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Answer to some open questions on covering dimension for finite lattices
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Answer to some open questions on covering dimension for finite lattices Hai-feng Zhang a,b, *, Meng Zhou b , Guang-jun Zhang a a
School of Instrument Science and Opto-electronics Engineering, Beihang University, XueYuan Road No. 37, HaiDian District, Beijing, 100191, China b School of Mathematics and Systems Science, Beihang University, XueYuan Road No. 37, HaiDian District, Beijing, 100191, China
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Article history: Received 23 February 2016 Received in revised form 24 October 2016 Accepted 28 October 2016 Available online xxxx Keywords: Lattice theory Covering dimension
a b s t r a c t Dube, Georgiou, Megaritis and Moshokoa (2015) characterized the covering dimension of a finite lattice. The covering dimension is completely different from the order dimension. The covering dimension of a finite lattice L will be denoted by dim(L) in this paper. Dube, Georgiou, Megaritis and Moshokoa (2015) posed three open questions. In this paper we answer two of these three questions. Question 1 is that whether the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . We give a positive answer for the above Question 1. Question 2 is that whether the relation dim(L1 ◦ L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . We give a negative answer to the above Question 2 by an example. © 2016 Elsevier B.V. All rights reserved.
1. Introduction Our references for elementary properties of lattices are [1,3,6]. In [4] the authors developed the theory of covering dimension for finite lattices. We now recall the definition of the covering dimension of a bounded lattice L and some important results in [4]. Let L be a bounded lattice. The top element and ⋁ the bottom element of L will be denoted by 1L and 0L , respectively. A cover of L is a subset C of L such that 0L ̸ ∈ C and C = 1L . A cover D of L is a refinement of a cover C , written D ≪ C , if for every d ∈ D there exists c ∈ C such that d ≤ c. Definition 1 (See [4]). Let L be a bounded lattice and k ∈ N0 , where N0 = {0, 1, 2, . . .}. The order of a subset C of L is defined as follows: (a) ord(C ) = k if the meet of any k + 2 distinct elements of C is 0L and there exist k + 1 distinct elements of C whose meet is not 0L . (b) ord(C ) = ∞ if for every k ∈ N0 there exist k distinct elements of C whose meet is not 0L . Definition 2 (See [2]). Let Blat be the class of bounded lattices. We define the covering dimension function dim : Blat → N0 as follows: (1) dim(L) ≤ k if every finite cover C of L has a finite refinement R with ord(R) ≤ k. (2) dim(L) = k, where k ≥ 1 if dim(L) ≤ k and dim(L) ̸ ≤ k − 1. (3) dim(L) = ∞ if the inequality dim(L) ≤ k does not hold for any k ∈ N0 .
*
Corresponding author. E-mail address: [email protected] (H.-f. Zhang).
http://dx.doi.org/10.1016/j.disc.2016.10.019 0012-365X/© 2016 Elsevier B.V. All rights reserved.
Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Fig. 1. The lattice K .
To make the definition of the covering dimension of a bounded lattice L more explicit, we give an equivalent definition as follows: Definition 3. The covering dimension of a bounded lattice L is the least k such that every finite cover of L has a finite refinement with order at most k, when there is a finite such k. And otherwise dim(L) = ∞. Definition 4 (See [5]). Let (X , ≤) be a poset. An order ≼ on X is a linear extension of ≤ if ≤⊆≼ and the poset (X , ≼) is a linearly ordered set. The order dimension dimo (X ) of a finite poset (X , ≤) is defined as the smallest cardinal number k such that the relation ≤ is the intersection of k linear extensions on X . Now we give an example to illustrate that the covering dimension differs from the order dimension. Example 1. Consider the lattice K represented by Fig. 1. Observe that dim(K ) = 0. By Definition 4, dimo (K ) = 2. Thus, we have dim(K ) ̸ = dimo (K ). The covering dimension of a finite lattice L will be denoted by dim(L) in this paper. Definition 5 (See [4]). A cover C of a finite lattice L is minimal if C ⊆ R for every refinement R of C . Definition 6. The product of two partially ordered sets L1 and L2 is the partially ordered set L1 × L2 where the relation ≤ is defined as follows: (x1 , x2 ) ≤ (y1 , y2 ) ⇐⇒ x1 ≤ y1 and x2 ≤ y2 , for every (x1 , x2 ), (y1 , y2 ) ∈ L1 × L2 . It is easy to see that the product of two finite lattices is a finite lattice. Definition 7. The lexicographic product L1 ◦ L2 of two lattices L1 and L2 is the partially ordered set L1 × L2 where the relation
≤ defined as follows: (x1 , x2 ) ≤ (y1 , y2 )
x1 < y1 x1 = y1 and x2 ≤ y2 .
