More about sharp and meager elements in Archimedean atomic lattice effect algebras

0.21 MB·English

by  Josef Niederle

More about sharp and meager elements in Archimedean
atomic lattice effect algebras
Josef Niederlea, Jan Pasekaa,∗
1
1 aDepartment of Mathematics and Statistics, Faculty of Science, Masaryk University,
0 Kotl´aˇrska´ 2, CZ-611 37 Brno, Czech Republic
2

n
a
J
Abstract
3
1
The aim of our paper is twofold. First, we thoroughly study the set of meager

elementsM(E),thecenterC(E)andthecompatibilitycenterB(E)inthesetting
]
O of atomic Archimedean lattice effect algebras E. The main result is that in
L this case the center C(E) is bifull (atomic) iff the compatibility center B(E) is
. bifull (atomic) whenever E is sharply dominating. As a by-product, we give
h
a new descriciption of the smallest sharp element over x ∈ E via the basic
t
a decomposition of x. Second, we prove the Triple Representation Theorem for
m
sharply dominating atomic Archimedean lattice effect algebras.
[
Keywords: lattice effect algebra, center, atom, MacNeille completion, sharp

1 element, meager element
v
0
8
5 Introduction
2
The history of quantum structures started at the beginning of the 20th
.
1
century. Observable events constitute a Boolean algebra in a classical physical
0
system. Becauseeventstructuresinquantummechanicscannotbedescribedby
1
1 Booleanalgebras,BirkhoffandvonNeumann introducedorthomodularlattices
: which were considered as the standard quantum logic. Later on, orthoalge-
v
braswereintroducedasthe generalizationsoforthomodularposets,whichwere
i
X considered as ”sharp” quantum logic.
r In the nineties of the twentieth century, two equivalentquantum structures,
a
D-posets andeffectalgebraswereextensivelystudied, whichwere consideredas
”unsharp” generalizations of the structures which arise in quantum mechanics,
in particular, of orthomodular lattices and MV-algebras.
In[16]PasekaandRieˇcanov´apublishedasopenproblemwhether the center
C(E) is a bifull sublattice of an Archimedean atomic lattice effect algebra E.
This question was answered by M. Kalina in [10] who proved that C(E) need
not be a bifull sublattice of E even if C(E) is atomic.
∗Correspondingauthor
Email addresses: [email protected] (JosefNiederle),[email protected]
(JanPaseka)
Preprint submitted toElsevier January 14, 2011
Theaimofourpaperistwofold. First,wethoroughlystudythesetofmeager
elementsM(E),thecenterC(E)andthecompatibilitycenterB(E)inthesetting
of atomic Archimedean lattice effect algebras E. The main result of Section 2
isthatinthis casethe centerC(E)isbifull(atomic)iffthe compatibilitycenter
B(E)is bifull (atomic)wheneverE is sharplydominating. As a by-product,we
give a new descriciption of the smallest sharp element over x∈E via the basic
decomposition of x. Second, in Section 3 we prove the Triple Representation
Theorem established by G. Jenˇca in [8] in the setting of complete lattice effect
algebras for sharply dominating atomic Archimedean lattice effect algebras.
1. Preliminaries and basic facts
Effect algebras were introduced by D.J. Foulis and M.K. Bennett (see [4])
for modelling unsharp measurements in a Hilbert space. In this case the set
E(H) of effects is the set of all self-adjoint operators A on a Hilbert space H
between the null operator 0 and the identity operator 1 and endowed with the
partialoperation+ defined iff A+B is in E(H), where + is the usual operator
sum.
Ingeneralform,aneffectalgebraisinfactapartialalgebrawithonepartial
binary operation and two unary operations satisfying the following axioms due
to D.J. Foulis and M.K. Bennett.
Definition 1.1. [22] A partial algebra (E;⊕,0,1) is called an effect algebra if
0, 1 are two distinct elements and ⊕ is a partially defined binary operation on
E which satisfy the following conditions for any x,y,z ∈E:
(Ei) x⊕y =y⊕x if x⊕y is defined,
(Eii) (x⊕y)⊕z =x⊕(y⊕z) if one side is defined,
(Eiii) for every x∈E there exists a unique y ∈E such that x⊕y =1 (we put
x′ =y),
(Eiv) if 1⊕x is defined then x=0.
