Gen. Math. Notes, Vol. 26, No. 1, January 2015, pp. 84-101
ISSN 2219-7184; Copyright © ICSRS Publication, 2015
www.i-csrs.org
Available free online at http://www.geman.in
Some Regular Elements, Idempotents and
Right Units of Semigroup BX (D) Defined by
X- Semilattices which is Union of a Chain and
Two Rhombus
Yasha Diasamidze1 and Giuli Tavdgiridze2
1,2
Sh. Rustaveli State University, Batumi, Georgia
1
E-mail: diasamidze_ya@mail.ru
2
E-mail: g.tavdgiridze@mail.ru
(Received: 28-9-14 / Accepted: 17-11-14)
Abstract
In
this
paper
we
take
Q = {T0 , T1 ,..., Tm−6 , Tm−5 , Tm−4 , Tm−3 , Tm−2 , Tm−1, Tm } ( m ≥ 6)
,
subsemilattice of X- semilattice of unions D where the elements Ti s are satisfying
the following properties,
T0 ⊂ T1 ⊂ ... ⊂ Tm−5 ⊂ Tm−4 ⊂ Tm−2 ⊂ Tm , T0 ⊂ T1 ⊂ ... ⊂ Tm−5 ⊂ Tm−3 ⊂ Tm−2 ⊂ Tm , T0 ⊂ T1 ⊂ ... ⊂ Tm−5 ⊂
⊂ Tm−3 ⊂ Tm−1 ⊂ Tm , Tm−4 \ Tm−3 ≠ ∅, Tm−3 \ Tm−4 ≠ ∅, Tm−4 \ Tm−1 ≠ ∅, Tm−1 \ Tm−4 ≠ ∅, Tm−2 \ Tm−1 ≠ ∅,
Tm−1 \ Tm−2 ≠ ∅, Tm−4 ∪ Tm−3 = Tm−2 , Tm−4 ∪ Tm−1 = Tm−2 ∪ Tm−1 = Tm .
We will investigate the properties of regular elements and idempotents of the
complete semigroup of binary relations BX (D) satisfying V (D, α)=Q. Also we
investigate right units of the semmigroup BX (Q). For the case where X is a finite
set we derive formulas by means of which we can calculate the numbers of
regular elements, idempotents and right units of the respective semigroup.
Some Regular Elements, Idempotents and…
85
Keywords: Semigroup, Semilatticce, Binary relations, Right units, Regular
elements.
1
Introduction
Let X be an arbitrary nonempty set, D be a X − semilattice of unions, i.e. a
nonempty set of subsets of the set X that is closed with respect to the set-theoretic
operations of unification of elements from D , f be an arbitrary mapping from X
into D . To each such a mapping f there corresponds 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
x∈ X
( f : X → D) is denoted by 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 a X − semilattice of unions
D (see [2,3 2.1 p. 34]).
By ∅ we denote an empty binary relation or 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 ) , T ∈ D , ∅ ≠ D′ ⊆ D and t ∈ D = ∪ Y . Then by symbols we denote the
Y ∈D
following sets:
yα = { x ∈ X | yα x} , Yα = ∪ yα , V ( D,α ) = {Yα | Y ∈ D} ,
y∈Y
X ∗ = {T | ∅ ≠ T ⊆ X } , Dt′ = {Z ′ ∈ D′ | t ∈ Z ′} , DT′ = {Z ′ ∈ D′ | T ⊆ Z ′} .
ɺɺ′ = {Z ′ ∈ D′ | Z ′ ⊆ T } , l ( D′, T ) = ∪ ( D′ \ D′ ) , Y α = { x ∈ X | xα = T } .
D
T
T
T
Under the symbol ∧ ( D, Dt ) we mean an exact lower bound of the set Dt in the
semilattice D .
Definition 1.1: An element α taken from the semigroup BX ( D ) called a regular
element of the semigroup BX ( D ) if in BX ( D ) there exists an element β such that
α β α =α .
Definition 1.2: We say that a complete X − semilattice of unions D is
an XI − semilattice of unions if it satisfies the following two conditions:
(a)
⌣
∧ ( D, Dt ) ∈ D for any t ∈ D ;
(b)
Z = ∪ ∧ ( D, Dt ) for any nonempty element Z of D (see [2,3 definition
t∈Z
1.14.2]).
Definition 1.3: Let D be an arbitrary complete X − semilattice of unions,
α ∈BX ( D) and YTα = {x ∈ X | xα = T } . If
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Yasha Diasamidze et al.
V ( X ∗ , α ), if ∅ ∉ D,

V [α ] = V ( X ∗ , α ), if ∅ ∈ V ( X ∗ , α ),
V ( X ∗ , α ) ∪ {∅}, if ∅ ∉ V ( X ∗ , α ) and ∅ ∈ D,
then it is obvious that any binary relation α of a semigroup BX ( D ) can always be
written in the form α = ∪ (YTα × T ) the sequel, such a representation of a binary
T ∈V [α ]
relation α will be called quasinormal.
Note that for a quasinormal representation of a binary relation α , not all sets YTα
(T ∈ V [α ]) can be different from an empty set. But for this representation the
following conditions are always fulfilled:
(a)
YTα ∩ YTα′ = ∅ , for any T , T ′ ∈ D and T ≠ T ′ ;
(b)
X=
∪
T ∈V [α ]
YTα
(See [2, 3 definition 1.11.1]).
Definition 1.4: We say that a nonempty element T is a non limiting element of the
set D ′ if T \ l ( D′, T ) ≠ ∅ and a nonempty element T is a limiting element of the set
D ′ if T \ l ( D′, T ) = ∅ (see [2, 3 definition 1.13.1 and definition 1.13.2]).
Definition 1.5: The one-to-one mapping ϕ between the complete X − semilattices
of unions φ ( Q, Q ) and D′′ is called a complete isomorphism if the condition
ϕ ( ∪ D1 ) =
∪ ϕ (T ′ )
T ′∈D1
is fulfilled for each nonempty subset D1 of the semilattice D′ (see [2, 3 definition
6.3.2]).
Definition 1.6: Let α be some binary relation of the semigroup BX ( D ) . We say
that the complete isomorphism ϕ between the complete semilattices of unions Q
and D′ is a complete α − isomorphism if
(a)
(b)
Q = V ( D, α ) ;
ϕ ( ∅ ) = ∅ for ∅ ∈ V ( D, α ) and ϕ (T ) α = T for any T ∈ V ( D, α ) (see [2,3
definition 6.3.3]).
Lemma 1.1: Let Y = { y1 , y2 ,..., yk } and D j = {T1 ,..., T j } be some sets, where k ≥ 1
and j ≥ 1 . Then the number s( k , j ) of all possible mappings of the set Y on any such
subset of the set D′j
that T j ∈ D ′j can be calculated by the formula
k
k
s ( k , j ) = j − ( j − 1) (see [2, 3 Corollary 1.18.1]).
Some Regular Elements, Idempotents and…
87
Lemma 1.2: Let D j = {T1 , T2 ,...T j } , X and Y − be three such sets, that ∅ ≠ Y ⊆ X .
