to the local homomorphism l : B! T by l(b) = l(b0) l(bk), where b = b0 + + bk is the canonical representation. Then g l = f k 1. Put h = l k. Corollary 2.3.16. All countably in nite nondiscrete regular left topological groups are locally isomorphic. Corollary 2.3.17. Let X and Y be countably in nite nondiscrete regular local left topological groups. Then Ult(X) is proper isomorphic to Ult(Y ). Proof. This follows immediately from Theorem 2.3.12 and Corollary 2.3.16. Corollary 2.3.18. For every countable nondiscrete regular left topological group X, there is a proper injective homomorphism f : Ult(X)! Ult(B). If X is rst countable, f can be chosen to be an isomorphism. 2.3 Local homomorphisms 82 Corollary 2.3.19. Let X be a countable nondiscrete regular local left topo- logical group and let S = Ult(X). Let T and Q be nite semigroups, let g : T ! Q be a surjective homomorphism, and let f : S ! Q be a proper homomorphism. Then there is a proper homomorphism h : S ! T such that f = g h. We have the following strengthened version of Theorem 2.3.13. Theorem 2.3.20. [[65], Theorem 2.1] Let G be a countable nondiscrete reg- ular left topological group and let D be a discrete subset of G such that e =2 D and c? D n D feg. Then there is a continuous bijection h : G ! B with h(e) = 0 such that 1. h(xy) = h(x)h(y) whenever max supp(h(x)) + 2 min supp(h(y)), 2. for every x 2 D, h(x) is basic. If G is rst countable then h can be chosen to be a homeomorphism. Proof. Let W be the set of words on the letters 0 and 1 with empty word ;. For any w = a0 an and v = b0 bm, de ne w + v = c0 ck by k = maxfn;mg and cl = al; if l n; bl; otherwise. Notice that each nonempty w 2 W has a unique canonical representation in the form = w0 + + wk, where (a) for every l k, wl = 0il1jl with il; jl < ! and il + jl > 0, (b) for every l k 1, jl > 0, (c) for every l k 1; il + jl < il+1. Words of the form 0i1j , where i; j < ! and j > 0, are called basic. Words of the form 0i , where i < !, are called zero. From now on when we write w = w0 + + wk, we mean that this is the canonical representation. For any w 2 F , jwj denotes the length of w. Since G is countable, enumerate G n feg as fxn : n < !g. We shall assign to each w 2 W a point x(w) 2 G and a clopen neighborhood X(w) of x(w) such that (1) x(0n) = e and X(;) = G, 2.3 Local homomorphisms 83 (2) X(wa0) \X(wa1) = ; and X(wa0) [X(wa1) = X(w), (3) x(w) = x(w0) x(wk) and X(w) = x(w0) x(wk 1)X(wk), where w = w0 + + wk , (4) X(wa0) \D fx(w)g for nonzero w, (5) xn 2 fx(w) : jwj = n+ 1g. We take as X(0) a clopen neighborhood of e such that x0 =2 X(0). Put X(1) = GnX(0), x(0) = e and x(1) = x0. Fix n 1 and suppose that X(w) and x(w) have been constructed for all w with jwj n such that conditions (1)-(5) hold. Notice that the subsets X(w), jwj = n, form a partition of X. So, one of them, say X(u), contains xn. Let u = u0 + +ur . Then X(u) = x(u0) x(ur 1)(X(ur)) and xn = x(u0) x(ur 1)yn for some yn 2 X(ur). If yn = x(ur), we take as X(0n) a clopen neighborhood of e such that for each basic w with jwj = n, X(w) n x(w)X(0n+1) 6= ; and, following (4), for each nonzero w with jwj = n, x(w) X(0n+1)\D fx(w)g. For each such w, put x(wa0) = x(w), X(wa0) = x(w)X(0n) and X(wa1) = X(w) nX(wa0), and take as x(wa1) any element of X(wa1). If yn 6= x(ur), we take X(0n) in addition such that yn =2 x(ur)X(0n), and put x(uar 1) = yn. For all nonbasic nonzero w with jwj = n+1, we de ne X(w) and x(w) by condition (3). Then x(w) = x(w0) x(wk) 2 x(w0) x(wk 1)X(wk) = X(w), and if xn =2 fx(w) : jwj = ng, then xn = x(u0) x(ur 1)x(u a r 1) 2 fx(w) : jwj = n+ 1g: To check (2), let jwj = n 1. Then x(wa0) = x(w0 + + wk + 0 n) = x(w0) x(wk)x(0 n) = x(w0) x(wk) = x(w); X(wa0) = x(w)X(0n) = x(w0) x(wk)X(0 n) = x(w0) x(wk 1)X(w a k 0); X(wa1) = x(w0) x(wk 1)X(w a k 1); 2.3 Local homomorphisms 84 so, X(wa0) [X(wa1) = x(w0) x(wk 1)[X(w a k 0) [ Y (w a k 1)] = x(w0) x(wk 1)X(wk) = X(w); X(wa0) \X(wa1) = ;: Now for every x 2 Gnfeg, there is w 2 W with nonzero last letter such that x = x(w), so fv 2 W : x = x(v)g = fwa0n : n < !g. It follows that we can de ne h : G! B by putting, for every w = a0 an 2 W , h(x(w)) = w = (a0; ; an; 0; 0; ): It is clear that h is a bijection with h(e) = 0. Since for every w = a0 an 2 W , X(w) consists of all elements x 2 G such that h(x) = (a0; ; an; ), h is continuous. To check that h(xy) = h(x)h(y) whenever max supp(h(x)) + 2 min supp(h(y), let x = x(w) and y = x(v) 2 X(0jwj+1). Then h(x(w)x(v)) = h(x(w + v)) = w + v = w + v = h(x(w)) + h(x(v)): To check that h(x) is basic for every x 2 D, let x = x(w) 2 D with w = w0 + + wk. Then x 2 x(w0)X(0 jwj+1) and x(w0)X(0 jwj+1) \D fx(w0)g; so x = x(w0). Hence h(x) = w0 is basic. If G is rst countable, we can choose fX(0n) : n < !g to be a neighborhood base of e, and then h will be a homeomorphism. Corollary 2.3.21. Let G be a countable regular left topological group and let D be a discrete subset of G such that e =2 D and c? D nD feg. Then every mapping from D into a semigroup S can be extended to a local homomorphism f : G! S. Proof. Let h : G ! B be a local homomorphism guaranteed by Theorem 2.3.16 and let f0 be a mapping from D into S. De ne a local homomorphism g : B ! S as follows. Let x 2 B. If h 1(x) 2 D, put g(x) = f0(h 1(x)). For any other basic x 2 B, pick g(x) arbitrarily. If x is nonbasic, put g(x) = g(x0) g(xk), where x = x0 + + xk is the canonical representation of x. We then de ne f = g h. Obviously, fjD = f0. To see that f is a local homomorphism, let x 2 G n feg. Put Ux = fy 2 G : minn supp(h(y)) max supp(h(x)) + 2g [ feg: 2.3 Local homomorphisms 85 Then Ux is a neighborhood of e because h is continuous, and for every y 2 Ux n feg, f(xy) = g(h(x) + h(y)) = g(h(x)) g(h(y)) = f(x)f(y): Corollary 2.3.22. Every countable regular left topological group containing a discrete subset with exactly one accumulation point admits a local homo- morphism onto N. Proof. Let G be a countable regular left topological group and let D be a discrete subset of G with c? DnD = feg. For every x 2 G, put f0(x) = 1 2 N. By Corollary 2.3.17, the mapping f0 : D ! N can be extended to a local homomorphism f : G! N. Since e 2 c? D and f(D) = f1g, f is onto. Corollary 2.3.16 suggests the question whether any two homeomorphic zero- dimensional local left topological groups are locally isomorphic. The follow- ing example shows that the answer to this question is negative. Example 2.3.23. The topological groups G = Q ! Z2 and H = Q ! Z3 are homeomorphic but not locally isomorphic. To show this, assume on the contrary that there is a local isomorphism f : G ! H. Construct inductively a decreasing sequence (Un)n 0 such that f(y0 + y) = f(y0) + f(y) for all y 2 Un. Then f(y0 + yn) = f(y0) + f(yn): 2.4 The ultra lter semigroup H 86 Since also f(y0) = f(x0 + + xn 1 + yn) = f(x0 + + xn 1) + f(yn); we obtain that f(y0 + yn) = f(x0 + + xn 1) + f(yn) + f(yn) 6= f(x0 + + xn 1): But on the other hand f(y0 + yn) = f(x0 + + xn 1 + yn + yn) = f(x0 + + xn 1); - a contradiction. 2.4 The ultra lter semigroup H Given a semigroup S, certain important subsemigroups of S depend very little on the algebraic structure of S. In fact, they can be de ned in terms of direct sums of sets with a distinguished element [56] (see also [29], p. 111). Let be an in nite cardinal and for every ordinal < , let A be a set with a distinguished element 0 2 A . Let X be the direct sum X = L < A . We recall that the direct sum (strong direct sum) topology on X is generated by taking as a neighborhood base at x 2 X the subsets of the form fy 2 X : y( ) = x( ) for all 2 Fg where F is a nite subset of (respectively, fy 2 X : y( ) = x( ) for all < g where < ). There is a natural partial semigroup operation + on X: for any x; y 2 X with supp(x) \ supp(y) = ;, x+ y is de ned by supp(x+ y) = supp(x)[ supp(y) and (x+ y)( ) = x( ) if 2 supp(x) y( ) if 2 supp(y): Lemma 2.4.1. (X;+) equipped with the direct sum (strong direct sum) topol- ogy is a local left topological group. Proof. We prove the case of the direct sum topology. The strong direct sum topology case is similar. For every x 2 X, let Ux = fy 2 X : y( ) = 0 for all 2 supp(x)g = fy 2 X : supp(y) \ supp(x) = ;g: Then Ux is a neighborhood of 0 2 X, x + Ux is de ned, with x + 0 = x, and the mapping Ux 3 y 7! x + y 2 x + Ux is clearly surjective. From the 2.4 The ultra lter semigroup H 87 de nition of +, it is easy to see that the mapping Ux 3 y 7! x + y 2 x + Ux is injective. It is also clear that the associative rule x+ (y+ z) = (x+ y) + z holds whenever the sums involved are de ned. To see that this mapping is continuous, let z = x + y for some y 2 Ux and let W X be open with z 2 (x + Ux) \W . We may assume that there is a nite subset F of such that W = fw 2 X : w( ) = z( ) for all 2 Fg. Take V = fv 2 X : v( ) = y( ) for all 2 Fg. Then V is open and y 2 Ux \ V . In addition, for every v 2 Ux \ V , (x + v)( ) = z( ) for every 2 F , i.e. x + v 2 (x + Ux) \W . It follows that the mapping Ux 3 y 7! x+ y 2 x+Ux is continuous. Now let V X be open and let z = x + y for some y 2 Ux \ V . Since Ux \ V X is open, there is a basic neighborhood W of y with W Ux \ V . Since W is a basic neighborhood of y, there exists a nite subset E of such that W = fw 2 X : w( ) = y( ) for all 2 Eg. Then z 2 x+W x+ (Ux \ V ) and (x+w)( ) = z( ) for every 2 E [ supp(x). Therefore, x+ (Ux \V ) is open. It follows that the mapping Ux 3 y 7! x + y 2 x + Ux is open. Hence Ux 3 y 7! x+ y 2 x+ Ux is a homeomorphism. Corollary 2.4.2. The operation + on X can be extended to X. Proof. This follows from Remark 2.3.5 (1). We will denote the ultra lter semigroup of X with respect to the discrete sum (respectively, the strong discrete sum) topology by Ult0(X) (respectively, Ult(X)). It is easy to see that Ult(X) Ult0(X). Corollary 2.4.3. For every p 2 X and q 2 Ult0(X), p + q has a base consisting of subsets of the form S x2A x + Bx where A 2 p and for every x 2 A, Bx 2 q with supp(x) \ supp(y) = ; for all y 2 Bx. Proof. This follows from Remark 2.3.5 (1). Alternatively, for each < , pick a semigroup operation on A with identity 0 , and let denote the corresponding semigroup operation on X. Then for all x; y 2 X with supp(x) \ supp(y) = ;, x + y = x y, and for all p 2 X and q 2 Ult0(X), p+ q = p q. Remark 2.4.4. Let D = fx 2 X : j supp(x)j = 1g, and let S be a semigroup. Then any mapping f0 : D ! S can be extended to a local homomorphism f : X ! S by f(x) = Y 2supp(x) f0(x( )): 2.4 The ultra lter semigroup H 88 It follows that if minfj L < A j : < g jSj, then Ult(X) admits a continuous homomorphism onto S. Remark 2.4.5. If for each < , A is a group, then the direct sum topology and the strong direct sum topology on X are group topologies (with respect to the componentwise group operation on X). Proof. We consider the direct sum topology. The case for the strong direct sum topology is similar. Let f : X X ! X, f(x; y) = xy. Let (a; b) 2 X X and let U be a basic neighborhood of ab. There is a nite subset F of such that U = fz 2 X : z( ) = ab( ) for all 2 Fg. Consider V = fz 2 X : z( ) = a( ) for all 2 Fg and W = fz 2 X : z( ) = b( ) for all 2 Fg. V is a basic neighborhood of a and W is a basic neighborhood of b and for any (x; y) 2 V W and for any 2 F , xy( ) = x( )y( ) = a( )b( ) = ab( ). Therefore, f(V W ) U . It follows that f is continuous. Now let h : X ! X, h(x) = x 1. Let a 2 X and let U be a basic neigh- borhood of a 1. Then U 1 is a basic neighborhood of a with h(U 1) U . Hence, h is continuous. Note that in Section 2.3 the direct sums were equipped with the strong direct sum topology. Again, from now on, unless otherwise stated, all direct sums are assumed to be equipped with the strong direct sum topology. Recall that for each n 2 N, the support of n is a nite subset of ! de ned by the unique expression of n as n = X i2supp(n) 2i: For each n 2 !, consider the set 2nN. Note the following. (i) x 2 2nN if and only if min supp(x) n. (ii) The family A := (2nN)n2! has the nite intersection property. (iii) For any n;m 2 N with n m and for any x 2 2nN, x+ 2mN 2nN. Therefore, the set H = \ n2! c? N(2 nN) 2.4 The ultra lter semigroup H 89 is a subsemigroup of N. Now let T be a compact right topological semigroup and let ? : N ! T with ?[N] (T ). Assume that there is some k 2 N such that whenever x; y 2 N and max supp(x) + k < min supp(y), one has ?(x+ y) = ?(x)?(y). Let n 2 N and let A = 2nN. For each x 2 A, take Bx = 2 (max supp(x)+k+1) N: Then, for each y 2 Bx, min supp(y) max supp(x)+k+1 > max supp(x)+ k. Therefore, for each n 2 N and each x 2 2nN, there is m 2 N, such that for all y 2 2mN, ?(xy) = ?(x)?(y). Therefore the restriction of the continuous extension ? : N! T to H is a homomorphism. Remark 2.4.6. Recall that T = T n2N c? N(nN). Note the following. (a) As in the case of H, it can easily be veri ed that T is a subsemigroup of N. (b) Since for any n 2 !, 2nN 2 fmN : m 2 Ng, it follows that T H. Proposition 2.4.7. [[29], Lemma 6.6] All idempotents of N are in T. Proof. For each n 2 N, let ?n be the natural homomorphism from N to the group of integers modulo n: ?n : N! Zn; ?n(x) = x(modn): Since Zn is nite, ?