# Topological Aspects of the Product of Lattices

Authors: Carmen Vlad
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

LetX be an arbitrary nonempty set and L a lattice of subsets of X such that ∅, X∈L. A(L) denotes the algebra generated by L, and M(L) denotes those nonnegative, finite, finitely additive measures on A(L). In addition, I(L) denotes the subset of M(L) which consists of the nontrivial zero-one valued measures. The paper gives detailed analysis of products of lattices, their associated Wallman spaces, and products of a variety of measures.


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## Body

## 1. Introduction

It is well known that given two measurable spaces and measures on them, we can obtain the product measurable space and the product measure on that space. The purpose of this paper is to give detailed analysis of product lattices and their associated Wallman spaces and to investigate how certain lattice properties carry over to the product lattices. In addition, we proceed from a measure theoretic point of view. We note that some of the material presented here has been developed from a filter approach by Kost, but the measure approach lends to a generalization of  measures and to an easier treatment of topological style lattice properties.


## 2. Background and Notations

In this section we introduce the notation and terminology that will be used throughout the paper. All is fairly standard, and we include it for the reader’s convenience.LetX be an arbitrary nonempty set and L a lattice of subsets of X such that ∅,  X∈L. A lattice L is a partially ordered set any two elements (x,y) of which have both sup(x,y)  and inf(x,y).A(L) denotes the algebra generated by L; σ(L) is the σ algebra generated by L; δ(L) is the lattice of all countable intersections of sets from L; τ(L) is the lattice of arbitrary intersections of sets from L; ρ(L) is the smallest class closed under countable intersections and unions which contains L.


### 2.1. Lattice Terminology

The latticeL is called:δ-lattice if L is closed under countable intersections; complement generated if L∈L implies L=∩Ln′, n=1,…,∞, Ln∈L (where prime denotes the complement); disjunctive if for x∈X and  L1∈L such that x∉L1 there exists L2∈L with x∈L2 and  L1∩L2=∅; separating (or T1)  if  x,y∈X and x≠y  implies there exists L∈L such that x∈L, y∉L; T2 if for x,y∈X and  x≠y there exist L1,L2∈L such that x∈L1′, y∈L2′, and L1′∩L2′=∅; normal if for any L1,L2∈L with  L1∩L2=∅ there exist L3,L4∈L with L1⊂L3′, L2⊂L4′, and L3′∩L4′=∅; compactif for any collection {Lα} of sets of  L with ∩αLα=∅, there exists a finite subcollection with empty intersection; countably compact if for any countable collection {Lα} of sets of L with ∩αLα=∅, there exists a finite subcollection with empty intersection.


### 2.2. Measure Terminology

M(L) denotes those nonnegative, finite, finitely additive measures on A(L).A measureμ∈M(L) is called: σ-smooth on L if for all sequences {Ln} of sets of L with Ln↓∅, μ(Ln)→0;σ-smooth on A(L) if for all sequences {An} of sets of A(L) with An↓∅, μ(An)→0, that is, countably additive.L-regularif for any A∈A(L), (2.1)μ(A)=sup⁡⁡{μ(L)∣L⊂A,L∈L}.We denote byMR(L) the set of L-regular measures of M(L); Mσ(L) the set of σ-smooth measures on L, of M(L); Mσ(L) the set of σ-smooth measures on A(L) of M(L);  MRσ(L) the set of  L-regular measures of Mσ(L).In addition,I(L), IR(L), Iσ(L), Iσ(L), IRσ(L) are the subsets of the corresponding M’s  which consist of  the nontrivial zero-one valued measures.Finally, letX,Y be abstract sets and L1 a lattice of subsets of X and L2 a lattice of subsets of Y.  Let μ1∈M(L1) and  μ2∈M(L2).Theproduct measureμ1×μ2∈M(L1×L2) is defined by
(2.2)(μ1×μ2)(L1×L2)=μ1(L1)μ2(L2)∀L1∈A(L1),L2∈A(L2).


### 2.3. Lattice-Measure Correspondence

Thesupportof μ∈M(L) is S(μ)=∩{L∈L/μ(L)=μ(X)}.In caseμ∈I(L) then the support is S(μ)=∩{L∈L/μ(L)=1}.With this notation and in light of the above correspondences, we now note:For anyμ∈I(L), there exists ν∈IR(L) such that μ≤ν on L (i.e., μ(L)≤ν(L) for all L∈L). For any μ∈I(L), there exists ν∈IR(L′) such that μ≤ν on L′.L is compact if and only if  S(μ)≠∅ for every μ∈IR(L). L is countably compact if and only if IR(L)=IRσ(L). L is normal if and only if for each  μ∈I(L), there exists a unique ν∈IR(L) such that μ≤ν on L. L is regular if and only if whenever μ1,μ2∈I(L) and μ1≤μ2 on L, then S(μ1)=S(μ2). L is replete if and only if for any μ∈IRσ(L), S(μ)≠∅. L is prime-complete if and only if for any μ∈Iσ(L), S(μ)≠∅.Finally, ifμx is the measure concentrated at x∈X, then μx∈IR(L), for all x∈X if and only if L is disjunctive.For further results and related matters see [1–3].


