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Generated sets of the complete semigroup binary relations defined by semilattices of the finite chains
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ScienceDirect Transactions of A. Razmadze Mathematical Institute 172 (2018) 378–387 www.elsevier.com/locate/trmi
Original article
Generated sets of the complete semigroup binary relations defined by semilattices of the finite chains Omari Givradze, Yasha Diasamidze, Nino Tsinaridze ∗ Department of Mathematics, Faculty of Physics, Mathematics and Computer Sciences, Shota Rustaveli Batumi State University, 35 Ninoshvili St., Batumi 6010, Georgia Received 20 April 2018; received in revised form 30 July 2018; accepted 18 August 2018 Available online 31 August 2018
Abstract In this article, we study generated sets of the complete semigroups of binary relations defined by X -semilattices unions of the finite chains. We found uniquely irreducible generating set for the given semigroups. c 2018 Ivane Javakhishvili Tbilisi State University. Published by Elsevier B.V. This is an open access article under the CC ⃝ BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Keywords: Semigroup; Semilattice; Binary relation
1. Introduction Let X be an arbitrary nonempty set, D be an X -semilattice of unions which is closed with respect to the settheoretic 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 correspondence a binary relation α f on the set X that satisfies the condition ⋃( ) αf = {x} × f (x) . x∈X
The set of all such α f ( f : X → D) is denoted by B X (D). It is easy to prove that B X (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 xαy. Further, let x, y ∈ X , Y ⊆ X , α ∈ B X (D), D˘ = Y ∈D Y and ∗ Corresponding author.
E-mail address: [email protected] (N. Tsinaridze). Peer review under responsibility of Journal Transactions of A. Razmadze Mathematical Institute. https://doi.org/10.1016/j.trmi.2018.08.004 c 2018 Ivane Javakhishvili Tbilisi State University. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND 2346-8092/⃝ license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
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T ∈ D. We denote by the symbols yα, Y α, V (D, α), X ∗ and V (X ∗ , α) the following sets: ⋃ { } yα, V (D, α) = {Y α | Y ∈ D}, X ∗ = Y | ∅ ̸= Y ⊆ X , yα = {x ∈ X | yαx}, Y α = y∈Y
{ } V (X ∗ , α) = Y α | ∅ ̸= Y ⊆ X ,
{ } DT = Z ∈ D | T ⊆ Z ,
{ } YTα = y ∈ X | yα = T .
It is well known the following statements: ˘ Z 1 , Z 2 , . . . , Z m−1 } be some finite X -semilattice of unions and C(D) = {P0 , P1 , P2 , . . . , Theorem 1.1. Let D = { D, 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˘ Z 1 Z 2 · · · Z m−1 ϕ= P0 P1 P2 · · · Pm−1 ˆZ = D \ D Z , then the following equalities are valid: and D D˘ = P0 ∪ P1 ∪ P2 ∪ · · · ∪ Pm−1 , ⋃ Z i = P0 ∪ ϕ(T ).
(1.1)
ˆZ T ∈D i
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, whereas sets P j (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, Chapter 11]). Definition 1.1. We say that an element α of the semigroup B X (D) is external if α ̸= δ ◦ β for all δ, β ∈ B X (D) \ {α} (see [1, Definition 1.15.1]). It is well known, that if B is all external elements of the semigroup B X (D) and B ′ is any generated set for the B X (D), then B ⊆ B ′ (see [1, Lemma 1.15.1]). ⋃ Definition 1.2. The representation α = T ∈D (YTα × T ) of binary relation α is called quasinormal, if ⋃ YTα = X and YTα ∩ YTα′ = ∅ for any T, T ′ ∈ D, T ̸= T ′ . T ∈D
Definition 1.3. Let α, β ⊆ X × X . Their product δ = α ◦ β is defined as follows: xδy (x, y ∈ X ) if there exists an element z ∈ X such that xαzβy (see [1, Chapter 1.3]). 2. Let Σm (X, m) be a class of all X -semilattices of unions whose every element is isomorphic to an X -semilattice of ˘ which satisfies the condition unions D = {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m = D}, Z 1 ⊂ Z 2 ⊂ Z 3 ⊂ · · · ⊂ Z m−2 ⊂ Z m−1 ⊂ Z m = D˘
(2.1)
(see Fig. 1). Let C(D) = {P0 , P1 , P2 , P3 , . . . , Pm−2 , Pm−1 ( }˘ be a family of sets, where P0), P1 , P2 , P3 , . . . , Pm−2 , Pm−1 are Z m−1 m−2 pairwise disjoint subsets of the set X and ϕ = PD0 ZP11 ZP22 ZP33 ·· ·· ·· ZPm−2 is a mapping of the semilattice Pm−1
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Fig. 1.
D onto the family of sets C(D). Then the formal equalities of the semilattice D have a form: D˘ = Z m = P0 ∪ P1 ∪ P2 ∪ P3 ∪ · · · ∪ Pm−1 , Z m = P0 ∪ P1 ∪ P2 ∪ P3 ∪ · · · ∪ Pm−2 , Z m−1 = P0 ∪ P1 ∪ P2 ∪ P3 ∪ · · · ∪ Pm−3 , ..........................................
(2.2)
Z 3 = P0 ∪ P1 ∪ P2 , Z 2 = P0 ∪ P1 , Z 1 = P0 , Here the elements P1 , P2 , P3 , . . . , Pm−2 , Pm−1 are bases sources, the element P0 is sources of completeness of the semilattice D. Therefore |X | ≥ m − 1, since |Pi | ≥ 1, i = 1, 2, . . . , m − 1 (see Theorem 1.1). In this paper we are learning irreducible generating sets of the semigroup B X (D) defined by semilattices of the class Σm (X, m). Definition 2.1. In the sequel, by symbol Σm.0 (X, m) we denote all semilattices D = {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m } of the class Σm (X, m) for which Z 1 is not empty. From the last inequality and from the formal equalities (2.2) it follows that Z 1 = P0 ̸= ∅, i.e. in this case |X | ≥ m. Note that, if D ∈ Σm,0 (X, m) and by symbol |X | is denoted the power of a set X , then the inequality |X | ≥ |D| always is true. ˜ = {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m } is irreducible generating set for the semilattice D. It is easy to see that D Lemma 2.1. If D ∈ Σm (X, m), then the following statements are true: (a)
P0 = Z 1 ;
(b)
Pi−1 = Z i \ Z i−1 , i = 2, 3, . . . , m − 2, m − 1, m.
(2.3)
Proof. From the formal equalities of the semilattice D immediately follows the following statements: P0 = Z 1 ,
Pi−1 = Z i \ Z i−1 , i = 2, 3, . . . , m − 2, m − 1, m.
The statements (a) and (b) of Lemma 2.1 are proved. □ It is well known, that the binary relation α, representation, which has a form ˘ × D) ˘ ε = (Z 1 × Z 1 ) ∪ ((Z 2 \ Z 1 ) × Z 2 ) ∪ · · · ∪ ((Z m−1 \ Z m−2 ) × Z m−1 ) ∪ ((Z m \ Z m−1 ) × Z m ) ∪ ((X \ D) is largest right unit of the semigroup B X (D) (see [1, Theorem 15.1.1]) and if by E (rX ) (D) is denoted all right units of the semigroup B X (D), then all right units of the semigroup are its external elements.
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Definition 2.2. We denote the following sets by symbol Ak and B(Ak ): { } Ak = V (X ∗ , α) | |V (X ∗ , α)| = k for anyα ∈ B X (D) , { } B(Ak ) = α ∈ B X (D) | V (X ∗ , α) ∈ Ak , where k = m, m − 1, m − 2, . . . , 3, 2, 1. Lemma 2.2. Let α ∈ B(Am ), then α is external element for the semigroup B X (D). Proof. Let α ∈ B(Am ) and α = δ ◦ β for some δ, β ∈ B X (D) \ {α}. If quasinormal representation of binary relation δ has a form m ⋃ δ= (Yiδ × Z i ), i=1
then α =δ◦β =
m m (⋃ ) ⋃ (Yiδ × Z i β). (Yiδ × Z i ) ◦ β =
(2.4)
i=1
i=1
It is easy to see that ˘ Z 1 β ⊆ Z 2 β ⊆ Z 3 β ⊆ · · · ⊆ Z m−2 β ⊆ Z m−1 β ⊆ Z m β = Dβ
(2.5)
since Z 1 ⊂ Z 2 ⊂ Z 3 ⊂ · · · ⊂ Z m−2 ⊂ Z m−1 ⊂ Z m = D˘ by definition of the semilattice D. By preposition { } { } Z 1 β, Z 2 β, Z 3 β, . . . , Z m−2 β, Z m−1 β, Z m β = Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m ( ) |V (D, α)| = |V (D, δ ◦ β)| = m , i.e. there exists one to one mapping ϕ of the set {Z 1 β, Z 2 β, Z 3 β, . . . , Z m−2 β, Z m−1 β, Z m β} on the set {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m } which satisfies the condition (2.1) and (2.5). Now, let ϕ(Z i β) = Z j for some Z i ⊆ Z j (i ≤ j). If Z i ̸= Z j , then there exist such p and q (1 ≤ p ̸= q ≤ m), that Z p β = Z q for which ( p > q), since D is finite. But equality Z p β = Z q and inclusion p > q contradicts the inclusions (2.5), i.e. i = j. Of this follows that Z i β = Z i for all i = 1, 2, 3, . . . , m. We have that α =δ◦β =
m ⋃ i=1
(Yiδ × Z i β) =
m ⋃ (Yiδ × Z i ) = δ. i=1
But equality α = δ contradicts the preposition δ ∈ B X (D) \ {α}.
