A metrizable semitopological semilattice with non-closed partial order
aa r X i v : . [ m a t h . GN ] F e b A METRIZABLE SEMITOPOLOGICAL SEMILATTICEWITH NON-CLOSED PARTIAL ORDER
TARAS BANAKH, SERHII BARDYLA, AND ALEX RAVSKY
Abstract.
We construct a metrizable semitopological semilattice X whose partial order P = { ( x, y ) ∈ X × X : xy = x } is a non-closed dense subset of X × X . As a by-product we findnecessary and sufficient conditions for the existence of a (metrizable) Hausdorff topology on aset, act, semigroup or semilattice, having a prescribed countable family of convergent sequences. Introduction
In this paper we shall construct an example of a metrizable semitopological semilattice withnon-closed partial order.A semilattice is a commutative semigroup X whose any element x ∈ X is an idempotent in thesense that xx = x . A typical example of a semilattice is any partially ordered set X in which anyfinite non-empty set F ⊂ X has the greatest lower bound inf( F ). In this case the binary operation X × X → X , ( xy ) inf { x, y } , turns X into a semilattice.Each semilattice X carries a partial order ≤ defined by x ≤ y iff xy = x . For this partial orderwe have xy = inf { x, y } .A ( semi ) topological semilattice is a semilattice X endowed with a topology such that the binaryoperation X × X → X , xy xy , is (separately) continuous.The continuity of the semilattice operation in a Hausdorff topological semilattice implies thefollowing well-known fact, see [5, VI-1.14]. Proposition 1.1.
For any Hausdorff topological semilattice X the partial order P = { ( x, y ) ∈ X × X : xy = x } is a closed subset of X × X . It is natural to ask whether this proposition remains true for Hausdorff semitopological semi-groups. The following example answers this question in negative.
Example 1.2.
There exists a metrizable countable semitopological semilattice X whose partialorder is dense and non-closed in X × X . This example will be constructed in Section 6 after some preliminary work, made in Sections 2–5. In Section 2 we establish necessary and sufficient conditions on a set X and function ℓ :dom( ℓ ) → X defined on a subset dom( ℓ ) ⊂ X ω ensuring that X admits a (metrizable) Hausdorfftopology in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ). In Section 3 we studythe analogous problem for acts, i.e. sets endowed with monoids of self-maps and in Section 4, 5we apply the obtained results about acts to constructing topologies with prescribed convergentsequences on semigroups and semilattices. More information on the closedness of the partial orderin semitopological semilattices can be found in [1, § Convergent sequences in topological spaces
Let X be a set and X ω be its countable power. Elements of X ω are sequences s = ( s n ) n ∈ ω .Let ℓ : dom( ℓ ) → X be a function defined on a subset dom( ℓ ) ⊂ X ω .A topology τ on the set X is called ℓ -admissible if each sequence s ∈ dom( ℓ ) converges to thepoint ℓ ( s ) in the topological space ( X, τ ). Mathematics Subject Classification.
Key words and phrases. semitopological semilattice, partial order, convergent sequence, act, semigroup.The second author was supported by the Austrian Science Fund FWF (Grant I 3709-N35).
Observe that the indiscrete topology {∅ , X } on X is ℓ -admissible. So, the family of ℓ -admissibletopologies is not empty. This family has the largest element. This is the topology τ ℓ consistingof all subsets U ⊂ X such that for any sequence s = ( s n ) n ∈ ω ∈ dom( ℓ ) with ℓ ( s ) ∈ U the set { n ∈ ω : s n / ∈ U } is finite. The topology τ ℓ will be referred to as the largest ℓ -admissible topology on X . In this section we discuss the following problem. Problem 2.1.
Under which conditions the largest ℓ -admissible topology τ ℓ on X is Hausdorff ? Below we define two necessary conditions of the Haudorffness of the topology τ ℓ .The function ℓ : dom( ℓ ) → X is defined to be • T -separating if for any sequence s ∈ dom( ℓ ) any any point x ∈ X with x = ℓ ( s ) the set { n ∈ ω : s n = x } is finite; • T -separating if ℓ is T -separating and for any sequences s, t ∈ dom( ℓ ) with ℓ ( s ) = ℓ ( t )there exists a finite set F ⊂ ω such that s n = t m for any n, m ∈ ω \ F .We say that a topology τ on a set T satisfies the separation axiom T if each finite subset of T is τ -closed in T . In this case we say that ( T, τ ) is a T -space . Lemma 2.2.
