The Lawson number of a semitopological semilattice

For a Hausdorff topologized semilattice $X$ its $Lawson\;\; number$ $\bar\Lambda(X)$ is the smallest cardinal $\kappa$ such that for any distinct points $x,y\in X$ there exists a family $\mathcal U$ of closed neighborhoods of $x$ in $X$ such that $|\mathcal U|\le\kappa$ and $\bigcap\mathcal U$ is a subsemilattice of $X$ that does not contain $y$. It follows that $\bar\Lambda(X)\le\bar\psi(X)$, where $\bar\psi(X)$ is the smallest cardinal $\kappa$ such that for any point $x\in X$ there exists a family $\mathcal U$ of closed neighborhoods of $x$ in $X$ such that $|\mathcal U|\le\kappa$ and $\bigcap\mathcal U=\{x\}$. We prove that a compact Hausdorff semitopological semilattice $X$ is Lawson (i.e., has a base of the topology consisting of subsemilattices) if and only if $\bar\Lambda(X)=1$. Each Hausdorff topological semilattice $X$ has Lawson number $\bar\Lambda(X)\le\omega$. On the other hand, for any infinite cardinal $\lambda$ we construct a Hausdorff zero-dimensional semitopological semilattice $X$ such that $|X|=\lambda$ and $\bar\Lambda(X)=\bar\psi(X)=cf(\lambda)$. A topologized semilattice $X$ is called (i) $\omega$-$Lawson$ if $\bar\Lambda(X)\le\omega$; (ii) $complete$ if each non-empty chain $C\subset X$ has $\inf C\in\overline{C}$ and $\sup C\in\overline{C}$. We prove that for any complete subsemilattice $X$ of an $\omega$-Lawson semitopological semilattice $Y$, the partial order $\le_X=\{(x,y)\in X\times X:xy=x\}$ of $X$ is closed in $Y\times Y$ and hence $X$ is closed in $Y$. This implies that for any continuous homomorphism $h:X\to Y$ from a compete topologized semilattice $X$ to an $\omega$-Lawson semitopological semilattice $Y$ the image $h(X)$ is closed in $Y$.


Introduction
In this paper we introduce a new cardinal invariantΛ(X) of a Hausdorff topologized semilattice X, called the Lawson number of X. Introducing of the Lawson number was motivated by studying the closedness properties of complete topologized semilattices. Complete topologized semilattices were studied by the first two authors in [1], [2], [3], [4], [5], [6]. It turns out that complete semitopological semilattices share many common properties with compact topological semilattices, in particular their continuous homomorphic images in Hausdorff topological semilattices are closed.
A semilattice is any commutative semigroup of idempotents (an element x of a semigroup is called an idempotent if xx = x).
Each semilattice X carries the natural partial order ≤ X defined by x ≤ y iff xy = x = yx. Many properties of semilattices are defined in the language of the natural partial order. In particular, for a point x ∈ X we can consider its upper and lower sets ↑x := {y ∈ X : xy = x} and ↓x := {y ∈ X : xy = y} in the partially ordered set (X, ≤ X ).
A subset C of a semilattice X is called a chain if xy ∈ {x, y} for any x, y ∈ C. A semilattice X is called chain-finite if each chain in X is finite. A semilattice is called linear if it is a chain in itself.
A semilattice endowed with a topology is called a topologized semilattice. A topologized semilattice X is called a (semi )topological semilattice if the semigroup operation X ×X → X, (x, y) → xy, is (separately) continuous.
In [17] Stepp proved that for any homomorphism h : X → Y from a chain-finite semilattice to a Hausdorff topological semilattice Y , the image h(X) is closed in Y . In [1], the first authors improved this result of Stepp proving the following theorem.
Theorem 1 (Banakh, Bardyla). For any homomorphism h : X → Y from a chain-finite semilattice to a Hausdorff semitopological semilattice Y , the image h(X) is closed in Y .
Topological generalizations of the notion of chain-finiteness are the notions of chain-compactness and completeness, discussed in [5].