(N) = Zn = (Zn) and, therefore, the continuous exten- sion ?n : N! Zn is a homomorphism. Let p 2 N. Then, p 2 ker(?n), ?n(p) = 0(modn), lim x!p ?n(x) = 0(modn), nN = [0] 2 p; where [0] is the equivalence class of 0 modulo n. It follows that the kernel of ?n is c? N(nN). Let p 2 N be any idempotent. Since ?n(p) = ?n(p + p) = ?n(p) + ?n(p), then ?n(p) = 0(mod n). Therefore, p 2 c? N(nN). Since this holds for every n 2 N, p 2 T n2N c? N(nN) = T. Lemma 2.4.8. De ne : N ! ! and : N ! ! by (n) = max(supp(n)) and (n) = min(supp(n)). Then, for any p 2 N and any q 2 H, (p+ q) = (q) and (p+ q) = (p): 2.4 The ultra lter semigroup H 90 Proof. Let p 2 N and let q 2 H. Given any m 2 N, choose r 2 N such that (m) < r. Let n 2 2rN. Then (m) < r (n). Therefore, supp(m + n) = supp(m) [ supp(n), (m+ n) = (n) and (m+ n) = (m). Since 2rN 2 q, (m+ q) = q- lim n22rN (m+ n) = q- lim n22rN (n) = (q): It follows that (p+ q) = (q). Similarly, (p+ q) = (p). Using Lemma 2.4.8, one can show that N is the center of ( N;+) and of ( N; ). It is known ( see [[27], Theorem 3.9]) that if p is an idempotent in K( N), then (p + N + p) \ H contains a copy of the free group on 22 ! generators. It is also known (see [[29], Theorem 6.32]) that copies of H can be found in S for any in nite cancellative semigroup S. Proposition 2.4.9. [[29], Theorem 6.4] Any compact right topological semi- group T with a countable dense topological center is an image of H under a continuous homomorphism. Proof. Enumerate (T ) as (T ) = fti : i 2 Ig, where I is either ! or f0; 1; 2; ; kg for some k 2 !. We then choose a disjoint partition fAi : i 2 Ig of f2n : n 2 !g, with each Ai being in nite. We de ne a mapping : N! T by rst de ning on f2n : n 2 !g by (2n) = ti if 2 n 2 Ai: We then extend to N by (n) = i2supp(n) (2 i); with the terms in this product occurring in the order of increasing i. It is easy to see that for any m;n 2 N, if max supp(x) < min supp(m), then (n + m) = (n) + (m). Therefore, the continuous extension : N ! T is a homomorphism on H. Now let i 2 I and let x 2 A i . Then x 2 H and (xi) = ti. It follows that (H) (T ) and thus (H) c?( (T )) = T . Corollary 2.4.10. Every nite discrete semigroup is the image of H under a continuous homomorphism. 2.4 The ultra lter semigroup H 91 Proof. A nite discrete semigroup is a compact right topological semigroup which is equal to its topological center. De nition 2.4.11. Given any cardinals ! and 2, H ; = Ult( M ) and H = H ;2: Given !, we will often denote the group L < 2 by B and for each < , e is that member of B such that supp(e ) = f g. Note that B! = B, one of the groups we have seen in section 2.3, and H! = Ult(B). Therefore, from Section 2.3, we make the following conclusion. Proposition 2.4.12. Let X be a countably in nite nondiscrete regular local left topological group. Then, Ult(X) is isomorphic to H!. Proposition 2.4.13. H! and H are topologically and algebraically isomor- phic. Proof. De ne ? : N! B by ?(x) = 0; if i =2 supp(x) 1; if i 2 supp(x): Clearly, for every x; y 2 N with max supp(x) < min supp(y), ?(x + y) = ?(x)?(y). It is easy to see that ? is injective and that for any nonzero a 2 B, a = ?(x), where x 2 N with supp(x) = supp(a). Hence ? maps N bijectively onto B n f0g. It is also easy to see that supp(?(x)) = supp(x). In particular for every n 2 N, since x 2 2nN if and only if min supp(x) n, ?(2nN) = Hn.Therefore the continuous extension ? satis es ?(H) = H!. Since, ? is injective, so is ?. Thus ?jH is a homeomorphism onto H!. Now let p; q 2 H. Then ?(p+ q) = ?(p - lim x2N q - lim y2N (x+ y)) = p - lim x2N q - lim y2N ?(x+ y): Given any x 2 N, in n = max(supp(x)) + 1, then for all y 2 2nN, ?(x+ y) = ?(x)?(y), so that, since q 2 H and ?(x) and ? are continuous, q - lim y2N ?(x+ y) = q - lim y22nN ?(x+ y) = q - lim y22nN (?(x)?(y)) = ?(x)?(y): Thus ?(p+ q) = p - lim x2N (?(x)?(q)) = ?(p)?(q). 2.4 The ultra lter semigroup H 92 Indeed this proposition is an example of the general statement as stated in [[29], Theorem 6.15]. Proposition 2.4.14. Let S be a countably in nite discrete group. Then S n S contains a topological and algebraic copy T of H such that K(T ) = K( S) \ T . Proof. [[30], Theorem 2.3]. Proposition 2.4.15. [[30], Theorem 2.5] Let be an in nite cardinal. Every compact right topological semigroup T containing a dense subset A such that jAj and A (T ) is a continuous homomorphic image of H . Proof. Enumerate A as ft : < g, with repetition if jAj < . Let fI : < g be a partition of into subsets of size . De ne h : B ! T by rst agreeing that for each < , h(e ) = t , where 2 I . Then for a nite F , de ne h X 2F e ! = 2F h(e ); where the product is taken in increasing order of indices. De ne h(0) arbi- trarily. Let h : B ! T be the continuous extension of h and let h be the restriction of h to H . To see that h(H ) = T , it su ces to show that A h(H ). Given < we have that jI j = . Pick a -uniform ultra lter p on fe : 2 I g. Then p 2 H and h(p) = t because f is constantly equal to t on fe : 2 I g. It is easy to see that whenever x; y are nonzero elements of B with max supp(x) < min supp(y), then h(x+ y) = h(x)h(y). Therefore h is a homomorphism. Proposition 2.4.16. [[30], Theorem 2.7] Let be an in nite cardinal and let S be an in nite cancellative discrete semigroup with cardinality . Then S n S contains a topological and algebraic copy of H . Proof. Choose a -sequence (t ) < in S with distinct nite products. Let T = FP ((t ) < ). For each < , let T = FP ((t ) < < ). Put eT = \ < c? S(T ) = \ < T : eT is a subsemigroup of S. De ne : T ! B by taking, for each nite F ( 2F t ) = X 2F e : 2.5 Local isomorphisms of direct sums 93 Since (t ) 2 has distinct nite products, is well de ned. Since the op- eration on T is that given by concatenation, it follows that the restric- tion of the continuous extension : eT :! B to eT is a homomor- phism. It is clear that and, hence, are injective. Since for each < , (T ) = fx 2 B : min supp(x) > g, maps eT onto H . 2.5 Local isomorphisms of direct sums Let be an in nite cardinal and for every ordinal < let A be a set with a distinguished element 0 , and let = minfj L < A j : < g. In this section we show that Ult( L < A ) is topologically and algebraically isomorphic either to H or to H ; . Furthermore, we show that H ; and H are topologically and algebraically isomorphic if either = = ! or < = or = and cf( ) = cf( ) < . This section forms the main part of [61]. De nition 2.5.1. Let X = L < A . Then (1) (X) = minfj L < A j : < g, (2) X is narrow if there is < such that j L < A j < (X) for all < , (3) X is broad if for every < there is < such that j L < A j = (X), and (4) X is reduced if (X) = jXj and either j L < A j < jXj for all < or for every < there is < such that j L < A j = jXj. Lemma 2.5.2. For every X = L < A there is < such that Y =L < A is reduced. Proof. Let = (X). Pick 0 < such that j L 0 < A j = . If for every 2 [ 0; ) there is < such that j L < A j = , put = 0. Otherwise pick 2 [ 0; ) such that j L < A j < for all < . Note that for every < , Ult( L < A ) is topologically and algebraically isomorphic to Ult( L < A ). 2.5 Local isomorphisms of direct sums 94 Lemma 2.5.3. Let X = L < A , let = cf( ), and let ( ) < be an increasing co nal sequence in such that 0 = 0 and = sup < if is a limit ordinal. Then X is locally isomorphic to Y = L < where = j L < +1 A j. Proof. For every < , let X = fx 2 X : ; 6= supp(x) [ ; +1)g and pick a bijection f : X ! n f0g. De ne f : X ! Y as follows. Put f(0) = 0. Now let 0 6= x 2 X. Then x can be written in the from x = x1 + + xn where, for each i 2 [1; n], xi 2 X i for some i < and 1 < < n. De ne y = f(x) by supp(y) = f 1; ; ng and y( i) = f i(xi): Clearly, f is bijective. Furthermore, (i) if x; y 2 X n f0g and max supp(x) < min supp(y) for some < , then f(x+ y) = f(x) + f(y), and (ii) for every < , f(fx 2 X : min supp(x) g) = fy 2 Y : min supp(y) g: Hence, f : X ! Y is a local isomorphism. Proposition 2.5.4. Let X = L < A be a reduced and let = jXj. Then X is locally isomorphic to L 2 if = ! or X is narrow and to L otherwise. Proof. If = !, then X is locally isomorphic to L 2 by Corollary 2.3.14. Suppose > ! and let = cf( ). Consider two cases. Case 1: X is narrow. Then cf( ) = . Construct inductively increasing co nal sequences ( ) < in and ( ) < in such that (i) 0 = 0 = 0, (ii) if < is limit, then = sup < and = sup < , and (iii) for every < , j L < +1 A j = j L < +1 2j. 2.5 Local isomorphisms of direct sums 95 Then by Lemma 2.5.3, both X and L 2 are locally isomorphic toL < +1. Case2: X is broad. Construct inductively an increasing co nal se- quence ( ) < in such that (i) 0 = 0, (ii) if < is limit, then = sup < and (iii) for every < , j L < +1 A j = . Then by Lemma 2.5.3, both X and L are locally isomorphic toL . Corollary 2.5.5. Let X = L < A and let = (X). Then Ult(X) is topologically and algebraically isomorphic to H if = ! or X is narrow and to H ; otherwise. It follows from the next lemma that if > ! is regular, then L 2 and L are not homeomorphic. Lemma 2.5.6. Let > ! be regular. Then every family of pairwise disjoint nonempty open subsets of L 2 has cardinality < . Proof. Let fU : < g be a family of nonempty open subsets of L 2. For every < , pick x 2 U . Applying the -system lemma [[36], Chapter II, Theorem 1.6], we obtain a subset S of cardinality and a nite subset F such that for any distinct ; 2 S, one has supp(x )\ supp(x ) = F and x ( ) = x ( ) for each 2 F . Now let 2 S. It then follows that jf 2 S : x =2 U gj < . Consequently, there is 2 S distinct from such that x 2 U , and so U \ U 6= ;. We now come to the main result of this section. Theorem 2.5.7. Let ; ! and let 2. Then the following statements are equivalent: (1) L and L 2 are locally isomorphic, (2) L and L 2 are homeomorphic, and 2.5 Local isomorphisms of direct sums 96 (3) either = = ! or < = or = and cf( ) = cf( ) < . The crucial moment in the proof of Theorem 2.5.7 is contained in the follow- ing proposition. Proposition 2.5.8. If and cf( ) = cf( ) < , then L and L 2 are locally isomorphic. The proof of Proposition 2.5.8 involves some auxiliary constructions. First of all, by Lemma 2.5.3, one may suppose that is regular, and so = cf( ) < : Pick a co nal -sequence ( ) < in such that (1) 0 > , and (2) for every > 0, > sup < . Then L 2 is locally isomorphic to L < . To see this, de ne inductively a -sequence ( ) < in by (i) 0 = 0, (ii) = sup < if is limit, and (iii) +1 = + , and apply Lemma 2.5.3. We shall prove that L and L < are locally isomorphic. De nition 2.5.9. (a) X = L . (b) For every < , X ; = fx 2 X : supp(x) = f g and x( ) 2 g. (c) For every < < , X ; is the subset of all x 2 X such that (i) supp(x) [ ; ], (ii) for each 2 supp(x), x( ) 2 , and (iii) for each 2 [ ; ), there is 2 [ ; ] such that x( ) =2 . (d) For every < and x 2 X ; , l(x) = and r(x) = . 2.5 Local isomorphisms of direct sums 97 Lemma 2.5.10. Every nonzero x 2 X can be uniquely written in the form x = x1 + + xn where for each i 2 [1; n], xi 2 X ; for some < , and for each i 2 [1; n 1], r(xi) < l(xi+1). Proof. Let = min supp(x) and = minf : x( ) 2 for all g. If supp(x) [ ; ], then x 2 X ; . Otherwise de ne x1; y 2 X by supp(x1) = f 2 supp(x) : g; supp(y) = f 2 supp(x) : > g; x1( ) = x( ) for all 2 supp(x1), and y( ) = x( ) for all 2 supp(y). Then x1 2 X ; , r(x1) < min supp(y) and x = x1 + y. Continuing this process, in j supp(x)j steps we obtain the required decomposition. To see the uniqueness, let x = x01+ +x 0 m be another such decomposition. It su ces to show that x01 = x1. We have that x 0 1 2 X 0; 0 for some 0 0 < . Then 0 = min supp(x) = and 0 = minf : x( ) 2 for all g = . Consequently, x01 = x1. De nition 2.5.11. (a) Y = L < . (b) For every < , Y ;0 = fy 2 Y : supp(y) = f g and y( ) 2 n g. (c) For every < and nonzero < , Y ; = fy 2 Y : supp(y) [ ; + ] and y( ) = g: (d) For every ; < and y 2 Y ; , l(y) = and r(y) = + . Lemma 2.5.12. Every nonzero y 2 Y can be uniquely written in the form y = y1 + + yn where for each i 2 [1; n], yi 2 Y ; for some ; < , and for each i 2 [1; n 1], r(yi) < l(yi+1). Proof. The proof is similar to that of Lemma 2.5.10. We restrict ourselves to showing the existence of such a decomposition. Let = min supp(y). If y( ) 2 n , put = 0. Otherwise = y( ). If supp(y) [ ; + ], then y 2 Y ; + . Otherwise de ne y1; z 2 Y by supp(y1) = f 2 supp(y) : + g; supp(z) = f 2 supp(y) : > + g; y1( ) = y( ) for all 2 supp(y1), and z( ) = y( ) for all 2 supp(z). Then y1 2 Y ; + , r(y1) < min supp(z) and y = y1 + z. Continuing this process, in j supp(y)j steps we obtain the required decomposition. 2.5 Local isomorphisms of direct sums 98 As in Example 2.3.4, we call the decompositions in Lemma 2.5.10 and Lemma 2.5.12 canonical. Proof of Proposition 2.5.8. We show that X and Y are locally isomorphic. For every ; < , both jX ; + j = + and jY ; j = + , so there is a bijection f ; : X ; + ! Y ; . De ne f : X ! Y as follows. Put f(0) = 0. Now let 0 6= x 2 X and let x = x1 + + xn be a canonical decomposition. For each i 2 [1; n], let i = l(xi), i = r(xi), and write i as i = i + i for some i < . Put f(x) = f 1; 1(x1) + + f n; n(xn): Note that this is a canonical decomposition. It then follows that f is a bijection. Furthermore, (i) if x; y 2 X n f0g, x = x1 + + xn is the canonical decomposition and r(xn) < min supp(y), then f(x+ y) = f(x) + f(y), and (ii) for every < , f(fx 2 X : min supp(x) g) = fy 2 Y : min supp(y) g: Hence f : X ! Y is a local isomorphism. Now we are in a position to prove Theorem 2.5.7. Proof of Theorem 2.5.7. (1)) (2) is obvious. (2) ) (3): If = !, then = !. Let > !. If < , then = . Suppose that . Then = and clearly cf( ) = cf( ). Since L can be partitioned into nonempty open subsets, it follows from Lemma 2.5.6 that cf( ) < . (3) ) (1): If either = = ! or < = , then L andL 2 are locally isomorphic by Proposition 2.5.4. And if = and cf( ) = cf( ) < , then they are locally isomorphic by Proposition 2.5.8. Corollary 2.5.13. Let ; ! and 2. Suppose that either = = ! or < = or = and cf( ) = cf( ) < . Then H ; and H are topologically and algebraically isomorphic. 2.6 The ultra lter semigroup 0+ 99 2.6 The ultra lter semigroup 0+ Unless, otherwise stated, we consider the semigroup (R;+). De nition 2.6.1. [25] Let S be a subsemigroup of (R;+). We de ne the following, where R is equipped with the usual topology. 1. : S ! [ 1;1] is the continuous extension of the identity function. 2. B(S) = 1( 1;1). 