States on Subtraction Algebras 1 Introduction
States on Subtraction Algebras 1 Introduction
States on Subtraction Algebras 1 Introduction
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Internati<strong>on</strong>al Mathematical Forum, Vol. 8, 2013, no. 24, 1155 - 1162<br />
HIKARI Ltd, www.m-hikari.com<br />
http://dx.doi.org/10.12988/imf.2013.3485<br />
<str<strong>on</strong>g>States</str<strong>on</strong>g> <strong>on</strong> Subtracti<strong>on</strong> <strong>Algebras</strong><br />
Sang Mo<strong>on</strong> Lee<br />
Department of Computer Science and Informati<strong>on</strong> Engineering<br />
Korea Nati<strong>on</strong>al University of Transportati<strong>on</strong><br />
Chungju 380-702, Korea<br />
smlee@ut.ac.kr<br />
Kyung Ho Kim<br />
Department of Mathematics<br />
Korea Nati<strong>on</strong>al University of Transportati<strong>on</strong><br />
Chungju 380-702, Korea<br />
ghkim@ut.ac.kr<br />
Copyright c○ 2013 Sang Mo<strong>on</strong> Lee and Kyung Ho Kim. This is an open access article<br />
distributed under the Creative Comm<strong>on</strong>s Attributi<strong>on</strong> License, which permits unrestricted<br />
use, distributi<strong>on</strong>, and reproducti<strong>on</strong> in any medium, provided the original work is properly<br />
cited.<br />
Abstract<br />
In this paper, we study the Bosbach state <strong>on</strong> subtracti<strong>on</strong> algebra and<br />
show that there exists a Bosbach state via X/θ where θ is a c<strong>on</strong>gruence<br />
relati<strong>on</strong> induced by an ideal of X.<br />
Mathematics Subject Classificati<strong>on</strong>: 06F35, 03G25, 08A30<br />
Keywords: Subtracti<strong>on</strong> algebra, Bosbach state, Ker (s), c<strong>on</strong>gruence class<br />
1 Introducti<strong>on</strong><br />
B. M. Schein [3] c<strong>on</strong>sidered systems of the form (Φ; ◦, \), where Φ is a set of<br />
functi<strong>on</strong>s closed under the compositi<strong>on</strong> “◦” of functi<strong>on</strong>s (and hence (Φ; ◦) is<br />
a functi<strong>on</strong> semigroup) and the set theoretic subtracti<strong>on</strong> “\” (and hence (Φ; \)<br />
is a subtracti<strong>on</strong> algebra in the sense of [1]. He proved that every subtracti<strong>on</strong><br />
semigroup is isomorphic to a difference semigroup of invertible functi<strong>on</strong>s. B.
1156 Sang Mo<strong>on</strong> Lee and Kyung Ho Kim<br />
Zelinka [4] discussed a problem proposed by B. M. Schein c<strong>on</strong>cerning the structure<br />
of multiplicati<strong>on</strong> in a subtracti<strong>on</strong> semigroup. He solved the problem for<br />
subtracti<strong>on</strong> algebras of a special type, called the atomic subtracti<strong>on</strong> algebras.<br />
In this paper, we study the Bosbach state <strong>on</strong> subtracti<strong>on</strong> algebra and show<br />
that there exists a Bosbach state via X/θ where θ is a c<strong>on</strong>gruence relati<strong>on</strong><br />
induced by an ideal of X.<br />
2 Subtracti<strong>on</strong> algebras<br />
By a subtracti<strong>on</strong> algebra we mean an algebra (X; −) with a single binary<br />
operati<strong>on</strong> “−” that satisfies the following identities: for any x, y, z ∈ X,<br />
(S1) x − (y − x) =x;<br />
(S2) x − (x − y) =y − (y − x);<br />
(S3) (x − y) − z =(x − z) − y.<br />
The last identity permits us to omit parentheses in expressi<strong>on</strong>s of the form<br />
(x − y) − z. The subtracti<strong>on</strong> determines an order relati<strong>on</strong> <strong>on</strong> X: a ≤ b ⇔<br />
a − b =0, where 0 = a − a is an element that does not depend <strong>on</strong> the choice<br />
of a ∈ X. The ordered set (X; ≤) is a semi-Boolean algebra in the sense of<br />
[6], that is, it is a meet semilattice with zero 0 in which every interval [0,a]is<br />
a Boolean algebra with respect to the induced order. Here a ∧ b = a − (a − b);<br />
the complement of an element b ∈ [0,a]isa − b; and if b, c ∈ [0,a], then<br />
b ∨ c =(b ′ ∧ c ′ ) ′ = a − ((a − b) ∧ (a − c))<br />
= a − ((a − b) − ((a − b) − (a − c))).<br />
In a subtracti<strong>on</strong> algebra, the following are true: for all x, y, z ∈ X,<br />
(p1) (x − y) − y = x − y.<br />
(p2) x − 0=x and 0 − x =0.<br />
(p3) x − y ≤ x.<br />
(p4) x − (x − y) ≤ y.<br />
(p5) (x − y) − (y − x) =x − y.<br />
(p6) x − (x − (x − y)) = x − y.<br />
(p7) (x − y) − (z − y) ≤ x − z.<br />
(p8) x ≤ y if and <strong>on</strong>ly if x = y − w for some w ∈ X.