□
˘ ≥ 1, then any element of set B(Ak ) is generated by elements of the set Lemma 2.3. Let D ∈ Σm.0 (X, m). If |X \ D| B(Ak+1 ) (k = m − 1, . . . , 4, 3, 2). Proof. Let D ∈ Σm.0 (X, m), then |X | ≥ |D| = m. By Lemma 2.2 it follows that if α ∈ B(Am ) , then α is external element for the semigroup B X (D). Now, let k < m and α ∈ B(Ak ), then quasinormal representation of a binary ⋃k relation has a form α = i=1 (Yiα × Ti ), where Y1α , Y2α , . . . , Ykα ̸∈ {∅}, T1 , T2 , . . . , Tk ∈ D and T1 ⊂ T2 ⊂ · · · ⊂ Tk . It ∗ is easy to see, that V (X , α) = {T1 , T2 , . . . , Tk }, i.e. |V (X ∗ , α)| = k. From the sets Y1α , Y2α , . . . , Ykα there exists such α α set Yqα , that satisfies the condition |Yqα | ≥ 2 since k < m. Therefore, we may suppose, that Yqα = Yq1 ∪ Yq2 , where α α α α Yq1 , Yq2 ̸∈ {∅} and Yq1 ∩ Yq2 = ∅. Further, there exist such element T ∈ D (T ̸∈ {T1 , T2 , . . . , Tk }). For the element T we consider the following cases: (1) (2) (3) (4)
T ⊂ T1 ; Tk ⊂ T ; T1 ⊂ T ⊂ Tq ; Tq ⊂ T ⊂ Tk .
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(1). Let T ⊂ T1 and T = T0 . If quasinormal representations of the binary relations δ and β have the forms δ = (Y1α × T0 ) ∪ (Y2α × T1 ) ∪ (Y3α × T2 ) ∪ · · · α α α α ∪ (Yq−1 × Tk−2 ) ∪ (Yq1 × Tq−1 ) ∪ (Yq2 × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ · · · ∪ (Ykα × Tk ); β = (T0 × T1 ) ∪ ((T1 \ T0 ) × T2 ) ∪ ((T2 \ T1 ) × T3 ) ∪ · · · ∪ ((Tq−2 \ Tq−3 ) × Tq−1 ) ∪ ((Tq \ Tq−2 ) × Tq ) ∪ ((Tq+1 \ Tq ) × Tq+1 ) ∪ · · · ∪ ((Tk \ Tk−1 ) × Tk ) ∪ ((X \ Tk ) × T0 ), α α α α where Y1α , Y2α , Y3α , . . . , Yq−1 Yq1 Yq2 Yq+1 , . . . , Ykα ̸∈ {∅}, then it is easy to see, that |V (X ∗ , δ)| = |V (X ∗ , β)| = k + 1, ∗ ∗ i.e. V (X , δ), V (X , β) ∈ Ak+1 and
T0 β = T1 , T1 β = T2 , . . . , Tq−2 β = Tq−1 , Tq−1 β = Tq β = Tq , Tq+1 β = Tq+1 , Tq+2 β = Tq+2 , . . . , Tk β = Tk . From the last equalities we obtain, that: δ ◦ β = (Y1α × T0 β) ∪ (Y2α × T1 β) ∪ (Y3α × T2 β) ∪ · · · α α α α ∪(Yq−1 × Tq−2 β) ∪ (Yq1 × Tq−1 β) ∪ (Yq2 × Tq β) ∪ (Yq+1 × Tq+1 β) ∪ · · · ∪ (Ykα × Tk β) α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ (Y3α × T3 ) ∪ v ∪ (Yq−1 × Tq−1 ) ∪ α α α ((Yq1 ∪ Yq2 ) × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ · · · ∪ (Ykα × Tk ) = α, α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k. Therefore, V (X ∗ , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ).
(2). Let Tk ⊂ T and Tk+1 = T . If quasinormal representations of the binary relations δ and β have the forms α α δ = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 × Tq−1 ) ∪ (Yq1 × Tq )∪ α α (Yq2 × Tq+1 ) ∪ (Yq+1 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk+1 );
β = (T1 × T1 ) ∪ ((T2 \ T1 ) × T2 ) ∪ ((T3 \ T2 ) × T3 ) ∪ · · · ∪ ((Tq−1 \ Tq−2 ) × Tq−1 ) ∪ ((Tq+1 \ Tq−1 ) × Tq ) ∪ ((Tq+2 \ Tq+1 ) × Tq+1 ) ∪ ((Tq+3 \ Tq+2 ) × Tq+2 ) ∪ · · · ∪ ((Tk+1 \ Tk ) × Tk ) ∪ ((X \ Tk+1 ) × Tk+1 ), α α α α where Y1α , Y2α , . . . , Yq−1 Yq1 Yq2 Yq+1 , . . . , Ykα ̸∈ {∅}. It is easy to see, that |V (X ∗ , δ)| = |V (X ∗ , β)| = k + 1, ∗ ∗ i.e. V (X , δ), V (X , β) ∈ Ak+1 and
T1 β = T1 , T2 β = T2 , . . . , Tq−1 β = Tq−1 , Tq β = Tq+1 β = Tq , Tq+2 β = Tq+1 , Tq+3 β = Tq+2 , . . . , Tk+1 β = Tk . From the last equalities we obtain, that: α δ ◦ β = δ = (Y1α × Tβ1 ) ∪ (Y2α × Tβ2 ) ∪ · · · ∪ (Yq−1 × Tq−1 β) α α α ∪(Yq1 × Tq β) ∪ (Yq2 × Tq+1 β) ∪ (Yq+1 × Tq+2 β) ∪ · · · ∪ (Ykα × Tk+1 β) α α × Tq ) ∪ (Yq2 × Tq ) ∪ · · · ∪ (Ykα × Tk ) = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq1 α α α α = (Y1 × T1 ) ∪ (Y2 × T2 ) ∪ · · · ∪ ((Yq1 ∪ Yq2 ) × Tq ) ∪ · · · ∪ (Ykα × Tk ) = α, α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k. ∗ Therefore, V (X , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ).
(3). Let T1 ⊂ T ⊂ Tq . Then there exists an element T which is covered by element Tl , where 2 ≤ l ≤ q, i.e. Tl−1 ⊂ T ⊂ Tl . So, we have the following chain: T1 ⊂ T2 ⊂ Tl−1 ⊂ T ⊂ Tl ⊂ Tl+1 ⊂ Tl+2 ⊂ · · · ⊂ Tq ⊂ Tq+1 ⊂ Tq+2 ⊂ · · · ⊂ Tk . If quasinormal representations of the binary relations δ and β have the forms α α α δ = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yl−1 × Tl−1 ) ∪ (Ylα × T ) ∪ (Yl+1 × Tl ) ∪ (Yl+2 × Tl+1 ) ∪ · · · α α α α α ∪ (Yq−1 × Tq−2 ) ∪ (Yq1 × Tq−1 ) ∪ (Yq2 × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk );
β = (T1 × T1 ) ∪ ((T2 \ T1 ) × T2 ) ∪ ((T3 \ T2 ) × T3 ) ∪ · · · ∪ ((Tl−1 \ Tl−2 ) × Tl−1 ) ∪ ((T \ Tl−1 ) × Tl ) ∪ ((Tl \ T ) × Tl+1 ) ∪ ((Tl+1 \ Tl ) × Tl+2 ) ∪ · · · ∪ ((Tq−2 \ Tq−3 ) × Tq−1 ) ∪ ((Tq \ Tq−2 ) × Tq ) ∪ ((Tq+1 \ Tq ) × Tq+1 ) ∪ · · · ∪ ((Tk \ Tk−1 ) × Tk ) ∪ ((X \ Tk ) × T ),
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α α α α α α α where Y1α , . . . , Yl−1 , Ylα , Yl+1 , . . . , Yq−1 , Yq1 , Yq2 , Yq+1 , Yq+2 , . . . , Ykα ̸∈ {∅}. It is easy to see, that |V (X ∗ , δ)| = ∗ ∗ ∗ |V (X , β)| = k + 1, i.e. V (X , δ), V (X , β) ∈ Ak+1 and
T1 β = T1 , T2 β = T2 , . . . , Tl−1 β = Tl−1 , Tβ = Tl , Tl β = Tl+1 , Tl+1 β = Tl+2 , . . . , Tq−2 β = Tq−1 , Tq−1 β = Tq β = Tq , Tq+1 β = Tq+1 , Tq+2 β = Tq+2 , . . . , Tk β = Tk . From the last equalities we obtain, that: α δ ◦ β = (Y1α × T1 β) ∪ (Y2α × T2 β) ∪ · · · ∪ (Yl−1 × Tl−1 β) ∪ (Ylα × Tβ) α α α α α ∪ (Yl+1 × Tl β) ∪ (Yl+2 × Tl+1 β) ∪ · · · ∪ (Yq−1 × Tq−2 β) ∪ (Yq1 × Tq−1 β) ∪ (Yq2 × Tq β) α α ∪ (Yq+1 × Tq+1 β) ∪ (Yq+2 × Tq+2 β) ∪ · · · ∪ (Ykα × Tk β) α α α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yl−1 × Tl−1 ) ∪ (Ylα × Tl ) ∪ (Yl+1 × Tl+1 ) ∪ (Yl+2 × Tl+2 ) ∪ · · · α α α α ∪ (Yq1 × Tq ) ∪ (Yq2 × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk ) α α α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yl−1 × Tl−1 ) ∪ (Ylα × Tl ) ∪ (Yl+1 × Tl+1 ) ∪ (Yl+2 × Tl+2 ) ∪ · · · α α α α ∪ ((Yq1 ∪ Yq2 ) × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk ) = α, α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k. ∗ Therefore, V (X , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ).