The function ℓ is T -separating if and only if the topology τ ℓ satisfies the separationaxiom T .Proof. Assume that ℓ is T -separating and take any finite set F ⊂ X . To show that the set U := X \ F belongs to the topology τ ℓ , it suffices to check that for every s ∈ dom( ℓ ) with ℓ ( s ) ∈ U the set { n ∈ ω : s ( n ) / ∈ U } is finite. By the T -separating property of ℓ , for every x ∈ F the setΩ x = { n ∈ ω : s n = x } is finite and so is the set { n ∈ ω : s n / ∈ U } = S x ∈ F Ω x .Now assuming that each finite subset F ⊂ X is closed in the topology τ ℓ , we shall prove thatthe function ℓ is T -separating. Given any point x ∈ X \ { ℓ ( s ) } , observe that ℓ ( s ) ∈ X \ { x } ∈ τ ℓ implies that the set { n ∈ ω : s n / ∈ X \ { x }} = { n ∈ ω : s n = x } is finite, which means that ℓ is T -separating. (cid:3) Lemma 2.3.
If the topology τ ℓ is Hausdorff, then the function ℓ is T -separating.Proof. By Lemma 2.2, the function ℓ is T -separating. To prove that ℓ is T -separating, take twosequences s, t ∈ dom( ℓ ) with ℓ ( s ) = ℓ ( t ). By the Hausdorff property of the topology τ ℓ , there aredisjoint open sets U, V ∈ τ ℓ such that ℓ ( s ) ∈ U and ℓ ( t ) ∈ V . By the definition of the topology τ λ , the sets F := { n ∈ ω : s n / ∈ U } and E := { m ∈ ω : t m / ∈ V } are finite. Then s n = t m for any n, m ∈ ω \ ( F ∪ E ). (cid:3) Now we shall prove that the largest ℓ -admissible topology τ ℓ is Hausdorff if the function ℓ is T -separating and dom( ℓ ) is at most countable.For a sequence s ∈ dom( ℓ ) and a subset I ⊂ ω let s [ I ] := { s n : n ∈ I } and s [ I ] ∗ := s [ I ] ∪ { ℓ ( s ) } . Lemma 2.4.
Assume that the function ℓ : dom( ℓ ) → X is T -separating. Let s ∈ dom( ℓ ) and I ⊂ ω . (1) the set s [ I ] ∗ is closed in ( X, τ ℓ ) ; (2) each point x ∈ s [ I ] ∗ \ { ℓ ( s ) } is isolated in s [ I ] ∗ ; (3) the subspace s [ I ] ∗ of ( X, τ ℓ ) is compact and Hausdorff.Proof.
1. The inclusion X \ s [ I ] ∗ ∈ τ ℓ will follow as soon as we show that for any sequence t ∈ dom( ℓ ) with ℓ ( t ) / ∈ s [ I ] ∗ the set { n ∈ ω : t n ∈ s [ I ] ∗ } is finite. By the T -separating propertyof ℓ , there exists a finite set Ω ⊂ ω such that s n = t m for any n, m ∈ ω \ Ω. Consider thefinite set E = { ℓ ( s ) } ∪ { s n : n ∈ Ω } \ { ℓ ( t ) } . By the T -separating property of ℓ , the setΛ = Ω ∪ { n ∈ ω : t n ∈ E } is finite. Then the set { n ∈ ω : t n ∈ s [ I ] ∗ } ⊂ Λ is finite, too.2. Given any point x ∈ s [ I ] ∗ \ { ℓ ( s ) } , observe that s [ I ] ∗ \ { x } = s [ J ] ∗ where J = I \ s − ( x ).By Lemma 2.4(1), the subspace s [ J ] ∗ is closed in ( X, τ ℓ ) and then the singleton { x } = s [ I ] ∗ \ s [ J ] ∗ is open in s [ I ] ∗ .3. The compactness of s [ I ] ∗ follows from the fact that each neighborhood U ∈ τ ℓ of ℓ ( s )contains all but finitely many points of the set s [ I ]. To see that s [ I ] ∗ is Hausdorff, take any two METRIZABLE SEMITOPOLOGICAL SEMILATTICE WITH NON-CLOSED PARTIAL ORDER 3 distinct points x, y ∈ s [ I ] ∗ . One of these points is distinct from the limit point ℓ ( s ) of the sequence s and we lose no generality assuming that x = ℓ ( s ). By Lemmas 2.2 and 2.4(2), the singleton U x = { x } closed-and-open in s [ I ] ∗ and so is its complement U y = s [ I ] ∗ \ U x . Then U x , U y aredisjoint neighborhoods of the points x, y in s [ I ] ∗ , witnessing that the subspace s [ I ] ∗ of ( X, τ ℓ ) isHausdorff. (cid:3) Corollary 2.5.
