Generated Sets of the Complete Semigroup Binary Relations Defined

Applied Mathematics, 2018, 9, 369-382 http://www.scirp.org/journal/am ISSN Online: 2152-7393 ISSN Print: 2152-7385

Generated Sets of the Complete Semigroup Binary Relations Defined by Semilattices of the Class Σ 8 ( X , n + k + 1) Yasha Diasamidze, Omari Givradze, Nino Tsinaridze, Giuli Tavdgiridze Faculty of Mathematics, Physics and Computer Sciences, Shota Rustaveli State University, Batumi, Georgia Email:

How to cite this paper: Diasamidze, Y., Givradze, O., Tsinaridze, N. and Tavdgiridze, G. (2018) Generated Sets of the Complete Semigroup Binary Relations Defined by Semilattices of the Class

Abstract

369-382. https://doi.org/10.4236/am.2018.94028

Keywords

Σ8 ( X , n + k + 1) . Applied Mathematics, 9,

find uniquely irreducible generating set for the given semigroups.

Semigroup, Semilattice, Binary Relation

Received: March 22, 2018 Accepted: April 24, 2018 Published: April 27, 2018 Copyright © 2018 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0). http://creativecommons.org/licenses/by/4.0/ Open Access

In this article, we study generated sets of the complete semigroups of binary relations defined by X-semilattices unions of the class Σ8 ( X , n + k + 1) , and

1. Introduction Let X be an arbitrary nonempty set, D is an X-semilattice of unions which is closed with respect to the set-theoretic union of elements from D, f be an arbitrary mapping of the set X in the set D. To each mapping f we put into corres-

pondence a binary relation α f on the set X that satisfies the condition = α f  ({ x} × f ( x ) ) . The set of all such α f ( f : X → D ) is denoted by x∈X

BX ( D ) . It is easy to prove that BX ( D ) is a semigroup with respect to the operation of multiplication of binary relations, which is called a complete semigroup of binary relations defined by an X-semilattice of unions D. We denote by ∅ an empty binary relation or an empty subset of the set X. The condition ( x, y ) ∈ α will be written in the form xα y . Further, let x, y ∈ X ,  Y ⊆ X , α ∈ BX ( D ) , D =  Y and T ∈ D . We denote by the symbols Y ∈D

yα , Y α , V ( D, α ) , X ∗ and V ( X ∗ , α ) the following sets: yα = {Y α | Y ∈ D} , { x ∈ X | yα x} , Y α =  yα , V ( D, α ) = y∈Y

∗

X= DOI: 10.4236/am.2018.94028

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DT = T }. {Z ∈ D | T ⊆ Z } , YTα = { y ∈ X | yα = It is well known the following statements:  Theorem 1.1. Let D = D, Z1 , Z 2 , , Z m−1

{

}

be some finite X-semilattice of unions and C ( D ) = { P0 , P1 , P2 , , Pm−1} be the family of sets of pairwise nonintersecting subsets of the set X (the set ∅ can be repeated several times). If ϕ is a mapping of the semilattice D on the family of sets C ( D ) which satisfies the conditions   D Z1 Z 2  Z m−1  ϕ =    P0 P1 P2  Pm−1  and Dˆ Z = D \ DZ , then the following equalities are valid:  (1.1) D = P0 ∪ P1 ∪ P2 ∪  ∪ Pm−1 , Z i = P0 ∪  ϕ (T ). T ∈Dˆ Zi

In the sequel these equalities will be called formal. It is proved that if the elements of the semilattice D are represented in the form (1.1), then among the parameters Pi ( 0 < i ≤ m − 1) there exist such parameters that cannot be empty sets for D. Such sets Pi are called bases sources, where sets Pj ( 0 ≤ j ≤ m − 1) , which can be empty sets too are called completeness sources. It is proved that under the mapping ϕ the number of covering elements of the pre-image of a bases source is always equal to one, while under the mapping ϕ the number of covering elements of the pre-image of a completeness source either does not exist or is always greater than one (see [1] [2] chapter 11). Definition 1.1. We say that an element α of the semigroup BX ( D ) is external if α ≠ δ  β for all δ , β ∈ BX ( D ) \ {α } (see [1] [2] Definition 1.15.1). It is well known, that if B is all external elements of the semigroup BX ( D ) and B′ is any generated set for the BX ( D ) , then B ⊆ B′ (see [1] [2] Lemma 1.15.1).

= α Definition 1.2. The representation

called quasinormal , if

 YTα = X

T ∈D

 (YTα × T )

T ∈D

of binary relation α is

∅ for any T , T ′ ∈ D , and YTα ∩ YTα′ =

T ≠ T ′ (see [1] [2] chapter 1.11). Definition 1.3. Let α , β ⊆ X × X . Their product δ = α  β is defined as follows: xδ y ( x, y ∈ X ) if there exists an element z ∈ X such that xα z β y (see [1], chapter 1.3).

2. Result Let Σ8 ( X , n + k + 1)

(3 ≤ k ≤ n)

be a class of all X-semilattices of unions

whose every element is isomorphic to an X-semilattice of unions  D = Z1 , Z 2 , , Z n+k , D , which satisfies the condition:   Z n+i ⊂ Z i ⊂ D= , ( i 1, 2, , k ) ; Z j ⊂ D,= ( j 1, 2, , n + k ) ; (see Figure 1). Z p \ Z q ≠ ∅ and Z q \ Z p ≠ ∅ (1 ≤ p ≠ q ≤ n + k ) .