### 2.4. The General Wallman Space and  Wallman Topology

TheWallman topology in IRσ(L) is obtained by taking all(2.3)Wσ(L)={μ∈IRσ(L)/μ(L)=1},L∈L
as a base for the closed sets in IRσ(L) and then IRσ(L) is called the general Wallman space associated with X and L. Assuming L is disjunctive, Wσ(L)={Wσ(L)/L∈L} is a lattice in IRσ(L), isomorphic to L under the map  L→Wσ(L), L∈L. Wσ(L) is replete and a base for the closed sets tWσ(L), all arbitrary intersections of sets of Wσ(L).IfA∈A(L), then Wσ(A)={μ∈IRσ(L)/μ(A)=1} and the following statements are true:(2.4)Wσ(A∪B)=Wσ(A)∪Wσ(B),Wσ(A∩B)=Wσ(A)∩Wσ(B),Wσ(A′)=Wσ(A)′,A⊃BiffWσ(A)⊃Wσ(B),A(Wσ(L))=Wσ(A(L)).The Induced Measure
Letμ∈IRσ(L) and consider the induced measureμ̲∈IRσ(Wσ(L)), defined by
(2.5)μ̲(Wσ(A))=μ(A),A∈A(L).
The map μ→μ̲  is a bijection between IRσ(L) and IRσ(Wσ(L)).


## 2.1. Lattice Terminology

The latticeL is called:δ-lattice if L is closed under countable intersections; complement generated if L∈L implies L=∩Ln′, n=1,…,∞, Ln∈L (where prime denotes the complement); disjunctive if for x∈X and  L1∈L such that x∉L1 there exists L2∈L with x∈L2 and  L1∩L2=∅; separating (or T1)  if  x,y∈X and x≠y  implies there exists L∈L such that x∈L, y∉L; T2 if for x,y∈X and  x≠y there exist L1,L2∈L such that x∈L1′, y∈L2′, and L1′∩L2′=∅; normal if for any L1,L2∈L with  L1∩L2=∅ there exist L3,L4∈L with L1⊂L3′, L2⊂L4′, and L3′∩L4′=∅; compactif for any collection {Lα} of sets of  L with ∩αLα=∅, there exists a finite subcollection with empty intersection; countably compact if for any countable collection {Lα} of sets of L with ∩αLα=∅, there exists a finite subcollection with empty intersection.


## 2.2. Measure Terminology

M(L) denotes those nonnegative, finite, finitely additive measures on A(L).A measureμ∈M(L) is called: σ-smooth on L if for all sequences {Ln} of sets of L with Ln↓∅, μ(Ln)→0;σ-smooth on A(L) if for all sequences {An} of sets of A(L) with An↓∅, μ(An)→0, that is, countably additive.L-regularif for any A∈A(L), (2.1)μ(A)=sup⁡⁡{μ(L)∣L⊂A,L∈L}.We denote byMR(L) the set of L-regular measures of M(L); Mσ(L) the set of σ-smooth measures on L, of M(L); Mσ(L) the set of σ-smooth measures on A(L) of M(L);  MRσ(L) the set of  L-regular measures of Mσ(L).In addition,I(L), IR(L), Iσ(L), Iσ(L), IRσ(L) are the subsets of the corresponding M’s  which consist of  the nontrivial zero-one valued measures.Finally, letX,Y be abstract sets and L1 a lattice of subsets of X and L2 a lattice of subsets of Y.  Let μ1∈M(L1) and  μ2∈M(L2).Theproduct measureμ1×μ2∈M(L1×L2) is defined by
(2.2)(μ1×μ2)(L1×L2)=μ1(L1)μ2(L2)∀L1∈A(L1),L2∈A(L2).


## 2.3. Lattice-Measure Correspondence

Thesupportof μ∈M(L) is S(μ)=∩{L∈L/μ(L)=μ(X)}.In caseμ∈I(L) then the support is S(μ)=∩{L∈L/μ(L)=1}.With this notation and in light of the above correspondences, we now note:For anyμ∈I(L), there exists ν∈IR(L) such that μ≤ν on L (i.e., μ(L)≤ν(L) for all L∈L). For any μ∈I(L), there exists ν∈IR(L′) such that μ≤ν on L′.L is compact if and only if  S(μ)≠∅ for every μ∈IR(L). L is countably compact if and only if IR(L)=IRσ(L). L is normal if and only if for each  μ∈I(L), there exists a unique ν∈IR(L) such that μ≤ν on L. L is regular if and only if whenever μ1,μ2∈I(L) and μ1≤μ2 on L, then S(μ1)=S(μ2). L is replete if and only if for any μ∈IRσ(L), S(μ)≠∅. L is prime-complete if and only if for any μ∈Iσ(L), S(μ)≠∅.Finally, ifμx is the measure concentrated at x∈X, then μx∈IR(L), for all x∈X if and only if L is disjunctive.For further results and related matters see [1–3].


## 2.4. The General Wallman Space and  Wallman Topology

TheWallman topology in IRσ(L) is obtained by taking all(2.3)Wσ(L)={μ∈IRσ(L)/μ(L)=1},L∈L
as a base for the closed sets in IRσ(L) and then IRσ(L) is called the general Wallman space associated with X and L. Assuming L is disjunctive, Wσ(L)={Wσ(L)/L∈L} is a lattice in IRσ(L), isomorphic to L under the map  L→Wσ(L), L∈L. Wσ(L) is replete and a base for the closed sets tWσ(L), all arbitrary intersections of sets of Wσ(L).IfA∈A(L), then Wσ(A)={μ∈IRσ(L)/μ(A)=1} and the following statements are true:(2.4)Wσ(A∪B)=Wσ(A)∪Wσ(B),Wσ(A∩B)=Wσ(A)∩Wσ(B),Wσ(A′)=Wσ(A)′,A⊃BiffWσ(A)⊃Wσ(B),A(Wσ(L))=Wσ(A(L)).The Induced Measure
Letμ∈IRσ(L) and consider the induced measureμ̲∈IRσ(Wσ(L)), defined by
(2.5)μ̲(Wσ(A))=μ(A),A∈A(L).
The map μ→μ̲  is a bijection between IRσ(L) and IRσ(Wσ(L)).