If the function ℓ : dom( ℓ ) → X is T -separating, then for every finite subset S ⊂ dom( λ ) the subspace S [ ω ] ∗ = S s ∈ S s [ ω ] ∗ of ( X, τ ℓ ) is compact, Hausdorff, and closed in ( X, τ ℓ ) . This corollary follows from Lemma 2.4 and the following known fact.
Lemma 2.6.
The union A ∪ B of two closed Hausdorff subspaces of a topological space T isHausdorff.Proof. We lose no generality assuming that T = A ∪ B . The Hausdorff property of the space T = A ∪ B will follow as soon as we check that its diagonal ∆ T = { ( x, x ) : x ∈ T } is closed in T × T . For this observe that ∆ T = ∆ A ∪ ∆ B . Since the space A is Hausdorff and closed in T , itsdiagonal ∆ A is closed in A × A and in T × T . By analogy, the diagonal ∆ B is closed in B × B andin T × T . Then the union ∆ T = ∆ A ∪ ∆ B is closed in T × T and the space T is Hausdorff. (cid:3) We say that the topology τ of a topological space X is generated by a family K of subspaces of X if a set U ⊂ X is open in X if any only if for every K ∈ K the intersection U ∩ K is open inthe subspace topology of K . A topology τ on a set X is called a k ω -topology if it is generated bya countable family of compact subsets of the topological space ( X, τ ). Lemma 2.7.
If the function ℓ : dom( ℓ ) → X is T -separating and dom( ℓ ) is at most countable,then the topology τ ℓ is Hausdorff and normal. Moreover, τ λ is a k ω -topology, generated by thecountable family K = { s [ ω ] ∗ : s ∈ dom( ℓ ) } of compact sets in ( X, τ ℓ ) .Proof. The definition of the topology τ ℓ ensures that it is generated by the countable family K .Now we show that the topology τ ℓ is Hausdorff and normal. By Lemma 2.2, the topology τ ℓ satisfies the separation axiom T . Now it suffices to check that this topology is normal. From nowon, we consider X as a topological space endowed with the topology τ ℓ .Given two disjoint closed sets A, B ⊂ X we should find two disjoint open sets V, W ⊂ X suchthat A ⊂ V and B ⊂ W .Let dom( ℓ ) = { s n } n ∈ ω be an enumeration of the countable set dom( ℓ ). By Corollary 2.5, forevery n ∈ ω the subspace K n := S i ≤ n s i [ ω ] ∗ of ( X, τ ℓ ) is compact, Hausdorff, and closed in ( X, τ ℓ ).Let A := A ∩ K and B := B ∩ K . By induction we shall construct sequences ( A n ) n ∈ ω , ( B n ) n ∈ ω ,( V n ) n ∈ ω , ( W n ) n ∈ ω of subsets in X such that for every n ∈ ω the following conditions are satisfied:(1) the sets A n , B n are disjoint and closed in K n ;(2) A n ⊂ V n ⊂ K n and B ⊂ W n ⊂ K n ;(3) the sets V n , W n are open in K n and V n ∩ W n = ∅ ;(4) A n +1 = V n ∪ ( K n +1 ∩ A ) and B n +1 = W n ∪ ( K n +1 ∩ B );(5) A n +1 ∩ B n +1 = ∅ .Assume that for some n ∈ ω disjoint closed sets A n , B n ⊂ K n with K n ∩ A ⊂ A n and K n ∩ B ⊂ B n have been constructed. By the normality of the compact Hausdorff space K n , there are open sets V n , W n ⊂ K n satisfying the conditions (2),(3). Define the sets A n +1 , B n +1 by the formula (4) andobserve that A n +1 ∩ B n +1 = ( A n +1 ∩ B n +1 ∩ K n ) ∪ ( A n +1 ∩ B n +1 ∩ ( K n +1 \ K n ) ⊂ ( V n ∩ W n ) ∪ ( A ∩ B ) = ∅ . After completing the inductive construction, observe that V := S n ∈ ω V n and W = S n ∈ ω W n aredisjoint open sets in X such that A ⊂ V and B ⊂ W . (cid:3) The following theorem is the main result of this section.
Theorem 2.8.
