# Research on Partial Ordered Sets That Can Be Constructed as Effect Algebras

Authors: Hai-feng Zhang; Meng Zhou
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this paper, we study effect algebra-induced partial ordered sets. All possible cases of effect algebras generated by bounded partial ordered setMI of height 2 are given. In addition, the structure of chain effect algebra is studied carefully and the corresponding results are obtained.


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## Body

## 1. Introduction

In the past few decades, algebraic structuremodels used to describe objective things have emerged in large numbers, providing effective tools for our scientific research. In the field of quantum mechanics and quantum logic, there are effects algebra [1], mv-algebra and bck-algebra [2]. In particular, effect algebra greatly promotes the rapid development of quantum theory and quantum logic.Effect algebra is an important concept introduced by Foulis and Bennett through algebraic abstraction when they studied quantum logic. Since 1994, the study of effect algebra has been favored by scholars.First, Foulis and Bennett gave a series of basic properties of effect algebra [1]. In 1996, Gudder proposed the concept of accurately measurable elements and pivot elements [3], and proved that the effect algebra in which all elements are pivot elements is an orthogonal modular lattice. In 1999, Riecanova removed the condition of existence of identity elements in the effect algebra and obtained the generalized effect algebra [4]. In 2002, Gudder and Greechie extracted some properties of sequence product operation in Hilbert space and proposed sequence effect algebra [5]. In 2019, Wu et al. proposed L-algebras in [6], which is a generalization of homogeneous effect algebra. The relationship between effect algebras and other algebraic structures has also been studied extensively.Generally speaking, effects algebra is an algebraic structure with a binary partial operation and a unary operation, and contains elements 0,1. The elements of effects algebra are events that are not sharp or clear, such as fuzzy events, quantum effects. Therefore, we can think of effect algebra as fuzzy and ambiguous quantum logic [7]. Effects algebra, of course, is also an algebraic abstraction of the various physical models of quantum mechanics. At the same time, effect algebra is in the category of partial order structure, it is MV-algebra.As we all know, effect algebra has been applied to quantum theory and quantum logic with great success. However, because effect algebra is a defect of partial algebra, its algebraic structure is not perfect, which brings inconvenience to the application research. The partial binary operation + of effect algebra are fully embodied in the partial order relation introduced. Therefore, it is a good idea to study effect algebra from the perspective of partial ordered sets.To avoid the difficulty of partial algebras, starting from the concept of effect algebras, we use effect algebras to induce the unique partial order and make it a partial order set. We study effect algebras from this perspective. The types, quantities and structures of partial ordered sets are discussed in effect algebras. The results show that this method can give some interesting theorems describing the structure of effect algebras perfectly.Below, we first give the basic definition and some properties used in the paper.


## 2. Preliminaries

First we introduce the concept of effects algebra and some properties that will be used.Definition 1.
[1] Aneffect algebra(EAfor short) is a systemX=X,+,0,1where0,1∈Xand+is a partial binary operation onXsatisfying the conditions:(E1) for allu,v∈X, u+v is defined ⇒v+u is defined, and u+v=v+u;(E2) for allu,v,w∈X,u+v+w is defined ⇒u+v+w is defined, and u+v+w=u+v+w;(E3) for allu∈X there is a unique u′∈E such that u+u′=1;(E4) for allu∈X, if 1+u is defined then u=0.The mappingx↦x′ is a total unary operation on X We can define the so-called induced order≤X on X by stipulation(1)≤X=a,b∈X×X|∃r∈Xs.t.a+r=b.Since0≤Xu≤X1 is true for any u∈X, X,≤X is a bounded poset, denoted by PX. When ≤X is a lattice-order relation, we call the effect algebra X a lattice-order effect algebra (LEA for short). When ≤X is a total-order relation, we call the effect algebra X a chain effect algebra (CEA for short).Definition 2.
Two effect algebrasX=X,+,′,0X,1XandY=Y,⊕,†,0Y,1Yareisomorphicif there is a bijectionφfromXtoYsuch that for everyu,vinXthe following four equations hold:(I1)u+v is defined ⇔φu⊕φv is defined and then φu+v=φu⊕φv;(I2)φ0X=0Y;(I3)φ1X=1Y;(I4)φu′=φu†.
Such aφis called anisomorphism, denoted byX≃φYX≃Yfor short).Definition 3.