{ ⇔
By checking the existence of meets and joins, it is easy to see that the lexicographic product of two lattices is a lattice. Definition 8 (See [1]). In a bounded poset P, a covers b (in notation, a ≻ b) if b < a and there exists no x such that b < x < a. An element a ∈ P is an atom of P if a ≻ 0 and a dual atom of P if 1 ≻ a. Corollary 1 (See [4]). If L is a finite bounded lattice and k = max{ord(C ) : C is a minimal cover of L}, then dim(L) = k. Theorem 1 (See [4]). If L is a finite lattice and k ∈ N0 , then dim(L) ≤ k if and only if every minimal cover C of L satisfies ord(C ) ≤ k. At the end of [4] the authors posed three questions. In this paper we solve two of these three questions. Question 1 is that whether the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . In Section 2 we give a positive answer for Question 1. We show that if L1 and L2 are finite lattices, then dim(L1 × L2 ) = max{dim(L1 ), dim(L2 )}. In Section 3 we give several results about the covering dimension of some special lattices. Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Question 2 in [4] is that whether the relation dim(L1 ◦ L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . At the end of this paper we give a negative answer for Question 2 by means of Example 2. Example 2 shows that there exist two finite lattices L1 and L2 such that dim(L1 ◦ L2 ) ̸ ≤ dim(L1 ) + dim(L2 ) + 1. 2. Main results In [4], the authors proposed the following question: Question 1. Does the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 hold for all finite lattices L1 and L2 ? Now, we answer Question 1 positively by the following results. These results show that the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . Theorem 2. If L1 and L2 are finite lattices, then dim(L1 × L2 ) = max{dim(L1 ), dim(L2 )}. Proof. Obviously L1 × L2 is a finite lattice. By Corollary 1, there exist two minimal covers C1 and C2 of L1 and L2 , respectively, such ord(C2 ) = dim(L2 ). We may assume C1 = {a1 , . . . , as } and C2 = {b1 , . . . , bt }. By definition, ⋁ that ord(C1 ) = dim(L1 ) and⋁ C2 = b1 ∨ · · · ∨ bt = 1L2 . C1 = a1 ∨ · · · ∨ as = 1L1 and Let C3 = {(a1 , 0⋁ L2 ), . . . , (as , 0L2 ), (0L1 , b1 ), . . . , (0L1 , bt )}. For all (a, b), (c , d) ∈ L1 ×L2 , it is easy to prove that (a, b)∨(c , d) = (a ∨ c , b ∨ d). So, C3 = (a1 ∨ · · · ∨ as ∨ 0L1 , b1 ∨ · · · ∨ bt ∨ 0L2 ) = (1L1 , 1L2 ). Thus, C3 is a cover of L1 × L2 . If C3 is not a minimal cover of L1 × L2 , then there exists a cover R of L1 × L2 that refines C3 and is such that C3 ̸ ⊆ R. Without loss we may assume that R = {(a′1 , 0L2 ), . . . , (a′n , 0L2 ), (0L1 , b′1 ), . . . , (0L1 , b′r )} and (am , 0L2 ) ∈ C3 \R. By ⋁of generality definition, R = (a′1 ∨ · · · ∨ a′n ∨ 0L1 , b′1 ∨ · · · ∨ b′r ∨ 0L2 ) = (1L1 , 1L2 ). We consider the set C ′ = {a′1 , . . . , a′n }. Note that C ′ is a cover of L1 . Since (am , 0L2 ) ̸ ∈ R, we have am ̸ ∈ C ′ . Since R is a refinement of C3 , for every (a′k , 0L2 ) ∈ {(a′1 , 0L2 ), . . . , (a′n , 0L2 )} there exists (ai(k) , 0L2 ) ∈ C3 such that (a′k , 0L2 ) ≤ (ai(k) , 0L2 ). Therefore, for every element a′k ∈ C ′ there exists an element ai(k) ∈ C1 such that a′k ≤ ai(k) . Thus, C ′ is a refinement of C1 . By am ̸ ∈ C ′ , we see that C1 ̸ ⊆ C ′ , which contradicts the fact that C1 is a minimal cover. Thus, C3 is a minimal cover of L1 × L2 . We may assume max{ord(C1 ), ord(C2 )} = ord(C1 ) = k1 . By ord(C1 ) = k1 , the meet of any k1 + 2 distinct elements of C1 is 0L1 , and there exist k1 + 1 distinct elements of C1 whose meet is not 0L1 . Without loss of generality we may assume that a1 ∧ · · · ∧ ak1 +1 ̸ = 0L1 . Thus, (a1 , 0L2 ) ∧ · · · ∧ (ak1 +1 , 0L2 ) ̸ = (0L1 , 0L2 ), and therefore ord(C3 ) ≥ k1 . Suppose that there exist k1 + 2 distinct elements of C3 whose meet is not (0L1 , 0L2 ). We will show that this leads to a contradiction. For every i ∈ {1, . . . , s} and j ∈ {1, . . . , t } we have (ai , 0L2 )∧(0L1 , bj ) = (0L1 , 0L2 ). Therefore, there exist distinct i1 , . . . , ik1 +2 ∈ {1, . . . , s} such that (ai1 , 0L2 ) ∧ · · · ∧ (aik +2 , 0L2 ) ̸ = (0L1 , 0L2 ) or there exist distinct j1 , . . . , jk1 +2 ∈ {1, . . . , t } 1 such that (0L1 , bj1 ) ∧ · · · ∧ (0L1 , bjk +2 ) ̸ = (0L1 , 0L2 ). 1 Thus, ai1 ∧ · · · ∧ aik +2 ̸ = 0L1 or bj1 ∧ · · · ∧ bjk +2 ̸ = 0L2 , which contradicts the fact that ord(C1 ) = k1 and ord(C2 ) ≤ k1 . 1 1 So, the meet of any k1 + 2 distinct elements of C3 is (0L1 , 0L2 ). Therefore, ord(C3 ) = k1 = max{ord(C1 ), ord(C2 )}. Since dim(L1 ) = ord(C1 ) and dim(L2 ) = ord(C2 ), we have ord(C3 ) = max{dim(L1 ), dim(L2 )}. By Corollary 1, dim(L1 × L2 ) ≥ max{dim(L1 ), dim(L2 )}. If D is a minimal cover of L1 × L2 , then for every (x1 , x2 ) ∈ D we have x1 = 0L1 or x2 = 0L2 . Indeed, let (x1 , x2 ) ∈ D such ⋁ that x1 ̸= 0L1 and x2 ̸= 0L2 . We prove that D0 = {(0L1 , x2 ), (x1 , 0L2 )} ∪ (D\{(x1 , x2 )}) is a cover of L1 × L2 . Suppose that D0 = (y1 , y2 ) ̸ = 1L1 ×L2 . Note that (y1 , y2 ) is an upper ⋁bound of {(0L1 , x2 ), (x1 , 0L2 )}. Since (x1 , x2 ) is the least upper bound of {(0L1 , x2 ), (x1 , 0L2 )}, we have (x1 , x2 ) ≤ (y1 , y2 ) and D = (y1 , y2 ), which is a contradiction. Thus, D0 is a refinement of D such that D ̸ ⊆ D0 , which contradicts the fact that D is minimal. Thus, we may assume D = {(c1 , 0L2 ), . . . , (cp , 0L2 ), (0L1 , d1 ), . . . , (0L1 , dr )}. Therefore, c1 ∨· · ·∨ cp = 1L1 and d1 ∨· · ·∨ dr = 1L2 . We observe that D1 = {c1 , . . . , cp } is a cover of L1 and D2 = {d1 , . . . , dr } is a cover of L2 . We will prove that D1 is a minimal cover of L1 and D2 is a minimal cover of L2 . Suppose that D1 is not a minimal cover of L1 . Note that there exists a cover D′ = {x1 , . . . , xq } of L1 , a refinement of D1 , such that D1 ̸ ⊆ D′ . We consider the family: D3 = {(x1 , 0L2 ), . . . , (xq , 0L2 ), (0L1 , d1 ), . . . , (0L1 , dr )}. Since D′ is a cover of L1 and D2 is a cover of L2 , also D3 is a cover of L1 × L2 . Also, since D′ is a refinement of D1 , for every element xk ∈ D′ there exists an element ci(k) ∈ D1 such that xk ≤ ci(k) . Therefore, for every element (xk , 0L2 ) ∈ {(x1 , 0L2 ), . . . , (xq , 0L2 )} there exists an element (ci(k) , 0L2 ) ∈ D such that (xk , 0L2 ) ≤ (ci(k) , 0L2 ). By definition, for every element (0L1 , dk ) ∈ D3 \{(x1 , 0L2 ), . . . , (xq , 0L2 )} we have (0L1 , dk ) ∈ D. Thus, D3 is a refinement of D. Since D1 ̸ ⊆ D′ , we have D ̸ ⊆ D3 , which is a contradiction. Thus, D1 is a minimal cover of L1 . Similarly, we can prove that D2 is a minimal cover of L2 . According to the previous proof, ord(D) = max{ord(D1 ), ord(D2 )}. By Corollary 1, max{ord(D1 ), ord(D2 )} ≤ max{dim(L1 ), dim(L2 )}. Thus, ord(D) ≤ max{dim(L1 ), dim(L2 )}. By Theorem 1, dim(L1 × L2 ) ≤ max{dim(L1 ), dim(L2 )}. Thus, dim(L1 × L2 ) = max{dim(L1 ), dim(L2 )}.
