Characterizing categorically closed commutative semigroups
aa r X i v : . [ m a t h . A C ] J a n CHARACTERIZING CATEGORICALLY CLOSED COMMUTATIVESEMIGROUPS
TARAS BANAKH, SERHII BARDYLA
Abstract.
Let C be a class of Hausdorff topological semigroups which contains all zero-dimensionalHausdorff topological semigroups. A semigroup X is called C -closed if X is closed in each topologicalsemigroup Y ∈ C containing X as a discrete subsemigroup; X is projectively C -closed if for eachcongruence ≈ on X the quotient semigroup X/ ≈ is C -closed. A semigroup X is called chain-finite iffor any infinite set I ⊆ X there are elements x, y ∈ I such that xy / ∈ { x, y } . We prove that a semigroup X is C -closed if it admits a homomorphism h : X → E to a chain-finite semilattice E such that forevery e ∈ E the semigroup h − ( e ) is C -closed. Applying this theorem, we prove that a commutativesemigroup X is C -closed if and only if X is periodic, chain-finite, all subgroups of X are bounded,and for any infinite set A ⊆ X the product AA is not a singleton. A commutative semigroup X isprojectively C -closed if and only if X is chain-finite, all subgroups of X are bounded and the union H ( X ) of all subgroups in X has finite complement X \ H ( X ). Introduction and Main Results
In many cases, completeness properties of various objects of General Topology or TopologicalAlgebra can be characterized externally as closedness in ambient objects. For example, a metricspace X is complete if and only if X is closed in any metric space containing X as a subspace. Auniform space X is complete if and only if X is closed in any uniform space containing X as a uniformsubspace. A topological group G is Ra˘ıkov complete if and only if it is closed in any topological groupcontaining G as a subgroup.On the other hand, for topological semigroups there are no reasonable notions of (inner) complete-ness. Nonetheless we can define many completeness properties of semigroups via their closedness inambient topological semigroups.A topological semigroup is a topological space X endowed with a continuous associative binaryoperation X × X → X , ( x, y ) xy . Definition 1.1.
Let C be a class of topological semigroups.A topological semigroup X is called C -closed if for any isomorphic topological embedding h : X → Y to a topological semigroup Y ∈ C the image h [ X ] is closed in Y .A semigroup X is called C -closed if so is the topological semigroup X endowed with the discretetopology. C -closed topological groups for various classes C were investigated by many authors includingArhangel’skii, Banakh, Choban, Dikranjan, Goto, Luka´sc and Uspenskij [1, 2, 5, 8, 15]. C -Closedtopological semilattices were investigated by Gutik, Repovˇs, Stepp and the authors in [3, 11, 12, 16,17]. For more information about complete topological semilattices and pospaces we refer to the recentsurvey of the authors [4].We shall be interested in the C -closedness for the classes: • T S of Hausdorff topological semigroups; Mathematics Subject Classification.
Key words and phrases. commutative semigroup, semilattice, group, C -closed semigroup, chain-finite semigroup,periodic semigroup.The second author was supported by the Austrian Science Fund FWF (Grant M 2967). • T z S of zero-dimensional Hausdorff topological semigroups; • T S of topological semigroups satisfying the separation axiom T .Recall that a topological space is zero-dimensional if it has a base of the topology consisting of clopen (= closed-and-open) subsets. A topological space X satisfies the separation axiom T if eachfinite subset is closed in X .Since T z S ⊆ T S ⊆ T S , for every semigroup we have the implications: T S -closed ⇒ T S -closed ⇒ T z S -closed . From now on we assume that C is a class of topological semigroups such that T z S ⊆ C ⊆ T S . Now we recall two known characterizations of C -closedness for semilattices and groups.Let us mention that a semigroup X is called a semilattice if it is commutative and each element x ∈ X is an idempotent , which means that xx = x . Each semilattice E carries a partial order ≤ defined by x ≤ y iff xy = x . Given two elements x, y of a semilattice we write x < y if x ≤ y and x = y .A subset C of a semigroup X is called a chain if xy ∈ { x, y } for any elements x, y ∈ C . A subset C of a semilattice is a chain if and only if any elements x, y ∈ C are comparable in the partial order ≤ .A semigroup X is called chain-finite if X contains no infinite chains. Observe that a commutativesemigroup X is chain-finite if and only if its maximal semilattice E ( X ) = { x ∈ X : xx = x } ischain-finite.The following characterization of C -closed semilattices was proved in [3]. Theorem 1.2.
A semilattice X is C -closed if and only if X is chain-finite. A semigroup X is defined to be • periodic if for every x ∈ X there exists n ∈ N such that the power x n is an idempotent; • bounded if there exists n ∈ N such that for every x ∈ X the power x n is an idempotent.It is clear that each bounded semigroup is periodic (but not vice versa).The following characterization of C -closed commutative groups was obtained by the first author in[2]. Theorem 1.3.
A commutative group X is C -closed if and only if X is bounded. In this paper we characterize C -closed commutative semigroups.Our principal tool for establishing the C -closedness is the following theorem. Theorem 1.4.
A semigroup X is C -closed if X admits a homomorphism h : X → E to a chain-finitesemilattice E such that for every e ∈ E the semigroup h − ( e ) is C -closed. Theorem 1.4 will be applied in the proof of the following theorem, which is one of the main resultsof this paper.
Theorem 1.5.
A commutative semigroup X is C -closed if and only if X is periodic, chain-finite, allsubgroups of X are bounded, and for every infinite subset A ⊆ X the set AA = { xy : x, y ∈ A } is nota singleton. Corollary 1.6.
Each subsemigroup of a C -closed commutative semigroup is C -closed. Example 1.7.
Take any infinite cardinal κ and endow it with the binary operation ∗ defined by x ∗ y = ( x = y and x, y ∈ κ \ { , } ;0 otherwise . The semigroup X = ( κ, ∗ ) was introduced by Taimanov [18]. Gutik [10] proved that the semigroup X is T S -closed but the quotient semigroup X/I by the ideal I = { , } is not T z S -closed. HARACTERIZING CATEGORICALLY CLOSED COMMUTATIVE SEMIGROUPS 3
Example 1.7 shows that the C -closedness is not preserved by taking quotient semigroups. Thisobservation motivates introducing the definitions of ideally and projectively C -closed semigroups.Let us recall that a congruence on a semigroup X is an equivalence relation ≈ on X such that forany elements x ≈ y of X and any a ∈ X we have ax ≈ ay and xa ≈ ya . For any congruence ≈ ona semigroup X , the quotient set X/ ≈ has a unique semigroup structure such that the quotient map X → X/ ≈ is a semigroup homomorphism. The semigroup X/ ≈ is called the quotient semigroup X by the congruence ≈ .A subset I of a semigroup X is called an ideal if IX ∪ XI ⊆ X . Each ideal I ⊆ X determines thecongruence ( I × I ) ∪ { ( x, y ) : x = y } ⊆ X × X . The quotient semigroup of X by this congruence isdenoted by X/I and called the quotient semigroup of X by the ideal I . If I = ∅ , then the quotientsemigroup X/ ∅ can be identified with the semigroup X .A semigroup X is called • projectively C -closed if for any congruence ≈ on X the quotient semigroup X/ ≈ is C -closed; • ideally C -closed if for any ideal I ⊆ X the quotient semigroup X/I is C -closed.It is easy to see that for every semigroup the following implications hold:projectively C -closed ⇒ ideally C -closed ⇒ C -closed.For a semigroup X the union H ( X ) of all subgroups of X is called the Clifford part of X .A semigroup X is called • Clifford if X = H ( X ); • almost Clifford if X \ H ( X ) is finite.Another principal result of this paper is the following characterization. Theorem 1.8.
For a commutative semigroup X the following conditions are equivalent: (1) X is projectively C -closed; (2) X is ideally C -closed; (3) the semigroup X is chain-finite, almost Clifford, and all subgroups of X are bounded. We do not know whether the equivalence (1) ⇔ (2) in Theorem 1.8 remains true for any (semi)group,see Question 9.4. Remark 1.9.
Theorem 1.8 implies that the projective C -closedness of commutative semigroups areinherited by subsemigroups and quotient semigroups. Remark 1.10.