For a set X and function ℓ : dom( ℓ ) → X defined on a countable subset dom( ℓ ) ⊂ X ω the following conditions are equivalent: TARAS BANAKH, SERHII BARDYLA, AND ALEX RAVSKY (1) X admits a metrizable topology τ in which every sequence s ∈ dom( ℓ ) converges to thepoint ℓ ( s ) . (2) X admits a Hausdorff topology τ in which every sequence s ∈ dom( ℓ ) converges to thepoint ℓ ( s ) . (3) The following two properties are satisfied: (3 a ) for any s ∈ dom( ℓ ) and x ∈ X with s = ℓ ( s ) the set { n ∈ ω : s n = x } is finite; (3 b ) for any sequences s, t ∈ dom( ℓ ) with ℓ ( s ) = ℓ ( t ) there exists a finite set F ⊂ ω suchthat s n = t m for any n, m ∈ ω \ F .Proof. The implications (1) ⇒ (2) ⇒ (3) are trivial.To prove that (3) ⇒ (1), assume the the condition (3) is satisfied. Then the function ℓ is T -separating and by Lemma 2.7, the largest ℓ -admissible topology τ ℓ is Hausdorff and normal.Consider the countable subset D = S s ∈ dom( ℓ ) s [ ω ] ∗ of X and observe that X \ D is a closed-and-open discrete subspace of the topological space X endowed with the topology τ ℓ . Being countableand Tychonoff, the closed-and-open subspace D of X is zero-dimensional. Then for any distinctpoints x, y ∈ D we can choose a closed-and-open subset U x,y ⊂ X such that x ∈ U x,y and y / ∈ U x,y .Let τ be the topology on D , generated by the countable subbase { U x,y , X \ U x,y : x, y ∈ D, x = y } .It is clear that the topology τ is second-countable, Hausdorff, zero-dimensional and hence regular.By Urysohn Metrization Theorem [4, 4.2.9], the topological space D τ = ( D, τ ) is metrizable. Thenthe topology of the topological sum on ( X \ D ) ⊕ D τ is also metrizable. Since τ ⊂ τ ℓ , the topology τ is ℓ -admissible, which means that each sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ). (cid:3) The countability of the domain dom( ℓ ) in Theorems 2.8 is essential as shown by the followingexample. Example 2.9.
There exists a T -separating function ℓ : dom( ℓ ) → { , } ⊂ ω defined on a subset dom( ℓ ) ⊂ [ ω ] ω of cardinality | dom( ℓ ) | = ω such that the largest ℓ -admissible topology τ ℓ is notHausdorff.Proof. To construct such function ℓ , take any Hausdorff ( ω , ω )-gap on ω , which is a pair (cid:0) ( A i ) i ∈ ω , ( B i ) i ∈ ω (cid:1) of families of infinite subsets of ω satisfying the following two conditions:(H1) for any i < j < ω we have A i ⊂ ∗ A j and B i ⊂ ∗ B j ;(H2) A i ∩ B j is finite for any i, j ∈ ω ;(H3) for any set C ⊂ ω one of the sets { i ∈ ω : A i ⊂ ∗ C } or { i ∈ ω : B i ⊂ ∗ ω \ C } is at mostcountable.Here the notation A ⊂ ∗ B means that the complement A \ B is finite.It is well-known [6, Ch.20] that Hausdorff ( ω , ω )-gaps do exist in ZFC.For every i ∈ ω choose any bijective functions α i : ω → A i and β i : ω → B i , and put dom( ℓ ) = { α i , β i : i ∈ ω } . Let ℓ : dom( ℓ ) → { , } ⊂ ω be the function such that ℓ − (0) = { α i } i ∈ ω and ℓ − (1) = { β i } i ∈ ω . The injectivity of the functions α i , β i and the condition (H2) ensure that thefunction ℓ is T -separating. Assuming that the ℓ -admissble topology on ω is Hausdorff, we couldfind two disjoint open sets U , U ∈ τ ℓ such that 0 ∈ U and 1 ∈ U . By condition (H3), thereexists i ∈ ω such that A i ∗ U or B i ∗ U . In the first case the set { n ∈ ω : α i ( n ) / ∈ U } isinfinite, which contradicts U ∈ τ ℓ . In the second case the set { n ∈ ω : β i ( n ) / ∈ U } is infinite,which contradicts U ∈ τ ℓ . (cid:3) Convergent sequences in topological acts
For a set X denote by X X the set of all self-maps X → X . The set X X endowed with theoperation of composition is a monoid whose unit is the identity map id X of X .An act is a pair ( X, A ) consisting of a set X and a submonoid A ⊂ X X . Elements of the set A are called the shifts of the act ( X, A ).A topology τ on the underlying set X of an act ( X, A ) is called an shift-continuous if each shift α ∈ A is a continuous self-map of the topological space ( X, τ ). Theorem 3.1.