A topologized semilattice X is called • chain-compact if each closed chain in X is compact; • complete if each non-empty chain C ⊂ X has inf C ∈ C and sup C ∈ C.
Here C stands for the closure of C in X. Chain-compact and complete topologized semilattices appeared to be very helpful in studying the closedness properties of topologized semilattices, see [1], [2], [3], [4], [5], [6], [12]. By Theorem 3.1 [1], a Hausdorff semitopological semilattice is chain-compact if and only if it is complete (see also Theorem 4.3 [5] for generalization of this characterization to topologized posets). In [1] the first two authors proved the following closedness property of complete topologized semilattices. In [4] the first two authors gave the following partial answer to Problem 1.
Theorem 3 (Banakh, Bardyla). For any continuous homomorphism h : X → Y from a complete topologized semilattice X to a sequential Hausdorff semitopological semilattice Y , the image h(X) is closed in Y .
Another partial result to Problem 1 was given in [6].
Theorem 4 (Banakh, Bardyla, Ravsky). For any continuous homomorphism h : X → Y from a complete topologized semilattice X to a functionally Hausdorff semitopological semilattice Y , the image h(X) is closed in Y .
Let us recall that a topological space X is functionally Hausdorff if for any distinct points x, y ∈ X there exists a continuous map f : X → R such that f (x) = f (y).
In fact, in [6] Theorem 4 was derived from the following closedness property of the partial order of a complete subsemilattice of a functionally Hausdorff semitopological semilattice.
Theorem 5 (Banakh, Bardyla, Ravsky). For any complete subsemilattice X of a functionally Hausdorff semitopological semilattice Y , the partial order ≤ X of X is a closed subset of Y × Y .
In this paper we shall show that the answer to Problem 1 is affirmative under the additional condition that the semitopological semilattice Y is ω-Lawson. We shall also prove a counterpart of Theorem 5 for complete subsemilattices of ω-Lawson semitopological semilattices.
We define a topologized semilattice X to be ω-Lawson if for any distinct points x, y ∈ X there exists a countable family U of closed neighborhoods of x such that U is a subsemilattice of X that does not contain y. A topologized semilattice X is ω-Lawson if and only if it is Hausdorff and has at most countable Lawson numberΛ(X).
The Lawson numberΛ(X) of a Hausdorff topologized semilattice X is defined as the smallest cardinal κ such that for any distinct points x, y ∈ X there exists a family U of closed neighborhoods of x such that |U| ≤ κ and U is a subsemilattice of X that does not contain y.
The Lawson number will be studied in more details in Section 1. In that section we shall prove that every Hausdorff topological semilattice X hasΛ(X) ≤ ω. On the other hand, for any infinite cardinal λ we shall construct a Hausdorff zero-dimensional semitopological semilattice X of cardinality |X| = λ and Lawson numberΛ(X) = cf(λ). In Section 3 we prove the main result of this paper: Theorem 6. For any complete subsemilattice X of a ω-Lawson semitopological semilattice Y , the natural partial order ≤ X of X is a closed subset of Y × Y . Consequently, X is closed in Y .
Since the completeness is preserved by continuous homomorphisms into Hausdorff semitopological semilattices (see Lemma 1), Theorem 6 implies the following corollary giving a partial answer to Problem 1.
Corollary 1. For any continuous homomorphism h : X → Y from a complete topologized semilattice to an ω-Lawson semitopological semilattice Y the image h(X) is closed in Y .
Problem 2. Let X be a complete subsemilattice of an ω 1 -Lawson semitopological semilattice Y . Is X closed in Y ? Is the natural partial order ≤ X of X closed in X × X?

The Lawson number of a Hausdorff topologized semilattice
It is easy to see that each ω-Lawson topologized semilattice is Hausdorff. In fact, ω-Lawson topologized semilattices can be equivalently defined as Hausdorff topologized semilattices with countable Lawson number. Definition 1. The Lawson numberΛ(X) of a Hausdorff topologized semilattice X is the smallest cardinal κ such that for any distinct points x, y ∈ X there exists a family U of closed neighborhoods of x such that |U| ≤ κ and the intersection U is a subsemilattice of X that does not contain y.