We often denote the effect algebra (E;⊕,0,1) briefly by E. On every effect
algebraE apartialorder≤andapartialbinaryoperation⊖canbe introduced
as follows:
x≤y and y⊖x=z iff x⊕z is defined and x⊕z =y.
If E with the defined partial order is a lattice (a complete lattice) then
(E;⊕,0,1) is called a lattice effect algebra (a complete lattice effect algebra).
Definition 1.2. Let E be an effect algebra. Then Q⊆E is called a sub-effect
algebra of E if
(i) 1∈Q
(ii) ifoutofelementsx,y,z ∈E withx⊕y =z twoareinQ,thenx,y,z ∈Q.
2
If E is a lattice effect algebra and Q is a sub-lattice and a sub-effect algebra of
E, then Q is called a sub-lattice effect algebra of E.
Note that a sub-effect algebra Q (sub-lattice effect algebra Q) of an effect
algebra E (of a lattice effect algebra E) with inherited operation ⊕ is an effect
algebra (lattice effect algebra) in its own right.
ForanelementxofaneffectalgebraE wewriteord(x)=∞ifnx=x⊕x⊕
···⊕x (n-times) exists for every positive integer n and we write ord(x)=n if
x
n is the greatest positive integer such that n x exists in E. An effect algebra
x x
E is Archimedean if ord(x)<∞ for all x∈E.
A minimal nonzero element of an effect algebra E is called an atom and E
is called atomic if below every nonzero element of E there is an atom.
For a poset P and its subposet Q ⊆ P we denote, for all X ⊆ Q, by X
WQ
the join of the subset X in the poset Q whenever it exists. Recall also Q ⊆ P
is densely embedded in P if for every element x ∈ P there exist S,T ⊆ Q such
that x= S = T.
WP VP
WesaythatafinitesystemF =(x )n ofnotnecessarilydifferentelements
k k=1
n
ofaneffectalgebra(E;⊕,0,1)isorthogonalifx ⊕x ⊕···⊕x (written x or
1 2 n k
L
k=1
F)existsinE. Herewedefinex1⊕x2⊕···⊕xn =(x1⊕x2⊕···⊕xn−1)⊕xn
L n−1 n−1
supposing that x is defined and x ≤ x′ . We also define ∅ = 0.
L k L k n L
k=1 k=1
An arbitrary system G = (xκ)κ∈H of not necessarily different elements of E
is called orthogonal if K exists for every finite K ⊆ G. We say that for a
L
orthogonal system G = (xκ)κ∈H the element G exists iff { K | K ⊆ G
L W L
is finite} exists in E and then we put G = { K | K ⊆ G is finite}. We
L W L
say that G is the orthogonal sum of G. (Here we write G ⊆ G iff there is
1
L
H1 ⊆H such that G1 =(xκ)κ∈H1).
An element u∈E is called finite if either u=0 or there is a finite sequence
{a ,a ,...,a } of not necessarily different atoms of E such that u=a ⊕a ⊕
1 2 n 1 2
···⊕a . NotethatanyatomofE isevidentlyfinite. Anelementv ∈E iscalled
n
cofinite if v′ ∈E is finite.
Elements x and y of a lattice effect algebra E are called compatible (x ↔y
for short) if x∨y =x⊕(y⊖(x∧y)) (see [13, 20]).
Remarkable sub-lattice effect algebras of a lattice effect algebra E are
(1) A block M of E, which is any maximal subset of pairwise compatible
elements of E (in fact M is a maximal sub-MV-algebra of E, see [20]).
(2) The set S(E)={x∈E |x∧x′ =0} of sharp elements of E (see [6], [7]),
which is an orthomodular lattice (see [9]).
(3) The compatibility center B(E) of E, B(E) = {M ⊆ E | M is a block
T
of E}={x∈E |x↔y for every y ∈E} which is in fact an MV-algebra
(MV-effect algebra).
(4) The center C(E) = {x ∈ E | y = (y ∧x)∨(y ∧x′) for all y ∈ E} of
E is a Boolean algebra (see [5]). In every lattice effect algebra it holds
C(E)=B(E)∩S(E)=S(B(E)) (see [18] and [19]).