□
Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Fig. 2. The lattice L that has only one dual atom x.
Fig. 3. The lattice L that has only one atom x.
Corollary 2. If L1 and L2 are finite lattices, then dim(L1 × L2 ) = 0 if and only if dim(L1 ) = 0 and dim(L2 ) = 0. Now we have a positive answer to the first question in [4]. Let L1 , . . . , Ln be finite chains. We know that every finite chain has covering dimension 0. It is easy to see that L = L1 × · · · × Ln is a finite lattice for which dim(L) = 0. The following corollary follows easily from the preceding theorem. Corollary 3. If L1 , L2 , . . . , Ls are finite lattices, then dim(L1 × L2 × · · · × Ls ) = max{dim(L1 ), dim(L2 ), . . . , dim(Ls )}. 3. The covering dimension of some special lattices We give several results about the covering dimension of some special lattices below. Theorem 3. If a finite lattice L has only one dual atom x, then dim(L) = 0. Proof. If a finite lattice L has only one dual atom x, then Fig. 2 is the order diagram of L. Since x is the only one dual atom of L, for every a ∈ L\{1L } we have a ≤ x. Let C be a cover of L. It is easy to prove that C contains 1L . Hence the cover {1L } is a refinement of every cover of L, and therefore ord({1L }) = 0. Thus, dim(L) = 0. □ Theorem 4. If a finite lattice L has only one atom x, then dim(L) = max{|C | − 1 | C is a minimal cover of L}. Proof. If a finite lattice L has only one atom x, then Fig. 3 is the order diagram of L. Let C be a minimal cover of L. Since ⋀ x is the only one atom of L, we have a ≥ x for every a ∈ L\{0L }. Also since C is a minimal cover of L, we have 0L ̸∈ C . Thus, C ≥ x ̸ = 0L , and therefore ord(C ) = |C | − 1. By Corollary 1, dim(L) = max{|C | − 1 : C is a minimal cover of L}. □ Theorem 5. Let (L1 , ≤1 ), (L2 , ≤2 ), . . . , (Ls , ≤s ) be finite lattices with the property that Li ∩ Lj = ∅, where i ̸ = j. If L = (L1 \{1L1 }) ∪ (L2 \{1L2 }) ∪ · · · ∪ (L(s−1) \{1L(s−1) }) ∪ Ls and the relation ≤ on L is defined as follows (see Fig. 4): a, b ∈ L i a ∈ Li , b ∈ Lj
{ a ≤ b ⇐⇒
and and
a≤i b i < j,
then L is a finite lattice and dim(L) = max{|C | − 1 : C is a minimal cover of L}. Proof. It is easy to see that 0Lk ≥ (Lk−1 \{1Lk−1 }) for every k ∈ {2, . . . , s} in L. By checking the existence of meets and joins, L is a finite lattice. Let C be a minimal cover of L. If C ̸ ⊆ Ls , then there exists a ∈ C such that a ̸ ∈ Ls .
⋁
Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Fig. 4. The lattice L.
Fig. 5. The lattices L1 , L2 and their lexicographic product L1 ◦ L2 .
