Download ROUGH SETS DETERMINED BY QUASIORDERS 1. Introduction

ROUGH SETS DETERMINED BY QUASIORDERS
JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
Abstract. In this paper, the ordered set of rough sets determined by
a quasiorder relation R is investigated. We prove that this ordered set
is a complete, completely distributive lattice, whenever the partially ordered set of the equivalence classes of R ∩ R−1 does not contain infinite
ascending chains. We show that on this lattice can be defined three
different kinds of complementation operations, and we describe its completely join-irreducible elements. We also characterize the case in which
this lattice is a Stone lattice. Our results generalize some results of
J. Pomykala and J. A. Pomykala (1988) and M. Gehrke and E. Walker
(1992) in case R is an equivalence.
1. Introduction
Rough set theory was introduced by Pawlak in [11]. His idea was to develop a formalism for dealing with vague concepts and sets. In rough set
theory it is assumed that our knowledge is restricted by an indiscernibility
relation. An indiscernibility relation is an equivalence E such that two elements of a universe of discourse U are E-equivalent if we cannot distinguish
these two elements by their properties known by us. By the means of an
indiscernibility relation E, we can partition the elements of U into three
disjoint classes with respect to any set X ⊆ U :
(1) The elements which are certainly in X. These are elements x ∈ U whose
E-class x/E is included in X.
(2) The elements which certainly are not in X. These are elements x ∈ U
such that their E-class x/E is included in X’s complement X c .
(3) The elements which are possibly in X. These are elements whose E-class
intersects with both X and X c . In other words, x/E is not included in
X nor in X c .
Based on this observation, Pawlak defined the lower approximation X H
of X to be the set of those elements x ∈ U whose E-class is included in
X. The upper approximation X N of X consists of elements x ∈ X whose
E-class intersects with X. Then, the sets X H and X N can be viewed as
sets of elements that belong certainly and possibly to X, respectively. The
difference X N \ X H can be viewed as the actual area of uncertainly.
2000 Mathematics Subject Classification. Primary 06A06; Secondary 06D10, 06D15,
68T37.
Key words and phrases. Rough set, rough approximations, quasiorder, Alexandrov
topology, de Morgan operation, pseudocomplement, completely distributive lattice, Stone
lattice, completely join-irreducible element.
The partial support by Hungarian National Research Found (Grant No. T049433/05
and T046913/04) is acknowledged by the second author.
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JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
Interestingly, we may define another indiscernibility relation, but now
between subsets of U . The relation ≡ is based on approximations and it
is called rough equality. The sets X and Y are ≡-related if both of their
approximations are the same, that is, X H = Y H and X N = Y N . The equivalence classes of ≡ are called rough sets. Each element in the same rough
set looks the same, when observed through the knowledge given by the indiscernibility relation E. Namely, if X ≡ Y , then exactly the same elements
belong certainly and possibly to X and Y .
Lattice-theoretical study of rough sets was initiated by Iwiński in [5]. He
noticed that rough sets can be represented simply by their approximations.
Hence, the set of rough sets can be defined as
RS = {(X H , X N ) | X ⊆ U }.
Iwiński also noted that RS may be canonically ordered by the coordinatewise
order:
(X H , X N ) ≤ (Y H , Y N ) ⇐⇒ X H ⊆ Y H and X N ⊆ Y N .
J. Pomykala and J. A. Pomykala showed in [12] that RS = (RS , ≤) is a
Stone lattice. This means that RS is a distributive lattice such that each
element (X H , X N ) has a pseudocomplement
(X Nc , X Nc ).
Later this result was improved by Comer [3] by showing that in fact RS is
a double Stone lattice.
Finally, in [4] Gehrke and Walker described the structure of RS precisely.
They showed that RS is isomorphic to 2I ×3J , where 2 and 3 are the chains
of two and three elements, I is the set of singleton E-classes, J is the set of
non-singleton equivalence classes of E, 2I is the pointwise ordered set of all
mappings from I to the two-element chain, and 3J is the pointwise ordered
set of all maps from J to the 3-element chain. Note that if each element of U
is indiscernible only with itself, then E is the identity relation, all E-classes
are singletons, and RS is isomorphic to 2U . This may be interpreted so that
rough sets really generalize “classical sets”.
In the literature can be found numerous studies on rough sets determined
by other types of indiscernibility relations than equivalences; see [8, 10] for
further references. The idea is that R may now be an arbitrary binary
relation, and rough lower and upper approximations are then defined in
terms of R. This means that x ∈ X N if there is y ∈ X such that x R y, and
x ∈ X H if x R y implies y ∈ X. Rough equality relation, the set of rough
sets RS , and its partial order are defined as before.
It is known that if R is reflexive and symmetric, then RS is not always
even a semilattice [6]. Similarly, if R is just transitive, RS is not necessarily
a semilattice [7]. However, if R is symmetric and transitive, RS is a complete
double Stone lattice [7]. Unfortunately, the structure of RS in case R is a
quasiorder, that is, R is reflexive and transitive, has been unknown. In this
paper, we prove that if R is a quasiorder such that the partially ordered
set of the equivalence classes of R ∩ R−1 does not contain infinite ascending
chains, then RS is a complete sublattice of ℘(U ) × ℘(U ).
ROUGH SETS DETERMINED BY QUASIORDERS
3
This paper is structured as follows. In the next section, we give the
definition of rough sets determined by arbitrary relations, and recall some
of their well-known properties. We also present a decomposition theorem
for rough sets that are defined by a relation that is at least serial. At the
end of the section, we concentrate on rough sets determined by quasiorders
and recall their essential connection to Alexandrov topologies.
Section 3 is devoted to our main result showing that if R is a quasiorder
such that the partially ordered set of the equivalence classes of R ∩ R−1
contains only finite ascending chains, then RS is a complete sublattice of
℘(U )×℘(U ). Note that this implies directly that RS is completely distributive.
Then, we study the lattice structure of RS more carefully. In Section 4,
we show that there can be defined three different kinds of complementation
operations. The completely join-irreducible and completely meet-irreducible
elements of RS are described in Section 5. In Section 6, we characterize the
case in which RS is a Stone lattice.
2. Rough Set Approximations
We begin by defining the rough set approximations based on arbitrary
binary relations. Let R be any binary relation on U . We denote for any x ∈
U , R(x) = {y ∈ R | x R y}. For any subset X ⊆ U , the lower approximation
of X is
X H = {x ∈ U | R(x) ⊆ X}
and the upper approximation of X is
X N = {x ∈ U | R(x) ∩ X 6= ∅}.