3. Given x 2 R, x+ = 1(x) \ (x;1) \ S and x = 1(x) \ ( 1; x) \ S. 4. U = S x2R x + and D = S x2R x It follows that an ultra lter p on S is in B(S) if and only if there is some n 2 N such that [ n; n] \ S 2 p. Lemma 2.6.2. [25] Let S be a subsemigroup of R. (a) Let x 2 R and let p 2 S. Then p 2 x+ if and only if for every > 0, (x; x + ) \ S 2 p. Also p 2 x if and only if for every > 0, (x ; x) \ S 2 p. (b) Let x 2 R. Then x+ 6= ; if and only if x 2 c?R((x;1)\S) and x 6= ; if and only if x 2 c?R(( 1; x) \ S). (c) Let p; q 2 B(S), let x = (p), and let y = (q). If p 2 x+, then p+ q 2 (x+ y)+. If p 2 x , then p+ q 2 (x+ y) . (d) B(S) n S = U [D. If U and D are nonempty, they are disjoint right ideals of B(S). In particular, B(S) is not commutative. (e) If 0+ 6= ;, then 0+ is a compact subsemigroup of B(S). (f) If x 2 S and p 2 S, then x+ p = p+ x. Proof. (a) Let p 2 x+ and let > 0. Then (x ; x+ ) is a neighborhood of x in R and, therefore, (x ; x + ) \ S 2 p. It follows that (x; x + ) \ S = ((x ; x + ) \ S) \ ((x;1) \ S) 2 p. Conversely, suppose (x; x + ) \ S 2 p for every > 0. Since (x; x + ) \ S (x;1) \ S, (x;1) \ S 2 p. In addition, since for every neighborhood U of x in [ 1;1], U \ S (x; x + ) \ S for some > 0, (p) = x. Therefore, p 2 x+ \ (x;1) \ S. Similarly the second statement follows. 2.6 The ultra lter semigroup 0+ 100 (b) Let x 2 R. Suppose x+ 6= ; . Pick p 2 x+ and let U R be a neighbor- hood of x in R. Then, (x+ ; x ) U for some > 0. Then, (x; x+ )\S 2 p. In particular (x; x+ )\S 6= ;. Therefore, ; 6= (x; x+ )\S U \ (x;1)\ S. It follows that x 2 c?R(( 1; x)\ S). Conversely, sup- pose x 2 c?R(( 1; x)\S). Then, there exists an ultra lter p on S con- taining the family fU \((x;1)\S) : U is a neighborhood of x in Rg. It then follows that (p) = x and, for example taking U = (x ;1) for some > 0, we see that (x;1) \ S 2 p. Similarly one establishes the second statement. (c) Assume rst that p 2 x+. To see that p + q 2 (x + y)+, let > 0 be given and let A = (x+ y; x+ y+ )\S. T see that A 2 p+ q, we show that (x; x+ )\S fz 2 S : z+A 2 qg. So let z 2 (x; x+ )\S. Let = minfz x; x+ zg. Since (q) = y, we have (y ; y+ )\S 2 q and (y ; y+ )\ S z +A, so z +A 2 q as required. The proof that p+ q 2 (x+ y) is nearly identical. (d) Let p 2 B(S) n S. Let x = (p). Then x 2 ( 1;1) and x =2 S. Since p is an ultra lter, ( 1; x) 2 p or (x;1) 2 p. Assume that (x;1) 2 p. Then p 2 x+. Suppose p 2 y for some y 2 R. Then, x = (p) = y and ( 1; x) \ S = ( 1; y) \ S 2 p. This implies that ; 2 p, which is a contradiction. It follows that D \ U = ;. It follows easily from (c) that D and U are right ideals of B(S). In particular, B(S) is not commutative. (e) Suppose 0+ 6= ;. It follows easily from (c) that 0+ is a subsemigroup of B(S). Since 0+ = 1(0) \ (0;1) \ S is closed in S, 0+ is compact. (f) It su ces to note that the left shift x and the right shift x are con- tinuous functions agreeing on S, hence on Sd. 0+ is a subsemigroup of R, while for any other x 2 R, x+ and x are not subsemigroups of R. In addition, the following theorem shows that 0+ holds all of the algebraic structure of B(R) not already revealed by R. Theorem 2.6.3. [25] Let S = R. (a) 0+ and 0 are isomorphic. (b) The function ? : Rd (f0g[0+[0 )! B(R) de ned by ?(x; p) = x+p is a continuous isomorphism onto B(R). 2.6 The ultra lter semigroup 0+ 101 Proof. (a) De ne : 0+ ! 0 by (p) = p, where p = f A : A 2 pg. Let p; q 2 0+ with (p) = (q). Then, p = ( p) = ( q) = q. Therefore is injective. Since S = R, p 2 0+ if and only if (p) = 0 and (0;1) 2 p if and only if (p) = 0 and ( 1; 0) 2 p. Since 0 has the neighborhood base f( ; ) : > 0g and since ( ; ) = ( ; ), i.e. 0 has a neighborhood base consisting of symmetric subsets of R, (p) = 0 if and only if ( p) = 0. It follows that p 2 0+ if and only if p 2 0 . Hence is surjective. It follows that is bijective. Let p; q 2 0+. To see that (p+q) = (p)+ (q) it su ces, since (p+q) and (p)+ (q) are both ultra lters, to show that (p+q) (p)+ (q). So let A 2 (p+ q). Then A 2 p+ q so B = fx 2 R : x+ ( A) 2 qg 2 p and hence B 2 (p). Then B fx 2 R : x + A 2 (q)g so A 2 (p) + (q) as required. (b) To see that ? is a homomorphism, let (x; p) and (y; q) be in Rd (f0g[ 0+[0 ). Since p+y = y+p, we have ?(x; p)+?(y; q) = ?(x+y; p+q). To see that ? is injective, assume we have ?(x; p) = ?(y; q). By Lemma 2.51 (c) we have x = (x + p) = (y + q) = y. Then x + p = x + q so p = x + x + p = x + x + q = q. To see that ? is surjective, let q 2 B(R) and let x = (q). Let p = x+ q. Then q = x+ p = ?(x; p). To see that ? is continuous, let (x; p) 2 Rd (f0g [ 0+ [ 0 ) and let A 2 x+p. Then x+A 2 p so fxg (c? S( x+A)) is a neighborhood of (x; p) contained in ? 1(c? SA). 0+ has yielded useful combinatorial results. Therefore the algebra of 0+ is of great interest. Since fx 2 R : x 0g does not contribute to 0+ at all, we will assume that 0+ is de ned in terms of some semigroup S (0;1). Since we want 0+ 6= ;, we need 0 2 c?RS. In addition, this su ces to imply that S is dense in (0;1). Furthermore, 0+ is far from being an ideal of B(S). In fact, 0+ is the inverse image of 0 under the homomorphism S ! [0;1] and, thus, 0+ is a prime subsemigroup of S. It follows that no general results apply to 0+ beyond those that apply to any compact right topological semigroup. 2.6 The ultra lter semigroup 0+ 102 Remark 2.6.4. Let T be the translation invariant topology on (R;+) with the following neighborhood base at 0: B = f[0; 1 n ) : n 2 Ng; i.e. T is the Sorgenfrey topology. Let S be a dense subsemigroup of (0;1). Let us still denote by T , the topology induced on S by the Sorgenfrey topol- ogy. Then 0+ is the ultra lter semigroup of ((0;1) \ S; T ), i.e. 0+ = Ult(T ) = \ n2N In; where In = (0; 1n) \ S. Proof. Let p 2 0+. Then (p) = 0 and (0;1) \ S 2 p. (p) = 0 implies that for every neighborhood U of 0 in T , U \ (0;1) \ S 2 p. In particular, (0; 1n) \ S 2 p for every n 2 N. Therefore 0 + Ult(T ). Conversely, let p 2 Ult(T ). Since f(0; 1n) : n 2 Ng is a base for the neighborhood lter at 0, it follows easily that p 2 0+. Remark 2.6.5. It is worth noting that (H; ) is an ideal of ( N; ) (see for example [[23], Theorem 7.5]) and, similarly, ((0+); ) is an ideal of ( (0; 1); ). Therefore, much is known about the structures of (H; ) and ((0+); ). Chapter 3 The smallest ideal and its closure 3.1 Combinatorial characterizations of K(T) and its closure In this section we present the extension of the combinatorial characteriza- tions of K( S) and its closure to K(T) and its closure. Most of the materials in this section come from [22]. Let S be a semigroup. Recall the following. (a) A subset A S is said to be syndetic if there exists a nite G S such that S = G 1A. (b) A subset A S is said to be piecewise syndetic if there exists a nite G S such that for every nite F S there exists x 2 S such that Fx G 1A. (c) K( S) can be characterized as : K( S) = fp 2 S : for all A 2 p; fx 2 S : x 1A 2 pg is syndeticg: (d) c? K( S) can be characterized as: c? K( S) = fp 2 S : for all A 2 p; A is piecewise syndeticg: 103 3.1 Combinatorial characterizations of K(T) and its closure 104 Considering the semigroup (N;+) we observe that statements (a) and (b) can be written as follow. (a0) A subset A N is syndetic if and only if there exists l 2 N such that for every x 2 N there is t 2 N with 1 t l such that t + x 2 A if and only if there exists l 2 N such that N = Sl t=1( t+ A). (b0) A subset A S is piecewise syndetic if and only if there exists l 2 N such that for each n 2 N there exists x 2 N with fx+ 1; ; x+ ng Sl t=1( t+ A). In statement (b0) su ciency follows easily from the fact that for every nonempty F 2 Pf (N) and every x 2 N, x+ F fx+ 1; x+ 2; ; x+ maxFg. Proposition 3.1.1. For every discrete semigroup S, c? K( S) is an ideal. Proof. Since S is a right topological semigroup and K( S) is a right ideal, c? K( S) is a right ideal (see Section 1.4). Since S is a right topological semigroup with a dense topological center, c? K( S) is a left ideal. It follows that c? K( S) is an ideal. T is a compact right topological semigroup and, therefore, T has the smallest ideal K(T). We would like to present the characterizations of K(T) and its closure and then show that c?T K(T) is not an ideal. Lemma 3.1.2. [[27], Lemma 1.5] If A is a left ideal, right ideal, a minimal left ideal, a minimal right ideal, or the smallest ideal of N, then A \ T is the corresponding object in T. Proof. Let A be a left ideal or right ideal of N. Then A contains an idem- potent. Therefore, A\ T 6= ;. It is clear that A\ T is a left ideal or a right ideal of T. Observe that if p 2 T, q 2 N n N, and p + q 2 T or q + p 2 T, then q 2 T. Consequently, A\T is minimal in T if A is minimal in N. As a consequence of Lemma 3.1.2, we have the following description of K(T). Corollary 3.1.3. K(T) = K( N) \ T. In fact, K(H) = K( N) \H, see [[29], Theorem 1.65] 3.1 Combinatorial characterizations of K(T) and its closure 105 De nition 3.1.4. [[22], De nition 2.1 (b)] Let A N. A is said to be piecewise T-syndetic if there exists a sequence (lm)N in N such that for each n 2 N there exists x 2 N such that whenever m 2 N with m n, fx+ 1; ; x+ ng lm[ t=1 ( t+ (A \mN)): Remark 3.1.5. Let A N. (a) If A is piecewise T-syndetic, then A is piecewise syndetic. (b) The converse of (a) is false. Proof. (a): Suppose A is piecewise T-syndetic. Pick a sequence (lm)N in N satisfying the condition in De nition 3.1.4 and take l = l1. Let n 2 N. By assumption, there exists x 2 N such that f1 + x; ; n+ xg l[ t=1 ( t+ (A \ 1N)) = l[ t=1 ( t+ A): Thus A is piecewise syndetic. (b): Let A be a set of odd positive integers, which is syndetic in N. Since A \ 2N = ;, then A can not satisfy the condition in De nition 3.1.4 for any sequence (lm)N in N. Hence A is piecewise syndetic but not piecewise T-syndetic. The following lemma gives a stronger version of De nition 3.1.4. Lemma 3.1.6. [[22], Lemma 2.6] Let A N. A is piecewise T-syndetic if and only if there exists a sequence (lm)N in N such that for each n 2 N there exists x 2 n!N such that whenever m 2 N with m n, then fx+ 1; ; x+ ng lm[ t=1 ( t+ (A \m!N)): Proof. (() Suppose A satis es the given condition. Since k!N N and k!N kN for every k 2 N, it follows easily that A is piecewise T- syndetic. 3.1 Combinatorial characterizations of K(T) and its closure 106 ()) Suppose A is piecewise T-syndetic. Pick a sequence (lm)N such that for each n 2 N there exists x 2 N such that whenever m 2 N with m n, then fx + 1; ; x + ng Slm t=1( t + (A \ mN)). Form a sequence (l0m)N in N by taking l 0 m = lm! for each m 2 N. Let n 2 N be given and let r = n! + n. Pick y 2 N such that for m r, fy + 1; ; y + rg Slm t=1( t+ (A \mN)). Pick x 2 fy + 1; ; y + n!g \ n!N. Let m n so that m! n! < n! + n = r. Then fx+ 1; ; x+ ng fy + 1; ; y + n! + ng = fy + 1; ; y + rg lm![ t=1 ( t+ (A \m!N)) = l0m[ t=1 ( t+ (A \m!N)): Lemma 3.1.7. [[22], Lemma 2.7] The following statements hold. (a) N is piecewise T-syndetic and ; is not piecewise T-syndetic. (b) If A B N and A is piecewise T-syndetic, then B is piecewise T-syndetic. (c) If A;B N such that A[B is piecewise T-syndetic then, A is piecewise T-syndetic or B is piecewise T-syndetic. Proof. (a) For every n 2 N and every x 2 N, fx + 1; ; x + ng 6= ;. On the other hand, for every t 2 N and every m 2 N, t + (; \ mN) = t+ ; = ;. It follows that ; is not piecewise T-syndetic. It is very easy to see that for everym 2 N and every n 2 N, f2; 3; ; 1+ ng Sm t=1( t + mN). Therefore, take (lm)N = (m)N and given any n 2 N, take x = 1. It follows that N is piecewise T-syndetic. (b) Since A B implies A \mN B \mN, the statement follows. (c) Let A;B N such that A[B is piecewise T-syndetic. Pick a sequence (lm)N as guaranteed by Lemma 3.1.6. Since the union Slm t=1( t+ ((A[ B) \ mN)) does not depend on the order of the terms l1; ; lm, we may assume that lm lm+1 for every m 2 N. For each n 2 N pick 3.1 Combinatorial characterizations of K(T) and its closure 107 xn 2 n!N such that whenever m 2 N with m n, fxn + 1; ; xn + ng Slm t=1( t+ (m!N \A)). As in [[24], Lemma 3.4] we may obtain a subsequence (yn)N of (xn)N such that (1) If t r n then yr + t 2 A if and only if yn + t 2 A: (2) yn 2 n!N and fyn + 1; ; yn + ng Slm t=1( t+ (m!N\ (A[B))) whenever m 2 N with m n. Let F = fn 2 N : yn + n 2 Ag. Then (3) for t 2 f1; 2; ; ng, yn + t 2 A if and only if t 2 F . We assume rst that there is some a 2 N such that for each k there is some z with fz+1; z+2 ; z+kg\F \a!N = ;. In this case we de ne km = lm if m a and km = la if m < a. Let s 2 N be given. We will produce x 2 N such that fx+1; x+2; ; x+sg Skm t=1( t+(B\m!N)) whenever m 2 N with m s. This will mean that B is piecewise T- syndetic. Let b = ks+s and pick r 2 N such that fr+1; ; r+bg\F \a!N = ;. Let n = maxfr+b; ag and let x = yn+r. Let m 2 f1; 2; ; sg. To see that fx+1; x+2; ; x+sg Skm t=1( t+(B\m!N)), let i 2 f1; 2; ; sg and note that x+ i = yn + r + i yn + r + b yn + n. Case 1: m < a. Then x+ i 2 fyn + 1; yn + 2; ; yn + ng and n a so x + i 2 Sla t=1( t + ((A [ B) \ a!N)). Pick t 2 f1; 2; ; lag such that x+ i+ t 2 (A[B)\ a!N. Now t la = km and x+ i+ t 2 a!N m!N so it su ces to show that x+ i+ t 2 B. Suppose instead x+ i+ t 2 A. Then yn+r+ i+ t 2 A and r+ i+ t r+s+ la r+s+ks = r+b n so, by (3), r+i+t 2 F . But also yn 2 n!N a!N and yn+r+i+t 2 a!N so r+ i+ t 2 a!N. But i+ t s+ la s+ks = b so r+ i+ t =2 F \a!N, a contradiction. Case 2: m a. Since m s n and x+i 2 fyn+1; yn+2; ; yn+ng, we have x+i 2 Slm t=1( t+((A[B)\m!N)). Pick t 2 f1; 2; ; lmg such that x+ i+ t 2 (A[B)\m!N. Then t lm = km and x+ i+ t 2 m!N so it su ces to show that x+ i+ t 2 B. Suppose instead x+ i+ t 2 A. Then yn+r+i+t 2 A and r+i+t r+s+km r+s+ks = r+b n so r+ i+ t 2 F . Also yn 2 n!N a!N and yn+r+ i+ t 2 m!N a!N so r+i+t 2 a!N. But i+t s+ks = b so r+i+t 2 F\a!N, a contradiction. Now assume that for each a, there exists k such that for each z, fz + 1; z + 2 ; z + kg \ F \ a!N 6= ;. Pick for each m some km such 3.1 Combinatorial characterizations of K(T) and its closure 108 that, for each r 2 N, fr + 1; r + 2; ; r + kmg \ F \ m!N 6= ;, and if m > 1, km km 1. Let s 2 N be given. We will produce x 2 N such that fx + 1; x + 2; x + sg Skm t=1( t + (A \ m!N)) for all m 2 f1; 2; ; sg. Let n = s+ks and let x = yn. Let m 2 f1; 2; ; sg. To see that fx + 1; x + 2; ; x + sg Skm t=1( t + (A \ m!N)), let i 2 f1; 2; ; sg. Pick t 2 f1; 2; ; kmg such that i + t 2 F \ m!N. Then i + t s + km s + ks = n. Thus yn + i + t 2 m!N. Thus yn + i + t 2 A, by (3). But also yn 2 n!N m!N so yn + i + t 2 m!N so x+ i+ t 2 A \m!N as required. Corollary 3.1.8. Let G P(N) such that G is closed under nite inter- sections and for every A 2 G, A is piecewise T-syndetic. Then there exists p 2 N such that G p and for every B 2 p, B is piecewise T-syndetic. Proof. This follows immediately from Lemma 3.1.7. Lemma 3.1.9. [[22], Lemma 2.2] K(T) fp 2 N : for all A 2 p; A is piecewise T-syndeticg: Proof. Let p 2 K(T) and let A 2 p. Then given m 2 N, A \mN 2 p, since p 2 T. Let Fm = fx 2 N : x+ (A \mN) 2 pg: Since p 2 K( N), Fm is syndetic. Therefore, pick lm 2 N such that N = lm[ t=1 ( t+ Fm): For each y 2 N and m 2 N pick t(y;m) 2 f1; 2; ; lmg such that y + t(y;m) 2 Fm. Now let n 2 N . Then for any y;m 2 f1; 2; ; ng, (y + t(y;m)) + (A \mN) 2 p. Pick x 2 n\ y=1 n\ m=1 (y + t(y;m)) + (A \mN): Let m n be given. To see that fx + 1; ; x + ng Tlm t=1, let y 2 f1; 2; ; ng. Then x+ y + t(y;m) 2 A \mN. So x+ y 2 ( t(y;m) + (A \ mN)) Slm t=1( t+ (A \mN)). 