If f is such mapping of the set X , in the set D j , for which f ( y ) = Tj for some
y ∈ Y , then the number s of all those mappings f of the set X in the set D j is
(
equal to s = j X \Y ⋅ j Y − ( j − 1)
Y
) (see [2,3 Theorem 1.18.2]).
Lemma 1.3: Let D be a complete X − semilattice of unions. If a binary relation
ε of the form
(
⌣
⌣
ε = ∪ ({t} × ∧ ( D, Dt ) ) ∪ ( X \ D ) × D
⌣
t ∈D
)
is a right unit of BX ( D ) , then it is largest right unit.
⌣
Theorem 1.1: Let D = {D, Z1 , Z 2 ,..., Z n −1 } be some finite X − semilattice of unions
and C( D) = { P0 , P1, P2 ,..., Pn−1} be the family of sets of pairwise nonintersecting subsets of
the set X . If ϕ is a mapping of the semilattice D on the family of sets C ( D )
⌣
which satisfies the condition ϕ ( D ) = P0 and ϕ ( Z i ) = Pi for any i = 1, 2,..., n − 1 and
Dˆ Z = D \ {T ∈ D | Z ⊆ T } ,
then the following equalities are valid:
⌣
D = P0 ∪ P1 ∪ P2 ∪ ... ∪ Pn −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
(*), then among the parameters Pi ( i = 0,1, 2,..., n − 1) there exist such parameters that
cannot be empty sets for D. Such sets Pi ( 0 < i ≤ n − 1) are called basis sources,
whereas sets Pj ( 0 ≤ j ≤ n − 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 preimage of a basis 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 [5]) .
Theorem 1.2: A binary relation ε ∈ BX ( D) is a right units of this semigroup iff ε
is idempotent and D = V ( D, ε ) (see Theorem 4.1.3).
Theorem 1.3: Let D be a finite X − semilattice of unions and α ∈ BX ( D ) ; D (α )
be the set of those elements T of the semilattice Q = V ( D,α ) \ {∅} which are
nonlimiting elements of the set QɺɺT . Then a binary relation α having a
quasinormal representation of the form α = ∪ (YTα × T ) is a regular element of
T ∈V ( D ,α )
the semigroup BX ( D ) iff V ( D, α ) is a XI − semilattice of unions and for some
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Yasha Diasamidze et al.
α − isomorphism ϕ from V ( D, α ) to some X − subsemilattice D′ of the semilattice
D the following conditions are fulfilled:
a)
b)
∪α
ɺɺ (
T ′∈D
)T
YTα′ ⊇ ϕ (T ) for any T ∈ D (α ) ;
ɺɺ (α ) (see [2,3
YTα ∩ ϕ ( T ) ≠ ∅ for any nonlimiting element T of the set D
T
Theorem 6.3.3]).
⌣
Lemma 1.4: Let D = {D, Z1 , Z 2 ,..., Z n −1 } and C ( D ) = {P0 , P1 , P2 ,..., Pn −1} are the finite
semilattice of unions and the family of sets of pairwise nonintersecting subsets of
⌣
Z1 Z 2 ...Z n −1  is a mapping of the semilattice D on the family of
the set X , ϕ =  D

 P0 P1 P2 ...Pn −1 
⌣
the sets C ( D ) , If ϕ (T ) = P ∈ C ( D ) for some D ≠ T ∈ D , then Dt = D \ DɺɺT for all t ∈ P
(see [***] Lemma 11.6.1).
Definition 1.8: Let Q and D′ be respectively some XI and X − subsemilattices of
the complete X − semilattice of unions D . Then Rϕ ( Q , D ′ ) is a subset of the
semigroup BX ( D ) such that α ∈ Rϕ ( Q, D′ ) only if the following conditions are
fulfilled for the elements α and ϕ :
a)
b)
c)
The binary relation α be regular element of the semigroup BX ( D ) ;
V ( D,α ) = Q ;
ϕ is a complete α − isomorphism between the complete semilattices of
unions Q and D′ satisfying the conditions a ) and b ) of the Theorem 1.4
(see [2,3] definition 6.3.4).
Further ε Q , Ω ( Q ) and Φ ( Q, D′ ) respectively are the identity mapping of the
semilattice Q , the set of all XI − subsemilattices of the complete X − semilattice of
unions D such that Q′ ∈ Ω ( Q ) if there exists a complete isomorphism between the
semilattices Q′ and Q and the set of all complete isomorphisms of the
XI − semilattice of unions into the semilattice D ′ such that ϕ ∈ Φ ( Q, D′ ) if ϕ is a
α − isomorphism for some α ∈ BX ( D ) and V ( D, α ) = Q .
Next, let
R ( Q, D ′ ) =
∪
ϕ∈Φ ( Q , D ′)
Rϕ ( Q, D ′ )
and R ( D ′ ) =
R ( Q, D′ ) is an arbitrary element of the set
∪
Q ′∈Ω ( Q )
R ( Q ′, D ′ ) .
{R ( Q, D′) | ϕ ∈ Φ ( Q, D′)}
ϕ
and a mapping
between the complete XI and X − semilattices of unions Q and D′ is a
complete isomorphism corresponding to the set R ( Q, D′ ) (see Theorem 6.3.5).
T →T
Some Regular Elements, Idempotents and…
89
Theorem 1.4: A regular element α of the semigroup B X ( D ) is idempotent iff the
mapping ϕ satisfying the condition ϕ (T ) = T α for any T ∈ V ( D, α ) is an identity
mapping of the semilattice V ( D, α ) (see[2,3 Theorem 6.3.7).
Theorem 1.5: If E X( r ) ( Q ) is the set of all right units of BX ( Q ) , then
E X( ) ( Q ) = Rε Q ( Q, Q ) (see Theorem 6.3.11).
r
Theorem 1.6: Let X be a finite set. if ϕ is a fixed element of the set Φ ( Q, D′ ) ,
Ω ( Q ) = m0 and q be number of all automorphisms of the semilattice then
R ( D ′ ) = m0 ⋅ q ⋅ Rϕ ( Q, D ′ )
(See Theorem 6.3.5).
2
Results
Theorem 2.1: Let Q = {T0 ,T1,..., Tm−6 , Tm−5 , Tm−4 ,Tm−3 ,Tm−2 , Tm−1, Tm } ( m ≥ 5)
subsemilattice of the semilattice D and
be
a
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm − 4 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm −1 ⊂ Tm ,
Tm − 4 \ Tm −3 ≠ ∅, Tm −3 \ Tm − 4 ≠ ∅, Tm − 4 \ Tm −1 ≠ ∅,
Tm −1 \ Tm − 4 ≠ ∅, Tm − 2 \ Tm −1 ≠ ∅, Tm −1 \ Tm − 2 ≠ ∅,
Tm − 4 ∪ Tm −3 = Tm − 2 , Tm − 4 ∪ Tm −1 = Tm − 2 ∪ Tm −1 = Tm .
Then Q is always an XI − semilattice of unions.
Proof: Let P0 , P1 ,..., Pm −1 and C be the pairwise nonintersecting subsets of the set
X and ϕ be a mapping of the semilattice Q onto the family of sets
{P0 , P1 ,..., Pm−1 , C } that has the form
Tm
Tm − 2
Tm − 6 Tm −5 Tm − 4 Tm −3 Tm − 2 Tm −1 Tm 
ϕ =  TP0 TP1 ...