3
All these sub-lattice effect algebras of a lattice effect algebra E are in fact
fullsub-latticeeffectalgebrasofE. Thismeansthattheyareclosedwithrespect
to all suprema and infima existing in E of their subsets [9, 21].
The MV-effect algebras E are precisely lattice effect algebraswith a unique
block (i.e., E =B(E)).
The following statements are well known.
Statement 1.3. Let E be a lattice effect algebra. Then
(i) [9,Theorem2.1] Assume b∈E, A⊆E are such that A exists in E and
W
b↔a for all a∈A. Then
(a) b↔ A.
W
(b) {b∧a:a∈A} exists in E and equals b∧( A).
W W
(ii) [9, Theorem 3.7], [21, Theorem 2.8] S(E), B(E) and C(E) are full sub-
lattice effect algebras of E.
(iii) [15, Lemma 3.3] Let x,y ∈E. Then x∧y = 0 and x ≤y′ iff kx∧ly =0
and kx≤(ly)′, whenever kx and ly exist in E.
(iv) [17, Proposition 1] Let {b | α∈ Λ} be a family of elements in E and let
α
a∈E with a≤b for all α∈Λ. Then
α
( {b |α∈Λ})⊖a= {b ⊖a|α∈Λ}
_ α _ α
if one side is defined.
(v) [25, Theorem 3.5] For every atom a ∈ E with ord(a) < ∞, n a is the
a
smallest sharp element over a.
(vi) [19, Corollary 4.3] Let x,y ∈E. Then x⊕y =(x∨y)⊕(x∧y) whenever
x⊕y exists.
(vii) [3, Proposition1.8.7] Let b∈E, A⊆E are such that A exists in E and
W
b⊕a exists for all a∈A. Then {b⊕a:a∈A}=b⊕ A.
W W
(viii) [24, Lemma 4.1] Assume that z ∈ C(E). Then, for all x,y ∈ E with
x≤y′, (x⊕y)∧z =(x∧z)⊕(y∧z).
Statement 1.4. [23, Theorem 3.3] Let E be an Archimedean atomic lattice
effect algebra. Then to every nonzero element x∈E there are mutually distinct
atoms a ∈E and positive integers k , α∈E such that
α α
x= {k a |α∈E}= {k a |α∈E},
M α α _ α α
and x∈S(E) iff k =n =ord(a ) for all α∈E.
α aα α
Statement 1.5. [14,Theorem8] Let E bean atomic Archimedean lattice effect
algebra and let M = {M |κ ∈ H} be a family of all atomic blocks of E. For
κ
each κ∈H let A be the set of all atoms of M . Then:
κ κ
(i) For each κ∈H, A is a maximal pairwise compatible set of atoms of E.
κ
(ii) For x∈E and κ∈H it holds x∈M iff x↔A .
κ κ
(iii) M ∈ M iff there exists a maximal pairwise compatible set A of atoms of
E suchthat A⊆M and if M is a block of E with A⊆M then M =M .
1 1 1
4
(iv) E = {M |κ∈H}.
κ
S
(v) B(E)= {M |κ∈H}.
κ
T
(vi) C(E)= {C(M )|κ∈H}= {S(M )|κ∈H}.
κ κ
T T
(vii) S(E)= {C(M )|κ∈H}= {S(M )|κ∈H}.
κ κ
S S
Lemma 1.6. Let E be a lattice effect algebra and let b ∈ E, A ⊆ E are such
that A exists in E and b⊕a exists for all a ∈ A. Then b⊕ A exists in E
W W
and b⊕ A=(b∨ A)⊕ {b∧a:a∈A}.
W W W
Proof. Clearlyb↔aforalla∈A. ByStatement1.3,(i)wehavethatb↔ A
and {b∧a:a∈A}=b∧( A). Furthermore, b≤a′ for all a∈A and heWnce
b ≤ W{a′ | a ∈ A} = ( A)′W. Therefore b⊕ A exists. In view of Statement
V W W
1.3, (vi)
b⊕ A=(b∨ A)⊕(b∧ A)=(b∨ A)⊕ {b∧a:a∈A}.