## 3. The Case of Finite Product of Lattices

### 3.1. Notations

LetX, Y  be abstract  sets and  L1 a lattice of subsets of X and L2 a lattice of subsets of Y.We denote: (1)
L*=L1×L2={L1×L2/L1∈L1,L2∈L2},(2)
L=L(L*), the lattice generated by L*.We have the following:(3)
A(L1)×A(L2)=A(L1×L2),(4)
A(L*)=A(L),(5)
SL(μ)=SL*(μ),(6)
Iσ(L*)=Iσ(L),(7)
IR(L*)=IR(L).


### 3.2. Results

Theorem 3.1 (the finite product of lattices/regular measures).
LetX, Y  be abstract  sets and let  L1,L2 be lattices of subsets of X and Y, respectively. Then IR(L1)×IR(L2)=IR(L).Proof.
ForA∈A(L1)×A(L2)=A(L1×L2), we have A=⋃i=1nA1i×A2i, disjoint union and A1i∈A(L1), A2i∈A(L2).
Letμ∈IR(L1) and  ν∈IR(L2)  and consider  μ×ν defined on A(L1)×A(L2).
Ifμ×ν(A)=1, then μ×ν(A1i×A2i)=1 for some i.
Thenμ(A1i)ν(A2i)=1, and since μ  and  ν  are zero-one valued measures, μ(A1i)=1 and ν(A2i)=1.  By the regularity of μ and ν there exist  L1⊂A1i, L1∈L1 with μ(L1)=1 and  L2⊂A2i, L2∈L2 with ν(L2)=1.
Thereforeμ×ν(L1×L2)=μ(L1)ν(L2)=1 and L1×L2∈L*.
If we letM=L1×L2⊂A1i×A2i⊂A, then
(3.1)μ×ν(A)=sup⁡⁡{μ×ν(M)/M⊂A,M∈L*}⟹μ×ν∈IR(L*).Conversely, letμ∈IR(L*)=IR(L) and define μ1 on A(L1) by μ1(A)=μ(A×Y), A∈A(L1). Since μ is a zero-one measure on A(L1×L2), it follows that μ1 is a zero-one measure on A(L1), that is, μ1∈IR(L1).Supposeμ1(A)=μ(A×Y)=1; there exists A×Y⊃L1×L2∈L* such that μ(L1×Y)=1 and  μ(L1×L2)=1. Then μ1(L1)=μ(L1×Y)=1 and L1⊂A which shows that μ1∈IR(L1).  Similarly take μ2 on A(L2) defined by μ2(B)=μ(X×B), B∈A(L2).Then, as beforeμ2 is regular on L2.Finally for anyA∈A(L1) and any B∈A(L2) we have μ1×μ2(A×B)=μ1(A)μ2(B)=μ(A×Y)μ(X×B)=μ[(A×Y)∩(X×B)]=μ[(A∩X)×(Y∩B)]=μ(A×B) which shows that μ=μ1×μ2, and thereforeIR(L1)×IR(L2)=IR(L).Theorem 3.2 (the product of lattices/σ-smooth regular measures).
LetX,Y be abstract  sets and let  L1,L2 be lattices of subsets of X and Y, respectively. Then IRσ(L1)×IRσ(L2)=IRσ(L1×L2).Proof.
Letμ∈IRσ(L1) and ν∈IRσ(L2). Hence for A1n∈A(L1) with A1n↓∅ we have μ(A1n)→0 and for A2n∈A(L2) with A2n↓∅ we have ν(A2n)→0, n=1,2,….
Consider the sequence{Bn} of sets from A(L1)×A(L2). As in Theorem 3.1  Bn=⋃i=1kA1ni×A2ni, disjoint union and A1ni∈A(L1), A2ni∈A(L2).
Suppose thatBn↓∅, that is, A1ni×A2ni↓∅  for all i. Therefore A1n↓∅ or A2n↓∅ or both:
(3.2)μ×ν(Bn)=μ×ν[⋃i=1k(A1ni×A2ni)]=∑i=1kμ×ν(A1ni×A2ni)=∑i=1kμ(A1ni)ν(A2ni)⟶0,thereforeμ×ν∈IRσ(L1×L2).Conversely, letμ∈IRσ(L1×L2) and define μ1 on A(L1) by(3.3)μ1(A)=μ(A×Y),A∈A(L1).If{An} is a sequence of sets with An∈A(L1) and An↓∅, then An×Y↓∅, and since μ∈Iσ(L1×L2) it follows that μ(An×Y)→0.Thereforeμ1∈Iσ(L1).Similarly, definingμ2 on A(L2) by μ2(B)=μ(X×B),  B∈A(L2) we get  μ2∈Iσ(L2).Henceμ=μ1×μ2∈Iσ(L1)×Iσ(L2)=Iσ(L1×L2).Theorem 3.3 (product of supports of measures).
LetX,Y be abstract sets and let L1,L2 be lattices of subsets of  X and Y,respectively. The following statements are true:(a)
ifμ=μ1×μ2∈I(L1)×I(L2)=I(L1×L2) then S(μ)=S(μ1)×S(μ2);(b)
ifL1 and L2 are compact lattices then L is compact.Proof.
We have(a)
S(μ)=S(μ1×μ2)=∩{L1×L2∈L1×L2/μ(L1×L2)=μ(X×Y)},
(3.4)S(μ1)×S(μ2)=∩{L1×L2/L1∈L1,L2∈L2,μ1(L1)=μ1(X),μ2(L2)=μ2(Y)},Butμ(L1×L2)=μ1×μ2(L1×L2)=μ1(L1)μ2(L2) and μ(X×Y)=μ1×μ2(X×Y)=μ1(X)μ2(Y),(b)
S(μ)=S(μ1)×S(μ2)≠∅, sinceS(μi)≠∅,Li being compact.Theorem 3.4 (product of Wallman spaces/Wallman topologies).
Consider the spacesIR(Li)  with the Wallman topologies  tWi(Li), i=1,2.
It is known that the topological spaces(IR(Li),tWi(Li))  are compact and T1. Then the topological space (IR(L1)×IR(L2),tW1(L1)×tW2(L2)) is also compact and T1.Proof.
SinceIR(Li) are compact topological spaces,
(3.5)SL1(μ̲)=∩{W1(L1)∈W1(L1)/μ̲(W1(L1))=1}≠∅,SL2(ν̲)=∩{W2(L2)∈W2(L2)/ν̲(W2(L2))=1}≠∅.
We have(3.6)μ̲(W1(A))=μ(A),μ∈IR(L1),A∈A(L1),μ̲∈IR(W1(L1)),ν̲(W2(B))=ν(B),ν∈IR(L2),B∈A(L2),ν̲∈IR(W2(L2)).
Therefore(3.7)μ×ν̲(W1(A)×W2(B))=μ×ν(A×B)=μ(A)ν(B),μ̲×ν̲(W1(A)×W2(B))=μ̲(W1(A))ν̲(W2(B))=μ(A)ν(B),
so that μ×ν̲=μ̲×ν̲∈IR(W1(L1))×IR(W2(L2)), and then SL1×L2(μ×ν̲)=SL1×L2(μ̲×ν̲)=SL1(μ̲)×SL2(ν̲)≠∅⇒IR(L1)×IR(L2) is compact.To show thatIR(L1)×IR(L2) is a T1-space, let μ,ν∈IR(L) and suppose μ≠ν. Since μ=μ1×μ2 with μ1,μ2∈IR(L1) and ν=ν1×ν2 with ν1,ν2∈IR(L2) we get μ1≠ν1 and μ2≠ν2. There exist L1,L̃1∈L1 and L2,L̃2∈L2 with (3.8)μ1∈W1(L1),ν1∈W1(L1)′;ν1∈W1(L̃1),μ1∈W1(L̃1)′,μ2∈W2(L2),ν2∈W2(L2)′;ν2∈W2(L̃2),μ2∈W2(L̃2)′.
Therefore μ1(L1)=μ2(L2)=1,  ν1(L1)=ν2(L2)=0, μ1(L̃1)=μ2(L̃2)=0, ν1(L̃1)=ν2(L̃2)=1 which implies μ∈W(L1×L2), ν∈W(L1×L2)′; ν∈W(L̃1×L̃2), μ∈W(L̃1×L̃2)′.Theorem 3.5 (product of normal lattices).
LetX,Y  be abstract  sets and let  L1,L2  be normal lattices of subsets of X and Y, respectively. Then  L is a normal lattice of subsets of X×Y.Proof.
Letμ∈I(L) and ν,ρ∈IR(L) such that μ≤ν,ρ on L.
Then, sinceμ=μ1×μ2∈I(L1)×I(L2), ν=ν1×ν2∈IR(L1)×IR(L2) and ρ=ρ1×ρ2∈IR(L1)×IR(L2), we obtain μi≤νi, ρi on Li, i=1,2.
Li normal lattices ⇒νi=ρi; therefore ν1×ν2=ρ1×ρ2, that is, ν=ρ.


