A metrizable Lawson semitopological semilattice with non-closed partial order
aa r X i v : . [ m a t h . GN ] S e p A METRIZABLE LAWSON SEMITOPOLOGICAL SEMILATTICEWITH NON-CLOSED PARTIAL ORDER
TARAS BANAKH, SERHII BARDYLA AND ALEX RAVSKY
Abstract.
We construct a metrizable Lawson semitopological semilattice X whose partial order ≤ X = { ( x, y ) ∈ X × X : xy = x } is not closed in X × X . This resolves a problem posed earlierby the authors. In this paper we shall construct an example of a metrizable Lawson semitopological semilatticewith non-closed partial order, thus answering a problem posed by the authors in [1].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 semilattice X is called ↑ -finite if for every element x ∈ X its upper set ↑ x := { y ∈ X : x ≤ y } is finite.A ( semi ) topological semilattice is a semilattice X endowed with a topology such that the binaryoperation X × X → X , xy xy , is (separately) continuous. A semitopological semilattice is Lawson if it has a base of the topology consisting of open subsemilattices.The continuity of the semilattice operation in a Hausdorff topological semilattice implies thefollowing well-known fact, see [4, VI-1.14].
Proposition 1.
For any Hausdorff topological semilattice X the partial order ≤ X := { ( x, y ) ∈ X × X : xy = y } is a closed subset of X × X . This proposition does not generalize to semitopological semilattices as shown by the followingexample constructed by the authors in [1].
Example 1.
There exists a metrizable countable semitopological semilattice X whose partial orderis dense and non-closed in X × X . At the end of the paper [1] the authors asked whether there exsists a Lawson Hausdorff semi-topological semilattice X with non-closed partial order. In this paper we shall give an affirmativeanswer to this question. Moreover, for any cardinal κ we can construct a Lawson semitopologicalsemilattice X with non-closed partial order such that the topological space of X is a P κ -space.A topological space ( X, τ ) is called a P κ -space if for any family U ⊂ τ of cardinality |U| ≤ κ the intersection T U belongs to the topology τ . Each topological space is a P κ -space for any finitecardinal κ .A topological space X is called zero-dimensional if it has a base of the topology, consisting ofclopen sets. A subset of a topological space is clopen if it is both closed and open. It is easy tosee that every regular P ω -space is zero-dimensional. The weight of a topological space ( X, τ ) isthe smallest cardinality of a base of the topology τ .The following example (answering Problem 1 in [1]) is the main result of this paper. Mathematics Subject Classification.
Key words and phrases. semitopological semilattice, Lawson semilattice, partial order, convergent sequence, act,semigroup.The second author was supported by the Austrian Science Fund FWF (Grant I 3709-N35).
Example 2.
For any infinite cardinal λ there exists a Lawson semitopological semilattice X havingthe following properties: (1) the partial order { ( x, y ) ∈ X × X : xy = x } of X is not closed in X × X ; (2) the semilattice X is ↑ -finite; (3) the cardinality and the weight of the space X both are equal to λ ; (4) X is a Hausdorff zero-dimensional space; (5) X is a P κ -space for any cardinal κ < cf( λ ) ; (6) if λ = ω , then the countable space X is metrizable.Proof. We identify the cardinal λ with the set [0 , λ ) of ordinals, smaller than λ . For two ordinals α < β in λ by [ α, β ) we denote the oerder-interval consisting of ordinals γ such that α ≤ γ < β .Consider the semilattice { , , } , endowed with the operation of minimum. In its power { , , } λ consider the subsemilattice X consisting of the functions x : λ → { , , } having fi-nite support supp( x ) := x − ( { , } ).For a function x ∈ X let k x k denote the smallest ordinal α ∈ λ such that supp( x ) ∩ [ α, λ ) = ∅ .If supp( x ) = ∅ , then k x k = max(supp( x )) + 1 and hence the ordinal k x k − x )).The following property of the semilattice X easily follows from the definition of X . Claim 1.