3.1 Combinatorial characterizations of K(T) and its closure 109 Theorem 3.1.10. [[22], Theorem 2.3] c?T K(T) = fp 2 N : for all A 2 p; A is piecewise T-syndeticg. Proof. Let H = fp 2 N : for all A 2 p; A is piecewise T-syndeticg. Let q 2 N n H and let B 2 q such that B is not piecewise T-syndetic. Then B \ H = ;. Therefore q 2 B N n H. It follows that H is closed. Since K(T) H, c? K(T) c? H = H. For the other inclusion, let p 2 H and let A 2 p. We would like to show that A\K(T) = ;. Choose a sequence (lm)N in N such that for each n 2 N there exists x 2 N such that whenever m 2 N with m n, fx+ 1; ; x+ ng lm[ t=1 ( t+ (A \mN)): For each n 2 N choose xn 2 N such that whenever m 2 N with m n, fxn + 1; ; xn + ng lm[ t=1 ( t+ (A \mN)): Let D = fxn+i : i; n 2 N with 2i ng and let E = fxn+i : i; n 2 N with i ng. We observe that for every m 2 N, fx2m + 1; ; x2m + mg D and, therefore, mN\D 6= ;. In addition, m1N\ \mkN = m1 mkN for every m1;m2; ;mk 2 N. Therefore the family D[fmN : m 2 Ng has the nite intersection property. Pick q 2 N such that D[fmN : m 2 Ng q. Then q 2 T. Next we observe that q + N E. Indeed, let r 2 N. To see that E 2 q + r, we show that fx 2 N : x + E 2 qg = N. Let m 2 N. To see that m + E 2 q, we show that fxn + i : i; n 2 N; 2i n; and 2m ng m + E. (Since all but nitely members of D are in m + E, we then have m + E 2 q.) Let i; n 2 N with 2i n and 2m n. Then i + m n so xn + i + m 2 E as required. Now q + T is a right ideal of T so (q + T) \ K(T) 6= ;. Pick r 2 (q + T) \ K(T). Since q + T E, E 2 r. We claim that ffx 2 N : x + A 2 rgg [ fmN : m 2 Ng has the nite intersection property. As before, it su ces to let m 2 N and show that fx 2 N : x+A 2 rg\mN 6= ;. Now E 2 r so, since it includes all but nitely members of E, fxn+ i : i; n 2 N and i n and m ng 2 r. By the choice of the x0ns, fxn + i : i; n 2 N and i n and m ng Slm t=1( t+ (A\mN)), soSlm t=1( t+(A\mN)) 2 r. Pick t 2 f1; 2; ; lmg such that t+(A\mN) 2 r. Now r 2 T so mN 2 r so pick x 2 ( t+ (A\mN))\mN. Then x 2 mN and x+t 2 mN so t 2 mN. Also t+A 2 r. Thus fx 2 N : x+A 2 rg\mN 6= ;. 3.1 Combinatorial characterizations of K(T) and its closure 110 Pick s 2 N such that ffx 2 N : x + A 2 rgg [ fmN : m 2 Ng s. Then s 2 T and A 2 r+s. Since r 2 K(T), r+s 2 K(T) and hence r+s 2 K(T)\A as required. Corollary 3.1.11. Let G P(N) such that G is closed under nite inter- sections and for every A 2 G, A is piecewise T-syndetic. Then there exists p 2 c?K(T) such that G p. Proof. This follows immediately from Corollary 3.1.8 and Theorem 3.1.10. Theorem 3.1.12. [[22], Theorem 2.8] c?T K(T) is not a left ideal of T. Proof. Take any bijection : N ! N N and de ne a doubly indexed sequence (u(a;b))(a;b)2N N recursively as follows. u( (n)) = 1 + maxfu(a;b) + a! + b+ a : (a; b) = (k) for some k < ng: Then (u(a;b))(a;b)2N N is a one-to-one sequence and for each (a; b) 2 N N we have u(a;b) + a! + b+ a < minfu(c; d) : u(c; d) > u(a; b)g: For each a 2 N let Ea = [ b2N fu(a; b) + 1; U(a; b) + 2; ; u(a; b) + bg \ a!N and for each k 2 N let Ak = [ a k Ea: Since Ak Ak+1 for every k 2 N, fAk : k 2 Ng is closed under nite intersections. We claim that for every k 2 N, Ak is not piecewise T-syndetic. Let k 2 N be given. De ne a sequence (lm)N in N as follows. lm = m! if m > k k! if m k: Let n 2 N be given and let d = maxfn; kg. Let x = u(k;d!+d) and let m n be given. We would like to show that fx+1; ; x+ng Slm t=1( t+(Ak\mN)). We look at two cases. Case 1: m k. Pick t 2 f1; 2; ; k!g such that x + i + t 2 k!N. Then x + i + t 2 mN. Also i + t n + k! d + d! so x + i + t 2 fu(k;d+d!) + 1; u(k;d+d!) + 2; ; u(k;d+d!) + d+ d!g\ k!N. Thus x+ i+ t 2 mN as required. 3.1 Combinatorial characterizations of K(T) and its closure 111 Case 2: m > k. Pick t 2 f1; 2; ;m!g such that x + i + t 2 m!N. Then x + i + t 2 mN \ k!N. Also i + t n + m! d + d! so x + i + t 2 Ak \mN as required. Therefore we can pick p 2 c? K(T) such that fAk : k 2 Ng p. Pick q 2 T such that fk! : k 2 Ng 2 q. Let B = S k2N( (k 1)!+Ak). Then B 2 q+p. We show that B is not piecewise T-syndetic, so that q+p =2 c? K(T). Suppose instead that B is piecewise T-syndetic. Pick a sequence (lm)N in N such that for each n 2 N there exists x 2 n!N such that whenever m 2 N with m n, then fx+1; ; x+ng Slm t=1( t+(B\m!N)). Pick m > l1 and pick 2 N such that ! ( 1)! > l1. Let n = lm + ( 1)l1 +m. Pick x 2 n!N such that, for each r 2 N with r n, fx+ 1; ; x+ ng Slr t=1( t+ (B \ r!N)). Pick t0 2 N with t0 lm such that x+ 1 + t0 2 B \m!N. Recursively given i 2 f1; 2; ; g, t0 2 f1; 2; ; lmg, and ftj : j 2 f1; 2; ; i 1gg f1; 2; ; l1g, we have x+ 1 + i 1X j=0 tj x+ 1 + lm + ( 1)l1 x+ n: Therefore pick ti 2 f1; 2; ; l1g such that x + 1 + Pi j=0 tj 2 B. For each i 2 f0; 1; ; g pick ki 2 N such that x+ 1 + Pi j=0 tj 2 Aki + (ki 1)! and, since x+ 1 + Pi j=0 tj (ki 1)! 2 Aki , pick ai ki and bi 2 N such that x+ 1 + iX j=0 tj (ki 1)! 2 fu(ai;bi) + 1; u(ai;bi) + 2; ; u(ai;bi) + big \ a!N: We show rst that a0 > m. Suppose instead a0 m. Then since x+ 1 + t0 2 m!N, a0! divides x+ 1 + t0. But also x+ 1 + t0 (k0 1)! 2 a0!N so that a0! divides (k 1)! while a0 k0, a contradiction. We now show that each i 2 f1; 2; ; g, (ai; bi) = (a0; b0). Suppose not and pick the rst i such that (ai; bi) 6= (a0; b0). Since (u(a;b))(a;b)2N N is a one-to-one sequence, u(ai;bi) 6= u(a0;b0). 3.1 Combinatorial characterizations of K(T) and its closure 112 Case 1: u(ai;bi) < u(a0;b0). Then u(a0;b0) > u(ai;bi) + ai! + bi + ai > u(ai;bi) + bi + (ki 1)! x+ 1 + iX j=0 tj (ki 1)! ! + (ki 1)! = x+ 1 + iX j=0 tj > x+ 1 + t0 > x+ i+ t0 (k0 1)! > u(a0;b0); a contradiction. Case 2: u(ai;bi) > u(a0;b0). Note that a0 > m > l1. Thus u(ai;bi) > u(a0;b0) + a0! + b0 + a0 > u(a0;b0) + b0 + a0! + l1 = u(ai 1;bi 1) + bi 1 + l1 + ai 1! > x+ 1 + i 1X j=0 tj (ki 1 1)! ! + ti + (ki 1 1)! = x+ 1 + iX j=0 tj > x+ 1 + iX j=0 tj (kj 1)! > u(ai;bi); a contradiction. Let (a; b) = (a0; b0), so that for all i 2 f0; 1; ; g, (a; b) = (ai; bi). We have established that a > m. Let i be any member of f1; 2; ; g. Then a! divides x+ 1 + Pi j=0 tj (ki 1)! and a! divides x+ 1 + Pi 1 j=0 tj (ki 1 1)!. Therefore, a! divides ti (ki 1)! + (ki 1 1)!. Pick an integer d such that da! = ti (ki 1)! + (ki 1 1)!. Then da! < ti + (ki 1 1)! ti + (a 1)! l1 + (ai 1)! < m+ (a 1)! < 2(a 1)!. Thus (a 1)!(da 2) < 0 so da < 2. Since a > m > l1, a > 2 so d 0. But also da! = ti (ki 1)! + (ki 1 1)! (ki 1)! (a 1)!. Thus (a 1)!(da+ 1) > 0 so da+ 1 > 0 so d 0 and hence d = 0. Consequently, for each i 2 f1; 2; : : : ; g, ti = (ki 1)! (ki 1 1)!. Since ti 1, ki ki 1. Since k0 1, ki i+1. Thus t = (k 1)! (k 1 1)! (k 1)! (k 2)! ! ( 1)! > l1 t , a contradiction. 3.2 Combinatorial characterizations of K(0+) and its closure 113 Corollary 3.1.13. K(T) is not closed. In particular, c?T K(T) 6= (c? N K( N)) \ T. Proof. Since c?T K(T) is not an ideal of T, c?T K(T) 6= K(T). In particular K(T) is not closed. Since c? N K( N) is an ideal of N, by Lemma 3.1.2 c? N K( N)\T is an ideal of T. It follows that c?T K(T) 6= (c? N K( N))\ T. 3.2 Combinatorial characterizations of K(0+) and its closure In this section we present the extension of the combinatorial characterizations of K( S) and its closure to K(0+) and its closure. Most of the materials in this section come from [25]. Since 0+ is a compact right topological semigroup, K(0+) exists. However, since 0+ is not an ideal of R, a description of K(0+) does not follow from known results about arbitrary discrete semigroups. We would like to present the characterization of the members of K(0+) and its closure. De nition 3.2.1. Let S be a dense subsemigroup of (0;1). A subset A S is said to be syndetic near 0 if and only if for every > 0 there exist some nite F (0; ) \ S and some > 0 such that S \ (0; ) S t2F ( t+ A). The following theorem characterizes K(0+). Theorem 3.2.2. [[25], Theorem 3.3] Let S be a dense subsemigroup of (0;1) and let p 2 0+. The following statements are equivalent. (a) p 2 K(0+). (b) For all A 2 p, fx 2 S : x+ A 2 pg is syndetic near 0. (c) For all r 2 0+, p 2 0+ + r + p. Proof. (a)) (b). Suppose p 2 K(0+), let A 2 p, let B = fx 2 S : x+A 2 pg and suppose B is not syndetic near 0. Pick > 0 such that for all F 2 Pf ((0; ) \ S) and all > 0, (S \ (0; )) \ (S n F 1B) 6= ;. Let G = f(S \ (0; )) \ (S n F 1B) : F 2 Pf ((0; ) \ S) and > 0g. 3.2 Combinatorial characterizations of K(0+) and its closure 114 Then G has the nite intersection property. Pick r 2 S with G r. Since fS \ (0; ) : > 0g r we have r 2 0+. Pick a minimal left ideal L of 0+ with L 0+ + r. Since K(0+) is the union of all the minimal right ideals of 0+, pick a minimal right ideal R of 0+ with p 2 R. Then L \ R is a group. Let q be the identity of L \R. Then R = q + 0+, so p 2 q + 0+ so p = q + p and therefore B 2 q. Also q 2 0+ + r so pick w 2 0+ such that q = w + r. Then (0; ) \ S 2 w and ft 2 S : t+B 2 rg 2 w so pick t 2 (0; ) \ S such that t+B 2 r. But (S \ (0; 1)) n ( t+B) 2 G r, a contradiction. (b)) (c). Let r 2 0+. For each A 2 p, let B(A) = fx 2 S : x+A 2 pg and let C(A) = ft 2 S : t+B(A) 2 rg. It is easy to see that for any A1; A2 2 p , one has B(A1 \ A2) = B(A1) \B(A2) and C(A1 \ A2) = C(A1) \ C(A2). We claim that for every A 2 p and every > 0, C(A) \ (0; ) 6= ;. To see this, let A 2 p and let > 0 be given and pick F 2 Pf ((0; ) \ S) and > 0 such that (0; ) \ S F 1B(A). Since (0; ) \ S 2 r we have F 1B(A) 2 r and hence there is some t 2 F with t + B(A) 2 r. Then t 2 C(A) \ (0; ) and, therefore, C(A) \ (0; ) 6= ;. Thus f(0; ) \ C(A) : > 0 and A 2 pg has the nite intersection property so pick q 2 S with f(0; )\C(A) : > 0 and A 2 pg q. Then q 2 0+. We claim that p = q + r + p. It su ces to show that p q + r + p. Let A 2 p be given. Then ft 2 S : t + B(A) 2 rg = C(A) 2 q so B(A) 2 q + r, so A 2 q + r + p as required. (c)) (a). Pick r 2 K(0+). De nition 3.2.3. Let S be a dense subsemigroup of (0;1). A subset A S is said to be piecewise syndetic near 0 if and only if there exist sequences (Fn)n2N and ( n)N such that (a) For each n 2 N, Fn is a nite subset of (0; 1n) \ S and 0 < n < 1 n and (b) for all nite G S and all > 0 there is some x 2 (0; )\S such that for all n 2 N, (G \ (0; n)) + x S t2Fn ( t+ A). Theorem 3.2.4. [[25], Theorem 3.5] Let S be a dense subsemigroup of (0;1) and let A S. The following statements are equivalent. (a) K(0+) \ A 6= ;. (b) A is piecewise syndetic near 0. 3.2 Combinatorial characterizations of K(0+) and its closure 115 Proof. (a) ) (b): Suppose K(0+) \ A 6= ;. Let p 2 K(0+) \ A and let B = fx 2 S : x+A 2 pg. Then B is syndetic near 0. Inductively for n 2 N pick Fn 2 Pf ((0; 1n) \ S) and n 2 (0; 1 n) (with n n 1 if n > 1) such that S \ (0; n) Fn +B. Let G 2 Pf (S) be given. If G \ (0; 1) = ;, the conclusion is trivial, so assume G \ (0; 1) 6= ; and let H = G \ (0; 1). For each y 2 H let m(y) = maxfn 2 N : y < ng. For each n 2 f1; 2; ;m(y)g, we have y 2 Fn + B so pick t(y; n) 2 Fn such that y 2 t(y; n) + B. Then given y 2 H and n 2 f1; 2; ;m(y)g, one has (t(y; n) + y) + A 2 p. Now let > 0 be given. Then (0; ) 2 p so pick x 2 (0; ) \ \ y2H m(y)\ n=1 ( (t(y; n) + y) + A)) : Then given n 2 N and y 2 G \ (0; n), one has y 2 H and n m(y) so t(y; n) + y + x 2 A so y + x 2 t(y; n) + A Fn + A (b) ) (a): Suppose A is piecewise syndetic near 0. Pick sequences (Fn)n2N and ( n)N such that 1. For each n 2 N, Fn is a nite subset of (0; 1n) \ S and 0 < n < 1 n and 2. for all nite G S and all > 0 there is some x 2 (0; )\S such that for all n 2 N, (G \ (0; n)) + x Fn + A. Given G 2 Pf (S) and > 0, let C(G; ) = fx 2 (0; ) \ S : for all n 2 N; (G \ (0; n)) + x Fn + Ag: By assumption each C(G; ) 6= ;. Further, given G1 and G2 in Pf (S) and 1; 2 > 0, one has C(G1 [G2; minf 1; 2g) C(G1; 1) \ C(G2; 2): Therefore, fC(G; ) : G 2 Pf (S) and > 0g has the nite intersection property. Pick p 2 S with fC(G; ) : G 2 Pf (S) and > 0g p. Since C(G; ) (0; ), one has p 2 0+. We claim that 0+ + p \ n2N ( Fn + A): 3.2 Combinatorial characterizations of K(0+) and its closure 116 Let n 2 N and let q 2 0+. To show that Fn + A 2 q + p, we show that (0; n) \ S fy 2 S : y + ( Fn + A) 2 pg: So let y 2 (0; n)\S. Then C(fyg; n) 2 p and C(fyg; n) y+( Fn+A). Now pick r 2 (0+ + p) \K (since 0+ + p is a left ideal of 0+). Given n 2 N, Fn + A 2 r so pick tn 2 Fn such that tn + A 2 r. Now for each n 2 N, tn 2 Fn (0; 1n) so limn!1 tn = 0 so pick q 2 0 + \ ftn : n 2 Ng. Then q + r 2 K(0+) and ftn : n 2 Ng ft 2 S : t+ A 2 rg so A 2 q + r. Therefore we have the following characterization of c?K(0+). Corollary 3.2.5. Let S be a dense subsemigroup of (0;1) and let p 2 S. The following statements are equivalent. (a) p 2 c?K(0+). (b) Every A 2 p is piecewise syndetic near 0. It follows that: c?0+ K(0+) = fp 2 S : for all A 2 p; A is piecewise syndetic near 0g: De nition 3.2.6. Let S be a dense subsemigroup of (0;1). (a) A subset A S is said to be central near 0 if and only if there is some idempotent p 2 K(0+) with A 2 p. (b) A collection A P(S) is said to be collectionwise piecewise syndetic near 0 if and only if there exist functions F : Pf (A)! Y n2N ((0; 1 n ) \ S) and : Pf (A)! Y n2N (0; 1 n ) such that for every > 0, every G 2 Pf (S), and every H 2 Pf (A), there is some t 2 (0; ) \ S such that for every n 2 N and every F 2 Pf (H), (G \ (0; (F)n)) + t [ x2F (F)n ( x+ \ F): Theorem 3.2.7. Let S be a dense subsemigroup of (0;1) and let A P(S). There exists p 2 K(0+) such that A p if and only if A is collectionwise piecewise syndetic near 0. 