SEQUENTIAL PRODUCTS
ON
EFFECT ALGEBRAS
Stan Gudder
Department of Mathematics
University of Denver
Denver Colorado 80208
[email protected]
Richard Greechie
Department of rvIathematics
Louisiana Tech University
Rusto且， Louisiana 71272
[email protected]
Abstract
A sequential effect algebra (SEA) is an effect algebra on which a
sequential product with natural properties is defined. The propertíes
of sequential products on Hi1 bert space effect algebras are discussed.
For a general SEA , relationships between. sequential independence.
coexistence and compatibility are given. It is shown that the sharp
elements of a SEA form an orthomodular poset. The sequential center
of a SEA is discussed and a characterization of when the sequential
center ís isomorphic to a fuzzy set system is presented. It is shown
that the existence of a sequential product is a strong restriction that
eliminates many effect algebras frorn being SEA's. For example , there
are no finite nonboolean SEA's. A measure of sharpness called he
sharpness index is studied. The existence of horizontal sums of SEA's
is characterized and exarnples of horizontal sums and tensor products
are presented.
1
Introd uction
Two measurement5 αand b cannot .be performed 5imultaneously in generaL
50 they are frequently executed 5equentially. 飞斗飞~ denote by α 。 b a sequential measurement in whichαis performed first and b second. \Ve ca l1 α 。 b
l
the sequential product of αand b. \Ve shal1 restrict our attention to yes-no
measurements. called effects , which have only two possible results. For generality, we do not 臼sume that effects are perfectly accurate me臼urements.
That is. they may be fuzzy or unsharp. :也5 we shall see , the sharp effects are
those that satisfy α 。 α=α.
A paradigm situation is an optic a1 bench in which abeam of partic1 es
prepared in a certain state is injected at the left and then impinge first upon
a filterαand then upon a filter b. Particles that p臼s through both filters
enter a detection device at the right of b. Because of quantum interference.
the order of placement of αand b usually makes a difference and we have
α 。 b#b 。 α. 1f it happens thatα 。 b = boa we say thatαand b are sequentially
independent and writeαI b.
1n recent years quantum effects have been studied within a general a1gebraic frame飞;\'ork called an effect algebra. 1n Section 2 we sh a11summarize the
basic de且 nitions concerning effect algebras and the properties of sequential
products on Hilbert space effect a1gebras. The simplest of these properties
are employed as a.xioms in Section 3 for a sequential effect algebra (SE.-\). A
SE:\. is an effect algebra on which a sequential product with natural properties is defined. \Ve be1ieve that the a.xioms for a SEA are physically motivated
and can be tested , for example , in the optical bench situation. Various properties of a SE人 are proved in Section 3. For instance , relationships between
sequential independence , coexistence and compatibility are given. 1t is also
sho\\'n that the sharp elements form an orthomodular poset.
The sequential center C(E) of a SEA E is the set of elements αε E such
thatαI b for every b ε E. 1n Section 4 it is shown that C(E) coincides
with the set of sharp central elements which has previously been studied.
:\loreover. a characterization is given for when C(E) is isomorphic to a fuzzy
set system. Section 5 shows that the existence of a sequentia1 product is a
strong restriction that eliminates many effect algebras from being SEA 's. It
is shown' that a Boolean algebra admits a unique sequenti a1 product and that
certain effect algebras admit a sequential product only if they are Boolean.
~Ioreover. it is prO\'ed that if a map preserves the sequential product then it
completely preseryes the effect algebra structure of the sharp elements.
Section 6 defines the sharpness index of an effect. 1t is demonstrated that
if E is isotropically 且ni
2
tions 8 and 9. The existence of horizontal sums is characterized and some
examples of horizontal sums and tensor products are given.
2
Hilbert Space Sequential Products
This section summarizes the b臼ic definitions concerning effect algebr臼 [1 ，
6, 7, 8, 14 , 15] and the properties of sequential products on Hilbert space
effect algebr出 [2. 3. 10 , 12 , 13]. If EÐ is a partial binary operation , we write
α -L b ifαEÐ b is defined. .\n effect algebra is a system (E , 0 , 1, EÐ) where
0, 1 are distinct elements of E and EÐ is a partial binary operation on E that
satisfies the following conditions.
(E1) If a • b, then b 1.
αand
bEÐ a =αEÐ b.
(E2) If α -L b and c -L (αEÐ b) , then b 1. c， α 1. (b EÐ c) andαEÐ(bEÐc)=
(αEÐ b) EÐ c.
(E3) For every αE E there exists a unique a' E E such that
and αEÐ a' = 1.
(E4)
1fα -L
1, then
α 1.
a'
α=0.
1n the sequel ，飞，yhenever we writeαEÐ b we are implicit!y assuming that
α -L b. 飞飞，~e define α ~ b if there exists a c E E such that αEÐ c = b. If such a
c E E exists then it is unique and we write c = b e α. It can be shown that
(E， 三，') is a partially ordered set with 0 ~二 α 三 1 for all αεE.α"α. and
α 三 b implies b' ~ a'. Moreover , we have α 1. b if and only ifα 三 b'. If α4α
we call αan isotropic element and when 0 is the only isotropic element of
E , then we call E an orthoalgebra. If the n-fold orthosum a EÐα@ … @α
is defined in E we denote this element of E by nα. If there is a largest ηEN
such that. na is defined , then ηis the isotropic index of αand if no such n
exists. then ηhas isotropic index ∞. An element αε E is sharp ifα ^a' = O.
Xotice that ifα 笋 o is sharp then a has isotropic index 1. \Ve say that E is
isotropically finite if everyα 笋 o in E has finite isotropic index.
飞飞'e now give some standard examples of effect algebr臼. For a Boolean algebra 8 , defineα 1. bifα^b = 0 and in this c臼eαEÐb=αvb. Then (8, 0, 1, EÐ)
is an effect algebra that happens to be an orthoalgebra. In particular if X is
a nonempty set , then (2 X ， 白， X , EÐ) is an effect algebra. These effect algebras
correspond to classicallogic and set theory. For the function space [0 , 1]''\
3
on the inten.al [0 , 1] ç lR define the functions fo , h by fo(x) = O. fI (x) = 1
for all x εX. For f , 9 ç [0 , 1]'\ , we define f 1. 9 if f(x) + g(x) 三 1 for all
Z εX and in this case (J æ g)(x) = f(x) + g(x). Then ([0 , l]X , fo , h , æ) is
the effect algebra of fuzzy subsets of X. A particularly simple effect algebra
is the interval [0 , 1] ç R. For α ， b ε[0 ， 1] we defineα 1. b ifα 十 b 三 1 and in
this case α æb= α +b.
In this section we are mainly concerned with the set ê(H) of all selfadjoint operators on a Hilbert space H that satisfy 0 三 (Ax ， x) 三位 ， x) for
all x E H. For .4., B ε ê(H) we define A 1. B if A + B E ê(H) and in
this case .4 EÐ B = A + B. Then (ê(H) , O,!, æ) is an effect algebra that we
call a Hilbert space effect algebra. This effect algebra is important in
studies of the foundations of quantum physics and quantum measurement
theory [2 , 3. 4，而， 17]. The quantum effects A ε ê (H) correspond to
yes-no measurements that may be unsharp. The set of projection operators
P(H) on H form an orthoalgebra that is a sub-effect algebra of ê(H). The
elements of P( H) correspond to sharp quantum effects.
F is additive ifα 1. b
If E and F are effect algebras , we say that rþ: E
implies rþ(α) 1.功(b) and rþ(αæ b) = φ(α) EÐ rþ(b). If rþ: E
F is additive
F is a morphism and
and 0(1) = 1, then φis a morphism. If 4>: E
φ(α) 1. rþ(b) implies that α 1. b, then φis a monomorphism. .\ surjective
monomorphism is an isomorphism. It is easy to see that a morphism d> is
an isomorphism if and only if rþ is bijective and φ-1 is a morphism. .\ state
on E is a morphism s: E → [O~ 1]. \Ve interpret s( α) as the probability that
the effectαis observed (has answer yes) when the system is in the state s.
We denote the set of states on E by n(E). A set of states S ç n(E) is order
determining if s(α) 三 s(b) for all s ε S implies thatα 三 b.