Let X c denote the complement U \ X of X. Then,
X Nc = X cH and X Hc = X cN ,
that is,
H
and
N
are dual. In addition,
[ N [
H = {X N | X ∈ H}
and
\
H
H
=
\
{X H | X ∈ H}
for all H ⊆ ℘(U ).
Rough sets may now be defined as in case of equivalences. Let us denote
for any X ⊆ U ,
A(X) = (X H , X N ),
and call it the rough set of X. Furthermore, we denote by
RS = {A(X) | X ⊆ U }
the set of all rough sets. The set RS can be ordered coordinatewise by
(X H , X N ) ≤ (Y H , Y N ) ⇐⇒ X H ⊆ Y H and X N ⊆ Y N ,
obtaining in this way a bounded partially ordered set RS = (RS , ≤) with
A(∅) = (∅H , ∅) as the least element and A(U ) = (U, U N ) as the greatest
element. A rough set A(X) is called an exact element of RS if X H = X =
X N.
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JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
Let us define the mapping:
c : RS → RS , A(X) 7→ A(X c ).
Since c((X H , X N )) = (X Nc , X Hc ) for any X ⊆ U , the mapping c is well
defined, and it is easy to see that the pair (c, c) is an order-reversing Galois
connection on RS. This implies directly the following proposition.
Proposition 2.1. The partially ordered set RS is self-dual, that is, RS is
order-isomorphic to its dual RS op .
For any binary relation R on U , a set C is called a connected component of
R, if C is an equivalence class of the smallest equivalence relation containing
R. Clearly, for any connected component C of R and any elements c ∈ C and
d ∈ U , R(c) ⊆ C and R(d) ∩ C 6= ∅ implies d ∈ C. Hence, C N ⊆ C ⊆ C H .
Recall that a binary relation R on U is serial, if R(x) 6= ∅ for every x ∈ U .
For a serial relation R, we have that X H = {x ∈ U | R(x) ⊆ X} ⊆ {x ∈ U |
R(x)∩X 6= ∅} = X N for all X ⊆ U . Therefore, for any connected component
C of R, C H = C N = C and thus A(C) = (C, C) is an exact element of RS .
Let {Ci | i ∈ I} stand for the set of the connected components of R and
denote for each i ∈ I by RS (Ci ) the set of rough sets on the component
Ci determined by the restriction of R to Ci . In addition, the corresponding
ordered set is denoted by RS(Ci ).
Next we present a decomposition theorem for rough sets determined by
serial relations. First, we prove the following lemma:
Lemma 2.2. Let R be a serial relation. Then, the following assertions hold.
(i) If {Ck | k ∈ K} is a subset of connected components
S of R,Hthen
S for any
N
X
,
family {(XkH , XkN ) ∈ RS (Ck ) | k ∈ K}, the pair
k∈K k
k∈K Xk
is a rough set on U .
(ii) If (X H , X N ) is a rough set on U , then the pair (X H ∩ Ci , X N ∩ Ci ) is
a rough set on Ci for any i ∈ I.
N
S
S
Xk
Proof. (i) Clearly,
= k∈K XkN . We will show that also
k∈K
H
H
S
S
S
Xk
= k∈K XkH . Since XkH ⊆
Xk
for all k ∈ K, we
k∈K
k∈K
H
H
S
S
S
H
.
. On the other
have k∈K Xk ⊆
k∈K Xk
k∈K Xk S
S hand, let x ∈
Because R(x) 6= ∅ and R(x) ⊆ k∈K Xk ⊆ k∈K Ck , there exist m ∈ K and
y ∈ Cm such that x R y. Hence, x ∈ Cm and R(x) ⊆ Cm . This implies
[
[ R(x) ⊆ Cm ∩
Xk =
Cm ∩ Xk = Xm ,
k∈K
k∈K
because Xm ⊆ Cm and Xk ∩ Cm = ∅ for all k ∈ K \ {m}. Thus, we obtain
H
S
S
S
H ⊆
H , which gives
H
x ∈ Xm
X
X
=
k
k∈K
k∈K
k∈K Xk . Therefore,
k
S
S
S
H
N
A k∈K Xk =
k∈K Xk , k∈K Xk is a rough set on U .
(ii) We prove that (X ∩ Ci )H = X H ∩ Ci and (X ∩ Ci )N = X N ∩ Ci for all
i ∈ I. It is easy to see that
(X ∩ Ci )H = X H ∩ Ci H = X H ∩ Ci .
Furthermore, (X ∩ Ci )N ⊆ X N and (X ∩ Ci )N ⊆ CiN = Ci imply (X ∩Ci )N ⊆
X N ∩ Ci . For the converse, suppose that x ∈ X N ∩ Ci . Then, R(x) ∩ X 6= ∅
and R(x) ⊆ Ci imply R(x) ∩ (X ∩ Ci ) 6= ∅, that is, x ∈ (X ∩ Ci )N . This
proves (X ∩ Ci )N = X N ∩ Ci .
ROUGH SETS DETERMINED BY QUASIORDERS
5
Corollary 2.3. If R is a serial relation, then for anySsubset {C
k | k ∈ K}
of the connected components of R, the rough set A k∈K Ck is an exact
element of RS .
Proof. By the proof of Lemma 2.2(i),
[
H
[
N
[
[
[
Ck
=
CkH =
Ck =
CkN =
Ck .
k∈K
k∈K
k∈K
k∈K
k∈K
Q
For any index set K, let P = k∈K Pk be the Cartesian product of the
partially ordered sets PkQ= (Pk , ≤k ), and let xk denote the k-th coordinate
of an element x ∈ P = k∈K Pk . We will also write x = (xk )k∈K . Recall
that the partial order ≤ of P is defined coordinatewise (see e.g [15]), that
is, for any x, y ∈ P we have x ≤ y if and only if xk ≤k yk for all k ∈ K.
Theorem
2.4. If R is a serial relation on U , then RS is order-isomorphic
Q
to i∈I RS(Ci ), where {Ci | i ∈ I} is the set of connected components of R.
Proof. Consider the maps
Y
Φ : RS →
RS (Ci )
and
Y
Ψ:
i∈I
RS (Ci ) → RS
i∈I
defined by
Φ(A(X)) = ((X H ∩ Ci , X N ∩ Ci ))i∈I
for any A(X) = (X H , X N ) ∈ RS , and
[
[
Ψ(((XiH , XiN ))i∈I ) =
XiH ,
XiN
i∈I
for any
((XiH , XiN ))i∈I
∈
Q
i∈I
i∈I
RS (Ci ).