3.3 Generalization of combinatorial characterizations 117 3.3 Generalization of combinatorial charac- terizations At this stage we would like to generalize even further the de nitions of synde- tic and piecewise syndetic subsets of a semigroup. This time we will consider an arbitrary closed subsemigroup of S. Recall that every closed subsemi- group T of S is of the form F for some lter F on S. The materials in this section and in the next section are the contents of [60]. De nition 3.3.1. Let S be a semigroup and let F and G be lters on S. A subset A S is (F ;G)-syndetic if for every V 2 F , there is a nite F V such that F 1A 2 G. If F = G, we say F-syndetic instead of (F ;G)-syndetic. Remark 3.3.2. Let S be a semigroup, let F and G be lters on S and let A S. (a) If F G and A is either F-syndetic or G-syndetic, then A is (F ;G)- syndetic. (b) A is syndetic if and only if A is fSg-syndetic. Hence, (F ;G)-syndetic is a generalization of syndetic. Proof. (a) Suppose F G. Let V 2 F . Since F G, V 2 G. If A is F-syndetic, there exists a nite F V such that F 1A 2 F G. If A is G- syndetic, there exists a nite E V such that E 1A 2 G. In both cases A is (F ;G)-syndetic. (b) Suppose A is syndetic. Let V 2 fSg. Then V = S. By assumption, there exists a nite F S such that F 1A = S 2 fSg. Therefore, A is fSg-syndetic. The converse is clear. Lemma 3.3.3. Let S be a semigroup, let T be a closed subsemigroup of S, let L be a minimal left ideal of T , let F and G be lters on S such that F = T and G = L, and let A S. The following statements are equivalent. (1) A \ L 6= ;. 3.3 Generalization of combinatorial characterizations 118 (2) A is G-syndetic. (3) A is (F ;G)-syndetic. Proof. (First note that since L is a nonempty closed subset of T (being a minimal left ideal), L is a nonempty closed subset of S. In particular G such that G = L exists.) (1) ) (2): Suppose A \ L 6= ;. Pick p 2 A \ L. For every q 2 L, one has p 2 L = Lq. Thus for each q 2 L, there exists rq 2 L such that p = rqq. Now to show that A is G-syndetic, let V 2 G. Since L = G, V 2 rq for all q 2 L. Since A 2 p = rqq, fx 2 S : x 1A 2 qg 2 rq for each q 2 L. Since rq is a lter, V \ fx 2 S : x 1A 2 qg 6= ; for every q 2 L. For each q 2 L, choose xq 2 V such that x 1q A 2 q, i.e. such that q 2 x 1q A. Then the set f x 1q A : q 2 L g is an open cover of L. Since L is compact, there exists a nite fq1; ; qng L such that L x 1q1 A [ [ x 1 qn A = x 1 q1 A [ [ x 1 qn A = F 1A; where F = fxq1 ; ; xqng V . G F 1A implies that F 1A 2 G. It follows that A is G-syndetic. (2) ) (3): Suppose A is G-syndetic. Since G = L T = F , F G. By Remark 3.3.2 (a), A is (F ;G)-syndetic. (3) ) (1): Suppose A is (F ;G)-syndetic. Pick q 2 L. For every V 2 F , take a nite F = fx1; ; xng V such that F 1A 2 G q. Then, x 11 A [ [ x 1 n A = F 1A 2 q. Since q is an ultra lter, x 1i A 2 q for some 1 i n. Therefore, for every V 2 F choose xV 2 V such that A 2 xV q. Pick r 2 T\c? SfxV : V 2 Gg. Since ft 2 S : t 1A 2 qg fxV : V 2 Gg 2 r, ft 2 S : t 1A 2 qg 2 r. Therefore A 2 rq. Since L is a left ideal of T , rq 2 L. It follows that A \ L 6= ;. The next theorem characterizes ultra lters from K(T ), where T is a closed subsemigroup of S. Theorem 3.3.4. Let S be a semigroup, let T be a closed subsemigroup of S, let F be a lter on S such that F = T , and let p 2 T . Then p 2 K(T ) if and only if for every A 2 p, fx 2 S : x 1A 2 pg is F-syndetic. Proof. Suppose that p 2 K(T ). Let L = Tp. Then L is a left ideal of T and, since T is a subsemigroup of S, L is a left ideal of S. Since K(T ) Tp, p 2 Tp. Let G be a lter on S such that is L = G. Now let A 2 p, let B = fx 2 S : x 1A 2 pg and let V 2 F . Then p 2 A \ L. It follows that 3.3 Generalization of combinatorial characterizations 119 A is G-syndetic. Pick a nite subset F V such that F 1A 2 G. Since L = Tp, there is W 2 F such that Wp f 1A. We claim that W F 1B. Indeed, let y W . Then there is x 2 F such that yp 2 x 1A. It follows that (xy) 1A 2 p. Hence, xy 2 B, and then y 2 x 1B. Conversely, suppose that p =2 K(T ). Pick q 2 K(T ). Then Tqp K(T ) and, therefore, p =2 Tqp. It follows that there is A 2 p such that A\Tqp = ;. Now let B = fx 2 S : x 1A 2 pg. We claim that B is not F-syndetic. To show this, pick a minimal left ideal L of T contained in Tp and let G be the lter on S such that G = L. Assume on the contrary that B is F-syndetic. Then B is also (F ;G)-syndetic. Hence by Lemma 3.3.3, B \L 6= ;. Consequently, B 2 rq for some r 2 T , and so fx 2 S : x 1A 2 pg 2 rq. But then A 2 rqp, which is a contradiction. Let S be a dense subsemigroup of ((0;1);+), let F be the lter on S with F = 0+ and let A S (equivalently let (R; T ) be the Sorgenfrey topological space, let F denote the neighborhood lter at 0 and let A R.). It follows form Theorem 3.2.2 and Theorem 3.3.4 that A is syndetic near 0 if and only if A is F-syndetic. Now we characterize ultra lters from c? K(T ). De nition 3.3.5. Let S be a semigroup and let F be a lter on S. A subset A S is piecewise F-syndetic if for every V 2 F , there is a nite FV V and WV 2 F such that the family f(x 1F 1V A) \ V : V 2 F ; x 2 WV g has the nite intersection property. Remark 3.3.6. 1. Let S be a semigroups and let A S. A is piecewise syndetic if and only if A is piecewise fSg-syndetic. Therefore piecewise F-syndetic is a generalization of piecewise syndetic. 2. The family f(x 1F 1V A)\V : V 2 F ; x 2 WV g has the nite intersec- tion property if and only if whenever V1; ; Vn 2 F and H1; ; Hn are nite subsets of WV1 ; ;WVn , respectively, there is y 2 V1\ \Vn such that Hiy F 1 Vi A for each i = 1; ; n. Proof. 3.3 Generalization of combinatorial characterizations 120 1. This is clear from the de nitions. 2. The necessity is easy to see, by taking each Hi to be a singleton. Now to prove the su ciency. Note that we may assume that each Hi is nonempty. For each 1 i i, pick xi 2 Hi WVi . Pick y 2 n\ i=1 (x 1i F 1 Vi A \ Vi): Then xiy 2 F 1 Vi A for every i 2 f1; 2; ; ng. Since xi was an arbitrary element of Hi, we conclude that Hiy F 1 Vi A. Theorem 3.3.7. Let T be a closed subsemigroup of S, let F be a lter on S such that T = F , and let A S. Then A \K(T ) 6= ; if and only if A is piecewise F-syndetic. Proof. Suppose A \K(T ) 6= ;. Pick p 2 A \K(T ) 6= ;. Let L = Tp and let G denote the lter on S such that G = L. Then by Lemma 3.3.3, A is (F ;G)- syndetic. Consequently, for every V 2 F , there is a nite FV V such that F 1V A 2 G. Since L = Tp, it follows that for every V 2 F , there is WV 2 F such that WV p F 1 V A, and so x 1F 1V A 2 p for all x 2 WV . Hence, the family f(x 1F 1V A)\V : V 2 F ; x 2 WV g has the nite intersection property. It follows that A is piecewise F-syndetic. Conversely, suppose that A is piecewise F-syndetic. Then for every V 2 F,there is a nite FV V and WV 2 F such that f(x 1F 1 V A) \ V : V 2 F ; x 2 WV g has the nite intersection property. Pick an ultra lter q on S. extending this family. Then q 2 T and for every V 2 F and x 2 WV , xq 2 F 1 V A. It follows that for every V 2 F , Tq F 1 V A. Let L be the minimal left ideal of T contained in Tq and let G denote the lter on S such that G = L. Then for every V 2 F , one has F 1V A 2 G. Hence, A is (F ;G)-syndetic. Therefore, by Lemma 3.3.3, A \ L 6= ;. Corollary 3.3.8. Let T be a closed subsemigroup of S, let F be a lter on S such that T = F , and let p 2 T . Then p 2 c? K(T ) if and only if every A 2 p is piecewise F-syndetic. From Theorem 3.1.10, Theorem 3.2.5 and Theorem 3.3.7 we can see that \piecewise T-syndetic" and \piecewise syndetic near 0" are examples of \piecewise F - syndetic". 3.4 c? K(T ) is not an ideal 121 3.4 c? K(T ) is not an ideal We have seen that in a right topological semigroup the closure of a right ideal is a right ideal, while the closure of a left ideal is not necessarily a left ideal. In this section we present our main result: c? K(Ult(T )) is not a left ideal for any nondiscrete regular left invariant topology T on an arbitrary in nite group G satisfying some extra condition as stated in Lemma 3.4.1 below. So, throughout this section G is an arbitrary in nite group. Recall the following, where X is a topological space. (1) Let x 2 X. The character of x, denoted x(X), is the minimum of cardinalities of neighborhood bases at x. (2) The character of X, denoted (X), is de ned as (X) = supf x(X) : x 2 Xg: It is easy to see that for a left topological group (G; T ) and any x 2 G, x(G) = e(G) = (G), where e is the identity of G. Lemma 3.4.1. Let T be a left invariant topology on G and let denote the character of T . Then the following conditions are equivalent: (1) there is a neighborhood base fW : < g at the identity of (G; T ) such that W W if < < , and W = T < W if is a limit ordinal, (2) there is a neighborhood base fW : < g at the identity of (G; T ) such that W W if < < , and (3) for every family A T of cardinality less that , T A 2 T . If T satis es these conditions, then is regular. Proof. suppose < !. Let N = fU1; ; U g be a neighborhood base at the identity of G. Then U := U1 \ \ U 2 N and, hence, fUg is a neighborhood base at the identity. It follows that = 1 and, in this case, all three statements are clearly equivalent. Therefore we may assume that !. 3.4 c? K(T ) is not an ideal 122 (1) ) (2): This implication is clear, since condition (2) is part of condition (1). (2) ) is regular: Suppose (2) holds. Let be an ordinal such that there exists an order-preserving function f : ! with supff( ) : < g = . Then, for each < there exists < such that < f( ) and, therefore, W Wf( ). It follows that fWf( ) : < g is a neighborhood base at the identity. Therefore, = jfWf( ) : < gj . Hence is a regular cardinal. (2)) (3): Suppose (2) holds. Let A T be a family of cardinality less than and let W = T A. Let x 2 W . Then fxW : < g is a neighborhood base at x. Therefore for every V 2 A choose (V ) < such that xW (V ) V . Let = supf (V ) : V 2 Ag. Then < , since is regular, and xW W . Hence W 2 T . (3) ) (1): Suppose (3) holds. Let fV : < g be a neighborhood base at the identity of (G; T ). For every < , put W = 8 < : T V if is not limit T < V if is limit: By (3), fW : < g T . Hence fW : < g is the required neighbor- hood base at the identity. Therefore, our main theorem is as follows. Theorem 3.4.2. Let T be a nondiscrete regular left invariant topology on G and let T = Ult(T ). Suppose that T satis es the conditions from Lemma 3.4.1. Then c? K(T ) is not a left ideal of T . As an immediate consequence we obtain that Corollary 3.4.3. Both c? K(0+) and c? K(H ) are not left ideals. In the rest of this section we prove Theorem 3.4.2. Let jGj = and let denote the character of T . Lemma 3.4.4. There is a function : G! such that (a) for every X G with jXj = , j (X)j = , (b) if = !, then for every x 2 G, there is a neighborhood V of e 2 G such that (xy) = (y) for all y 2 V n feg, 3.4 c? K(T ) is not an ideal 123 (c) if > !, then for every x 2 G, there is a subset Y G with jGnY j < such that (xy) = (yx) = (y) for all y 2 Y , and (d) if < , then for every < , e 2 c?T 1( + 1). Proof. Consider two cases. Case 1 : = !. Then !. Since our topologies are assumed to satisfy the T1 separation axiom and T is nondiscrete, 6< !. Therefore = !. In particular, G is rst-countable. Recall that we denote by B the countably in nite Boolean group L ! Z2 endowed with the topology with a neighborhood base at zero consisting of the subgroups Hn = fx 2 L ! Z2 : x(m) = 0 for all m < ng, where n < ! (see Example 2.3.4). By Theorem 2.3.13, there is a homeomorphism h : G! B such that h(e) = 0 and h(xy) = h(x)+h(y) whenever max supp(h(x))+2 min supp(h(y)). De ne : G! ! by (x) = 8 < : max supp(h(x)) if x 6= e 0 if x = e: Since for any x 2 G and for any 0 i max supph(x), (h(x))(i) has only a nite number of possible values (two in this case), is a nite-to- one function. In addition, h is injective. Therefore, for any X G with jXj = !, j (X)j = !. Thus (a) is satis ed. To check (b), let x 2 G. If x = e, then xy = ey = y and, therefore, (xy) = (y) for all y 2 G. Therefore, we may assume that x 6= e. Let m = max supp(h(x)) + 2 and let V = fy 2 G : m min supp(h(y))g [ feg: Then V = h 1(Hm) [ feg = h 1(Hm [ f0g) 2 T . It follows that V is an open neighborhood of e. Furthermore, for every y 2 V nfeg one has (xy) = max supp(h(xy)) = max supp(h(x) + h(y)) = max supp(h(y)) = (y): Case 2 : > !. Construct inductively a -sequence (G ) < of sub- groups of G with G0 = feg such that (i) for every < , jG j < , 3.4 c? K(T ) is not an ideal 124 (ii) for every < , G G +1, (iii) for every limit ordinal < , G = S < G , (iv) S < G = G, and (v) if < , then for every < , e 2 c?T (G +1 nG ). Note that G = S < (G +1 nG )[G0 is a disjoint union of sets. De ne : G! by (x) = 8 < : + 1 if x 2 G +1 nG 0 if x 2 G0: Clearly satis es (a) and (d). To check (c), let x 2 G. One may assume that x 6= e. Then x 2 G +1nG for some < . Now let Y = GnG +1 and let y 2 Y . Then y 2 G +1nG for some +1. Assume xy 2 G . Then y = x 1(xy) 2 G , a contradiction. Therefore xy 2 G +1 n G . Similarly, yx 2 G +1 nG . Hence, (xy) = (yx) = + 1 = (y). Let : G! be a function guaranteed by Lemma 3.4.4 and let : G! denote the continuous extension of . Recall that for any p 2 G, (p) = fE : 1(E) 2 pg. Corollary 3.4.5. (1) If = !, then (qp) = (p) for all q 2 G and p 2 T . (2) If > !, then (qp) = (p) for all q 2 G and p 2 U(G). Proof. (1) Let q 2 G. To see that (qp) = (p), let R 2 p. For every x 2 G, there is a neighborhood Vx of e 2 G such that (xy) = (y) for all y 2 Vx n feg. Since p 2 T , Vx n feg 2 p. Put Px = R \ Vx n feg. Then S x2G xPx 2 qp and, whenever x 2 G and y 2 Px, one has (xy) = (y) 2 (R). It follows that for every A 2 q, every x 2 A and every y 2 Px, (xy) = (y). Therefore, for every A 2 q, ( S x2A xPx) (R) 2 (p). Since the sets S x2A xPx, where A 2 q, form a base for qp, the sets ( S x2A xPx), where A 2 q, form a base for (qp). It follows that (qp) (p) and, hence, (qp) = (p). 