(see[8]). LetF,≤be a poset, andf,g∈P.(1)
gcoversfinF, denoted byf≺g, iff<gand∀t∈P,f≤t≤g⇒f=torg=t.In a posetF,≤ with a smallest element 0, the element covering the 0 is called an atom. Let AF=a∈F|a is an atom }.(2)
f,g∈F are called comparableiff≤gorg≤f. Otherwisefandgareincomparable, which denoted byf‖g.The following are the basic properties of effect algebra given by Foulis and Bennett in 1994, which we will use.Lemma 1 (see[1]).
LetX=X,+,0,1be an EA andg1,g2,k,l∈X. Then(1)
g1+g2is defined⇔g2≤Xg1′⇔g1≤Xg2′;(2)
ifg1+g2is defined thenk+lis defined for allk≤Xg1andl≤Xg2;(3)
g1≤Xg2⇔g2′≤Xg1′;(4)
k′′=k;(5)
ifg1≤Xg2 and g2+k is defined then g1+k is defined and g1+k≤Xg2+k;(6)
g1+g2=k⇔g1′=g2+k′⇔g1=g2+k′′.Theorem 1.
The necessary and sufficient condition for the EAX=X,+,′,0E,1EandY=Y,⊕,†,0F,1Fto be isomorphic is that there exists a bijectiveφfromXtoYsuch that∀g,h∈Xthe following condition (I1) is true.(I1)g+h is defined ⇔φg⊕φh is defined and φg+h=φg⊕φh,Proof.
⇐: Since φ:X⟶Y is a bijection, then ∃j∈X, such that φj=1Y. Hence(2)φ1X=φj+j′=φj⊕φj′=1Y⊕φj′.
By (E3), we haveφj′=0Y. Therefore, φ1X=1Y , j=1X and j′=0X, which shows that (I2) and (I3) holds.
For everyl∈X , Since l+l′=1X, we have(3)φl⊕φl′=φl+l′=φ1X=1Y.
Henceφl′=φl†, i.e. (I4) holds. Therefore, X≃φY.⇒: By Definition 2, it is trivial.Definition 4.
By anisomorphismbetween two posetsS,≤andT,≤, is meant a one-one correspondenceψbetweenSandTsuch that(4)s≤t⇔ψs≤ψt.
Two posets are calledisomorphic iff there exists an isomorphism between them, we writeψ:S≃ψpoTor justS≃T; an isomorphism of a partly ordered set with itself is called an automorphism. A many-one corregiondence satisfying (4) is called isotone.By theconverse of a relation R is meant the relation Rc such that xRcy if and only if yRx.Definition 5.
By thedualPcof a posetPis meant that poset defined by the converse relation on the same elements.According to Definitions2 and 4, we can obtain the following lemma. It gives the conclusion that the isomorphism of EA isomorphism can imply the order isomorphism.Lemma 2.
LetX=X,+,′,0X,1XandY=Y,⊕,†,0Y,1Ybe EAs,X≃φY. ThenX,≤X≃φpoY,≤Y.The condition (I1) can be further simplified by the following theorem.Theorem 2.
Two effect algebrasX=X,+,′,0X,1XandY=Y,⊕,†,0Y,1Yare isomorphic iff∃φ:X⟶Y, and for allw,v∈X(I1)′ if w+v is defined then φw⊕φv is defined and φw+v=φw⊕φv,(li)X,≤X≃φpoY,≤YProof.
⇐: First of all, since X,≤X≃φpoY,≤Y, φ is bijective.
For allu∈Y, 0X+φ−1u is defined, then so does φ0X⊕φφ−1u and(5)φ0X⊕φφ−1u=φ0X+φ−1u=φφ−1u=u.
Thusφ0X=0Y.
Sinceφ:X⟶Y is a bijection, then ∃h∈X, such that φh=1Y. Hence(6)φ1X=φh+h′=φh⊕φh′=1Y⊕φh′.
By (E3), we haveφh′=0Y. Thus h′=0X and h=1Y.Therefore, which shows that (I2) and (I3) holds.
For everyw∈X, Since w+w′=1X, we have(7)φw⊕φw′=φw+w′=φ1X=1Y.
Henceφw′=φw†, i.e. (I4) holds.
The following is the proof of condition (I1)′ and (li) implication condition (I1).
Letk,l,r∈X and φk⊕φl is defined, φk⊕φl=φr in Y. Then(8)φk≤Yφr.
therefore(9)k≤Xr.
∃o∈X, such that k+o=r. By (I1), we have φk⊕φo=φr. Then φl=φo and l=o. i.e. (I1) holds.
Therefore,X≃fY.
⇒:By Definition 2 and Lemma 2, it is trivial.Remark 1.
The condition (I1)′ in Theorem 1does not imply the condition (I1), see Example 1.Example 1.