In view of Lemma 2.2, the maps Φ and Ψ are well-defined, and it is easy
to check that both Φ and Ψ are order-preserving. Hence, to prove that Φ
and Ψ are order-isomorphisms, it is enough to show that they are mutually
inverse maps.
For any A(X) = (X H , X N ) ∈ RS , we obtain:
Ψ (Φ(A(X))) = Ψ(((X H ∩ Ci , X N ∩ Ci ))i∈I )
[
[
=
(X H ∩ Ci ), (X N ∩ Ci )
i∈I
i∈I
[
= XH ∩
Ci , X N ∩
i∈I
H
[
Ci
i∈I
N
= (X , X )
= A(X).
Q
Furthermore, for any ((XiH , XiN ))i∈I ∈ i∈I RS (Ci ), we have:
[
[
Φ(Ψ(((XiH , XiN ))i∈I )) = Φ
XjH ,
XjN
j∈I
=
[
j∈I
j∈J
H
Xj
∩ Ci ,
[
j∈I
XjN ∩ Ci
i∈I
.
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JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
Because (XjH , XjN ) ∈ RS (Cj ) for every j ∈ I, XjH and XjN are subsets of Cj .
Additionally, for any i, j ∈ I such that i 6= j, Ci ∩ Cj = ∅. These facts imply
that
[
[
XjH ∩ Ci =
XjH ∩ Ci = XiH
j∈I
and
[
j∈I
j∈I
[
XjN ∩ Ci =
XjN ∩ Ci = XiN ,
j∈I
for each i ∈ I. Thus, we obtain Φ(Ψ(((XiH , XiN ))i∈I )) = ((XiH , XiN ))i∈I .
In view of the above equalities, the maps Φ and Ψ are order-preserving
inverse mappings
of each other and hence they are order-isomorphisms. So,
Q
RS and i∈I RS(Ci ) are order-isomorphic.
We may also determine rough set approximations in terms of the inverse
R−1 of R, that is,
X O = {x ∈ U | R−1 (x) ⊆ X}
and
X M = {x ∈ U | R−1 (x) ∩ X 6= ∅}.
Interestingly, the pairs (N ,O ) and (M ,H ) are order-preserving Galois connections on ℘(U ). The end of this section is devoted to approximations determined by quasiorders. First, we recall the notion of Alexandrov topologies
that is closely connected to rough sets determined by quasiorders – for further details see [9], for example.
An Alexandrov topology [1] is a topology which is closed also under arbitrary intersections. Let T be an Alexandrov topology on U . Then, for each
X ⊆ U , there exists the smallest neighbourhood
\
N (X) = {Y ∈ T | X ⊆ Y }
of X, which is the smallest open set containing X. Let us denote by N (x)
the smallest neighbourhood of the point x. Then,
BT = {N (x) | x ∈ U }
is the smallest base of T . This means that for every X ∈ T ,
[
X = {N (x) | x ∈ X}.
Furthermore, BT is the family of completely join-irreducible elements of T as
a lattice. Recall that for a complete lattice L, anWelement x ∈ L is completely
join-irreducible if for every subset S of L, x = S implies that x ∈ S.
For a quasiorder R, the rough approximations satisfy for all X ⊆ U :
X NO = X N , X MH = X M , X HM = X H , X ON = X O .
These approximations determine two Alexandrov topologies on U :
T N = {X N | X ⊆ U } = {X O | X ⊆ U }
and
T H = {X H | X ⊆ U } = {X M | X ⊆ U }.
Clearly, these topologies are dual, that is, for all X ⊆ U ,
X ∈ T N ⇐⇒ X c ∈ T H
ROUGH SETS DETERMINED BY QUASIORDERS
7
For the Alexandrov topology T N :
(i) N : ℘(U ) → ℘(U ) is the smallest neighbourhood operator.
(ii) M : ℘(U ) → ℘(U ) is the closure operator. Note that the family of closed
sets for the topology T N is T H .
O
(iii) : ℘(U ) → ℘(U ) is the interior operator, that is, it maps each set to
the greatest open set contained into the set in question.
(iv) The set { {x}N | x ∈ U } = {R−1 (x) | x ∈ U } is the smallest base.
Similarly, for the topology T H :
(i) M : ℘(U ) → ℘(U ) is the smallest neighbourhood operator.
(ii) N : ℘(U ) → ℘(U ) is the closure operator.
(iii) H : ℘(U ) → ℘(U ) is the interior operator.
(iv) The set { {x}M | x ∈ U } = {R(x) | x ∈ U } is the smallest base.
3. Lattices of Rough Sets Determined by Quasiorders
In this section, we present a finitary condition under which the partially
ordered set of quasiorder-based rough sets forms a complete lattice.
Let R ⊆ U × U be a quasiorder and denote by U/R ∩ R−1 the set of
equivalence classes of R ∩ R−1 . It is well-known that R defines a partial
order ≤R on the quotient set U/R ∩ R−1 by setting any A, B ∈ U/R ∩ R−1 ,
A ≤R B if there exist a ∈ A and b ∈ B with a R b.
Notice that in fact A ≤R B if and only if a R b for all a ∈ A and b ∈
B. Let us denote the set of maximal elements of the partially ordered set
(U/R ∩ R−1 , ≤R ) by max.
Lemma 3.1. Let R be a quasiorder and let M = (R ∩ R−1 )(a) for some
a ∈ U . Then, the following assertions are equivalent:
(i) M ∈ max;
(ii) M = R(b) for all b ∈ M ;
(iii) R(x) ⊆ R(a) implies R(x) = R(a) for all x ∈ U .
Proof. (i)⇒(ii): Clearly, M = (R ∩ R−1 )(b) ⊆ R(b) for any b ∈ M . On the
other hand, if x ∈ R(b), then b R x and M ≤R (R ∩ R−1 )(x). Suppose M is
maximal. Then x ∈ (R ∩ R−1 )(x) = M . So, R(b) ⊆ M and M = R(b).
(ii)⇒(iii): If R(x) ⊆ R(a), then x ∈ R(a) = (R ∩ R−1 )(a). Suppose
y ∈ R(a). Then, x R y by transitivity and so y ∈ R(x). Hence, R(x) = R(a).
(iii)⇒(i): If M ≤R (R ∩ R−1 )(x) for some x ∈ U , then a R x. Therefore,
y ∈ R(x) implies y ∈ R(a) which is equivalent to R(x) = R(a) by our
assumption. In particular, we have x R a, which gives (R ∩ R−1 )(x) ≤R M .