The sequential product on ê(H) is defined by A 0 B = .-1 1/ 2 BA 1/2
where .-1 1/ 2 is the unique positive square root of A. [2 , 3, 10 , 12 , 13]. We have
that .-1 0 B ε ê (H) because
•
•
•
。三 (.-1 1 / 2 BA 1 / 2 X ， X) = (B川2 X ， 川2X) 三 (A 1/2ι .4 1 / 2 x)
= (Ax , x)
空 (x ， x)
for all x E H. :"J' otice that B 叶 AoB is additive on ê(H) and that IoB = B
for all B εε (H). We now present some of the important properties of the
sequential product on ê(H). If.-1 0 B = B 0 A we say that .4. and B are
sequentially independent and write A I B. The following result is proved
in [12. 13]
4
Theorem 2. 1. (i) For A. B E E(H) ， 矿.-l o B E P(H) then A. B
(ii) For .4, B E E(H ), .4 I B 扩 αη d only if AB = BA.
= B .-l.
Applying Theorem 2.1 we obtain the following properties ofthe sequential
product A 0 B.
Corollary 2.2. (i) 扩 A 0 B = 0, then B 0 A = O.
αnd .-l o (B 0 C) = (A 0 B) 0 C for αII C E E(H).
then C I A 0 B αnd C I (A E9 B).
(ii) 扩 A
(iii)
B , then .-l 1 B'
扩 CIA αndC I B
I
The next three results are also proved in [13]. Notice that Theorem 2.3
gives the converse of the second part of Corollary 2.2(ii).
Theorem 2.3. For .4, B E E(H) , A
C E E(H) if αnd only if A I B.
Theorem 2.4. For f1 , B
(i) A 0 B = B. (ii) B 0 A
0
(B
0
(A
C)
0
B)
0
C for every
ε E (H)
the following stαtements are equivalent.
= B. (iii) AB = BA = B.
Theorem 2.5. For A , B E
.-1 1 B.
E(H) ωe
have B
=A
0
B E9 A' 0 B if and only if
\Ve denote the set of positive trace class operators with trace 1 on H by
The normal states on E(H) have the form Pw( .4) = tr(vV .4) for
some n'ε 1J (H). \Ve say that A , B E E(H) are stochastically independent relative to ~V E 万 (H) if Pw( .4 0 B) = ~以 A)Pw(B). The next result
is proved in [13].
1J (H).
Theorem 2.6. For .4, B E E(H) the following statements α陀 equivalent.
(i) .-l o (C 0 B) = (.-l 0 C) 0 B for every C E E(H). (ii) C 0 (.4 0 B) = (C 0 .4.) 0 B
for every C E E(H). (iii) Pw(AoB) = Pw(A)Pw(B) for every lV E 1J (H).
(iv) A. = CI or B = CI for some 0 三 C 三1.
\Ve close this section 飞，vith an application of the interesting work in [18]
This result shows that the sequential product determines the effect algebra
structure of E(H) when dim H 之 3.
•
Theorem 2.7. Suppose that dim H ~二 3. 扩 φ: E(H)
E(H) is α bijection
sαtisfying cþ( .-l 0 B) = 4>(.4) o 4> (B) for αII A , B E E(H) , then ø is αn effect
algebra isomorphism.
3
Proof. For A E ê(H) we have
4> (A2) = 4> (A 0 A) = 4> (A) o 4> (A) = 4> (A)2
Hence , for .4, B E ê(H) we have
4> (ABA) = 4> (A 2 0 B) = 4> (A 2) o 4> (B) = 4>(A)2 0 4> (B) = rþ(A)rþ(B)rþ(A)
Applying Theorem 2 [18] , 4> has the form rþ(A) = U AU. where U is either a
unitary or an anti-unitary operator on H. It easily follows that rþ is an effect
algebra isomorphism.
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3
Sequential Effect Algebras
This section summarizes the basic definitions and results for sequential effect
algebras. For a binary operation 0 , ifα 。 b=b。 αwe write a I b. A sequential
effect algebra (SEA) is a system (E , 0, L EÐ , 0) where (E , 0, 1, EÐ) is an effect
E is a binary operation that satisfies the following
algebra and 0: E x E
conditions.
•
(S1)
b叶 α。b
(S2)
1 。 α=αfor
is additive for
allαε E.
all a E E.
(S3) Ifα 。 b = 0, then αI b.
(S4) IfαI b, then αI b' and α 。 (boc) = (α 。 b)
0
c for all c E E.
(S5) If c Iαand c I b, then c Iα 。 b and c I (αEÐ b).
飞毛￥
call an operation that satisfies (S1)-(S5) a sequentia1 product on E.
IfαI b for all α . b E E we call E a commutative SEA.
The effect algebra (0 , 1] ç lR is a SEA with sequential product a 0 b =
αb. Corollary 2.2 shows that ê(H) is a SEA under the operation A 0 B =
A. 1/2 B A 1/2. NIore generally, this operation makes any von Neumann alge bra
a SEA. A Boolean algebra is a SEA under the operation α 。 b= α^ b. Let
X be a nonempty set and let :F ç [0 , 1]X. We call :F a fuzzy set system
on X if 10 ， /1 ε :F， if 1 E :F then !t - f E :F, if f , 9 E :F with 1 + 9 三 l
then 1 + 9 E :F and if 1, 9 E :F then f 9 E :F. Then :F becomes a SEA when
6
f $ 9 == f + 9 for f 十 g 三 1 and f 0 9 == f9. Except for E(H) , all of these
examples are commutative. The following lemma summarizes some of the
properties of a SEA E.
= 1 。 α=α for αII αE E.
(ii)α 。 b 三 α for all α， b E E. (iii) 牙 α 三 b， then c 。 α 三 co b for αllc ε E.
(iv) lf α 三 b， then co (be α)=cobec 。 α. (v) If α 三 b， clα and c I b, then
c 1 (be α) .
Lemma 3. 1.
(i)α 。 0=0 。 α=
0
and α 。 1
Proof. (i) By additivity we have
α 。 0$0=α 。 0=α 。 (0
$ 0)
=α 。 0$α 。。
and by cancellation we have α 。 o = O. Applying (S3) gives
01α ， applyíng (S2) and (S4) gives α 。 1=1 。 α=α.
(ii) Applying (S1) gives
α=α 。 α=α 。 (b
(iii)
If α 三 b
0 。 α=
O. Since
$ b') = α 。 b$ α 。 b' 主 α 。 b
there exists a d E E such that
d = b. Hence ,
α$
co b = c 。 α $cod~co α
(iv) If α 三 b， then by (iii) c 。 α 三 c 0 b. Since α $(be α) = b we have that
coa$co(be α) == c 0 b. Hence, c 0 (b e α ) =cobeco α.
(v) This follows from (S4) , (S5) and the identity b e α==(α$ b')'.
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We denote the set of sharp elements of E by Es. It is clear that 0, 1 E Es
and that a' E E s wheneverαε Es.
Lemma 3.2. The fol1owi叼
α， = O. (iii)α 。 α=α.
statements a陀 equivalen t. (i)αE
Proof. To show that (i) imp1ies (ii) suppose that
we have
α ^a'
Es.
(ii)α 。
= O. By Lemma 3.1(ii)
α 。 a' =α' 。 α 三 α， α'
Hence.α 。 a'
= O. To show that (ii) implies (iii) suppose
Then
α=α 。 α@α 。 α-α 。 α
7
thatα 。 α.'
= O.
To show that (iii) implies (i) suppose that
α 。 α=α.
Then
α=α 。 α@α 。 α-α@α 。 d
so by cancellation α 。 α， = O. If b 三 α， a' , then by Lemma 3.1(iii) we have
that a 0 b ~二 α 。 a' = O. Hence ， α 。 b = 0 and similarly a' 0 b = O. Hence ,
b 。 α =b 。 α， = 0 so that
b=b 。 αæb 。 α'=0
Hence.α 〈 α，
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= O.
Lemma 3.3. (i) lf α 。 b = 0 , then α .L b. (ii) For αε E， b E Es ， α 。 b=O
if and only if α .L b. (iii) For α~ b E Es , wíth α .L b we hαve αæ b E Es.
(iv) For a , b E Es wi的 α 三 b we have b e αε Es. (v) For α ， b ε Es wi仇
αI b we have α 。 b E Es.
