=Paper=
{{Paper
|id=Vol-1624/paper3
|storemode=property
|title=Partial Duplication of Convex Sets in Lattices
|pdfUrl=https://ceur-ws.org/Vol-1624/paper3.pdf
|volume=Vol-1624
|authors=Laurent Beaudou,Lhouari Nourine
|dblpUrl=https://dblp.org/rec/conf/cla/BeaudouN16
}}
==Partial Duplication of Convex Sets in Lattices==
https://ceur-ws.org/Vol-1624/paper3.pdf
Partial Duplication of Convex Sets in Lattices
Laurent Beaudou and Lhouari Nourine
Clermont-Université, Université Blaise Pascal, LIMOS, CNRS, France
{labeaudo@univ-bpclermont.fr, nourine@isima.fr}
Abstract. In this paper, we generalize the classical duplication of in-
tervals in lattices. Namely, we deal with partial duplication instead of
complete convex subsets. We characterize those subsets that guarantee
the result to still be a lattice. Moreover, we show that semi-distributive
and extremal lattices can be encompassed by such construction where
classical duplication fails.
Introduction
The aim of this paper is to give a characterization of several classes of lattices
obtained by doubling suborder (not necessary convex) in lattices. This construc-
tion generalizes the one that uses convex duplication introduced by Day [1] and
followed by several results on the characterizations and algorithmic aspects of
these classes of lattices such as: Bounded, Upper Bounded and normal classes
of lattices. These results have been obtained by Day on his own [2, 3] or with
Nation and Tschantz [4], Bertet and Caspard [5–7] and Geyer [8].
In the opposite of constructing a lattice, decomposing a lattice using prop-
erties of duplication to small lattices has been also considered in the litera-
ture. Markowsky [9, 10] has shown that extremal lattices can be factorized using
prime/coprime property which correspond to the double arrow or perspective
relation as introduced in [11]. Janssen and Nourine [12] have given a procedure
to decompose a semidistributive lattice according to a simplicial elimination
scheme. Others decomposition related to subdirect product construction and
congruence can be found in [13–15].
In this paper we give a necessary and sufficient condition for duplications that
maintain the lattice structure. We also give other properties that guarantee some
combinatorial properties of lattices such as ∧-semidistributivity and extremality.
As a by-product of our results and existing ones, we obtain characterizations of
some classes of lattices.
1 Preliminaries
In this paper, all considered lattices are finite. For classic definitions of lattices,
we refer the reader to the celebrated monograph of Birkhoff [16]. Still, we adress
some specific definitions that are of special interest for this documenr. Let (X, 6)
be a lattice (denoted L) with ∨ and ∧ the usual join and meet operations. An
element j in X is called join-irreducible in L if x = z ∨ t implies x = z or x = t.
The set of all join-irreducible elements is denoted by J(L). The set M (L) of all
meet-irreducible elements is defined dually. The height h(L) of a lattice L is the
length of the longest chain from ⊥ to > (the least and greatest elements of L).
Given an element x in X, the set ↑ (x, L) is the subset of X containing every
element y such that x 6 y. Set ↓ (x, L) is defined dually.
Given two elements x and y in any lattice L, we use relations ., % and . %
defined in [11] as follows,
x . y if x is minimal in L− ↓ (y, L),
x % y if y is maximal in L− ↑ (x, L),
x.% y if x . y and x % y.
Note that whenever x . y, x needs to be a join-irreducible element. Similarly,
if x % y, y needs to be a meet-irreducible element.
A lattice is said meet-semidistributive, if for all x, y, z ∈ L, x ∧ y = x ∧ z
implies x ∧ y = x ∧ (y ∨ z). It is said semidistributive if it is meet-semidistributive
and join-semidistributive. We may use . %L to denote the set of pairs (j, m) in L
such that j .% m. The subscript may be omitted when the context is clear.
2 Doubling construction
We study the possibilities of copying a part of a lattice so that it remains in
certain classes of well-known lattices.