{

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}

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Figure 1. Diagram of the semilattice D.

It is easy to see that D = {Z1 , Z 2 , , Z n+k } is irreducible generating set of the semilattice D. Let C ( D ) = { P0 , P1 , P2 , , Pn+k } be a family of sets, where P0 , P1 , P2 , , Pn+k   D Z1 Z 2  Z n+k  are pairwise disjoint subsets of the set X and ϕ =   is a map  P0 P1 P2  Pn+k  ping of the semilattice D onto the family of sets C ( D ) . Then the formal equalities of the semilattice D have a form:

 D =

n+k

n+k

Pi , j  P= = i; Z j

=i 0=i 0, i≠ j

1, 2, , n= ; Zn+q

n+k

Pi , q =

=i 0, i ≠ q,n+ q

1, 2, , k .

(2.0)

= Pi ( i 1, 2, , n + k ) are bases sources, the element P0 Here the elements are sources of completeness of the semilattice D. Therefore X ≥ n + k (by 1( i 1, 2, , n + k ) symbol X we denoted the power of a set X), since Pi ≥= (see [1] [2] chapter 11). In this paper we are learning irreducible generating sets of the semigroup BX ( D ) defined by semilattices of the class Σ8 ( X , n + k + 1) . Note, that it is well known, when k = 2 , then generated sets of the complete semigroup of binary relations defined by semilattices of the class Σ8 ( X , 2 + 2 + 1) =Σ8 ( X ,5 ) . In this paper we suppose, that 3 ≤ k ≤ n . Remark, that in this case (i.e. k ≥ 3 ), from the formal equalities of a semilattice D follows, that the intersections of any two elements of a semilattice D is not empty. Lemma 2.0 If D ∈ Σ8 ( X , n + k + 1) , then the following statements are true: a)

n+k

 Zi = P0 ; i =1

\ Z j P= b) Z j +1= j , j 1, 2, , n − 1; \ Z n+q P= c) Z q = n+ q , q 1, , k . Proof. From the formal equalities of the semilattise D immediately follows the following statements:

 n+k   n+k  , \ 1, 2, , n − 1; Z P Z Z Pi  P= = =  i 0 j+1 j   Pi  \  = j, j  =i 1 =  i≠i j 0,+1=   ii≠0,j       n+k   n+ k  \ Z n+q   Pi  \   , q 1, , k . Z q= Pi  P= =  i 0,=    n+q =  i≠q   i≠qi ,n0,+q      n+k

The statements a), b) and c) of the lemma 2.0 are proved. DOI: 10.4236/am.2018.94028

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Lemma 2.0 is proved. We denoted the following sets by symbols D1 , D2 and D3 :

= D1

Z1 , Z 2 , , Z k } , D2 {Z k += {= {Z n+1 , Z n+2 , , Z n+k } . 1 , Z k + 2 , , Z n } , D3

Lemma 2.1. Let D ∈ Σ8.0 ( X , n + k + 1) and α ∈ BX ( D ) . Then the following

statements are true: 1) Let T , T ′ ∈ D2 ∪ D3 , T ≠ T ′ . If T , T ′ ∈ V ( D, α ) , then α is external element of the semigroup BX ( D ) ; 2) Let T ∈ D1 , T ′ ∈ D2 ∪ D3 . If T ′ ⊄ T and T , T ′ ∈ V ( D, α ) , then α is external element of the semigroup BX ( D ) . 3) Let T , T ′ ∈ D1 and T ≠ T ′ . If T , T ′ ∈ V ( D, α ) and k ≥ 3 , then α is external element of the semigroup BX ( D ) ;  Proof. Let Z 0 = D and α = δ  β for some δ , β ∈ BX ( D ) \ {α } . If quasinormal representation of binary relation δ has a form

 (YTδ × T ) ,

= δ

T ∈V ( D ,δ )

then

= α δ= β

 (YTδ × T β ) .

(2.1)

T ∈V ( D ,δ )

From the formal equalities (2.0) of the semilattice D we obtain that: n+k

n+k

= Pi β , j ; Z jβ   Pi β=

= Z0 β

=i 0 =i 0, i≠ j

= Z n+q β

1, 2, , n; (2.2)

n+k

=  Pi β , q 1, 2, , k.

i =0, i ≠ q ,n + q

where Pi β ≠ ∅ for any Pi = ≠ ∅ ( i 0,1, 2, , n + k ) and β ∈ BX ( D ) by definition of a semilattice D from the class Σ8.0 ( X , n + k + 1) . Now, let Z m β = T and Z j β = T ′ for some T ≠ T ′ , T , T ′ ∈ D2 ∪ D3 , then from the equalities (2.3) follows that = T P= T ′ since T and T ′ are minim0β al elements of the semilattice D and P0 ≠ ∅ by preposition. The equality T = T ′ contradicts the inequality T ≠ T ′ . The statement a) of the Lemma 2.1 is proved. Now, let Z m β = T and Z j β = T ′ , for some T ∈ D1 , T ′ ∈ D2 ∪ D3 and T ′ ⊄ T , then from the equalities 2.3 follows, that

= T ′ Z= Z= jβ 0β = T ′ Z= n+q β

n+k



i =0, i ≠ q ,n + q

n+k

 Pi β i =0

, if

j = 0 , or = T ′ Z= jβ

n+k

 Pi β

i =0, i≠ j

, 1 ≤ j ≤ n , or

Pi β

where j= n + q . For the Z j β = T ′ we consider the following cases:

T ′ Z= 1) If = 0β

n+k

 Pi β , then we have i =0

P0 β= P1β=  = Pn+k β= T ′ , since T ′ is a minimal element of a semilattice D. On the other hand, DOI: 10.4236/am.2018.94028

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But the equality T = T ′ contradicts the inequality T ≠ T ′ . Thus we have, that j ≠ 0 .