Proof. (i)
If α 。 b
= 0, then b 。
α=α 。 b
= O. Hence ,
α=α 。 bæα 。 b' = α 。 b'
(ii) Ifα 三
(iii) Since
b'
then b 。 α 三 b 0 b' = O.
= 0 we have
= b' 。 α 三 b'
Hencè ， α 。 b=b 。 α=0.
α。 b
(αæ
b) 0 (αæ b) = (α æb) 。 αæ(α æb) ob
=α 。 (α æb) æbo (α æb)=
αæb
(i\') Thi5 follow5 from iii) and the identity b e α=(αe b')'.
(v) Since αIb 飞ve have
(α 。 b)
0 (α 。 b) = α 。 [b 0 (α 。 b)] = α 。 [b 0 (b 。 α)]
=α 。 [(bob) 。 α]=α 。 (b 。 α)=α 。 (α 。 b)
=
(α 。 α)
ob
=α。b
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It follows from Lemma 3.3(iii) that Es is a sub-~ffect algebra of E that
i5 an orthoalgebra. In general , ifα . b ε Es then α 。 b ~ E s 50 E S i5 not a
sub-SE人 of
E.
Theorem 3.4. Let αεE αnd b E Es. (i)α 三 b if and only if α。 b=b。 α=α
αnd b 三 α if αnd only if α 。 b = b 。 α = b. (ii) 1f αI b, then α ^b= α 。 b.
( iii) 扩 α .L b. then α æb= αvb=(α， 0 b')'.
8
Prool (i) If b 。 α=α. then α = b 。 α 三 b. Similarly: ifα 。 b = b then
b 三 α. Conversely, suppose thatα 三 b. Then α 1.. b' so by Lemma 3.3(ii)
α 。 b' =
b' 0 a = O. Hence ， αI b and we have
α=α 。 bæα 。 b' = α 。 b
If b
三 αthen
a'
:三 b'
so from our previous
b。
α，
= b0
(b' 。 α')
work ， α， 0
= (b
0
b' = b'
0
a'
= α，
Hence ,
b') 。 α'=0
and we have
b=b 。 αæb 。 α'=b 。 α=α 。 b
(ii) We have α 。 b=b 。 α 三 α ， b. Suppose that c 三 α， b. Then there exists a
αand by (i) we have b 0 c = c. Hence ,
d E E such that c æ d =
b。
α =
b 0 (c æ d) = b 0 c æ b 0 d = c æ b 0 d > c
飞~e
conclude that α ^b= α 。 b.
(iii) It is clear that α ， b 三 αæ b. Suppose that α . b ~二 c. Then there exists
a d E E such that αæ d = c and by (i) we have c 0 b = b 0 c = b. By
Lemma 3.3(iiLα 。 b=b 。 α= 0 so that
b=boc=bo( α æd)
= b 。 α æbod=bod
b I (c e α). Hence , b I d and b = d 0 b 三 d.
d = c. Hence , by (ii) we have that
主 pplying Lemma 3.1(v) we have
It follows that a æ
b 三 αæ
α æb= αvb=(α，
Corollary 3.5. E A
^ b')' =
is α sub-effect α1gebr，α 01 E
Proof. This follows from Theorem 3.4 (iii).
(α，
0
b')'
口
that is an orthomodular poset.
口
叭;e
say that α ， b E E coexist if there exist c, d. e E E such that c æ d æ e
is de自 ned and α =cæd， b=cffie [16 , 17].
Theorem 3.6. (i) 厅 αI b then ααnd b coexist. (ii) For αε E ， b E Es ,
αIb 矿 and only if α and b coexist.
9
ProoJ. (i)
IfαI
b, then
α=α 。 b E9 α 。 b'
and
b=b 。 α E9 b 。 α-α 。 b E9 a' 0 b
Now
1=α@α-α E9 (α，
0
b E9 α， 0 b') =
(α@α，
0
E9 α，
b)
0
b'
Hence.
(α' 。的 -α@α，
0
b = α 。 b E9 α 。 b' E9 α， 0 b
It follows thatαand b coexist. (ii) If a and b coexist , thenα = c E9 d, b = c Eg e
for some c, d , e E E where c E9 d E9 e is defìned. Since c 三 b we have b I c.
Since d .1 b we have d 三 b'. Hence, b I d and it follows that b Iα.
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It is well-known that the converse of Theorem 3.6(i) does not hold [3].
We say thatα ， b E Es 缸e compatible if there exist mutually orthogonal
elements c, d , e E Es such thatα= c v d and b = c V e.
Corollary 3.7. Fora , b E ES1
α Ib 矿 αnd only 矿 α and
b α陀
compαtible.
ProoJ. If a I b, then by the proof of Theorem 3.6 and Lemma 3.3(v) ， α 。 b，
α 。 b' and a' 0 b are mutually orthogonal elements of Es. By Theorem 3.4 (iii)
we have
α=α 。 b E9 α 。 b' = α 。 bv α 。扩
and
b= α 。 b
E9 a' 0 b = α 。 b V a' 0 b
Hence , a and b are compatible. Conversely, suppose thatαand b are compatible and α= c V d, b = c V e where c, d , e E Es are mutually orthogonal.
By Theorem 3.4(iii) ， α= c E9 d, b = c E9 e. Since e .1 c, e .1 d , we have e .1 α­
Hence , c E9 d E9 e is defined. Thus ， αand b coexist so by Theorem 3.6(ii) we
have thatαI b.
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Corollary 3.8. A SEA is α commutαtive 0付hoα1gebra if and only if it is
Boolean algebra. 1n a comm u. tative SEA E , Es is α Booleαnα1gebra.
The next result shows that for certain special cases ,
related.
0
α
and E9 are closely
Corollary 3.9. (i) 1f αI b and α .1 b, then a E9 b = α 。 b E9 (α， 0 b')'. (ii) 1f
b E Es and α .1 b, then α E9 b=(α， 0 b')'. (iii) 1f α， b E Es and a' .1 b':
then α 。 b = (α， E9 b')'.
α.
10
4
马rve
"The Sequential Center of a SEA
define the sequential center of a SEA E as
C(E) =
{αε E: αI
b for all b E E}
The next result follows from our previous work.
Theorem 4. 1. (i) C(E) is a sub-SEA of E which is α commutative monoid
under o. (ii) C(E) n Es is a Boolean algebra.
An element
αε
E is principal 证
i f b, c 三
αand
b 1. c imply 也
t ha
剖tbEÐc~三二
α.
It 丽
i s 51
conver5e hold5 in a SEA.
Lemma 4.2. An element αε
E
is p时ncipal if and only if a E Es.
Proof. Suppose αis principal and b ~二
αEÐ
b ~二
α， a'.
Then b 1.
αand b， α 三 α.
Hence ,
α=αEÐO
By cancellation b = 0 so that α^ a' = O. Conversely, suppose that a E Es
and b, c ~二 αwith b .L c. By Theorem 3 .4 (i)α 。 b = b and a 0 c = c. Hence ,
α 三 α 。 (bEÐc) = α 。 bæ α 。 c=bEÐc
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The next result holds for an arbitrary effect algebra [9]. However, the
proof is much simpler for a SEA.
Corollary 4.3. (i) If α， b E Es and α ^b
α ， bE Es αnd αv b exists , then αvb ε Es.
臼ists，
then
α 八 b
E Es.
(ii) 厅
Proof. (i) By Lemma 4.2 ， αand b are principal. Supp05e that c, d 三 α^band
c .L d. Then c 三 α ， b and d ~二 α ， b so that c EÐ d ~二 α， b. Hence , c æd ~二 α^ b.
Thus ， α^ b is principal so by Lemma 4.2 ， α^ b E Es. (ii) Since αv b exist5 ,
a' /\ b':: ( αv b)' exists and is sharp. Hence , a v b E Es.
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.-\.ccording to [9 ], an element αE E is central if a , a' are principal and
for every p E E tÈere exist q, r E E such that q 三 α r 三 a' and p = q æ r.
In [9] the center C(E) is defined to be the set of all central elements.
Theorem 4.4. (i) If αε Es ， then αIp 扩 and only 矿 there exist q, r E E
such that q 三 α.r 三 dαnd p = q ær. (ii) C(E) = C(E) n Es.