It is easy to prove that C \{a} is a cover of L. And obviously C \{a} is a refinement of C . Since C is a minimal cover of L, this is a contradiction. Thus, C ⊆⋀ Ls . ⋀ In Fig. 4 we observe that Ls = xs ̸ = 0L . So C ̸ = 0L , and therefore ord(C ) = |C | − 1. By Corollary 1, we have dim(L) = max{|C | − 1 : C is a minimal cover of L}.
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In [4], the authors posed another open question as follows: Question 2. Does the relation dim(L1 ◦ L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 hold for all finite lattices L1 and L2 ? Now, we give a negative answer to Question 2 by the following example: Example 2. Consider the finite lattices L1 and L2 shown on the left in Fig. 5. Consider their lexicographic product L1 ◦ L2 , shown on the right in Fig. 5. Let C = {(1L1 , x1 ), (1L1 , x2 ), (1L1 , x3 )}. We observe that C is a minimal cover of L1 ◦ L2 and ord(C ) = 2. Thus, dim(L1 ◦ L2 ) ≥ 2. We also observe that dim(L1 ) = dim(L2 ) = 0. Thus, dim(L1 ◦ L2 ) ̸ ≤ dim(L1 ) + dim(L2 ) + 1. Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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References [1] [2] [3] [4]
Garrett Birkhoff, Lattice theory, Mathematical Gazette (34) (1948). Charalambous, G. M, Dimension theory for sigma-frames, J. Lond. Math. Soc. (1974) 149–160. B.A. Davey, H.A. Priestley, Introduction to Lattices and Order, Cambridge University Press, 2002. Themba Dube, Dimitris N. Georgiou, Athanasios C. Megaritis, Seithuti P. Moshokoa, A study of covering dimension for the class of finite lattices, Discrete Math. 338 (7) (2015) 1096–1110. [5] Ben Dushnik, E.W. Miller, Partially ordered sets, Amer. J. Math. 63 (1941) 600–610. [6] Jorge Picado, Ales Pultr, Frames and Locales, Springer, Basel, 2012.
Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
)
–
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Note
Answer to some open questions on covering dimension for finite lattices Hai-feng Zhang a,b, *, Meng Zhou b , Guang-jun Zhang a a
School of Instrument Science and Opto-electronics Engineering, Beihang University, XueYuan Road No. 37, HaiDian District, Beijing, 100191, China b School of Mathematics and Systems Science, Beihang University, XueYuan Road No. 37, HaiDian District, Beijing, 100191, China
article
info
Article history: Received 23 February 2016 Received in revised form 24 October 2016 Accepted 28 October 2016 Available online xxxx Keywords: Lattice theory Covering dimension
a b s t r a c t Dube, Georgiou, Megaritis and Moshokoa (2015) characterized the covering dimension of a finite lattice. The covering dimension is completely different from the order dimension. The covering dimension of a finite lattice L will be denoted by dim(L) in this paper. Dube, Georgiou, Megaritis and Moshokoa (2015) posed three open questions. In this paper we answer two of these three questions. Question 1 is that whether the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . We give a positive answer for the above Question 1. Question 2 is that whether the relation dim(L1 ◦ L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . We give a negative answer to the above Question 2 by an example. © 2016 Elsevier B.V. All rights reserved.
1. Introduction Our references for elementary properties of lattices are [1,3,6]. In [4] the authors developed the theory of covering dimension for finite lattices. We now recall the definition of the covering dimension of a bounded lattice L and some important results in [4]. Let L be a bounded lattice. The top element and ⋁ the bottom element of L will be denoted by 1L and 0L , respectively. A cover of L is a subset C of L such that 0L ̸ ∈ C and C = 1L . A cover D of L is a refinement of a cover C , written D ≪ C , if for every d ∈ D there exists c ∈ C such that d ≤ c. Definition 1 (See [4]). Let L be a bounded lattice and k ∈ N0 , where N0 = {0, 1, 2, . . .}. The order of a subset C of L is defined as follows: (a) ord(C ) = k if the meet of any k + 2 distinct elements of C is 0L and there exist k + 1 distinct elements of C whose meet is not 0L . (b) ord(C ) = ∞ if for every k ∈ N0 there exist k distinct elements of C whose meet is not 0L . Definition 2 (See [2]). Let Blat be the class of bounded lattices. We define the covering dimension function dim : Blat → N0 as follows: (1) dim(L) ≤ k if every finite cover C of L has a finite refinement R with ord(R) ≤ k. (2) dim(L) = k, where k ≥ 1 if dim(L) ≤ k and dim(L) ̸ ≤ k − 1. (3) dim(L) = ∞ if the inequality dim(L) ≤ k does not hold for any k ∈ N0 .