 0 1 ... Pm − 6 Pm −5 Pm − 4 Pm −3 Pm − 2 Pm −1 C 
Tm −1
Tm − 4
Tm −3
i
i
i
Tm −5
Tm −6
T1
T0
Fig. 1
Then the formal equalities corresponding to the semilattice Q are written as
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Yasha Diasamidze et al.
Tm =C∪P0 ∪P1 ∪...∪Pm−6 ∪Pm−5 ∪Pm−4 ∪Pm−3 ∪Pm−2 ∪Pm−1,
Tm−1 = C∪P0 ∪P1 ∪...∪Pm−6 ∪Pm−5 ∪Pm−4 ∪Pm−3 ∪Pm−2,
Tm−2 = C∪P0 ∪P1 ∪...∪Pm−6 ∪Pm−5 ∪Pm−4 ∪Pm−3 ∪Pm−1,
Tm−3 =C∪P0 ∪P1 ∪...∪Pm−6 ∪Pm−5 ∪Pm−4,
Tm−4 = C∪P0 ∪P1 ∪...∪Pm−6 ∪Pm−5 ∪Pm−3 ∪Pm−1,
Tm−5 =C∪P0 ∪P1 ∪...∪Pm−6,
Tm−6 =C∪P0 ∪P1 ∪...∪Pm−7,
−−−−−−−−−−−−
T1 = C∪P0,
T0 =C,
(2.1)
where C ≥ 0 , Pm −5 ≥ 0 , Pm −3 ≥ 0 and P0 , P1 ,..., Pm − 6 , Pm − 4 , Pm − 2 , Pm −1 ∉ {∅} . Further, let
t ∈ Tm . Then from equalities (2.1) and from the Lemma 1.2 we have
T0 , if t ∈ C ,
T1 , if t ∈ P0 ,
− − − − − −
Tm − 5 , if t ∈ Pm − 6 ,
∧ ( Q, Qt ) = Tm − 5 , if t ∈ Pm − 5 ,
Tm − 3 , if t ∈ Pm − 4 ,
Tm − 5 , if t ∈ Pm − 3 ,
Tm −1 , if t ∈ Pm − 2 ,
Tm − 4 , if t ∈ Pm −1 .
Then We have obtained that ∧ ( Q, Qt ) ∈ D for any t ∈ Tm . Furthermore, if
Q ∧ = {∧ ( Q, Qt ) t ∈ Tm } , then Q ∧ = {T0 , T1 ,..., Tm − 5 , Tm − 4 , Tm − 3 , Tm −1 } and it is easy to verify
that any nonempty element of the semilattice Q is the union of some elements of
the set Q ∧ . Now, taking into account Definition 1.2, we obtain that Q is an
XI − semilattice of unions.
For the largest right unit ε of the semigroup BX ( D ) we have:
ε = ( Pm − 2 × Tm −1 ) ∪ ( Pm − 4 × Tm −3 ) ∪ ( Pm −1 × Tm − 4 ) ∪ ( ( Pm − 6 ∪ Pm −5 ∪ Pm −3 ) × Tm −5 ) ∪ ...
∪ ( P0 × T1 ) ∪ ( C × T0 ) ∪ ( ( X \ Tm ) × Tm ) .
(See Lemma 1.3).
Of the formal equalities (2.1) follows that:
Pm − 2 = Tm −1 \ Tm − 2 , Pm −1 = Tm − 4 \ Tm −1 , Pm − 4 = Tm − 3 \ Tm − 4 ,
Pm − 6 ∪ Pm − 5 ∪ Pm − 3 = (Tm −1 ∩ Tm − 4 ) \ Tm − 6 , P0 = T1 \ T0 , C = T0
Then we have
ε = ( (Tm −1 \ Tm − 2 ) × Tm −1 ) ∪ ( (Tm −3 \ Tm − 4 ) × Tm −3 ) ∪ ( (Tm − 4 \ Tm −1 ) × Tm − 4 ) ∪ ( ( (Tm −1 ∩ Tm − 4 ) \ Tm − 6 ) × Tm −5 ) ∪
∪ ( (Tm − 6 \ Tm − 7 ) × Tm − 6 ) ∪ ... ( (T1 \ T0 ) × T1 ) ∪ (T0 × T0 ) ∪ ( ( X \ Tm ) × Tm ) .
Theorem 2.2: Let Q = {T0 ,T1,..., Tm−6 , Tm−5 , Tm−4 ,Tm−3 ,Tm−2 , Tm−1, Tm } ( m ≥ 5)
subsemilattice of the semilattice D such that
be
a
Some Regular Elements, Idempotents and…
91
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm − 4 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm −1 ⊂ Tm ,
Tm − 4 \ Tm −3 ≠ ∅, Tm −3 \ Tm − 4 ≠ ∅, Tm − 4 \ Tm −1 ≠ ∅,
Tm −1 \ Tm − 4 ≠ ∅, Tm − 2 \ Tm −1 ≠ ∅, Tm −1 \ Tm − 2 ≠ ∅,
Tm − 4 ∪ Tm −3 = Tm − 2 , Tm − 4 ∪ Tm −1 = Tm − 2 ∪ Tm −1 = Tm .
(see Fig. 1). A binary relation α of the semigroup BX ( D ) that has a quasinormal
m
representation of the form α = ∪ (Yiα × Ti ) , where Q = V ( D, α ) , is a regular
i=0
element of the semigroup BX ( D) iff for some α − isomorohism ϕ of the
semilattice Q on some X − subsemilattice D′ = {ϕ (T1 ) ,ϕ (T2 ) ,...,ϕ (Tm )} of the
semilattice D satisfies the conditions.
Proof: To begin with, we recall that Q is an XI − semilattice of unions (see
Theorem 2.1). Now we are to find the nonlimiting element of the sets Qɺɺq∗ of the
semilattice Q∗ = Q \ {∅} (see definition 1.4). Indeed, let Tq ∈ Q∗ , where
Then for
q = 0,1, 2,..., m
(
(
(
ɺɺ
l (Q
ɺɺ
l (Q
ɺɺ
l (Q
ɺɺ
l (Q
)
(
(
)
)
q = 0,1, 2,..., m .