11
Proof. (i) Suppose p = q 6 r where q 三 α ， r 三 a'. Since αI q and αI r we
have that αI p. Conversely, suppose that αI p. Then p = p 。 αæ p 0 a' and
we have that p 。 α=α 。 p 三 αand po a' =α， 0 p 三 a'. (ii) IfαE C(E ), then
by Lemma 4.2.αε Es. Moreover, by (i) we have thatαε C(E). Hence ,
C(E) ç C(E) n Es. Conversely, ifαε C(E) n Es , then by (i) we have that
αε C(E). Hence , C(E) n Es ç C(E).
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•
Example. Let F be the set of differentiable functions f: [0 , 1]
[0 , 1].
Then F is a fuzzy set system under the previously defined partial operation
f æ 9 = 1 + 9 if 1 + 9 ~二 1 and the operation 1 0 9 = 1g. Hence , F is a
commutative SEA 50 that C(F) = F. However , F is not lattice order
because 1 八 9 does not exist in F in general. This shows that a
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commutative SEA need not be an MV-algebra.
This example showed that C(E) need not be lattice ordered. If E and
F are SEA、 a map <þ: E
F is an isomorphism if <þ is an effect algebra
isomorphism satisfying Ø( α 。 b) = <Þ( α) 0 Ø(b) for all 矶 b E E. \Ve shall give
an example later which shows that C(E) need not be isomorphic to a fuzzy
set system. Our next result characterizes when C(E) is indeed isomorphic to
a fuzzy set system. In contrast to C(E) , it is interesting to note that C(E)
is always a Boolean algebra even for an arbitrary effect algebra [9]. A state
s on a is called multiplicative if s(α 。 b) = s(α )s(b) for all α ， b.
•
Theorem 4.5. The sequential center C(E) is isomorphic to α fuzzy set system if and only 矿 C(E) α dmits an order deterrnining set 01 multiplicative
stα tes.
•
Proof. Suppose F is a fuzzy set system on 0 and <þ: C(E)
F is an isomorphism. For ωε0 ， αε C(E) defineω(α) = <Þ(α)(ω). Then 叫 1) = <þ(1)(ω)=
1.If α .1 b we have
马~(α 6b)= φ(α æb)(ω)
= [<þ(α) æ <þ (b)] (ω)=φ(α)(ω) + <þ (b)(ω)
=ω(α)+ω (b)
Hence , {山':ωεn} forms a set of states on C(E). These states are multiplicative because
ω(α 。 b) = φ(α 。 b)(ω)
=
[4> (α ) 4> (b)] (ω)
=ω(α)ω (b)
12
=
ø(α)(ω)φ (b)(ω)
To show that this set of states is order determining, suppose that w( α) 三
ω (b) for every :..J E f2. Then Ø( α)(ω) 三 Ø(b)(ω) for every ωE f2 so that
Ø( α) 三伊 (b). Since φis an isomorphism , we have that α 三 b.
Conversely, suppose f2 is an order determining set of multiplicative states
on C(E). Define
C(E)
[0 , 1]0 by Ø(α)(ω)=ω(α) and let :F ç [O , l]n
be the range of ø. Since Ø(1) = h and tþ(O) = 10 , 11 ， /0 ε :F. For Ø( α) E :F
we have
•
ø:
[!t -
Ø(α)] (ω)=1-ω(α)=ω(α') = Ø(α')(ω)
so that !t一 φ(α) E :F and 伊 (α') =
11 -
Ø(α). If α J..
币(α)(ω)=ω(α) 三 ω (b')
b, then
= 1 - Ø(b)(ω)
Hence , Ø(α) J.. Ø(b) and we have
功(α Ei1 b)(ω)=ω(α Ei1
b)
= ω(α)+ω (b)
= Ø( α)(ω) + Ø(b)(ω)
for every ωε f2.飞，ve conclude that tþ(α Ei1 b) = tþ(α ) EBtþ(b). For 功(α) ， Ø(b) E :F
with φ(α) + ó(b) 三 1 we have ω(α) 三 ω (b'). Since f2 is order determining ,
we have thatα J.. b. Hence , ø is a monomorphism and Ø(α) + tþ(b) ε :F. For
Ø( α) ，仿 (b) E :F we have
[tþ( α) 功 (b)] (ω) =φ(α)(ω )tþ(b)(ω)=ω(α)ω (b) = ω(α 。 b)
= Ø(α 。 b)(ω)
for everωε Sl. Hence , Ø(α)tþ(b) E :F and Ø(α 。 b) = Ø(α)Ø(b). We conclude
that :F is a fuzzy set system and
C(E)
:F is an isomorphism.
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•
ø:
order determining set 01 multiplicative
states n if αnd only if E is 臼 omorphic to a luzzy set system f2. Under this
isomorphism. Es is a Booleαn α1gebra 01 subsets 01 n.
Corollary 4.6. A SEA E
Proof. For
αE
α dmits αn
Es we have
tþ(α)2 = tþ(α 。 α) = Ø(α).
Hence ， φ(α) is a
characteristic function which can be considered to be a subset of f2. The
result now follows.
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5
Existence of Sequential Products
This section shows that the existence of a sequential product is a strong
restriction that eliminates many effect algebvras from being SE 人 's.
13
Lemma 5. 1. For an effect algebra E that is a Boolean algebra there is a
unique sequential product α 。 b= α 八 b.
Proof. Since E = Es and any two element5 of E are compatible, by Corollary 3. ï we have thatαI b for every α， b E E and any 5equential product o.
It follow5 from Theorem 3.4 (ii) thatα 。 b= α 八 b.
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Lemma 5.2. Let E be"a SEA.
(i) 厅 αE
either α 三 b orα 三扩. (ii) 厅 α， b
E is αn atom then for every b E E
E E are distinct atoms, then α J.. b.
Proof. (i) Since α 。 b 三 αwe have a 0 b = 0 or α 。 b= α. By Lemma 3.3(i) , if
α 。 b = 0, then α 三扩.If α=α 。 b then α 。 b' = 050 again by Lemma 3.3(i)
we have that α 三 b. (ii) By (i) we have that α J.. b orα 三 b. In the second
case ， α = b which i5 a contradiction.
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Theorem 5.3. An atomistic orthoalgebra E admits a sequential product if
and only if E is Boolean.
Proof. If E i5 Boolean , we have seen that it admits a sequential product.
Conversely, suppose that E admits a sequential product. Since E = Es , by
Corollary 3.5 , E is an orthomodular poset. Let c , d E E. By Lemma 5.2 we
have that c = (V l;) V (Ved , d = (Vl;) V (Vdi ) where l;， 吨，向 are distinct
mutually orthogonal atoms. Since (Vei) J.. (Vdi ), c and d are compatible. It
follows that E is Boolean.
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Corollary 5.4. There does not exist a sequential product on P(H) , dim H 2:
Proof. For dim H 2: 2, P(H) is a nonboolean , atomistic orthoalgebra.
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Theorem 5.5. (i) An orthoalgebra E is a commutative SEA 矿 αnd only if
E is Boolean. (ii) If E is a chain finite SEA , then E is Boolean.
Proof. (i) It is clear that a Boolean algebra is a commutative SEA. Coníf an orthoalgebra E is commutative then by Corollary 3. ï any two
elements of E are compatible. It follows that E is Boolean.
(ii) Ifαε E is an atom , then a =α 。 α@α 。 α， implies thatα 。 α= 0 or
α 。 α， == o. Suppose that a 。 α= O. By Lemma 3.3(i)α J.. α50 that 2αeXÏsts.
Now
versely~
α 。 (2α) ==α 。 (α@α) =α 。 α@α 。 α=0
14
Hence. 2α 1. α50 that 3αexi5t5. Continuing by induction ， παexi5t5 for
all n E N. Since α< 2α< 3α 〈… form5 an infìnite chain , thi5 i5 a
contradiction. Hence ， α 。 α， = 0 50 by Lemma 3.2 a i5 5harp. Since the
orth05um of 5harp element5 i5 5harp , we have E = Es. Hence, E i5 an
atomi5tic orthoalgebra and the re5ult follow5 from Theorem 5.3.