The general framework is the following. Let L be a lattice on some set X
with partial order 6. Let C be any subset of X that will be copied. We call C 0
the copy of C (meaning there is a bijection ϕ from C to C 0 ). The convex closure
of C in L is the set H(C) = {y : ∃x, z ∈ C with x 6 y 6 z}. We may now
consider the partial order (X ∪ C 0 , 4) where the relation 4 is defined as follows
for any pair (x, y) of elements of X ∪ C 0 :
x ∈ X, y ∈ X and x 6 y
x ∈ X − H(C), y ∈ C 0 and x 6 ϕ−1 (y)
x 4 y if
x ∈ C 0 , y ∈ X and ϕ−1 (x) 6 y
x ∈ C 0 , y ∈ C 0 and ϕ−1 (x) 6 ϕ−1 (y).
Note that if x is in H(C), y is in C and x 6 y, we do not have x 4 ϕ(y). It
is routine to check that 4 defines a partial order on X ∪ C 0 . We shall denote
this partial order L[C]. If C is the empty set, then this process does not alter L
(L[∅] = L). We shall distinguish two specific subsets of C, namely the minimal
elements L = {l1 , l2 , . . . , ln } and the maximal elements U = {u1 , u2 , . . . , um }.
Figure 1 depicts an example of a copy with two minimal and two maximal
elements in C.
In order to guarantee that the resulting partial order remains a lattice, we
need to enforce two properties about C. The first one says that if the join of two
13 13
12 12
H(C) 11 10 11 10
8 7 9 8 7 9
5 6 5 6
4 4
2 3 2 3
1
1
Fig. 1. Depiction of a copy with U = {11, 12} and L = {2, 3}.
copied elements is in the convex closure of C, then it must be copied. The second
says that if an element x in H(C) covers an element which is not in H(C), then
x must be copied.
∀(x, y) ∈ C 2 , x ∨ y ∈ H(C) ⇒ x ∨ y ∈ C, (P1 )
∀x ∈ H(C), ∀y ∈ X − H(C), x covers y ⇒ x ∈ C. (P2 )
Remark 1. When U and L are singletons, Property (P1 ) says that (C, 6) is a
join-sublattice of L.
We shall first prove that properties (P1 ) and (P2 ) are necessary and sufficient
conditions for the resulting partial order to be a lattice.
Proposition 1. Given a lattice L = (X, 6) and a subset C of X, L[C] is a
lattice if and only if (P1 ) and (P2 ) are satisfied.
Proof. We first show that (P1 ) and (P2 ) are necessary. Suppose that L[C] is a
lattice. We shall prove both properties separately.
– (P1 ). Let x and y be two elements of C such that their join z in L is in
H(C). There is some element u in U such that z 6 u. By hypothesis, L[C]
is a lattice, so there is a join of ϕ(x) and ϕ(y) in L[C] let us call it t0 . By the
definition of 4, we have ϕ(x) 4 ϕ(u) and ϕ(y) 4 ϕ(u). This ensures that t0
is between ϕ(x) and ϕ(u). But those elements can only be in C 0 and there
must be an element t in C such that t0 = ϕ(t). This element t is then larger
than both x and y so that z 6 t and by definition of 4, z 4 t. But we also
have ϕ(x) 4 z and ϕ(y) 4 z so that t 4 z. Finally, t = z and thus the join
of x and y is in C.
– (P2 ). Let x be an element of H(C) and y be an element of X − H(C) which
is covered by x in L. Since x is in H(C), there are elements l and u in L and
U such that l 6 x 6 u. In turn, ϕ(l) 4 x. We also know that y 4 x, thus
the join of ϕ(l) and y in L[C] is less than or equal to x. For a contradiction,
suppose that x is not in C. Then x covers y in L[C] so that the join of y and
ϕ(l) in L[C] must be exactly x. Now y 4 ϕ(u) and ϕ(l) 4 ϕ(u) so the join
of y and ϕ(l) must be below ϕ(u). This is a contradiction since x 64 ϕ(u).