By [2, Proposition 10], the semidirect product
Z ⋊ {− , } (endowed with the binaryoperation h x, i i ∗ h y, j i = h x + i · y, i · j i ) is a C -closed group which is not bounded. This exampleshows that Theorem 1.3 and Corollary 1.6 can not be generalized to non-commutative semigroups.Theorems 1.5 and 1.8 will be proved in Sections 6, 8 after a preliminary work made in Sections 2–5and 7. 2. The topological semigroup of filters on a semigroup
In this section for every semigroup X we define the topological semigroup ϕ ( X ) of filters on X ,containing X as a dense discrete subsemigroup. This construction is our principal tool in the proofsof non- C -closedness of semigroups.We recall that a filter on a set X is any family F of nonempty subsets of X , which is closed underfinite intersections and taking supersets in X . A filter F is • free if T F = ∅ ; • principal if { x } ∈ F for some x ∈ X . TARAS BANAKH, SERHII BARDYLA
A subfamily
B ⊆ F is called a base of a filter F if F = { A ⊆ X : ∃ B ∈ B ( B ⊆ A ) } . By ϕ ( X )we denote the set of all filters on X . The set ϕ ( X ) is partially ordered by the inclusion relation.Maximal elements of the partially ordered set ϕ ( X ) are called ultrafilters . It is well-known that afilter F on X is an ultrafilter if and only if for any partition X = U ∪ V of X either U or V belongsto F . By β ( X ) ⊆ ϕ ( X ) we denote the set of all ultrafilters on X .Each point x ∈ X will be identified with the principal ultrafilter U x = { U ⊆ X : x ∈ U } ∈ β ( X ) ⊆ ϕ ( X ). So, X can be identified with the subset of β ( X ) consisting of all principal ultrafilters. Thuswe get the chain of inclusions X ⊆ β ( X ) ⊆ ϕ ( X ).The set ϕ ( X ) carries the canonical topology generated by the base consisting of the sets h U i = {U ∈ ϕ ( X ) : U ∈ U } where U ⊆ X runs over subsets of X . It can be shown that this topology satisfies the separationaxiom T . The set X of principal ultrafilters is dense in ϕ ( X ) and for each x ∈ X the singleton { x } = { F ∈ ϕ ( X ) : { x } ∈ F } = h{ x }i is an open set in ϕ ( X ). So, X is a dense discrete subspace of ϕ ( X ). The subspace β ( X ) of ultrafilters is compact, Hausdorff, zero-dimensional, and dense in ϕ ( X ).Consequently, each subspace of β ( X ) is zero-dimensional and Tychonoff.If X is a (commutative) semigroup, then ϕ ( X ) has a natural structure of a (commutative) topo-logical semigroup: for any filters U , V ∈ ϕ ( X ) their product U V is the filter generated by the base { U V : U ∈ U , V ∈ V} , where U V = { uv : u ∈ U, v ∈ V } . Each neighborhood of U V in ϕ ( X )contains a basic neighborhood h U V i for some U ∈ U and V ∈ V . Then h U i and h V i are basicneighborhoods of the filters U , V in ϕ ( X ) such that h U i · h V i ⊆ h U V i , which means that ϕ ( X ) is atopological semigroup, containing X as a dense discrete subsemigroup. Observe that the product oftwo ultrafilters is not necessarily an ultrafilter, so β ( X ) is not necessarily a subsemigroup of ϕ ( X ).3. Some properties of periodic semigroups
In this section we establish some properties of periodic semigroups. Let us recall that a semigroup S is periodic if for every x ∈ S there exists n ∈ N such that x n is an idempotent of S .For a subset A of a semigroup S , let ∞ √ A = { x ∈ X : ∃ n ∈ N ( x n ∈ A ) } . For an element e ∈ S , the set ∞ p { e } will be denoted by ∞ √ e . Observe that a semigroup S is periodicif and only if S = S e ∈ E ( S ) ∞ √ e , where E ( S ) = { e ∈ S : ee = e } is the set of idempotents of S .For an element a of a semigroup S the set H a = { x ∈ S : ( xS = aS ) ∧ ( S x = S a ) } is called the H -class of a . Here S = S ∪ { } where 1 is an element such that 1 x = x = x x ∈ S . We shall assume that x = 1 for every x ∈ S .By Corollary 2.2.6 [13], for every idempotent e ∈ E ( S ) its H -class H e coincides with the maximalsubgroup of S , containing the idempotent e . The union H ( S ) = [ e ∈ E ( S ) H e is the Clifford part of S .For two subsets A, B of a semigroup S their product in S is defined as A · B = { ab : a ∈ A, b ∈ B } . The set A · B will be also denoted by AB . Lemma 3.1.
For any idempotent e of a semigroup S we have ( ∞ p H e · H e ) ∪ ( H e · ∞ p H e ) ⊆ H e . HARACTERIZING CATEGORICALLY CLOSED COMMUTATIVE SEMIGROUPS 5
Proof.
First we prove that xe ∈ H e . Since x ∈ ∞ √ H e , there exists n ∈ N such that x n ∈ H e and hence x n ∈ H e . Observe that xeS = xx n S ⊆ x n S = eS and eS = x n S ⊆ x n +1 S = xeS and hence xeS = eS . By analogy we can prove that S xe = S e . Then xe ∈ H e by the definition of the H -class H e .Fix any elements x ∈ ∞ √ H e and y ∈ H e . Since H e is a group with neutral element e , we obtain xy = x ( ey ) = ( xe ) y ∈ H e H e = H e . By analogy we can prove that yx ∈ H e . (cid:3) For a periodic semigroup S , by π : S → E ( S ) we denote the map assigning to each x ∈ S theunique idempotent in the semigroup x N = { x n : n ∈ N } .For a semigroup S let Z ( S ) = { z ∈ S : ∀ x ∈ S ( xz = zx ) } be the center of S . Proposition 3.2. If S is a periodic semigroup with E ( S ) ⊆ Z ( S ) , then π : S → E ( S ) is a homo-morphism and H ( S ) is a subsemigroup of S .Proof. Since E ( S ) ⊆ Z ( S ), the set E ( S ) is a semilattice and hence E ( S ) carries the partial order ≤ defined by x ≤ y iff xy = x . Claim 3.3.
For any x ∈ S and y ∈ Z ( S ) we have π ( xy ) = π ( x ) π ( y ) .Proof. Since S is periodic, there exist numbers n, m ∈ N such that π ( x ) = x n and π ( y ) = y m . Since xy = yx , ( xy ) nm = x nm y mn = π ( x ) m π ( y ) n = π ( x ) π ( y ) ∈ E ( S ) and hence π ( xy ) = π ( x ) π ( y ). (cid:3) The following claim implies that H ( S ) is a subsemigroup of S . Claim 3.4.
For any x, y ∈ E ( S ) we have H x H y ⊆ H xy .Proof. Take any elements a ∈ H x and b ∈ H y and observe that since x, y ∈ E ( S ) ⊆ Z ( S ) weget abS = ayS = yaS = yxS = xyS and S ab = S xb = S bx = S yx = S xy and hence ab ∈ H xy . (cid:3) Claim 3.5.
For any x, y ∈ S we have π ( x ) π ( y ) ≤ π ( xy ) .Proof. By Lemma 3.1, xπ ( x ) ∈ ∞ p π ( x ) π ( x ) ∈ H π ( x ) and yπ ( y ) ∈ H π ( y ) . Then xyπ ( x ) π ( y ) = xπ ( x ) yπ ( y ) ∈ H π ( x ) H π ( y ) ⊆ H π ( x ) π ( y ) according to Claim 3.4. Hence π ( xπ ( x ) yπ ( y )) = π ( x ) π ( y ).By Claim 3.3, π ( x ) π ( y ) = π ( xπ ( x ) yπ ( y )) = π ( xyπ ( x ) π ( y )) = π ( xy ) π ( x ) π ( y ) , which means that π ( x ) π ( y ) ≤ π ( xy ). (cid:3) Claim 3.6.