_ _ _ _ _
2. Bifull sub-lattice effect algebras of lattice effect algebras
Definition 2.1. For a poset L and a subset D ⊆L we say that D is a -bifull
W
sub-poset of L iff, for any X ⊆D , X exists iff X exists, in which case
WL WD
X = X. Dually, the notionof -bifull sub-poset of Lis defined. We call
WL WD V
a subset D ⊆L to be a bifull sub-poset of L if it is both -bifull and -bifull.
W V
Remark 2.2. Clearly, if L is a complete lattice then D ⊆ L is a complete
sub-lattice of L (i.e., D inherits all suprema and infima of its subsets existing
in L) iff D is a bifull sub-poset of L. Moreover, if E is a lattice effect algebra
then a sub-lattice effect algebra D of E is a bifull sub-lattice effect algebra of E
iff it is -bifull.
W
AnimportantclassofeffectalgebraswasintroducedbyS.Gudderin[6]and
[7]. Fundamental example is the standard Hilbert spaces effect algebra E(H).
For an element x of an effect algebra E we denote
x = {s∈S(E)|s≤x} if it exists and belongs to S(E)
WE
x = {s∈S(E)|s≥x} if it exists and belongs to S(E).
e VE
Defibnition 2.3. ([6], [7].) An effect algebra (E,⊕,0,1) is called sharply dom-
inating if for every x ∈ E there exists x, the smallest sharp element such that
x≤x. That is x∈S(E) and if y ∈S(E) satisfies x≤y then x≤y.
b
Rbecall that ebvidently an effect algebra E is sharply dominbating iff for every
x ∈ E there exists x ∈ S(E) such that x ≤ x and if u ∈ S(E) satisfies u ≤ x
thenu≤x iff foreveryx∈E there existasmallestsharpelementxoverxand
e e
a greatest sharp element x below x.
e b
In what follows set (see [8, 25])
e
M(E)={x∈E | if v ∈S(E) satisfies v ≤x then v =0}.
An element x ∈ M(E) is called meager. Moreover, x ∈ M(E) iff x = 0.
Recall that x∈M(E), y ∈E, y ≤x implies y ∈M(E) and x⊖y ∈M(E).
e
5
Lemma 2.4. Let E be an effect algebra in which S(E) is a sub-effect algebra of
E and let x∈M(E) such that x exists. Then
(i) x⊖x∈M(E). b
(ii) Ibf y ∈ M(E) such that x⊕y exists and x⊕y = z ∈S(E) then x = z.
\
Moreover, if E is a lattice effect algebra then y exists and y =x⊖xb=z.
b b b
Proof. (i): Let u ∈ S(E) such that u ≤ x⊖x. Then x ≤ x⊖u ∈ S(E) which
yields that x≤x⊖u. Hence u=0, i.e., x⊖x∈M(E).
b b
(ii): Since x ≤ z and hence x ≤ z we have x ⊕ y = z = x ⊕ (z ⊖ x) and
b b b
x=x⊕(x⊖x). This yields x⊕y =x⊕(x⊖x)⊕(z⊖x). By the cancellation
b b b
law we get y =(x⊖x)⊕(z⊖x). Hence z⊖x=0, i.e., z =x.
b b b b
b |∈S{(zEb)} b b
Now, assume that E is a lattice effect algebra. Let u ∈ S(E), u ≥ y. Then
also u∧z ≥ y, u∧z ∈ S(E) and z ⊖(u ∧z) ∈ S(E). Then x⊕y = z =
y⊕((u∧z)⊖y)⊕(z⊖(u∧z)). Therefore x = ((u∧z)⊖y)⊕(z⊖(u∧z)).
Since z⊖(u∧z)∈S(E) this yields z⊖(u∧z)=0, i.e., z =u∧z ≤u.
Lemma 2.5. Let E be an effect algebra in which S(E) is a sub-effect algebra of
E and let x∈E such that x exists. Then
(i) x⊖x ∈ M(E) andex = x⊕(x⊖x) is the unique decomposition x =
x ⊕x , where x ∈S(E) and x ∈M(E). Moreover, x ∧x =0 and
S Me S e M e S M
if E is a lattice effect algebra then x=x ∨x .