we obtain respectively
ɺɺ∗ , T = ∪ ({T , T ,..., T } \ {T } ) = ∪{T , T ,..., T } = T ,
l Q
Tm m
0 1
m
m
0 1
m−1
m
∗
ɺɺ
l QTm−1 , Tm−1 = ∪ ({T0 , T1 ,..., Tm−5 , Tm−3 , Tm−1} \ {Tm−1} ) = ∪{T0 , T1 ,..., Tm−5 , Tm−3 } = Tm−3 ,
ɺɺ∗ , T
l Q
= ∪ ({T , T ,..., T , T , T , T } \ {T } ) = ∪{T , T ,..., T , T , T
)
)
) = ∪ ({T ,T ,..., T
) = ∪ ({T ,T ,..., T
) = ∪ ({T , T ,...,T
) = ∪ ({T ,T ,...,T
Tm−2 m− 2
∗
Tm−3 , Tm−3
∗
Tm−4 , Tm− 4
∗
Tm−5 , Tm−5
∗
Tm−6 , Tm−6
0
1
0
1
0
1
0
1
0
1
m −5
m −4
m −3
m−2
m−2
0
1
m −5
} \ {Tm−3}) = ∪{T0 , T1 ,..., Tm−5 } = Tm−5 ,
m−5 , Tm − 4 } \ {Tm− 4 } ) = ∪ {T0 , T1 ,..., Tm −5 } = Tm−5 ,
m−5 } \ {Tm−5 } ) = ∪ {T0 , T1 ,..., Tm−6 } = Tm −6 ,
m− 6 } \ {Tm−6 } ) = ∪{T0 , T1 ,..., Tm− 7 } = Tm−7 ,
m−5 , Tm −3
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
ɺɺ∗ , T = ∪ ({T , T } \ {T } ) = ∪{T } = T ,
l Q
T1 1
0 1
1
0
0
∗
ɺɺ
l Q , T = ∪ ({T } \ {T } ) = ∪{∅} = ∅,
T0
0
0
0
Therefore
(
)
(
)
ɺɺ∗ , T = T \ T = ∅, T \ l Q
ɺɺ∗ , T
Tm \ l Q
Tm m
m
m
m −1
Tm−1 m −1 = Tm −1 \ Tm − 3 ≠ ∅,
∗
ɺɺ
ɺɺ∗ , T
Tm − 2 \ l QTm−2 , Tm − 2 = Tm − 2 \ Tm − 2 = ∅, Tm −3 \ l Q
Tm−3 m −3 = Tm − 3 \ Tm − 5 ≠ ∅,
∗
ɺɺ , T
ɺɺ∗
Tm − 4 \ l Q
Tm−4 m − 4 = Tm − 4 \ Tm −5 ≠ ∅, Tm − 5 \ l QTm−5 , Tm − 5 = Tm − 5 \ Tm − 6 ≠ ∅,
ɺɺ∗ , T
T
\l Q
=T
\T
≠ ∅,
m−6
(
(
(
Tm−6
m−6
)
)
)
m −6
(
(
m −7
)
)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
ɺɺ∗ , T = T \ T ≠ ∅, T \ l Q
ɺɺ∗ , T = T \ ∅ ≠ ∅, if T ≠ ∅,
T1 \ l Q
T1 1
1
0
0
T0 0
0
0
(
)
(
)
m −4
} = Tm−2 ,
m −3 ,
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Yasha Diasamidze et al.
(
)
i.e. Tq \ l QɺɺTq , Tq ≠ ∅ , where
q = 1, 2,..., m − 6, m − 5, m − 4, m − 3, m − 1 .
Thus we have
obtained that Tm , Tm− 2 are the limiting elements of the sets QɺɺT∗m , QɺɺT∗m−2 and the Tq
are
the
nonlimiting
elements
of
the
set
ɺɺ∗ ,
Q
Tq
where
Now, in view of Theorem (1.3) a binary
relation α of the semigroup B X ( D ) is a regular element of this semigroup iff there
exists an α − isomorphism ϕ of the semilattice Q on some X − subsemilattice
D ′ = {ϕ (T0 ) ,...,ϕ (Tm )} of the semilattice Q such that
q = 1, 2,..., m − 6, m − 5, m − 4, m − 3, m − 1 .
( )
Y0α ∪ Y1α ∪ ... ∪ Y pα ⊇ ϕ T p , Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ⊇ ϕ (Tm −3 ) ,
( )
Y0 ∪ Y1 ∪ ... ∪ Ym −5 ∪ Ymα−3 ∪ Ymα−1 ⊇ ϕ ( Tm −1 ) , Yqα ∩ ϕ Tq ≠ ∅
α
for any
α
α
p = 0,1,..., m − 4, m
and
q = 1, 2,..., m − 4, m − 3, m − 1 .
It is clearly understood that the inclusion Y0α ∪ Y1α ∪ ... ∪ Ymα = X ⊇ ϕ (Tm ) is always
valid. Therefore
( )
Y0α ∪ Y1α ∪ ... ∪ Y pα ⊇ ϕ T p , Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ⊇ ϕ (Tm −3 ) ,
( )
Y0 ∪ Y1 ∪ ... ∪ Ym −5 ∪ Ymα−3 ∪ Ymα−1 ⊇ ϕ ( Tm −1 ) , Yqα ∩ ϕ Tq ≠ ∅
α
for any
α
α
p = 0,1,..., m − 4, m
and
q = 1, 2,..., m − 4, m − 3, m − 1 .
Theorem is proved.
Corollary 2.1: Let Q = {T0 ,T1,..., Tm−6 , Tm−5 , Tm−4 ,Tm−3 ,Tm−2 , Tm−1, Tm } ( m ≥ 5)
subsemilattice of the semilattice D such that
be
a
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm − 4 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm −1 ⊂ Tm ,
Tm − 4 \ Tm −3 ≠ ∅, Tm −3 \ Tm − 4 ≠ ∅, Tm − 4 \ Tm −1 ≠ ∅,
Tm −1 \ Tm − 4 ≠ ∅, Tm − 2 \ Tm −1 ≠ ∅, Tm −1 \ Tm − 2 ≠ ∅,
Tm − 4 ∪ Tm −3 = Tm − 2 , Tm − 4 ∪ Tm −1 = Tm − 2 ∪ Tm −1 = Tm .
A binary relation α
of the semigroup BX ( D ) , which has a quasinormal
m
representation of the form α = ∪ (Yiα × Ti ) , such that где Q = V ( D, α ) is an
i =0
idempotent element of the semigroup BX ( D) iff
Y0α ∪ Y1α ∪ ... ∪ Y pα ⊇ T p , Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ⊇ Tm −3 ,
Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ∪ Ymα−1 ⊇ Tm −1 , Yqα ∩ Tq ≠ ∅
Some Regular Elements, Idempotents and…
for any
p = 0,1,..., m − 4
and
93
q = 1, 2,..., m − 4, m − 3, m − 1 .
Proof: As has been shown in the proof of Theorem 2.2, all elements of the set Q,
different from the elements T3, T5 are the nonlimiting elements of the semilattice
Q. Now the corollary immediately follows from Theorem 1.4. Corollary is
proved.
Corollary 2.2: Let Q = {T0 ,T1,..., Tm−6 , Tm−5 , Tm−4 ,Tm−3 ,Tm−2 , Tm−1, Tm } ( m ≥ 5) be a semilattice
such that
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm − 4 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm −1 ⊂ Tm ,
Tm − 4 \ Tm −3 ≠ ∅, Tm −3 \ Tm − 4 ≠ ∅, Tm − 4 \ Tm −1 ≠ ∅,
Tm −1 \ Tm − 4 ≠ ∅, Tm − 2 \ Tm −1 ≠ ∅, Tm −1 \ Tm − 2 ≠ ∅,
Tm − 4 ∪ Tm −3 = Tm − 2 , Tm − 4 ∪ Tm −1 = Tm − 2 ∪ Tm −1 = Tm .