(4). Let Tq ⊂ T ⊂ Tk . Then there exists an element T which is covered by element Tr , where q + 1 ≤ r ≤ k, i.e. Tr −1 ⊂ T ⊂ Tr . So, we have the following chain: T1 ⊂ T2 ⊂ · · · ⊂ Tq ⊂ Tq+1 ⊂ · · · ⊂ Tr −1 ⊂ T ⊂ Tr ⊂ Tr +1 ⊂ · · · ⊂ Tk . If quasinormal representations of the binary relations δ and β have the forms α α δ = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 × Tq−1 ) ∪ · · · ∪ (Yq1 × Tq ) α α α ∪ (Yq2 × Tq+1 ) ∪ (Yq+1 × Tq+2 ) ∪ (Yq+2 × Tq+3 ) ∪ · · · ∪ (Yrα−2 × Tr −1 )
∪ (Yrα−1 × T ) ∪ (Yrα × Tr ) ∪ (Yrα+1 × Tr +1 ) ∪ · · · ∪ (Ykα × Tk ); β = (T1 × T1 ) ∪ ((T2 \ T1 ) × T2 ) ∪ ((T3 \ T2 ) × T3 ) ∪ · · · ∪ ((Tq−1 \ Tq−2 ) × Tq−1 ) ∪ ((Tq+1 \ Tq−1 ) × Tq ) ∪ ((Tq+2 \ Tq+1 ) × Tq+1 ) ∪ · · · ∪ ((Tr −1 \ Tr −2 ) × Tr −2 ) ∪ ((T \ Tr −1 ) × Tr −1 ) ∪ ((Tr \ T ) × Tr ) ∪ ((Tr +1 \ Tr ) × Tr +1 ) ∪ · · · ∪ ((Tk \ Tk−1 ) × Tk ) ∪ ((X \ Tk ) × T ), α α α α α where Y1α , Y2α , . . . , Yq−1 , . . . , Yq1 , Yq2 , Yq+1 , Yq+2 , . . . , Yrα−2 , Yrα−1 , Yrα , Yrα+1 , . . . , Ykα ∗ ∗ ∗ |V (X , δ)| = |V (X , β)| = k + 1, i.e. V (X , δ), V (X ∗ , β) ∈ Ak+1 and
̸∈ {∅}. It is easy to see, that
T1 β = T1 , T2 β = T2 , . . . , Tq−1 β = Tq−1 , Tq β = Tq+1 β = Tq , Tq+2 β = Tq+1 , Tq+3 β = Tq+2 , . . . , Tr −1 β = Tr −2 , Tβ = Tr −1 , Tr β = Tr , Tr +1 β = Tr +1 , . . . , Tk β = Tk . From the last equalities we obtain, that: α α α × Tq−1 β) ∪ · · · ∪ (Yq1 × Tq β) ∪ (Yq2 × Tq+1 β) δ ◦ β = (Y1α × T1 β) ∪ (Y2α × T2 β) ∪ · · · ∪ (Yq−1 α α × Tq+2 β) ∪ (Yq+2 × Tq+3 β) ∪ · · · ∪ (Yrα−2 × Tr −1 β) ∪ (Yq+1
∪ (Yrα−1 × Tβ) ∪ (Yrα × Tr β) ∪ (Yrα+1 × Tr +1 β) ∪ · · · ∪ (Ykα × Tk β) α α α × Tq−1 ) ∪ · · · ∪ (Yq1 × Tq ) ∪ (Yq2 × Tq ) = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 α α ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Yrα−2 × Tr −2 )
∪ (Yrα−1 × Tr −1 ) ∪ (Yrα × Tr ) ∪ (Yrα+1 × Tr +1 ) ∪ · · · ∪ (Ykα × Tk ) α α α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 × Tq−1 ) ∪ · · · ∪ ((Yq1 ∪ Yq2 ) × Tq ) α α ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Yrα−2 × Tr −2 )
∪ (Yrα−1 × Tr −1 ) ∪ (Yrα × Tr ) ∪ (Yrα+1 × Tr +1 ) ∪ · · · ∪ (Ykα × Tk ) = α,
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α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, since α α Yqα = Yq1 ∪ Yq2 ,
V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k.
Therefore, V (X ∗ , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ). So, any element of set B(Ak ) is generated by elements of the set B(Ak+1 ), where k = m − 1, . . . , 4, 3, 2.
□
˘ ≥ 1, then any element of the set B(A1 ) is generated by elements of the Lemma 2.4. Let D ∈ Σm.0 (X, m). If |X \ D| set B(A2 ). Proof. Let T is any element of the semilattice D and α = X × T ∈ B(A1 ). If quasinormal representations of the binary relations δ and β (δ, β ∈ B X (D)) have forms ˘ × T ′ ), δ = (Y1δ × T ) ∪ (Y2δ × T ′ ), β = ( D˘ × T ) ∪ ((X \ D) where Y1δ , Y2δ ̸∈ {∅} (T ′ ∈ D, T ̸= T ′ ), then |V (X ∗ , δ)| = |V (X ∗ , β)| = 2, i.e. δ, β ∈ B(A2 ) and Tβ = T ′ β = T, δ ◦ β = (Y1δ × Tβ) ∪ (Y2δ × T ′ β) = (Y1δ × T ) ∪ (Y2δ × T ) = X × T = α, since the representation of a binary relation δ is quasinormal. ˘ ≥ 1, then any element of the set B(A1 ) is generated by elements of the set B(A2 ). □ Therefore, if |X \ D| ˘ ≥ 1 and Theorem 2.1. Let D ∈ Σm.0 (X, m) and D is a finite set. If |X \ D| { } { } ∗ ∗ Am = V (X , α) | |V (X , α)| = m for any α ∈ B X (D) , B(Am ) = α ∈ B X (D) | V (X ∗ , α) ∈ Am , then the set B(Am ) is irreducible generating set for the semigroup B X (D). ˘ ≥ 1. First, we proved that every element of the semigroup B X (D) is generated by elements of Proof. Let |X \ D| the set B(Am ). Indeed, let α be an⋃arbitrary element of the B X (D). Then quasinormal representation of a ⋃msemigroup m binary relation α has a form α = i=1 (Yiα × Z i ), where i=1 Yiα = X and Yiα ∩ Y jα = ∅ for any 1 ≤ i ̸= j ≤ m. For the number |V (X ∗ , α)| we consider the following cases: (1) If k = m − 1, . . . , 4, 3, 2, then from Lemma 2.3 immediately follows that any element of the set B(Ak ) is generated by elements of the set B(Ak+1 ). (2) If k = 1, then from Lemma 2.4 immediately follows that any element of the set B(A1 ) is generated by elements of the set B(A2 ). ˘ ≥ 1, then every element of the semigroup B X (D) are generated by elements of the set B(Am ). So, if |X \ D| From Lemma 2.2 it follows that, every element of the set B(Am ) is external element for the semigroup B X (D). Therefore, we have that B(Am ) is irreducible generating set for the semigroup B X (D). □ ˘ ≥ 1 and Theorem 2.2. Let D ∈ Σm.0 (X, m) and D is a finite set. If |X \ D| { } { } Am = V (X ∗ , α) | |V (X ∗ , α)| = m for any α ∈ B X (D) , B(Am ) = α ∈ B X (D) | V (X ∗ , α) ∈ Am . Then the paver of a set B(Am ) is equal to the paver surjection of the set X on the set D. Proof. Let X and D be finite sets. If α ∈ B(Am ), then by definition of a semigroup B X (D) there exists such a mapping f of a set X on the set D = {Z 1 , Z 2 , . . . , Z m }, since |X | ≥ |D| and α = α f , i.e. mapping f is some surjection of the set X on the set D. □ ˘ ≥ 1 and Corollary 2.1. Let D ∈ Σm.0 (X, m). If |X \ D| { } Am = V (X ∗ , α) | |V (X ∗ , α)| = m for any α ∈ B X (D) ,
{ } B(Am ) = α ∈ B X (D) | V (X ∗ , α) ∈ Am .