For an act ( X, A ) with countable set A of shifts and a function ℓ : dom( ℓ ) → X defined on a countable subset dom( ℓ ) ⊂ X ω the following conditions are equivalent: METRIZABLE SEMITOPOLOGICAL SEMILATTICE WITH NON-CLOSED PARTIAL ORDER 5 (1) X admits a shift-continuous metrizable topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) ; (2) X admits a shift-continuous Hausdorff topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) ; (3) the following two properties hold: (3a) for any s ∈ dom( ℓ ) , α ∈ A and x ∈ X with x = α ◦ ℓ ( s ) the set { n ∈ ω : α ◦ s n = x } is finite; (3b) for any sequences s, t ∈ dom( ℓ ) and any shifts α, β ∈ A with α ◦ ℓ ( s ) = β ◦ ℓ ( t ) thereexists a finite set F ⊂ ω such that α ◦ s n = β ◦ t m for any n, m ∈ ω \ F .Proof. The implications (1) ⇒ (2) ⇒ (3) are trivial. To prove that (3) ⇒ (1), assume that thecondition (3) is satisfied. Consider the set dom( A ℓ ) = { α ◦ s : α ∈ A , s ∈ dom( ℓ ) } ⊂ X ω and thefunction A ℓ : dom( A ℓ ) → X defined by the formula A ℓ ( α ◦ s ) = α ( ℓ ( s )). The conditions (3a) and(3b) ensure that the function A ℓ is well-defined and is T -separating. By Lemma 2.7, the largest A ℓ -admissible topology τ A ℓ on X is Hausdorff and normal. Since the monoid A contains theidentity map of X , the topology τ A ℓ is ℓ -admissible, which implies that each sequence s ∈ dom( ℓ )converges to ℓ ( s ) in the topological space ( X, τ A ℓ ).We claim that the topology τ A ℓ is shift-continuous. Given any shift α ∈ A and open set U ∈ τ A ℓ ,we need to check that α − ( U ) ∈ τ A ℓ . The latter inclusion holds if any only if for any sequence s ∈ dom( A ℓ ) with A ℓ ( s ) ∈ α − ( U ) the set { n ∈ ω : s n / ∈ α − ( U ) } = { n ∈ ω : α ◦ s n / ∈ U } is finite. Since A is a monoid, the sequence α ◦ s belongs to dom( A ℓ ). Since U ∈ τ A ℓ and A ℓ ( α ◦ s ) = α ( A ℓ ( s )) ∈ U , the set { n ∈ ω : α ◦ s n / ∈ U } = { n ∈ ω : s n / ∈ α − ( U ) } is finite and weare done.The countability of the sets dom( ℓ ) and A imply the countability of the sets dom( A ℓ ) and D = { s [ ω ] ∗ : s ∈ dom( A ℓ ) } . Observe that for every α ∈ A we have α [ D ] ⊂ D .By the definition of the topology τ A ℓ , the set D is open-and-closed in ( X, τ A ℓ ) and the comple-ment X \ D is discrete. Being Tychonoff, the countable subspace D of ( X, τ A ℓ ) is zero-dimensional.This allows us to choose a countable family B ⊂ τ A ℓ of open-and-closed sets that separate pointsof the countable set D in the sense that for any distinct points x, y ∈ D there exists a set B ∈ B such that x ∈ B and y / ∈ B . Since τ A ℓ is an act topology on ( X, A ), for every α ∈ A and B ∈ B the set α − ( B ) is closed-and-open. Then the topology τ D on D generated by the subbase { D ∩ α − ( B ) , D \ α − ( B ) : α ∈ A , B ∈ B} is second-countable, Hausdorff and zero-dimensional.By the Urysohn metrization Theorem [4, 4.2.9], the topological space D τ = ( D, τ D ) is metrizable.Then the topology τ of topological sum D τ ⊕ ( X \ D ) of D τ and the discrete topological space X \ D is metrizable. By the definition of the topology τ D , for every α ∈ A the restriction α ↾ D is a continuous self-map of the topological space D τ . Since X \ D is a closed-and-open discretesubspace of ( X, τ ), the continuity of α ↾ D implies that α is a continuous self-map of the metrizabletopological space ( X, τ ). This means that the topology τ is shift-continuous. Since τ ⊂ τ A ℓ , themetrizable topology τ is ℓ -admissible. (cid:3) Theorem 3.1 can be compared with the following result proved in [2, 3.4].