LetX1=01,11,p1,q1,,X2=02,12,p2,q2,. It is easy to verify thatX1=X1,+,′,01,11andX2=X2,⊕,†,02,I2are effect algebras, where+,′,⊕,†see the follows.
Letf:X1⟶X2,01↦02,11↦12,p1↦p2,q1↦q2. Then f is a bijection from X1 to X2, and satisfies condition (I1)′. But the effect algebras X1 and X2 are not isomorphic.
For each effect algebraE, a unique partially ordered set PE can be obtained under (1). Then, for a given bounded poset P,≤, can we introduce partial binary operation + and unary operation′such that P,+,′,0,1 is an EA and the induced order relation on P is exactly ≤?
The following counterexamples answers this question.Example 2.
LetV5=0,1,v1,v2,r,0<r<v1<1,0<r<v2<1, (see Figure 1).Figure 1
LatticeV5.IfEV5=V5,+,′,0,1 is an EA. Since 0<r<v1<1, we have r+r=v1. Similarly, since 0<r<v2<1, we have r+r=v2, hence v1=v2. This is a contradiction. Thus, V5,≤ cannot be constructed as an EA.Example 3.
LetV4=0,1,v1,v2and partial binary operations+,⊕and unaryoperations′,†are defined by
It is easy to prove thatE1V4=V4,+,′,0,1 and E2V4=V4,⊕,†,0,1 are two completely different effect algebras, but they both induce partial ordered sets of V4 (see Figure 2). In face, We have more general examples.Figure 2
LatticeV4.Example 4.
LetE0=0,1⊆ℝℝis a set of real numbers) and define⊕nand′nas follows:(10)x⊕ny=xn+yn1/nxn+yn≤1−xn+yn>1x′n=1−xn1/n.n∈Z+.ThenEn=E0,⊕n,′n,0,1 is a LEA with the induced order ≤n. For all u1,u2∈0,1, since(11)u1≤nu2⇔∃h∈0,1s.t.u1⊕nd=u2.Thus, ifu1,u2∈≤n, then u2=u1⊕nd=u1n+dnn⩾u1, i.e. u1,u2∈⩽, where ⩽ is usual orders on ℝ. Hence ≤n⊆⩽.Conversely, lets,t∈⩽, i.e. 0⩽s⩽t⩽1, then we have 0⩽sn⩽tn⩽1 and tn−snn∈0,1, s⊕ntn−snn=t, hence s≤nt, i.e. ≤⊆≤n. Thus ≤n=⩽n=1,2,⋯.Theorem 3.
LetX,≤be a bounded poset withX=n(n∈ℤ+). Ifn≤4, then posetX,≤can be used to construct an EA.Proof.
(1)
Whenn=1, the statement is clearly true.(2)
Ifn=2, then X=0,1, 0+0=0,0+1=1+0=1, 0′=1,1′=0, and X=X,+,0,1 is an EA.(3)
SinceX,≤ is a bounded poset, when n=3, we have X=0,a,1, and 0<a<1, i.e. poset X,≤ is a 3-element chain. Define + and′ as follows:
therefor,X=X,+,0,1 is an EA.(4)
For a bounded posetX,≤ , there are two cases when n= 4, one is a chain of four elements and the other is V4 (see Example 1). So by Theorem 2 and Example 1, we get X,+, which can be converted into EA. The proof is complete.Remark 2.
These results can be summarized in the following table.
We useEn to denote n-element effect algebras n=1,2,3.Remark 3.
Example2shows latticeV5with the least number of elements in non-effect algebra.Lemma 3.
LetK=K,+,0,1be an EA. Then′:K≃Kc, that is.K,≤Kis automorphic.Proof.
for allk1,k2∈K, if k1≤k2, then k2′≥k1′ by Lemma 1 (3). And if k2′≥k1′, then(12)k1=k1′′=k1′≤k2′,=k2′′=k2.
by Lemma 1 (4). Thus′:K≃Kc.Theorem 4.
In the isomorphism sense, there are only four types of effect algebra for five elements, which are:(1)
C5=0,u,v,w,1. Define + and′ as follows:(2)
N5=0,u,v,w,1. Define + and ′ as follows:(3)
M3=0,u,v,w,1. Define + and ′ as follows:(4)
M3=0,u,v,w,1. Define + and ′ as follows:Proof.
Since there are only three kinds of automorphic bounded five-element partial ordered sets:C5,N5, and M3, the theorem holds.Remark 4.
The five-element effect algebras whose induced poset isM3are not unique. There are altogether four of them. We’ll look at this in the next section.Corollary 1.