Thus, M is maximal.
Remark 3.2. (i) Let us suppose that any ascending chain of the partially
ordered set (U/R ∩ R−1 , ≤R ) is finite. Then, for any A ∈ U/R ∩ R−1 , there
exists M ∈ max such that A ≤R M . Hence, for any a ∈ U , there exists a
maximal M such that
(R ∩ R−1 )(a) ≤R M = R(b)
for each b ∈ M . Therefore, a R b for all b ∈ M and so M ⊆ R(a). This
means that for any a ∈ U , the set R(a) includes a maximal class M .
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JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
(ii) Notice that for any subset X ⊆ U and any M ∈ max,
X H ∩ M 6= ∅ implies M ⊆ X H
X N ∩ M 6= ∅ implies M ⊆ X N .
and
Indeed, if X H ∩ M 6= ∅, then there exists a ∈ M with R(a) ⊆ X. Hence, we
get M = (R ∩ R−1 )(a) = R(a) ⊆ X. Thus, we obtain R(m) = R(a) ⊆ X,
for all m ∈ M , and this implies M ⊆ X H . Analogously, if X N ∩ M 6= ∅, then
R(a) ∩ X 6= ∅ for some a ∈ M . So, for each m ∈ M , we get R(m) ∩ X =
R(a) ∩ X 6= ∅ and M ⊆ X N .
Recall that since R is a quasiorder, N is a closure operator and
interior operator. Thus, for any X ⊆ U ,
X H ⊆ X ⊆ X N,
X HH = X H ,
and
H
is an
X NN = X N .
These facts will be needed in the proof of our next theorem.
Theorem 3.3. Let U be a non-empty set and let R ⊆ U × U be a quasiorder
such that any ascending chain in the partially ordered set (U/R ∩ R−1 , ≤R )
is finite. Then, RS is a complete sublattice of ℘(U ) × ℘(U ).
Proof. To prove that RS is a complete sublattice of ℘(U ) × ℘(U ), it suffices
H
N
to show that forTany family
I} = {(XS
i , Xi ) |i ∈ I} of rough
i ) | i ∈ S
T {A(X
H
N also are rough
H
N
and
sets, the pairs
i∈I Xi , i∈I Xi
i∈I Xi , i∈I Xi
sets.
T
(1) First,
we construct a set W ⊆ U that satisfies W H = i∈I XiH and
T
W N = i∈I XiN . Let us consider the set
\ N
\
Z=
XiN \
Xi ,
i∈I
i∈I
and observe that for any a ∈ Z, we have |R(a)| ≥ 2. Indeed, by T
the definition
of Z, a ∈ Z means that R(a) ∩ Xi 6= ∅ for all i ∈ I and R(a) ∩ i∈I Xi = ∅.
If R(a) has the form R(a)
T = {a}, then R(a) ∩ Xi 6= ∅ implies a ∈ Xi for
each i ∈ I, that is, a ∈ i∈ Xi , a contradiction.
Let us consider now those classes M ∈ max which are included in the set
Z. As for any a ∈ M , we have R(a) = M , we get also that |M | = |R(a)| ≥ 2.
Thus, for all M ∈ max with M ⊆ Z, we can select two elements rM , sM ∈ M
with rM 6= sM . Since for any M, N ∈ max such that M 6= N , we have
M ∩ N = ∅, we also obtain that {rM , sM } ∩ {rN , sN } = ∅ for all M, N ⊆ Z
with M 6= N . Thus, we now select the two elements rM , sM ∈ M with
rM 6= sM for all possible M ∈ max satisfying M ⊆ Z. Note that this
selection of the elements for all classes M ∈ max, where M ⊆ Z, requires
the use of the Zermelo’s Axioms of Choice.
We define next the sets Y and W by setting:
Y = Z \ {rM | M ∈ max and M ⊆ Z}
and
W =
\
Xi ∪ Y.
i∈I
T
T
We will show that W H = i∈I XiH and W N = i∈I XiN .
H
T
T
T
Clearly, Y, W ⊆ i∈I XiN and i∈I XiHT =
⊆ W H . In order
i∈I Xi
H
H
to prove the converse inclusion W ⊆ i∈I Xi , suppose that a ∈ W H .
ROUGH SETS DETERMINED BY QUASIORDERS
9
T
Then, R(a) ⊆ W =
i∈I Xi ∪ Y , and next we will show that necessarily
R(a) ∩ Y = ∅.
Indeed, if R(a) ∩ Y 6= ∅, then there exists an element b ∈ R(a) ∩ Y .
So, R(b) ⊆ R(a) and, in view of Remark 3.2(i), R(b) contains an element
−1
−1
c ∈ R(b) with (R
=
T ∩ R )(c) ∈ max. Therefore, we have (RT∩ R )(c)
R(c) ⊆ R(b) ⊆
X
∪
Y
.
As
b
∈
Y
⊆
Z
implies
R(b)
∩
X
i
i = ∅,
i∈I
i∈I
we get R(c) ⊆ R(b) ⊆ Y ⊆ Z.
On the other hand, since R(c) = (R ∩R−1 )(c) ∈ max, by the construction
of Y , there exists rM ∈ R(c) with rM ∈
/ Y , where M = R(c). As this result
contradicts R(c) ⊆ Y , we deduce R(a) ∩ Y = ∅.
T
Now, R(a) ∩ Y = ∅ and R(a) ⊆ W imply R(a) ⊆ i∈I Xi , which means
H T
T
T
= i∈I XiH . Therefore, we obtain T
W H = i∈I XiH .
that a ∈
i∈I Xi
(1), let usT
show the equality W NT= i∈I XiN . Obviously,
T To conclude
T part
T
N
N
N Since
N
i∈I Xi ⊆
i∈I Xi and Y ⊆
i∈I Xi imply W ⊆
i∈I Xi . T
i∈I Xi
N
N
N
is closed with respect to T: ℘(U ) → ℘(U ), we obtain W ⊆ i∈I Xi .
Conversely, we prove i∈I XiN ⊆ W N showing first that Z ⊆ Y N . For
that, take any z ∈ Z. Clearly, we may now assume z ∈
/ Y . Then, according
to the construction of Y , we have z = rM , for some M ∈ max with M ⊆ Z.