If X and D are finite sets |X | = n and |D| = m, then the number |B(Am )| elements of the set B(Am ) is equal to |B(Am )| = n m +
n−1 ∑ (−1)i · Cni · (n − i)m . i=1
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Proof. Let X and D be finite sets, |X | = n and |D| = m. In our case we have |X | ≥ |D|. It is well known that, the ∑n−1 number surjection of the set X on the set D is equal to the number n m + i=1 (−1)i · Cni · (n − i)m . Now, Corollary 2.1 immediately follows from Theorem 2.2. □ Example 2.1. X = {1, 2, 3} and D = {{1}, {1, 2}}. Then 3 = |X | > |D| = 2 and { } { } { } α1 = (1, 1), (2, 1), (3, 1) , α2 = (1, 1), (2, 1), (3, 1), (3, 2) , α3 = (1, 1), (2, 1), (2, 2)(3, 1) , { } { } α4 = (1, 1), (1, 2)(2, 1), (3, 1) , α5 = (1, 1), (1, 2), (2, 1), (2, 2)(3, 1) , { } α6 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2) , { } { } α7 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2) , α8 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2) . In this case we have B X (D) = {α1 , α2 , . . . , α7 , α8 };
B(A2 ) = {α2 , α3 , α4 , α5 , α6 , α7 },
B(A1 ) = {α1 , α8 }
and { } { } { } α3 ◦ α2 = (1, 1), (2, 1), (2, 2)(3, 1) ◦ (1, 1), (2, 1), (3, 1), (3, 2) = (1, 1), (2, 1), (3, 1) = α1 , { } { } α2 ◦ α4 = (1, 1), (2, 1), (3, 1), (3, 2) ◦ (1, 1), (1, 2)(2, 1), (3, 1) { } = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2) = α8 , i.e. the set B(A2 ) is irreducible generating set for the semigroup B X (D). Example 2.2. X = {1, 2, 3, 4} and D = {{1}, {1, 2}, {1, 2, 3}}. Then 4 = |X | > |D| = 3 and { } { } α1 = (1, 1), (2, 1), (3, 1), (4, 1) , α2 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) , { } α3 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } { } α4 = (1, 1), (1, 2), (2, 1), (3, 1), (4, 1) , α5 = (1, 1), (2, 1), (2, 2), (3, 1), (4, 1) , { } { } α6 = (1, 1), (2, 1), (3, 1), (3, 2), (4, 1) , α7 = (1, 1), (2, 1), (3, 1), (4, 1), (4, 2) , { } { } α8 = (1, 1), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2) , α9 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (4, 1) , { } { } α10 = (1, 1), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2) , α11 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (4, 1) , { } { } α12 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1) , α13 = (1, 1), (1, 2), (2, 1), (3, 1), (4, 1), (4, 2) , { } α14 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) , { } α15 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2) , { } α16 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2) , { } α17 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1) , { } { } α18 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (4, 1) , α19 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1) , { } { } α20 = (1, 1), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1) , α21 = (1, 1), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3) , { } α22 = (1, 1), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α23 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1) , { } α24 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2), (4, 3) , { } α25 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1) , { } α26 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1) , { } α27 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3) , { } α28 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α29 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α30 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2), (4, 3) , { } α31 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1) , { } α32 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) ,
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{ } α33 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2) , { } α34 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α35 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α36 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2) , { } α37 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α38 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α39 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α40 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α41 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α42 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α43 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α44 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α45 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α46 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1) , { } α47 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α48 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α49 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) , { } α50 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2) , { } α51 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α52 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1) , { } α53 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α54 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α55 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1) , { } α56 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2) , { } α57 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2), (4, 3) , { } α58 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (4, 1) , { } α59 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2) , { } α60 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α61 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (4, 1) , { } α62 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2) , { } α63 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2), (4, 3) , { } α64 = (1, 1), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2), (4, 3) , { } α65 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)v , { } α66 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2) , { } α67 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α68 = (1, 1), (1, 2), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3) , { } α69 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α70 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) , { } α71 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1) , { } α72 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (4, 1), (4, 2) , { } α73 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) ,
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{ } α74 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1) , { } α75 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1) , { } α76 = (1, 1), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α77 = (1, 1), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α78 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2), (4, 3) , { } α79 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1) , { } α80 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2) , { } α81 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) . In this case B X (D) = {α1 , α2 , α3 , . . . , α80 , α81 }, B(A2 ) = {α4 , α5 , . . . , α44 , α45 },
B(A3 ) = {α46 , α47 , . . . , α80 , α81 },
B(A1 ) = {α1 , α2 , α3 }
and α4 = α72 ◦ α76 ,
α5 = α49 ◦ α76 ,
α6 = α46 ◦ α76 ,
α8 = α54 ◦ α76 ,
α9 = α80 ◦ α76 ,
α10 = α51 ◦ α76 ,
α7 = α76 ◦ α76 , α11 = α52 ◦ α64 ,
α12 = α46 ◦ α64 ,
α13 = α60 ◦ α76 ,
α14 = α47 ◦ α64 ,
α15 = α53 ◦ α64 ,
α16 = α56 ◦ α64 ,
α17 = α79 ◦ α65 ,
α18 = α72 ◦ α77 ,
α19 = α49 ◦ α77 ,
α20 = α46 ◦ α77 ,
α21 = α76 ◦ α77 ,
α22 = α48 ◦ α77 ,
α23 = α80 ◦ α77 ,
α24 = α51 ◦ α77 ,
α25 = α52 ◦ α49 ,
α26 = α46 ◦ α49 ,
α27 = α60 ◦ α77 ,
α28 = α47 ◦ α49 ,
α29 = α53 ◦ α49 ,
α30 = α78 ◦ α67 ,
α31 = α79 ◦ α67 ,
α32 = α72 ◦ α79 ,
α33 = α49 ◦ α53 ,
α34 = α46 ◦ α52 ,
α35 = α76 ◦ α79 ,
α36 = α80 ◦ α79 ,
α37 = α48 ◦ α52 ,
α38 = α52 ◦ α55 ,
α39 = α51 ◦ α79 ,
α40 = α60 ◦ α79 ,
α42 = α65 ◦ α71 ,
α43 = α53 ◦ α55 ,
α44 = α78 ◦ α71 ,
α45 = α79 ◦ α71 ,
α41 = α67 ◦ α79 , α1 = α7 ◦ α7 ,
α2 = α7 ◦ α17 ,
α3 = α31 ◦ α31 .
References [1] Ya. Diasamidze, Sh. Makharadze, Complete Semigroups of Binary Relations, Kriter, Turkey, 2013.
Further Reading [1] Ya. Diasamidze, N. Aydin, A. Erdoˇgan, Generating set of the complete semigroups of binary relations, Appl. Math. Irvin 7 (2016) 98–107. [2] Ya.I. Diasamidze, Sh.I. Makharadze, Complete semigroups of binary relations defined by elementary and nodal X -semilattices of unions. Algebra, 19, J. Math. Sci. (New York) 111 (2002), no. 1, 3171–3226. [3] Ya.I. Diasamidze, T.T. Sirabidze, Complete semigroups of binary relations determined by three-element X -chains. Semigroups of binary relations, J. Math. Sci. (N.Y.) 117 (4) (2003) 4320–4350. [4] O. Givradze, Irreducible generating sets of complete semigroups of unions Bx (D) defined by semilattices of the class Σ2 (X, 4). (Russian), Sovrem. Mat. Prilozh. (2011) 74 translation in J. Math. Sci. (N.Y.) 186 (2012) no. 5, 745–750. [5] O. Givradze, Irreducible generating sets of complete semigroups of unions. (Russian), Sovrem. Mat. Prilozh. (2012) 84 translation in Part 3. J. Math. Sci. (N.Y.) 197 (2014) no. 6, 755–760. [6] A. Bakuridze, Generated sets of the complete semigroup binari relations defined by semilattices of the class Σ1 (X, 2), Int. J. Eng. Sci. Innovative Technol. (IJESIT) 5 (6) (2016) 17–26. [7] Ya. Diasamidze, O. Givradze, A. Bakuridze, Generated sets of the complete semigroup binari relations defined by semilattices of the class Σ1 (X, 3), Int. J. Eng. Sci. Innovative Technology (IJESIT) 5 (6) (2016) 52–69.