Theorem 3.2 (Banakh, Protasov, Sipacheva) . Let κ be an infinite cardinal, ( X, A ) be an act, x ∈ X , and ( x i ) i ∈ κ be a transfinite sequence of points in X . Assume that there exists a (notnecessarily bijective) enumeration A = { α i } i ∈ κ of the set A such that for each ordinal m ∈ κ andordinals i, j, k < m the following conditions are satisfied: (1) if α i ( x ) = α j ( x ) , then α i ( x m ) = α j ( x m ) ; (2) if α i ( x ) = α j ( x k ) , then α i ( x m ) = α j ( x k ) .Then X admits a shift-continuous hereditarily normal topology τ in which the transfinite sequence ( x i ) i ∈ λ converges to the point x in the sense that for every neighborhood O x ∈ τ of x there exists n ∈ κ such that x i ∈ O x for all i ≥ n in κ . TARAS BANAKH, SERHII BARDYLA, AND ALEX RAVSKY Convergent sequences in semigroups
Let X be a semigroup and X be the semigroup X with attached unit. A topology τ on X iscalled shift-continuous if for every a, b ∈ X the two-sided shift X → X, x axb, is a continuous self-map of the topological space ( X, τ ).Each semigroup X has the structure of an act ( X, A ) endowed with the family of shifts A = { s a,b : a, b ∈ X } . Applying Theorem 3.1 to this act, we obtain the following theorem, which is amain result of this section. Theorem 4.1.
For a countable semigroup X and a function ℓ : dom( ℓ ) → X defined on acountable subset dom( ℓ ) ⊂ X ω , the following conditions are equivalent: (1) The semigroup X admits a shift-continuous metrizable topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) ; (2) The semigroup X admits a shift-continuous Hausdorff topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) ; (3) the following two properties hold: (3a) for any s ∈ dom( ℓ ) , a, b ∈ X and x ∈ X with x = a · ℓ ( s ) · b the set { n ∈ ω : a · s n · b = x } is finite; (3b) for any s, t ∈ dom( ℓ ) and a, b, c, d ∈ X with a · ℓ ( s ) · b = c · ℓ ( t ) · d there exists a finiteset F ⊂ ω such that a · s n · b = c · t m · d for all n, m ∈ ω \ F . For commutative semigroups, Theorem 4.1 has a bit simpler form.
Theorem 4.2.
For a countable commutative semigroup X and a function ℓ : dom( ℓ ) → X definedon a countable subset dom( ℓ ) ⊂ X ω , the following conditions are equivalent: (1) The semigroup X admits a shift-continuous metrizable topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) . (2) The semigroup X admits a shift-continuous Hausdorff topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) . (3) The following two properties hold: (3a) for any s ∈ dom( ℓ ) , a ∈ X and x ∈ X with x = a · ℓ ( s ) the set { n ∈ ω : a · s n = x } isfinite; (3b) for any s, t ∈ dom( ℓ ) and a, b ∈ X with a · ℓ ( s ) = b · ℓ ( t ) there exists a finite subset F ⊂ ω such that a · s n = b · t m for any n, m ∈ ω \ F . Convergent sequences in semilattices
Applying Theorem 4.2 to semilattices we obtain the following characterization.
Theorem 5.1.