For any Hausdorff topologized semilattice X the Lawson numberΛ(X) is well-defined and does not exceed the closed pseudocharacterψ(X) of X, defined as the smallest cardinal κ such that for any point x ∈ X there exists a family U of closed neighborhoods of x such that |U| ≤ κ and U = {x}. Therefore,Λ(X) ≤ψ(X) for any Hausdorff topologized semilattice X.
Observe that a topologized semilattice X is ω-Lawson if and only if X is Hausdorff and has Λ(X) ≤ ω.

Definition 2.
A topologized semilattice X is defined to be κ-Lawson for a cardinal κ if X is Hausdorff andΛ(X) ≤ κ.
The Lawson number admits the following simple characterization. Proposition 1. The Lawson number of a Hausdorff topologized semilattice X is equal to the smallest cardinal κ such that for any distinct points x, y ∈ X there exist a closed subsemilattice L in X and a family V of closed neighborhoods of x such that |V| ≤ κ and V ⊂ L ⊂ X \ {y}.
Proof. We should prove thatΛ(X) = Λ(X) where Λ(X) is the smallest cardinal κ such that for any distinct points x, y ∈ X there exist a closed subsemilattice L in X and a family V of closed neighborhoods of x such that |V| ≤ κ and V ⊂ L ⊂ X \ {y}.
To see that Λ(X) ≤Λ(X), for any distinct points x, y ∈ X use the definition ofΛ(X) and find a family U of closed neighborhoods of x such that |U| ≤Λ(X) and L := U is a closed subsemilattice of X that does not contain y.
To see thatΛ(X) ≤ Λ(X), for any distinct points x, y ∈ X use the definition of Λ(X) and find a closed subsemilattice L of X and a family V of closed neighborhoods of x such that |V| ≤ Λ(X) and V ⊂ L ⊂ X \ {y}. Then U := {V ∪ L : V ∈ V} is a family of closed neighborhoods of x such that |U| ≤ |V| ≤ Λ(X) and U = ( V) ∪ L = L is a closed subsemilattice of X that does not contain y and witnesses thatΛ(X) ≤Λ(X).
The notion of a 1-Lawson semilattice extends the well-known notion of a Lawson semilattice (or else a topologized semilattice with small subsemilattices), introduced and studied by Lawson [14] (see also [10,Chapter 2]). Following [10, p.12], we define a topologized semilattice X to be • Lawson if it has a base of the topology consisting of open subsemilattices; • a V -semilattice if for any point x ∈ X and y / ∈ ↑x there exists a point v ∈ X \ ↓y such that ↑v is a neighborhood of x in X; • I-separated if for any distinct points x, y ∈ X there exists a continuous homomorphism f : Here by I we denote the unit interval [0, 1] endowed with the semilattice operation min.
Proof. To prove that X is 1-Lawson, fix any distinct points x, y ∈ X. We need to find a closed subsemilattice L in X that contains x in its interior but does not contain y.
Since X is a semitopological semilattice, the closure V of the semilattice V is a closed subsemilattice that contains x in its interior but does not contain y. Therefore, the semilattice X is 1-Lawson.
(3) ⇒ (1) Assume that X is a V -semilattice. If x / ∈ ↓y, then y / ∈ ↑x and there exists an element v ∈ X \ ↓y such that the upper set ↑v contains x in its interior. Since X is a Hausdorff semitopological semilattice the upper set ↑v = {z ∈ X : zv = v} is closed. Then the closed subsemilattice ↑v is a neighborhood of x that does not contain y. If x ∈ ↓y, then x / ∈ ↑y. Since X is a V -semilattice, there exists an element u ∈ X \ ↓x such that the upper set ↑u contains y in its interior. Observe that the complement U := X \ ↑u is an open subsemilattice of X, containing x. Then U is a closed subsemilattice of X, which is a neighborhood of x that does not contain y.