ScienceDirect Transactions of A. Razmadze Mathematical Institute 172 (2018) 378–387 www.elsevier.com/locate/trmi
Original article
Generated sets of the complete semigroup binary relations defined by semilattices of the finite chains Omari Givradze, Yasha Diasamidze, Nino Tsinaridze ∗ Department of Mathematics, Faculty of Physics, Mathematics and Computer Sciences, Shota Rustaveli Batumi State University, 35 Ninoshvili St., Batumi 6010, Georgia Received 20 April 2018; received in revised form 30 July 2018; accepted 18 August 2018 Available online 31 August 2018
Abstract In this article, we study generated sets of the complete semigroups of binary relations defined by X -semilattices unions of the finite chains. We found uniquely irreducible generating set for the given semigroups. c 2018 Ivane Javakhishvili Tbilisi State University. Published by Elsevier B.V. This is an open access article under the CC ⃝ BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Keywords: Semigroup; Semilattice; Binary relation
1. Introduction Let X be an arbitrary nonempty set, D be an X -semilattice of unions which is closed with respect to the settheoretic 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 correspondence a binary relation α f on the set X that satisfies the condition ⋃( ) αf = {x} × f (x) . x∈X
The set of all such α f ( f : X → D) is denoted by B X (D). It is easy to prove that B X (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 xαy. Further, let x, y ∈ X , Y ⊆ X , α ∈ B X (D), D˘ = Y ∈D Y and ∗ Corresponding author.
E-mail address: [email protected] (N. Tsinaridze). Peer review under responsibility of Journal Transactions of A. Razmadze Mathematical Institute. https://doi.org/10.1016/j.trmi.2018.08.004 c 2018 Ivane Javakhishvili Tbilisi State University. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND 2346-8092/⃝ license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
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T ∈ D. We denote by the symbols yα, Y α, V (D, α), X ∗ and V (X ∗ , α) the following sets: ⋃ { } yα, V (D, α) = {Y α | Y ∈ D}, X ∗ = Y | ∅ ̸= Y ⊆ X , yα = {x ∈ X | yαx}, Y α = y∈Y
{ } V (X ∗ , α) = Y α | ∅ ̸= Y ⊆ X ,
{ } DT = Z ∈ D | T ⊆ Z ,
{ } YTα = y ∈ X | yα = T .
It is well known the following statements: ˘ Z 1 , Z 2 , . . . , Z m−1 } be some finite X -semilattice of unions and C(D) = {P0 , P1 , P2 , . . . , Theorem 1.1. Let D = { D, 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˘ Z 1 Z 2 · · · Z m−1 ϕ= P0 P1 P2 · · · Pm−1 ˆZ = D \ D Z , then the following equalities are valid: and D D˘ = P0 ∪ P1 ∪ P2 ∪ · · · ∪ Pm−1 , ⋃ Z i = P0 ∪ ϕ(T ).
(1.1)
ˆZ T ∈D i
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, whereas sets P j (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, Chapter 11]). Definition 1.1. We say that an element α of the semigroup B X (D) is external if α ̸= δ ◦ β for all δ, β ∈ B X (D) \ {α} (see [1, Definition 1.15.1]). It is well known, that if B is all external elements of the semigroup B X (D) and B ′ is any generated set for the B X (D), then B ⊆ B ′ (see [1, Lemma 1.15.1]). ⋃ Definition 1.2. The representation α = T ∈D (YTα × T ) of binary relation α is called quasinormal, if ⋃ YTα = X and YTα ∩ YTα′ = ∅ for any T, T ′ ∈ D, T ̸= T ′ . T ∈D
Definition 1.3. Let α, β ⊆ X × X . Their product δ = α ◦ β is defined as follows: xδy (x, y ∈ X ) if there exists an element z ∈ X such that xαzβy (see [1, Chapter 1.3]). 2. Let Σm (X, m) be a class of all X -semilattices of unions whose every element is isomorphic to an X -semilattice of ˘ which satisfies the condition unions D = {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m = D}, Z 1 ⊂ Z 2 ⊂ Z 3 ⊂ · · · ⊂ Z m−2 ⊂ Z m−1 ⊂ Z m = D˘
(2.1)
(see Fig. 1). Let C(D) = {P0 , P1 , P2 , P3 , . . . , Pm−2 , Pm−1 ( }˘ be a family of sets, where P0), P1 , P2 , P3 , . . . , Pm−2 , Pm−1 are Z m−1 m−2 pairwise disjoint subsets of the set X and ϕ = PD0 ZP11 ZP22 ZP33 ·· ·· ·· ZPm−2 is a mapping of the semilattice Pm−1
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Fig. 1.
D onto the family of sets C(D). Then the formal equalities of the semilattice D have a form: D˘ = Z m = P0 ∪ P1 ∪ P2 ∪ P3 ∪ · · · ∪ Pm−1 , Z m = P0 ∪ P1 ∪ P2 ∪ P3 ∪ · · · ∪ Pm−2 , Z m−1 = P0 ∪ P1 ∪ P2 ∪ P3 ∪ · · · ∪ Pm−3 , ..........................................
(2.2)
Z 3 = P0 ∪ P1 ∪ P2 , Z 2 = P0 ∪ P1 , Z 1 = P0 , Here the elements P1 , P2 , P3 , . . . , Pm−2 , Pm−1 are bases sources, the element P0 is sources of completeness of the semilattice D. Therefore |X | ≥ m − 1, since |Pi | ≥ 1, i = 1, 2, . . . , m − 1 (see Theorem 1.1). In this paper we are learning irreducible generating sets of the semigroup B X (D) defined by semilattices of the class Σm (X, m). Definition 2.1. In the sequel, by symbol Σm.0 (X, m) we denote all semilattices D = {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m } of the class Σm (X, m) for which Z 1 is not empty. From the last inequality and from the formal equalities (2.2) it follows that Z 1 = P0 ̸= ∅, i.e. in this case |X | ≥ m. Note that, if D ∈ Σm,0 (X, m) and by symbol |X | is denoted the power of a set X , then the inequality |X | ≥ |D| always is true. ˜ = {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m } is irreducible generating set for the semilattice D. It is easy to see that D Lemma 2.1. If D ∈ Σm (X, m), then the following statements are true: (a)
P0 = Z 1 ;
(b)
Pi−1 = Z i \ Z i−1 , i = 2, 3, . . . , m − 2, m − 1, m.
(2.3)
Proof. From the formal equalities of the semilattice D immediately follows the following statements: P0 = Z 1 ,
Pi−1 = Z i \ Z i−1 , i = 2, 3, . . . , m − 2, m − 1, m.
The statements (a) and (b) of Lemma 2.1 are proved. □ It is well known, that the binary relation α, representation, which has a form ˘ × D) ˘ ε = (Z 1 × Z 1 ) ∪ ((Z 2 \ Z 1 ) × Z 2 ) ∪ · · · ∪ ((Z m−1 \ Z m−2 ) × Z m−1 ) ∪ ((Z m \ Z m−1 ) × Z m ) ∪ ((X \ D) is largest right unit of the semigroup B X (D) (see [1, Theorem 15.1.1]) and if by E (rX ) (D) is denoted all right units of the semigroup B X (D), then all right units of the semigroup are its external elements.
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Definition 2.2. We denote the following sets by symbol Ak and B(Ak ): { } Ak = V (X ∗ , α) | |V (X ∗ , α)| = k for anyα ∈ B X (D) , { } B(Ak ) = α ∈ B X (D) | V (X ∗ , α) ∈ Ak , where k = m, m − 1, m − 2, . . . , 3, 2, 1. Lemma 2.2. Let α ∈ B(Am ), then α is external element for the semigroup B X (D). Proof. Let α ∈ B(Am ) and α = δ ◦ β for some δ, β ∈ B X (D) \ {α}. If quasinormal representation of binary relation δ has a form m ⋃ δ= (Yiδ × Z i ), i=1
then α =δ◦β =
m m (⋃ ) ⋃ (Yiδ × Z i β). (Yiδ × Z i ) ◦ β =
(2.4)
i=1
i=1
It is easy to see that ˘ Z 1 β ⊆ Z 2 β ⊆ Z 3 β ⊆ · · · ⊆ Z m−2 β ⊆ Z m−1 β ⊆ Z m β = Dβ
(2.5)
since Z 1 ⊂ Z 2 ⊂ Z 3 ⊂ · · · ⊂ Z m−2 ⊂ Z m−1 ⊂ Z m = D˘ by definition of the semilattice D. By preposition { } { } Z 1 β, Z 2 β, Z 3 β, . . . , Z m−2 β, Z m−1 β, Z m β = Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m ( ) |V (D, α)| = |V (D, δ ◦ β)| = m , i.e. there exists one to one mapping ϕ of the set {Z 1 β, Z 2 β, Z 3 β, . . . , Z m−2 β, Z m−1 β, Z m β} on the set {Z 1 , Z 2 , Z 3 , . . . , Z m−2 , Z m−1 , Z m } which satisfies the condition (2.1) and (2.5). Now, let ϕ(Z i β) = Z j for some Z i ⊆ Z j (i ≤ j). If Z i ̸= Z j , then there exist such p and q (1 ≤ p ̸= q ≤ m), that Z p β = Z q for which ( p > q), since D is finite. But equality Z p β = Z q and inclusion p > q contradicts the inclusions (2.5), i.e. i = j. Of this follows that Z i β = Z i for all i = 1, 2, 3, . . . , m. We have that α =δ◦β =
m ⋃ i=1
(Yiδ × Z i β) =
m ⋃ (Yiδ × Z i ) = δ. i=1
But equality α = δ contradicts the preposition δ ∈ B X (D) \ {α}.