<str<strong>on</strong>g>States</str<strong>on</strong>g> <strong>on</strong> subtracti<strong>on</strong> algebras 1157<br />
(p9) x ≤ y implies x − z ≤ y − z and z − y ≤ z − x for all z ∈ X.<br />
(p10) x, y ≤ z implies x − y = x ∧ (z − y).<br />
(p11) (x ∧ y) − (x ∧ z) ≤ x ∧ (y − z).<br />
(p12) (x − y) − z =(x − z) − (y − z).<br />
A n<strong>on</strong>-empty subset I of a subtracti<strong>on</strong> algebra X is called a subalgebra if<br />
x − y ∈ I for all x, y ∈ I. A mapping f from a subtracti<strong>on</strong> algebra X to a<br />
subtracti<strong>on</strong> algebra Y is called a homomorphism if f(x − y) =f(x) − f(y)<br />
for all x, y ∈ X. A homomorphism f from a subtracti<strong>on</strong> algebra X to itself<br />
is called an endomorphism of X. Ifx ≤ y implies f(x) ≤ f(y), fis called an<br />
isot<strong>on</strong>e map.<br />
A n<strong>on</strong>empty subset I of a subtracti<strong>on</strong> algebra X is called an ideal of X if<br />
it satisfies the following:<br />
(I1) 0 ∈ I,<br />
(I2) for any x, y ∈ X, y ∈ I and x − y ∈ I imply x ∈ I.<br />
Lemma 2.1. Let (X; −) be a subtracti<strong>on</strong> algebra and I a n<strong>on</strong>empty subset<br />
of X. Then I is an ideal of X if and <strong>on</strong>ly if it satisfies the following properties:<br />
(1) x ∈ I and y ≤ x imply y ∈ I,<br />
(2) if x, y ∈ I and x ∨ y exists, then x ∨ y ∈ I.<br />
3 <str<strong>on</strong>g>States</str<strong>on</strong>g> <strong>on</strong> subtracti<strong>on</strong> algebras<br />
Definiti<strong>on</strong> 3.1. Let X be a subtracti<strong>on</strong> algebra. A Bosbach state <strong>on</strong> X is a<br />
functi<strong>on</strong> s : X → [0, 1] such that the following axioms hold:<br />
(i) s(0) = 0,<br />
(ii) s(x)+s(y − x) =s(y)+s(x − y) for all x, y ∈ X.<br />
For a Bosbach state s : X → [0, 1], define a set Ker(s) by<br />
Ker(s) ={x ∈ X | s(x) =0}.<br />
A functi<strong>on</strong> s : X → [0, 1] is called a trivial Bosbach state <strong>on</strong> X if s(x) = 0 for<br />
all x ∈ X. Also, the functi<strong>on</strong><br />
{<br />
0 if x =0,<br />
s(x) =<br />
t otherwise .<br />
is called a c<strong>on</strong>stant Bosbach state for some t ∈ [0, 1].
1158 Sang Mo<strong>on</strong> Lee and Kyung Ho Kim<br />
Example 3.2.<br />
Let X = {0,a,b} be a set in which “−” is defined by<br />
− 0 a b<br />
0 0 0 0<br />
a a 0 a<br />
b b b 0<br />
Define a functi<strong>on</strong> s : X → [0, 1] by<br />
Then s is a Bosbach state <strong>on</strong> X.<br />
s(0) = 0,s(a) =0.3,s(b) =0.7.<br />
Example 3.3.<br />
Let X = {0,x,y,1} be a set in which “−” is defined by<br />
Define a functi<strong>on</strong> s : X → [0, 1] by<br />
Then s is a Bosbach state <strong>on</strong> X.<br />
− 0 x y 1<br />
0 0 0 0 0<br />
x x 0 x 0<br />
y y y 0 0<br />
1 1 y x 0<br />
s(0) = 0,s(x) =0.3,s(y) =0.5,s(1) = 0.8.<br />
Propositi<strong>on</strong> 3.4. If s : X → [0, 1] is a Bosbach state <strong>on</strong> X, s a : X →<br />
[0, 1],s a (x) =s(x − a) for every x ∈ X is also a Bosbach state <strong>on</strong> X.<br />
Proof.<br />
Let X be a subtracti<strong>on</strong> algebra and a ∈ X. Then we have<br />
s a (0) = s(0 − a) =s(0) = 0.<br />
Also, we have for all x, y ∈ X,<br />
s a (x)+s a (y − x) =s(x − a)+s((y − x) − a))<br />
= s(x − a)+s((y − a)+(x − a))<br />
= s(y − a)+s((x − a)+(y − a))<br />
= s(y − a)+s((x − y) − a)<br />
= s a (y)+s a (x − y).<br />
This completes the proof.