T ′ Z= 2) Let 1 ≤ j ≤ n , i.e. = jβ

n+k

 Pi β , then we have, that

i =0, i≠ j

= Pj −1β= Pj +1β=  = Pn+k β= T ′ , P0 β= P1β=  since T ′ is a minimal element of a semilattice D. On the other hand:  n+k  n+k   P β  = P β  ∪ P β = T ′ ∪ Pj β , if m =0; i  j    i0, i  i 0= =      i≠ j      n+k  n + k      Pi β  =   Pi β  ∪ Pj β = T ′ ∪ Pj β , if 1 ≤ m ≤ n, m ≠ j; 0, 0, i i = =   i ≠ m, j   i ≠ m     T Z= =  mβ    n+k   Pi β = T ′, if m= n + j;   i = 0,  i ≠ j ,n+ j      n+k   n + k   Pi β  = Pi β  ∪ Pj β = T ′ ∪ Pj β , if m =+ n q , q ≠ j.     i = 0,   i = 0,      i ≠ q,n+ q, j   i ≠ q , n + q

The equality T = T ′ contradicts the inequality T ≠ T ′ . Also, the equality T= T ′ ∪ Pj β ( Pj β ∈ D ) contradicts the inequality T ≠ T ′ ∪ Z for any Z ∈ D and T ′ ⊄ T ( T ′ ⊄ T , by preposition) by definition of a semilattice D. 3) If j= n + q

T′ (1 ≤ q ≤ k ) , i.e.=

Z= n+q β

n+k



i =0, i ≠ q ,n + q

Pi β , then we have, that

P0 β= P1β=  = Pq−1β= Pq+1β=  = Pn+q−1β= Pn+q+1β=  = Pn+k β= T ′ , since T ′ is a minimal element of a semilattice D. On the other hand:

 n+k  n+k   P β  =  β P   i    i  ∪ Pq β ∪ Pn + q β =T ′ ∪ Pq β ∪ Pn + q β , if m =0;  =   i ≠ qi , n0,+ q   i 0,=      n+k  + n k       Pi β  =  Pi β  ∪ Pn + q β =T ′ ∪ Pn + q β , if 1 ≤ m = q ≤ n; =   i ≠ qi , n0,+ q   ii ≠ 0,q=     = T Z=  n + k    n+k mβ  =  T ′ ∪ Pq β ∪ Pn + q β , if 1 ≤ m ≠ q ≤ n; P β P β i  i  ∪ Pq β ∪ Pn + q β =  i   = 0, = i 0,      i ≠ m    i ≠ m,q ,n + q       n+k n+k  n + k =  =   P β P β P β T ∪ Pq β ∪ Pn + q ,    i    i   i  ∪ Pq β ∪ Pn + q = = = i 0, i 0,       ii ≠ 0,m= ,   i ≠ q ′, n + q ′   i ≠ , q ′, n + q ′, q , n + q   ′ + ≤ = + ≤ + k , q ≠ q ′ since j ≠ m. if n 1 m n q n  DOI: 10.4236/am.2018.94028

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The equality T = T ′ contradicts the inequality T ≠ T ′ . Also, the equality T =T ′ ∪ Pq β ∪ Pn+q β , or T= T ′ ∪ Pn+q β Pq β , Pn+q β ∈ D contradicts the inequality T ≠ T ′ ∪ Z for any Z ∈ D and T ′ ⊄ T by definition of a semilattice

(

)

D. The statement 2) of the Lemma 2.1 is proved.

Let T , T ′ ∈ D1 and T ≠ T ′ . If k ≥ 3 and Z j β = T ′ , Z m β = T , then from the formal equalities (2.0) of a semilattice D there exists such an element, that

Pq ⊆ Z j and Pq ⊆ Z m , where 0 ≤ q ≤ m + k . So, from the equalities (2.3) folT . Of from this and from the T ′ and Pq β ⊆ Z m β = lows that Pq β ⊆ Z j β = equalities (2.3) we obtain that there exists such an element Z ∈ D , for which the equalities T =′ Z ∪ Z ′ and T= Z ∪ Z ′′ , where Z ′, Z ′′ ∈ D . But such elements by definition of a semilattice D do not exist. The statement c) of the Lemma 2.1 is proved. Lemma 2.1 is proved. Lemma 2.2. Let D ∈ Σ8 ( X , n + k + 1) and α ∈ BX ( D ) . Then the following