However, in this case R(z) = R(rM ) = M contains also a fixed element
sM 6= rM with sM ∈ Y . Hence, we get R(z) ∩ Y 6= ∅, proving z ∈ Y N . Thus,
Z ⊆ Y N and we can write:
\
XiN =
\
=
\
i∈I
Xi
N
∪
\
XiN \
i∈I
i∈I
Xi
N
\
Xi
N i∈I
∪Z
i∈I
⊆
\
Xi
N
∪YN
i∈I
N
⊆W .
This proves W N =
rough set.
T
i∈I
XiN . Hence, (W H , W N ) =
T
i∈I
XiH ,
T
i∈I
XiN is a
S
S
H
N is a rough set. This is done
(2) We will prove that
i∈I Xi , i∈I Xi S
S
by constructing a set V ⊆ U such that V H = i∈I XiH and V N = i∈I XiN .
Let us first consider the set
S=
[
i∈I
Xi
N [
\
XiH
i∈I
and observe that for each b ∈ S, |R(b)| ≥ 2. Indeed, if we suppose that R(b)
N S
S
has only one element, that is, R(b) = {b}, then b ∈
X
= i∈I XiN
i
i∈I
implies R(b) ∩ Xm 6= ∅ for some m ∈ I, that is b ∈ Xm . However, in this
H , a contradiction.
case {b} = R(b) ⊆ Xm implies b ∈ Xm
10
JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
Let us now choose rM , sM ∈ M with rM 6= sM for all M ∈ max. Then,
we define the following three sets by setting:
n
[ N o
H1 = a ∈ S | R(a) *
Xi
,
i∈I
H2 = {rM | M ∈ max and M ⊆ S},
[
V =
XiH ∪ H1 ∪ H2 .
i∈I
N
S
S
S
Xi . This is because i∈I XiH ⊆ i∈I Xi ⊆
Observe that V ⊆
i∈I
N
N
S
S
by definition. Furthermore, if a ∈ H2 ,
i∈I Xi
i∈I Xi , and H1 ⊆
then a = rM for some M = R(rM ) ∈ max and M ⊆ S. This means
N
S
that a = rM ∈ R(a) = M ⊆ S and so a ∈
Xi . Thus, we have
i∈I
N
S
H2 ⊆
i∈I Xi .
Now, we will prove that
[
VH =
XiH
and
VN =
i∈I
[
XiN .
i∈I
Indeed, we have XiH ⊆ V for all i ∈ I by
all i ∈ I, XiH = XiHH ⊆ V H , which gives
[
the definition of V . Therefore, for
XiH ⊆ V H .
i∈I
To show the converse, suppose that a ∈ V H . Then,
[
a ∈ R(a) ⊆ V =
XiH ∪ H1 ∪ H2 .
i∈I
Observe that a ∈ H1 is not possible. Namely, if a ∈ H1 , then R(a) *
N
S
i∈I Xi , which implies R(a) * V , a contradiction. Thus, we get
[
a∈
XiH ∪ H2
i∈I
S
Next we will show that R(a) ⊆ V and a ∈ H2 imply a ∈ i∈I XiH .
Indeed, if a ∈ H2 , then, by definition, there exists a class M ∈ max with
M ⊆ S, and the corresponding elements rM , sM ∈ M , such that a = rM 6=
sM . As M = (R ∩ R−1 )(a), we get sM ∈ (R ∩ R−1 )(a) = R(a) ⊆ V , that is,
[
sM ∈
XiH ∪ H1 ∪ H2 .
i∈I
However, sM ∈
/ H2 by the construction of H2 . The case sM ∈ H1 is also
S
N
impossible, since
i∈I Xi . Therefore, we
S we Hhave R(sM ) = R(a) ⊆ V ⊆
obtain sM S
∈ i∈I Xi , which gives R(a) = R(sM ) ⊆ Xm , for some m ∈ I.
Thus, a ∈ i∈I XiH .
Since we deduced that R(a) ⊆ V implies
[
a∈
XiH ,
i∈I
we conclude that V
H
=
S
H
i∈I Xi .
ROUGH SETS DETERMINED BY QUASIORDERS
11
N
S
S
Let us prove now the equality V N = i∈I XiN . Because V ⊆
i∈I Xi ,
we have
[ NN [ N [
VN ⊆
Xi
=
Xi
=
XiN .
i∈I
i∈I
i∈I
S
To prove the reverse inclusion, suppose that a ∈ i∈I XiN . Then, there
N , and this implies R(a) ∩ X
exists m ∈ I such that a ∈ Xm
m 6= ∅. If
S
S
H
H N ⊆ V N , and the proof is
R(a) ∩
X
=
6
∅
holds,
then
a
∈
X
i
i
i∈I
i∈I S
H = ∅, that is,
completed. Therefore, we now assume R(a) ∩
i∈I Xi
[ N [
a∈
Xi \
XiH = S.
i∈I
i∈I
N
S
N and the proof is again
If R(a) *
i∈I Xi , then a ∈ H1 ⊆ V ⊆ V
N
S
completed. Hence, we restrict ourselves to the case R(a) ⊆
i∈I Xi .
in this case we get R(a) ⊆ S, because we have R(a) ∩
SHowever,
H = ∅ by our hypothesis. Now, by Remark 3.2(i), there exists
X
i
i∈I
a class M ∈ max with M ⊆ R(a). Thus, M ⊆ S and hence by construction of H2 , there exists an element rM ∈ M such that rM ∈ H2 ⊆ V . As
rM ∈ R(rM ) = M ⊆ R(a), we obtain R(a) ∩ V 6= ∅, that is, a ∈ V N . This
result implies that
[
XiN = V N .
i∈I
Hence,
S
i∈I
XiH ,
S
i∈I
N
Xi
is a rough set, which completes the proof.
The next corollary describes the meets and the joins in the complete
lattice RS.
Corollary 3.4. If R ⊆ U × U is a quasiorder with the property that any
ascending chain in the partially ordered set (U/R ∩ R−1 , ≤R ) is finite, then
RS is a completely distributive complete lattice such that
\
[
^
\
_
[
A(Xi ) =
XiH ,
XiN
and
A(Xi ) =
XiH ,
XiN
i∈I
i∈I
i∈I
i∈I
i∈I
i∈I
for all {A(Xi ) | i ∈ I} ⊆ RS .
Proof. As RS is a complete sublattice of the completely distributive lattice
℘(U ) × ℘(U ), it is always completely distributive. Because the meet and
the join of {A(Xi ) | i ∈ I} in RS coincide with their meet and join in the
lattice ℘(U ) × ℘(U ), we obtain the required formulas.