### 3.3. Examples

(1)
LetX, Y be topological spaces and let O1, O2 be the lattices of open sets of X and Y, respectively. Consider the product space X×Y with a base of open sets given by
(3.9){O1×O2/O1∈O1,O2∈O2}.We have(3.10)(O1×O2)′={(x,y)∈X×Y/(x,y)∉(O1×O2)}={(x,y)/(x,y)∈(X×O2′)or(x,y)∈(O1′×Y)}=(X×O2′)∪(O1′×Y)=(X×F2)∪(F1×Y).HenceF=t(L(F1×F2)) where F1,F2 are the lattices of closed sets of X and Y,respectively.(2)
LetX, Y be topological T3.5-spaces and let Z1, Z2 be the lattices of zero sets of continuous functions of X and Y, respectively. Then for the product space X×Y we consider a base of open sets given by
(3.11){Z1′×Z2′/Z1∈Z1,Z2∈Z2}
such that any open set from X×Y is of the form O=⋃αZ1α′×Z2α′ and any closed set is
(3.12)F=O′=⋂α(Z1ε×Y)∪(X×Z2α)∈t(L(Z1×Z2))
and then F=t(L(Z1×Z2)).


## 3.1. Notations

LetX, Y  be abstract  sets and  L1 a lattice of subsets of X and L2 a lattice of subsets of Y.We denote: (1)
L*=L1×L2={L1×L2/L1∈L1,L2∈L2},(2)
L=L(L*), the lattice generated by L*.We have the following:(3)
A(L1)×A(L2)=A(L1×L2),(4)
A(L*)=A(L),(5)
SL(μ)=SL*(μ),(6)
Iσ(L*)=Iσ(L),(7)
IR(L*)=IR(L).