statements are true: ∅,V ( D, α ) ∩ D3 = ∅ . If 1) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 = V ( D, α ) ∩ D1 ≥ 2 , then α is external element of the semigroup BX ( D ) ; ∅,V ( D, α ) ∩ D2 ≠ ∅,V ( D, α ) ∩ D3 = ∅ . If 2) Let V ( D, α ) ∩ D1 = V ( D, α ) ∩ D2 ≥ 2 , then α is external element of the semigroup BX ( D ) ; ∅,V ( D, α ) ∩ D2 = ∅,V ( D, α ) ∩ D3 ≠ ∅ . If 3) Let V ( D, α ) ∩ D1 = V ( D, α ) ∩ D3 ≥ 2 , then α is external element of the semigroup BX ( D ) ; 4) Let V ( D, α ) ∩ D1 = ∅, V ( D, α ) ∩ D2 ≠ ∅, V ( D, α ) ∩ D3 ≠ ∅ , then α is external element of the semigroup BX ( D ) ; 5) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 = ∅, V ( D, α ) ∩ D3 ≠ ∅ . If 1, 1 , or V ( D, α ) ∩ D1 = V ( D, α ) ∩ D1 ≥ 2 , V ( D, α ) ∩ D3 = V ( D, α ) ∩ D3 ≥ 2 then α is external element of the semigroup BX ( D ) ; 6) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 ≠ ∅, V ( D, α ) ∩ D3 = ∅ , then α is external element of the semigroup BX ( D ) ; 7) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 ≠ ∅, V ( D, α ) ∩ D3 ≠ ∅ , then α is external element of the semigroup BX ( D ) . Proof. Let α be any element of the semigroup BX ( D ) . It is easy that V ( D, α ) ∈ D . We consider the following cases: ∅,V ( D, α ) ∩ D2 = ∅,V ( D, α ) ∩ D3 = ∅ , then Let V ( D, α ) ∩ D1 =  V ( D, α ) ∈ { D} since V ( D, α ) is subsemilattice of the semilattice D. ∅,V ( D, α ) ∩ D3 = ∅. 1) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 =  V D , α ∩ D = 1 V D , α ∈ Z V D , α ∈ If ( ) 1 , then ( ) { j } , or ( ) {Z j , D} , where j = 1, 2, , k , since V ( D, α ) is subsemilattice of the semilattice D. If V ( D, α ) ∩ D1 ≥ 2 , then by statement c) of the Lemma 2.1 follows that α is external element of the semigroup BX ( D ) . ∅,V ( D, α ) ∩ D2 ≠ ∅,V ( D, α ) ∩ D3 = ∅. 2) Let V ( D, α ) ∩ D1 =  1 , then V ( D, α ) ∈ {Z j } , or V ( D, α ) ∈ {Z j , D} , where If V ( D, α ) ∩ D2 = j =k + 1, k + 2, , n , since V ( D, α ) is a subsemilattice of the semilattice D. DOI: 10.4236/am.2018.94028

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If V ( D, α ) ∩ D2 ≥ 2 , then by statement a) of the Lemma 2.1 follows that α is external element of the semigroup BX ( D ) .

∅, V ( D, α ) ∩ D2 = ∅,V ( D, α ) ∩ D3 ≠ ∅ . 3) Let V ( D, α ) ∩ D1 =  1 , then V ( D, α ) ∈ Z j , or V ( D, α ) ∈ Z j , D If V ( D, α ) ∩ D3 =

{

{ }

}

,

j =n + 1, n + 2, , n + k , since V ( D, α ) is subsemilattice of the semilattice D. If V ( D, α ) ∩ D3 ≥ 2 , then by statement a) of the Lemma 2.1 follows that α is external element of the semigroup BX ( D ) . 4) Let V ( D, α ) ∩ D1 = ∅, V ( D, α ) ∩ D2 ≠ ∅, V ( D, α ) ∩ D3 ≠ ∅ , then by the

statement a) of the Lemma 2.1 follows that α is external element of the semi-

group BX ( D ) .

5) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 = ∅, V ( D, α ) ∩ D3 ≠ ∅ .

∩ D1 1, V ( D, α ) = ∩ D3 1 , then V ( D, α ) = {Z n + q , Z q } , or V ( D, α ) =   V ( D, α ) = Z n+q , Zq , D , or V ( D, α ) = Z n+q , Z j , D where Z1 ≤ Z j ≤ Z k and If

{

}

q = 1, 2, , k .

{

 If V ( D, α ) = Z n+q , Z j , D

{

}

}

where j ≠ q, q = 1, 2, , k , then by the statement

2) of the Lemma 2.1 follows that α is external element of the semigroup

BX ( D ) ; ∩ D1 1, V ( D, α ) ∩ D3 ≥ 2 , or If V ( D, α ) = V ( D, α ) ∩ D1 ≥ 2, V ( D, α ) ∩ D3 = 1 , then from the statement 1) and 3) of the Lemma 2.1 follows that α is external element of the semigroup BX ( D ) respectively. 6) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 ≠ ∅, V ( D, α ) ∩ D3 = ∅ . Then from the statement b) of the Lemma 2.1 follows that α is external element of the semigroup BX ( D ) . 7) Let V ( D, α ) ∩ D1 ≠ ∅, V ( D, α ) ∩ D2 ≠ ∅, V ( D, α ) ∩ D3 ≠ ∅ , then by the statement a) of the Lemma 2.1 follows that α is external element of the semigroup BX ( D ) . Lemma 2.2 is proved. Now we learn the following subsemilattices of the semilattice D:  = Z n+ j , Z j , D} , where j 1, 2, , k ; A1 {=  j 1, 2, , n + k ; = = A 2 {Z j , D} , where = A3 = A4

{ } { } Z , Z }} , where j {{=  = j {{Z } ,{D}} , where n+ j

j

j

1, 2, , k ; 1, 2, , n + k .