For any x ∈ S and y ∈ H ( S ) we have π ( xy ) = π ( x ) π ( y ) .Proof. It follows from y ∈ H ( S ) that y ∈ H π ( y ) and hence y = yπ ( y ). Let y − be the inverse elementof y in the group H π ( y ) . By Claims 3.5 and 3.3, π ( x ) π ( y ) ≤ π ( xy ) = π ( xyπ ( y )) = π ( xy ) π ( y ) = π ( xy ) π ( y − ) ≤ π ( xyy − ) = π ( xπ ( y )) = π ( x ) π ( y )and hence π ( xy ) = π ( x ) π ( y ). (cid:3) Claim 3.7.
For every x, y ∈ S we have π ( xy ) = π ( yx ) .Proof. Since S is periodic, there exists n ∈ N such that ( xy ) n and ( yx ) n are idempotents. Takinginto account that E ( S ) ⊆ Z ( S ), we conclude that( xy ) n = (( xy ) n ) n +1 = (( xy ) n +1 ) n = ( x ( yx ) n y ) n = (( yx ) n ) n ( xy ) n = ( yx ) n ( xy ) n =( yx ) n (( xy ) n ) n = ( y ( xy ) n x ) n = (( yx ) n +1 ) n = (( yx ) n ) n +1 = ( yx ) n and hence π ( xy ) = ( xy ) n = ( yx ) n = π ( yx ). (cid:3) TARAS BANAKH, SERHII BARDYLA
Claim 3.8.
For every x, y ∈ S we have π ( xy ) = π ( x ) π ( y ) .Proof. By Claim 3.7, π ( xy ) = π ( yx ). Let e = π ( xy ) = π ( yx ). By Claim 3.5, π ( x ) π ( y ) ≤ π ( xy ) = e .Since S is periodic, there exists n ∈ N such that ( xy ) n = e = ( yx ) n . Observe that xeS = exS ⊆ eS and eS = eeS = ( xy ) n eS ⊆ xeS and hence xeS = eS . Similarly, S xe ⊆ S e and S e = S ee = S e ( yx ) n ⊆ S ex = S xe and hence xe ∈ H e . By analogy we can prove that ye ∈ H e . By Claim 3.4and the inequality π ( x ) π ( y ) ≤ e , we finally have π ( x ) π ( y ) = π ( x ) π ( y ) e = π ( xe ) π ( ye ) = π ( xeye ) = e = π ( xy ) . (cid:3)(cid:3) Sufficient conditions of C -closedness In this section we prove some sufficient conditions of the C -closedness of a semigroup. We startwith the following lemma that implies Theorem 1.4, announced in the introduction. Lemma 4.1.
A subsemigroup X of a topological semigroup Y is closed in Y if X admits a continuoushomomorphism h : X → E to a chain-finite discrete topological semilattice E such that for every e ∈ E the set h − ( e ) is closed in Y .Proof. To derive a contradiction, assume that X is not closed in Y . So, we can fix an element y ∈ X \ X ⊆ Y . Replacing Y by X , we can assume that X is dense in Y . Claim 4.2.
If for some a ∈ X , e ∈ E , and n ∈ N we have ay n ∈ h − ( e ) , then ay ∈ h − ( e ) .Proof. Since h − ( e ) is an open subspace of X , there exists an open subset W ⊆ Y such that W ∩ X = h − ( e ). Assuming that ay n ∈ h − ( e ) ⊆ W , we can find a neighborhood V ⊆ Y of y such that aV n ⊆ W . Then for every v ∈ X ∩ V we have h ( av ) = h ( a ) h ( v ) = h ( a ) h ( v ) n = h ( av n ) = h ( ay n ) = e and hence ay ∈ a ( X ∩ V ) ⊆ a ( X ∩ V ) ⊆ h − ( e ) = h − ( e ). (cid:3) Claim 4.3. If ay ∈ h − ( e ) for some a ∈ X and e ∈ E , then the point y has a neighborhood U ⊆ Y such that aU ⊆ h − ( e ) .Proof. Since the set h − ( e ) is open in X , there exists an open set W in Y such that h − ( e ) = X ∩ W .Since ay ∈ h − ( e ) ⊆ W , there exists a neighborhood U ⊆ Y of y such that aU ⊆ W . Then a ( U ∩ X ) ⊆ W ∩ X = h − ( e ) and aU ⊆ a ( U ∩ X ) ⊆ a ( U ∩ X ) ⊆ h − ( e ) = h − ( e ). (cid:3) Let T y be the family of all neighborhoods of y in Y . In the semilattice E consider the subset E y = { e ∈ E : ∃ U ∈ T y h [ X ∩ U ] ⊆ ↑ e } , where ↑ e = { f ∈ E : e ≤ f } . The set E y contains the smallest element of the semilattice E and hence is not empty. Let e be amaximal element of E y (which exists, because E is chain-finite) and W ∈ T y be a neighborhood of y such that h [ X ∩ W ] ⊆ ↑ e .By induction we shall construct a sequence ( v n ) n ∈ ω of points of W ∩ X such that for every n ∈ ω the following conditions are satisfied:(1) v · · · v n y / ∈ h − ( e ) ∪ h − ( h ( v · · · v n ));(2) e < h ( v · · · v n +1 ) < h ( v · · · v n ).To start the inductive construction, observe that Claim 4.2 implies that y / ∈ h − ( e ), so we can find aneighborhood V ⊆ W of y such that V V ∩ h − ( e ) = ∅ . Choose any element v ∈ V ∩ X and observe that v y ∈ V V ⊆ Y \ h − ( e ). Taking into account that h ( v ) = h ( v ) h ( v ) = h ( v v ) ∈ h [ X ∩ V V ] ⊆ E \{ e } and v ∈ V ⊆ W ⊆ h − [ ↑ e ], we conclude that h ( v ) > e .We claim that v y / ∈ h − ( h ( v )). Assuming that v y ∈ h − ( h ( v )) ⊆ X , we can apply Claim 4.3and find a neighborhood U ⊆ V of y such that v U ⊆ h − ( h ( v )). By the maximality of e , the set HARACTERIZING CATEGORICALLY CLOSED COMMUTATIVE SEMIGROUPS 7 h [ X ∩ U ] is not contained in h − [ ↑ h ( v )], so we can find an element u ∈ X ∩ U such that h ( u ) / ∈ ↑ h ( v ).Then h ( v ) = h ( v ) · h ( u ) = h ( v u ) = h ( v ), which is a desired contradiction.Now assume that for some n ∈ N the points v , . . . , v n − ∈ W ∩ X with v . . . v n − y / ∈ h − ( e ) ∪ h − ( h ( v · · · v n − )) and h ( v · · · v n − ) > e have been constructed. Claim 4.2 implies that v . . . v n − y / ∈ h − ( e ) ∪ h − ( h ( v · · · v n − )).Since the semigroups h − ( e ) and h − ( h ( v · · · v n − )) are closed in Y , we can find a neighborhood V ⊆ W of y such that the set v . . . v n − V V is disjoint with the closed set h − ( e ) ∪ h − ( h ( v · · · v n − )).Choose any element v n ∈ V ∩ X . Observe that v . . . v n − v n y ∈ v . . . v n − V V and hence v · · · v n y does not belong to the set h − ( e ) ∪ h − ( h ( v · · · v n − )). Observe also that the idempotent h ( v · · · v n ) = h ( v · · · v n − ) h ( v n ) ∈ h [ X ∩ v · · · v n − V V ] does not belong to the set { e } ∪ { h ( v · · · v n − ) } , whichimplies that e < h ( v · · · v n ) < h ( v · · · v n − ).Finally, we show that v · · · v n y / ∈ h − ( h ( v · · · v n )). Assuming the opposite and using Claim 4.3, wecan find a neighborhood U ⊆ W of y such that v · · · v n U ⊆ h − ( h ( v · · · v n )). Since h ( v · · · v n ) > e ,the maximality of the element e guarantees that h [ X ∩ U ]
6⊆ ↑ h ( v , . . . v n ), so we can choose anelement u ∈ X ∩ U \ h − [ ↑ h ( v · · · v n )] and conclude that h ( v · · · v n u ) = h ( v · · · v n ) h ( u ) < h ( v · · · v n ),which contradicts h ( v · · · v n u ) ∈ h ( v · · · v n U ) ⊆ h − ( h ( v . . . v n )). This contradiction completes theinductive step.After completing the inductive construction, we obtain a strictly decreasing sequence (cid:0) h ( v · · · v n ) (cid:1) n ∈ ω of idempotents in E , which is not possible as E is chain-finite. (cid:3) Next we prove a sufficient condition of the C -closedness of a bounded semigroup. Lemma 4.4.