<str<strong>on</strong>g>States</str<strong>on</strong>g> <strong>on</strong> subtracti<strong>on</strong> algebras 1159<br />
Propositi<strong>on</strong> 3.5. Let s : X → [0, 1] be a Bosbach state <strong>on</strong> X. Then If x ≤ y,<br />
then s(x) ≤ s(y) and s(x)+s(y − x) =s(y).<br />
Proof. Let s be a Bosbach state <strong>on</strong> X and x ≤ y. Then x − y =0. Since<br />
s(x)+s(y−x) =s(y)+s(x−y), we have 0 ≤ s(y−x) =s(x−y)+s(y)−s(x)=<br />
0+s(y) − s(x) =s(y) − s(x), and so s(y) ≥ s(x).<br />
Propositi<strong>on</strong> 3.6. Let s : X → [0, 1] be a Bosbach state <strong>on</strong> X. Then Ker(s)<br />
is a subalgebra of X.<br />
Proof. Let x, y ∈ Ker(s). Then s(x) =s(y) =0. Since x − y ≤ x, we have<br />
s(x − y) ≤ s(x) by Propositi<strong>on</strong> 3.5, and so 0 ≤ s(x − y) ≤ s(x) =0. This<br />
implies s(x − y) =0, i.e., x − y ∈ Ker(s).<br />
Propositi<strong>on</strong> 3.7.<br />
is an ideal of X.<br />
Let s : X → [0, 1] be a Bosbach state <strong>on</strong> X. Then Ker(s)<br />
Proof. Since s(0) = 0, we have 0 ∈ Ker(s). Let x − y ∈ Ker(s) and y ∈<br />
Ker(s). So, s(x − y) =s(y) =0. Since y − x ≤ y, we have s(y − x) ≤ s(y) by<br />
Propositi<strong>on</strong> 3.5, which implies 0 ≤ s(x − y) ≤ s(y) =0. Hence s(y − x) =0.<br />
Now, since s(x)+s(y − x) =s(y)+s(x − y), we obtain s(x)+0=0+0. This<br />
implies s(x) =0, i.e., x ∈ Ker(s).<br />
Theorem 3.8. Let f : X → Y be a homomorphism and s : Y → [0, 1] be a<br />
Bosbach state <strong>on</strong> Y. Then there exists an unique Bosbach state t : X → [0, 1]<br />
such that the following diagram is commutative (i.e., t = s ◦ f).<br />
X f <br />
Y<br />
<br />
s<br />
∃t <br />
[0, 1]<br />
Proof. Let t := s ◦ f. It is clear that t is well-defined. Now we show that t is<br />
a Bosbach state. We know that t(0) = (s ◦ f)(0) = s(f(0)) = s(0) = 0. Now,
1160 Sang Mo<strong>on</strong> Lee and Kyung Ho Kim<br />
let x, y ∈ X. Then<br />
t(x)+t(y − x) =(s ◦ f)(x)+(s ◦ f)(y − x)<br />
= s(f(x)) + s(f(y) − f(x))<br />
= s(f(y)) + s(f(x) − f(y))<br />
= s(f(y)) + s(f(x − y))<br />
=(s ◦ f)(y)+(s ◦ f)(x − y)<br />
= t(y)+t(x − y).<br />
Hence t is a Bosbach state. Finally, we prove that t is unique. Suppose that<br />
there exists an another Bosbach state r : X → [0, 1] such that r = s ◦ f. Then<br />
r(x) =(s ◦ f)(x) =t(x) for all x ∈ X. Hence we have r = t.<br />
Theorem 3.9. Let f : X → Y be a bijecti<strong>on</strong> homomorphism and s : X →<br />
[0, 1] be a Bosbach state <strong>on</strong> X. Then there exists an unique Bosbach state<br />
t : Y → [0, 1] such that s = t ◦ f.<br />
X f Y<br />
s<br />
<br />
∃t<br />
[0, 1]<br />
Proof. Let y ∈ Y. Since f is <strong>on</strong>to, there exists x ∈ X such that f(x) =y. Put<br />
t(y) =s(x). Then t(y) =t(f(x)) = (t◦f)(x) =s(x), and so t◦f = s. Since f is<br />
<strong>on</strong>e to <strong>on</strong>e, then Ker(f) =0, and so t(0) = t(f(0)) = (t◦f)(0) = s(0) = 0. Let<br />
y 1 ,y 2 ∈ Y. Then there exists x 1 ,x 2 ∈ X such that f(x 1 )=y 1 and f(x 2 )=y 2 .<br />
Then we have<br />
t(y 1 )+t(y 2 − y 1 )=t(f(x 1 )+t(f(x 2 ) − f(x 1 ))<br />
= t(f(x 1 )+t(f(x 2 − x 1 ))<br />