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Lemma 5.6. Let E
•
be α SEA
and let <jJ: E
E be an additive function
sαt吨fying (i)α= <jJ (1) E C(E) , (ii) if b E E with b 三 α ， then <jJ (b) = b.
Then a E Es αnd fþ(b) = α 。 b for every b E E.
Proof. By (ii)
fþ(α)=α.
Hence ,
α= <jJ (1) = <jJ (α) EÐ <jJ (a') =αEÐ <jJ (α')
50 by cancellation we have that
c E E. Hence ,
<jJ (α')
= O. If b
功 (b) EÐ <þ (c)
三 α"
then b EÐ c = a' for 50me
= <jJ(a') = 0
50 that fþ(b) = O. If d E E then a' 0 d 三 a' 50 that <jJ (α， 0 d) = O. A150 ,
α 。 d~二 α50 that <1>(α 。 d) = α 。 d. Thu5 , for every d E E we have that
<jJ (d)
Since a =
= <jJ (d 。 αEÐ d 0 a') = <jJ (α 。 d) EÐ </J(α， 0 d) = α 。 d
<jJ (α)=α 。 αwe
conclude
thatαE
Es.
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Lemma 5.6 can be used to give another proof of Lemma 5. 1. Indeed , let
α 。 b = α^ b be the 5equential product on E. For αE E = C(E) define
4>a: E
E by 仇 (b) = α • b where α • b i5 another 5equential product on E.
Then fþa 5ati5fie5 the condition5 of Lemma 5.6 50 thatα .b = 仇 (b) = α 。 b.
Let X be a finite nonempty 5et and let .:F(X) be the fuzzy 5et 5y5tem
F(X) = [0 , l]X. A 5equential product 0 on .:F (X) is homogeneous ifλ fl 。
f=f 。 λfl = Ãf for all f E F(X) ， λE [0 , 1].
•
Lemma 5.1.
fg.
.:F(X) αdmits α unique
homogeneo 'W3 sequential product f 0 9 =
Proof. It i5 clear that f 0 9 = f 9 is a homogeneou5 sequential product on
F(X). Let f. 9 be another homogeneou5 sequential product on .:F(X). Since
any two element5 of .:F(X)s are compatible , it follow5 from Corollary 3.7
and Lemma 3.3(v) that :F (X)s is a commutative sub-SEA of (.:F (X) , o).
15
For f E F(X )s define 衍: F(X)s
Lemma 5.6 , we have that
•
J.g = 告I (g)
F(X)s by 衍 (g) = f • g. Applying
=f
0
9 = J9
Hence , J. 9 = Jg for every J , g E F(X )s. Since J. 9 is homogeneous , we
have (λ J) • (μg) = (μg) • (入J) for every J, 9 E F(X )s， 人 με[0 ， 1]. For
any 民 v E F(X) we have U = ~二 λdi ， V = 汇的gj where λi ， 的 ε[0， 1] ,
Ji ， 岛 ε F(X )s，
i = 1, . .. ， π ， j = 1,... , m. It follows that U. 岛 = gj.u and
v • Ji = Ji • v for every i = 1,. . . , n , j = 1, . . . , m. Hence ,
u.V = 汇 μ川 gj = 汇 μjgj. U = 汇 μjÀigj • Ji
=
L >"i
J.l.
=U 0 V
jfi 0 gj
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The next result shows that if a map preserves the sequential product then
it completely preserves the 5tructure of the sharp elements.
•
Theorem 5.8. Let E. F be SEA 's and let cþ: E
F be α bijection sαtisfying
ø( α 。 b) = cþ( α) 。 φ (b). Then cþ: Es
Fs is an isomorphism. Moreover， 矿
αε Es or b E Es αnd α 三 b then cþ(α) 三 cþ( b).
•
Proof. If αε Es ， then φ(α) = cþ(α 。 α) = cþ(α) 0 cþ(α) 50 that
b ε Fs then <þ(α) = b for some αE E. But then
cþ(α)
= b0 b =
cþ(α)
0 <þ(α) =
cþ(α)ε Fs .
cþ(α 。 α)
Since φis injective , we have thatα=α 。 αsoαE Es. Hence , cþ: Es
i5 a bijection. Now cþ(α) = 1 for some αE E. Thus ,
cþ(l) = <þ (1) 0 1 = cþ (1) 0 cþ(α) =
Similarly.φ(b)
If αε Es
cþ (1 。 α)
=
cþ(α)
= 1
= 0 for some b E E and we have
φ(0)
or b E Es
= cþ(O 0 b)
withα 三 b
cþ( α)
=
then
cþ(b 。 α)
= φ(0)
0 cþ(b) = 0
α 。 b=b 。 α=α.
= rþ(b) 0 cþ( α)
16
If
Hence.
三 cþ(b)
•
Fs
Therefore , 1>: Es
•
Fs preserves order. If αE Es ,
。(α)
0
1> (a') =
1> (α 。 α') =
1> (0) = 0
Applying Lemma 3.3(ii) we have that 1> (α') 三1>(α)'. Now 1> (b) = 1> (α)' for
sorne b E E s. Since
1>(α 。 b) = 1> (α)
0
1> (b) =
1> (α) 。币(α)' = 0
we have that α 。 b = O. By Lernrna 3.3(ii) we have that b 三 a' 50 that
<Þ(α)' = 1> (b) 三 φ(α'). Hence ，币 (α') = 1>(α)'. Ifα， b E Es and α 1. b then
applying Theorern 3 .4 (iii) gives αEÐb=(α， 0 b')'. Since 1> (α) 1. 1> (b) we have
that
1> (αEÐ b) = 1> ((α，
0
b')') = [1> (α' 。扩)]' = [1> (α') o 1> (b')]'
= [1>(α)'
0
1>( b)']'
Also , 1>-1 : Fs
•
Es preserves order because 1> (α) 三 1> (b) irnplies that
1> (b 。 α)
Hence:α =b 。 α 三 b. Thus ，币:
is an
= 1>(α) EÐ 1>(b)
= 1> (b) 0 1> (α) = 1> (α)
Es
•
Fs is a monornorphisrn and therefore
isornorphisrn.
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The following exarnple show that Theorem 5.8 cannot be strengthened.
Let E = [0 , l]X be a fuzzy set systern. Defìne 1>: E
E by 1>(f) = j2. Then
1> is a bijection and o (f 0 g) = 1>(f) o 1> (g). But if f f/. Es then
•
。 (f') = (1 - 1) 2
:vIoreover, if f , 9 f/. Es with
6
For
#
1 - ]2 =φ (f)'
f 1. g , then 1>(f EÐ g) # </>(f) EÐ 1> (g).
Sharpness Index
ηεN， αε E
we
defìneα n_ α 。 α 。.
..。 α(ηfactors).
Lemma 6. 1. (i) We have that αn E Es if αnd only if an + 1 = ♂. (ii) Also ,
απε Es if αnd only if α m_ αn for all m 之 η.
17
Proof. (i) If αn E Es. then α2n = αn. But α2n
Conversely, ifαn+l = αn then we have that
n+l _
<二 αn+l <二 αn
_ _2n _
α=α= … =α
=
so that αn+l = αn
I_n 、 2
la'"
so thatαn E Es. Moreover, (ii) follows from (i).
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If there exists an m such
thatαm E Es , then the smallestn E N such
E E s is the sharpness index of a. If no such m exists , then the
sharpness index of αis ∞. We denote the sharpness index of αby s(α). Of
course， αE Es if 姐d only if s(α) = 1.
thatα n
Lemma
6.2. 扩 n
=
s(α) <∞ ，
then ♂
is
the largest
Proof. Clearly， αn E Es and ♂三 α. Suppose that
by Theorem 3叫ii) b 。 α=α 。 b. Hence ,
sha叩 element
b 三 αand
below a.
b E Es. Then
boa2 = α2 ob = α 。 b=b
and continuing by induction we have that b 。
αn = αn
0
b = b. Hence ,
b< αn
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Aσ-effect
that 八αi
a1gebra is an effect algebra such that 副主 α2> … implies
exists. A. σ-SEA is a SEA that is a σ-effect algebra E satisfying:
(1) If 副主 α2 主…， then b 0 (八向)=八 (b 。向) for every b E E;
(2) If 副主 α2 主… and b I 句， i = 1, 2, . . ., then b I 八αi.