The semilattice X has cardinality | X | = λ and is ↑ -finite. More precisely, for every x ∈ X its upper set ↑ x has cardinality |↑ x | ≤ | supp( x ) | . Now we shall define a topology τ on X turning the semilattice X into a Lawson semitopologicalsemilatice with non-closed partial order.Write the semilattice X as the union X ∪ X ∪ X of the subsemilattices X i := { x ∈ X : x (0) = i } for i ∈ { , , } . For every function x ∈ X and ordinal α ∈ [ k x k , λ ) we shall define a subsemilattice V α ( x ) of X as follows.If x ∈ X , then we put V α ( x ) := { x } .If x ∈ X , then we put V α ( x ) := (cid:8) y ∈ X : y ↾ [1 , α ) = x ↾ [1 , α ) and y ([ α, λ )) ⊂ { , } (cid:9) ∪ (cid:8) y ∈ X : y ↾ [1 , α ) = x ↾ [1 , α ) , y ([ α, λ )) ⊂ { , } and k y k > α (cid:9) . If x ∈ X , then we put V α ( x ) := (cid:8) y ∈ X : y ↾ [1 , α ) = x ↾ [1 , α ) (cid:9) ∪ (cid:8) y ∈ X : y ↾ [1 , α ) = x ↾ [1 , α ) , k y k > α and y ( k y k − (cid:9) . It is easy to see that x ∈ V β ( x ) ⊂ V α ( x ) for any x ∈ X and ordinals α, β ∈ λ with k x k ≤ α ≤ β .Endow X with the topology τ consisting of the sets U ⊂ X such that for any x ∈ U there existsan ordinal α ∈ [ k x k , λ ) such that V α ( x ) ⊂ U . Claim 2.
For every x ∈ X and α ∈ [ k x k , λ ) , the set V α ( x ) is open in the topological space ( X, τ ) .Proof. Given any element y ∈ V α ( x ), we should find an ordinal β ∈ [ k y k , λ ) such that V β ( y ) ⊂ V α ( x ). Choose any ordinal β ∈ [ α, λ ) such that β ≥ k y k . If y ∈ X , then V β ( y ) = { y } ⊂ V α ( x )and we are done. So, we assume that y / ∈ X . In this case x / ∈ X .If x ∈ X , then y ∈ V α ( x ) \ X ⊂ X and hence y ↾ [1 , α ) = x ↾ [1 , α ) and y ([ α, λ )) ⊂ { , } . Toprove that V β ( y ) ⊂ V α ( x ), take any z ∈ V β ( y ). The definition of V β ( y ) for y ∈ X ensures that z ∈ X ∪ X . If z ∈ X , then z ↾ [1 , β ) = y ↾ [1 , β ) and z ([ β, λ )) ⊂ { , } . Taking into account that α ≤ β , we conclude that z ↾ [1 , α ) = y ↾ [1 , α ) = x ↾ [1 , α ) and z ([ α, λ )) = z ([ α, β )) ∪ z ([ β, λ )) = y ([ α, β )) ∪ z ([ β, λ ) ⊂ y ([ α, λ )) ∪ z ([ β, λ )) ⊂ { , } , witnessing that z ∈ V α ( x ) ∩ X .If z ∈ X , then the definition of V β ( y ) ensures that z ↾ [1 , β ) = y ↾ [1 , β ), z ([ β, λ )) ⊂ { , } and k z k > β . Then z ↾ [1 , α ) = y ↾ [1 , α ) = x ↾ [1 , α ), z ([ α, λ )) = z ([ α, β )) ∪ z ([ β, λ )) = y ([ α, β )) ∪ z ([ β, λ )) ⊂ { , } and k z k > β ≥ α , witnessing that z ∈ V α ( x ) ∩ X . METRIZABLE LAWSON SEMITOPOLOGICAL SEMILATTICE WITH NON-CLOSED PARTIAL ORDER 3
Next, assume that x ∈ X . In this case y ∈ V α ( x ) \ X ⊂ X and y ↾ [1 , α ) = x ↾ [1 , α ). To provethat V β ( y ) ⊂ V α ( x ), take any z ∈ V β ( y ). The definition of V β ( y ) ensures that z ∈ X ∪ X . If z ∈ X , then z ↾ [1 , α ) = y ↾ [1 , α ) = x ↾ [1 , α ) and z ∈ V α ( x ) by the definition of V α ( x ). If z ∈ X , then z ↾ [1 , β ) = y ↾ [1 , β ), k z k > β and z ( k z k −
1) = 1. Then z ↾ [1 , α ) = y ↾ [1 , α ) = x ↾ [1 , α ), k z k > β ≥ α and z ( k z k −
1) = 1, witnessing that z ∈ V α ( x ). (cid:3) Claim 2 implies
Claim 3.
The family B := { V α ( x ) : x ∈ X, α ∈ [ k x k , λ ) } is a base of the topology τ . Claim 4.
The weight w ( X, τ ) of the topological space ( X, τ ) equals λ .Proof. Claim 3 implies that w ( X, τ ) ≤ |B| ≤ λ . Taking into account that X is a discrete subspaceof ( X, τ ), we conclude that w ( X, τ ) ≥ | X | = λ . (cid:3) The following claim implies that the topological space (
X, τ ) is zero-dimensional.