## 3.2. Results

Theorem 3.1 (the finite product of lattices/regular measures).
LetX, Y  be abstract  sets and let  L1,L2 be lattices of subsets of X and Y, respectively. Then IR(L1)×IR(L2)=IR(L).Proof.
ForA∈A(L1)×A(L2)=A(L1×L2), we have A=⋃i=1nA1i×A2i, disjoint union and A1i∈A(L1), A2i∈A(L2).
Letμ∈IR(L1) and  ν∈IR(L2)  and consider  μ×ν defined on A(L1)×A(L2).
Ifμ×ν(A)=1, then μ×ν(A1i×A2i)=1 for some i.
Thenμ(A1i)ν(A2i)=1, and since μ  and  ν  are zero-one valued measures, μ(A1i)=1 and ν(A2i)=1.  By the regularity of μ and ν there exist  L1⊂A1i, L1∈L1 with μ(L1)=1 and  L2⊂A2i, L2∈L2 with ν(L2)=1.
Thereforeμ×ν(L1×L2)=μ(L1)ν(L2)=1 and L1×L2∈L*.
If we letM=L1×L2⊂A1i×A2i⊂A, then
(3.1)μ×ν(A)=sup⁡⁡{μ×ν(M)/M⊂A,M∈L*}⟹μ×ν∈IR(L*).Conversely, letμ∈IR(L*)=IR(L) and define μ1 on A(L1) by μ1(A)=μ(A×Y), A∈A(L1). Since μ is a zero-one measure on A(L1×L2), it follows that μ1 is a zero-one measure on A(L1), that is, μ1∈IR(L1).Supposeμ1(A)=μ(A×Y)=1; there exists A×Y⊃L1×L2∈L* such that μ(L1×Y)=1 and  μ(L1×L2)=1. Then μ1(L1)=μ(L1×Y)=1 and L1⊂A which shows that μ1∈IR(L1).  Similarly take μ2 on A(L2) defined by μ2(B)=μ(X×B), B∈A(L2).Then, as beforeμ2 is regular on L2.Finally for anyA∈A(L1) and any B∈A(L2) we have μ1×μ2(A×B)=μ1(A)μ2(B)=μ(A×Y)μ(X×B)=μ[(A×Y)∩(X×B)]=μ[(A∩X)×(Y∩B)]=μ(A×B) which shows that μ=μ1×μ2, and thereforeIR(L1)×IR(L2)=IR(L).Theorem 3.2 (the product of lattices/σ-smooth regular measures).
LetX,Y be abstract  sets and let  L1,L2 be lattices of subsets of X and Y, respectively. Then IRσ(L1)×IRσ(L2)=IRσ(L1×L2).Proof.
Letμ∈IRσ(L1) and ν∈IRσ(L2). Hence for A1n∈A(L1) with A1n↓∅ we have μ(A1n)→0 and for A2n∈A(L2) with A2n↓∅ we have ν(A2n)→0, n=1,2,….
Consider the sequence{Bn} of sets from A(L1)×A(L2). As in Theorem 3.1  Bn=⋃i=1kA1ni×A2ni, disjoint union and A1ni∈A(L1), A2ni∈A(L2).
Suppose thatBn↓∅, that is, A1ni×A2ni↓∅  for all i. Therefore A1n↓∅ or A2n↓∅ or both:
(3.2)μ×ν(Bn)=μ×ν[⋃i=1k(A1ni×A2ni)]=∑i=1kμ×ν(A1ni×A2ni)=∑i=1kμ(A1ni)ν(A2ni)⟶0,thereforeμ×ν∈IRσ(L1×L2).Conversely, letμ∈IRσ(L1×L2) and define μ1 on A(L1) by(3.3)μ1(A)=μ(A×Y),A∈A(L1).If{An} is a sequence of sets with An∈A(L1) and An↓∅, then An×Y↓∅, and since μ∈Iσ(L1×L2) it follows that μ(An×Y)→0.Thereforeμ1∈Iσ(L1).Similarly, definingμ2 on A(L2) by μ2(B)=μ(X×B),  B∈A(L2) we get  μ2∈Iσ(L2).Henceμ=μ1×μ2∈Iσ(L1)×Iσ(L2)=Iσ(L1×L2).Theorem 3.3 (product of supports of measures).
LetX,Y be abstract sets and let L1,L2 be lattices of subsets of  X and Y,respectively. The following statements are true:(a)
ifμ=μ1×μ2∈I(L1)×I(L2)=I(L1×L2) then S(μ)=S(μ1)×S(μ2);(b)
ifL1 and L2 are compact lattices then L is compact.Proof.
We have(a)
S(μ)=S(μ1×μ2)=∩{L1×L2∈L1×L2/μ(L1×L2)=μ(X×Y)},
(3.4)S(μ1)×S(μ2)=∩{L1×L2/L1∈L1,L2∈L2,μ1(L1)=μ1(X),μ2(L2)=μ2(Y)},Butμ(L1×L2)=μ1×μ2(L1×L2)=μ1(L1)μ2(L2) and μ(X×Y)=μ1×μ2(X×Y)=μ1(X)μ2(Y),(b)
S(μ)=S(μ1)×S(μ2)≠∅, sinceS(μi)≠∅,Li being compact.Theorem 3.4 (product of Wallman spaces/Wallman topologies).
Consider the spacesIR(Li)  with the Wallman topologies  tWi(Li), i=1,2.
It is known that the topological spaces(IR(Li),tWi(Li))  are compact and T1. Then the topological space (IR(L1)×IR(L2),tW1(L1)×tW2(L2)) is also compact and T1.Proof.
SinceIR(Li) are compact topological spaces,
(3.5)SL1(μ̲)=∩{W1(L1)∈W1(L1)/μ̲(W1(L1))=1}≠∅,SL2(ν̲)=∩{W2(L2)∈W2(L2)/ν̲(W2(L2))=1}≠∅.
We have(3.6)μ̲(W1(A))=μ(A),μ∈IR(L1),A∈A(L1),μ̲∈IR(W1(L1)),ν̲(W2(B))=ν(B),ν∈IR(L2),B∈A(L2),ν̲∈IR(W2(L2)).
Therefore(3.7)μ×ν̲(W1(A)×W2(B))=μ×ν(A×B)=μ(A)ν(B),μ̲×ν̲(W1(A)×W2(B))=μ̲(W1(A))ν̲(W2(B))=μ(A)ν(B),
so that μ×ν̲=μ̲×ν̲∈IR(W1(L1))×IR(W2(L2)), and then SL1×L2(μ×ν̲)=SL1×L2(μ̲×ν̲)=SL1(μ̲)×SL2(ν̲)≠∅⇒IR(L1)×IR(L2) is compact.To show thatIR(L1)×IR(L2) is a T1-space, let μ,ν∈IR(L) and suppose μ≠ν. Since μ=μ1×μ2 with μ1,μ2∈IR(L1) and ν=ν1×ν2 with ν1,ν2∈IR(L2) we get μ1≠ν1 and μ2≠ν2. There exist L1,L̃1∈L1 and L2,L̃2∈L2 with (3.8)μ1∈W1(L1),ν1∈W1(L1)′;ν1∈W1(L̃1),μ1∈W1(L̃1)′,μ2∈W2(L2),ν2∈W2(L2)′;ν2∈W2(L̃2),μ2∈W2(L̃2)′.
Therefore μ1(L1)=μ2(L2)=1,  ν1(L1)=ν2(L2)=0, μ1(L̃1)=μ2(L̃2)=0, ν1(L̃1)=ν2(L̃2)=1 which implies μ∈W(L1×L2), ν∈W(L1×L2)′; ν∈W(L̃1×L̃2), μ∈W(L̃1×L̃2)′.Theorem 3.5 (product of normal lattices).
LetX,Y  be abstract  sets and let  L1,L2  be normal lattices of subsets of X and Y, respectively. Then  L is a normal lattice of subsets of X×Y.Proof.
Letμ∈I(L) and ν,ρ∈IR(L) such that μ≤ν,ρ on L.
Then, sinceμ=μ1×μ2∈I(L1)×I(L2), ν=ν1×ν2∈IR(L1)×IR(L2) and ρ=ρ1×ρ2∈IR(L1)×IR(L2), we obtain μi≤νi, ρi on Li, i=1,2.
Li normal lattices ⇒νi=ρi; therefore ν1×ν2=ρ1×ρ2, that is, ν=ρ.