Since any completely distributive complete lattice is both pseudocomplemented and dually pseudocomplemented, we may write the following corollary.
Corollary 3.5. If any ascending chain in the partially ordered set (U/R ∩
R−1 , ≤R ) is finite, then both RS and its dual RS op are pseudocomplemented
lattices.
12
JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
4. Complementation in the Lattice of Rough Sets
In the previous section, we showed that if any ascending chain in the
partially ordered set (U/R ∩ R−1 , ≤R ) is finite, then RS is a completely
distributive complete lattice. In this section, we describe different types of
complementation operations in RS.
Let us first consider the mapping
c : RS → RS , A(X) 7→ A(X c )
introduced already in Section 2. Now, for all α, β ∈ RS , we have
c(α ∨ β) = c(α) ∧ c(β)
c(α ∧ β) = c(α) ∨ c(β)
c(c(α)) = α
This means that c is so-called de Morgan operation on the lattice RS (for the
notion, see [2], for instance). Note that α∨c(α) = (U, U ) and α∧c(α) = (∅, ∅)
do not generally hold. However, X H ∩ X cH = (X ∩ X c )H = ∅H = ∅ and
X N ∪ X cN = (X ∪ X c )N = U N = U for any X ⊆ U .
As noted in the end of the previous section, RS is a pseudocomplemented
lattice. Next we will describe the pseudocomplementation operation in RS.
Suppose that L is a lattice with a least element 0. An element x∗ is a
pseudocomplement of x ∈ L, if x ∧ x∗ = 0 and for all a ∈ L, x ∧ a = 0 implies
a ≤ x∗ . An element can have at most one pseudocomplement. A lattice is
pseudocomplemented if each element has a pseudocomplement.
Proposition 4.1. In the lattice RS, for each X ⊆ U , we have
A(X)∗ = A(X NMc ).
Proof. A(X) ∧ A(X NMc ) = (∅, ∅), because X NMcN = X NMHc = X NMc and
X N ∩ X NMc = ∅ if and only if X N ⊆ X NM , and the latter holds trivially. For
the other part, X H ∩ X NMcH ⊆ X N ∩ X NMc = ∅.
On the other hand, if A(X) ∧ A(Y ) = (∅, ∅), then X N ∩ Y N = ∅ and
N
Y ⊆ X Nc . This implies Y ⊆ Y NO ⊆ X NcO = X NMc , from which we get
A(Y ) ≤ A(X NMc ). Thus, A(X)∗ = A(X NMc ).
Notice that RS is not necessarily a Stone lattice. In the final section, we
give a condition under which RS is a Stone lattice.
We conclude this section by describing also dual pseudocomplements in
RS. A dual pseudocomplement x+ of x ∈ L in a lattice L with a greatest
element 1 is such that x ∨ x+ = 1 and x ∨ y = 1 implies x+ ≤ y for all y ∈ L.
Proposition 4.2. In the lattice RS, for each X ⊆ U , we have
A(X)+ = A(X HOc ).
Proof. A(X) ∨ A(X HOc ) = (U, U ), because X HOcH = X HONc = X HOc , and
X HO ⊆ X H implies X Hc ⊆ X HOc and X H ∪X HOc ⊇ X H ∪X Hc = U . Similarly,
X N ∪ X HOcN ⊇ X H ∪ X HOc = U .
If A(X) ∨ A(Y ) = (U, U ), then X H ∪ Y H = U and X Hc ⊆ Y H . This
implies X HOc = X HcM ⊆ Y HM ⊆ Y . From this we directly obtain A(X HOc ) ≤
A(Y ).
ROUGH SETS DETERMINED BY QUASIORDERS
13
5. Completely Irreducible Elements
In this section, we find the set of completely join-irreducible elements of
RS, and show how all elements can be represented as a join of these. Also
completely meet-irreducible elements are characterized.
Let us define the set of rough sets
J = {(∅, {x}N ) | |R(x)| ≥ 2} ∪ {({x}M , {x}MN ) | x ∈ U }.
We can write the following lemma.
Lemma 5.1. The members of J are completely join-irreducible.
Proof. If |R(x)| ≥ 2, then R(x) 6⊆ {x} and {x}H = ∅, which implies
A({x}) = (∅, {x}N ) ∈ RS. The pair must be completely join-irreducible
in RS, because {x}N is completely join-irreducible in the topology T N .
It is clear that for each x ∈ U , A({x}M ) = ({x}M , {x}MN ) ∈ RS, because
{x}MH = {x}M . We show that ({x}M , {x}MN ) is completely join-irreducible.
Suppose that there exists a family {A(Xi )}i∈I = {(XiH , XiN )}i∈I such that
_
({x}M , {x}MN ) =
A(Xi ).
i∈I
{x}M
Because
is completely join-irreducible in T H , we have that {x}M = XiH
for some i ∈ I. Since XiH ⊆ Xi , this also implies {x}MN ⊆ XiN . The converse,
XiN ⊆ {x}MN , holds trivially. Thus, {x}MN = XiN and
({x}M , {x}MN ) = (XiH , XiN ).
Our next theorem shows that J is the set of all completely join-irreducible
elements.
Theorem 5.2. J is the set of completely join-irreducible elements of the
complete lattice RS. Any nonzero element of RS is a join of some completely join-irreducible elements of J .
Proof. Next, we will prove that each (X H , X N ) ∈ RS can be expressed as
the join of some elements in J . We begin by showing that for all X ⊆ U ,
[
X H = {{x}M | {x}M ⊆ X}.
Because X H ∈ T H and M : ℘(U ) → ℘(U ) is the smallest neighbourhood
operator of the topology T H , we get
[
XH =
{{x}M | x ∈ X H }
[
=
{{x}M | R(x) ⊆ X}
[
=
{{x}M | {x}M ⊆ X}.
S
Since {{x}M | {x}M ⊆ X} ⊆ X and N distributes over unions, we have
[
{{x}MN | {x}M ⊆ X} ⊆ X N .
Hence,
_
{({x}M , {x}MN ) | {x}M ⊆ X} ≤ (X H , X N ).
14
JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
Obviously,
_
{(∅, {x}N ) | x ∈ X and |R(x)| ≥ 2} ≤ (X H , X N )
Next we will show that
(X H , X N ) =
_
{(∅, {x}N ) | x ∈ X and |R(x)| ≥ 2}
_
∨
{({x}M , {x}MN ) | {x}M ⊆ X}.