A binary relation α of the semigroup BX ( Q ) that has a quasinormal representation
m
of the form α = ∪(Yiα × Ti ) , such that Q = V ( Q,α ) , is a right unit of the semigroup
BX ( Q ) iff
i =0
Y0α ∪ Y1α ∪ ... ∪ Y pα ⊇ T p , Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ⊇ Tm −3 ,
Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ∪ Ymα−1 ⊇ Tm −1 , Yqα ∩ Tq ≠ ∅
for any
p = 0,1,..., m − 4
and
q = 1, 2,..., m − 4, m − 3, m − 1 .
Proof: By assumption, Q = V ( Q, α ) . Now the validity of the corollary immediately
follows from Corollary 2.2 and from Theorem 1.2.
Corollary is proved.
Theorem 2.3: Let Q = {T0 ,T1,..., Tm−6 , Tm−5 , Tm−4 ,Tm−3 ,Tm−2 , Tm−1, Tm } ( m ≥ 5)
subsemilattice of the semilattice D such that
be
a
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm − 4 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm −1 ⊂ Tm ,
Tm − 4 \ Tm −3 ≠ ∅, Tm −3 \ Tm − 4 ≠ ∅, Tm − 4 \ Tm −1 ≠ ∅,
Tm −1 \ Tm − 4 ≠ ∅, Tm − 2 \ Tm −1 ≠ ∅, Tm −1 \ Tm − 2 ≠ ∅,
Tm − 4 ∪ Tm −3 = Tm − 2 , Tm − 4 ∪ Tm −1 = Tm − 2 ∪ Tm −1 = Tm .
If the
XI −
semilattices
Q
and D ′ = {T0 , T1 ,..., Tm − 6 , Tm −5 , Tm − 4 , Tm −3 , Tm − 2 , Tm −1 , Tm } are
α − isomorphic and Ω ( Q ) = m0 , then the following equality is valid:
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Yasha Diasamidze et al.
Tm
Tm − 2
Tm −1
Tm − 4
Tm −3
i
i
i
Tm −5
Tm −6
T1
T0
Fig. 2
(
R ( D ′ ) = m0 ⋅ 2
⋅ ( m − 4 ) ((
(
⋅ (( m − 2)
⋅ ( m − 3)
) ( − 2 ) ⋅⋅⋅ (( m − 5) − ( m − 6 )
)
) ⋅ ( m − 4)
− ( m − 5)
)⋅
(
− ( m − 4)
) ⋅ ( ( m − 3) − ( m − 4 ) ) ⋅
− ( m − 3)
) ⋅ ( m + 1) .
T1 \T0
−1 ⋅ 3
Tm−1 ∩Tm−4 \Tm−5
Tm−4 \ Tm−1
Tm−1 \Tm−2
T2 \T1
T2 \T1
Tm−6 \ Tm −7
Tm−5 \Tm−6
Tm −4 \Tm−1
Tm −1 \Tm−2
Tm−6 \ Tm−7
)⋅
Tm−5 \Tm −6
Tm −3 \ Tm −4
Tm−3 \Tm−4
X \Tm
Proof: In the first place, we note that the semilattice Q has only one
automorphisms (i.e. Φ( Q,Q) =1 ). Let α ∈ R ( Q , D ′ ) and a quasinormal
representation of a regular binary relation α have the form
α=
m
∪ (Y α × T ) .
i
i
(2.2)
i =1
Then according to Theorem 1.2 the condition α ∈ R ( Q , D ′ ) is fulfilled if
Y0α ∪Y1α ∪ ... ∪Ypα ⊇ Tp , p = 0,1,..., m − 4; Y0α ∪ Y1α ∪... ∪Ymα−5 ∪Ymα−3 ⊇ Tm−3 , ( 2.3)
Y0α ∪Y1α ∪ ... ∪Ymα−5 ∪ Ymα−3 ∪ Ymα−1 ⊇ Tm−1; Yqα ∩Tq ≠ ∅,
( 2.4)
q = 1,2,..., m − 4, m − 3, m −1 .
( 2.5)
Now, assume that fα is a mapping of the set X in D such that fα ( t ) = tα for any
t ∈ X . f 0α , f m − 5 , f m − 4 , f m − 3 , f m −1 , f kα ( k = 1, 2,..., m − 6 ) and f m +1 are respectively the
restrictions of the mapping fα on the sets
T0 , (Tm −1 ∩ Tm − 4 ) \ Tm − 6 , Tm − 4 \ Tm −1 , Tm − 3 \ Tm − 4 , Tm −1 \ Tm − 2 ,
T1 \ T0 ,..., Tm − 6 \ Tm − 7 , X \ Tm
We have, by assumption, that these sets do not intersect pair wise and the settheoretic union of these sets is equal to X .
Let us establish the properties of the mappings t ∈ X . f0α , f m −5 , f m − 4 , f m −3 , f m −1 , f kα
( k = 1, 2,..., m − 6 ) and f m +1 .
Some Regular Elements, Idempotents and…
95
1) t ∈ T0 . Hence by virtue of the inclusions ( 2.3) we have t ∈ Y0α , i.e., tα = T0 by the
definition of the set Y0α . Thus f0α ( t ) = T0 for any t ∈ T0 .
2) t ∈ (Tm −1 ∩ Tm − 4 ) \ Tm − 6 , . In that case, by virtue of inclusion ( 2.4 ) , ( 2.3) we have
t ∈ (Tm −1 ∩ Tm − 4 ) \ Tm −1 ∩ Tm − 4 ⊆
(
) (
)
⊆ Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ∪ Ymα−1 ∩ Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα− 4 = Y0α ∪ Y1α ∪ ... ∪ Ymα−5 .
Therefore tα ∈ {T0 , T1 ,..., Tm −5 } by the definition of the sets Y0α , Y1α ,..., Ymα−5 . Thus
f m − 5α ( t ) ∈ {T0 , T1 ,..., Tm −5 }
for any t ∈ (Tm −1 ∩ Tm − 4 ) \ Tm − 6 .
On the other hand, the inequality Ymα−5 ∩ Tm−5 ≠ ∅ is true. Therefore tm−5 ∈ Ymα−5 for
some element tm −5 ∈ Tm −5 . Hence it follows that tm − 5α = Tm − 5 . Furthermore, if
tm −5 ∈ Tm − 6 , then tm −5α ∈ {T0 , T1 ,..., Tm − 6 } . However the latter condition contradicts the
equality tm − 5α = Tm − 5 . The contradiction obtained shows that tm −5 ∈ Tm −5 \ Tm − 6 . Thus
f m − 5α ( tm − 5 ) = Tm −5 for some tm −5 ∈ Tm −5 \ Tm − 6 .
3)
t ∈ Tm − 4 \ Tm −1 .
In that case, by virtue of inclusion
α
α
α
( 2.3)
we have
α
t ∈ Tm − 4 \ Tm −1 ⊆ Tm − 4 ⊆ Y0 ∪ Y1 ∪ ... ∪ Ym −5 ∪ Ym −4 . Therefore tα ∈ {T0 , T1 ,..., Tm − 5 , Tm − 4 } by
the definition of the sets Y0α , Y1α ,...Ymα−5 , Ymα−4 . Thus f m − 4α ( t ) ∈ {T0 , T1 ,..., Tm −5 , Tm − 4 } for
any t ∈ Tm − 4 \ Tm −1 .