S M
\ \
(ii) If E is a lattice effect algebra such that x exists then x⊖x and x⊖x
\ \ \ \ \
exist, x⊖x=x⊖x=x⊖x, x=x⊕x⊖x=b x∨x⊖x and x∧ex⊖xb=0.
\
Moreobver,ex⊖x=e(x⊖bx)⊖(bx⊖ex). e e e e e
b e b e b
Proof. (i): Let v ∈S(E), v ≤x⊖x. Then v⊕x≤x and v⊕x∈S(E). Hence
v⊕x≤x, i.e., v =0, x ∈M(E) and x=x⊕(x⊖x).
M e e e
Assumethatthereisadecompositionx=x ⊕x suchthatx ∈S(E)and
e e e S Me S
x ∈M(E). Thenx ≤xandx =x⊖x =(x⊖x)⊕(x⊖x )≥x⊖x ∈S(E).
M S M S S S
It follows that x⊖x = 0 because x ∈ M(E). Therefore, x = x⊖x and
S e M e e Me
x =x.
S e e
′ ′
We have x =1⊖x ≥x⊖x =x . Hence x ∧x ≤x ∧x =0.
e S S S M S M S S
Let E be a lattice effect algebra. By Statement 1.3, (vi) we have that
x=x ⊕x =(x ∨x )⊕(x ∧x )=x ∨x .
S M S M S M S M
(ii): We have that x⊖ x ≥ x ⊖x, x ⊖x ∈ S(E). Let z ∈ S(E) such that
z ≥ x⊖x. Let us put w = z∧(x⊖x). Then w ≤ x⊖x hence w⊕x ∈ S(E)
b e e b e
existsandw⊕x≥x. Thisyieldsthatw⊕x≥x,i.e.,z ≥w ≥x⊖x. Therefore,
e b e b e e
\
x⊖x=x⊖x.e e b b e
Since x⊖x≤1⊖x=(x)′ we obtain that x∧x\⊖x≤x∧(x)′ =0.
b e e
\
We prboceeedsimilarelytoeprovethatx⊖x=ex⊖x. Eevideentlye, x⊖x≥x⊖x.
Let z ∈ S(E) such that z ≥ x⊖x. We put w = z ∧(x⊖x). Then x⊖x ≤
b e b b e b
w ≤ x⊖x ≤ x. It follows that x⊖w≤x and x⊖w ∈S(E) which yields that
b b e b
x⊖w≤x. Hence x⊖x≤w ≤z.
b e b b b
b e b e
6
\ \
Moreover, (x⊖x)⊖(x⊖x)=x⊖x=x⊖x.
b e b e b e
As provedin[1],S(E)isalwaysasub-effectalgebrainasharplydominating
effect algebra E.
Corollary2.6. [8,Proposition15]LetE beasharplydominatingeffectalgebra.
Then every x∈ E has a unique decomposition x = x ⊕x , where x ∈S(E)
S M S
and x ∈M(E), namely x=x⊕(x⊖x).
M
Moreover,the following staetement heolds.
Statement 2.7. Let E be a lattice effect algebra. Then
(i) [12,Corollary1] If E is a sharply dominating then S(E) is bifull in E.
(ii) [16, Lemma 2.7] If E is Archimedean and atomic then S(E) is bifull
in E.
First, we shall need an extension of Statement 1.3, (iii).
Lemma 2.8. Let E be a lattice effect algebra, x ,...,x ∈ E, k ,...,k ∈ N,
1 n 1 n
n≥2 such that k x exist in E for all 1≤i≤n. Then
i i
x ∧x =0 and x ≤x′ for all 1≤i<j ≤n
i j i j
iff
n k x exists and n k x = n k x ,
Lj=1 j j Lj=1 j j Wj=1 j j
k x ∧ k x =0 and k x ≤( k x )′
Li∈I i i Lj∈J j j Lj∈J j j Li∈I i i
for all ∅6=I ⊂{1,...,n}, J ={1,...,n}\I.