Claim 5.
For every x ∈ X and a non-zero ordinal α ∈ [ k x k , λ ) the set V α ( x ) is closed in thetopological space ( X, τ ) .Proof. Given any y ∈ X \ V α ( x ), we should find an ordinal β ∈ [ k y k , λ ) such that V β ( y ) ∩ V α ( x ) = ∅ .We claim that the ordinal β = max { α, k y k} has the desired property.Six cases are possible.1) If y ∈ X , then V α ( x ) ∩ V β ( y ) = V α ( x ) ∩ { y } = ∅ and we are done.2) If y ∈ X ∪ X and x ∈ X , then V α ( x ) ∩ V β ( y ) = { x } ∩ V β ( y ) = ∅ as k x k ≤ α ≤ β .3) x, y ∈ X . In this case y ∈ X \ V α ( x ) implies y ↾ [1 , α ) = x ↾ [1 , α ) and hence V α ( x ) ∩ V β ( y ) ⊂ V α ( x ) ∩ V α ( y ) = ∅ .4) x, y ∈ X . In this case y ∈ X \ V α ( x ) implies y ↾ [1 , α ) = x ↾ [1 , α ) or y ([ α, λ ))
6⊂ { , } .If y ↾ [1 , α ) = x ↾ [1 , α ), then V α ( x ) ∩ V β ( y ) ⊂ V α ( x ) ∩ V α ( y ) = ∅ . If y ↾ [1 , α ) = x ↾ [1 , α ), then y ([ α, λ ))
6⊂ { , } and the choice of the ordinal β ≥ k y k guarantees that y ([ α, β ))
6⊂ { , } . Thenfor any z ∈ V β ( y ) we get z ([ α, β )) = y ([ α, β ))
6⊂ { , } and hence z / ∈ V α ( x ), which implies V α ( x ) ∩ V β ( y ) = ∅ .5) x ∈ X and y ∈ X . In this case for any z ∈ V α ( x ) ∩ V β ( y ), we have z ∈ X . The inclusion z ∈ X ∩ V β ( y ) implies k z k > β and z ([ β, λ )) ⊂ { , } . On the other hand, z ∈ X ∩ V α ( x ) implies1 = z ( k z k− ∈ z ([ β, λ )) ⊂ { , } , which is a desired contradiction implying that V α ( x ) ∩ V β ( y ) = ∅ .6) x ∈ X and y ∈ X . In this case for any z ∈ V α ( x ) ∩ V β ( y ), we have z ∈ X . The inclusion z ∈ X ∩ V α ( x ) implies z ([ α, λ )) ⊂ { , } and the inclusion z ∈ X ∩ V β ( y ) implies k z k > β and1 = z ( k z k − ∈ z ([ β, λ )) ⊂ z ([ α, λ )) ⊂ { , } , which is a desired contradiction implying that V α ( x ) ∩ V β ( y ) = ∅ . (cid:3) Claim 6.
The topological space ( X, τ ) is Hausdorff.Proof. Given two distinct points x, y ∈ X , put α := max {k x k , k y k} and consider four possiblecases.1) If x ∈ X , then by Claims 2, 5, O x := { x } and O y := X \{ x } are disjoint clopen neighborhoodsof the points x, y , respectively.2) If y ∈ X , then O x := X \ { y } and O y := { y } are disjoint clopen neighborhoods of the points x, y , respectively.3) If x, y ∈ X or x, y ∈ X , then x = y impies x ↾ [1 , α ) = y ↾ [1 , α ). Consequently, V α ( x ) and V α ( y ) are disjoint clopen neighborhoods of the points x, y , respectively.4) If the doubleton { x, y } intersects both sets X and X , then O x := V α ( x ) and O y := X \ V α ( x )are disjoint clopen neighborhoods of the points x, y , respectively. (cid:3) Claim 7.
For any cardinal κ < cf( λ ) , the space ( X, τ ) is a P κ -space. TARAS BANAKH, SERHII BARDYLA AND ALEX RAVSKY
Proof.