On the other hand, the inequality Ymα−4 ∩ Tm− 4 ≠ ∅ is true. Therefore tm−4 ∈ Ymα− 4 for
some element tm −3 ∈ Tm − 4 . Hence it follows that tm − 4α = Tm − 4 . Furthermore, if
tm − 4 ∈ Tm −1 , then tm − 4α ∈ {T0 , T1 ,..., Tm − 5 , Tm − 3 , Tm −1} . However the latter condition
contradicts the equality tm − 4α = Tm − 4 . The contradiction obtained shows that
tm − 4 ∈ Tm − 4 \ Tm −1 , . Thus f m − 4α ( tm − 4 ) = Tm − 4 for some tm − 4 ∈ Tm − 4 \ Tm −1 .
4)
t ∈ Tm − 3 \ Tm − 4 , .
In that case, by virtue of inclusion
( 2.3)
we have
t ∈ Tm −3 \ Tm − 4 , ⊆ Tm −3 ⊆ Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 . Therefore tα ∈ {T0 , T1 ,..., Tm − 5 , Tm − 3 } by
the definition of the sets Y0α , Y1α ,...Ymα−5 , Ymα−3 . Thus f m −3α ( t ) ∈ {T0 , T1 ,..., Tm −5 , Tm −3 } for
any t ∈ Tm −3 \ Tm − 4 .
On the other hand, the inequality Ymα−3 ∩ Tm−3 ≠ ∅ is true. Therefore tm−3 ∈ Ymα−3 for
some element tm −3 ∈ Tm −3 . Hence it follows that tm − 3α = Tm − 3 . Furthermore, if
tm −3 ∈ Tm − 4 , then tm −3α ∈ {T0 , T1 ,..., Tm − 5 , Tm − 4 } . However the latter condition
contradicts the equality tm − 3α = Tm − 3 . The contradiction obtained shows that
tm −3 ∈ Tm −3 \ Tm − 4 , . Thus f m − 3α ( tm − 3 ) = Tm − 3 for some tm −3 ∈ Tm −3 \ Tm − 4 .
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Yasha Diasamidze et al.
t ∈ Tm −1 \ Tm − 2 .
In that case, by virtue of inclusion
( 2.4 )
we have
t ∈ Tm −1 \ Tm − 2 ⊆ Tm −1 ⊆ Y0α ∪ Y1α ∪ ... ∪ Ymα−5 ∪ Ymα−3 ∪ Ymα−1 .
Therefore
tα ∈ {T0 , T1 ,..., Tm − 5 , Tm −3 , Tm −1 }
by
the
definition
of
the
sets
Y0α , Y1α ,...Ymα−5 , Ymα−3 , Ymα−1 . Thus f m −1α ( t ) ∈ {T0 , T1 ,..., Tm −5 , Tm − 3 , Tm −1 } for any t ∈ Tm −1 \ Tm − 2 .
On the other hand, the inequality Ymα−1 ∩ Tm−1 ≠ ∅ is true. Therefore tm−2 ∈ Ymα−1 for
some element tm − 2 ∈ Tm −1 . Hence it follows that tm − 2α = Tm −1 . Furthermore, if
tm − 2 ∈ Tm − 2 , then tm − 2α ∈ {T0 , T1 ,..., Tm −5 , Tm − 4 , Tm − 3 , Tm − 2 } . However the latter condition
contradicts the equality tm − 2α = Tm −1 . The contradiction obtained shows that
tm − 2 ∈ Tm −1 \ Tm − 2 , . Thus f m −1α ( tm − 2 ) = Tm −1 for some tm − 2 ∈ Tm −1 \ Tm − 2 .
6) t ∈ Tk \ Tk −1 ( k = 1, 2,..., m − 6 ) . In that case, by virtue of inclusion ( 2.3) we have
t ∈ Tk \ Tk −1 ⊆ Tk ⊆ Y0α ∪ Y1α ∪ Y2α ∪ ... ∪ Ykα .
Therefore tα ∈ {T0 , T1 ,..., Tk } by the definition of the sets Y0α , Y1α ,..., Ykα . Thus
f kα ( t ) ∈ {T0 , T1 ,..., Tk } for any t ∈ Tk \ Tk −1 .
On the other hand, the inequality Ykα ∩ Tk ≠ ∅ is true. Therefore tk ∈ Ykα for some
element tk ∈ Tk . Hence it follows that tk α = Tk . Furthermore, if tk ∈ Tk −1 , then
tk α ∈ {T0 , T1 ,..., Tk −1 } . However the latter condition contradicts the equality t k α = Tk .
The contradiction obtained shows that tk ∈ Tk \ Tk −1 . Thus f kα ( tk ) = Tk for some
tk ∈ Tk \ Tk −1 .
m
m
i =0
i =0
7) t ∈ X \ Tm . Then by virtue of the condition X = ∪ Yiα we have t ∈ ∪ Yiα . Hence
we obtain tα ∈{T0 ,T1,T2 ,...,Tm} . Thus f mα ( t ) ∈ {T0 , T1 , T2 ,..., Tm } for any t ∈ X \ Tm .
Therefore for a binary relation α ∈ R ( Q, D′ ) that has a representation of form 2.2
there always exists a uniquely defined system
( f0α , fm−5α , fm−4α , fm−3α , fm−1α , fkα , fm+1α ) ,
(2.6)
where ( k = 1, 2,..., m − 6 ) . It is obvious that to different elements of the set R ( Q, D′ )
there correspond different ordered sets of form ( 2.6 ) .
Now let
f1 : T0 →{T0} , fm−5 : (Tm−1 ∩Tm−4 ) \ Tm−6 →{T0 ,T1,...,Tm−5} , fm−4 : Tm−4 \ Tm−1 →{T0 ,T1,...,Tm−5,Tm−4} ,
fm−3 : Tm−3 \ Tm−4 →{T0,T1,...,Tm−5,Tm−3} , fm−1 : Tm−1 \ Tm−2 →{T0 ,T1,...,Tm−5 ,Tm−3 ,Tm−1} ,
fk : Tk \ Tk −1 → {T0 , T1 ,..., Tk } k = 1, 2,..., m − 6 , f m +1 : X \ Tm → {T0 , T1 ,..., Tm }
Some Regular Elements, Idempotents and…
97
be the mappings satisfying the following conditions:
8) f1 ( t ) = T0 for any t ∈ T0 ;
9) f m −5 ( t ) ∈ {T0 , T1 ,..., Tm − 5 } for any t ∈ (Tm −1 ∩ Tm − 4 ) \ Tm −6 and f m −5α ( tm −5 ) = Tm −5 for
some tm −5 ∈ Tm −5 \ Tm − 6 ;
10) f m − 4 ( t ) ∈ {T0 , T1 ,..., Tm −5 , Tm − 4 } for any t ∈ Tm − 4 \ Tm −1 and f m − 4α ( tm − 4 ) = Tm − 4 for some
tm −3 ∈ Tm − 4 \ Tm −1 , ;
11) f m −3 ( t ) ∈ {T0 , T1 ,..., Tm −5 , Tm − 3 } for any t ∈ Tm − 3 \ Tm − 4 and f m −3α ( tm −3 ) = Tm −3 for some
tm −3 ∈ Tm −3 \ Tm − 4 , ;
12) f m −1 ( t ) ∈ {T0 , T1 ,..., Tm −5 , Tm −3 , Tm −1} for any t ∈ Tm −1 \ Tm − 2 , and f m −1α ( tm − 2 ) = Tm −1 for
some tm − 2 ∈ Tm −1 \ Tm − 2 ;
13) f k ( t ) ∈ {T0 , T1 ,..., Tk } for any t ∈ Tk \ Tk −1 and f k ( tk ) = Tk for some tk ∈ Tk \ Tk −1 ;
14) f m +1 ( t ) ∈ {T0 , T1 , T2 ,..., Tm } for any t ∈ X \ Tm .