Proof. Assume that x ∧x = 0 and x ≤ x′ for all 1 ≤ i < j ≤ n. Let k x
i j i j i i
exist in E for all 1 ≤ i ≤ n. If n = 2 then from Statement 1.3, (iii) we know
that k x ∧k x = 0, k x ≤ (k x )′ and k x ≤ (k x )′. Since k x ↔ k x
1 1 2 2 1 1 2 2 2 2 1 1 1 1 2 2
we have that k x ∨k x = k x ⊕(k x ⊖(k x ∧k x )) = k x ⊕k x . We
1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2
shall proceed by induction. Let n∈N be arbitrary, n≥3 and assume that the
statement holds for every m < n. Let us take ∅ 6= I ⊂ {1,...,n} arbitrarily
and put J = {1,...,n}\I. Hence |I| < n and |J| < n. Then we have (again
by Statement 1.3, (iii)) that k x ∧k x =0 andk x ≤(k x )′ for all i∈I and
i i j j j j i i
j ∈J. Thisandthe inductionassumptionyieldthat k x = k x ≤
(k x )′ for all i∈I. This is equivalent to k x ≤ ( Lj∈kJxj)′jfor Wallj∈iJ∈jI ij.e.,
i i k x = k x ≤ ( k x )′. FurithiermoLrej,∈kJ xj ↔j k x for all i ∈ I
Li∈I i i Wi∈I i i Lj∈J j j j j i i
and j ∈J implies by Statement 1.3, (i) that
k x ∧ k x = k x ∧ k x = (k x ∧k x )=0.
M i i M j j _ i i _ j j _ _ i i j j
i∈I j∈J i∈I j∈J i∈Ij∈J
=0
| {z }
Similarly by the induction assumption and Statement 1.3, (iii) and (vii),
n k x = n−1k x ⊕k x = n−1k x ⊕k x
Lj=1 j j (cid:16)Lj=1 j j(cid:17) n n (cid:16)Wj=1 j j(cid:17) n n
= n−1(k x ⊕k x )= n−1(k x ∨k x )= n k x .
Wj=1 j j n n Wj=1 j j n n Wj=1 j j
The converse implication is evident.
7
Corollary 2.9. Let E be an Archimedean lattice effect algebra and a ,...,a
1 n
mutually compatible different atoms from E, 1 ≤ k ≤ n for all 1 ≤ i ≤ n.
i ai
Thenk a ⊕···⊕k a existsandk a ⊕···⊕k a =k a ∨···∨k a . Moreover,
1 1 n n 1 1 n n 1 1 n n
n a ⊕···⊕n a = n a ∨···∨n a is the smallest sharp element over
a1 1 an n a1 1 an n
k a ⊕···⊕k a .
1 1 n n
Theorem 2.10. Let E be an atomic Archimedean lattice effect algebra and let
x ∈ M(E). Let us denote A = {a | a an atom of E, a ≤ x} and, for any
x
a∈A , we shall put kx =max{k∈N|ka≤x}. Then
x a
(i) For any a∈A we have kx <n .
x a a
(ii) The set F ={kxa| a∈A } is orthogonal and
x a x
x= {kxa|a an atom of E, a≤x}= F = F .
M a M x _ x
Moreover, for all B ⊆A and all naturalnumbersl <n ,b∈B suchthat
x b b
x= {l b|b∈B} we have that B =A and l =kx for all a∈A i.e.,
L b x a a x
F is theuniqueset of multiples of atoms from A suchthat its orthogonal
x x
sum is x.
(iii) For every atomic block M of E, x∈M implies that [0,x]⊆M.
(iv) x∈B(E) implies that [0,x]⊆B(E).
(v) If x exists then
b
\
x=x⊖x= {n a|a an atom of E, a≤x}= {n a| a∈A }
M a _ a x
b b
and
x⊖x= {(n −kx)a| a∈A }= {(n −kx)a| a∈A }.
M a a x _ a a x
b
(vi) If x is finite then [0,x] is a finite lattice, x = n k a = n k a
for a suitable finite set A = {a ,...,a } of atoLmsi=o1f iEiandW[i0=,1x]i∼=i
x 1 n
n [0,k a ].
Qi=1 i i
Proof. (i): Let a∈A . Since E is Archimedeanwe havekx ≤n . Assume that
x a a
kx =n . Then 0<n a≤x and n a∈S(E) by Statement 1.4, i.e., x6∈M(E),
a a a a
a contradiction.