Letℒ=L,+,′,0,1be an EA withE=n(n∈ℤ+). Ifn<6, thenℒ=L,+,′,0,1is a LEA.Example 5.
[6] TheP6=P6,+,′,0,1is an EA, where+,′see the follows.LetF,≤ be a poset, r,s∈F and r<s. Here are the definitions of the intervals:(13)r,s=d∈F|r≤d≤s,r,s=d∈F|r<d<s.Note thatP6 is not a lattice, next, we will consider lattice-ordered effect algebras.


## 3. Homo-Ordered Effect Algebras

Next we will study the poset induced by the effect algebra with the same property, and first give the definition of the same order effect algebra.Definition 6.
Two effect algebrasX=X,+,0X,1XandY=Y,+,0Y,1Yare calledHomo-orderedif the posetsX,≤XandY,≤Yare isomorphic, denoted byX≅poY.In Example3, E1V4≅poE2V4 and in Example 4, En≅poE1,n=1,2,⋯ holds. Below, we have more general results.Theorem 5.
LetZ=Z,+,0Z,1Zbe an EA,K,≤be a poset. IfPZ≃hK,≤, thenZ≅poK, whereK=K,⊕,†,0K,1K,∀l1,l2∈K:(14)l1⊕l2=hh−1l1+h−1l2h−1l1+h−1l2is defined,−otherwise,l1†=hh−1l1.Proof.
First, we prove thatK=K,⊕,†,0K,1K is an EA.
For alll1,l2∈K, if l1⊕l2 is defined, then h−1l1+h−1l2 is defined, and h−1l1+h−1l2=h−1l2+h−1l1 is defined, hence l2⊕l1 is defined and(15)l2⊕l1=hh−1l2+h−1l1=hh−1l1+h−1l2=l1⊕l2.
i.e. (E1) holds. Similarly, we can prove that (E2) holds as well.(16)l1⊕l1†=l1⊕hh−1l1′=hh−1l1+h−1hh−1l1,=hh−1l1+h−1l1′=h1Z=1K.
Then (E3) holds.
Let1K⊕l is defined l∈K, then h−11K+h−1l=1Z+h−1l is defined. Thus, h−1l=0Z,(17)l=hh−1l=h0Z=0K.
i.e. (E4) holds. HenceK=K,⊕,†,0K,1K is an EA.
Next, we show thatPK=K,≤. i.e. ≤K=≤.
Letk1,k2∈K and k1≤k2. Then we have h−1k1≤h−1k2 and ∃y∈Z, h−1k1+y=h−1k2.i.e.(18)h−1k1+h−1hy=h−1k2.
Thereforek1⊕hy=hh−1k1+h−1hy=hh−1k2=k2, i.e. k1≤Kk2.
Sincek1≤Kk2, then ∃r∈K, k1⊕r=k2. therefore we have h−1k1+h−1r=h−1k2. Hence h−1k1≤Zh−1k2. Since PZ≃hK,≤, we have k1≤k2. Thus, we conclude that PK=K,≤ holds as well. Therefore, PZ and PK are isomorphic.
Thus,Z≅poK, the proof is complete.Remark 5.
(1)
This theorem gives a way to construct a new EA from the poset of an EA.(2)
This method is not sufficient, seeExample 3, E1V4≅poE2V4holds, but+and⊕do not satisfy the relationship of Theorem 5.Definition 7.
LetK1=K1,+K1,′K1,0,1andK2=K2,+K2,′K2,0,1are effect algebras andK1∩K2=0,1,M=K1∪K2. If we put(19)r+t≔r+K1t,r+K1tis definedr,t∈K1,r+K2t,r+K2tis definedr,t∈K2,−otherwise,r′≔r′K1,r∈K1,r′K2,r∈K2,.
for allr,t∈Mthenℳ=M,+,′,0,1is EA, we callℳaunion effect algebraofK1andK2, denoted byℳ=K1⊔K2(see Figure 3.Figure 3
Union effect algebraK1⊔K2.In Example3, if we put G=0,a,1,H=0,b,1, then E1V4=G⊔ℋ.Definition 8.
LetK=K,+K,′K,0K,1Kandℒ=L,+L,′L,0L,1Lare EA. If we put(20)k1,h1+k2,h2≔k1+Kk2,h1+Lh2,k1+Kk2andh1+Lh2are defined,−,otherwise,k1,h1′≔k1′K,h1′L.
for allk1,h1,k2,h2∈K×L, obviously K⊗ℒ=K×L,+,′,0K,0L,1K,1Lis EA, we callK⊗ℒ a direct product effect algebraofKandℒ.In Example3, if we put G=H=0,1,G=G,+,′,0,1,ℋ=H,+,′,0,1 and 0+0=0,0+1=1+0=1,0′=1,1′=0, then E2V4=G⊗ℋ≅E2⊗E2.If all sub-chains in a posetP=P,≤ contain at most m+1 element m∈N, then we say that the height of the poset P=P,≤ is m, denoted by hP=m.Lemma 4.