(4) ⇒ (1) Assuming that X is I-separated, we can find a continuous homomorphism f : X → I such that f (x) = f (y). Choose any closed neighborhood N ⊂ I of f (x) such that f (y) / ∈ N . Then f −1 (N ) is a closed subsemilattice in X that contains x in its interior but does not contain y.
Now assume that X is compact. In this case the conditions (1)-(4) are equivalent by Theorem 7.1 in [2]. In fact, the equivalence of the conditions (2) and (4) is a classical result of Lawson [14] and [15]. Example 1. The topological semilattice Z ω with the Tychonoff product topology and coordinatewise operation of minimum is Lawson and I-separated but not a V -semilattice.
By [10, Example 2.21], there exists a metrizable compact topological semilattice, which is not Lawson and hence is not 1-Lawson. However, such a semilattice necessarily is ω-Lawson as shown by the following simple proposition.
Proof. Given two distinct points x, y ∈ X, choose a decreasing sequence (U n ) n∈ω of open neighborhoods of x such that y / ∈ U 0 and U n · U n ⊂ U n−1 and hence U n · U n ⊂ U n−1 for all n ∈ N. The choice of U 0 is possible by the Hausdorff property of X, and the choice of the neighborhoods U n is possible by the continuity of the semilattice operation at (x, x). It follows that the intersection n∈ω U n is a closed subsemilattice of X containing x but not y. Corollary 2. Each compact Hausdorff semitopological semilattice is ω-Lawson.
Let us also notice the following trivial (but useful) fact. Let us recall that a topological space X is Urysohn if any distinct points in X have disjoint closed neighborhoods. It is clear that each Urysohn space is Hausdorff.
We define a topological space X to be κ-Urysohn for a cardinal κ if for any distinct points x, y ∈ X there are families U x and U y of open sets on X such that max{|U x |, |U y |} ≤ κ, x ∈ U x , y ∈ U y and the sets U∈Ux U and V ∈Uy V are disjoint.
It is easy to see that a topological space is Urysohn if and only if it is 1-Urysohn.
Example 2. For every infinite cardinal κ, there exists a Hausdorff Lawson topological semilattice, which is not κ-Urysohn.
Proof. Take any ordinal λ of cofinality cf(λ) > κ (for example, put λ := κ + ). Consider the set L = {x α } α≤λ ∪ {z} ∪ {y α } α≤λ of pairwise distinct points endowed with the linear order in which x α < x β < z < y β < y α for any ordinals α < β ≤ λ. LetL := L \ {x λ , y λ }. On the set consider the semilattice operation Endow X with the topology τ consisting of all sets U ⊂ X satisfying the following three conditions: Taking into account that cf(λ) > κ, we can show that (X, τ ) is a required Hausdorff Lawson topological semilattice which is not κ-Urysohn.  Proof. Consider the set X := {A ⊂ λ : A = λ or A is finite} endowed with the semilattice operation of union. This semilattice has cardinality |X| = λ. Here we identify the cardinal λ with the set [0, λ) of all ordinals smaller than λ. Now the trick is to introduce an appropriate topology on the semilattice X. For this we define several kinds of sets in λ.
A finite subset A ⊂ λ is defined to be sparse if |A ∩ [α, α + ω)| ≤ 1 for any ordinal α ∈ λ. For a set A and an ordinal α ∈ λ consider the set and observe that λ / ∈ S[A; α]. Let α ∈ λ be an ordinal, n be a finite ordinal and ε be a positive real number. A subset A ⊂ λ is called (α, n, ε)-fat if there exists a limit ordinal β ∈ [α, λ) and a finite ordinal m > n that (iii) the set [0, β) ∩ A is finite and has cardinality < ε · m. The conditions (i),(ii) ensure that the ordinal β is unique.
Now we define a topology τ on the semilattice X. This topology consists of the sets U ⊂ X satisfying the following two conditions: (a) for any finite subset A ∈ U of λ there exists an ordinal α ∈ λ such that A ∈ S[A, α] ⊂ U ; (b) if λ ∈ U , then there exist ordinals α ∈ λ, k ∈ ω, and a positive real number ε such that F [α, k, ε] ⊂ U . It is easy to see that τ is a well-defined topology on X. Now we show that this topology is Hausdorff and zero-dimensional.