## 3.3. Examples

(1)
LetX, Y be topological spaces and let O1, O2 be the lattices of open sets of X and Y, respectively. Consider the product space X×Y with a base of open sets given by
(3.9){O1×O2/O1∈O1,O2∈O2}.We have(3.10)(O1×O2)′={(x,y)∈X×Y/(x,y)∉(O1×O2)}={(x,y)/(x,y)∈(X×O2′)or(x,y)∈(O1′×Y)}=(X×O2′)∪(O1′×Y)=(X×F2)∪(F1×Y).HenceF=t(L(F1×F2)) where F1,F2 are the lattices of closed sets of X and Y,respectively.(2)
LetX, Y be topological T3.5-spaces and let Z1, Z2 be the lattices of zero sets of continuous functions of X and Y, respectively. Then for the product space X×Y we consider a base of open sets given by
(3.11){Z1′×Z2′/Z1∈Z1,Z2∈Z2}
such that any open set from X×Y is of the form O=⋃αZ1α′×Z2α′ and any closed set is
(3.12)F=O′=⋂α(Z1ε×Y)∪(X×Z2α)∈t(L(Z1×Z2))
and then F=t(L(Z1×Z2)).


## 4. The General Case of Product of Lattices

Let{Xα}α∈Λ be a collection of abstract sets (Λ an arbitrary index set) and let Lα be the lattice of subsets of Xα for all α.We denote(4.1)L*=∏α∈ΛLα={∏α∈ΛLα/Lα∈Lα,Lα=Xαfor almost allα}.