LetH=H,≤be a bounded poset withhH=2, thenH=MI, whereMI=MI,≤M,MI=0,1∪I,I≠∅,0≤Ma≤M1, for alla∈I(see Figure 4).Figure 4
Bounded posetMI.Proof.
The proof can be obtained directly from the boundedness and height of the posetH,≤.Theorem 6.
LetH=H,≤be a bounded poset withhH=2, then there is an EAX=H,+,′,0,1such thatPX=H,≤.Proof.
Let0,1 be the smallest and largest element of a bounded poset H,≤, that is: 0≤x≤1, for any x∈H.
SincehH=2, H=I∪0,1,I=x,y,⋯ by Lemma 4 (see Figure 4). Obviously, X=H,+,′,0,1 is an EA and PX=H,≤, where +,′ see the follows.(21)0+j=j+0≔j,j+j≔1;j′=j,for allj∈I,0+1=1+0=1,0+0=0;0′=1,1′=0.Theorem 7.
LetX=X,+,′,0,1be an EA andI=X/0,1. Then the following are equivalent:(1)
PX=MI;(2)
For allu,v∈I, if u+v is defined then u+v=1.Proof.
(1)⇒ (2). For all u,v∈I, if u+v is defined, then u,v≤u+v and u∨v≤u+v. In MI, u∨v=1 for all u,v∈I. Thus we have u+v=1.(2)⇒ (1). For all u,v∈I, when u≠v, we have u<u∨v or v<u∨v. Then ∃s,t∈I, such that(22)u∨v=u+s,oru∨v=v+t.andu∨v=1 by (2).
SinceX≃Xc by Lemma 3, u∧v=0 for all u,v∈I. Hence PX=MI, the proof is complete.Theorem 8.
LetZ=Z,+,′,0,1be an EA withPZ=MI,I=Z/0,1. Then(23)Z=⊔Za|a∈I,where Za=Za,+,′,0,1, Za=0,a,a′,1,a∈I.Proof.
For alla∈I, Za=0,a,a′,1,+,′,0,1 and(24)Za=0,a,a′,1=0,a,1,a=a′;0,a,a′,1,a≠a′..
It is easy to verify thatZa is an EA and(25)Zr∩Zs=0,1,r≠sandr≠s′;Zr=Zs,r=sorr=s′..
for allr,s∈I. Thus, ⊔Za|a∈I is defined and ∪a∈IZa=Z, therefore, u+v is defined iff v∈Zu by Theorem 7. Hence Z=⊔Za|a∈I, the proof is complete.Corollary 2.
LetI=n∈ℤ+. In the isomorphism sense, there are altogethern/2+1different homo-ordered effect algebras withMIas the induced partial ordered set.Remark 6.
(1)
In Theorem8, when a≠a′, Ea≃E2⊗E2. Therefore, the effect algebra E with PE=MI is obtained by some 2-element effect algebras and 3-element effect algebras through ⊗ and ⊔ operations.(2)
We find out the structure of the EA of height 2 of its partial ordered set.The structure of the EA of height 3 of its partial ordered set. Here are some examples.(1)
+,′ of C1=E2⊗E2⊗E2:(2)
+,′ of C2=E2⊔ℰ2⊗E2:Example 6.
P6,C1=E2⊗E2⊗E2andC2=E2⊔ℰ2⊗E2are effect algebras whose partial ordered sets have height 3. But the posetPP6ofP6is not a lattice, andC1≅poC2,PC1=PC2is a cubeC2×C2×C2(see Figure 5(a)).Figure 5
LatticeC2×C2×C2 and poset P8. (a) C2×C2×C2 (b) P8.
(a)(b)Here is another example of an EAX whose poset PX is not a lattice.Example 7.
It is easy to verify thatP8=P8,+,′,0,1is an EA, where+,′see the follows.P8,≤ is not a lattice (see Figure 5(b)).Example 8.
The poset in Figure 6is not an induced poset of any lattice effect algebran≥4. At the same time, we notice that atn≠2,Znis all lattice, and we callZncrown lattice.Figure 6
The crown latticeZn.Figure7 below shows the crown lattice Z1,Z2,Z3, and Z4 with n=1,2,3, and 4.Figure 7
The crown latticesZ1,Z2,Z3,Z4.