We denoted the following sets by symbols A 0 and B ( A 0 ) :

{V ( D, α ) ⊆ D | V ( D, α ) ∉ A1 ∪ A 2 ∪ A3 ∪ A 4 } , B ( A0 ) = {α ∈ BX ( D ) | V ( D, α ) ∈ A0 }.

= A0

By definition of a set B ( A 0 ) follows that any element of the set is external element of the semigroup BX ( D ) . Lemma 2.3. Let D ∈ Σ8 ( X , n + k + 1) . If quasinormal representation of a bi-

nary relation α has a form DOI: 10.4236/am.2018.94028

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 α = (Ynα+ j × Z n+ j ) ∪ (Y jα × Z j ) ∪ (Y0α × D ) ,

where Ynα+ j , Y jα , Y0α ∉ {∅} and j = 1, 2, , k , then α is generated by elements of the elements of set B ( A 0 ) . Proof. 1). Let quasinormal representation of binary relations δ and β have a form



δ = (Ynδ+ j × Z n+ j ) ∪ (Y jδ × Z j ) ∪ (Yqδ × Z q ) ∪ (Y0δ × D ) , β=

(Z

n+ j

 × Z n+ j ) ∪ ( Z j \ Z n+ j ) × Z j ∪ D \ Z j × Z q ∪

) ((

(

)





) (( X \ D ) × D ) ,

α α α where Yn+ j , Y j , Yq ∉ {∅} , Z1 ≤ Z q ≤ Z k , q ≠ j , j =1, , k .   Z n+ j ∪ ( Z j \ Z n+ j ) ∪ D \ Z j ∪ X \ D

(

) (

)

  , n+k     = P0 ∪  Pi  ∪ Pn+ j ∪ Pj ∪ X \ D = D∪ X \D = X   i =1,   i ≠ j ,n + j  

(

)

(

)

since the representation of a binary relation β is quasinormal and by statement 3) of the Lemma 2.1 binary relations δ and β are external elements of the semigroup BX ( D ) . It is easy to see, that:

Z n+ j β = Z n+ j ,      n+k n+k  P β = ∪ P β ∪ Z j β =∪ P P P  i  n+ j  i  0  0 i = i 1, 1,=      i ≠ j i ≠ j ,n + j      = Z n+ j β ∪ Pn+ j β = Z n+ j ∪ Z j = Z j ,

  n+k   Z q β = P0 ∪  Pi  β = Z n+ j ∪ Z j ∪ Z q = D,   i =1,   i≠q   n+k   Dβ =  Pi β = Z n+ j ∪ Z j ∪ Z q = D i =0

Z q ∩ Z n + j ≠ ∅, Z q ∩ ( Z j \ Z n + j = ) Pn+ j ≠ ∅, Z q ∩ ( D \ Z j =) Pj ≠ ∅ equality (2.0))  α = δ  β = (Ynδ+ j × Z n+ j β ) ∪ (Y jδ × Z j β ) ∪ (Yqδ × Z q β ) ∪ (Y0δ × Dβ )   = (Ynδ+ j × Z n+ j ) ∪ (Y jδ × Z j ) ∪ (Yqδ × D ) ∪ (Y0δ × D )  = (Ynδ+ j × Z n+ j ) ∪ (Y jδ × Z j ) ∪ (Yqδ ∪ Y0δ ) × D = α ,

since

(

(see

)

δ α δ α δ δ Y0α . Last equalities are possible since if Yn+ j = Yn+ j , Y j = Y j and Yq ∪ Y0 = Yqδ ∪ Y0δ ≥ 1 ( Y0δ ≥ 0 , by preposition).

Lemma 2.3 is proved. Lemma 2.4. Let D ∈ Σ8 ( X , n + k + 1) . If quasinormal representation of a bi nary relation α has a form α = Y jα × Z j ∪ Y0α × D , where Y jα , Y0α ∉ {∅} ,

(

) (

)

= j 1, 2, , n + k , then binary relation α is generated by elements of the elements of set B ( A 0 ) . Proof. Let quasinormal representation of the binary relations δ and β have a DOI: 10.4236/am.2018.94028

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form:



δ = (Y jδ × Z j ) ∪ (Yqδ × Z q ) ∪ (Y0δ × D ) ,

(





) (



)

β = ( Z j × Z j ) ∪ ( D \ Z j ) × Zq ∪ ( X \ D ) × D , δ δ where Y j , Yq ∉ {∅} and Z1 ≤ Z j ≠ Z q ≤ Z n+k . Then from the statements a), b)

and c) of the Lemma 2.1 follows, that δ and β are generated by elements of the

set B ( A 0 ) and

Z jβ = Z j ,  Z q β = D , since Z q ∩ ( D \ Z j = ) Pj ≠ ∅,   Dβ = D; 

δ  β = (Y jδ × Z j β ) ∪ (Yqδ × Z q β ) ∪ (Y0δ × Dβ )   = Y jδ × Z j ∪ Yqδ × D ∪ Y0δ × D  = Y jδ × Z j ∪ Yqδ ∪ Y0δ × D = α ,

( (

) ( ) ((

) ( ) ) )

δ α δ δ Y0α = if Y j = Y j , Yq ∪ Y0 = and q 1, 2, , n + k . Last equalities are possible

since Yqδ ∪ Y0δ ≥ 1 ( Y0δ ≥ 0 by preposition). Lemma 2.4 is proved.