Proof. Given any element
To see that this set is closed in (X, τ ), choose any set A ∈ X \ F [α, n, ε]. It follows that A is a finite subset of λ, which is not (α, n, ε)-fat. Let β is the smallest limit ordinal such that the intersection A ∩ [β + ω, λ) is sparse. Let m ∈ ω be the smallest finite ordinal such that Since A is not (α, n, ε)-fat, one of the following conditions holds: (1) β < α; (2) m ≤ n; (3) β ≥ α and m > n but [β, β + m] ⊂ A;   The case A = λ can be considered by analogy.
Claims 1-4 show that the topology τ is Hausdorff and zero-dimensional.
Proof. Given any element a ∈ X, we should prove that the shift s a : X → X, s a : x → ax, is continuous. If a = λ, then s a (X) = {λ} is a singleton, so the continuity of s a is trivial. So, we assume that a is a finite subset of λ. To check the continuity of the shift s a at a point x ∈ X, fix any neighborhood O ax ∈ τ of the point ax = x ∪ a. If x = λ, then ax = λ and by the definition of the topology τ , there exists an ordinal α ∈ λ such that ax ∈ S[ax; α] ⊂ O ax . Replacing α by a larger ordinal, we can assume that ax ⊂ [0, α).
If x = λ, then ax = λ and by the definition of the topology τ , there exist α ∈ λ, k ∈ ω and ε > 0 such that F [α, k, ε] ⊂ O ax . Replacing α by a larger ordinal, we can assume that a ⊂ [0, α). Replacing k by a larger number, we can assume that |a| ≤ 1 2 εk. In this case O Claim 6. Let U ⊂ τ be a family of open sets and L be a τ -closed subsemilattice in X such that |U| < cf(λ) and ∅ = U ⊂ L. Then λ ∈ L.
Proof. Fix any element x ∈ U. If x = λ, then λ = x ∈ U ⊂ L and we are done. So, we assume that x is a finite subset of λ. To show that λ ∈ L, take any neighborhood O λ ∈ τ of λ ∈ X. By Claim 2, there exist ordinals α ∈ λ, k ∈ ω and a positive real number ε such that F [α, k, ε] ⊂ O κ .

Complete topologized semilattices
In this section we recall some known properties and characterizations of complete topologized semilattices.
By a poset we understand a set endowed with a partial order. A topologized poset is a poset endowed with a topology. So, each topologized semilattice is a topologized poset.
A subset D of a poset (X, ≤) is called • a chain if any elements x, y ∈ D are comparable in the sense that x ≤ y or y ≤ x; • up-directed if for any x, y ∈ D there exists z ∈ D such that x ≤ z and y ≤ z; • down-directed if for any x, y ∈ D there exists z ∈ D such that z ≤ x and z ≤ y.
It is clear that each chain in a poset is both up-directed and down-directed. A topologized posed X is defined to be • up-complete if any nonempty up-directed subset U ⊂ X has the smallest upper bound sup U ∈ U in X; • down-complete if any nonempty down-directed subset D ⊂ X has the greatest lower bound inf D ∈ D in X.
Proposition 5. For a topologized poset X the following conditions are equivalent: (1) X is up-complete; (2) each non-empty chain C ⊂ X has the smallest upper bound sup C ∈ C in X.
Here for a subset A of a topological space X by A we denote the closure of A in X. Proposition 5 implies the following useful characterization of completeness in topologized semilattices.

Corollary 3. A topologized semilattice X is complete if and only if it is up-complete and downcomplete.
This corollary implies that each closed subsemilattice of a complete topologized semilattice has the smallest element.
A topologized semilattice Y is called ↑-closed if for every y ∈ Y the upper set ↑y = {x ∈ Y : xy = y} is closed in Y . It is easy to see that each T 1 semitopological semilattice is ↑-closed.