□
˘ ≥ 1, then any element of set B(Ak ) is generated by elements of the set Lemma 2.3. Let D ∈ Σm.0 (X, m). If |X \ D| B(Ak+1 ) (k = m − 1, . . . , 4, 3, 2). Proof. Let D ∈ Σm.0 (X, m), then |X | ≥ |D| = m. By Lemma 2.2 it follows that if α ∈ B(Am ) , then α is external element for the semigroup B X (D). Now, let k < m and α ∈ B(Ak ), then quasinormal representation of a binary ⋃k relation has a form α = i=1 (Yiα × Ti ), where Y1α , Y2α , . . . , Ykα ̸∈ {∅}, T1 , T2 , . . . , Tk ∈ D and T1 ⊂ T2 ⊂ · · · ⊂ Tk . It ∗ is easy to see, that V (X , α) = {T1 , T2 , . . . , Tk }, i.e. |V (X ∗ , α)| = k. From the sets Y1α , Y2α , . . . , Ykα there exists such α α set Yqα , that satisfies the condition |Yqα | ≥ 2 since k < m. Therefore, we may suppose, that Yqα = Yq1 ∪ Yq2 , where α α α α Yq1 , Yq2 ̸∈ {∅} and Yq1 ∩ Yq2 = ∅. Further, there exist such element T ∈ D (T ̸∈ {T1 , T2 , . . . , Tk }). For the element T we consider the following cases: (1) (2) (3) (4)
T ⊂ T1 ; Tk ⊂ T ; T1 ⊂ T ⊂ Tq ; Tq ⊂ T ⊂ Tk .
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(1). Let T ⊂ T1 and T = T0 . If quasinormal representations of the binary relations δ and β have the forms δ = (Y1α × T0 ) ∪ (Y2α × T1 ) ∪ (Y3α × T2 ) ∪ · · · α α α α ∪ (Yq−1 × Tk−2 ) ∪ (Yq1 × Tq−1 ) ∪ (Yq2 × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ · · · ∪ (Ykα × Tk ); β = (T0 × T1 ) ∪ ((T1 \ T0 ) × T2 ) ∪ ((T2 \ T1 ) × T3 ) ∪ · · · ∪ ((Tq−2 \ Tq−3 ) × Tq−1 ) ∪ ((Tq \ Tq−2 ) × Tq ) ∪ ((Tq+1 \ Tq ) × Tq+1 ) ∪ · · · ∪ ((Tk \ Tk−1 ) × Tk ) ∪ ((X \ Tk ) × T0 ), α α α α where Y1α , Y2α , Y3α , . . . , Yq−1 Yq1 Yq2 Yq+1 , . . . , Ykα ̸∈ {∅}, then it is easy to see, that |V (X ∗ , δ)| = |V (X ∗ , β)| = k + 1, ∗ ∗ i.e. V (X , δ), V (X , β) ∈ Ak+1 and
T0 β = T1 , T1 β = T2 , . . . , Tq−2 β = Tq−1 , Tq−1 β = Tq β = Tq , Tq+1 β = Tq+1 , Tq+2 β = Tq+2 , . . . , Tk β = Tk . From the last equalities we obtain, that: δ ◦ β = (Y1α × T0 β) ∪ (Y2α × T1 β) ∪ (Y3α × T2 β) ∪ · · · α α α α ∪(Yq−1 × Tq−2 β) ∪ (Yq1 × Tq−1 β) ∪ (Yq2 × Tq β) ∪ (Yq+1 × Tq+1 β) ∪ · · · ∪ (Ykα × Tk β) α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ (Y3α × T3 ) ∪ v ∪ (Yq−1 × Tq−1 ) ∪ α α α ((Yq1 ∪ Yq2 ) × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ · · · ∪ (Ykα × Tk ) = α, α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k. Therefore, V (X ∗ , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ).
(2). Let Tk ⊂ T and Tk+1 = T . If quasinormal representations of the binary relations δ and β have the forms α α δ = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 × Tq−1 ) ∪ (Yq1 × Tq )∪ α α (Yq2 × Tq+1 ) ∪ (Yq+1 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk+1 );
β = (T1 × T1 ) ∪ ((T2 \ T1 ) × T2 ) ∪ ((T3 \ T2 ) × T3 ) ∪ · · · ∪ ((Tq−1 \ Tq−2 ) × Tq−1 ) ∪ ((Tq+1 \ Tq−1 ) × Tq ) ∪ ((Tq+2 \ Tq+1 ) × Tq+1 ) ∪ ((Tq+3 \ Tq+2 ) × Tq+2 ) ∪ · · · ∪ ((Tk+1 \ Tk ) × Tk ) ∪ ((X \ Tk+1 ) × Tk+1 ), α α α α where Y1α , Y2α , . . . , Yq−1 Yq1 Yq2 Yq+1 , . . . , Ykα ̸∈ {∅}. It is easy to see, that |V (X ∗ , δ)| = |V (X ∗ , β)| = k + 1, ∗ ∗ i.e. V (X , δ), V (X , β) ∈ Ak+1 and
T1 β = T1 , T2 β = T2 , . . . , Tq−1 β = Tq−1 , Tq β = Tq+1 β = Tq , Tq+2 β = Tq+1 , Tq+3 β = Tq+2 , . . . , Tk+1 β = Tk . From the last equalities we obtain, that: α δ ◦ β = δ = (Y1α × Tβ1 ) ∪ (Y2α × Tβ2 ) ∪ · · · ∪ (Yq−1 × Tq−1 β) α α α ∪(Yq1 × Tq β) ∪ (Yq2 × Tq+1 β) ∪ (Yq+1 × Tq+2 β) ∪ · · · ∪ (Ykα × Tk+1 β) α α × Tq ) ∪ (Yq2 × Tq ) ∪ · · · ∪ (Ykα × Tk ) = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq1 α α α α = (Y1 × T1 ) ∪ (Y2 × T2 ) ∪ · · · ∪ ((Yq1 ∪ Yq2 ) × Tq ) ∪ · · · ∪ (Ykα × Tk ) = α, α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k. ∗ Therefore, V (X , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ).
(3). Let T1 ⊂ T ⊂ Tq . Then there exists an element T which is covered by element Tl , where 2 ≤ l ≤ q, i.e. Tl−1 ⊂ T ⊂ Tl . So, we have the following chain: T1 ⊂ T2 ⊂ Tl−1 ⊂ T ⊂ Tl ⊂ Tl+1 ⊂ Tl+2 ⊂ · · · ⊂ Tq ⊂ Tq+1 ⊂ Tq+2 ⊂ · · · ⊂ Tk . If quasinormal representations of the binary relations δ and β have the forms α α α δ = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yl−1 × Tl−1 ) ∪ (Ylα × T ) ∪ (Yl+1 × Tl ) ∪ (Yl+2 × Tl+1 ) ∪ · · · α α α α α ∪ (Yq−1 × Tq−2 ) ∪ (Yq1 × Tq−1 ) ∪ (Yq2 × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk );
β = (T1 × T1 ) ∪ ((T2 \ T1 ) × T2 ) ∪ ((T3 \ T2 ) × T3 ) ∪ · · · ∪ ((Tl−1 \ Tl−2 ) × Tl−1 ) ∪ ((T \ Tl−1 ) × Tl ) ∪ ((Tl \ T ) × Tl+1 ) ∪ ((Tl+1 \ Tl ) × Tl+2 ) ∪ · · · ∪ ((Tq−2 \ Tq−3 ) × Tq−1 ) ∪ ((Tq \ Tq−2 ) × Tq ) ∪ ((Tq+1 \ Tq ) × Tq+1 ) ∪ · · · ∪ ((Tk \ Tk−1 ) × Tk ) ∪ ((X \ Tk ) × T ),
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α α α α α α α where Y1α , . . . , Yl−1 , Ylα , Yl+1 , . . . , Yq−1 , Yq1 , Yq2 , Yq+1 , Yq+2 , . . . , Ykα ̸∈ {∅}. It is easy to see, that |V (X ∗ , δ)| = ∗ ∗ ∗ |V (X , β)| = k + 1, i.e. V (X , δ), V (X , β) ∈ Ak+1 and
T1 β = T1 , T2 β = T2 , . . . , Tl−1 β = Tl−1 , Tβ = Tl , Tl β = Tl+1 , Tl+1 β = Tl+2 , . . . , Tq−2 β = Tq−1 , Tq−1 β = Tq β = Tq , Tq+1 β = Tq+1 , Tq+2 β = Tq+2 , . . . , Tk β = Tk . From the last equalities we obtain, that: α δ ◦ β = (Y1α × T1 β) ∪ (Y2α × T2 β) ∪ · · · ∪ (Yl−1 × Tl−1 β) ∪ (Ylα × Tβ) α α α α α ∪ (Yl+1 × Tl β) ∪ (Yl+2 × Tl+1 β) ∪ · · · ∪ (Yq−1 × Tq−2 β) ∪ (Yq1 × Tq−1 β) ∪ (Yq2 × Tq β) α α ∪ (Yq+1 × Tq+1 β) ∪ (Yq+2 × Tq+2 β) ∪ · · · ∪ (Ykα × Tk β) α α α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yl−1 × Tl−1 ) ∪ (Ylα × Tl ) ∪ (Yl+1 × Tl+1 ) ∪ (Yl+2 × Tl+2 ) ∪ · · · α α α α ∪ (Yq1 × Tq ) ∪ (Yq2 × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk ) α α α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yl−1 × Tl−1 ) ∪ (Ylα × Tl ) ∪ (Yl+1 × Tl+1 ) ∪ (Yl+2 × Tl+2 ) ∪ · · · α α α α ∪ ((Yq1 ∪ Yq2 ) × Tq ) ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Ykα × Tk ) = α, α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k. ∗ Therefore, V (X , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ).