Let x ∈ X N . If |R(x)| = 1, then clearly also x ∈ X H . In this case, {x}M =
{x} ⊆ X, x ∈ {x}M , and x ∈ {x}MN .
If |R(x)| ≥ 2, then we consider three different cases: (i) x ∈ X H , (ii)
x ∈ X \ X H , and (iii) x ∈ X N \ X.
(i) If x ∈ X H , then x ∈ {x}M = R(x) ⊆ X, and trivially x ∈ {x}MN .
(ii) If x ∈ X, but x ∈
/ X H , then {x}M = R(x) 6⊆ X. However, x ∈ {x}N
and x ∈ X. In addition, x ∈ X and x ∈
/ X H imply |R(x)| ≥ 2 and {x}H = ∅.
(iii) If x ∈ X N , but x ∈
/ X, then necessarily x ∈ {y}N for some y ∈ X such
that x 6= y. We have now two possibilities, either R(y) ⊆ X or R(y) 6⊆ X. If
{y}M = R(y) ⊆ X, then necessarily x ∈
/ {y}M . However, x ∈ {y}N ⊆ {y}MN .
M
If R(y) 6⊆ X, then R(y) = {y} contains at least two elements, because
y ∈ X. This implies {y}H = ∅. Thus, (∅, {y}N ) ∈ RS and recall that
x ∈ {y}N and y ∈ X.
Clearly, the image of the set J under the mapping c : A(X) 7→ A(X c ) is
M = {A({x}c ) | |R(x)| ≥ 2} ∪ {A({x}Mc ) | x ∈ U }
= {({x}Nc , U ) | |R(x)| ≥ 2} ∪ {({x}MNc , {x}Mc ) | x ∈ U }.
Since the completely meet-irreducible elements of the lattice RS are just
the completely join-irreducible elements of its dual RS op , and because RS
is order-isomorphic to RS op via the mapping A(X) 7→ A(X c ), we obtain
the following corollary.
Corollary 5.3. M is the set of completely meet-irreducible elements of
the complete lattice RS. Any nonunit element of RS is a meet of some
completely meet-irreducible elements of M.
6. Characterization of the Stonean Case
In the case of a quasiorder R ⊆ U ×U , the smallest equivalence containing
R is R ∨ R−1 , where ∨ denotes the join in the lattice of all quasiorders on
U ordered with the set-inclusion relation ⊆. Furthermore, R ∨ R−1 is equal
to the transitive closure of the relation R ∪ R−1 . Hence, the connected
components of R are just the equivalence classes of R ∨ R−1 .
Proposition 6.1. Let R ⊆ U × U be a quasiorder such that the partially
ordered set (U/R ∩ R−1 , ≤R ) contains only finite ascending chains. Then,
the following assertions are equivalent for any X ⊆ U :
(i) A(X) is a complemented element of RS;
(ii) A(X) is an exact element of RS ;
(iii) X is a union of some equivalence classes of R ∨ R−1 .
ROUGH SETS DETERMINED BY QUASIORDERS
15
Proof. (i)⇒(ii): Assume that there exists a set Y ⊆ U such that A(Y ) is the
complement of A(X). Then, X N ∩Y N = ∅ and X H ∪Y H = U . Thus, we have
Y c ⊆ Y Hc ⊆ X H ⊆ X ⊆ X N ⊆ Y Nc ⊆ Y c , proving X H = X = X N = Y c .
(ii)⇒(iii): Clearly, (ii) implies X H = X and X O = X NO = X N = X.
Hence, for any x ∈ X, we have (R ∪ R−1 )(x) ⊆ X, that is, X is closed with
respect to the relation R ∪ R−1 . Then, X must also be closed with respect
−1
to the transitive closure R ∨ R
R−1 , that is, (R ∨ R−1 )(x) ⊆ X for
S of R ∪−1
all x ∈ X. This implies X = {(R
S ∨ R )(x) | x ∈ X}.
(iii)⇒(i): Suppose that X = k∈K Ck , where {Ck | k ∈ K} is a set of
some equivalence classes of R ∨ R−1 . Because for each k ∈ K, the set Ck
is a connected component of R, X H = X N = X by Corollary 2.3. Then,
X cH = X Nc = X c = X Hc = X cN , that is, also A(X c ) is an exact element
of RS . Since A(X) ∧ A(X c ) = (X ∩ X c , X ∩ X c ) = (∅, ∅) = A(∅) and
A(X) ∨ A(X c ) = (X ∪ X c , X ∪ X c ) = (U, U ) = A(U ), we have that A(X)
is a complemented element of RS.
Remark 6.2. If the assumption of Proposition 6.1 is satisfied, then from the
arguments of the above proof it follows also that A(X) is exact if and only
if A(X c ) is exact.
An element a of a bounded lattice L = (L, ≤) is called a central element
of L if a is complemented and for all x, y ∈ L the sublattice generated by
{a, x, y} is distributive. Notice that the complement of a central element is
unique and it is also a central element of L. Clearly, the least element 0 and
the greatest element 1 of L are always central elements. It is known that L ∼
=
L1 × L2 for some nontrivial bounded lattices L1 and L2 if and only if there
exists a pair of central elements c1 , c2 ∈ L \ {0, 1} such that (c1 ] ∼
= L1 , (c2 ] ∼
=
L2 and c1 and c2 are complements of each other (for details, see e.g. [13, 14]).
A lattice L is directly irreducible if there are no nontrivial lattices L1 and
L2 satisfying L ∼
= L1 × L2 . Clearly, this is equivalent to the fact that L has
no nontrivial central elements. It is also obvious that the central elements
of a bounded distributive lattice are exactly its complemented elements.
Proposition 6.3. If R is a quasiorder such that the partially ordered set
(U/R ∩ R−1 , ≤R ) contains only finite ascending chains, then the following
assertions are true.
(i) For all i ∈ I, the lattices RS(Ci ) are directly irreducible.
(ii) The lattice RS is directly irreducible if and only if R is a connected
quasiorder, that is, R has a single connected component.
Proof. (i) Assume that there exists k ∈ I such that the lattice RS(Ck ) is
directly reducible. Then, RS(Ck ) has at least one nontrivial central element.
This means that in RS(Ck ) exists a complemented element A(X) for some
X ⊆ Ck such that A(X) 6= (∅, ∅) and A(X) 6= (Ck , Ck ). Then, according
to Proposition 6.1, X is a join of some equivalence classes of the restriction
of R ∨ R−1 to Ck . However (x, y) ∈ R ∨ R−1 is satisfied for all x, y ∈ Ck ,
because Ck is an equivalence class of R ∨ R−1 . This fact implies X = Ck ,
that is, A(X) = (Ck , Ck ), a contradiction.