*
Corresponding author. E-mail address: [email protected] (H.-f. Zhang).
http://dx.doi.org/10.1016/j.disc.2016.10.019 0012-365X/© 2016 Elsevier B.V. All rights reserved.
Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Fig. 1. The lattice K .
To make the definition of the covering dimension of a bounded lattice L more explicit, we give an equivalent definition as follows: Definition 3. The covering dimension of a bounded lattice L is the least k such that every finite cover of L has a finite refinement with order at most k, when there is a finite such k. And otherwise dim(L) = ∞. Definition 4 (See [5]). Let (X , ≤) be a poset. An order ≼ on X is a linear extension of ≤ if ≤⊆≼ and the poset (X , ≼) is a linearly ordered set. The order dimension dimo (X ) of a finite poset (X , ≤) is defined as the smallest cardinal number k such that the relation ≤ is the intersection of k linear extensions on X . Now we give an example to illustrate that the covering dimension differs from the order dimension. Example 1. Consider the lattice K represented by Fig. 1. Observe that dim(K ) = 0. By Definition 4, dimo (K ) = 2. Thus, we have dim(K ) ̸ = dimo (K ). The covering dimension of a finite lattice L will be denoted by dim(L) in this paper. Definition 5 (See [4]). A cover C of a finite lattice L is minimal if C ⊆ R for every refinement R of C . Definition 6. The product of two partially ordered sets L1 and L2 is the partially ordered set L1 × L2 where the relation ≤ is defined as follows: (x1 , x2 ) ≤ (y1 , y2 ) ⇐⇒ x1 ≤ y1 and x2 ≤ y2 , for every (x1 , x2 ), (y1 , y2 ) ∈ L1 × L2 . It is easy to see that the product of two finite lattices is a finite lattice. Definition 7. The lexicographic product L1 ◦ L2 of two lattices L1 and L2 is the partially ordered set L1 × L2 where the relation
≤ defined as follows: (x1 , x2 ) ≤ (y1 , y2 )
x1 < y1 x1 = y1 and x2 ≤ y2 .
{ ⇔
By checking the existence of meets and joins, it is easy to see that the lexicographic product of two lattices is a lattice. Definition 8 (See [1]). In a bounded poset P, a covers b (in notation, a ≻ b) if b < a and there exists no x such that b < x < a. An element a ∈ P is an atom of P if a ≻ 0 and a dual atom of P if 1 ≻ a. Corollary 1 (See [4]). If L is a finite bounded lattice and k = max{ord(C ) : C is a minimal cover of L}, then dim(L) = k. Theorem 1 (See [4]). If L is a finite lattice and k ∈ N0 , then dim(L) ≤ k if and only if every minimal cover C of L satisfies ord(C ) ≤ k. At the end of [4] the authors posed three questions. In this paper we solve two of these three questions. Question 1 is that whether the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . In Section 2 we give a positive answer for Question 1. We show that if L1 and L2 are finite lattices, then dim(L1 × L2 ) = max{dim(L1 ), dim(L2 )}. In Section 3 we give several results about the covering dimension of some special lattices. Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Question 2 in [4] is that whether the relation dim(L1 ◦ L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . At the end of this paper we give a negative answer for Question 2 by means of Example 2. Example 2 shows that there exist two finite lattices L1 and L2 such that dim(L1 ◦ L2 ) ̸ ≤ dim(L1 ) + dim(L2 ) + 1. 