(4). Let Tq ⊂ T ⊂ Tk . Then there exists an element T which is covered by element Tr , where q + 1 ≤ r ≤ k, i.e. Tr −1 ⊂ T ⊂ Tr . So, we have the following chain: T1 ⊂ T2 ⊂ · · · ⊂ Tq ⊂ Tq+1 ⊂ · · · ⊂ Tr −1 ⊂ T ⊂ Tr ⊂ Tr +1 ⊂ · · · ⊂ Tk . If quasinormal representations of the binary relations δ and β have the forms α α δ = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 × Tq−1 ) ∪ · · · ∪ (Yq1 × Tq ) α α α ∪ (Yq2 × Tq+1 ) ∪ (Yq+1 × Tq+2 ) ∪ (Yq+2 × Tq+3 ) ∪ · · · ∪ (Yrα−2 × Tr −1 )
∪ (Yrα−1 × T ) ∪ (Yrα × Tr ) ∪ (Yrα+1 × Tr +1 ) ∪ · · · ∪ (Ykα × Tk ); β = (T1 × T1 ) ∪ ((T2 \ T1 ) × T2 ) ∪ ((T3 \ T2 ) × T3 ) ∪ · · · ∪ ((Tq−1 \ Tq−2 ) × Tq−1 ) ∪ ((Tq+1 \ Tq−1 ) × Tq ) ∪ ((Tq+2 \ Tq+1 ) × Tq+1 ) ∪ · · · ∪ ((Tr −1 \ Tr −2 ) × Tr −2 ) ∪ ((T \ Tr −1 ) × Tr −1 ) ∪ ((Tr \ T ) × Tr ) ∪ ((Tr +1 \ Tr ) × Tr +1 ) ∪ · · · ∪ ((Tk \ Tk−1 ) × Tk ) ∪ ((X \ Tk ) × T ), α α α α α where Y1α , Y2α , . . . , Yq−1 , . . . , Yq1 , Yq2 , Yq+1 , Yq+2 , . . . , Yrα−2 , Yrα−1 , Yrα , Yrα+1 , . . . , Ykα ∗ ∗ ∗ |V (X , δ)| = |V (X , β)| = k + 1, i.e. V (X , δ), V (X ∗ , β) ∈ Ak+1 and
̸∈ {∅}. It is easy to see, that
T1 β = T1 , T2 β = T2 , . . . , Tq−1 β = Tq−1 , Tq β = Tq+1 β = Tq , Tq+2 β = Tq+1 , Tq+3 β = Tq+2 , . . . , Tr −1 β = Tr −2 , Tβ = Tr −1 , Tr β = Tr , Tr +1 β = Tr +1 , . . . , Tk β = Tk . From the last equalities we obtain, that: α α α × Tq−1 β) ∪ · · · ∪ (Yq1 × Tq β) ∪ (Yq2 × Tq+1 β) δ ◦ β = (Y1α × T1 β) ∪ (Y2α × T2 β) ∪ · · · ∪ (Yq−1 α α × Tq+2 β) ∪ (Yq+2 × Tq+3 β) ∪ · · · ∪ (Yrα−2 × Tr −1 β) ∪ (Yq+1
∪ (Yrα−1 × Tβ) ∪ (Yrα × Tr β) ∪ (Yrα+1 × Tr +1 β) ∪ · · · ∪ (Ykα × Tk β) α α α × Tq−1 ) ∪ · · · ∪ (Yq1 × Tq ) ∪ (Yq2 × Tq ) = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 α α ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Yrα−2 × Tr −2 )
∪ (Yrα−1 × Tr −1 ) ∪ (Yrα × Tr ) ∪ (Yrα+1 × Tr +1 ) ∪ · · · ∪ (Ykα × Tk ) α α α = (Y1α × T1 ) ∪ (Y2α × T2 ) ∪ · · · ∪ (Yq−1 × Tq−1 ) ∪ · · · ∪ ((Yq1 ∪ Yq2 ) × Tq ) α α ∪ (Yq+1 × Tq+1 ) ∪ (Yq+2 × Tq+2 ) ∪ · · · ∪ (Yrα−2 × Tr −2 )
∪ (Yrα−1 × Tr −1 ) ∪ (Yrα × Tr ) ∪ (Yrα+1 × Tr +1 ) ∪ · · · ∪ (Ykα × Tk ) = α,
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α α since Yqα = Yq1 ∪ Yq2 , i.e. α = δ ◦ β, since α α Yqα = Yq1 ∪ Yq2 ,
V (X ∗ , α) = {T1 , T2 , . . . , Tk } and |V (X ∗ , α)| = k.
Therefore, V (X ∗ , α) ∈ Ak , i.e. element α is generated by elements of the set B(Ak+1 ). So, any element of set B(Ak ) is generated by elements of the set B(Ak+1 ), where k = m − 1, . . . , 4, 3, 2.
□
˘ ≥ 1, then any element of the set B(A1 ) is generated by elements of the Lemma 2.4. Let D ∈ Σm.0 (X, m). If |X \ D| set B(A2 ). Proof. Let T is any element of the semilattice D and α = X × T ∈ B(A1 ). If quasinormal representations of the binary relations δ and β (δ, β ∈ B X (D)) have forms ˘ × T ′ ), δ = (Y1δ × T ) ∪ (Y2δ × T ′ ), β = ( D˘ × T ) ∪ ((X \ D) where Y1δ , Y2δ ̸∈ {∅} (T ′ ∈ D, T ̸= T ′ ), then |V (X ∗ , δ)| = |V (X ∗ , β)| = 2, i.e. δ, β ∈ B(A2 ) and Tβ = T ′ β = T, δ ◦ β = (Y1δ × Tβ) ∪ (Y2δ × T ′ β) = (Y1δ × T ) ∪ (Y2δ × T ) = X × T = α, since the representation of a binary relation δ is quasinormal. ˘ ≥ 1, then any element of the set B(A1 ) is generated by elements of the set B(A2 ). □ Therefore, if |X \ D| ˘ ≥ 1 and Theorem 2.1. Let D ∈ Σm.0 (X, m) and D is a finite set. If |X \ D| { } { } ∗ ∗ Am = V (X , α) | |V (X , α)| = m for any α ∈ B X (D) , B(Am ) = α ∈ B X (D) | V (X ∗ , α) ∈ Am , then the set B(Am ) is irreducible generating set for the semigroup B X (D). ˘ ≥ 1. First, we proved that every element of the semigroup B X (D) is generated by elements of Proof. Let |X \ D| the set B(Am ). Indeed, let α be an⋃arbitrary element of the B X (D). Then quasinormal representation of a ⋃msemigroup m binary relation α has a form α = i=1 (Yiα × Z i ), where i=1 Yiα = X and Yiα ∩ Y jα = ∅ for any 1 ≤ i ̸= j ≤ m. For the number |V (X ∗ , α)| we consider the following cases: (1) If k = m − 1, . . . , 4, 3, 2, then from Lemma 2.3 immediately follows that any element of the set B(Ak ) is generated by elements of the set B(Ak+1 ). (2) If k = 1, then from Lemma 2.4 immediately follows that any element of the set B(A1 ) is generated by elements of the set B(A2 ). ˘ ≥ 1, then every element of the semigroup B X (D) are generated by elements of the set B(Am ). So, if |X \ D| From Lemma 2.2 it follows that, every element of the set B(Am ) is external element for the semigroup B X (D). Therefore, we have that B(Am ) is irreducible generating set for the semigroup B X (D). □ ˘ ≥ 1 and Theorem 2.2. Let D ∈ Σm.0 (X, m) and D is a finite set. If |X \ D| { } { } Am = V (X ∗ , α) | |V (X ∗ , α)| = m for any α ∈ B X (D) , B(Am ) = α ∈ B X (D) | V (X ∗ , α) ∈ Am . Then the paver of a set B(Am ) is equal to the paver surjection of the set X on the set D. Proof. Let X and D be finite sets. If α ∈ B(Am ), then by definition of a semigroup B X (D) there exists such a mapping f of a set X on the set D = {Z 1 , Z 2 , . . . , Z m }, since |X | ≥ |D| and α = α f , i.e. mapping f is some surjection of the set X on the set D. □ ˘ ≥ 1 and Corollary 2.1. Let D ∈ Σm.0 (X, m). If |X \ D| { } Am = V (X ∗ , α) | |V (X ∗ , α)| = m for any α ∈ B X (D) ,
{ } B(Am ) = α ∈ B X (D) | V (X ∗ , α) ∈ Am .