For a countable semilattice X and a function ℓ : dom( ℓ ) → X defined on acountable subset dom( ℓ ) ⊂ X ω the following conditions are equivalent: (1) The semilattice X admits a shift-continuous metrizable topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) . (2) The semilattice X admits a shift-continuous Hausdorff topology τ in which every sequence s ∈ dom( ℓ ) converges to the point ℓ ( s ) . (3) The following two conditions hold: (3a) for any s ∈ dom( ℓ ) , a ∈ X and x ∈ X with x = a · ℓ ( s ) the set { n ∈ ω : a · s n = x } isfinite; (3b) for any s, t ∈ dom( ℓ ) and a, b ∈ X with a · ℓ ( s ) = b · ℓ ( t ) there exists a finite set F ⊂ ω such that a · s n = b · t m for any n, m ∈ ω \ F .Proof. The implications (1) ⇒ (2) ⇒ (3) are trivial. The implication (3) ⇒ (1) will follow fromTheorem 4.2 as soon as we check that the condition (3) of Theorem 5.1 implies condition (3) ofTheorem 4.2. So, assume that condition (3) of Theorem 5.1 is satisfied.To check the condition (3a) of Theorem 4.2, take any sequence s ∈ dom( ℓ ) and element a ∈ X and x ∈ X with x = a · ℓ ( s ). If a ∈ X , then the set { n ∈ ω : a · s n = x } is finite by the METRIZABLE SEMITOPOLOGICAL SEMILATTICE WITH NON-CLOSED PARTIAL ORDER 7 condition (3a) of Theorem 5.1. So, we assume that a is an external unit for X . In this case { n ∈ ω : a · s n = x } = { n ∈ ω : s n = x } ⊂ { n ∈ ω : x · s n = x · x = x } . Assuming that the set { n ∈ ω : s n = x } is infinite, we conclude that the set { n ∈ ω : x · s n = x } is infinite, which impliesthat x · ℓ ( s ) = x . Since ℓ ( s ) · ℓ ( s ) = ℓ ( s ) = a · ℓ ( s ) = x = ℓ ( s ) · x , the set { n ∈ ω : ℓ ( s ) · s n = x } = { n ∈ ω : ℓ ( s ) · s n = ℓ ( s ) · x } ⊃ { n ∈ ω : s n = x } is finite, which contradicts our assumption.Next, we check the condition (3b) of Theorem 4.2. Given any sequences s, t ∈ dom( ℓ ) andelements a, b ∈ X with a · ℓ ( s ) = b · ℓ ( t ), we need to find a finite set F ⊂ ω such that a · s n = b · t m for any n, m ∈ ω \ F . The condition (3a) of Theorem 5.1 ensures that a or b does not belong to X . We lose no generality assuming that a / ∈ X and hence a is the external unit to X . In this casethe inequality a · ℓ ( s ) = b · ℓ ( t ) transforms into the inequality ℓ ( s ) = b · ℓ ( t ). We claim that thereexists an element c ∈ X such that c · ℓ ( s ) = cb · ℓ ( t ). If ℓ ( s ) = ℓ ( s ) · b · ℓ ( t ), then put c = ℓ ( s ). If ℓ ( s ) = ℓ ( s ) · b · ℓ ( t ), then put c = b · ℓ ( t ) and conclude that c · ℓ ( s ) = ℓ ( s ) = b · ℓ ( t ) = cb · ℓ ( t ).In both cases we get c · ℓ ( s ) = cb · ℓ ( t ). By condiction (3b) of Theorem 5.1, there exists a finiteset F ⊂ ω such that c · s n = cb · t m and hence s n = b · t m for any n, m ∈ ω \ F . (cid:3) A topology on a semilattice X is called Lawson if it has a base consisting of open subsemilattices.
Example 5.2.
There exists a countable semilattice X and a function ℓ : { s, t } → X defined on asubset { s, t } ⊂ X ω such that (1) The semilattice X admits a shift-continuous metrizable topology τ in which the sequence s converges to ℓ ( s ) and the sequence t converges to ℓ ( t ) . (2) The semilattice X admits no Lawson Hausdorff topology τ in which the sequence s con-verges to ℓ ( s ) and the sequence t converges to ℓ ( t ) .Proof. By [3, 2.21], there exists a compact metrizable topological semilattice K containing twopoints x, y ∈ K such that for any neighborhoods O x , O y ⊂ K of the points x, y there exists a finitesubset F ⊂ O x such that inf F ∈ O y . Fix countable neighborhood bases { V n } n ∈ ω and { W n } n ∈ ω at the points x, y , respectively. For every n ∈ ω choose a finite subset F n ⊂ T k ≤ n V k such thatinf F n ∈ T k ≤ n W k . Let s ∈ X ω be a sequence such that s n = inf F n for every n ∈ ω and t ∈ X ω be a sequence such that F n = { t k : P i In this section we shall apply Theorem 5.1 to construct an example of a metrizable semitopo-logical semilattice with dense non-closed partial order.Consider the semilattice { , , } endowed with