3.4 c? K(T ) is not an ideal 125 (2) Let q 2 G and p 2 U(G). To see that (qp) = (p), let R 2 p. For every x 2 G, there is Yx G with jG n Yxj < such that (xy) = (y) for all y 2 Yx. Since p is a uniform ultra lter, Yx 2 p. Put Px = R\Yx. Then S x2G xPx 2 qp and, whenever x 2 G and y 2 Px, one has (xy) = (y) 2 (R). As in (1), it follows that (qp) = (p). Corollary 3.4.6. There is D with jDj = j j such that whenever D0 D and jD0j = j j, 1 (U(D0)) \ T is a left ideal of T . Proof. Consider two cases. Case 1: = . Let fV : < g be a decreasing neighborhood base at e in T . Inductively, for each < , pick x 2 V n feg such that (x ) =2 f (x ) : < g. We have that (x ) < is a - sequence converging to e and such that (x ) 6= (x ) if 6= . Put D = f (x ) : < g. To see that D is as required, let D0 D and jD0j = . Then fx : (x ) 2 D0g is a -subsequence of (x ) < , and so is converging to e. It follows that L = 1 (U(D0))\T is nonempty. To check that L is a left ideal of T , let p 2 L and q 2 T . Since T is a subsemigroup of G, qp 2 T . By Corollary 3.4.5, (qp) = (p) 2 U(D0). Hence, qp 2 L. Case 2: < . Put D = f + 1 : < g. To see that D is as required, let D0 D and jD0j = . For every 2 D0, e 2 c?T 1( ). It follows that L = 1 (U(D0))\T is nonempty. Then, applying Corollary 3.4.5, we obtain that L is a left ideal of T . Construct inductively an open neighborhood base fW : < g at the identity of G and a -sequence (a ) < of elements of G such that W0 = G and for every < , the following conditions are satis ed: (i) a 2 W , (ii) c?TW +1 W , (iii) a W +1 W nW +1, (iv) for every y 2 W +1 n feg, one has (a y) = (y), and 3.4 c? K(T ) is not an ideal 126 (v) W = T < W if is a limit ordinal. Let D be a subset of guaranteed by Corollary 3.4.6. Pick any family fD : < g of pairwise disjoint subsets of D of cardinality . For every < , put E = [ < D ; A = 1(D ) \W and B = [ < A : De ne N ; L T and M L by N = \ < E ; L = 1 (N) \ T; and M = \ < B : Lemma 3.4.7. M \ c? K(T )) 6= ;. Proof. It follows from Corollary 3.4.6 that for each < , 1 (U(D )) \ T is a left ideal of T . Consequently, there is p 2 1 (U(D )) \ K(T ), so p 2 A \K(T ). Let p 2 \ < c? Gfp : < g: Then p 2M \ (c?K(T )). Lemma 3.4.8. For every q 2 G n feg and r 2 L, qr =2 B0. Proof. It su ces to show that for every x 2 G n feg, xr =2 B0. There is < such that x 2 W n W +1. Since r 2 T and c?TW +2 W +1, xr 2 W nW +2. Assume that xr 2 B0. Then xr 2 S +1 A , so (r) = (xr) 2 S +1D . Hence, (r) =2 E +2 , a contradiction. Corollary 3.4.9. M \ T 2 = ; and so M \K(T ) = ;. Proof. Assume on the contrary that p = qr for some p 2 M and q; r 2 T . Then (p) = (qr) = (r), so r 2 1 (N) \ T = L. But by Lemma 3.4.8, p = qr =2 B0, which is a contradiction. Therefore, M \ T 2 = ;. Since K(T )2 = K(T ) T 2, M \K(T ) = ;. Let A = fa : < g. Since (a ) < converges to e in T , fW : < g[fAg has the nite intersection property. Therefore there exists p 2 G with fW : < g[fAg p . It follows that p 2 A\T and, therefore, A\T 6= ;. 3.4 c? K(T ) is not an ideal 127 On the other hand, whenever q 2 U(A), one has fx 2 G : x 1A 2 qg = feg, consequently by Theorem 3.3.4, A \K(T ) = ;. Now let C = [ < a B +1: Note that (C \ T ) (M). Indeed, if x 2 C \W , then x = a y for some and y 2 W +1, so (x) = (a y) = (y) 2 W +1. It is clear also that AM C. Since M \ (c? K(T )) 6= ; (Lemma 3.4.7), it follows that (T (c? K(T ))) \ C 6= ;. Therefore (since C is open) in order to prove that c? K(T ) is not a left ideal, it su ces to show that C \ K(T ) = ;. To this end, let q 2 K(T ) and let r be the identity of the maximal subgroup in K(T ) containing q. Then q = qr and, since K(T ) \ A = ;, q =2 A. Consequently by the next lemma, q =2 C. Lemma 3.4.10. Let qr 2 C for some q; r 2 T . Then q 2 A and r 2M . Proof. Since (r) = (qr) 2 and (C \ T ) (M), one has r 2 L. To see that q 2 A, assume the contrary. Then there is x 2 G n (A [ feg) such that xr 2 C. It follows that xr 2 a B +1 for some < , and so a 1 xr 2 B +1, which contradicts Lemma 3.4.8. Now to see that r 2 M , pick Q 2 q such that Q A and Qr C. Then for every < such that a 2 Q, one has B +1 2 r, and the set of all such is co nal in . Hence r 2M . Chapter 4 Finite ultra lter semigroups Given a (left) topological group G, Ult(G) is either nite or of cardinality greater than or equal to 22 ! . In fact, this is true for any closed subsemigroup of G . Ult(G) re ects quite good topological properties of G. In [66] it was shown that every countable regular homogeneous space admits a structure of a left topological group. This allows us to de ne Ult(X) for any countable regular space X. In [19], Hewitt studied maximal spaces. He gave a general way of constructing them and he used this construction to produce irresolv- able spaces. If the topology of a (left) topological group is maximal, then Ult(G) is nite. Projectives in the category of nite semigroups were studied in [67, 73, 62]. It turned out that for every countable regular left topological group G, if Ult(G) is nite, then Ult(G) is projective. In Section 4.1 we look at resolvable spaces, maximal spaces and spaces X satisfying the condition that for every x 2 X, there is only a nite number of ultra lters converging to x. In Section 4.2, we look at projectives. In section 4.3 we present results on topological groups G with Ult(G) nite. We denote by FinS the category of nite semigroups and semigroup homomorphisms between them, and we denote by KHausRS the category of compact Hausdor right topological semigroups and semigroup homomorphisms between them. 4.1 Almost maximal spaces and almost max- imal left topological groups De nition 4.1.1. Let X be a topological space and let x 2 X. The disper- sion character (x;X) ofX at the point x is the minimum of the cardinalities 128 4.1 Almost maximal spaces and almost maximal left topological groups 129 of open subsets of X containing x. The cardinal number (X) := minf (x;X) : x 2 Xg is called the dispersion character of X. Lemma 4.1.2. Let X be a topological space. (1) If (X) is in nite, then X is dense-in-itself (i.e. X has no isolated point). (2) If X is T0 and dense-in-itself, then (X) is in nite. Proof. (1) Suppose (X) is in nite. Then every nonempty open subset of X is in nite. In particular X has no isolated point. (2) Suppose X is T0 and dense-in-itself. Assume (X) is nite. Pick a subset U X such that jU j = (X) and pick a 2 U . Since X is dense- in-itself, there exists x 2 U n fag. Pick an open subset V X such that jV \ fa; xgj = 1. Then, U \ V is an open subset of X with jU \ V j < (X), a contradiction. It follows that (X) is in nite. We call two subsets, A and B, of a set X complementary if A \ B = ; and A [B = X. De nition 4.1.3. Let X be a nonempty topological space. X is said to be resolvable if X can be partitioned into two dense subsets. If X = A [ B where A and B are disjoint dense subsets of X, we say that fA;Bg is a resolution of X. Remark 4.1.4. Let X be a topological space with an isolated point c and let A and B be dense subsets of X. Then c 2 A \ B. Therefore A \ B 6= ;. In particular X has no resolution. Therefore, when we want to know which spaces are resolvable, we need only to consider spaces without isolated points. Theorem 4.1.5. [[19], Theorem 19] The following statements about a topo- logical space X are equivalent. (a) X is resolvable; (b) X contains two disjoint dense subsets; (c) X contains a dense set with empty interior; 4.1 Almost maximal spaces and almost maximal left topological groups 130 (d) X contains two complementary subsets with empty interiors. Proof. These implications are obvious. De nition 4.1.6. Let X be a topological space which is dense-in-itself. X is called irresolvable if X is not resolvable. Remark 4.1.7. It is clear from Theorem 4.1.5 (c) that a space X which is dense-in-itself is irresolvable if and only if every dense subset of X has a nonempty interior. Lemma 4.1.8. Let X be a topological space. Let S be the family of all ordered pairs fA;Bg of disjoint subsets of X. S has the partial order de ned by fA1; B1g fA2; B2g if and only if A1 A2 and B1 B2, and maximal elements of S are exactly the pairs fA;X n Ag, where A X. Proof. It is clear that is a partial order. Since fX; ;g 2 S, S 6= ;. Every chain (fAi; Big)i2I is S has an upper bound f S i2I Ai; S i2I Big. By Zorn?s Lemma S has a maximal element. It is easy to see that maximal elements of S are of the form fA;X n Ag, where A X. Corollary 4.1.9. Let X be a resolvable space. The maximal resolutions of X with respect to the order in Lemma 4.1.8 are exactly pairs of the form fA;X n Ag, where A X with c? A = X = c? (X n A). Proof. Immediate from Lemma 4.1.8. Let X be a nonempty topological space. Following the concept of Hewitt [19] , we will call a dense subset of X whose complement is also dense a CD-set. Thus a maximal resolution of X is a pair consisting of a CD-set and its complement. Theorem 4.1.10. [[19], Theorem 21] A topological space X containing a dense subset A which is resolvable in its relative topology is also resolvable. Proof. LetA be a dense subset ofX which is resolvable in its relative topology with a resolution of CD-sets fB;Cg. It is easy to see that fB;C [ (X nA)g is a resolution of X. Theorem 4.1.11. A topological space X is resolvable if and only if every nonempty open subset of X contains a nonempty subset which is resolvable in its relative topology. 4.1 Almost maximal spaces and almost maximal left topological groups 131 Proof. Necessity. Suppose X is resolvable. Let U be a nonempty subset of X and let fA;Bg be a resolution of X. It is easy to see that fA\U;B \Ug is a resolution of U . Su ciency. Let Uo be an arbitrary nonempty open subset of X and let A0 be a subset of U0 which is resolvable in its relative topology with a pair of CD-sets fD0; E0g. Let U1 = X n c?(A0). If U1 = ;, then c?(A0) = X and fD0; E0 [ (X nA0)g is a resolution of X. If U1 6= ;, it must, as an open subset of X, contain a set A1 which has a resolution of CD-sets fD1; E1g in its relative topology. Suppose that sets A with resolutions of CD-sets fD ; E g have been selected for every ordinal < , also being an ordinal. Take U = X n c? ( S < A ). Then U is an open subset of X. If U 6= ;, we select a subset A of U with a resolution, in its relative topology, of CD-sets fD ; E g. By induction, it follows that there is some ordinal number 0 such that c? ( S < 0 A ) = X. It can then be shown that the sets S < 0 D and S < 0 E are disjoint and dense in X. By Theorem 4.1.5 (b), X is resolvable. Proposition 4.1.12. [[62], Lemma 2.3] If a homogeneous space X has a resolvable subspace, then X itself is also resolvable. Proof. Suppose X has a resolvable subspace Y0. Let fA0; B0g be a partition of Y0 into dense subsets. Consider the family S of all pairs fA;Bg of disjoint subsets of X such that A0 A, B0 B, and c? A = c? B. Since fA0; B0g 2 S, S 6= ;. De ne a relation on S by fA1; B1g fA2; B2g if and only if A1 A2 and B1 B2. It is easy to see that is a partial order on S. Let (fAi; Big)i2I be a chain in S. Let A = S i2I Ai and let B = S i2I Bi. Then fA;Bg is an upper bound for (fAi; Big)i2I . By Zorn?s Lemma S has a maximal element. Let fA;Bg be a maximal element of S. We claim that fA;Bg is a partition of X into dense subsets. Already A \ B = ;. It is clear that Y = A [ B is closed. Assume that Y 6= X. Pick x 2 X n Y and y 2 Y . Let f : X ! Y be a homeomorphism with f(y) = x. Choose an open neighborhood U of y such that f(U) \ Y = ;. Put A1 = A [ f(U \ A) and B1 = B [ f(U \ B). Then fA1; B1g 2 S and fA;Bg < fA1; B1g, a contradiction. It follows that Y = X. Since c? A = c? B and A [ B = X, c? A = c? B = c? A [ c? B = c? (A [B) = c? X = X. Theorem 4.1.13. [[19], Theorem 7] Let X be a set. Any family of topologies on X is partially ordered by inclusion. The family of all topologies on X form a complete lattice under this partial order. 