A bounded semigroup X is C -closed if E ( X ) is a C -closed semigroup and for everyinfinite set A ⊆ X the set AA is not a singleton.Proof. Assuming that the semigroup X is not C -closed, we can find an isomorphic topological embed-ding h : X → Y of X endowed with the discrete topology to a topological semigroup ( Y, τ ) ∈ C ⊆ T S .By our assumption, the set h [ E ( X )] is closed in Y , being a C -closed semigroup. Being discrete, thesubspace h [ X ] is open in its closure h [ X ]. Identifying X with its image h [ X ] and replacing Y by h [ X ], we conclude that X is a dense open discrete subsemigroup of a topological semigroup Y ∈ T S such that E ( X ) is closed in Y .Since X is bounded, there exists n ∈ N such that for every x ∈ X the power x n is an idempotentof X . Pick any point a ∈ Y \ X . Note that a n ∈ { x n : x ∈ X } ⊆ E ( X ) = E ( X ). Let e = a n ∈ E ( X ).By the continuity of the semigroup operation, the point a has a neighborhood O a ⊆ Y such that O na = { e } . Let H e be the maximal subgroup of Y , containing e . Then X ∩ H e is the maximal subgroupof X containing e . For every x ∈ O a ∩ X we get x n = e and hence x m ∈ X ∩ H e for all m ≥ n (seeLemma 3.1). We claim that for any m ≥ n the element a m belongs to the semigroup X . Taking intoaccount that ( a m ) n = ( a n ) m = e m = e , we can find a neighborhood U ⊆ Y of a m such that U n = { e } .Next, find a neighborhood V ⊆ O a of a such that V m ⊆ U . It follows that a m is contained in theclosure of the set W := { v m : v ∈ V ∩ X } ⊆ X ∩ H e . Assuming that a m / ∈ X , we conclude that theset W ⊆ U ∩ X ∩ H e is infinite. Since W is a subset of the group X ∩ H e , the product W n ⊆ U n isinfinite and cannot be equal to the singleton U n = { e } . This contradiction shows that a m ∈ X forall m ≥ n . Then there exists a number k ∈ ω such that a k / ∈ X but a k +1 ∈ X . By the continuityof the semigroup operation, the point b = a k has a neighborhood O b ⊆ Y such that O b = { b } ⊆ X .Since b ∈ X \ X , the set A = O b ∩ X is infinite and AA ⊆ O b = { b } is a singleton. (cid:3) Finally we establish a sufficient condition of the C -closedness of a periodic commutative semigroupwith a unique idempotent. Lemma 4.5.
A periodic commutative semigroup X with a unique idempotent e is T S -closed if themaximal subgroup H e of X is bounded and for every infinite set A ⊆ X the set AA is not a singleton. TARAS BANAKH, SERHII BARDYLA
Proof.
Assume that the maximal subgroup H e of X is bounded and for every infinite set A ⊆ X theset AA is not a singleton. To derive a contradiction, assume that X is not T S -closed and hence X isa non-closed discrete subsemigroup of some topological semigroup ( Y, τ ) ∈ T S . Replacing Y by theclosure of X , we can assume that X is dense and hence open in Y . Claim 4.6.
The semigroup X is an ideal in Y .Proof. Given any elements x ∈ X and y ∈ Y , we should prove that xy ∈ X . Since X is periodic,there exists n ∈ N such that x n = e . Consider the set n √ H e = { b ∈ X : b n ∈ H e } . We claim that n √ H e is an ideal in X . Indeed, for any b ∈ n √ H e and z ∈ X we have ( bz ) n = b n z n ∈ H e z n ⊆ H e as H e isan ideal in X (see Lemma 3.1). Since the group H e is bounded, the semigroup n √ H e is bounded, too.By Lemma 4.4, the bounded semigroup n √ H e is T S -closed and hence closed in Y .Taking into account that n √ H e is an ideal in X and x ∈ n √ H e , we conclude that xY = xX ⊆ xX ⊆ n p H e · X = n p H e = n p H e ⊆ X. (cid:3) Take any point y ∈ Y \ X and consider its orbit y N = { y n : n ∈ N } . Claim 4.7. y N ∩ X = ∅ .Proof. To derive a contradiction, assume that y n ∈ X for some n ∈ N . We can assume that n isthe smallest number with this property. It follows that n ≥ n − ≥ n + ( n − ≥ n .Then y n − / ∈ X and y n − ∈ y n y n − ⊆ XY ⊆ X , because X is an ideal in Y . Since X is an opendiscrete subspace of the topological semigroup Y , there exists a neighborhood U ∈ τ of y such that U n − = { y n − } . Consider the set A = ( U ∩ X ) n − and observe that AA ⊆ U n − = { y n − } is asingleton. On the other hand, y n − ∈ A \ X which implies that the set A is infinite. But the existenceof such set A contradicts our assumptions. (cid:3) By our assumption, the maximal subgroup H e of X is bounded and hence there exists p ∈ N suchthat x p = e for every x ∈ H e . Consider the subsemigroup P = { x p : x ∈ X } in the semigroup X . Itfollows from y ∈ X that the element y p belongs to the closure of the set P in Y . Claim 4.8.
P e = { e } .Proof. Given any element x ∈ P , find an element z ∈ X such that x = z p . By Lemma 3.1, thesubgroup H e is an ideal in X , which implies ze ∈ H e . The choice of p ensures that xe = z p e p =( ze ) p = e . (cid:3) Claim 4.9.
For any x ∈ P there exists n ∈ N such that x ( y p ) m = e for all m ≥ n .Proof. Since X is an ideal in Y we have xy p ∈ X . Since X is an open discrete subspace of thetopological semigroup ( Y, τ ), there exists a neighborhood U ⊆ Y of y such that xU p = { xy p } . Chooseany element u ∈ U ∩ X and observe that xU p = { xu p } = { xy p } . Let V = { v p : v ∈ U ∩ X } ⊆ P andobserve that xV = { xu p } . Then xV V = xu p V = u p xV = u p xu p = xu p . Proceeding by induction,we can show that xV n = xu np for every n ∈ N . Since the semigroup X is periodic, there exists n ∈ N such that u pn = e . Then for every m ≥ n , we obtain xV m = xu mp = xu np u p ( m − n ) ∈ xeP = { e } by Claim 4.8. Then x ( y p ) m ∈ xV m ⊆ xV m = { e } = { e } . (cid:3) Now we are able to finish the proof of Lemma 4.5. Inductively we shall construct sequences ofpoints ( x k ) k ∈ N in X , positive integer numbers ( n k ) k ∈ N , ( m k ) k ∈ N and open neighborhoods ( U k ) k ∈ ω of y in Y such that for every k ∈ N the following conditions are satisfied: HARACTERIZING CATEGORICALLY CLOSED COMMUTATIVE SEMIGROUPS 9 (i) x k ∈ U k − ;(ii) x pm k k / ∈ { e } ∪ { x pm i i : i < k } , x pm k k = e , and m k > n k − ;(iii) x pk ( y p ) n k = e , x pk U pn k k = { e } , and n k > n k − ;(iv) y ∈ U k ⊆ U k − .To start the inductive construction, choose any neighborhood U ⊆ Y of y such that e / ∈ U p .Such neighborhood exists since e = y p by Claim 4.7. Also put n = 2. Now assume that for some k ∈ N and all i < k we have constructed a point x i , a neighborhood U i of y and two numbers n i , m i satisfying the inductive conditions. Since y N ∩ X = ∅ , there exists a neighborhood W ⊆ Y of y suchthat e / ∈ W l for any natural number l ≤ p (2 + 2 k + n k − ). Choose any point x k ∈ U k − ∩ W ∩ X .Then e = x lk for any natural number l ≤ p (2 + 2 k + n k − ). Since the semigroup X is periodic and hasa unique idempotent e , there exists a number l k such that ( x pk ) l k + k +1 = e . We can assume that l k isthe smallest number with this property. Then ( x pk ) l k + k = e . The choice of the neighborhood W ∋ x k ensures that l k > n k − + k + 1. Claim 4.10.