### 4.1. Results

Theorem 4.1 (the product of lattices/regular measures).
One has(4.2)∏α∈ΛIR(Lα)=IR(L)=IR(∏α∈ΛLα).Proof.
We note that∏α∈ΛA(Lα)=A(∏α∈ΛLα)=A(L) and that ∏α∈ΛA(Lα) is the collection of all finite cylinder sets which means that if B∈∏α∈ΛA(Lα) then B is a cylinder set for which there exists a nonempty finite subset F={α1,α2,…,αn} of Λ and a subset EF∈∏α∈FA(Lα) such that B=PF-1(EF) with
(4.3)PF:∏α∈ΛXα⟶∏α∈ΛXα=Xα1×Xα2×⋯×Xαn,Pα:∏α∈ΛXα⟶Xα.
Letμα∈IR(Lα) for all α∈Λ with μα:A(Lα) and define
(4.4)μ=∏α∈Λμα∈∏α∈ΛIR(Lα),μ:∏α∈ΛA(Lα).
LetA∈∏α∈ΛA(Lα) with μ(A)=1. Then (∏α∈Λμα)(A)=(∏α∈Λμα)(PF-1(EF))=1 that is
(4.5)∏α∈FA(Lα)→PF-1∏αA(Lα)→∏αμα{0,1}
and for EF∈∏α∈FA(Lα) we have
(4.6)(∏α∈ΛμαPF-1)(EF)=(∏α∈Λμα)(PF-1(EF))=(∏α∈F)(EF)=(μα1×μα2×⋯×μαn)(EF)=1.
As in the finite case, we getEF⊃Lα1×Lα2×⋯Lαn where Lαi∈Lαi and μαi(Lαi)=1 for all i=1,2,…,n. Then A=PF-1(EF)⊃PF-1(Lα1×Lα2×⋯Lαn) and PF-1(Lα1×Lα2×⋯Lαn)=1, which shows that μ(A)=sup{μ(PF-1(Lα1×Lα2×⋯Lαn))/PF-1(Lα1×Lα2×⋯Lαn)⊂A and PF-1(Lα1×Lα2×⋯Lαn)∈∏α∈ΛLα=L*}; hence μ is L-regular.
Conversely, letμ∈IR(L)=IR(∏α∈ΛLα) and define μα on A(Lα) by
(4.7)μα(A)=μ(A×∏β∈Λ-{α}Xβ),A∈A(Lα),that is,μα(A)=μ(Pα-1(A)).
Sinceμ is a zero-one valued measure on A(∏α∈ΛLα) it follows from the above definition that μα∈I(Lα). If μα(A)=1, then μ(Pα-1(A))=1, and since μ is L-regular, there exists ∏β∈ΛLβ such that Pα-1(A)⊃∏β∈ΛLβ∈L* and μ(∏β∈ΛLβ)=1.
ThenPα-1(Lα)⊂Pα-1(A) and μα(Lα)=μα(Pα-1(Lα))=1.
Thereforeμα(A)=sup{μα(Lα)/Lα⊂A,Lα∈Lα}, that is, μα∈IR(Lα). Next, if B∈L*, we may consider B=PF-1(Lα1×Lα2×⋯Lαn) and then ∏α∈Λμα(B)=∏α∈Λμα(PF-1(Lα1×Lα2×⋯Lαn))=∏α∈Fμα(Lα1×Lα2×⋯Lαn)=(μα1×μα2×⋯×μαn)(Lα1×Lα2×⋯Lαn)=μα1(Lα1)μα2(Lα2)⋯μαn(Lαn)=μ(Pα1-1(Lα1))⋯μ(Pαn-1(Lαn)). If ∏α∈Λμα(B)=1, then μ(Pαi-1(Lαi)) for all i; hence μ(⋂i=1nPαi-1(Lαi))=1 and μ(PF-1(Lα1×Lα2×⋯Lαn))=μ(∏αLα)=1, that is, μ(B)=1. Thus μ=∏αμα∈IR(L*), and then μ=∏αμα on ∏αA(Lα).Theorem 4.2 (the product of normal lattices).
LetLα be a lattice of subsets of Xα. Then(a)
ifμ=∏αμα∈I(∏α∈ΛLα)=∏α∈ΛI(Lα) we have S(μ)=∏α∈ΛS(μα);(b)
ifLα disjunctive for all α∈Λ, then L=L(∏α∈ΛLα) is a disjunctive lattice of subsets of ∏α∈ΛXα;(c)
suppose thatLα is a normal lattice of subsets of Xα for all α∈Λ; then L=L(∏α∈ΛLα) is a normal lattice of subsets of ∏α∈ΛXα.Proof.
(a)
We haveS(μα)=∩{Lα∈Lα/μα(Lα)=μα(Xα)=1} and S(μ)=S(∏α∈Λμα)=∩{∏α∈ΛLα∈∏α∈ΛLα/μ(∏αLα)=μ(∏αXα)=1}. But μ(∏αLα)=1 implies ∏αμα(∏α∈ΛLα)=1. Then S(μ)=∏α∈ΛS(μα).(b)
Letx=(xα)α∈Λ∈∏α∈ΛXα Since ∏α∈Λμxα(∏α∈ΛAα)=μx(∏α∈ΛAα) we get    ∏α∈Λμxα=μx.Lα disjunctive implies μxα∈IR(Lα) for all α∈Λ and then ∏α∈Λμxα∈∏α∈ΛIR(Lα)=IR(∏α∈ΛLα); therefore μx∈IR(∏α∈ΛLα) which proves that L=L(∏α∈ΛLα) is disjunctive.(c)
Letμ∈I(L) and ν,ρ∈IR(L) such that μ≤ν,ρ on L.
Butμ=∏αμα∈∏α∈ΛI(Lα) and both ν=∏α∈να, ρ=∏α∈ρα∈∏α∈ΛIR(Lα) and  then  ∏αμα≤∏α∈να and ∏αμα≤∏α∈ρα on L with μα(Aα)=μ(Pα-1(Aα)),   να(Aα)=ν(Pα-1(Aα)) for Aα∈A(Lα). By the previous work we get μα≤να and μα≤ρα on Lα.Since eachLα is normal it follows that να=ρα for all α∈Λ, and therefore ν=(να)α∈Λ=(ρα)α∈Λ=ρ which proves that L is a normal lattice.


### 4.2. Examples

(3)
LetXα be a topological T3.5-spaces and let Lα=Zα be the replete lattices of zero sets of continuous functions of Xα for all α∈A.Then eachXα is said to be realcompact.Consider a latticeZ of subsets of ∏α∈AXα   such that
(4.8)∏α∈AZα⊂t(∏α∈AZα).ThenZ is replete, and ∏α∈AXα is realcompact.(4)
LetXα be a T2 and 0-dimensional space and let Lα=Cα be the replete lattice of clopensets for all α∈A. Then each Xα is said to be N-compact. Consider any lattice C of subsets of ∏α∈AXα such that ∏α∈ACα⊂t(∏α∈ACα)⊂t(∏α∈AZα)=F and C is replete and ∏α∈AXα is N-compact.