Lemma 2.5. Let D ∈ Σ8 ( X , n + k + 1) . If quasinormal representation of a bi-

nary relation α has a form α = (Ynα+ j × Z n+ j ) ∪ (Y jα × Z j ) , where Ynα+ j , Y jα ∉ {∅} , j = 1, 2, , k , then binary relation α is generated by elements of the elements of set B ( A 0 ) . Proof. Let quasinormal representation of a binary relations δ, β have a form 

δ = (Ynδ+ j × Z n+ j ) ∪ (Yqδ × Z q ) ∪ (Y0δ × D ) , β=

(Z

n+ j

 × Z n+ j ) ∪ D \ Z n+ j × Z j ∪

((





) (( X \ D ) × D ) ,

)

δ δ where Yn+ j , Yq ∉ {∅} , j ≠ q and j = 1, 2, , k . Then from the Lemma 2.2

follows that β is generated by elements of the set B ( A 0 ) , δ ∈ B ( A 0 ) and

Z n+ j β = Z n+ j , Z q β = Z n+ j ∪ Z j = Z j , since  ≠ ∅ ≠ Z q ∩ D \ Z n+= P , j q (see equality (2.0)) j n+ j    Dβ = Z j since D ∩ X \ D = ∅,

(

)

Z q ∩ Z n+ j ≠ ∅

,

( )  δ  β = (Y × Z β ) ∪ (Y × Z β ) ∪ (Y × D ) = (Y × Z ) ∪ (Y × Z ) ∪ (Y × Z ) = (Y × Z ) ∪ ( (Y ∪ Y ) × Z ) = α , δ

n+ j

δ

n+ j

δ

n+ j

δ

n+ j n+ j n+ j

δ

q

q

δ

0

δ

q

j

j

0

δ

δ

q

0

j

Y jα . Last equalities are possible since Ynδ+ j = Ynα+ j and Yqδ ∪ Y0δ = Yqδ ∪ Y0δ ≥ 1 ( Y0δ ≥ 0 by preposition).

if

Lemma 2.5 is proved.

Lemma 2.6. Let D ∈ Σ8 ( X , n + k + 1) . Then the following statements are true:

1) If quasinormal representation of a binary relation α has a form α= X × Z j DOI: 10.4236/am.2018.94028

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( j = 1, 2, , k ) B ( A0 ) .

, then binary relation α is generated by elements of the set

 2) If quasinormal representation of a binary relation α has a form α= X × D ,

then binary relation α is generated by elements of the set B ( A 0 ) .

Proof. 1) Let T ∈ D \ ( D2 ∪ D3 ) . If quasinormal representation of a binary re-

lations δ, β have a form



δ = (Y jδ × Z j ) ∪ (Y0δ × D ) , β=

(Z

n+ j





) (( X \ D ) × D ) ,

(

× Z n+ j ) ∪ ( Z j \ Z n+ j ) × Z j ∪

δ δ where Y j , Y0 ∈ {∅} , j = 1, 2, , k

 Z n+ j ∪ ( Z j \ Z n+ j ) ∪ X \ D

(

)

 n+k      ∪ P ∪ P ∪ X \ D = = P D∪ X \D = X ( j n+ j ) i    i≠ ij=,n0,+ j   

(

)

(

)

(see equalities (2.0) and (2.1)), then from the Lemma 2.4 follows that δ is generated by elements of the set B ( A 0 ) and from the Lemma 2.3 element β is gen-

erated by elements of the set B ( A 0 ) and

Z j β = Z n+ j ∪ Z j = Z j ,    Dβ = Z j , since D ∩ ( X \ D ) = ∅,  δ  β = (Y jδ × Z j β ) ∪ (Y0δ × Dβ ) = (Y jδ × Z j ) ∪ (Y0δ × Z j ) = X × Z j = α , since representation of a binary relation δ is quasinormal. The statement a) of the lemma 2.6 is proved. 2) Let quasinormal representation of a binary relation δ have a form  δ = ( Z n+ j × Z q ) ∪ ( X \ Z n+ j ) × D ,

)

(

where j ≠ q , then from the Lemma 2.4 follows that δ is generated by elements of the set B ( A 0 ) and  n+k   n+k      Z qδ =   Pi  δ =   Piδ  = Z q ∪ D = D, Dδ = D , since  i 0,=   i 0,  =  i≠q   i≠q      j ≠ q, Z qδ ∩ Z n+1 ≠ ∅ and Z qδ ∩ ( X \ Z n+1 ) ≠ ∅;  δ  δ = ( Z n+ j × Z qδ ) ∪ ( X \ Z n+ j ) × Dδ ,    = Z n+ j × D ∪ ( X \ Z n+ j ) × D = X × D = α

(

) (

(

)

)

since representation of a binary relation δ is quasinormal. The statement b) of the lemma 2.6 is proved. Lemma 2.6 is proved.