(ii): From Statement 1.4, (i) we know that there is a subset B ⊆ A and
x
natural numbers l <n ,b∈B such that
b b
x= {l b|b∈B}= {l b|b∈B}.
M b _ b
Let us show that F = {l b | b ∈ B}. Evidently, l ≤ kx < n and l b ≤ x for
x b b b b b
all b ∈ B. Hence, for any finite subset D ⊆ B and for any c ∈ B, we have by
Corollary2.9that c⊕ {l b|b∈D} exists. This yields that {l b|b∈D}≤
b b
c′ and therefore x≤ c′Lfor all c ∈ B. Now, let a ∈ A . ThenLa ≤ x≤ c′ for all
x
c∈B i.e., a↔c.
8
We then have
06=kxa = kxa∧x=kxa∧ {l b|b∈B}
a a a W b
= {kxa∧l b|b∈B}=kxa∧l a.
W a b a a
ThethirdequationfollowsfromStatement1.3,(i) andthelastequationfollows
from the fact that a 6=b, a ↔b implies kxa∧l b =0. Hence kx ≤l ≤ kx i.e.,
a b a a a
a ∈ B and lx = k . Therefore A = B and F = {l b | b ∈ B}. The remaining
a a x x b
part of the statement is evident.
(iii): Let y ≤ x. Then y ∈ M(E), A ⊆ A and ky ≤ kx for all a ∈ A . Recall
y x a a y
thatby[16,Lemma2.7(i)]weknowthatM isabifullsub-latticeeffectalgebra
of E. Since M is atomic we have that x = {l b | b ∈ AM}= {l b | b ∈
LM b x LE b
AM}; here AM = {a | a an atom of M, a ≤ x}. This immediately implies by
x x
(ii) that the sets AM and A coincide. Therefore, A ⊆ M. Note also that M
x x x
is closed under arbitrary joins existing in E. Hence y = {kya| a≤y}∈M.
W a
(iv): Itfollowsimmediatelyfrom(iii)andbyB(E)= {M ⊆E |M is an atomic
T
block of E} (see 1.5).
(v): We have that x = {kxa | a an atom of E, a ≤ x}. Let a ∈ A . Then
L a x
a ≤ x ≤ x ∈ S(E). Therefore n a ≤ x. Assume that z ∈ S(E), n a ≤ z for all
a a
a ∈ A . Then kxa ≤ z for all a ∈ A , i.e., x ≤ z. This yields that x ≤ z, i.e.
x b a xb
x= {n a| aan atom ofE, a≤x}. ByStatement2.7,(ii)weobtainthat
WS(E) a b
xb= WE{naa | a an atom of E, a ≤x}. Let G ⊆Ax, G finite. Then L{naa |
a∈G}= {n a| a∈G}≤x. Hence x= {n a| a an atom of E, a≤x}.
b a a
W L
Further, we have
b b
x⊖x = ( {n a|a∈A })⊖( {kxa|a∈A })
L a x L a x
b ≥ ( {n a|a∈A ,a6=b}⊕n b)⊖( {n a|a∈A ,a6=b}⊕kxb)
L a x b L a x b
= (n −kx)b
b b
for all b ∈ A . Now, let z ∈ E such that z ≥ (n −kx)b for all b ∈ A .
x b b x
Then also z ∧ (x ⊖ x) ≥ (n −kx)b. Hence (z∧(x⊖x)) ⊕ kxb ≥ n b i.e.,
b b b b
(z∧(x⊖x))⊕ {kxa|a∈A }≥ {n a|a∈A }. This yields
Lb a x W a x b
zb≥z∧(x⊖x)≥( {n a|a∈A })⊖( {kxa|a∈A })=x⊖x.
L a x L a x
Therefore x⊖bx= {(n −kx)a| a∈A }= {(n −kx)a| a∈Ab }.
L\a a x W a a x
The equbality x=x⊖x follows from Lemma 2.5, (ii).