Now we write the mapping f : X → D as follows:
 f 0 ( t ) , if t ∈ T0 ,
f
 m −5 ( t ) , if t ∈ (Tm −1 ∩ Tm − 4 ) \ Tm − 6 ,
 f m − 4 ( t ) , if t ∈ Tm − 4 \ Tm −1 ,

f ( t ) =  f m −3 ( t ) , if t ∈ Tm −3 \ Tm − 4 ,
 f m −1 ( t ) , if t ∈ Tm −1 \ Tm − 2 ,

 f k ( t ) , if t ∈ Tk \ Tk −1 , k = 1, 2,..., m − 6,
 f m +1 ( t ) , if t ∈ X \ Tm .

To the mapping f we put into correspondence the relation β = ∪ ({t} × f ( t ) ) .
t∈ X
Now let Yi β = {t ∈ X | tβ = Ti } , where i = 0,1, 2,..., m . With this notation, the binary relation β is represented as β = ∪ (Yi β × Ti ) . Moreover, from the defnition of the binary
m
i =0
relation β we immediately obtain
Y0β ∪ Y1β ∪ ... ∪ Ypβ ⊇ T p , p = 0,1,..., m − 6,
Y0β ∪ Y1β ∪ ... ∪ Ymβ−5 ∪ Ymβ−3 ⊇ Tm −3 ,
Y0β ∪ Y1β ∪ ... ∪ Ymβ−5 ∪ Ymβ−3 ∪ Ymβ−1 ⊇ Tm −1 ,
Yqβ ∩ Tq ≠ ∅, q = 1, 2,..., m − 4, m − 3, m − 1.
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Yasha Diasamidze et al.
since f m −5α ( tm −5 ) = Tm −5 for some tm −5 ∈ Tm −5 \ Tm − 6 ; f m − 4α ( tm − 4 ) = Tm − 4 for some
tm −3 ∈ Tm − 4 \ Tm −1 ; f m − 3α ( tm − 3 ) = Tm − 3 for some tm −3 ∈ Tm −3 \ Tm − 4 ; f m −1α ( tm − 2 ) = Tm −1 for
some tm − 2 ∈ Tm −1 \ Tm − 2 ; f k ( tk ) = Tk for some tk ∈ Tk \ Tk −1 , k = 1, 2,..., m − 6 .
Hence by virtue of Theorem 2.2 we conclude that the binary relation β is a
regular element of the semigroup BX ( D ) that belongs to the set R ( Q, D′ ) .
Therefore there exists a one-to-one correspondence between the binary relations
α of the set R ( Q, D′ ) that have representations of form ( 2.2 ) and ordered systems
of the form ( 2.6 ) .
The numbers of all mappings of the form f0α , f m −5 , f m − 4 , f m −3 , f m −1 , f kα
( k = 1, 2,..., m − 6 ) and
( m − 4 ) ( (T
(m − 3)T
m −1 ∩ T m − 4
m−4
\ T m −1
(α ∈ R ( Q , D ′ ) )
f m +1
) \ Tm − 6 ) \ (Tm − 5 \ Tm − 6 )
− (m − 4 )
(m − 2 )T
T m − 4 \ T m −1
m −1 \ T m − 2
(( m − 4 )
are equal respectively to 1,
Tm − 5 \ Tm − 6
− (m − 5)
Tm − 5 \ Tm − 6
),
, ( m − 3 ) T m − 3 \ T m − 4 − ( m − 4 ) Tm − 3 \ T m − 4 ,
− (m − 3)
T m −1 \ T m − 2
, ( k + 1) T \T − k T \T , ( k = 1, 2,..., m − 6 )
k
k −1
k
k −1
( m + 1) X \T .
m
Therefore the equality
(
R (Q, D ′) = 2
⋅ ( m − 4 ) ((
(
⋅ ( m − 3)
) ( − 2 ) ⋅ ⋅ ⋅ (( m − 5 )
)
) ⋅ (m − 4)
− (m − 5)
(
− (m − 4)
) ⋅ (( m − 2 )
T1 \ T0
−1 ⋅ 3
T2 \ T1
Tm −1 ∩Tm − 4 \ Tm −5
Tm − 3 \ Tm − 4
T2 \ T1
Tm −5 \ Tm − 6
Tm −3 \ Tm − 4
Tm −6 \ Tm − 7
Tm −5 \ Tm − 6
Tm −1 \ Tm − 2
− (m − 6)
) ⋅ (( m − 3)
− ( m − 3)
Tm −6 \ Tm − 7
Tm − 4 \ Tm −1
Tm −1 \ Tm − 2
)⋅
− (m − 4)
) ⋅ ( m + 1)
X \ Tm
Tm − 4 \ Tm −1
.
is valid.
Now, using the equalities Ω ( Q ) = m0 , Φ( Q, D′) =1 and Theorem 1.6, we Obtain
(
R ( D ′ ) = m0 ⋅ 2
⋅ ( m − 4 ) ((
(
⋅ ( m − 3)
) ( − 2 ) ⋅⋅⋅ (( m − 5) − ( m − 6) ) ⋅
)
) ⋅ ( m − 4)
− ( m − 5)
) ⋅ ( ( m − 3) − ( m − 4 )
(
− ( m − 4)
) ⋅ (( m − 2) − ( m − 3) ) ⋅ ( m + 1) .
T1 \T0
−1 ⋅ 3
Tm−1 ∩Tm−4 \Tm−5
Tm−3 \Tm−4
Theorem is proved.
T2 \T1
T2 \T1
Tm−5 \Tm−6
Tm −3 \Tm −4
Tm−6 \Tm −7
Tm −5 \Tm−6
Tm−1 \Tm−2
Tm−6 \Tm −7
Tm−4 \Tm −1
Tm−1 \Tm−2
X \Tm
Tm−4 \ Tm−1
)⋅
)⋅
Some Regular Elements, Idempotents and…
99
Corollary 1.4. Let Q = {T0 ,T1,..., Tm−6 , Tm−5 , Tm−4 ,Tm−3 ,Tm−2 , Tm−1, Tm } ( m ≥ 5)
subsemilattice of the semilattice D such that
be
a
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm − 4 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm − 2 ⊂ Tm ,
T0 ⊂ T1 ⊂ ... ⊂ Tm −5 ⊂ Tm −3 ⊂ Tm −1 ⊂ Tm ,
Tm − 4 \ Tm −3 ≠ ∅, Tm −3 \ Tm − 4 ≠ ∅, Tm − 4 \ Tm −1 ≠ ∅,
Tm −1 \ Tm − 4 ≠ ∅, Tm − 2 \ Tm −1 ≠ ∅, Tm −1 \ Tm − 2 ≠ ∅,
Tm − 4 ∪ Tm −3 = Tm − 2 , Tm − 4 ∪ Tm −1 = Tm − 2 ∪ Tm −1 = Tm .