1t can be shown that
&(H)
is a
σ-SEA.
Theorem 6.3. Let E be aσ -SEA.l1 αε E， then there exist b, c E Es s1J. ch
that b is the largest sharp element below ααnd c is the smallest shα叩 element
above a.
Proof. If n = s(α) <∞， then by Lemma 6.2 ，♂
Suppose that s(α) = 00. Since a 2三 α2
by (1) we have that
belowα.
amob
is the largest sharp element
2:…， b= 八a n exists. No\v
= αm 0 (八αn) = 八αn+m
n
n
18
= b
Since αmlαn for n = 1, 2,. . ., applying (2) gives a m I b. Hence, for all m E N
we have that
b= αmob
= b 。 αm
Therefore ,
b2 = b 0 (^α") = ^(b 。 α") = b
so that b E Es. Clearly, b 三 a. Suppose that d 三 αwith d E Es. Then
d=d 。 α=α 。 d so that d 三 α11 for all n E N. Hence , d 三〈α11 = b. 巩'e
conclude that b is the largest sharp element belowα. Let e be the largest
sharp element below a' and let c =ιThen α 三 e' = c and c E Es. If a ~二 f
where f E Es , then f' :三 a' . Hence, f' :三 e so that c = e' 二二 f. Hence, c is
the smal1 est sharp element above α.
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Theorem 6.3 shows that a u-SEA is sharply dominating [6]. 1叼articular ，
is sharply dominating. We denote the isotropic index of an element a
by i(α) .
ε (H)
1f 1 <
s(α) <∞ ， then there exists an m E N such that
m
♂。 (αm)' 笋 o and i [a 0 (αm)']= ∞.
Theorem 6.4.
Proof. Suppose that s(α) = 2. Since α~ Es we have that
Lemma 6.1(ii) , a 3 =α2. Then
2 _ _3 "" _2 _ _, _ _2 '" _2
α=α 国 α 。 α=α 时 α 。 α
α 。 αF 笋 O.
By
implies that a 2 0 a' = O. Now
α'=(α')2 æα' 。 α
imp 1ies that
α' 。 α=α 。 α-α 。 (α')2 æα2 。 α-α 。 (α')2 = (α')2 。 α
Hence.
(a')3 。 α = (a'? 。 α=α' 。 α
Continuing by induction , we have that
(α')2 =
(a')n 。 α=α' 。 αfor
(a')3 EÐ (a')2 。 α = (a')3 ea ' 。 α
19
every n E N. Now
lt follows that
α， = [(α')3 号 α' 。 α] EÐα' 。 α
Hence. 2(αF 。 α) is defìned and a' = (α')3 EÐ 2(α1 。 α). ln a similar way
(a')3 == (a')4 EÐ
(α') :i 。 α=(α')4
EÐ a' 。 α
50 that
α， == [(a')4
e (['。 α]
Hence ， 3(α'0α) i5 defìned and a' ==
EÐ 2(α' 。 α)
(a')4 EÐ 3(αF 。 α). Continuing by induction ,
we conclude that n(α' 。 α) is defìnecl for all n E N. Hence , i(α 。 α') ==∞.
For the general c臼e ，槌sume that s(α) = n where 1 <η< ∞. Suppose
that n is even and n = 2m. Then (αm)2 == a2m E Es and since m 三
2m - 1 == n - 1 we have αm rt Es. Hence , s(αm) == 2. It follow5 frorn
our previous work thatαm 0 (αm)' 弄 o and i [am 0 (αm) ，] =∞. Finally,
suppose that n i5 odd and n == 2m - 1. Then (αm)2 = α2m E Es. Since
tn
m 三 2m - 2 == n - 1, we have a rt Es. Hence , s(αm) == 2 and again
αm 0 (α勺'弄 o and i [a m 0 (αm)'] == x.
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Corollary 6.5. If E is isotropically finite , then for every a E E \ Es we
have
s(α)== ∞.
lt is easy to see that ê(H) is isotropically fìnite. However , the nonstandard unit interval * [0 , 1] with its usual product is a SEA that is not
isotropically fìnite. ln fact , every infìllitesimal in *[0 , 1] has infìnite isotropic
index. It follows frorn Corollary 6.5 that every unsharp element of E(H) hω
infìnite sharpness index. The converse of Corollary 6.5 does not hold. For
exarnple , in *[0 , 1] every unsharp element has infìnite sharpness index. ln
the next section we 5hall show that there exists a SEA with an elernent α
satisfyi吨 1
< s( α) < 810.
Corollary 6.6. Let E be an
is
shα叩. (ii) 扩 E is αtomic
isotropic1αlly
finite SEA. (i) Every atom of E
then E is an orthoalgebra.
Proof. (i) lfαis an atorn , since a 2 三 α we have a 2 =αor α2 == O. ln the
latter case s(α) = 2 which contradicts Corollary 6.5. Hence.αis sharp.
(ii) Suppose that E is not an orthoalgebra. Then there exists a b E E with
b 八 b' 笋 O. Since E is atornic. there exists an atorn αwithα~ b, b'. But then
α 三 b 三 α， so that α rt Es. This contradicts (i).
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20
Theorem 6.7. (i)扩 αE E s and b E E with b 1. α ， then (a æ b)n = α æb
(ii) If E is ασ -SEA αnd α . b ε E with b 1. α ， then ^(αæ bn) = α æ^b n .
n
n
.
n
Proof. (i) Since b 三 a we have b 0 a' = a' 0 b = b. Hence.α 。 b = O. We
now prove the result by induction n. The result certainly holds for η= 1.
Assuming the result holds for ηwe have that
l
(αæ b) π+1 = (α 67
b)n
0
(α æb) = (αæ
bn) 0 (αæ b)
n
=(αæ b ) 。 αæ(α æbn)ob= α 。 (αæ
bn) æ b 0
n
(α æb )
=αæ bn -r 1
(ii) Since a æ 八俨三 αæ bn , we have that α@ 八 bn 三 ^(αæ bn). Since
α æb n 主 α ， we have that 八 (α æb n ) ~二 α.From 八 (α EÐbn ) :三 α æbn it follows that
八 (α æbn )8 α 三 bn . Hence，八 (α æbn )8a 三八bn so that 八 (α@ 俨)三 α@ 八bn . 口
If E is a σ-SE.-\. we have seen that for any αε E there exists a largest
sharp element belowα. We denote this element by LaJ .
Corollary 6.8. Let E be a σ -SEA
b 1. d and αæ c = b æ d, then
叨的 α ， b
E Es and c. d E E. (i)
lf α 1.
c.
lαæcJ =αæ lcJ = bEÐ ldJ = lbædJ
(ii) 扩 lcJ
= LdJ = 0 ， α 1. c, b 1. d αnd αEÐ c = b 67 d. then α = b and c = d.
(iii )扩 b 1. d , l dJ = 0 and α 三 b æ d , theηα 三 b.
Proof. (i) By Theorem 6.7(i) we have
αEÐ
cn = (a EÐ ct = (b EÐ dt = b æ d"
By Theorem 6.ï(ii) and the proof of Theorem 6.3 we have
n
αEÐ LcJ =α@ 八 c = 八 (αEÐ
cn ) = 八 (bEÐd ) =bEÐ d" :=bæ ldJ
n
~Joreover.
laEÐcJ= 八 (α 67c) π= 八 (αEÐ
cn ) = α@ 八 c = αe lcJ
n
(ii) By (i) we have
α=αEÐlcJ :=bEÐldJ
21
=b
Hence , c = d by cancellation.
(iii) Since α 三 b EÐ d, there exists a c E E such that αEÐ c = b EÐ d. Applying
(i) gives
α 三 αEÐl叫 = b EÐ ldJ = b
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Theorem 6.9. 扩 E is a σ -SEA then any αE E has a unique 陀P陀sentation
a=bEÐc 叫 ere b E Es αnd l叫= O.
Proof It is clear thatα= laJ EÐ (αe La J ). Ií b E E s and b 三 αe 例， then
b 三 α. Hence , b 三 l~J and b 三凶， so that b = O. Th erefore, la e laJJ = O.
This gives the desired representation. To show uniqueness , suppose we have
two such representationsα = b EÐ c = b1 EÐ CI, b, b1 E Es , lcJ = lctJ = o.