The set { ( x pk ) l k + i : 0 ≤ i ≤ k } has cardinality k + 1 .Proof. Assuming that this set has cardinality smaller than k + 1, we can find two numbers i, j suchthat l k ≤ i < j ≤ l k + k and x pik = x pjk . The equality x pik = x pjk = x pik x p ( j − i ) k implies x pik = x pik x np ( j − i ) k for all n ∈ N . Find a (unique) number n ∈ N such that i ≤ n ( j − i ) < j . Then x pn ( j − i ) k x pn ( j − i ) k = x pn ( j − i ) − pik x pik x pn ( j − i ) k = x pn ( j − i ) − pik x pik = x pn ( j − i ) k and hence x pn ( j − i ) k is an idempotent. Since the semigroup X contains a unique idempotent, x pn ( j − i ) k = e and hence j > n ( j − i ) ≥ l k + k + 1, which contradicts the choice of j . (cid:3) By Claim 4.10, there exists a number j such that 0 ≤ j ≤ k and ( x pk ) l k + j / ∈ { e }∪ { x pm i i : 1 ≤ i < k } .Put m k = l k + j and observe that( x pk ) m k = ( x pk ) l k + k +1 ( x pk ) l k +2 j − k − ∈ eP = { e } by Claim 4.8. By Claim 4.9, there exists a number n k > n k − such that x pk ( y p ) n k = e . Since X is anopen discrete subspace of the topological semigroup Y , there exists a neighborhood U k ⊆ U k − of y such that x pk ( U k ) pn k = { e } . This completes the inductive construction.Now consider the set A = { x pm k k : k ∈ N } of P . The inductive condition (ii) guarantees that A isinfinite and a = e for every a ∈ A . Also for any i < j we have x pm i i x pm j j = x pi x pn i j x p ( m i − i x p ( m j − n i ) j ∈ x pi ( U i ) pn i P = eP = { e } . Therefore, AA = { e } is a singleton. But the existence of such set A is forbidden by our assumption. (cid:3) Some properties of T z S -closed semigroups Lemma 5.1.
For each T z S -closed semigroup X , its center Z ( X ) is chain-finite.Proof. To derive a contradiction, assume that the semigroup Z ( X ) contains an infinite chain C . Takeany free ultrafilter U ∈ β ( X ) ⊆ ϕ ( X ) containing the set C . Since C is a chain, for every set U ⊆ C in U we have U U = U , which implies that U U = U . Let Y be the smallest subsemigroup of thesemigroup ϕ ( X ), containing the set X ∪ {U } . Since the set C is contained in the center of thesemigroup X and U U = U , the semigroup Y is equal to the set X ∪ { x U : x ∈ X } ⊆ β ( X ). Then X is not T z S -closed, being a proper dense subsemigroup of the Hausdorff zero-dimensional topologicalsemigroup Y . (cid:3) Corollary 5.2.
For a semilattice X the following conditions are equivalent: (1) X is projectively C -closed; (2) X is C -closed; (3) X is chain-finite.Proof. The implication (1) ⇒ (2) is trivial, the implication (2) ⇒ (3) follows from Lemma 5.1. Theimplication (3) ⇒ (1) follows from Lemma 4.1 and the observation that each homomorphic image ofa chain-finite semilattice is chain-finite. (cid:3) Lemma 5.3.
If a semigroup X is T z S -closed, then for any infinite subset A ⊆ Z ( X ) the set AA isnot a singleton.Proof. Assume that for some infinite set A ⊆ Z ( X ) the product AA is a singleton. Choose anyultrafilter U ∈ β ( X ) containing the set A and observe that U U is a principal ultrafilter (containingthe singleton AA ). Then the subsemigroup Y ⊆ ϕ ( X ) generated by the set X ∪ {U } is equal to X ∪ { x U : x ∈ X } and hence is contained in β ( X ). Consequently, X is a non-closed subsemigroupof Hausdorff zero-dimensional topological semigroup Y , which means that X is not T z S -closed. (cid:3) Lemma 5.4.
The center Z ( X ) of any T z S -closed semigroup X is periodic.Proof. Assuming that Z ( X ) is not periodic, find x ∈ Z ( X ) such that the powers x n , n ∈ N , arepairwise distinct. On the set X consider the free filter F generated by the base consisting of the sets x n ! N = { x n ! k : k ∈ N } , n ∈ N .Taking into account that ( n + 1)! N ⊆ n ! N + n ! N ⊆ n ! N for all n ∈ N , we conclude that F F = F , so F is an idempotent of the semigroup ϕ ( X ). Let Y = X ∪ { a F : a ∈ X } be the smallest subsemigroupof ϕ ( X ) containing the set X ∪ {F } . We endow Y with the subspace topology inherited from ϕ ( X ).Then Y is a topological semigroup, containing X as a proper dense discrete subsemigroup. Since thespace ϕ ( X ) is T it is sufficient to show that the space Y is zero-dimensional, because zero-dimensional T spaces are Hausdorff.By I denote the set of all elements a ∈ X such that the function N → X , n ax n , is injective.It is clear that for every a ∈ I the filter a F is free and hence does not belong to the set X ⊂ Y ofprincipal ultrafilters. Claim 5.5.
For any a ∈ X \ I the filter a F is principal.Proof. By the definition of the set I , there are two numbers n, k ∈ N such that ax n = ax n + k = ax n x k and hence ax n = ax n x ki for all i ∈ N . Find a number j ∈ N such that 0 ≤ kj − n < k and observe thatfor every integer number i > j we get ax ki = ax kj − n x n x k ( i − j ) = x kj − n ax n x k ( i − j ) = x kj − n ax n = ax kj .Consequently, for the set F = { x ki : i > j } ∈ F the set aF = { ax kj } is a singleton, which impliesthat the filter a F is principal. (cid:3) Claim 5.6.
The topological semigroup Y is zero-dimensional.Proof. We need to show that for any point y ∈ Y , any neighborhood O y ⊆ Y of y contains a clopenneighborhood of y . If y ∈ X , then y is an isolated point of the space y and { y } is an open neighborhoodof y , contained in O y . Using Claim 5.5, one can show that { y } is a closed set in Y .Next, assume that y / ∈ X and hence y = a F for some a ∈ X . By Claim 5.5, a ∈ I . Find a set F = x k ! N ∈ F such that h aF i ⊆ O y . We claim that the basic open set h aF i is closed in Y . Given anypoint t ∈ Y \ h aF i , we should find a neighborhood O t ⊆ Y , which is disjoint with h aF i . If t ∈ X , thenthe neighborhood O t = { t } of t is disjoint with h aF i and we are done. So, we assume that t / ∈ X . Inthis case t = b F for some b ∈ I , according to Claim 5.5.We claim that aF ∩ bF = ∅ . To derive a contradiction, assume that aF ∩ bF contains somecommon element ax k ! n = bx k ! m where n, m ∈ N . Then ax k !( n + i ) = bx k !( m + i ) for all i ∈ N and hencethe complement ax k ! N \ bx k ! N is finite, which implies that the set aF = ax k ! N belongs to the free filter b F = t and hence t ∈ h aF i , but this contradicts the choice of t .This contradiction shows that aF ∩ bF = ∅ and hence h bF i is a neighborhood of the filter t , disjointwith the set h aF i , which implies that h aF i is clopen and the space Y is zero-dimensional. (cid:3) HARACTERIZING CATEGORICALLY CLOSED COMMUTATIVE SEMIGROUPS 11
Therefore, X is not T z S -closed. (cid:3) Lemma 5.7.
Assume that X is a periodic T z S -closed semigroup with H ( X ) ⊆ Z ( X ) . If X containsan unbounded subgroup, then for some e ∈ E ( X ) and x ∈ X there exists an infinite set A ⊆ x · H e such that AA is a singleton.Proof. To derive a contradiction, assume that X contains an unbounded subgroup but for any e ∈ E ( X ), x ∈ X and an infinite set A ⊆ x · H e the set AA is not a singleton.Since E ( X ) ⊆ H ( X ) ⊆ Z ( X ), the set of idempotents E ( X ) is a semilattice. Let π : X → E ( X )be the map assigning to each x ∈ X the unique idempotent in the set x N . By Proposition 3.2, π is ahomomorphism.Since X contains an unbounded subgroup, for some idempotent e ∈ E ( X ) the maximal subgroup H e containing e is unbounded. By Lemma 5.1, the semilattice E ( X ) is chain-finite. Consequently,we can find an idempotent e whose maximal group H e is unbounded but for every idempotent f < e the group H f is bounded.In the semigroup X , consider the set T = S (cid:8) ∞ p H f : f ∈ E ( X ) , f e < e (cid:9) . Claim 5.8.