The following lemma (that can be derived from Corollary 3) is proved in [5,Lemma 5.3].
Lemma 1. Let h : X → Y be a continuous surjective homomorphism between topologized semilattices. If X is complete and Y is ↑-closed, then the topologized semilattice Y is complete.

Proof of Theorem 6 and Corollary 1
The proof of Theorem 6 is based on the following lemma.
Lemma 2. Let X be a complete subsemilattice of a semitopological semilattice Y . Let a pair (x, y) ∈ Y × Y belong to the closure of the natural partial order ≤ X of X in Y × Y , and let {U n } n∈ω , {V n } n∈ω be sequences of closed neighborhoods of the points x and y in Y , respectively. Then there exist points x ′ ∈ X ∩ n∈ω U n and y ′ ∈ X ∩ n∈ω V n such that x ′ ≤ y ′ .
Proof. Replacing each set U n by i≤n U i , we can assume that U n+1 ⊂ U n for all n ∈ ω. By the same reason, we can assume that the sequence (V n ) n∈ω is decreasing. For every n ∈ ω denote by U • n and V • n the interiors of the sets U n and V n in Y . By induction we shall construct sequences (x n ) n∈ω and (y n ) n∈ω of points of X such that for every n ∈ ω the following conditions are satisfied: (1 n ) x n ≤ y n ; To choose the initial points x 0 , y 0 , use the separate continuity of the semilattice operation and find neighborhoods U ′ 0 ⊂ U • 0 and V ′ 0 ⊂ V • 0 of x and y in Y such that U ′ 0 x ⊂ U • 0 and V ′ 0 y ⊂ V • 0 . By our assumption, there are points x 0 ∈ X ∩ U ′ 0 and y 0 ∈ X ∩ V ′ 0 such that x 0 ≤ y 0 . The choice of the neighborhoods U ′ 0 and V ′ 0 ensures that the conditions (2 0 ) and (3 0 ) are satisfied. Now assume that for some n ∈ N points x 0 , . . . , x n−1 and y 0 , . . . , y n−1 of X are chosen so that the conditions (1 n−1 )-(3 n−1 ) are satisfied. The condition (2 n−1 ) implies that for every i ≤ n we have the inclusion x i · · · x n−1 xx = x i · · · x n−1 x ∈ U • i (if i = n, then we understand that x i · · · x n−1 x = x). Using the continuity of the shift s x : Y → Y , s x : z → xz, we can find a neighborhood U ′ n ⊂ Y of x such that x i · · · x n−1 · (U ′ n ∪ U ′ n x) ⊂ U • i for every i ≤ n. By analogy, we can find a neighborhood V ′ n ⊂ Y of y such that y i · · · y n−1 · (V ′ n ∪ V ′ n y) ⊂ V • i for every i ≤ n. By our assumption, there are points x n ∈ X ∩ U ′ n and y n ∈ X ∩ V ′ n such that x n ≤ y n . The choice of the neighborhoods U ′ 0 and V ′ 0 ensures that the conditions (2 n ) and (3 n ) are satisfied. This completes the inductive step. Now for every i ∈ ω consider the chain C i = {x i · · · x n : n ≥ i} ⊂ U • i in X. By the completeness of X, this chain has inf x n for all i > n, we see that inf C i is a lower bound of the chain C i+1 and hence inf C i ≤ inf C i+1 . By the completeness of X, for every i ∈ ω the chain D i : By analogy, for every k ∈ ω consider the chain E i = {y i · · · y n : n ≥ i} ⊂ V • i in X. By the completeness of X, this chain has inf To finish the proof of Lemma 2, it suffices to show that sup D 0 ≤ sup F 0 . The inductive conditions (1 n ), n ∈ ω, imply that inf C i ≤ inf E i for all i ∈ ω and sup D 0 = sup{inf C i : i ∈ ω} ≤ sup{inf E i : i ∈ ω} = sup F 0 .
The following two lemmas imply Theorem 6. Lemma 3. Let Y be an ω-Lawson semitopological semilattice. For any complete subsemilattice X ⊂ Y the natural partial order ≤ X of X is closed in Y × Y .