Let us now prove that (P1 ) and (P2 ) are sufficient conditions for L[C] to be a
lattice. For this, it suffices to prove that any pair of elements have a least upper
bound. From the definition of 4, one may check that for any x in X ∪ C 0 we
have
↑ (x, L)
if x ∈ H(C)
↑ (x, L[C]) = ↑ (x, L) ∪ ϕ(↑ (x, L)) if x ∈ X − H(C)
↑ (ϕ (x), L) ∪ ϕ(↑ (ϕ (x), L)) if x ∈ C 0 ,
−1 −1
where ϕ(A) denotes all elements that can be written as ϕ(a) for some a in A. In
all three cases, there exists an element a in X such that ↑ (x, L[C]) is exactly
↑ (a, L) or ↑ (a, L) ∪ ϕ(↑ (a, L)).
Now, notice that for any two subsets of X, A and B, the intersection of A
and ϕ(B) is always empty. From this and the distributivity of set operations, we
may derive that for any pair (x, y) of elements in X ∪ C 0 , there are two elements
a and b in X such that
↑ (a, L)∩ ↑ (b, L)
↑ (x, L[C])∩ ↑ (y, L[C]) = or
(↑ (a, L)∩ ↑ (b, L)) ∪ ϕ(↑ (a, L)∩ ↑ (b, L)).
In the first case, ↑ (a, L)∩ ↑ (b, L) is a subset of X and since L is a lattice,
we know there is a least element. For the second case, let c be the join of a and
b in L. We distinguish three subcases.
– If c is in C, then ϕ(c) is definitely less than any element of both ↑ (a, L)∩ ↑ (b, L)
and ϕ(↑ (a, L)∩ ↑ (b, L)). Therefore, ϕ(c) is the least upper bound of a and
b.
– If c is not in H(C), then c is a least element of ↑ (a, L)∩ ↑ (b, L). Consider
any element x of ϕ(↑ (a, L)∩ ↑ (b, L)). Then there ϕ−1 (x) is an element of
C which is greater than a and b. Therefore it is also bigger than c. By the
definition of 4, x is greater than c in L[C]. Element c is then the least upper
bound of a and b.
– If c is in H(C) − C, then a and b cannot be both elements of C by (P1 ).
Property (P2 ) basically tells us that for any chain from an element out of
H(C) to some element in H(C), there is an element of C. As a consequence,
there is a0 (respectively b0 ) in C such that a 6 a0 6 c (respectively b 6 b0 6 c).
But then the join of a0 and b0 must be c and since c is not in C, this leads
to a contradiction of (P1 ) so that this third subcase can never occur.
Thus any pair of elements in L[C] has a least upper bound. Since there is also a
bottom element in L[C], we conclude that L[C] is a lattice. t
u
As a useful side result, we get that (P1 ) and (P2 ) imply that the copy of any
join-irreducible element in L is a join-irreducible element of L[C].
Proposition 2. Given a lattice L, and a subset C satisfying, (P1 ) and (P2 ),
then for any element j in J(L) ∩ C, its copy ϕ(j) is a join-irreducible element
of L[C].
Proof. Let j be such an element and j∗ its unique predecessor in L. For a contra-
diction, suppose that ϕ(j) is not a join-irreducible element in L[C]. This implies
that j∗ is in H(C) but has not been copied. Consider two different elements a
and b covered by ϕ(j). In particular, ϕ(j) is the join of a and b in L[C]. Further-
more, a and b are less than j in L[C]. Whether they are in C 0 or in X, this means
they are also less than j∗ . But j∗ is not comparable to ϕ(j) the join of a and b,
which is a contradiction since Proposition 1 ensures that L[C] is a lattice. t
u
We may first notice that the copying process creates only m new and pairwise
distinct meet-irreducible elements.
Remark 2. Given a lattice L and a subset C satisfying (P1 ) and (P2 ),
M (L[C]) = M (L) ∪ {ϕ(u1 ), ϕ(u2 ), . . . , ϕ(um )}.