For every a ∈ T , the set G a = { x ∈ H e : ax = ae } is a subgroup of H e such that thequotient group H e /G a is bounded.Proof. Observe that for any x, y ∈ G a we have axy = aey = ay = ae , which means that G a is asubsemigroup of the group H e . Since the group H e is periodic, the subsemigroup G a is a subgroupof H e . It remains to prove that the quotient group H e /G a is bounded. To derive a contradiction,assume that H e /G a is unbounded.Let f = π ( a ). It follows from a ∈ T that f e < e . Now the minimality of e ensures that the group H fe is bounded. Then there exists p ∈ N such that x p = f e for any x ∈ H fe . Claim 5.9.
For every x ∈ H fe and h ∈ H e we have xh p = x .Proof. By Proposition 3.2, π ( f eh ) = f eπ ( h ) = f ee = f e and by Lemma 3.1, f eh = f eh · f e = f eh · π ( f eh ) ∈ H π ( feh ) = H fe . Then ( f eh ) p = f e and xh p = ( xf e ) h p = x ( f eh ) p = xf e = x. (cid:3) In the group H e consider the subgroup G = { h p : h ∈ H e } . By Proposition 3.2, π ( ae ) = π ( a ) π ( e ) = f e and hence ( ae ) n ∈ H fe for some n ∈ N . Claim 5.9 ensures that a n G = a n ( e n G ) = ( ae ) n G = { ( ae ) n } = { a n e } is a singleton and hence G ⊆ G a n . Let k ≤ n be the smallest number such that thesubgroup G ∩ G a k has finite index in G . We claim that k = 1. Assuming that G ∩ G a has finite indexin G , we conclude that the quotient group G/ ( G ∩ G a ) is finite and hence bounded. Since the quotientgroup H e /G is bounded, the quotient group H e / ( G ∩ G a ) is bounded and so is the quotient group H e /G a . But this contradicts our assumption. This contradiction shows that k = 1. The minimalityof k ensures that the subgroup G ∩ G a k − has infinite index in G . Since the group G ∩ G a k has finiteindex in G , the subgroup G ∩ G a k − has infinite index in the group G ∩ G a k . So, we can find an infiniteset I ⊆ G ∩ G a k such that x ( G ∩ G a k − ) ∩ y ( G ∩ G a k − ) = ∅ for any distinct elements x, y ∈ I . Observethat for any distinct elements x, y ∈ I we have a k x = a k e = a k y and a k − x = a k − y (assuming that a k − x = a k − y , we obtain that a k − e = a k − xx − = a k − yx − and hence yx − ∈ G ∩ G a k − whichcontradicts the choice of the set I ).Then the set A = a k − I is infinite. We claim that AA is a singleton. Indeed, for any x, y ∈ I we have a k − xa k − y = a k xa k − y = a k ea k − y = a k − ea k y = e k − ea k e = a k − e . Therefore, AA = { a k − e } .But the existence of such set A contradicts our assumption. (cid:3) Let Q ∞ = { z ∈ C : ∃ n ∈ N ( z n = 1) } be the quasi-cyclic group, considered as a dense subgroup ofthe compact Hausdorff group T = { z ∈ C : | z | = 1 } . Claim 5.10.
There exists a homomorphism h : H e → Q ∞ whose image h [ H e ] is infinite.Proof. By Lemma 5.4, the group H e is periodic, so for every x ∈ H e we can choose the smallestnumber p ( x ) ∈ N such that x p ( x ) = e . Since H e is unbounded, there is a sequence ( x n ) n ∈ N of elementssuch that p ( x n ) > n Q k
If a T z S -closed periodic semigroup X has X · H ( X ) ⊆ Z ( X ) , then each subgroup of X is bounded.Proof. Assuming that X contains an unbounded subgroup, we can apply Lemma 5.7 and find elements e ∈ E ( X ), x ∈ X , and an infinite subset A ⊆ x · H e such that the set AA is a singleton. Since A ⊆ X · H ( X ) ⊆ Z ( X ), we can apply Lemma 5.3 and conclude that the semigroup X is not T z S -closed. But this contradicts our assumption. (cid:3) Lemma 5.12.
Let X be a T z S -closed semigroup and e ∈ E ( X ) be an idempotent such that thesemigroup H e ∩ Z ( X ) is bounded. Then for any sequence ( x n ) n ∈ ω in ∞ √ e ∩ Z ( X ) \ H e there exists n ∈ ω such that x n / ∈ { x pn +1 : p ≥ } .Proof. To derive a contradiction assume that there exists a sequence ( x n ) n ∈ ω in ∞ √ e ∩ Z ( X ) \ H e suchthat for every n ∈ N there exists p n ≥ x n − = x p n n . Since the semigroup H e ∩ Z ( X ) isbounded, there exists n e ∈ N such that x n e = e for each x ∈ H e ∩ Z ( X ).Consider the additive subsemigroup Q + = (cid:8) kp ··· p n : k, n ∈ N (cid:9) of the semigroup of positive rationalnumbers endowed with the binary operation of addition of rational numbers. Let h : Q + → ∞ √ e ∩ Z ( X )be the (unique) homomorphism such that h ( p ··· p n ) = x n for all n ∈ N . Then h (1) = h ( p p ) = x p = x / ∈ H e . By Lemma 3.1, the subgroup H e is an ideal in ∞ √ e . Consequently, the preimage h − [ H e ] = h − [ H e ∩ Z ( X )] is an upper set in Q + , which means that for any points q < r in Q + with q ∈ h − [ H e ] we get r ∈ h − [ H e ]. Then L = h − [ ∞ √ e \ H e ] is a lower set, which contain 1 andhence contains the interval Q + ∩ (0 , h ↾ L injective. Assuming that h ( a ) = h ( b ) for some distinct points a < b in L , we can find natural numbers k and n < m such that a = np ··· p k and b = mp ...p k . Then x nk = h ( a ) = h ( b ) = x mk and hence x nk ∈ H e and a = np ··· p k ∈ h − [ H e ].But this contradicts the choice of a ∈ L ⊆ Q + \ h − [ H e ].Let s = sup L ∈ (0 , + ∞ ) and W = { q ∈ n e Q + : s < q < s } ⊆ L . The injectivity of h ↾ L guaranteesthat the set h [ W ] is infinite. Observe that for every points a, b ∈ W we get a + b > s = s andhence h ( a + b ) ∈ H e ∩ Z ( X ) and thus h ( a + b ) = h ( a + b ) e . Find z ∈ Q + such that a + b = n e z .Then h ( a + b ) = h ( n e z ) e = h ( z ) n e e = ( h ( z ) e ) n e = e by the choice of n e and the inclusion h ( z ) e ∈ ∞ √ H e · H e ⊆ H e (see Lemma 3.1). This implies that the infinite set A = h [ W ] ⊆ Z ( X ) has AA = { e } .Applying Lemma 5.3, we conclude that the semigroup X is not T z S -closed which contradicts ourassumption. (cid:3) Proof of Theorem 1.5
We should prove that a commutative semigroup X is C -closed if and only if X is periodic, chain-finite, all subgroups of S are bounded and for every infinite set A ⊆ X the set AA is not a singleton.The “only if” part follows from Lemmas 5.1, 5.3, 5.4 and 5.11. To prove the “if” part, assumethat X is periodic, chain-finite, all subgroups of S are bounded and for every infinite set A ⊆ X theset AA is not a singleton. By the periodicity, X = S e ∈ E ( X ) ∞ √ e . Consider the map π : X → E ( X )assigning to each x ∈ X the unique idempotent in the orbit x N . By Proposition 3.2, the map π isa semigroup homomorphism. By Lemma 4.5, for every idempotent e ∈ E ( X ) the semigroup ∞ √ e is C -closed. Since X is chain-finite, so is the semilattice E ( X ). Applying Lemma 4.1, we conclude thatthe semigroup X is C -closed.7. C -closedness of quotient semigroups In this section we prove some lemmas that will be used in the proof of Theorem 1.8.