## 4. Chain Effect Algebra (CEA)

In the previous section we obtained the complete structure of a class of effect algebras.They are constructed from 2-element and 3-element effect algebra by⊗ and ⊔ operations. Since both 2-element and 3-element effect algebras are chain effect algebras, we will discuss chain effect algebras in this section.Lemma 5.
LetX=X,+,′,0,1be an EA andu,v,p,q∈X. Then(1)
ifu+p=u+qthenp=q;(2)
ifu+p=pthenu=0;(3)
u≺v iff there exists a atomp∈Xsuch thatu+p=v;(4)
ifu≺vthenv′≺u′.Proof.
(1)
Letw=u+p=u+q, then p′=u+w′,q′=u+w′ by Lemma 1 (4). Thus p′=q′, that is p=q.(2)
Sinceu+p=p=0+p, Hence p=0 by (1).(3)
Ifu≺v, then ∃d∈X, u+d=v. Let y∈X,0≤y<d, since u+d=v, we have u+y is defined and u≤u+y<u+d=b by (1), then u=u+y and y=0 by (2). Hence d is atom of X,≤. Conversely, let d be an atom of X, and u+d=v, then u≤v. If there ∃e∈Xu≤e<v. that is e=u+y<v=u+d for some y∈X. Then 0≤y<d, Since d is an atom, we have y=0, i.e. u=e. Hence u≺v.(4)
Sinceu≺v , then ∃w∈X , u+w=v and w is atom of X by (3). We have u′+w=v′ by Lemma 1 (4), Then v′≺u′, the proof is complete.Theorem 9.
IfL,≤is ann-element chain0=l1<l2<⋯<ln−1<ln=1, there is and only one effect algebra constructed by posetP,≤, and its+and′operations are as follows:Proof.
Obviously, the+ and ′ operations given in the theorem satisfy the condition (E1) – (E4), that is, ℒ=L,+,′,0,1 is an EA. The order relation induced on ℒ is ≤.
The following shows that the effect algebra constructed byL,≤ is unique.
LetL,⊕,†,0,1 be an EA and the induced poset is the L,≤. Since l1=0, we have(26)l1⊕lr=lr,r=1,2,…,n.
Sincelr≺lr+1, then ∃x∈L, lr⊕x=lr+1r=1,2,⋅n−1, then x=l2 by Lemma 5 (1). Hence(27)l2⊕lr=lr+1r=1,2,…,n−1,andl2⊕lnis not defined.
Therefore,lr+2=l2⊕lr+1=l2⊕l2⊕lr=l2⊕l2⊕lr=l3⊕lr, that is(28)l3⊕lr=lr+2r=1,…,n−2,−r=n−1,n.
Similarly, we can show that(29)l4⊕lr=lr+3r=1,…,n−3,−r=n−2,n−1,n,⋯⋯⋯⋯⋯⋯⋯⋯,li⊕lr=lr+i−1r=1,…,n+1−i,−r=n+2−i,…,n−1,n,i=1,2,3,…,n−1.
Considering the above mentioned, we can get:⊕=+. And, according to the above equation, lk⊕ln−k+1=1, hence lk†=ln−k+1, k=1,2,…,n, i.e. †=′.
Thus, the effect algebra constructed by posetL,≤ is unique, the proof is complete.Definition 9.
(see[9]). LetP,≤be a partial ordered set.(1)
P,≤has theascending chain condition(ACC) if it has no infinite strictly ascending sequences, that is, for any ascending sequence(30)a1≤a2≤a3≤⋯.∃m∈N , am+r=am for all r≥0.(2)
P,≤has thedescending chain condition(DCC) if it has no infinite strictly descending sequences, that is, for any descending sequence(31)a1≥a2≥a3≥⋯.∃m∈N , am+r=am for all r≥0.(3)
An effect algebraX=X,+,′,0,1has the ACC(DCC) ifX,≤has the ACC(DCC).where≤ is induced order of X.Definition 10.
(see[9]). A posetP,≤is said to have amaximal conditionif each non-empty subset ofP,≤contains a maximal element. Dually, the posetP,≤can be defined to haveminimal conditions.Lemma 6 (see[9]).
LetX,≤be a poset, then(1)
The sufficient and necessary condition forX,≤to satisfy ACC is thatX,≤has the maximum condition.(2)
The sufficient and necessary condition forX,≤to satisfy DCC is thatX,≤has the minimal condition.Theorem 10.
LetX=X,+,0,1be an EA, thenXhas the ACC iff it has the DCC.Proof.