Join-irreducible elements can be of four types. Any join-irreducible element
of L which is not copied remains a join-irreducible element in L[C]. By Propo-
sition 2, any join-irreducible element of L which is copied has its image as a
join-irreducible element of L[C]. In addition, any element l in L becomes a
join-irreducible element in L[C]. And finally, some elements of C 0 might be join-
irreducible in L[C] even though their pre-image by ϕ is not join-irreducible in
L. This paragraph is summed up in the following remark.
Remark 3. Given a lattice L and a subset C satisfying (P1 ) and (P2 ),
J(L[C]) = (J(L) − C) ∪ ϕ(J(L) ∩ C) ∪ {l1 , l2 , . . . , ln } ∪ R,
where R denotes the join-irreducible elements of C 0 which are not the copy of a
join-irreducible element.
3 Preserving combinatorial properties
In this paper we want to keep control on the number of join-irreducible elements
in order to guarantee that the lattice L[C] satisfies several combinatorial prop-
erties. To this end, we would like the sizes of J(L[C]) and J(L) to differ by only
one. Remark 3 ensures that |J(L[C])| − |J(L)| = |R| + n. Since n is at least 1, we
need to enforce that R is empty and L is a singleton. Having R as an empty set
means that any join-irreducible element of L[C] in C 0 is the image of a former
join-irreducible of L in C. We thus states the additional property,
∀j ∈ J(L[C]) ∩ C 0 , ϕ−1 (j) ∈ J(L)
. (P0 )
L = {l}
Remark 4. Given a lattice L and a subset C satisfying (P0 ), (P1 ) and (P2 ),
|J(L[C])| = |J(L)| + 1.
In addition we shall consider three properties that will allow us to circum-
scribe the type of lattice that we want to obtain.
L = {⊥} (⊥)
U is a singleton, (U )
∀x ∈ C, ∀y ∈ X, ϕ(x) . y in L[C] ⇒ x . y in L. (.)
Each of these properties allows us to control some combinatorial parameter
of L[C]. Namely, (⊥) controls the height of the lattice, (U ) controls the number
of its meet-irreducible elements and (.) controls the number of pairs related
through relation .
%. These are formalized in the following theorem.
Theorem 1. Given L a lattice and C a subset satisfying (P0 ), (P1 ) and (P2 ),
we have the following implications:
(i) if (⊥), then h(L[C]) = h(L) + 1,
(ii) if (U ), then |M (L[C])| = |M (L)| + 1,
(iii) if (.), then | .%L[C] | = | .
%L | + |U |.
Proof. Fact (i) is trivial and (ii) is obtained by considering Remark 2. Let us
focus on (iii). By Property (P0 ), we know that L is a singleton. Let l denote its
single element.
Claim 1.1. For any u in U , l .
% ϕ(u).
The only predecessor of l in L[C] is ϕ(l) and for any u in U , the only successor
of ϕ(u) in L[C] is u itself. Furthermore l 4 u, ϕ(l) 4 ϕ(u) and l 64 ϕ(u). This
concludes the proof of Claim 1.1.
Claim 1.2. Reciprocally, for any meet-irreducible element m of L[C], if l .
% m,
then there is u in U such that m = ϕ(u).
A stronger statement is that when l . m, then there is u in U such that
m = ϕ(u).
We prove the stronger statement for a later use. Let m be an element of
M (L[C]) − ϕ(U ), thus m is in X. We shall prove that l . m cannot occur
in L[C]. For a contradiction, suppose that l . m in L[C]. This means that
ϕ(l) 4 m and by the definition of 4, we get that l 6 m in L and in turn that
ϕ(l) 4 m which is a contradiction. This concludes the proof of Claim 1.2.
Claim 1.3. Similarily, for any join-irreducible element j of L[C] and any u in U ,
if j .
% ϕ(u), then j = l.
Once again, a stronger statement is obtained when . % is replaced by ..
We also prove the stronger statement. Let u be some element of U and
suppose for a contradiction that ϕ(u) is in relation . with some join-irreducible
element j distinct from l. Then j 64 ϕ(u) and j 4 u. Thus j cannot be in C 0 (it
would be less than both u and ϕ(u) or not less than both of them). Therefore, j
is a join-irreducible element of L with a single predecessor j∗ . In L[C], j has also
a single predecessor j∗ which is not in C 0 . Since we assumed that j∗ 4 ϕ(u), it
cannot be in H(C) (they would be non-comparable). Since j has not been copied,
Property (P2 ) guarantees that j is not in H(C) either. But by the definition of
4, j must be either less than both u and ϕ(u) or not less than both of them.