=(t ◦ f)(x 1 )+(t ◦ f)(x 2 − x 1 )<br />
= s(x 1 )+s(x 2 − x 1 )<br />
= s(x 2 )+s(x 1 − x 2 )<br />
=(t ◦ f)(x 2 )+(t ◦ f)(x 1 − x 2 )<br />
=(t(f(x 2 )+t(f(x 1 − x 2 ))<br />
= t(f(x 2 )) + t(f(x 1 ) − f(x 2 ))<br />
= t(y 2 )+t(y 1 − y 2 ).<br />
Hence t is a Bosbach state. Suppose that there exists an another Bosbach<br />
state r : X → [0, 1] such that s = r ◦ f. Let y ∈ Y. Then there exists x ∈ X
<str<strong>on</strong>g>States</str<strong>on</strong>g> <strong>on</strong> subtracti<strong>on</strong> algebras 1161<br />
such that f(x) =y. Now, r(y) =r(f(x)) = (r ◦ f)(x) =s(x), but by definiti<strong>on</strong><br />
of t, we get r(y) =s(x) =t(y). Hence we have r = t.<br />
Let I be an ideal of X. C<strong>on</strong>sider relati<strong>on</strong> θ <strong>on</strong> X by<br />
(x, y) ∈ θ ⇔ x − y ∈ I and y − x ∈ I.<br />
Then θ is a c<strong>on</strong>gruence relati<strong>on</strong> <strong>on</strong> X. For x ∈ X, we denote by [x] the c<strong>on</strong>gruence<br />
class of x and let<br />
X/θ = {[x] | x ∈ X}.<br />
Then X/θ is a subtracti<strong>on</strong> algebra where [x]−[y] =[x−y] for all x, y ∈ X. Also,<br />
we denote p θ : X → X/θ by p θ (x) =[x]. If x ∈ I, we have [x] = [0] = 0. We<br />
introduce a relati<strong>on</strong> “ ≤ ”<strong>on</strong>X/θ by [x] ≤ [y] if and <strong>on</strong>ly if [x] − [y] = [0] = 0.<br />
Propositi<strong>on</strong> 3.10. Let s : X → [0, 1] be a Bosbach state <strong>on</strong> X. Then<br />
(i) [x] ≤ [y] ⇔ s(x − y) =0.<br />
(ii) [x] =[y] ⇔ s(x − y) =s(y − x) =0.<br />
Proof. It is easy to prove from Propositi<strong>on</strong> 3.5.<br />
Theorem 3.11. Let I be an ideal of X and s : X → [0, 1] be a Bosbach state<br />
<strong>on</strong> X. Then there exists an unique Bosbach state t : X/θ → [0, 1] such that the<br />
following diagram is commutative (i.e., s = t ◦ p θ ), in fact, θ is a c<strong>on</strong>gruence<br />
relati<strong>on</strong> induced by ideal I.<br />
X p θ<br />
<br />
X/θ<br />
<br />
s<br />
∃t<br />
<br />
[0, 1]<br />
Corollary 3.12. Let X be a subtracti<strong>on</strong> algebra and s : X → [0, 1] be a<br />
Bosbach state <strong>on</strong> X. Then there exists an unique Bosbach state t : X/θ → [0, 1]<br />
such that the following diagram is commutative (i.e., s = t ◦ p θ ), in fact, θ is<br />
a c<strong>on</strong>gruence relati<strong>on</strong> induced by ideal Ker(s).<br />
X p θ<br />
<br />
X/θ<br />
<br />
s<br />
∃t<br />
<br />
[0, 1]
1162 Sang Mo<strong>on</strong> Lee and Kyung Ho Kim<br />
References<br />
[1] J. C. Abbott, Sets, Lattices and Boolean <strong>Algebras</strong>, Allyn and Bac<strong>on</strong>,<br />
Bost<strong>on</strong> 1969.<br />
[2] J. C. Abbott, <strong>Algebras</strong> of implicati<strong>on</strong> and semi-lattices, Séminarire<br />
Dubreil (Algèbre et théorie des nombres) 20(2) (1966-1967), exp. n 0 20,<br />
1–8.<br />
[3] B. M. Schein, Difference Semigroups, Comm. in Algebra 20 (1992), 2153–<br />
2169.<br />
[4] B. Zelinka, Subtracti<strong>on</strong> Semigroups, Math. Bohemica, 120 (1995), 445–<br />
447.<br />
Received: April 21, 2013