4.1 Almost maximal spaces and almost maximal left topological groups 132 Proof. It is fairly obvious that inclusion is a partial order. Let (Ti)i2I be a nonempty family of topologies on X. Let G = S i2I Ti and let T be the topology on X generated by G. Then Ti T for every i 2 I and T is the smallest topology on X with this property. Thus T is the least upper bound for (Ti)i2I . Now let = T i2I Ti. Then is the greatest lower bound for (Ti)i2I . Let X be a set and let T1 and T2 be topologies on X. We will call T2 an expansion of T1 (and T1 a contraction of T2 ) if T2 is ner than T1, i.e. T1 T2 . An expansion or contraction is said to be proper or improper as the inclusion T1 T2 is proper or improper. Lemma 4.1.14. [[19], Theorem 2] Let X be a set and let T1 and T2 be topologies on X with T1 T2. Let A X and let x 2 X. Then intT1(A) intT2(A) and c?T2(A) c?T1(A). If x is an accumulation point of A in T2, then x is an accumulation point of A in T1. Proof. The statements follow immediately from the fact that T1 T2. It is also an easy observation that any expansion of a T0, T1, Hausdor , or Urysohn topological space satis es the same separation axiom. Lemma 4.1.15. Every nonempty nite open set in a T0 space contains an isolated point. In particular, every nonempty nite T0 space has an isolated point. Proof. Let X be a T0 space. It is clear that any open singleton contains an isolated point. Let n 2 N and suppose every nonempty open subset consisting of at most n elements contains an isolated point. Let U = fx1; ; xn+1g X be open. Pick an open subset V X such that jV \ fx1; x2gj = 1. Then U \ V is an open subset of X consisting of at most n elements. By assumption, U \ V contains an isolated point. Hence U contains an isolated point. By induction, every nonempty nite open subset of X contains an isolated point. The second statement follows at once. It follows from Theorem 4.1.15 that every T0-space has dispersion character in nite or equal to 1. Theorem 4.1.16. [[19], Theorem 9] Every T0-space X has an expansion Y which is a T1 space and which has the same dispersion character as X. 4.1 Almost maximal spaces and almost maximal left topological groups 133 Proof. Let (X; T ) be a T0-space and let T 0 = fU X : jX n U j < !g [ f;g; i.e. the co nite topology on X. Let T 00 be the topology generated by T [T 0. Since T 0 is a T1-space, T 00 is a T1-space. Suppose (X; T ) = 1. Since T T 00, (X; T 00) = 1. On the other hand, if (X; T ) is in nite, the intersection of any open set U 2 T with an open set V 2 T 0 must have the same cardinal number as U , since the removal of a nite number of points from an in nite set does not change the cardinal number of that set. Hence (X; T ) = (X; T 00). Theorem 4.1.17. Every Hausdor space X has an expansion Y which is a Urysohn space, whose isolated points are exactly the isolated points in X, and which has the dispersion character equal to the dispersion character of X. Proof. [[19], Theorem 10]. Theorem 4.1.18. Let X be a set and let (Ti)i2I be a nonempty family of topologies on X completely ordered by inclusion. If Tj is T0, T1, Hausdor or Urysohn, for some j 2 I, then the least upper bound of (Ti)i2I satis es the same separation property. If Ti is regular (completely regular) for every i 2 I, then the least upper bound of (Ti)i2I is regular (completely regular). Proof. [[19], Theorem 11]. De nition 4.1.19. Let be a cardinal number and let P be a property intelligible for topological spaces. Let (X; T ) be a topological space. (X; T ) is called -maximal with respect P if T is maximal among topologies on X satisfying the following properties. (a) (X; T ) , (b) (X; T ) has the property P . De nition 4.1.20. Let be a cardinal number and let (X; T ) be a topo- logical space. (X; T ) is called -maximal if T is maximal among topologies on X with dispersion character greater than or equal to . An !-maximal space is simply said to be maximal. 4.1 Almost maximal spaces and almost maximal left topological groups 134 Let P be a property intelligible of topological spaces. If every expansion of a space with the property P has the property P , then P is called expansion- invariant. In [[19], Theorems 12-16], Hewitt showed that certain topolo- gies has -maximal expansions with respect to some properties. Further, he proved that every maximal space is irresolvable. In [14], van Douwen proved that there exists a countable regular maximal space in ZFC. In [38], Ma- lykhin proved, under Martin?s Axiom (to be stated later), that there exists a nondiscrete Hausdor topological group whose underlying space is maximal. In particular, maximal spaces exist. In the sequel our maximal topological spaces are Hausdor . The follow- ing lemma gives two characterizations of maximal spaces among Hausdor spaces. Proposition 4.1.21. [[62], Proposition 2.1] Let (X; T ) be a Hausdor topo- logical space. The following statements are equivalent. (1) X is maximal, (2) X is without isolated points, extremally disconnected and has no resolv- able subspaces and no nonclosed nowhere dense subsets, (3) for each x 2 X there is exactly one nonprincipal ultra lter on X con- verging to x. Proof. Let X = (X; T ). (1) ) (2): It is clear that X has no isolated points. Assume that X is not extremally disconnected. Let V be an open subset of X with a nonopen closure c? V . Then T [ fU \ c? V : U 2 T g is a base for a topology T1 on X which re nes T . Notice that c? V X has no isolated points in T . Consequently, if U \ c? V 6= ;, then U \ c? V is in nite. Hence, (X; T1) has no isolated points, a contradiction. Assume that X has a resolvable subspace Y . Let A be a subset of Y such that A and Y nA are dense in Y . Then A X is a nonopen subset without isolated points. Hence, one can re ne T to the topology T1 without isolated points by taking as a base T [ fU \ A : U 2 T g, a contradiction. Assume now that X has a nonclosed nowhere dense subset B. Let x 2 c? B n B and let Z = (X n c? B) [ fxg. Then Z X is a nonopen subset without isolated points. Hence, one can re ne T to the topology T1 without isolated points by taking as a base T [ fU \ Z : U 2 T g, a contradiction. 4.1 Almost maximal spaces and almost maximal left topological groups 135 (2) ) (3): First we prove that every nonprincipal ultra lter p on X con- verging to a point x 2 X has a base of open subsets. To this end, let A 2 p and suppose that int A =2 p. Let B = A n ((int A) [ fxg). Then B 2 p and int B = ;. Put U = int c? B. We claim that U = ;. Indeed, otherwise B is nowhere dense and, since x 2 c? B, nonclosed. Put C = U \ B. Then C is dense in U and int C = ;, hence U n C is also dense in U , so U is resolv- able, a contradiction. Now suppose that there are two di erent nonprincipal ultra lters p and q on X converging to a point x 2 X. Choose disjoint open subsets Up 2 p and Uq 2 q. Then x 2 c? Up \ c? Uq, so X is not extremally disconnected, a contradiction. (3) ) (1) Let X = (X; T ) and let T1 be any re nement of T . Then there exists a nonprincipal ultra lter p on X which is convergent to a point x 2 X in T but is not convergent in T1. But then x is an isolated point in T1. Hence, T is maximal. Corollary 4.1.22. A homogeneous space is maximal if and only if it is nondiscrete, extremally disconnected, irresolvable and has no nonclosed nowhere dense subsets. Proof. It is immediate from Proposition 4.1.21 and Proposition 4.1.12. De nition 4.1.23. A topological (left topological) groupG is called maximal if its underlying space is maximal. Proposition 4.1.24. Let G be an in nite group with identity e and let p 2 G be an idempotent. Then G(p) is maximal. Proof. Recall that the topology T of G(p) is T = fU G : a 1U 2 p for every a 2 Ug with the neighborhood lter at e equal to the family fA [ feg : A 2 pg, see Section 2.2. Since pp = p, T is Hausdor . Suppose there exists a topology T 0 on G such that T ( T 0 and G has no isolated points in T 0. Pick V 2 T 0 n T and pick a 2 V such that a 1V =2 p. Then G n a 1V 2 p. Therefore (Gna 1V )[feg is a neighborhood of e in T , and thus a((Gna 1V )[feg) = (GnV )[fag is a neighborhood of a in T . Then (GnV )[fag is a neighborhood of a in T 0. But then (G n V ) [ fag) \ V = fag is a neighborhood of a in T 0, a contradiction. Conversely we have the following. 4.1 Almost maximal spaces and almost maximal left topological groups 136 Proposition 4.1.25. Let (G; T ) be an in nite maximal left topological group. There exists an idempotent p 2 G such that (G; T ) = G(p). Proof. [[53], Theorem 1.1] Corollary 4.1.26. Let G be an in nite group. There is a bijection between the maximal left invariant topologies on G and the idempotents in G . We recall that in a semigroup S, an idempotent p 2 S is said to be strongly right maximal if the equation xp = p has the unique solution x = p. Theorem 4.1.27. Let G be an in nite group with identity e and let p 2 G . Let T be the topology for G(p). Then the following are equivalent. 1. T is regular. 2. The idempotent p is strongly right maximal in G . 3. The map ( p)jG is a homeomorphism from G(p) onto Gp. 4. T is regular and extremally disconnected. 5. T is zero dimensional. Proof. [[29], Theorem 9.15] Corollary 4.1.28. Let G be an in nite group with identity e and let p 2 G . The topologies of the groups G(p) and G[p] coincide if and only if the underlying space of G(p) is regular. Corollary 4.1.29. Let G be an in nite group with identity e and let p 2 G be an idempotent. If p is strongly right maximal, then G(p) is homogeneous, zero dimensional, Hausdor , extremally disconnected, and maximal among all topologies without isolated points. Distinct strongly right maximal idem- potents q and r give rise to distinct topologies G(q) and G(r). Proof. [[29], Corollary 9.17] De nition 4.1.30. A topological space X is called almost maximal if it is without isolated points and for every x 2 X there are only nitely many nonprincipal ultra lters on X converging to x. 4.1 Almost maximal spaces and almost maximal left topological groups 137 Note that as a consequence of Proposition 4.1.21, every maximal space is almost maximal. By an almost maximal topological (left topological) group we will mean a topological (respectively, left topological) group whose un- derlying space is almost maximal. Note that every almost maximal left topological group G gives a nite semi- group Ult(G). Conversely, by Proposition 2.1.13, if every nite semigroup can be realized as a subsemigroup of G for some left topological group G, then every nite semigroup is an ultra lter semigroup. We will return to this remark in the next section. Proposition 4.1.31. [[62], Proposition 2.6] Let G be an Abelian torsion free group. Then every almost maximal left invariant topology on G is Hausdor . Proof. Let T be an almost maximal left invariant topology on G and let S = Ult(T ). Then S is nite. Assume on the contrary that T is not Hausdor . Then there is a 2 G n f0g such that (a + S) \ S 6= ;. It follows that fq 2 S : a+ q 2 Sg is a nonempty and, obviously, a subsemigroup of S (and even a right ideal), and consequently, as any nite semigroup, has an idempotent. Thus, there is an idempotent p 2 S with a + p 2 S. Since p is an idempotent and G is Abelian, (a+ p) + + (a+ p) = (a+ + a) + p. Hence na + p 2 S for all n 2 N. Since G is torsion free, all elements na are di erent. But then all elements na + p are di erent as well, so S is in nite, a contradiction. It follows that T is Hausdor . Corollary 4.1.32. Let G be an Abelian torsion free group and let T be an almost maximal left invariant topology on G. Then jUlt(G)j = 1 if and only if (G; T ) is maximal. Proof. By Proposition 4.1.31, (G; T ) is Hausdor . Suppose jUlt(G)j = 1. Since (G; T ) is homogeneous, for every x 2 G there is exactly one nonprinci- pal ultra lter on G converging to x. Therefore, by Proposition 4.1.21, (G; T ) is maximal. The converse follows directly from Proposition 4.1.21. In [63] it was proved that there are no nontrivial nite groups in N. In fact, it was proved that for any countable group G and any nite group F in G, if the subgroup fx 2 G : xF = Fg is trivial, the group F is trivial as well. The following theorem follows from this result and Proposition 2.1.9. Theorem 4.1.33. [[62], Proposition 2.7] For every countable Hausdor left topological group G, the ultra lter Ult(G) has no nontrivial nite subgroups. 4.1 Almost maximal spaces and almost maximal left topological groups 138 Corollary 4.1.34. Let G be a countable Abelian group and let T be an almost maximal left invariant topology on G. Then T is Hausdor if and only if Ult(T ) has no nontrivial subgroups. Proof. ): This implication follows directly from Theorem 4.1.33. (: Let S = Ult(T ). Suppose S has no nontrivial nite subgroups and assume that T is not Hausdor . Then there exists a 2 G n f0g such that (a + S) \ S 6= ;. Then, as in the proof of Proposition 4.1.31, there is an idempotent p 2 S such that hai+ p S, where hai = fna : n 2 Ng. Clearly hai + p is a subsemigroup isomorphic to hai. Since S is nite, hai + p is a nite subgroup, a contradiction. Let ? be a lter on a topological space X. We recall the following. ? is called open (respectively, closed) if ? has a base consisting of open (respectively, closed) subsets. We will denote by c? ? the largest closed lter on X con- tained in ? and by ? the largest open lter (possibly improper) containing ?. Lemma 4.1.35. [[62], Lemma 2.10] Let G be an almost maximal left topo- logical group, let S = Ult(G), and let be a lter on G. Then (a) is closed if and only if for every p 2 G n , pS \ = ;, (b) is open if and only if S . Proof. (a) Suppose is closed. Let p 2 Gn . Then * p. Pick a closed F 2 n p and let U = G n F . Then p 2 U and U \ = ;. Since U is open, U S U . Since pS \ U S \ = ;, pS \ = ;. Conversely, assume for every p 2 G n , pS \ = ; and assume that is not closed. Then there is U 2 such that U is not closed. Given V 2 , pick xV 2 c? V n U . Then e 2 c? x 1 V V . Then the family fN : N is a neighborhood of the identity g [ fx 1V V g has the nite in- tersection property and, therefore, is contained in some ultra lter qV . Then qV 2 S and x 1 V V 2 qV , or equivalently, V 2 xV qV . Since S is nite, one may suppose that qV = q is the same lter. Pick an ultra l- ter p on G such that for every V 2 , fxW : W 2 and W V g 2 p. Then pq 2 and G n U 2 p, so p 2 G n . (b) Suppose is open. Then by Corollary 2.1.11, S . Conversely, suppose S and let U 2 . For every p 2 and q 2 S, U 2 pq, so there exists Ap;q 2 p such that Ap;q q U . Denote 4.1 Almost maximal spaces and almost maximal left topological groups 