This is a contradiction. This concludes the proof of Claim 1.3.
Claim 1.4. For any j in J(L) − C and m in M (L), j .
% m in L if and only if
j.
% m in L[C].
Let j be an element of J(L) − C then j is a join-irreducible element of L[C]
and its only predecessor in L[C] is the same as in L, say j∗ . Let m be a meet-
irreducible from L. It remains a meet-irreducible element in L[C]. But its only
successor in L[C] can be the same as in L, say m∗ or its copy ϕ(m∗ ). In any case,
the comparability of j and m is the same in both L and L[C]. Same stands for j∗
and m. Now if the only successor of m is the same in L and L[C], j . % m in L if
and only if j .
% m in L[C]. In the case where the only successor of m in L[C] is
ϕ(m∗ ), if j 4 ϕ(m∗ ), we also have j 6 m∗ . Reciprocally, if j 6 m∗ , we only need
to prove that j is not in H(C) to conclude that j 4 ϕ(m∗ ). Suppose that j is in
H(C). Since m∗ is in C, there is an element u in U such that m 4 ϕ(m∗ ) 6 ϕ(u).
This implies that m is not in H(C) (otherwise it would not be comparable with
ϕ(m∗ )) so that j 66 m. If j .% m in L, it means that j∗ 6 m thus j∗ is not in
H(C) either. In the end, since j has not been copied, Property (P2 ) allows us to
say that j is not in H(C). So j .% m in L if and only if j . % m in L[C], ending
the proof of Claim 1.4.
Claim 1.5. For any j in J(L) ∩ C and m in M (L), j .
% m in L if and only if
ϕ(j) .
% m in L[C].
We still have to study the case when the join-irreducible element is a copy
of a former join-irreducible element. Let j be in J(L) ∩ C and m be a meet-
irreducible element of L. We want to prove that ϕ(j) .% m in L[C] if and only
if j .
% m in L. In this case, we know that the only predecessor of ϕ(j) is some
element between ϕ(l) and ϕ(j). This element can then be written ϕ(x) for some
x in C between l and j. Thus, x 6 j∗ . By the definition of 4, ϕ(j) 64 m if
and only if j 66 m. Clearly, if j∗ 6 m, we have that ϕ(x) 4 m. Conversely, if
ϕ(x) 4 m, and ϕ(j) 64 m it means that ϕ(j) . m. By Property (.), we have
that j . m, thus j∗ 6 m. Let m∗ be the only successor of m in L. In L[C]
the only successor of m is either m∗ or ϕ(m∗ ). In the latter case, if j 6 m∗ ,
ϕ(j) 4 ϕ(m∗ ) and reciprocally. In the former case, we also have that ϕ(j) 4 m∗
if and only if j 6 m∗ . Therefore ϕ(j) .
% m in L[C] if and only if j .% m in L.
This concludes the proof of Claim 1.5
Summarising the previous results, we obtain that
%L[C] ={(l, ϕ(u)) : u ∈ U }
.
∪ {(j, m) ∈ J(L) × M (L) : j .
% m in L and j ∈
/ C}
∪ {(ϕ(j), m) ∈ C 0 × M (L) : j ∈ J(L) ∩ C and j .
% m in L}.
In terms of cardinality, we get that | .
%L[C] | = | .
%L | + |U |. t
u
We may notice that Property (.) is actually only needed for Claim 1.5.
Indeed, if this property is not satisfied, we may have new relations between the
image of a join-irreducible element and some old meet-irreducible element (see
Figure 2).
The proof of the third implication of Theorem 1 can be adapted to prove a
fourth implication. We prove separately for an easier reading.
Theorem 2. Given L a lattice and C a subset satisfying (P0 ), (P1 ) and (P2 ),
we have
(.) ⇒ | .L[C] | = | .L | + |U |
Proof. We use the same ideas as in the proof of Theorem 1.
Claim 2.6. For any u in U , l . ϕ(u) where L = {l}.
This a direct consequence of Claim 1.1.