If X and D are finite sets |X | = n and |D| = m, then the number |B(Am )| elements of the set B(Am ) is equal to |B(Am )| = n m +
n−1 ∑ (−1)i · Cni · (n − i)m . i=1
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Proof. Let X and D be finite sets, |X | = n and |D| = m. In our case we have |X | ≥ |D|. It is well known that, the ∑n−1 number surjection of the set X on the set D is equal to the number n m + i=1 (−1)i · Cni · (n − i)m . Now, Corollary 2.1 immediately follows from Theorem 2.2. □ Example 2.1. X = {1, 2, 3} and D = {{1}, {1, 2}}. Then 3 = |X | > |D| = 2 and { } { } { } α1 = (1, 1), (2, 1), (3, 1) , α2 = (1, 1), (2, 1), (3, 1), (3, 2) , α3 = (1, 1), (2, 1), (2, 2)(3, 1) , { } { } α4 = (1, 1), (1, 2)(2, 1), (3, 1) , α5 = (1, 1), (1, 2), (2, 1), (2, 2)(3, 1) , { } α6 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2) , { } { } α7 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2) , α8 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2) . In this case we have B X (D) = {α1 , α2 , . . . , α7 , α8 };
B(A2 ) = {α2 , α3 , α4 , α5 , α6 , α7 },
B(A1 ) = {α1 , α8 }
and { } { } { } α3 ◦ α2 = (1, 1), (2, 1), (2, 2)(3, 1) ◦ (1, 1), (2, 1), (3, 1), (3, 2) = (1, 1), (2, 1), (3, 1) = α1 , { } { } α2 ◦ α4 = (1, 1), (2, 1), (3, 1), (3, 2) ◦ (1, 1), (1, 2)(2, 1), (3, 1) { } = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2) = α8 , i.e. the set B(A2 ) is irreducible generating set for the semigroup B X (D). Example 2.2. X = {1, 2, 3, 4} and D = {{1}, {1, 2}, {1, 2, 3}}. Then 4 = |X | > |D| = 3 and { } { } α1 = (1, 1), (2, 1), (3, 1), (4, 1) , α2 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) , { } α3 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } { } α4 = (1, 1), (1, 2), (2, 1), (3, 1), (4, 1) , α5 = (1, 1), (2, 1), (2, 2), (3, 1), (4, 1) , { } { } α6 = (1, 1), (2, 1), (3, 1), (3, 2), (4, 1) , α7 = (1, 1), (2, 1), (3, 1), (4, 1), (4, 2) , { } { } α8 = (1, 1), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2) , α9 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (4, 1) , { } { } α10 = (1, 1), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2) , α11 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (4, 1) , { } { } α12 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1) , α13 = (1, 1), (1, 2), (2, 1), (3, 1), (4, 1), (4, 2) , { } α14 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) , { } α15 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2) , { } α16 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2) , { } α17 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1) , { } { } α18 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (4, 1) , α19 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1) , { } { } α20 = (1, 1), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1) , α21 = (1, 1), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3) , { } α22 = (1, 1), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α23 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1) , { } α24 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2), (4, 3) , { } α25 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1) , { } α26 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1) , { } α27 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3) , { } α28 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α29 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α30 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2), (4, 3) , { } α31 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1) , { } α32 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) ,
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{ } α33 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2) , { } α34 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α35 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α36 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2) , { } α37 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α38 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α39 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α40 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α41 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α42 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α43 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α44 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α45 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α46 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1) , { } α47 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α48 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α49 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) , { } α50 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2) , { } α51 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α52 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1) , { } α53 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α54 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) , { } α55 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1) , { } α56 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2) , { } α57 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2), (4, 3) , { } α58 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (4, 1) , { } α59 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2) , { } α60 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α61 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (4, 1) , { } α62 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2) , { } α63 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2), (4, 3) , { } α64 = (1, 1), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2), (4, 3) , { } α65 = (1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)v , { } α66 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2) , { } α67 = (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α68 = (1, 1), (1, 2), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3) , { } α69 = (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α70 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) , { } α71 = (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1) , { } α72 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (4, 1), (4, 2) , { } α73 = (1, 1), (1, 2), (1, 3), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) ,
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{ } α74 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1) , { } α75 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1) , { } α76 = (1, 1), (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) , { } α77 = (1, 1), (2, 1), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) , { } α78 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (4, 1), (4, 2), (4, 3) , { } α79 = (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1) , { } α80 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (4, 1), (4, 2) , { } α81 = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) . In this case B X (D) = {α1 , α2 , α3 , . . . , α80 , α81 }, B(A2 ) = {α4 , α5 , . . . , α44 , α45 },
B(A3 ) = {α46 , α47 , . . . , α80 , α81 },
B(A1 ) = {α1 , α2 , α3 }
and α4 = α72 ◦ α76 ,
α5 = α49 ◦ α76 ,
α6 = α46 ◦ α76 ,
α8 = α54 ◦ α76 ,
α9 = α80 ◦ α76 ,
α10 = α51 ◦ α76 ,
α7 = α76 ◦ α76 , α11 = α52 ◦ α64 ,
α12 = α46 ◦ α64 ,
α13 = α60 ◦ α76 ,
α14 = α47 ◦ α64 ,
α15 = α53 ◦ α64 ,
α16 = α56 ◦ α64 ,
α17 = α79 ◦ α65 ,
α18 = α72 ◦ α77 ,
α19 = α49 ◦ α77 ,
α20 = α46 ◦ α77 ,
α21 = α76 ◦ α77 ,
α22 = α48 ◦ α77 ,
α23 = α80 ◦ α77 ,
α24 = α51 ◦ α77 ,
α25 = α52 ◦ α49 ,
α26 = α46 ◦ α49 ,
α27 = α60 ◦ α77 ,
α28 = α47 ◦ α49 ,
α29 = α53 ◦ α49 ,
α30 = α78 ◦ α67 ,
α31 = α79 ◦ α67 ,
α32 = α72 ◦ α79 ,
α33 = α49 ◦ α53 ,
α34 = α46 ◦ α52 ,
α35 = α76 ◦ α79 ,
α36 = α80 ◦ α79 ,
α37 = α48 ◦ α52 ,
α38 = α52 ◦ α55 ,
α39 = α51 ◦ α79 ,
α40 = α60 ◦ α79 ,
α42 = α65 ◦ α71 ,
α43 = α53 ◦ α55 ,
α44 = α78 ◦ α71 ,
α45 = α79 ◦ α71 ,
α41 = α67 ◦ α79 , α1 = α7 ◦ α7 ,
α2 = α7 ◦ α17 ,
α3 = α31 ◦ α31 .
References [1] Ya. Diasamidze, Sh. Makharadze, Complete Semigroups of Binary Relations, Kriter, Turkey, 2013.
Further Reading [1] Ya. Diasamidze, N. Aydin, A. Erdoˇgan, Generating set of the complete semigroups of binary relations, Appl. Math. Irvin 7 (2016) 98–107. [2] Ya.I. Diasamidze, Sh.I. Makharadze, Complete semigroups of binary relations defined by elementary and nodal X -semilattices of unions. Algebra, 19, J. Math. Sci. (New York) 111 (2002), no. 1, 3171–3226. [3] Ya.I. Diasamidze, T.T. Sirabidze, Complete semigroups of binary relations determined by three-element X -chains. Semigroups of binary relations, J. Math. Sci. (N.Y.) 117 (4) (2003) 4320–4350. [4] O. Givradze, Irreducible generating sets of complete semigroups of unions Bx (D) defined by semilattices of the class Σ2 (X, 4). (Russian), Sovrem. Mat. Prilozh. (2011) 74 translation in J. Math. Sci. (N.Y.) 186 (2012) no. 5, 745–750. [5] O. Givradze, Irreducible generating sets of complete semigroups of unions. (Russian), Sovrem. Mat. Prilozh. (2012) 84 translation in Part 3. J. Math. Sci. (N.Y.) 197 (2014) no. 6, 755–760. [6] A. Bakuridze, Generated sets of the complete semigroup binari relations defined by semilattices of the class Σ1 (X, 2), Int. J. Eng. Sci. Innovative Technol. (IJESIT) 5 (6) (2016) 17–26. [7] Ya. Diasamidze, O. Givradze, A. Bakuridze, Generated sets of the complete semigroup binari relations defined by semilattices of the class Σ1 (X, 3), Int. J. Eng. Sci. Innovative Technology (IJESIT) 5 (6) (2016) 52–69.