2. Main results In [4], the authors proposed the following question: Question 1. Does the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 hold for all finite lattices L1 and L2 ? Now, we answer Question 1 positively by the following results. These results show that the relation dim(L1 × L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 holds for all finite lattices L1 and L2 . Theorem 2. If L1 and L2 are finite lattices, then dim(L1 × L2 ) = max{dim(L1 ), dim(L2 )}. Proof. Obviously L1 × L2 is a finite lattice. By Corollary 1, there exist two minimal covers C1 and C2 of L1 and L2 , respectively, such ord(C2 ) = dim(L2 ). We may assume C1 = {a1 , . . . , as } and C2 = {b1 , . . . , bt }. By definition, ⋁ that ord(C1 ) = dim(L1 ) and⋁ C2 = b1 ∨ · · · ∨ bt = 1L2 . C1 = a1 ∨ · · · ∨ as = 1L1 and Let C3 = {(a1 , 0⋁ L2 ), . . . , (as , 0L2 ), (0L1 , b1 ), . . . , (0L1 , bt )}. For all (a, b), (c , d) ∈ L1 ×L2 , it is easy to prove that (a, b)∨(c , d) = (a ∨ c , b ∨ d). So, C3 = (a1 ∨ · · · ∨ as ∨ 0L1 , b1 ∨ · · · ∨ bt ∨ 0L2 ) = (1L1 , 1L2 ). Thus, C3 is a cover of L1 × L2 . If C3 is not a minimal cover of L1 × L2 , then there exists a cover R of L1 × L2 that refines C3 and is such that C3 ̸ ⊆ R. Without loss we may assume that R = {(a′1 , 0L2 ), . . . , (a′n , 0L2 ), (0L1 , b′1 ), . . . , (0L1 , b′r )} and (am , 0L2 ) ∈ C3 \R. By ⋁of generality definition, R = (a′1 ∨ · · · ∨ a′n ∨ 0L1 , b′1 ∨ · · · ∨ b′r ∨ 0L2 ) = (1L1 , 1L2 ). We consider the set C ′ = {a′1 , . . . , a′n }. Note that C ′ is a cover of L1 . Since (am , 0L2 ) ̸ ∈ R, we have am ̸ ∈ C ′ . Since R is a refinement of C3 , for every (a′k , 0L2 ) ∈ {(a′1 , 0L2 ), . . . , (a′n , 0L2 )} there exists (ai(k) , 0L2 ) ∈ C3 such that (a′k , 0L2 ) ≤ (ai(k) , 0L2 ). Therefore, for every element a′k ∈ C ′ there exists an element ai(k) ∈ C1 such that a′k ≤ ai(k) . Thus, C ′ is a refinement of C1 . By am ̸ ∈ C ′ , we see that C1 ̸ ⊆ C ′ , which contradicts the fact that C1 is a minimal cover. Thus, C3 is a minimal cover of L1 × L2 . We may assume max{ord(C1 ), ord(C2 )} = ord(C1 ) = k1 . By ord(C1 ) = k1 , the meet of any k1 + 2 distinct elements of C1 is 0L1 , and there exist k1 + 1 distinct elements of C1 whose meet is not 0L1 . Without loss of generality we may assume that a1 ∧ · · · ∧ ak1 +1 ̸ = 0L1 . Thus, (a1 , 0L2 ) ∧ · · · ∧ (ak1 +1 , 0L2 ) ̸ = (0L1 , 0L2 ), and therefore ord(C3 ) ≥ k1 . Suppose that there exist k1 + 2 distinct elements of C3 whose meet is not (0L1 , 0L2 ). We will show that this leads to a contradiction. For every i ∈ {1, . . . , s} and j ∈ {1, . . . , t } we have (ai , 0L2 )∧(0L1 , bj ) = (0L1 , 0L2 ). Therefore, there exist distinct i1 , . . . , ik1 +2 ∈ {1, . . . , s} such that (ai1 , 0L2 ) ∧ · · · ∧ (aik +2 , 0L2 ) ̸ = (0L1 , 0L2 ) or there exist distinct j1 , . . . , jk1 +2 ∈ {1, . . . , t } 1 such that (0L1 , bj1 ) ∧ · · · ∧ (0L1 , bjk +2 ) ̸ = (0L1 , 0L2 ). 1 Thus, ai1 ∧ · · · ∧ aik +2 ̸ = 0L1 or bj1 ∧ · · · ∧ bjk +2 ̸ = 0L2 , which contradicts the fact that ord(C1 ) = k1 and ord(C2 ) ≤ k1 . 