Claim 2.7. For any j in J(L) − C and m in M (L), j . m in L if and only if
j . m in L[C].
Let j be an element of J(L) − C then j is a join-irreducible element of L[C]
and its only predecessor in L[C] is the same as in L, say j∗ . Let m be a meet-
irreducible from L. It remains a meet-irreducible element in L[C]. But its only
successor in L[C] can be the same as in L, say m∗ or its copy ϕ(m∗ ). In any
case, the comparability of j and m is the same in both L and L[C]. Same stands
for j∗ and m since j is not copied. Then j . m in L if and only if j . m in
L[C], ending the proof of Claim 2.7.
Claim 2.8. For any j in J(L) ∩ C and m in M (L), j . m in L if and only if
ϕ(j) . m in L[C].
By Property (.), we have for any j in J(L) ∩ C and m in M (L), ϕ(j) . m
in L[C] imply j . m in L.
For the converse, let j be in J(L) ∩ C and m be a meet-irreducible element
of L such that j . m in L. We want to prove that ϕ(j) . m in L[C]. First,
by definition of 4, ϕ(j) is incomparable to m. If j∗ ∈
/ H(C) then j∗ = ϕ(j)∗
by definition of 4, and then ϕ(j) . m in L[C]. Now suppose that j∗ ∈ H(C).
In this case, we know that the only predecessor ϕ(j)∗ of ϕ(j) is some element
between ϕ(l) and ϕ(j). This element can then be written ϕ(x) = ϕ(j)∗ for some
x in C between l and j. Then x 6 j∗ and thus ϕ(x) 4 m. So ϕ(j) . m in L[C].
Summarising the previous results (and the strong versions of Claims 1.2 and 1.3),
we obtain that
.L[C] ={(l, ϕ(u)) : u ∈ U }
∪ {(j, m) ∈ J(L) × M (L) : j . m in L and j ∈
/ C}
∪ {(ϕ(j), m) ∈ C 0 × M (L) : j ∈ J(L) ∩ C and j . m in L}.
In terms of cardinality, we get that | .L[C] | = | .L | + |U |. t
u
4 4
3 2 3 2
ϕ(2)
1
1
0 0
ϕ(0)
L L[C]
C = {0, 2}
Fig. 2. ϕ(2) . 3 in L[C] while we do not have 2 . 3 in L.
Lattices characterizations given in the following theorem can be found in
several papers (see for example [10, 17, 18]).
Theorem 3. Let L be a finite lattice. Then L is
– meet-semidistributive if and only if | .
% | = |J(L)| [18].
– semidistributive if and only if | .
% | = |J(L)| = |M (L)| [18].
– meet-extremal if and only if h(L) = |M (L)| [10].
– extremal if and only if h(L) = |J(L)| = |M (L)| [10].
– distributive if and only if | . | = |J(L)| = |M (L)| [17].
Remark 5. Notice that a lattice that is semidistributive and extremal does not
imply that is distributive (see Figure 3). In fact it is not graded. This explain
the property (.).
As a corollary of Theorems 1 2 and 3, we obtain a wider range of possibilities
to build specific types of lattices by preserving some combinatorial characteriza-
tions.
Corollary 1. Given a lattice L and a subset C verifying (P0 ), (P1 ) and (P2 )
the following implications are true:
1. L is distributive, (.), (⊥) and (U ) imply that L[C] is distributive.
2. L is semidistributive, (.), and (U ) imply that L[C] is semidistributive
3. L is meet-semidistributive and (.) imply that L[C] is meet-semidistributive
4. L is extremal, (⊥) and (U ) imply that L[C] is extremal
5. L is meet-extremal and (⊥) imply that L[C] is meet-extremal
One challenging problem is the characterization of contexts where their con-
cepts lattices satisfy the considered properties. For doubling convex sets, there
are nice FCA characterization and algorithms that recognize bound, lower (up-
per) bounded, semidistributive and convex lattices [5–7, 1–4, 8].
Acknowledgment The authors are very grateful to the referee for their construc-
tive comments. This work has been funded by the french national research agency
(ANR Graphen, 2015-2018).
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