IfX has the ACC and let p1,p2,p3,⋯⊆X be descending sequence, i.e.(32)p1≥p2≥p3≥⋯.
Thenp1′≤p2′≤p3′≤⋯, therefor, ∃m∈N , am+k′=am′ for all k≥0 by ACC. Thus an+k=an for all k≥0, and X=X,+,′,0,1 has the DCC, i.e. ACC ⇒ DCC and vice versa. The proof is complete.
Using the above two theorems, we get the following result.Theorem 11.
A chain effect algebraC=C,+,′,0,1must is one of the following:(1)
Cis a finite set0=p1,p2,…,pn−1,pn=1and(33)0=p1<p2<⋯<pn−1<pn=1.(2)
Chave an infinite strictly ascending chain(34)1>q1>q2>q3>⋯,and an infinite strictly descending chain(35)0<q1′<q2′<q3′<⋯.Proof.
IfC is a finite set. Obviously, (1) is true.
Now let’s assume that C is an infinite set, and let’s prove thatC can only be (2). In face, C fails to have the DCC and ACC (if not, C has the ACC, then C has the DCC by Theorem 10, hence C is a finite set. This is a contradiction.). Hence C,≤ have an infinite strictly ascending chain(36)1>q1>q2>q3>⋯.
Obviously,(37)0<q1′<q2′<q3′<⋯.
is an infinite strictly descending chain inC,≤. The proof is complete.
Here is the simplest example of an infinite chain effect algebra.Example 9.
LetC0=0=a0,a1,…,an,…,bn,bn−1,…,b1,b0=1, and define+and′as follows:(38)as+at=as+t,bs+bt=−,as+bt=bt−ss≤t,−s>t,∀s,t=0,1,2,⋯,as′=bs,bs′=as∀s=0,1,2,⋯.
ThenC0=C0,+,′,0,1 is an infinite CEA. And(39)0=a0<a1<⋯<an<⋯<bn<bn−1<⋯<b1<b0=1.Theorem 12.
LetX=X,+,0,1be an EA,≤its induced order,t,u,w∈X. Ift<uandu+wis defined thent,u,≤≃t+w,u+w,≤.Proof.
Sinceu+w is defined, t<u, we have z+w is defined and z+w∈t+w,u+w∀w∈t,u by Lemma 1 (2) and (5).
Letf:t,u,≤⟶t+w,u+w,≤,z↦z+w,∀z∈t,u.
For everyl,k∈t,u(40)l≤k⇔l+w≤k+w⇔fl≤fk.
Thus,t,u,≤≃t+w,u+w,≤.Corollary 3.
Letℒ=L,+,0,1be a CEA,p∈C. Ifnp=p+p+⋯+p︷nis defined then we have:(41)0,p,≤≃p,2p,≤≃2p,3p,≤≃⋯≃n−1p,np,≤.Theorem 13.
LetX=X,+,0,1be an EA withX,≤has no atoms. Ifl<k,l,k∈Xthen∃w∈Xsuch thatl<w<k.Proof.
Considerl,k∈X,l<k. So ∃s∈X,s≠0 such that l+s=k by Definition 1 (E1). Since X,≤ has no atoms, we have: ∃y∈X such that 0<y<s, therefore l<l+y<l+s=k, the result holds.Corollary 4.
Letℳ=M,+,0,1be a finite EA. IfM,≤has a atomp∈M, such that∀t∈M/0,p≤t, thenM,≤is a chain.Proof.
For the sequenced,2d,3d,⋯ in M,≤, since M is finite, we have: ∃k∈N, kd∈M but k+1d is undefined.
Sincekd≤1, we have kd+m=1 for some m∈M. If m≠0, then d≤b and k+1d=kd+d is defined by Lemma 1 (2). This is a contradiction, hence m=0 . Thus kd=1. Now let’s drove that M is equal to 0,d,2d,…,k−1d,1.
Assume thatc∈M and c∉0,d,2d,…,k−1d,1. Since d<c and 0≺d≺2d≺⋯≺k−1d≺1, then ∃t1≤t<k, td<c, but t+1d<c. Hence(42)c=td+y∃y∈M.
Obvious,y≠0, and y≥d, thus c=td+y≥td+d. This is a contradiction. Hence(43)M=0,d,2d,…,k−1d,1,M=k+1,
andM,≤ is a chain.
The following example shows that Corollary4 fails when L is an infinite EA.Example 10.