Lemma 2.7. Let D ∈ Σ8 ( X , n + k + 1) . Then the following statements are true:  a) If X \ D ≥ 1 and T ∈ D2 ∪ D3 , then binary relation α= X × T is gener-

ated by elements of the elements of set B ( A 0 ) ; DOI: 10.4236/am.2018.94028

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

b) If X = D and T ∈ D2 ∪ D3 , then binary relation α= X × T is external

element for the semigroup BX ( D ) . Proof. 1) If quasinormal representation of a binary relation δ has a form n+k  δ = (Y0δ × D ) ∪  (Y jδ × Z j ) , j= k +1

δ where Y j ≠ ∅ for all j =k + 1, k + 2, , n + k , then δ ∈ B ( A 0 ) \ {α } . Let qua-

sinormal representation of a binary relations β have a form   β = D × T ∪   ({t ′} × f ( t ′ ) ) , where f is any mapping of the set X \ D in the

(

)

t ′∈X \ D

( D2 ∪ D3 ) \ {T } . It is easy to see, that β ≠ α D2 ∪ D3 belong to the semilattice V ( D, β ) , i.e.

set

and two elements of the set

δ ∈ B ( A 0 ) \ {α } . In this case

we have that Z j β = T for all j =k + 1, k + 2, , n + k . 

δ  β == δ (Y0δ × Dβ ) ∪ =

(Y

δ

0

n+k

n+k

j = k +1

)  (Y

×T ∪

j = k +1

n+k  =   Y0δ ∪  Y jδ  j = k +1 

 (Y jδ × Z j β )

δ j

×T

)

   × T  = X × T = α ,  

since the representation of a binary relation δ is quasinormal. Thus, the element

α is generated by elements of the set B ( A 0 ) .

The statement a) of the lemma 2.7 is proved.



2) Let X = D , α= X × T , for some T ∈ D2 ∪ D3 and α = δ  β for some

δ , β ∈ BX ( D ) \ {α } . Then we obtain that Z j β = T since T is a minimal element of the semilattice D. Now, let subquasinormal representations β of a binary relation β have a form   n+k   = β    Pi  × T  ∪   {t ′} × β 2 ( t ′ ) , =   i 0   t′∈X \ D

(

)

 P0 P1 P2  Pn+k  where β1 =   is normal mapping. But complement mapping T T T  T   β 2 is empty, since X \ D = ∅ , i.e. in the given case, subquasinormal representation β of a binary relation β is defined uniquely. So, we have that β = β = X × T = α (see property 2) in the case 1.1), which contradict the condition, that β ∉ BX ( D ) \ {α } .



Therefore, if X = D and α= X × T , for some T ∈ D2 ∪ D3 , then α is ex-

ternal element of the semigroup BX ( D ) .

The statement 2) of the Lemma 2.7 is proved. Lemma 2.7 is proved.

Theorem 2.1. Let D ∈ Σ8 ( X , n + k + 1) , k ≥ 3 , and

= D1

Z1 , Z 2 , , Z k } , D2 {Z k += {= {Z n+1 , Z n+2 , , Z n+k } ; 1 , Z k + 2 , , Z n } , D3

= A1

DOI: 10.4236/am.2018.94028

379



Z , Z , D}} , where q {{= n+q

q

1, 2, , k ;

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

= j 1, 2, , n + k ; {{Z , D}} , where A {{= = Z , Z }} , where j 1, 2, , k ;  = A {{Z } , { D}} , where = j 1, 2, , n + k ; = A2

j

n+ j

3

4

j

j

{V ( D, α ) ⊂ D | V ( D, α ) ∉ A1 ∪ A 2 ∪ A3 ∪ A 4 } ,

= A0

B ( A0 ) = {α ∈ BX ( D ) | V ( D, α ) ∈ A0 } ,

B0 = { X × T | T ∉ D2 ∪ D3 }

Then the following statements are true:  1) If X \ D ≥ 1 , then the S0 = B ( A0 ) is irreducible generating set for the semigroup BX ( D ) ;  B0 ∪ B ( A0 ) is irreducible generating set for the 2) If X = D , then the S= 1 semigroup BX ( D ) .  Proof. Let D ∈ Σ8 ( X , n + k + 1) , k ≥ 3 and X \ D ≥ 1 . First, we proved that every element of the semigroup BX ( D ) is generated by elements of the set S0 . Indeed, let α be an arbitrary element of the semigroup BX ( D ) . Then quasinormal representation of a binary relation α has a form n+k  α = (Y0α × D ) ∪  (Yiα × Z i ) , i =1

where

n+k

 Yiα = X

α α ∅ and Yi ∩ Y j =

i =0

(0 ≤ i ≠

(

j ≤ n + k ) . For the V X ∗ , α

)

we consider the following cases:

(

)

1) If V X ∗ , α ∉ A1 ∪ A 2 ∪ A3 ∪ A 4 , then α ∈ B ( A0 ) ⊆ S0 by definition of

a set S0 .

(

)

Now, let V X ∗ , α ∈ A1 ∪ A 2 ∪ A3 ∪ A 4 .

(

) (

2) If V X , α ∈ A1 , then quasinormal representation of a binary relation α  α α α has a form α = Ynα+ j × Z n+ j ∪ Y jα × Z j ∪ Y0α × D , where Yn+ j , Y j , Y0 ∉ {∅} ∗

( j = 1, 2, , k )

) (

) (

)

and from the Lemma 2.3 follows that α is generated by elements

of the elements of set B ( A 0 ) ⊆ S0 by definition of a set S0 .