If E X( r ) (Q ) is the set of all right units of the semigroup BX (Q ) , then
(
E X( r ) ( Q ) = 2
⋅( m − 4)
T1 \ T0
( (Tm−1 ∩Tm−4 ) \ Tm−5 )
(
⋅ ( m − 3)
Tm −3 \ Tm− 4
)(
−1 ⋅ 3
T2 \ T1
(
−2
⋅ ( m − 4)
− ( m − 4)
T2 \ T1
) ⋅⋅⋅ ( ( m − 5)
Tm −5 \ Tm−6
Tm −3 \ Tm −4
− ( m − 5)
) ⋅ (( m − 2)
Tm −6 \ Tm −7
Tm−5 \ Tm− 6
Tm−1 \ Tm −2
− ( m − 6)
) ⋅ ( ( m − 3)
− ( m − 3)
Tm−6 \ Tm− 7
Tm− 4 \ Tm−1
)⋅
− ( m − 4)
) ⋅ ( m + 1)
Tm−1 \ Tm −2
X \ Tm
Tm− 4 \ Tm−1
)⋅
.
Proof: By virtue of Theorem 6.3.1 we have E X( r ) ( Q ) = RεQ ( Q, Q ) , where ε Q is the
identity mapping of the semilattice
Q
. Hence, taking into account the equalities
Q = D ′ RεQ (Q, Q) = R ( Q, D′) and
(
R ( D ′ ) = m0 ⋅ 2
⋅ ( m − 4 ) ((
(
⋅ ( m − 3)
) ( − 2 ) ⋅⋅⋅ (( m − 5) − ( m − 6) ) ⋅
)
) ⋅ ( m − 4)
− ( m − 5)
) ⋅ ( ( m − 3) − ( m − 4 )
(
− ( m − 4)
) ⋅ (( m − 2) − ( m − 3) ) ⋅ ( m + 1) .
T1 \T0
−1 ⋅ 3
T2 \T1
Tm−1 ∩Tm−4 \Tm−5
Tm−3 \Tm−4
T2 \T1
Tm−6 \Tm −7
Tm−5 \Tm−6
Tm −3 \Tm −4
Tm−6 \Tm −7
Tm −5 \Tm−6
Tm−4 \Tm −1
Tm−1 \Tm−2
Tm−1 \Tm−2
Tm−4 \ Tm−1
)⋅
X \Tm
We obtain
(
EX( r ) ( Q ) = 2
⋅( m − 4)
(
T1 \ T0
( (Tm−1 ∩Tm−4 ) \ Tm−5 )
⋅ ( m − 3)
Tm −3 \ Tm− 4
)(
−1 ⋅ 3
(
T2 \ T1
−2
⋅ ( m − 4)
− ( m − 4)
T2 \ T1
) ⋅⋅⋅ ( ( m − 5)
Tm −5 \ Tm−6
Tm −3 \ Tm −4
− ( m − 5)
) ⋅ (( m − 2)
Tm −6 \ Tm −7
Tm−5 \ Tm− 6
Tm−1 \ Tm −2
− ( m − 6)
) ⋅ ( ( m − 3)
− ( m − 3)
Tm−6 \ Tm− 7
Tm− 4 \ Tm−1
Tm−1 \ Tm −2
)⋅
− ( m − 4)
) ⋅ ( m + 1)
X \ Tm
Tm− 4 \ Tm−1
)⋅
.
Corollary 1.5: Let Q = {T0 ,T1,T2 ,...,T6} be a subsemilattice of the semilattice D and
T0 ⊂T1 ⊂ T2 ⊂T4 ⊂ T6 , T0 ⊂ T1 ⊂ T3 ⊂ T4 ⊂ T6
T0 ⊂T1 ⊂ T3 ⊂ T5 ⊂T6, T2 \ T3 ≠∅, T3 \ T2 ≠∅,
T2 \ T5 ≠∅, T5 \ T2 ≠∅, T4 \ T5 ≠∅, T5 \ T4 ≠∅,
T2 ∪T3 = T4, T2 ∪T5 = T4 ∪T5 = T6.
100
Yasha Diasamidze et al.
T6
T4
T5
T2
T3
T1
T0
Fig. 4
XI −
(see Fig. 13.6.4). If the
semilattices
D ′ = {T1 , T2 ,..., T6 }
and
Q
are
α − isomorphic and Ω ( Q ) = m0 , the equality is valid.
R ( D ′ ) = m0 ⋅ 2
(
⋅ 3
( (T5 ∩T2 ) \T1 )
T2 \T5
−2
(
⋅ 2
T2 \ T5
T1 \ T0
) ⋅ (3
)
−1 ⋅
T3 \ T2
−2
T3 \T2
)⋅(4
T5 \T4
−3
T5 \T4
)⋅7
X \T6
.
Proof: The corollary immediately follows from Theorem 1.3.
Corollary 1.6: Let Q = {T0 , T1 , T2 , T3 , T4 , T5 , T6 , T7 } is a subsemilattice of the semilattice
D such that
T0 ⊂ T1 ⊂ T2 ⊂ T3 ⊂ T5 ⊂ T7 , T0 ⊂ T1 ⊂ T2 ⊂ T4 ⊂ T5 ⊂ T7 ,
T0 ⊂ T1 ⊂ T2 ⊂ T4 ⊂ T6 ⊂ T7 , T4 \ T3 ≠ ∅, T3 \ T4 ≠ ∅,
T6 \ T3 ≠ ∅, T3 ∪ T6 ≠ ∅, T6 \ T5 ≠ ∅, T5 ∪ T6 ≠ ∅,
T4 ∪ T3 = T5 , T6 ∪ T3 = T6 ∪ T5 = T7
T7
T6
T5
T3
T4
T2
T1
T0
Fig. 5
(see Fig. 13.3.7). If the
D′ = {T0 , T1, T2 , T3 , T4 , T5 , T6 , T7 }
statement is true:
XI −
semilattices
Q = {T0 , T1 , T2 , T3 , T4 , T5 , T6 , T7 }
and
are α − isomorphic and Ω ( Q ) = m0 , then the following
(
⋅(4
R ( D′) = m0 ⋅ 2
T1 \T0
T3 \T6
)
−1 ⋅ 3
−3
( (T6 ∩T3 ) \T2 )
T3 \T6
) ⋅(4
T4 \T3
(
T2 \T1
−3
T4 \T3
⋅ 3
−2
T2 \T1
) ⋅ (5
)⋅
T6 \T5
−4
T6 \T5
) ⋅8
Proof: The corollary immediately follows from Theorem 1.3. □
X \T7
.
Some Regular Elements, Idempotents and…
101
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