Proof. By Corollary 3, the complete semitopological semilattice X is both up-complete and downcomplete. To show that the partial order ≤ X := {(x, y) ∈ X × X : x ≤ y} is closed in Y × Y , take any pair (y 1 , y 2 ) in the closure of the set ≤ X in Y × Y . For every i ∈ {1, 2}, let U i be the set of all countable families U of closed neighborhoods of y i in Y such that U is a subsemilattice of Y . By Lemma 2, for any U 1 ∈ U 1 and U 2 ∈ U 2 there are points x 1 ∈ X ∩ U 1 and x 2 ∈ X ∩ U 2 such that x 1 ≤ x 2 . In particular, the closed subsemilattice X ∩ U 1 is not empty and has the smallest element inf(X ∩ U 1 ) ∈ X (by the down-completeness of X). Denote this smallest element by x(U 1 ). It follows that x(U 1 ) := inf(X ∩ U 1 ) ≤ x 1 ≤ x 2 . Consequently, the closed subsemilattice (↑x(U 1 )) ∩ (X ∩ U 2 ) ∋ x 2 is not empty and has the smallest element (by down-completeness of X), which will be denoted by y(U 1 , U 2 ). Observe that x(U 1 ) ∈ X ∩ U 1 , y(U 1 , U 2 ) ∈ X ∩ U 2 and x(U 1 ) ≤ y(U 1 , U 2 ). For any families U 1 ∈ U 1 and U 2 , U ′ 2 ∈ U 2 with U 2 ⊆ U ′ 2 we have y(U 1 , U 2 ) ≤ y(U 1 , U ′ 2 ). Therefore, the set {y(U 1 , U 2 ) : U 2 ∈ U 2 } ⊂ X is up-directed and by the up-completeness of X, it has the smallest upper bound in X, which will be denoted by y(U 1 ). It follows that x(U 1 ) ≤ y(U 1 ). We claim that y(U 1 ) = y 2 . In the opposite case we can use the ω-Lawson property of Y and choose a countable family U ′ 2 ∈ U 2 such that y(U 1 ) / ∈ U ′ 2 . Taking into account that the set {y(U 1 , U 2 ∩ U ′ 2 ) : U 2 ∈ U 2 } is cofinal in {y(U 1 , U 2 ) : U 2 ∈ U 2 }, we conclude that y(U 1 ) = sup{y(U 1 , U 2 ) : U 2 ∈ U 2 } = sup{y(U 1 , U 2 ∩ U ′ 2 ) : U 2 ∈ U 2 } ∈ U ′ 2 , which contradicts the choice of the family U ′ 2 . This contradiction shows that y 2 = y(U 1 ) ∈ X. Now we see that x(U 1 ) ≤ y(U 1 ) = y 2 for every U 1 ∈ U 1 . By the up-completeness of the semitopological semilattice X, the up-directed subset {x(U 1 ) : U 1 ∈ U 1 } has the smallest upper bound x ∈ X. It follows from x(U 1 ) ≤ y 2 for all U 1 ∈ U 1 that x ≤ y 2 .
Proof. By Lemma 3, the partial order ≤ X := {(x, y) ∈ X × X : xy = x} is a closed subset of Y × Y . By Corollary 3, the complete semilattice X has the smallest element min X ∈ X. Consider the continuous map f : Y → Y × Y , f : y → (min X, y), and observe that X = f −1 (≤ X ) is a closed subset of X, being the preimage of the closed set ≤ X under the continuous map f . Finally, we prove Corollary 1.
Lemma 5. For every continuous homomorphism h : X → Y from a complete topologized semilattice X to an ω-Lawson semitopological semilattice Y , the image h(X) is closed in Y .
Proof. Observe that the ω-Lawson property of Y implies that the semitopological semilattice Y is Hausdorff and hence ↑-closed. By Lemma 1, the semitopological semilattice h(X) of Y is complete and by Lemma 4, h(X) is closed in Y .