## 4.1. Results

Theorem 4.1 (the product of lattices/regular measures).
One has(4.2)∏α∈ΛIR(Lα)=IR(L)=IR(∏α∈ΛLα).Proof.
We note that∏α∈ΛA(Lα)=A(∏α∈ΛLα)=A(L) and that ∏α∈ΛA(Lα) is the collection of all finite cylinder sets which means that if B∈∏α∈ΛA(Lα) then B is a cylinder set for which there exists a nonempty finite subset F={α1,α2,…,αn} of Λ and a subset EF∈∏α∈FA(Lα) such that B=PF-1(EF) with
(4.3)PF:∏α∈ΛXα⟶∏α∈ΛXα=Xα1×Xα2×⋯×Xαn,Pα:∏α∈ΛXα⟶Xα.
Letμα∈IR(Lα) for all α∈Λ with μα:A(Lα) and define
(4.4)μ=∏α∈Λμα∈∏α∈ΛIR(Lα),μ:∏α∈ΛA(Lα).
LetA∈∏α∈ΛA(Lα) with μ(A)=1. Then (∏α∈Λμα)(A)=(∏α∈Λμα)(PF-1(EF))=1 that is
(4.5)∏α∈FA(Lα)→PF-1∏αA(Lα)→∏αμα{0,1}
and for EF∈∏α∈FA(Lα) we have
(4.6)(∏α∈ΛμαPF-1)(EF)=(∏α∈Λμα)(PF-1(EF))=(∏α∈F)(EF)=(μα1×μα2×⋯×μαn)(EF)=1.
As in the finite case, we getEF⊃Lα1×Lα2×⋯Lαn where Lαi∈Lαi and μαi(Lαi)=1 for all i=1,2,…,n. Then A=PF-1(EF)⊃PF-1(Lα1×Lα2×⋯Lαn) and PF-1(Lα1×Lα2×⋯Lαn)=1, which shows that μ(A)=sup{μ(PF-1(Lα1×Lα2×⋯Lαn))/PF-1(Lα1×Lα2×⋯Lαn)⊂A and PF-1(Lα1×Lα2×⋯Lαn)∈∏α∈ΛLα=L*}; hence μ is L-regular.
Conversely, letμ∈IR(L)=IR(∏α∈ΛLα) and define μα on A(Lα) by
(4.7)μα(A)=μ(A×∏β∈Λ-{α}Xβ),A∈A(Lα),that is,μα(A)=μ(Pα-1(A)).
Sinceμ is a zero-one valued measure on A(∏α∈ΛLα) it follows from the above definition that μα∈I(Lα). If μα(A)=1, then μ(Pα-1(A))=1, and since μ is L-regular, there exists ∏β∈ΛLβ such that Pα-1(A)⊃∏β∈ΛLβ∈L* and μ(∏β∈ΛLβ)=1.
ThenPα-1(Lα)⊂Pα-1(A) and μα(Lα)=μα(Pα-1(Lα))=1.
Thereforeμα(A)=sup{μα(Lα)/Lα⊂A,Lα∈Lα}, that is, μα∈IR(Lα). Next, if B∈L*, we may consider B=PF-1(Lα1×Lα2×⋯Lαn) and then ∏α∈Λμα(B)=∏α∈Λμα(PF-1(Lα1×Lα2×⋯Lαn))=∏α∈Fμα(Lα1×Lα2×⋯Lαn)=(μα1×μα2×⋯×μαn)(Lα1×Lα2×⋯Lαn)=μα1(Lα1)μα2(Lα2)⋯μαn(Lαn)=μ(Pα1-1(Lα1))⋯μ(Pαn-1(Lαn)). If ∏α∈Λμα(B)=1, then μ(Pαi-1(Lαi)) for all i; hence μ(⋂i=1nPαi-1(Lαi))=1 and μ(PF-1(Lα1×Lα2×⋯Lαn))=μ(∏αLα)=1, that is, μ(B)=1. Thus μ=∏αμα∈IR(L*), and then μ=∏αμα on ∏αA(Lα).Theorem 4.2 (the product of normal lattices).
LetLα be a lattice of subsets of Xα. Then(a)
ifμ=∏αμα∈I(∏α∈ΛLα)=∏α∈ΛI(Lα) we have S(μ)=∏α∈ΛS(μα);(b)
ifLα disjunctive for all α∈Λ, then L=L(∏α∈ΛLα) is a disjunctive lattice of subsets of ∏α∈ΛXα;(c)
suppose thatLα is a normal lattice of subsets of Xα for all α∈Λ; then L=L(∏α∈ΛLα) is a normal lattice of subsets of ∏α∈ΛXα.Proof.
(a)
We haveS(μα)=∩{Lα∈Lα/μα(Lα)=μα(Xα)=1} and S(μ)=S(∏α∈Λμα)=∩{∏α∈ΛLα∈∏α∈ΛLα/μ(∏αLα)=μ(∏αXα)=1}. But μ(∏αLα)=1 implies ∏αμα(∏α∈ΛLα)=1. Then S(μ)=∏α∈ΛS(μα).(b)
Letx=(xα)α∈Λ∈∏α∈ΛXα Since ∏α∈Λμxα(∏α∈ΛAα)=μx(∏α∈ΛAα) we get    ∏α∈Λμxα=μx.Lα disjunctive implies μxα∈IR(Lα) for all α∈Λ and then ∏α∈Λμxα∈∏α∈ΛIR(Lα)=IR(∏α∈ΛLα); therefore μx∈IR(∏α∈ΛLα) which proves that L=L(∏α∈ΛLα) is disjunctive.(c)
Letμ∈I(L) and ν,ρ∈IR(L) such that μ≤ν,ρ on L.
Butμ=∏αμα∈∏α∈ΛI(Lα) and both ν=∏α∈να, ρ=∏α∈ρα∈∏α∈ΛIR(Lα) and  then  ∏αμα≤∏α∈να and ∏αμα≤∏α∈ρα on L with μα(Aα)=μ(Pα-1(Aα)),   να(Aα)=ν(Pα-1(Aα)) for Aα∈A(Lα). By the previous work we get μα≤να and μα≤ρα on Lα.Since eachLα is normal it follows that να=ρα for all α∈Λ, and therefore ν=(να)α∈Λ=(ρα)α∈Λ=ρ which proves that L is a normal lattice.


## 4.2. Examples

(3)
LetXα be a topological T3.5-spaces and let Lα=Zα be the replete lattices of zero sets of continuous functions of Xα for all α∈A.Then eachXα is said to be realcompact.Consider a latticeZ of subsets of ∏α∈AXα   such that
(4.8)∏α∈AZα⊂t(∏α∈AZα).ThenZ is replete, and ∏α∈AXα is realcompact.(4)
LetXα be a T2 and 0-dimensional space and let Lα=Cα be the replete lattice of clopensets for all α∈A. Then each Xα is said to be N-compact. Consider any lattice C of subsets of ∏α∈AXα such that ∏α∈ACα⊂t(∏α∈ACα)⊂t(∏α∈AZα)=F and C is replete and ∏α∈AXα is N-compact.



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*Source: 920737-2011-09-28.xml*