the operation of taking minimum. In the semi-lattice { , , } ω consider the countable subsemilattice X consisting of functions f : ω → { , , } having non-empty finite support supp( f ) := f − ( { , } ).It is easy to see that the partial order on { , , } induced by the semilattice operation (ofminimum) coincides with the usual linear order on { , , } . Then the semilattice operation (ofcoordinatewise minimum) on X ⊂ { , , } ω induces the natural partial order on X . TARAS BANAKH, SERHII BARDYLA, AND ALEX RAVSKY For every n ∈ ω consider the functions n , n ∈ X defined by n ( i ) = ( i = n n ( i ) = ( i = n . It is clear that n · n = n and hence n ≤ n for all n ∈ ω . Theorem 6.1. The semilattice X admits a metrizable shift-continuous topology τ such that theset { ( n , n ) : n ∈ N } is dense in the square X × X of the semitopological semilattice ( X, τ ) . Since { ( n , n ) } n ∈ ω ⊂ P := { ( x, y ) ∈ X × X : xy = x } 6 = X × X the partial order P of X is dense and non-closed in X × X .Proof. Write the countable set X × X as X × X = { ( x k , y k ) : k ∈ ω } . For every k, n ∈ ω considerthe elements z k,n = k n and u k,n := k n of the set X . These elements form sequences ~z k =( z k,n ) n ∈ ω and ~u k = ( u k,n ) n ∈ ω , which are elements of the set X ω . Let dom( ℓ ) = { ~z k , ~u k : k ∈ ω } and ℓ : dom( ℓ ) → X be the function defined by ℓ ( ~z k ) = x k and ℓ ( ~u k ) = y k for k ∈ ω .We claim that for the function ℓ the condition (3) of Theorem 5.1 is satisfied. In fact, thecondition (3a) is satisfied in the stronger form: for any s ∈ dom( ℓ ) and a, b ∈ X the set { n ∈ ω : a · s n = b } ⊂ { n ∈ ω : ∃ k ∈ ω with 2 k n ∈ supp( b ) } is finite.Now we check the condition (3b). Fix elements a, b ∈ X and sequences s, t ∈ dom( λ ) such that a · ℓ ( s ) = b · ℓ ( t ). It is easy to see that a · s n = b · t m for any n, m ∈ ω \ F where F = { n ∈ ω : ∃ k ∈ ω such that 2 k n ∈ supp( a ) ∪ supp( b ) } .By Theorem 5.1, the semilattice X admits a metrizable shift-continuous topology τ in whichevery sequence s ∈ dom( ℓ ) converges to ℓ ( s ). In particular, for every k ∈ ω the sequence( k n ) n ∈ ω = ~z k converges to x k = ℓ ( ~z k ) and the sequence ( k n ) n ∈ ω = ~z k converges to y k = ℓ ( ~u k ). Consequently, the set { ( k n , k n ) : k, n ∈ ω } ⊂ { ( m , m ) : m ∈ ω } is densein X × X = { ( x k , y k ) : k ∈ ω } .Since { ( n , n ) } n ∈ ω ⊂ P := { ( x, y ) ∈ X × X : xy = x } 6 = X × X, the partial order P is a dense non-closed subset of X × X . (cid:3) Problem 6.2. Does the semilattice X admit a Lawson Hausdorff shift-continuous topology suchthat the partial order P = { ( x, y ) ∈ X × X : xy = x } is not closed (and dense) in X × X ? References [1] T. Banakh, S. Bardyla, Complete topologized posets and semilattices , preprint(https://arxiv.org/abs/1806.02869).[2] T. Banakh, I. Protasov, O. Sipacheva, Topologization of sets endowed with an action of a monoid , TopologyAppl. (2014) 161–174.[3] J.H. Carruth, J.A. Hildebrant, R.J. Koch, The theory of topological semigroups , Vol. 2. Monographs and Text-books in Pure and Applied Mathematics, 100. Marcel Dekker, Inc., New York, 1986.[4] R. Engelking, General topology , Heldermann Verlag, Berlin, 1989.[5] G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M. Mislove, D.S. Scott, Continuous lattices and domains ,Cambridge University Press, Cambridge, 2003.[6] W. Just, M. Weese, Discovering modern set theory. II. Set-theoretic tools for every mathematician , GraduateStudies in Mathematics, 18. American Mathematical Society, Providence, RI, 1997. T.Banakh: Ivan Franko National University of Lviv (Ukraine) and Jan Kochanowski University inKielce (Poland) E-mail address : [email protected] S. Bardyla: Institute of Mathematics, Kurt G¨odel Research Center, Vienna (Austria) E-mail address : [email protected] A.Ravsky: Department of Analysis, Geometry and Topology, Pidstryhach Institute for AppliedProblems of Mechanics and Mathematics National Academy of Sciences of Ukraine E-mail address ::