(ii) If R is a connected quasiorder on U , then R ∨ R−1 has just one equivalence class U . Therefore, by applying Proposition 6.1, the complemented
elements are just A(∅) and A(U ). This means that RS contains only the
16
JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
trivial central elements A(∅) and A(U ) that are the least and the greatest
elements of RS, respectively. Thus, the lattice RS is directly irreducible.
The other
Q part is an obvious consequence of (i) and the isomorphism of
RS and i∈I RS(Ci ) established in Theorem 2.4.
Let L be a pseudocomplemented bounded distributive lattice. If x∗ ∨x∗∗ =
1 holds for all x ∈ L, then L is called a Stone lattice. Obviously, this
is equivalent to the fact that x∗ is a complemented element of L for each
x ∈ L.
Theorem 6.4. Let R ⊆ U ×U be a quasiorder such that the partially ordered
set (U/R ∩ R−1 , ≤R ) contains only finite ascending chains. Then, RS is a
Stone lattice if and only if R−1 ◦ R = R ∨ R−1 .
Proof. Assume that RS is a Stone lattice. Then, for any X ⊆ U , the rough
set A(X)∗ = A(X NMc ) is a complemented element of RS. This implies
by Proposition 6.1 and Remark 6.2 that A(X NMc ) and also its complement
A(X NM ) are exact elements of RS . Let x be an arbitrary element of U and
let us set X = {x}. Then, it is easy to see that X NM = (R−1 ◦ R)(x).
By Proposition 6.1, (R−1 ◦ R)(x) is equal to the union of some equivalence
classes of R ∨ R−1 . Because R−1 ◦ R ⊆ R ∨ R−1 , this implies (R−1 ◦ R)(x) =
(R ∨ R−1 )(x). As this equality is satisfied for all x ∈ U , we obtain R−1 ◦ R =
R ∨ R−1 .
Conversely, assume the condition of Theorem 6.4 and the equality R−1 ◦
R = R ∨ R−1 are satisfied. Then, according to Corollaries 3.4 and 3.5, RS is
a distributive pseudocomplemented complete lattice. We have to show that
A(X)∗ is complemented for any X ⊆ U .
Assume that X ⊆ U . Let x ∈ X NM and y ∈ (R ∨ R−1 )(x). Then, by
the definition of X NM , there exist z ∈ R−1 (x) ∩ X N and v ∈ R(z) ∩ X.
These mean x R−1 z and z R v, from which we obtain (x, v) ∈ R−1 ◦ R with
v ∈ X. Therefore, (v, x) ∈ R ∨ R−1 , and so (x, y) ∈ R ∨ R−1 implies
(v, y) ∈ R ∨ R−1 . This means that y ∈ (R ∨ R−1 )(v) = (R−1 ◦ R)(v) =
{v}NM ⊆ X NM . This result proves that (R ∨ R−1 )(x) ⊆ X NM for all x ∈ X NM .
From this we get that X NM is the union of some classes of R ∨ R−1 . So,
by Proposition 6.1, A(X NM ) is a complemented and exact element of RS.
Then, by Remark 6.2, also A(X)∗ = A(X NMc ) is an exact and complemented
element of RS. Therefore, RS is a Stone lattice.
A partially ordered set P = (P, ≤) is called down-directed, if for any
a, b ∈ P there exists c ∈ P with c ≤ a, b. In what follows, we deduce
some corollaries of the above theorem in cases R is a partial order or an
equivalence, respectively. Notice that case (i) of the next corollary shows
that the result of [4] concerning the direct product decomposition of the
lattice RS also follows from our Theorem 6.4. We note that the details on
the direct decomposition of complete Stone lattices can be found in [13].
Corollary 6.5. Let R be a binary relation on U .
(i) If R is an equivalence, then RS is a completely distributive Stone lattice
which is isomorphic to a direct product of chains of two and three
elements.
ROUGH SETS DETERMINED BY QUASIORDERS
17
(ii) If R is a partial order such that any ascending chain in the partially
ordered set (U, R) is finite, then RS is a Stone lattice if and only if
any connected component of (U, R) is down-directed.
Proof. (i) Since R is an equivalence, R ∩ R−1 = R and the fact that RS is a
completely distributive lattice follows also from Corollary 3.4. Additionally,
R−1 ◦ R = R ∨ R−1 because R is an equivalence. Hence, Theorem 6.4
implies that
Q RS is a Stone lattice. In view of Theorem 2.4, we have that
RS and i∈I RS(Ci ) are isomorphic, where {Ci | i ∈ I} is the set of the
equivalence classes of R. If Ci consists of a single element a, then RS (Ci ) =
{(∅, ∅), ({a}, {a})} and RS(Ci ) is a chain of two elements. Similarly, if
|Ci | ≥ 2, then RS (Ci ) = {(∅, ∅), (∅, Ci ), (Ci , Ci )} and RS(Ci ) is a chain of
three elements. Therefore, RS is of the form 2J × 3K , where J ∪ K = I and
J ∩ K = ∅. Note that here J stands for singleton equivalence classes and K
denotes non-singleton R-classes.
(ii) If R is a partial order, then R ∩ R−1 is the identity relation on U
and hence (U/R ∩ R−1 , ≤R ) and (U, R) can be identified. Thus, by applying
Corollary 3.5, we obtain that RS is a completely distributive lattice whenever any ascending chain in (U, R) is finite. As the connected components
{Ci | i ∈ I} are the equivalence classes of R ∨ R−1 , it is easy to check that
R−1 ◦ R = R ∨ R−1 if and only if for any i ∈ I and a, b ∈ Ci , there exists
c ∈ Ci with c R a and c R b. This means that all partially ordered sets (Ci ,≤)
are down-directed.
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18
JOUNI JÄRVINEN, SÁNDOR RADELECZKI, AND LAURA VERES
[15] William T. Trotter, Combinatorics and partially ordered sets: Dimension theory,
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Jouni Järvinen: Department of Information Technology, FI-20014 University of Turku, Finland
E-mail address: [email protected]
Sándor Radeleczki: Institute of Mathematics, University of Miskolc,
3515 Miskolc-Egyetemváros, Hungary
E-mail address: [email protected]
Laura Veres: Institute of Mathematics, University of Miskolc, 3515 Miskolc-Egyetemváros, Hungary
E-mail address: [email protected]