Then by Corollary 6圳i) we have b = b1 and c = Cl'
口
7
An Example
This section presents an example of a SEA that has an element a satisfying
s(α) = 2. Let E = ω+ω.. be the set 飞;vi th elements
{O ， l ， α， 2α， . . . ， α" (2αt...
By convention , we defìne
0α=
O. Defining EÐ on E by
(mα) EÐ (ηα) = (m
and
}
+ n)α
whenη<m
(mα)'
æ(na) =
(na)
æ(ma)' =
((m-π)α)'
it is easy to see that (E , 0 , 1, EÐ) is an effect algebra [1.]. Moreover , ja 三 (ka)'
for every j , k E N because
(jα) EÐ ((k
+ j)α)' =
(kα)'
Theorem 7. 1. There is a unique sequential product on the
E= ω+ω ..
22
effect α1gebra
ProoJ. For x , y E E define
-
<
if x
u+
飞
m
= ma , y = nα
ifx= mα ， y =
n a.• if x = (ma)' , y
、‘
，，E
、El·
Nud
rl--
。
z
OZHU
F
(ηa)'
or x =
(mα)' ，
y = na
= (nα)'
It is clear that 10 x = x for every x E E. Since E is commutative under 0 ,
in order to show that 0 is a sequential product we only need to verify
(D) xo (yEÐz)
(A)
X 0
= (xoy) EÐ (xoz) , y 1. z
(y 0 z) = (x 0 y)
z
0
There are eight possibilities for x , y , z.
(1)
(2)
(3)
(-1)
(5)
(6)
(7)
(8)
y
ma
x
ra
ra
ra
(mα)'
Tα
(mα)'
mα
z
nα
(nα)'
nα
(na)'
(rα)'
mαm
(ra)'
(ra)'
mα
(na)'
(ma)'
(ma)'
na
(rα)'
(nα)'
To verify (D) we only need to consider c出es (1) , (2) , (5) and (6) because
(2) and (3) a.s well a.s (6) and (7) are symmetric and (4) and (8) never occur
because y 1- z in these ca.ses. In Case (1) both the left and right hand sides
are O. In Ca.se (2) both the left and right hand sides are x. In Case (5) both
the left and right hand sides are y æz. In Case (6) both the left and right
hand sides are ((n - m + r)α)' .
To verify (A) , we check all eight cases.
(1) ra 0 (ma
(2) ra 0
(3)
0
na) = ra 0
(mα 。 (nα)')
0
= ra 0
=
0
=
0 0
mα= 0
(nα)
=
00
= (ra
(nα)'
0
=
mα) 0
(rα 。 ma) 0 (nα)'
rα 。 ((mα)' 。 ηα )=raona=O=(rα 。 (mα)') 0
23
na
na
(4) rα 。 ((mα)' 0 (ηα)') = rα 。 ((m + η)α)' = rα=rα 。 (nα)'
=
(5)
(rα)'
0
(mα 。 ηα)
(6)
(rα)'
0
(mα 。 (ηα)')
=
(rα 。 (mα)') 0 (ηα)'
(rα)'oO=O=mα 。 nα = ((rα)' omα)
=
(rα)'
0
mα=mα=
= ((rα)' 0 mα) 0
ma 0
0
na
(nα)'
(ηα)'
(7) (rα)' 0 ((ma)' 。 ηα) = (rα)' 。 ηα=ηα = ((r + m)a)' 。 πα
=
((rα)'
0
(mα)') 。 ηα
(8) (rα)'o((mα)' 0 (nα)') = (rα)' 0 ((m + η)α)' = ((r +m +η)α)'
= ((r 十 m)α)' 0 (na)' =
((rα)'
0
(mα)')
0
(ηα)'
It follows that (E , 0) is a SEA.
For uniqueness , suppose that .: E
E is another sequential product on
E. Since αis an atom and α·α 三 α， we have α·α= 0 or α·α=α. But
α 三 a' so thatα~ Es. Hence， α·α= O. Therefore， ηα.mα=ηmα·α=0
for every n , m E N. Since every x E E h臼 the form nαor (nα)' ， it ís clear
that E is commutative under •. For x = mα ， y = (ηα)' we have that
•
x.y =mα • (na)' = (mα·πα) EÐ (mα. (ηα)') = ma = x ^ y
Applying Theorem 3.4 (iii) we have that
((na)' • (ma)')' = nαEÐmα=(η +m) α
Hence , (nα)'. (mα)' = ((η +m) α))'. \Ve conclude that x.y = xoy for every
x , y ε E.
口
The elements nαεω+ω 气 n "# 0, satisfy s(ηα) = 2 while the elements
(nα)' ， n "# 0, satisfy s ((nα)') =∞. For this example , ifx 1. y , then s(xEÐy) =
max (s(x) , s(y)). Of course , this equation does not hold for an arbitrary SEA.
Notice that ω+ω· is not isotropically finite and is not a σ-SEA. Since ω+ω­
does not admit an order determining set of states ， ω+ω· is a commutative
SEA that is not isomorphic to a fuzzy set system.
24
8
Horizontal Sums
Let (Ei , Oi , l i , æi) , i E 1 , be a co11ection of effect algebras 时th card. (1) > 1.
Their horizonta1 sum E = HS (丘， i ε1) is defined 臼 fo11ows. Identify all
the Oi with a single element 0 and all the 1i with a single element 1. Let
EI = Ei \{Oi! 1d , form the disjoint union 山E: and let E = {O, l}uúE:. For
α ， b ε Ei for 50me i ε1 defineα æb= αEÐi b and no other orthosums are
define on E. It is then easy to check that (E , 0, 1, EÐ) is an effect algebra. If
the Ei' i E 1 , are SEA's we now investigate whether HS(Ei , i E 1) admits
a sequential product and is thus a SEA. For an arbitrary effect algebra E
we use the notation E' = E \{O , 1} and we denote the trivial effect algebra
{Oll} by 2.
Tbeorem 8. 1. Let El' E2 笋 2 be effect algebras. (i) 扩 El hωαn atom , then
H S (E 1 , E2 ) does not admit α sequential product. (ii) 11 El is an orthoalgebra,
then HS(E1 , E2 ) does not admit α sequential product.
Proof. (i) Suppose that H S(E l! E 2 ) admits a sequential product o. Let a E
El be an atom and let b E E~. Then α 。 b = 0 orα 。 b' = o. If α 。 b = 0, then
α=α 。 b' = b' 。 α <
b'
which is a contradiction. Similarly, ifα 。 b' = 0, then
α=α 。 b=b 。 α <b
which is again a contradiction.
(ii) Suppose that H S(Eb E2 ) admits a sequential product o. Let aε E~ ，
b E E~. Then α 。 b =1= 0 because otherwise we would again obtain α 三 b' which
is a contradiction. Similarly， α 。 H 笋 o so that 0 <α 。 b， αo b' < 1. Since El
is an orthoalgebra and α 。 b 三 αwe have that
α 。 (α 。 b)
= (α 。 b) 。 α=α 。 b
and αI (α 。 b)'. Hence ,
((α 。 b)' 。 α)
0 b = (α 。 b)' 0 (α 。 b) = 0
But
α=[α 。 (α 。 b)] æ [a 0 (α 。 b)'] = α 。 b EÐ [(α 。 b)' 。 αl
æ (α 。 b') and α 。 b' :并 0 ， we have α 笋 α 。 b. Hence, 0 <
(α 。 b)' 。 α< 1. It fo11ows that ((α 。 b)' 。 α)ob 笋 o which is a contradiction. 口
Since
α= (α 。 b)
25
Applying Theorem 8.1 , if HS(E11 E2 ) is a SEA then neither El nor E2
can be a nontrivi a1 Boolean a1 gebra orω+ω.. The next resul t characterizes
horizont a1 sums of SEA's that admit a sequential product and hence form a
F is positive if
SEA. If E ， F 缸e effect a1 gebras , an additive map 4J: E
φ(α) = 0 implies α= o.