Given any family
U ⊂ τ of cardinality |U| ≤ κ < cf( λ ), we should prove that the intersection T U belongs to the topology τ . Fix any point x ∈ T U . By the definition of the topology τ , for everyset U ∈ U ⊂ τ there exists an ordinal α U ∈ [ k x k , λ ) such that V α U ( x ) ⊂ U . Since |U| ≤ κ < cf( λ ),the ordinal α = sup { α U : U ∈ U} is strictly smaller than λ . Since V α ( x ) ⊂ T U , the set T U belongs to the topology τ by the definition of τ . (cid:3) Claim 8. If λ = ω , then the countable space ( X, τ ) is metrizable.Proof. Being Hausdorff and zero-dimensional, the space (
X, τ ) is regular. If λ = ω , then byClaim 4, the space ( X, τ ) is second-countable. By the Urysohn Metrization Theorem 4.2.9 [3], thespace (
X, τ ) is metrizable. (cid:3)
Claim 9. ( X, τ ) is a Lawson semitopological semilattice.Proof. Given any element a ∈ X , we first prove the continuity of the shift s a : X → X , s a : x ax ,at any point x ∈ X . Given any neighborhood O ax ∈ τ of ax , we need to find a neighborhood O x ∈ τ of x such that aO x ⊂ O ax . Using Claim 2, find an ordinal α ∈ λ such that α ≥ k ax k =max {k a k , k x k} and V α ( ax ) ⊂ O ax . It remains to prove that aV α ( x ) ⊂ V α ( ax ) ⊂ O ax .Four cases are possible.1) x ∈ X . In this case aV α ( x ) = a { x } = { ax } ⊂ V α ( ax ).2) a ∈ X or x ∈ X . In this case ax ∈ X and for any z ∈ V α ( x ) we have z ↾ [1 , α ) = x ↾ [1 , α ),which implies az ↾ [1 , α ) = ax ↾ [1 , α ) and finally az ∈ V α ( ax ) ∩ X . Therefore, aV β ( x ) ⊂ V α ( ax ).3) a ∈ X and x ∈ X . In this case ax ∈ X and for any z ∈ V α ( x ) we have z ↾ [1 , α ) = x ↾ [1 , α )and z ([ α, λ )) ⊂ { , } . Taking into account that k a k ≤ α , we conclude that az ↾ [1 , α ) = ax ↾ [1 , α )and ( az )([ α, λ )) = z ([ α, λ )) ⊂ { , } , which implies az ∈ V α ( ax ) ∩ X and aV α ( x ) ⊂ V α ( ax ) ⊂ O ax .4) a ∈ X and x ∈ X . In this case ax ∈ X and V α ( x ) ⊂ X ∪ X . Observe that for any z ∈ V α ( x ) we have z ↾ [1 , α ) = x ↾ [1 , α ) and z ([ α, λ )) ⊂ { , } . Taking into account that k a k ≤ α , weconclude that az ↾ [1 , α ) = ax ↾ [1 , α ) and ( az )([ α, λ )) = z ([ α, λ )) ⊂ { , } . If z ∈ X , then az ∈ X and az ∈ V α ( ax ) ∩ X . If z ∈ X , then z ∈ V α ( x ) implies k z k > α and then k az k ≥ k z k > α and az ∈ V α ( az ) ∩ X . In both cases we get az ∈ V α ( ax ) and hence aV α ( x ) ⊂ V α ( ax ) ⊂ O ax .Therefore, ( X, τ ) is semitopological semilattice. Since the base B = { V α ( x ) : x ∈ X, α ∈ [ k x k , λ ) } of the topology τ consists of open subsemilattices of X , the semitopological semilattice( X, τ ) is Lawson. (cid:3)
Claim 10.
The partial order ≤ X := { ( x, y ) ∈ X × X : xy = x } of X is not closed in X × X .Proof. For every α ∈ ω consider the elements α : λ → { , } and α : λ → { , } of X , uniquelydetermined by the conditions − α (0) = { α } = − α (1). It is clear that α ≤ α and hence( , ) / ∈≤ X . On the other hand, for any ordinal α ∈ [1 , λ ) we have { ( β , β ) : β ∈ [ α, λ ) } ⊂ (cid:0) V α ( ) × V α ( ) (cid:1) ∩ (cid:0) ≤ X (cid:1) , which means that the pair ( , ) / ∈≤ X belongs to the closure of ≤ X . Consequently, the partialorder ≤ X of X is not closed in X × X . (cid:3) Claims 1–10 imply that (
X, τ ) is a Lawson semitopological semilattice possessing the properties(1)–(6) of Example 2. (cid:3)
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A metrizable Lawson semitopological semilattice with non-closed partialorder , preprint (https://arxiv.org/abs/1902.08760).[2] 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.[3] R. Engelking,
General topology , Heldermann Verlag, Berlin, 1989.[4] G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M. Mislove, D.S. Scott,
Continuous lattices and domains ,Cambridge University Press, Cambridge, 2003.
METRIZABLE LAWSON SEMITOPOLOGICAL SEMILATTICE WITH NON-CLOSED PARTIAL ORDER 5
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
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