Lemma 7.1.
Let X be a periodic semigroup, e ∈ E ( X ) and Z n := { z ∈ Z ( X ) : z n ∈ H e } for n ∈ N .If for some ℓ ∈ N the set Z ℓ \ H e is infinite, then there exist a finite set F ⊆ Z ℓ and an infinite set A ⊆ Z ℓ \ F X such that AA ⊆ F ∪ H e ⊆ F X .Proof. Lemma 3.1 implies that Z n ⊆ Z n +1 for all n ∈ ω . Let Z ∞ = S n ∈ N Z n = Z ( X ) ∩ ∞ √ H e . By ourassumption, there exists a number ℓ ∈ N such that the set Z ℓ \ H e is infinite. We can assume that ℓ is the smallest number with this property. The obvious equality Z = Z ( X ) ∩ H e implies that ℓ ≥ Z ℓ − \ H e is finite by the minimality of ℓ .Choose any sequence ( z n ) n ∈ ω of pairwise distinct points in the infinite set Z ℓ \ Z ℓ − . Claim 7.2.
For every z ∈ Z ℓ we have z ∈ Z ℓ − .Proof. Since ℓ ≥
2, we have 2 ℓ − ≥ ℓ and hence( z ) ℓ − = z ℓ − = z ℓ z ℓ − ∈ H e ∪ ( H e · ∞ p H e ) ⊆ H e as H e is an ideal in ∞ √ H e by Lemma 3.1. Then z ∈ Z ℓ − according to the definition of Z ℓ − . (cid:3) Claim 7.3.
For every n ∈ N we have Z ∞ ∩ Z n X ⊆ Z n .Proof. For any z ∈ Z n and x ∈ X , with zx ∈ Z ∞ , we should prove that ( zx ) n ∈ H e . The inclusion z ∈ Z n implies z n ∈ H e and π ( z ) = e . By Proposition 3.2, e = π ( zx ) = π ( z ) π ( x ) = eπ ( x ) and hence π ( ex n ) = π ( e ) π ( x ) = e and thus ex n ∈ ∞ √ H e . Then( zx ) n = z n x n = ( z n e ) x n = z n ( ex n ) ∈ H e · ∞ p H e ⊆ H e . (cid:3) If for some z ∈ Z ℓ , the set A = ( Z ℓ \ Z ℓ − ) ∩ ( zX ) is infinite, then for the finite set F =( Z ℓ − \ H e ) ∪ { e } we have A ∩ F X ⊆ ( Z ℓ \ Z ℓ − ) ∩ Z ℓ − X = ∅ by Claim 7.3. On the other hand, AA ⊆ Z ℓ ∩ z X ⊆ Z ℓ ∩ Z ℓ − X ⊆ Z ℓ − = F ∪ H e ⊆ F X . Therefore, the finite set F and the infinite set A have the properties required in Lemma 7.1.So, we assume that for every z ∈ Z ℓ , the set ( Z ℓ \ Z ℓ − ) ∩ ( zX ) is finite.Let T = {h i, j, k i ∈ ω × ω × ω : i < j < k } and χ : T → { , , } be the function defined by theformula χ ( i, j, k ) = z i z j ∈ Z ℓ − ;1 if z i z j / ∈ Z ℓ − and z i z j = z i z k ;2 if z i z j / ∈ Z ℓ − and z i z j = z i z k . HARACTERIZING CATEGORICALLY CLOSED COMMUTATIVE SEMIGROUPS 15
By the Ramsey Theorem 5 in [9], there exists an infinite set Ω ⊆ ω such that χ [ T ∩ Ω ] = { c } forsome c ∈ { , , } .If c = 0, then the infinite set A = { z n : n ∈ Ω } has AA ⊆ Z ℓ − = F ∪ H e ⊆ F X for the finite set F = ( Z ℓ − \ H e ) ∪ { e } . On the other hand, A ∩ F X ⊆ ( Z ℓ \ Z ℓ − ) ∩ ( Z ℓ − X ) = ∅ by Claim 7.3.If c = 1, then for any i ∈ Ω the set { z i z j : i < j ∈ Ω } is an infinite subset of the set z i Z ℓ \ Z ℓ − ⊆ ( Z ℓ \ Z ℓ − ) ∩ ( z i X ), which is finite by our assumption. Therefore, the case c = 1 is impossible.If c = 2, then z i z j = z i z k / ∈ Z ℓ − for any numbers i < j < k in Ω. Then for every i < k in Ω wehave z i z k = z i z i + where i + = min { j ∈ Ω : i < j } .Now consider two cases.1) The set A = { z i z i + : i ∈ Ω } ⊆ Z ℓ \ Z ℓ − is infinite. Observe that for any numbers i < j in Ω wehave z i z i + z j z j + = z i z j z j z j + ∈ z j Z ℓ ⊆ Z ℓ − Z ℓ ⊆ Z ℓ − . Also z i z i + z i z i + ∈ z i Z ℓ ⊆ Z ℓ − Z ℓ ⊆ Z ℓ − . Then AA ⊆ Z ℓ − = F ∪ H e for the finite set F = Z ℓ − ∪ { e } .Also A ∩ F S ⊆ ( Z ℓ \ Z ℓ − ) ∩ Z ℓ − X = ∅ by Claim 7.3.2) The set C = { z i z i + : i ∈ Ω } is finite. In this case there exists an element c ∈ C such that theset Λ = { i ∈ Ω : c = z i z i + } is infinite. By our assumption, the set ( Z ℓ \ Z ℓ − ) ∩ cX is finite. Thenthe infinite set A = { z i : i ∈ Λ } \ cX has AA = { z i : i ∈ Λ } ∪ { z i z j : i, j ∈ Λ , i < j } ⊆ Z ℓ − ∪ { z i z i + : i ∈ Λ } = Z ℓ − ∪ { c } ⊆ F ∪ H e ⊆ F X for the finite set F = ( Z ℓ − \ H e ) ∪ { c, e } . Also A ∩ F X ⊆ (( Z ℓ \ Z ℓ − ) ∩ ( Z ℓ − X )) ∪ ( A ∩ cX ) = ∅ . In all cases we have constructed an infinite set A ⊆ Z ℓ and a finite set F ⊆ Z ℓ such that AA ⊆ F ∪ H e ⊆ F X and A ∩ F X = ∅ . (cid:3) Remark 7.4.
The finite set F from the previous lemma can be forced to have at most two elements.For this we need to use one more time the Ramsey Theorem. Let A be the infinite set constructedin the previous lemma. Recall that AA ⊆ F ∪ H e and A ∩ F X = ∅ . Write the set F ∪ { e } as { f , f , . . . , f n } where f = e . The Pigeonhole Principle implies that there exists k ≤ n and aninfinite subset B ⊆ A such that { x : x ∈ B } ⊆ H e if k = 0 and { x : x ∈ B } = { f k } if k >
0. By[ B ] we denote the set of all two-element subsets of B . Consider the function χ : [ B ] → { , . . . , n } defined by the formula: χ ( { a, b } ) = ( ab ∈ H e ; i if ab = f i for some i ∈ { , . . . , n } .By the Ramsey Theorem, there exist a number i ∈ { , . . . , n } and an infinite subset A ′ ⊆ B suchthat χ ( { x, y } ) = i for any distinct elements x, y ∈ A ′ . Then for the set F ′ = { f k , f i } we have A ′ A ′ ⊆ F ′ ∪ H e and A ′ ∩ F ′ X ⊆ A ∩ F X = ∅ . Lemma 7.5.
