A metrizable Lawson semitopological semilattice with non-closed partial order

We construct a metrizable Lawson semitopological semilattice $X$ whose partial order $\le_X=\{(x,y)\in X\times X:xy=x\}$ is not closed in $X\times X$. This resolves a problem posed earlier by the authors.


Introduction
In this paper we shall construct an example of a metrizable semitopological semilattice with non-closed partial order.
A semilattice is a commutative semigroup X whose any element x ∈ X is an idempotent in the sense that xx = x. A typical example of a semilattice is any partially ordered set X in which any finite 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 order we have xy = inf{x, y}.
A (semi)topological semilattice is a semilattice X endowed with a topology such that the binary operation X × X → X, xy → xy, is (separately) continuous.
The continuity of the semilattice operation in a Hausdorff topological semilattice implies the following well-known fact, see [5,.
Observe that the indiscrete topology {∅, X} on X is ℓ-admissible. So, the family of ℓ-admissible topologies is not empty. This family has the largest element. This is the topology τ ℓ consisting of 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 1 -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 2 -separating if ℓ is T 1 -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 1 if each finite subset of T is τ -closed in T . In this case we say that (T, τ ) is a T 1 -space.
Lemma 2.2. The function ℓ is T 1 -separating if and only if the topology τ ℓ satisfies the separation axiom T 1 .
Proof. Assume that ℓ is T 1 -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 1 -separating property of ℓ, for every x ∈ F the set Ω x = {n ∈ ω : s n = x} is finite and so is the set {n ∈ ω : s n / ∈ U } = x∈F Ω x . Now assuming that each finite subset F ⊂ X is closed in the topology τ ℓ , we shall prove that the function ℓ is T 1 -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 1 -separating. Lemma 2.3. If the topology τ ℓ is Hausdorff, then the function ℓ is T 2 -separating.
Proof. By Lemma 2.2, the function ℓ is T 1 -separating. To prove that ℓ is T 2 -separating, take two sequences s, t ∈ dom(ℓ) with ℓ(s) = ℓ(t). By the Hausdorff property of the topology τ ℓ , there are disjoint 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). Now we shall prove that the largest ℓ-admissible topology τ ℓ is Hausdorff if the function ℓ is T 2 -separating and dom(ℓ) is at most countable.
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 2 -separating property of ℓ, there exists a finite set Ω ⊂ ω such that s n = t m for any n, m ∈ ω \ Ω. Consider the finite set E = {ℓ(s)} ∪ {s n : n ∈ Ω} \ {ℓ(t)}. By the T 1 -separating property of ℓ, the set Λ = Ω ∪ {n ∈ ω : t n ∈ E} is finite. Then the set {n ∈ ω : t n ∈ s[I] * } ⊂ Λ is finite, too. 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 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 are disjoint neighborhoods of the points x, y in s[I] * , witnessing that the subspace s[I] * of (X, τ ℓ ) is Hausdorff.
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 is Hausdorff.
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 Since the space A is Hausdorff and closed in T , its diagonal ∆ A is closed in A × A and in T × T . By analogy, the diagonal ∆ B is closed in B × B and in T × T . Then the union ∆ T = ∆ A ∪ ∆ B is closed in T × T and the space T is Hausdorff.
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 in the subspace topology of K. A topology τ on a set X is called a k ω -topology if it is generated by a countable family of compact subsets of the topological space (X, τ ).
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 1 . Now it suffices to check that this topology is normal. From now on, 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 such that 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, for every n ∈ ω the subspace K n := i≤n s i [ω] * of (X, τ ℓ ) is compact, Hausdorff, and closed in (X, τ ℓ ). Let A 0 := A∩K 0 and B 0 := B ∩K 0 . 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 ; 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) and observe that After completing the inductive construction, observe that V := n∈ω V n and W = n∈ω W n are disjoint open sets in X such that A ⊂ V and B ⊂ W .
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: (1) X admits a metrizable topology τ in which every sequence s ∈ dom(ℓ) converges to the point ℓ(s). (2) X admits a Hausdorff topology τ in which every sequence s ∈ dom(ℓ) converges to the point ℓ(s). (3) The following two properties are satisfied: (3a) for any s ∈ dom(ℓ) and x ∈ X with s = ℓ(s) the set {n ∈ ω : s n = x} is finite; (3b) 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 .
To prove that (3) ⇒ (1), assume the the condition (3) is satisfied. Then the function ℓ is T 2 -separating and by Lemma 2.7, the largest ℓ-admissible topology τ ℓ is Hausdorff and normal. Consider the countable subset D = s∈dom(ℓ) s[ω] * of X and observe that X \ D is a closed-andopen discrete subspace of the topological space X endowed with the topology τ ℓ . Being countable and Tychonoff, the closed-and-open subspace D of X is zero-dimensional. Then for any distinct points 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. Then the 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).
The countability of the domain dom(ℓ) in Theorems 2.8 is essential as shown by the following example.
Proof. To construct such function ℓ, take any Hausdorff (ω 1 , ω 1 )-gap on ω, which is a pair (A i ) i∈ω1 , (B i ) i∈ω1 of families of infinite subsets of ω satisfying the following two conditions: (H1) for any i < j < ω 1 we have A i ⊂ * A j and B i ⊂ * B j ; (H2) A i ∩ B j is finite for any i, j ∈ ω 1 ; (H3) for any set C ⊂ ω one of the sets {i ∈ ω 1 : A i ⊂ * C} or {i ∈ ω 1 : B i ⊂ * ω \ C} is at most countable. Here the notation A ⊂ * B means that the complement A \ B is finite. It is well-known [6, Ch.20] that Hausdorff (ω 1 , ω 1 )-gaps do exist in ZFC. For every i ∈ ω 1 choose any bijective functions α i : ω → A i and β i : ω → B i , and put dom(ℓ) = The injectivity of the functions α i , β i and the condition (H2) ensure that the function ℓ is T 2 -separating. Assuming that the ℓ-admissble topology on ω is Hausdorff, we could find two disjoint open sets U 0 , U 1 ∈ τ ℓ such that 0 ∈ U 0 and 1 ∈ U 1 . By condition (H3), there exists i ∈ ω 1 such that A i ⊂ * U 0 or B i ⊂ * U 1 . In the first case the set {n ∈ ω : α i (n) / ∈ U 0 } is infinite, which contradicts U 0 ∈ τ ℓ . In the second case the set {n ∈ ω :