LetK=0,1,kp,kp′|k=1,2,⋯∪at|t=0,±1,⋯∪bt|t=0,±1,⋯,and(44)sp+tp=s+tp,sp+tp′=t−sp′s<t,1s=t,−s>t,s,t∈Z+,np+ai=an+i,ai+aj=1i+j=0,−i−jp′i+j<0,−i+j>0,i,j∈Z,np+bi=bn+i,bi+bj=1i+j=0,−i−jp′i+j<0,−i+j>0,i,j∈Z.
thenK=K,+,0,1 is an EA, but K,≤ is not chain (See Figure 8). In face, K,≤ is not even a lattice (a0,b0 has no least upper bound in k,≤).Figure 8
PosetK,≤.Example 11.
Let(45)K0=0,p,2p,…,np,…,np′,…,2p′,p′,1∪a0,a1,a−1,⋯,(46)sp+tp=s+tp,sp+tp′=t−sp′s<t,1s=t,−s>t,s,t∈Z+,np+ai=an+i,ai+aj=1i+j=0,−i−jp′i+j<0,−i+j>0,i,j∈Z.
thenK0=K0,+,0,1 is an EA, and K0,≤ is a chain (See Figure 9).Figure 9
ChainK,≤.We naturally ask the question: in Corollary4, if ℒ=L,+,0,1 is a LEA must L,≤ be a chain?The following theorem answers this question.Theorem 14.
Letℒ=L,+,0,1be a LEA. IfL,≤has a atomp∈L, such that∀l∈L/0,p≤l, thenL,≤is a chain.Proof.
If∃m∈N such that mp∈L but m+1p is undefined. Then by Corollary 4, we know that the theorem is true. The theorem will be proved in the case where np is defined ∀n∈N.
Obvious, for allk∈N, 0<kp<1. According to the proof of Corollary 4, similarly, we can get(47)x∈L|x≤np=0,p,2p,…,n−1p,np,0<p<2p<⋯<np,
and its dual(48)x∈L|x≥np′=1,p′,2p′,…,np′,np′<⋯<2p′<p′<1.
Sincep<p′, we have np<np′n∈N.
Next, we will prove thatL is a chain. Since(49)0,p,2p,…,np,…,np′,…,2p′,p′,1⊆C.
Assume thatu,v∈L are incomparable. Then(50)w,v∉0,p,2p,…,np,…,np′,…,2p′,p′,1,and, ∀s,t∈N , sp<w,v<tp′. Since L,≤ is a lattice, we have w∧v∈L, let d=w∧v. Since d<w,v, we have ∃x,y∈L,x,y≥p such that d+x=w,d+y=v. Then(51)d=w∧v=d+x∧d+y=d+x∧y,hence x∧y=0, This is a contradiction, thus L,≤ is a chain.Definition 11.
LetLbe a lattice,j,m∈L,(1)
jis join-irreducible ifj=u∨v⇒j=uorj=v, (u,v∈L).(2)
m is meet-irreducible ifm=s∧t⇒m=sorm=t, (s,t∈L).
JL=j∈L|j is join-irreducible} and ML=m∈L|m is meet-irreducible}.Theorem 15.
LetC=C,+,0,1be a LEA. Then the following conditions are equivalent:(1)
C,≤is a chain.(2)
1is join-irreducible element ofC,≤.(3)
0is meet-irreducible element ofC,≤.Proof.
(1)⇒ (2): Let s∨t=1,s,t∈C. Since C,≤ is a chain, we have: x and y are comparable. Hence s∨t=s or s∨t=t, that is. 1=s or 1=t. Thus, 1 is join-irreducible.(2)⇒ (3): By Lemma 2.(3)⇒ (1): Let C=C,+,0,1 be a LEA. Suppose that C,≤ is not a chain. then ∃p,q∈C have p‖q holds.
SinceC,≤ is a lattice, p∧q∈C, let d=p∧q. Since p‖q, we have: d<p,q, then ∃x,y∈C,x,y>0 such that d+x=p,d+y=q. Hence(52)d=p∧q=d+x∧d+y=d+x∧y,
that is.x∧y=0, That contradicts the fact that 0 is meet-irreducible element. Thus C,≤ is a chain.Corollary 5.
Letℒ=L,+,0,1be a LEA. If 0 is meet-irreducible element ofL,≤, thenL=JL=ML.


## 5. Conclusion

The main content of this paper is to study the properties and structures of LEAs from the perspective of partial ordered sets. We study the characterization of original effect algebras by partial ordered sets induced by EAs. The structure and number of effect algebras generated byMI bounded partially ordered sets of height 2 are solved.We study the chain effect algebra and give some necessary and sufficient conditions for determining the LEA as a CEA. It is proved that a finite EA is a CEA if and only if it has only one atom, and some counterexamples are given.



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