Let X be an ideally T z S -closed semigroup and for some e ∈ E ( X ) ∩ Z ( X ) the semigroup H e ∩ Z ( X ) is bounded. Then the set ∞ √ H e ∩ Z ( X ) \ H e is finite.Proof. To derive a contradiction, assume that the set ∞ √ H e ∩ Z ( X ) \ H e is infinite. For every n ∈ N consider the set Z n = { z ∈ Z ( X ) : z n ∈ H e } and let Z ∞ = S n ∈ N Z n = Z ( X ) ∩ ∞ √ H e .If for some ℓ ∈ N the set Z ℓ \ H e is infinite, then we can apply Lemma 7.1 and find a finite set F ⊂ Z ℓ ⊆ Z ( X ) and infinite set A ⊆ Z ℓ such that AA ⊆ F ∪ H e ⊆ F X and A ∩ ( F X ) = ∅ . Consider the ideal I = F X and the quotient semigroup X/I . Then the quotient image q [ A ] of A in X/I is aninfinite set in Z ( X/I ) such that q [ A ] q [ A ] = q [ I ] is a singleton. By Lemma 5.3, the semigroup X/I isnot T z S -closed, which contradicts our assumption. This contradiction shows that for every n ∈ N theset Z n \ H e is finite.Since the semigroup Z ( X ) ∩ H e is bounded, there exists p ∈ N such that x p = e for all x ∈ Z ( X ) ∩ H e .Consider the subsemigroup P = { z p : z ∈ Z ∞ } in Z ∞ . Claim 7.6. P ∩ H e = { e } .Proof. Given any element x ∈ P ∩ H e , find z ∈ Z ∞ such that x = z p . Lemma 3.1 implies that ze ∈ ∞ √ e · e ⊆ H e and hence ( ze ) p = e by the choice of p . Then x = xe = z p e = ( ze ) p = e . (cid:3) Claim 7.7.
For every n ∈ ω the set P \ Z n is not empty.Proof. Assuming that P \ Z n = ∅ , we conclude that P ⊆ Z n and hence Z ∞ ⊆ Z pn . Then the set Z ∞ ⊆ Z pn \ H e is finite, which contradicts our assumption. (cid:3) Consider the tree T = [ n ∈ ω (cid:8) ( t k ) k ∈ n ∈ P n : t = e ∧ (cid:0) ∀ k ∈ n \ { } ( t ( k ) ∈ Z k \ H e ∧ t ( k ) = t ( k − (cid:1)(cid:9) endowed with the partial order of inclusion of functions. Since the sets Z n \ H e are finite, the tree T has finitely many branching points at every vertex. On the other hand, this tree has infiniteheight. This follows from the fact that for every element z ∈ P \ Z k , there exists n > k such that z n ∈ P ∩ H e = { e } but z n − / ∈ H e . Then the sequence ( z n − i ) i ∈ k belongs to the tree T . By K˝onig’sLemma 5.7 [14], the tree T has an infinite branch which is a sequence ( z n ) n ∈ ω in P such that z ∈ H e and z n = z n − , z n ∈ Z n \ H e , for all n ∈ N . But the existence of such a sequence contradictsLemma 5.12. (cid:3) Proof of Theorem 1.8
Given a commutative semigroup X , we should prove the equivalence of the following conditions:(1) X is projectively C -closed;(2) X is ideally C -closed;(3) X is chain-finite, almost Clifford, and all subgroups are bounded.The implication (1) ⇒ (2) is trivial.(2) ⇒ (3) Assume that X is ideally C -closed. Then X is C -closed and by Lemmas 5.1, 5.4, 5.11, X = Z ( X ) is chain-finite, periodic, and all subgroups of X are bounded. It remains to prove that X is almost Clifford. By Lemma 7.5, for every e ∈ E ( X ) the set ∞ √ H e \ H e is finite. Assuming thatthe semigroup X is not almost Clifford, we conclude that the set B = { e ∈ E ( X ) : ∞ √ H e = H e } is infinite. Since the semilattice E ( X ) is chain-finite, we can apply the Ramsey Theorem and findan infinite antichain C ⊆ B (the latter means that xy / ∈ { x, y } for any distinct elements x, y ∈ C ).Let R = { e ∈ E ( X ) : ∃ c ∈ C ( e < c ) } . It is straightforward to check that R is an ideal in E ( X ).Proposition 3.2 implies that J = S e ∈ R ∞ √ H e is an ideal in X . By Lemma 3.1, H e is an ideal in ∞ √ H e for each e ∈ E ( X ). Using this fact, it is easy to check that I = J ∪ S e ∈ C H e is an ideal in X . Since X is ideally C -closed, the quotient semigroup X/I is C -closed. By the choice of the set C , for every e ∈ C the set ∞ √ H e \ H e is not empty and by the periodicity contains an element a e such that a e ∈ H e .Then A = { a e : e ∈ C } ⊆ X \ I is an infinite set such that AA ⊆ I . Moreover, the image q [ A ] of A in the quotient semigroup X/I is an infinite subset of
X/I such that q [ A ] q [ A ] = q ( I ) is a singleton,which contradicts Lemma 5.3.(3) ⇒ (1) Assume that X is chain-finite, periodic, almost Clifford and all subgroups of X arebounded. By Theorem 1.5, the projectively C -closedness of X will be proved as soon as we check that HARACTERIZING CATEGORICALLY CLOSED COMMUTATIVE SEMIGROUPS 17 for any congruence ≈ on S the quotient semigroup Y = X/ ≈ is periodic, chain-finite, all subgroupsof X/ ≈ are bounded and for any infinite subset A ⊆ X/ ≈ the set AA is not a singleton.Let q : X → Y = X/ ≈ be the quotient homomorphism. The periodicity of X implies the periodicityof Y . To see that Y is chain-finite, observe that for every e ∈ E ( Y ) the (periodic) semigroup q − ( e )contains an idempotent. This implies that E ( Y ) = q [ E ( X )]. Since X is chain-finite, its maximalsemilattice E ( X ) is chain-finite. By Corollary 5.2, E ( X ) is projectively C -closed and then so is itshomomorphic image E ( Y ). Using Corollary 5.2 one more time, we obtain that the semilattice E ( Y )is chain-finite. The following claim implies that the semigroup Y is almost Clifford and all subgroupsof Y are bounded. Claim 8.1.
For any idempotent e ∈ E ( Y ) there exists an idempotent s ∈ E ( X ) such that h [ H s ] = H e .Proof. Since X is chain-finite, the semilattice E ( X ) ∩ q − ( e ) contains the smallest idempotent s . Weclaim that H e = q [ H s ]. In fact, the inclusion q [ H s ] ⊆ H e is trivial. To see that H e ⊆ q [ H s ], takeany element y ∈ H e ⊆ Y and find x ∈ X with q ( x ) = y . Find n ∈ N such that x n ∈ E ( X ) and y n = e . It follows from y = q ( x ) that e = y n = q ( x n ) and hence s = sx n by the minimality of s . ByProposition 3.2, π ( sx ) = π ( s ) π ( x ) = sx n = s and then sx = ( ss ) x = s ( sx ) = π ( sx ) sx ∈ H π ( sx ) = H s by Lemma 3.1. Finally, y = ey = h ( s ) h ( x ) = h ( sx ) ∈ h [ H s ] and hence H e = h [ H s ]. (cid:3) It remains to prove that for any infinite subset A ⊆ Y the set AA is not a singleton. This followsfrom the next lemma. Lemma 8.2.
For any infinite set A in an almost Clifford semigroup S the set AA is infinite.Proof. Since S is almost Clifford, the set A ∩ H ( S ) is infinite. If for some idempotent e ∈ E ( S ) theintersection A ∩ H e is infinite, then for any a ∈ A ∩ H e the set a ( A ∩ H e ) ⊆ ( AA ) ∩ H e ⊆ AA isinfinite (since shifts in groups are injective) and hence AA is not a singleton. So, assume that forevery e ∈ E ( S ) the intersection A ∩ H e is finite.Then the set E = { e ∈ E ( S ) : A ∩ H e = ∅} is infinite. For every e ∈ E , fix an element a e ∈ A ∩ H e and observe that a e ∈ AA ∩ H e , which implies that the set AA ⊇ { a e : e ∈ E } is infinite. (cid:3) Some open problems
In this section we collect some open problems, motivated by the results, obtained in this paper.
Question 9.1.
Does there exist a T z S -closed semigroup which is not T S -closed? A semigroup S is a band if S = E ( S ). Question 9.2.
Is each T S -closed band chain-finite? A semigroup S is inverse if for any x ∈ S there exists a unique element x − ∈ S such that xx − x = x and x − xx − . Question 9.3.
Is any T S -closed inverse semigroup chain-finite? Question 9.4.
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Email address : [email protected] S. Bardyla: University of Vienna, Institute of Mathematics, Kurt G¨odel Research Center (Austria)
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