(vi): Let x = n k a . By (ii) we have that the only atoms below x are
a ,...,a . HenLcbei=x1b=i i n k a = n k a . From the proof of (iii) we
k1now thant any element ofL[0i,=x1] isioif theWfoi=rm1 i ni l a for uniquely determined
Wi=1 i i
natural numbers 0 ≤ l < n , 1 ≤ i ≤ n and conversely, for any system of
i ai
natural numbers 0 ≤ l < n , 1 ≤ i ≤ n, n l a ∈ [0,x]. This yields the
required isomorphism bietweeani[0,x] and nW[i0=,1kiai].
Qi=1 i i
Note that Theorem 2.10 (ii), (iv) immediately yields that the set of meager
(finite meager) elements of an atomic Archimedean lattice effect algebra is a
dual of a weak implication algebra introduced in [2].
Motivated by [8, Proposition15] we have the following proposition.
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Proposition 2.11. Let E be an atomic Archimedean MV-effect algebra. Then:
(i) Let x ∈ M(E) and y ∈ E such that x∧y = 0 and x exists. Then
x∧y =0.
b
(ibi) M(E) is a -bifull sub-poset of E.
W
(iii) M(E) is a lattice ideal of E.
Proof. (i): As in Theorem 2.10 let us put A = {a | a an atom of E, a ≤ x}.
x
Evidently, a ∧ y = 0 and y ≤ a′ for all a ∈ A . Therefore by Statement
x
1.3, (iii) n a ∧ y = 0 for all a ∈ A . Then Theorem 2.10, (v) yields that
a x
x∧y = {n a| a∈A }∧y = {n a∧y | a∈A }=0.
a x a x
W W
(ii): Let X ⊆M(E). Assume that z = X exists. Let u∈E be an upper
b WM(E)
boundofX. Hencealsou∧z isanupperboundofX andclearlyu∧z ismeager.
Therefore z =u∧z ≤u, i.e., z = X.
WE
Now, assume that z = X exists. It is enough to check that z ∈ M(E).
WE
Let t ∈ S(E), t ≤ z, t 6= 0. Then there exists an atom b ∈ E such that b ≤ t.
Let us put kx = max{k | kb ≤ x} < n (since any x ∈ X is meager) and
b b
k = max{kx | x ∈ X}< n . Hence also n b ≤ t ≤ z and n b = n b∧ X =
b b b b b b WE
{n b∧x | x ∈ X}= {kxb | x ∈ X}=k b < n b, a contradiction. Hence
WE b WE b b b
z =0 and z ∈M(E).
(iii): It follows immediately from (ii) because M(E) is a downset in E and E is
e
a lattice.
Moreover we have
Proposition2.12. LetE beanatomicArchimedeanlatticeeffectalgebra. Then
(i) For all X ⊆B(E)∩M(E), X exists iff X exists, in which
WE WB(E)
case X = X ∈M(E).
WE WB(E)
(ii) B(E)∩M(E) is a -bifull sub-poset of E.
W
Proof. (i): Let X ⊆B(E)∩M(E). Assume first that z = X exists. Any
WB(E)
x ∈ X is by Theorem 2.10 of the form x = {kxa | a ∈ A } = {kxa |
WE a x WB(E) a
a∈A },A ⊆B(E)∩M(E). Hencez = { {kxa| a∈A }|x∈X}.
x x WB(E) WB(E) a x
Letusputk =max{kx |x∈X}<n . Thenz = {k a|a∈A ,x∈X}.
a a a WB(E) a x
First, we shall show that z ∈ M(E). Assume that there is y 6= 0, y ≤ z,
y ∈ S(E). Then there is an atom c ∈ E such that c ≤ y i.e., also n c ≤ y ≤ z.
c
Either c∈A for some x∈X or c∧a=0 for all a∈A ,x∈X. Letc∈A for
x x x
some x∈X. Then n c∈B(E). Therefore
c
n c = n c∧z =n c∧ {k a|a∈A ,x∈X}
c c c WB(E) a x
= {n c∧k a|a∈A ,x∈X}=k c<n c,
WB(E) c a x c c
acontradiction. Now,letc∧a=0foralla∈A ,x∈X. Thenc↔ayieldsthat
x
k a ≤ (n c)′ ∈ C(E). Hence z ≤ (n c)′. But n c = n c∧z ≤ n c∧(n c)′ = 0
a c c c c c c
and we have a contradiction again. Hence z ∈M(E).
10