139 Vq = S p2 Ap;q. Then Vq 2 and V q q U . Let V = T q2S Vq. Then V 2 and V S U . Finally, put W = fx 2 U : xS Ug. Clearly, W is open and, since V W , W 2 . Hence, is open. Corollary 4.1.36. Let G be an almost maximal left topological group, let S = Ult(G), and let be a lter on G. Then (a) c? = [ fp 2 G : pS \ 6= ;g, (b) = fp 2 : pS g. Proposition 4.1.37. [[62], Proposition 2.12] Let (G; T ) be an almost max- imal left topological group and let S = Ult(T ). Then (1) T is regular if and only if S1 is left saturated in G, (2) if T is Hausdor , T is regular if and only if S is left saturated in G . Proof. (1) ()): This implication is Lemma 2.1.15. ()): Note that T is regular if and only if the neighborhood lter of the identity is closed. Now suppose S1 is left saturated in G. Let N be the neighborhood lter at the identity, so that S1 = N . Let p 2 GnN . Since G is cancellative, pS G . Therefore, pS\N = pS\S1 = pS\S pS1\S1 = ;. It follows that pS \ N = ;. By Lemma 4.1.13 (a), N is closed. Hence T is regular. (2) Assume T is regular. Then, by (1), S1 is left saturated in G and, thus, S is left saturated in G . Conversely, suppose S is left saturated in G. By (1) it su ces to show that S1 is left saturated in G. Let p 2 G n S1. If p =2 G, then pS1 \ S1 = (pS [ S) \ S1 = (pS [ S) \ S = ;. Suppose p 2 G. Since T is Hausdor , pS \ S = ; by Proposition 2.1.9. Then pS1 \ S1 = (pS [ fpg) \ S1 = pS \ S1 = pS \ S = ;. The following is a stronger version of Proposition 4.1.37. Proposition 4.1.38. [[62], Proposition 2.13] Let G be an almost maximal left topological group, let X be an open neighborhood of the identity, and let S = Ult(G). Then (a) X is regular i for every x 2 X and p 2 X n xS1, pS \ xS = ; , 4.1 Almost maximal spaces and almost maximal left topological groups 140 (b) if G is Hausdor , X is regular i for every x 2 X and p 2 X n xS, pS \ xS = ;, where Xd = X nX, (c) if G is Hausdor and X is a subsemigroup such that for every x 2 X, jx 1X nXj < !, then X is regular i S is left saturated in X . Proof. First we note the following. For every x 2 G, the neighborhood lter at x is given x = xN , where N is the neighborhood lter at the identity. Thus x = xS 1. X is regular if and only if for every x 2 X, x is closed. (a) ): Suppose X is regular. Then xS1 is closed. Let x 2 X and let p 2 X nxS1. By Lemma 4.1.35 (a), pS\xS1 = ;. Since pS\xS pS\xS1, pS \ xS = ;. (: Conversely, suppose the condition is satis ed. Let x 2 X and let p 2 G n x. By assumption, pS \ xS = ;. It remains to show that x =2 pS. Since G is cancellative, G is an ideal of G. Therefore, pS G and, thus, x =2 pS. It follows that x is closed and, hence, X is regular. (b) Suppose G is Hausdor . Then for every x 2 G n feg, xS \ S = ;. Then the result follows from (a). (c) We can rewrite condition (b) as follows: for every x 2 X and q 2 x 1X n S, qS \ S = ;. Note that in this case x 1X = X . Then the results follows. Corollary 4.1.39. Let S be a nite subsemigroup in N Z and let T be a left invariant topology on Z with Ult(T ) = S. Then X = N[ f0g is an open neighborhood of 0 2 Z in T and X is regular if and only if S is left saturated in N . Proof. Let be the neighborhood lter of 0 in T . Then = S1. Since S N , N 2 p for every p 2 S. It follows that X is a neighborhood of 0 in T . For every x 2 X, x+X = fn 2 N : n xg X and x+X is a neighborhood of x. Therefore, X is open. Since (Z;+) is Abelian, T is Hausdor , by Proposition 4.1.31. For any x 2 N, j( x + X) n Xj = jf x; x + 1; ; 1gj = x < ! and j( 0 +X) nXj = jX nXj = 0 < ! . Therefore, the remaining statement follows from Proposition 4.1.38 (c). Proposition 4.1.40. [[62], Proposition 2.15] Let (G; T ) be an almost max- imal left topological group and let S = Ult(G). Then (a) T is extremally disconnected if and only if K(S) consists of only one minimal right ideal, (b) T is irresolvable if and only if K(S) is a left zero semigroup, 4.2 Projectives in the category of nite semigroups 141 (c) for every p 2 S, p does not contain nowhere dense subsets if and only if p 2 K(S), (d) (G; T ) has no nonclosed nowhere dense subsets if and only if S = K(S). Proof. (a) We note that the implication from the right to the left is Lemma 2.1.14, but we give a complete proof here. T is extremally disconnected if and only if there are no two disjoint open sets having e as an accumulation point, consequently by Lemma 4.1.35 (b), if and only if there are no two disjoint right ideals in S, so if and only if K(S) has only one minimal right ideal (since distinct minimal right ideals are disjoint). (b) T is irresolvable if and only if there is an open ultra lter on G converging to e, consequently, by Lemma 4.1.35 (b), if and only if there is a singleton right ideal in S, so if and only if K(S) is a left zero semigroup. (c) p does not contain nowhere dense subsets if and only if (c? p) 6= ;. By Corollary 4.1.36, c?p = fpg [ fq 2 G : p 2 qSg and (c? p) = fq 2 c? p : qS c? pg. If p =2 K(S), c? p \K(S) = ; and then (c? p) = ;. If p 2 K(S), c? p pS and then p 2 ( c? p) . (d) (G; T ) has no nonclosed nowhere dense subsets if and only if for every p 2 S, p does not contain nowhere dense subsets (if A 2 p is nowhere dense, then A n feg is nonclosed nowhere dense), consequently, by (c), if and only if S = K(S). 4.2 Projectives in the category of nite semi- groups We will call an object S in some category a projective if for every morphism f : S ! Q and every surjective morphism g : T ! Q there exists a morphism h : S ! Q such that the diagram S f h T g // // Q commutes, i.e. g h = f . We will call S an absolute coretract if for every surjective morphism f : T ! S there exists a morphism g : S ! T such that f g = idS. It is clear from the following diagram that if S is a projective, 4.2 Projectives in the category of nite semigroups 142 then S is an absolute coretract. S idS g T f // // S In some categories these notions coincide. Theorem 4.2.1. For every countable regular almost maximal left topological group G, Ult(G) is a projective in the category of nite semigroups. Proof. Proof. Let S = Ult(G), let Q be a nite semigroup, and let f : S ! Q be a homomorphism. Since S is nite, it su ces to prove that f is proper by Corollary 2.3.15. For each p 2 S, choose Cp 2 p such that Cp \ Cq = ; if p 6= q, and choose Ap 2 p such that Ap Cp, X = S p2U Ap [ feg is open in G, and Ap q Cpq for all p; q 2 S. Then for every p 2 S and x 2 Ap, there exists a neighborhood Vx of e 2 X such that x(Vx \ Aq) Apq for all q 2 S. De ne f0 : X ! Q putting for every p 2 S and x 2 Ap, f0(x) = f(p). Then f0 is a local homomorphism inducing f , i.e. f0jS = f . Projectives in FinS were described in [67, 73, 62]. In [67] the author came to projectives in FinS from projectives in KHausRS. As we remarked in the previous section, if every nite semigroup can be realized as a subsemigroup of G for some left topological group G, then every nite semigroup is an ultra lter semigroup. The absolute coretracts in KHausRS arose in solving the following two questions. The rst question is: Which nite semigroups are ultra lter semigroups of topological groups? The second question is con- cerned with the semigroup N. Until now it is not known whether N has elements of nite order other than idempotents. Which nite bands exist in N? It is well known that N contains closed subsemigroups admitting a continuous homomorphism onto any nite semigroup. Hence, N contains isomorphic copies of any nite absolute coretract in KHausRS. In [67] the author described nite idempotent absolute coretracts in KHausRS and proved that these are precisely idempotent absolute coretracts in FinS. All projectives in FinS turned out to be bands decomposing into a certain chain of rectangular components. Denote by W the semigroup of words of the form a1 anun u1, where 4.2 Projectives in the category of nite semigroups 143 ak; uk 2 !, 1 k n < !, with the operation a1 anvn v1 b1 bmvm v1 = 8 < : a1 anun u1 if n = m; a1 anun um+1vm v1 if n > m; a1 anbn+1 bmvm v1 if n < m: It is clear that W is a band. In fact, it is clear that W is a rectangu- lar band. For every n 2 N, denote by Wn the subset of W consisting of all words of length 2n. Then Wn is clearly a rectangular band. Let x = a1 anun u1 2 Wn and let y = b1 bmvm v1 for some m 6= n. Then xy 6= a1 anvm v1. In particular Wn is not contained in another proper rectangular subsemigroup of W . Hence Wn is a rectangular compo- nent. In addition, it is easy to see that for any n 2 N, Wn+1 Wn+1Wn. Hence W decomposes into a decreasing chain of the rectangular components Wn. (See the paragraph just after the proof of Remark 1.2.13 for the partial order on the set of rectangular components of a band). Next denote by M the set of all matrices M = (mi;j)ni;j=0 without the main diagonal (mi;i), where n 2 N and mi;j 2 !, satisfying the following conditions for every k 2 N with 1 k n: (a) m0;k m1;k mk 1;k 2 N and mk;0 mk;1 mk;k 1 2 N, (b) either mk 1;k = 1 and mk 1;k+1 = = mk 1;n = 0 or mk;k 1 = 1 and mk+1;k 1 = = mn;k 1 = 0. These are precisely matrices of the form: 4.2 Projectives in the category of nite semigroups 144 0 B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B @ 1 0 0 0 + 1 0 0 + 0 0 0 0 0 1 + + 1 + 0 1 0 0 0 0 0 0 + 0 0 0 1 1 0 0 0 + 1 0 0 + 0 0 0 0 0 1 + + 1 + 0 1 0 0 0 1 C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C A and their transposes, where + is a positive integer, is a nonnegative integer, and all rows and columns are nondecreasing up to the main diagonal. For every M = (mi;j)ni;j=0 2 M, denote by W (M) the subsemigroup ofSn i=1Wi which consists of all words a1 akuk u1 2 Wk, 1 k n, satisfying the following conditions: (a) both ak 6= 0 and uk 6= 0, (b) for every j < i k, if aj+i = = ai 1 = 0, then ai mj;i, and dually, if uj+1 = = ui 1 = 0, then ui mi;j. It can be shown that for di erent matrices M 2 M, the semigroups W (M) are nonisomorphic. The following theorem has been obtained. Theorem 4.2.2. [[62], Theorem 4.2] For every nite semigroup S the fol- lowing statements are equivalent. 1. S is isomorphic to W (M) for some M 2M, 2. S is a projective in FinS, 4.3 Almost maximal topological groups 145 3. S is an absolute coretract in FinS, 4. S is a projective in KHausRS,, 5. S is an absolute coretract in KHausRS. Proof. See [62, 73] As we will see in the next section, under Martin?s Axiom each of the state- ments in Theorem 4.2.2 is equivalent to the following statement. S is isomorphic to Ult(G) for some countable almost maximal topolog- ical group G. Corollary 4.2.3. For each countable regular almost maximal left topological group G, the semigroup Ult(G) is isomorphic to W (M) for some M 2M. Proof. This is immediate from Theorem 4.2.1 and Theorem 4.2.2. Theorem 4.2.4. Let G be a countable group and let S be a nite subsemi- group in G such that S1 is left saturated in G. Then S is isomorphic to W (M) for some M 2M. Proof. Since S is a nite subsemigroup of G , S = Ult(T ) for some left invariant topology T on G. Since Ult(T ) is nite and (G; T ) is homogeneous, (G; T ) is almost maximal. Since S1 is left saturated in G, T is regular. Therefore, by Theorem 4.2.1, S is a projective in FinS. The result follows by Theorem 4.2.2. 4.3 Almost maximal topological groups We start this section by stating Martin?s Axiom (MA) and some related concepts. De nition 4.3.1. Let (P; ) be a partially ordered set. (a) Two elements a; b 2 P are said to be compatible if there exists c 2 P such that c a and c b. Otherwise a and b are incompatible. 4.3 Almost maximal topological groups 146 (b) P is called a c.c.c. partially ordered set if whenever D is a subset of P consisting of pairwise incompatible elements (i.e. D is an antichain), one has that D is countable. (c) A subset D P is dense in P if for every a 2 P , there exists d 2 D with d a. (d) A lter on P is a nonempty subset G P satisfying the following conditions. (i) For every elements a; b 2 G, there is c 2 G with c a and c b. (ii) For every a 2 G and b 2 P , if a b, then b 2 G. (c.c.c. stands for countable chain condition.) When we say that a topological space X is c.c.c., we mean the topology on X with set inclusion is a c.c.c. partially ordered set. Thus a topological space is c.c.c. if and only if there is no uncountable family of pairwise disjoint nonempty open subsets of X. Lemma 4.3.2. Every separable topological space is c.c.c. Proof. Let X be separable. Pick a countable dense subset D of X. Let G be a family of pairwise disjoint nonempty open subsets of X. For each U 2 G, pick xU 2 U \D. Then xU 6= xV for U 6= V . Therefore, the correspondence G 3 U 7! xU 2 D is injective. It follows that G is countable. Theorem 4.3.3. Let (Xi)i2I be a family of c.c.c. topological spaces such that for any nite F I, the product Q i2F Xi is c.c.c.. Then Q i2I Xi is c.c.c. Proof. [[36], Theorem 1.9]. De nition 4.3.4. Let be an in nite cardinal. (a) MA( ) is the assertion that whenever P is a c.c.c. partially ordered set and D is a family of at most dense subsets of P , there is a lter G on P such that G \D 6= ; for all D 2 D. (b) Martin?s Axiom is the assertion that for all cardinals with ! < 2!, MA( ). 4.3 Almost maximal topological groups 147 (c) P( ) is the following assertion: Whenever A is a family of subsets of N such that jAj < and jA0 \ \ Anj = ! for any A0; ; An 2 A, then there is an in nite subset I N such that jI n Aj < ! for every A 2 A. Theorem 4.3.5. (a) If < 0 and MA( 0) is true, then MA( ) is true. (b) MA(!) is true. (c) P((2!)+) is false. Proof. (a) This statement is obvious. (b) Let P be a c.c.c. partially ordered set and let D be a family of at most ! dense subsets of P . Enumerate D as D = fD0; D1; D2; g. Pick any d0 2 D0. Since D1 is dense, we can pick d1 2 D1 such that d1 d0. Since D2 is dense, pick d2 2 D2 such that d2 d1. Continuing in this way, we obtain a decreasing sequence (dn)n