(

)

3) If V X ∗ , α ∈ A 2 , then quasinormal representation of a binary relation α  α α ∅} , j 1, 2, , n + k has a form α = Y jα × Z j ∪ Y0α × D , where Y j , Y0 ∉ {=

(

) (

)

and from the Lemma 2.4 follows that α is generated by elements of the elements of set B ( A 0 ) ⊆ S0 by definition of a set S0 .

(

) (

4) If V X ∗ , α ∈ A 3 , then quasinormal representation of a binary relation α

) (

)

α α has a form α = Ynα+ j × Z n+ j ∪ Y jα × Z j , where Yn+ j , Y j ∉ {∅} , j = 1, 2, , k

and from the Lemma 2.5 follows that α is generated by elements of the elements of set B ( A 0 ) ⊆ S0 by definition of a set S0 .

(

)

Now, let V X ∗ , α ∈ A 4 , then quasinormal representation of a binary rela tion α has a form α= X × D , or α= X × Z j , where = j 1, 2, , n + k .  5) If α= X × D , then from the statement b) of the Lemma 2.6 follows that

binary relation α is generated by elements of the set B ( A 0 ) . DOI: 10.4236/am.2018.94028

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6) If α= X × Z j , where = j 1, 2, , n + k , then from the statement a) of the Lemma 2.6 and 2.7 follows that binary relation α is generated by elements of the set B ( A 0 ) .

Thus, we have that S0 is a generating set for the semigroup BX ( D ) .  If X \ D ≥ 1 , then the set S0 is an irreducible generating set for the semigroup BX ( D ) since, S0 is a set external elements of the semigroup BX ( D ) . The statement a) of the Theorem 2.1 is proved.



Now, let X = D . First, we proved that every element of the semigroup BX ( D ) is generated by elements of the set S1 . The cases 1), 2), 3), 4) and 5) are proved analogously of the cases 1), 2), 3), 4) and 5 given above and consider case, when V X ∗ , α ∈ A1 .

(

( )

(

)

)

If V X ∗ , α = Z j , where j = 1, 2, , k , then from the statement a) of the Lemma 2.7 follows that binary relation α is generated by elements of the set B ( A0 ) . If V X ∗ , α = Z j , where Z j ∈ D2 ∪ D3 , then from the statement b) of the

Lemma 2.6 follows that binary relation α= X × T is external element for the semigroup BX ( D ) .

Thus, we have that S1 is a generating set for the semigroup BX ( D ) .  If X = D , then the set S1 is an irreducible generating set for the semigroup BX ( D ) since S1 is a set external elements of the semigroup BX ( D ) . The statement b) of the Theorem 2.1 is proved. Theorem 2.1 is proved. Corollary 2.1. Let D ∈ Σ8 ( X , n + k + 1)

= D1

( k ≥ 3) and {Z k += {Z n+1 , Z n+2 , , Z n+k } ; 1 , Z k + 2 , , Z n } , D3

Z1 , Z 2 , , Z k } , D2 {=



Z , Z , D}} , where q 1, 2, , k ; {{=  = A {{Z , D}} , where = j 1, 2, , n + k ; = A {{= Z , Z }} , where j 1, 2, , k ;  = A {{Z } , { D}} , where = j 1, 2, , n + k ;

= A1

n+q

2

j

n+ j

3

4

= A0

q

j

j

{V ( D, α ) ⊂ D | V ( D, α ) ∉ A1 ∪ A 2 ∪ A3 ∪ A 4 } , B ( A0 ) = {α ∈ BX ( D ) | V ( D, α ) ∈ A0 } ,

B0 = { X × T | T ∉ D2 ∪ D3 }

Then the following statements are true:  1) If X \ D ≥ 1 , then S0 = B ( A0 ) is the uniquely defined generating set for the semigroup BX ( D ) ;  B0 ∪ B ( A0 ) is the uniquely defined generating set 2) If X = D , then S= 1 for the semigroup BX ( D ) . Proof. It is well known, that if B is all external elements of the semigroup BX ( D ) and B′ is any generated set for the BX ( D ) , then B ⊆ B′ (see [1] [2] Lemma 1.15.1). From this follows that the sets S0 = B ( A0 ) and DOI: 10.4236/am.2018.94028

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S= B0 ∪ B ( A0 ) are defined uniquely, since they are sets external elements of 1 the semigroup BX ( D ) . Corollary 2.1 is proved. It is well-known, that if B is all external elements of the semigroup BX ( D ) and B′ is any generated set for the BX ( D ) , then B ⊆ B′ (Definition 1.1). In this article, we find irredusible generating set for the complete semigroups of binary relations defined by X-semilattices of unions of the class Σ8 ( X , n + k + 1) ( k ≥ 3) . This generating set is uniquely defined, since they are defined by elements of the external elements of the semigroup BX ( D ) .

References

DOI: 10.4236/am.2018.94028

[1]

Diasamidze, Y. and Makharadze, S. (2013) Complete Semigroups of Binary Relations. Kriter, Turkey, 1-519.

[2]

Я. И. Диасамидзе, Ш. И. Махарадзе (2017) Полные полугруппы бинарных отношений. Lambert Academic Publishing, 1-692.

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