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 the operation 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 ).
The countability of the sets dom(ℓ) and A imply the countability of the sets dom(Aℓ) and By the definition of the topology τ Aℓ , the set D is open-and-closed in (X, τ Aℓ ) and the complement X \D is discrete. Being Tychonoff, the countable subspace D of (X, τ Aℓ ) is zero-dimensional. This allows us to choose a countable family B ⊂ τ Aℓ of open-and-closed sets that separate points of 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 Aℓ is an act topology on (X, A), for every α ∈ A and B ∈ B the set α −1 (B) is closed-and-open. Then the topology τ D on D generated by the subbase {D ∩ α −1 (B), D \ α −1 (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 discrete subspace of (X, τ ), the continuity of α↾D implies that α is a continuous self-map of the metrizable topological space (X, τ ). This means that the topology τ is shift-continuous. Since τ ⊂ τ Aℓ , the metrizable topology τ is ℓ-admissible.  x ∈ X, and (x i ) i∈κ be a transfinite sequence of points in X. Assume that there exists a (not necessarily bijective) enumeration A = {α i } i∈κ of the set A such that for each ordinal m ∈ κ and ordinals i, j, k < m the following conditions are satisfied: 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.

Convergent sequences in semigroups
Let X be a semigroup and X 1 be the semigroup X with attached unit. A topology τ on X is called shift-continuous if for every a, b ∈ X 1 the two-sided shift 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 1 }. Applying Theorem 3.1 to this act, we obtain the following theorem, which is a main result of this section.
Theorem 4.1. For a countable semigroup X and a function ℓ : dom(ℓ) → X defined on 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, b ∈ X 1 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 1 with a·ℓ(s)·b = c·ℓ(t)·d there exists a finite set 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 defined on 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). (3a) for any s ∈ dom(ℓ), a ∈ X 1 and x ∈ X with x = a·ℓ(s) the set {n ∈ ω : a·s n = x} is finite; (3b) for any s, t ∈ dom(ℓ) and a, b ∈ X 1 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 a countable 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} is finite; (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 . To check the condition (3a) of Theorem 4.2, take any sequence s ∈ dom(ℓ) and element a ∈ X 1 and x ∈ X with x = a·ℓ(s). If a ∈ X, then the set {n ∈ ω : a·s n = x} is finite by the 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 implies that 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) and elements a, b ∈ X 1 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 case the inequality a·ℓ(s) = b·ℓ(t) transforms into the inequality ℓ(s) = b·ℓ(t). We claim that there exists 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 finite set F ⊂ ω such that c·s n = cb·t m and hence s n = b·t m for any n, m ∈ ω \ F .
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 a subset {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 converges 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 two points x, y ∈ K such that for any neighborhoods O x , O y ⊂ K of the points x, y there exists a finite subset 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 ⊂ k≤n V k such that inf F n ∈ 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 : i<n |F i | < k ≤ i≤n |F i |} for every n ∈ ω. Let ℓ(s) = y and ℓ(t) = x. Let X be the countable semilattice generated by the countable set {x, y} ∪ s[ω] ∪ t[ω]. The metrizable topology on X inherited from K witnesses that the condition (1) is satisfied.
It remains to prove that X admits no Lawson Hausdorff topology τ in which the sequence s converges to ℓ(s) = y and the sequence t converges to ℓ(t) = x. To derive a contradiction, assume that such topology τ exists. Then the points x, y have disjoint open neighborhoods O x , O y ∈ τ such that O x is a subsemilattice of X. Since the sequences s and t converge to y and x, respectively, there exists n ∈ ω such that s k ∈ O y and t k ∈ O x for all k ≥ n. Then F n ⊂ {t k } k≥n ⊂ O x and hence inf F n ∈ O x (as O x is a subsemilattice of X). On the other hand, inf F n = s n ∈ O y . Then inf F n ∈ O x ∩ O y , which contradicts the choice of the neighborhoods O x and O y .

The example
In this section we shall apply Theorem 5.1 to construct an example of a metrizable semitopological semilattice with dense non-closed partial order.
It is easy to see that the partial order on {0, 1, 2} induced by the semilattice operation (of minimum) coincides with the usual linear order on {0, 1, 2}. Then the semilattice operation (of coordinatewise minimum) on X ⊂ {0, 1, 2} ω induces the natural partial order on X.
For every n ∈ ω consider the functions 0 n , 1 n ∈ X defined by 0 n (i) = 0 if i = n 2 otherwise and 1 n (i) = 1 if i = n 2 otherwise.
It is clear that 0 n ·1 n = 0 n and hence 0 n ≤ 1 n for all n ∈ ω.
Theorem 6.1. The semilattice X admits a metrizable shift-continuous topology τ such that the set {(0 n , 1 n ) : n ∈ N} is dense in the square X × X of the semitopological semilattice (X, τ ). Since