•
Theorem 8.2. Let 马， i E 1 , be $EA 垣 and let E = HS(且 ， i ε 1). Th en
E admits a sequential product 矿 and only if for every αε E; there exists a
positive additive map 4J生 :EU → Ei such that 4J ji (1) =α for every i , j E 1 四th
i =/; j and if α ， b E E; , c ε 乌四协 α 。 b = bo a =/; 0, then 币;产 (c) = α 。 ejt(c)-
Proof Suppose that E admits a sequenti a1 product o. For a E E; , define
￠飞 :Ej → EiI j =/; i , by 愕'i(b) = α 。 b. Notice thatα 。 bε Ei becauseα 。 b 三 α­
Now 愕'i is clear1 y additive and 愕'i(1) = α. Suppose thatα 。 b= 愕:i(b) = 0
and b =/; o. ThenαI b so thatα111. Hence ,
α=α 。矿 = b' 。 αE
Ei n Ej
But then αε{0 ， 1} which is a contradiction. Hence，句 is positive. If
α ， b E E: , c E 乌 with a 0 b = b 。 α=/; 0, then
哝b(C) = (α 。 b)
0
c=α 。 (b 0 c) =α 。币~i(C)
Conversely, suppose 币2: 马→马， Z,J E 1 , i =/; j , satis马r the given
conditions. Define the operation 0 on E by
Iα 。 b
α 。 b=<
L4Jji(b)
ifα， b E
ifαE
Ei for some i E 1
E; , b E Ej , i =/; j E 1
飞鸟飞~
now show that 0 is a sequenti a1 product on E. If 矶的， ~ε 马 for some
i E 1 with b1 ..L ~ then c1early α 。 (b 1 EÐ~) =α 。 b 1 EÐα 。 b 2 • Otherwise ,
αε E: ， 仇， ~ε Ej ， i 笋 j ，皿d we have that
α 。 (b 1
EÐ b2 ) = 4J ji(b 1 EÐ b2 ) = 币'ji(bd EÐ 币'ji(b2 ) = α 。 b 1 EÐα 。 b2
Hence , (S1) holds. It is clear that (S2) holds. To show that (S3) holds ,
suppose thatα 。 b = o. Ifα， b E Ei for some í ε1， then b 。 α= 0 so suppose
that a E E: , b E 马， i =/; j. Then 币1i(b) = 0 so by positivity, b = o. Hence ,
b 。 α= o. To veri作 (S4) suppose thatα 。 b = b 。 αwhereαε E: ， b E Ej ,
i =/; j. Then
α 。 b= 得j(b) = b 。 α = 4J~i(α)ε Ein 马
26
Hence.α 。 b ε{0 ， 1}.
contradiction. If α 。 b
If a 0 b == 0, then CÞji(b) == 0 50 that b == 0 which is a
== 1, then cþ主 (b) == 1. Hence,
α= 饨 (1) 主伟 (b) == 1
so thatα= 1 which is again a contradiction. We conclude that α ， b ε 马
50me i ε!. Hence， αlli. Moreover, ifα， b E E: and c E E斗， i =/: j , then
α 。 (b 0 d) = α 。 CÞ~i(C) = CÞjib(c) = (α 。 b)
0
for
c
To verify (S5) , 5upp05e th剖 clαand c I b. As before， α， b， c E 马 or some
口
i E ! and the result follow5.
The next resu 1t gives a useful method for con5tructing a SEA from a
horizontal sum of SEA's E == HS( 马 I i E I). Suppose there exist effect
algebra morphisms 向:j: Ei →马 ， i=/:j ε!. Define the operation 0 on E by
lα 。 b
α 。 b==<
ifα ，
lα 。 cþji (b)
b E Ei for some
ifαε E: I b E Ej , i
iεI
i: j
E !
Corollary 8.3. We have that (E ,o) is a SEA if αnd only if for
b ε Ej ， i 笋 j E! ， α 。 b = 0 implies that a = 0 or b == O.
Proof. .-\s in the proof of Theorem 8.2 , if (E , 0) is a SEA and
α=α 。 b' == b' 0 a E Ei
eveηαε Ei1
ωb
= 0, then
n Ej
Hence， αε{0 ， 1}. If α=
1, then b = O.
Conversely, assume that 0 satisfies the given condition. For a E E; define
the map CÞji: Ej →瓦， t 笋 j E !, by 句j(b) ==α 。 CÞji(b). Then Øji is
additi\'e and CÞji(1) =α. If CÞji(b) = 0, then α 。 b = O. Since α 笋 0 ， by
assumption b == O. Hence , v is positive. Suppose thatα ， b E E: , c ε Ej with
aob = b 。 α 笋 O. Then
。'j:b(C) = (α 。 b)
0
CÞji(C) = α 。 (b 0 cþjj(c)) = α 。司i( c)
It fo11ows from the proof of Theorem 8.2 that (E , 0) is a SEA.
27
口
Theorem 9. 1. 11 X i= ø is a set and E is α SEA ， then the SEA tensor
product 012 x and E exists.
Proof. \Ve call a function
values and we define
1:
T=
X
•
E simple if
1 h凶
a
finite number of
{I E E X : 1 is simple}
On T define 1 1. 9 if I(x) 1. g(x) for a11 x E X and if 1 1. 9 define
(f $ g)(x) = I(x) $ g(x). Defining O(x) = 0 and l(x) = 1 for all x E X ,
it is easy to check that (T, 0, 1, EÐ) is an effect algebra. For 1, 9 E T define
10 g(x) = I(x) 0 g(x). Ag出n， it is easy to check that (T, 0, 1, $, 0) is a SEA.
Define r: 2x x E
T by
•
=
αo
Z
、·
rtt411
、、
-aJ
α
，，，.飞
A
Tra--
ifx E A
if x ~ A
飞飞马
use the notation r(A， α) = χAα. It is clear that r is an effect algebra
bimorphism. Since
r(A. a) 0 r(B , b) = (χAα) 0 (XBb) = XAnBa 0 b = r(A 0 B ， α 。 b)
we see that r is a SEA-bimorphism. Moreover ,
representatlon
1=
where 向#句: .4 i 内 Aj
eve巧r
1E T
has the unique
EÐ X. jaj = EÐ 咐i ， aj)
= 0,
-l
i 笋 j and uA i
= X.
Let ß: 2x x E
•
F
be a SEA-bimorphism. Define
T
F as follows. If 1 = æXA‘ αi is the
unique representation of 1, then ø(f) = æß(A ù 向 ). Notice that æß(A ， 向) is
defined because æß(Ai , 1) = 1 and ß(Aj ， 向)三 ß(A il 1). It is straightforward
to show that ø is a SEA-morphism. Moreover ,
ß(A， α)
ø:
•
= ø(χAα)=φ 。 r(A ， α)
口
A slightly more delicate argument than that used in Theorem 9.1 can be
employed to show that the SEA tensor product of a Boolean algebra with an
arbitrarv SEA alwavs exists.
30
Let X 笋的 be a set , Q be the rational numbers and define
:F(X)
Then
= {u: X •
Qn[O , 1]}
.r(X) is a fuzzy set system and thus forms a SEA.
Theorem 9.2. The SEA tensor product
ol .r(X) αnd
E(H) exists.
Proof. Let E = E(H) and defìne the SEA T 槌 in the proof of Theorem 9. 1.
Define r: :F (X) x E
T by r(吼叫 (x) = u(x) α. Then r is clearly an effect
algebra bimorphism. Since
•
r(u ， α)
0
r(v , b) = (uα)
0
(vb) = ωα 。 b=r(uov ， α 。 b)
r is a SEA-bimorphism. As in Theorem 9.1 , every
1E T
h槌 the
unique repF be a SEA-bimorphism.
resentation 1 = EÐr(XAi , ai). Let ß: .r(X) x E
As in Theorem 9.1 , define r/J: T
F by r/J(f) = EÐß(χ岛，向). Again , cþ is a
SE.-\.-morphism. Moreover , if u E :F(X) then u h槌 the unique representation u = 艺入iXAi where 沁笋 >"j ， Ai n Aj = , i # j and UA i = X. It 臼
iS
strai地
ghtti
岛
orwar时
dtωo s1
then ha毛.ve that
•
•
ø
ß(u ， α)
= ß(EÐ入iXAi' α) = EÐß( 入iXAi' a) = 9ß(XAi' 沁α)
=创(λ队， α) =
r/J
(2:: >"iX川) =φ 。 r(u， a)
口
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32