1 1 So, the meet of any k1 + 2 distinct elements of C3 is (0L1 , 0L2 ). Therefore, ord(C3 ) = k1 = max{ord(C1 ), ord(C2 )}. Since dim(L1 ) = ord(C1 ) and dim(L2 ) = ord(C2 ), we have ord(C3 ) = max{dim(L1 ), dim(L2 )}. By Corollary 1, dim(L1 × L2 ) ≥ max{dim(L1 ), dim(L2 )}. If D is a minimal cover of L1 × L2 , then for every (x1 , x2 ) ∈ D we have x1 = 0L1 or x2 = 0L2 . Indeed, let (x1 , x2 ) ∈ D such ⋁ that x1 ̸= 0L1 and x2 ̸= 0L2 . We prove that D0 = {(0L1 , x2 ), (x1 , 0L2 )} ∪ (D\{(x1 , x2 )}) is a cover of L1 × L2 . Suppose that D0 = (y1 , y2 ) ̸ = 1L1 ×L2 . Note that (y1 , y2 ) is an upper ⋁bound of {(0L1 , x2 ), (x1 , 0L2 )}. Since (x1 , x2 ) is the least upper bound of {(0L1 , x2 ), (x1 , 0L2 )}, we have (x1 , x2 ) ≤ (y1 , y2 ) and D = (y1 , y2 ), which is a contradiction. Thus, D0 is a refinement of D such that D ̸ ⊆ D0 , which contradicts the fact that D is minimal. Thus, we may assume D = {(c1 , 0L2 ), . . . , (cp , 0L2 ), (0L1 , d1 ), . . . , (0L1 , dr )}. Therefore, c1 ∨· · ·∨ cp = 1L1 and d1 ∨· · ·∨ dr = 1L2 . We observe that D1 = {c1 , . . . , cp } is a cover of L1 and D2 = {d1 , . . . , dr } is a cover of L2 . We will prove that D1 is a minimal cover of L1 and D2 is a minimal cover of L2 . Suppose that D1 is not a minimal cover of L1 . Note that there exists a cover D′ = {x1 , . . . , xq } of L1 , a refinement of D1 , such that D1 ̸ ⊆ D′ . We consider the family: D3 = {(x1 , 0L2 ), . . . , (xq , 0L2 ), (0L1 , d1 ), . . . , (0L1 , dr )}. Since D′ is a cover of L1 and D2 is a cover of L2 , also D3 is a cover of L1 × L2 . Also, since D′ is a refinement of D1 , for every element xk ∈ D′ there exists an element ci(k) ∈ D1 such that xk ≤ ci(k) . Therefore, for every element (xk , 0L2 ) ∈ {(x1 , 0L2 ), . . . , (xq , 0L2 )} there exists an element (ci(k) , 0L2 ) ∈ D such that (xk , 0L2 ) ≤ (ci(k) , 0L2 ). By definition, for every element (0L1 , dk ) ∈ D3 \{(x1 , 0L2 ), . . . , (xq , 0L2 )} we have (0L1 , dk ) ∈ D. Thus, D3 is a refinement of D. Since D1 ̸ ⊆ D′ , we have D ̸ ⊆ D3 , which is a contradiction. Thus, D1 is a minimal cover of L1 . Similarly, we can prove that D2 is a minimal cover of L2 . According to the previous proof, ord(D) = max{ord(D1 ), ord(D2 )}. By Corollary 1, max{ord(D1 ), ord(D2 )} ≤ max{dim(L1 ), dim(L2 )}. Thus, ord(D) ≤ max{dim(L1 ), dim(L2 )}. By Theorem 1, dim(L1 × L2 ) ≤ max{dim(L1 ), dim(L2 )}. Thus, dim(L1 × L2 ) = max{dim(L1 ), dim(L2 )}.
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Fig. 2. The lattice L that has only one dual atom x.
Fig. 3. The lattice L that has only one atom x.
Corollary 2. If L1 and L2 are finite lattices, then dim(L1 × L2 ) = 0 if and only if dim(L1 ) = 0 and dim(L2 ) = 0. Now we have a positive answer to the first question in [4]. Let L1 , . . . , Ln be finite chains. We know that every finite chain has covering dimension 0. It is easy to see that L = L1 × · · · × Ln is a finite lattice for which dim(L) = 0. The following corollary follows easily from the preceding theorem. Corollary 3. If L1 , L2 , . . . , Ls are finite lattices, then dim(L1 × L2 × · · · × Ls ) = max{dim(L1 ), dim(L2 ), . . . , dim(Ls )}. 3. The covering dimension of some special lattices We give several results about the covering dimension of some special lattices below. Theorem 3. If a finite lattice L has only one dual atom x, then dim(L) = 0. Proof. If a finite lattice L has only one dual atom x, then Fig. 2 is the order diagram of L. Since x is the only one dual atom of L, for every a ∈ L\{1L } we have a ≤ x. Let C be a cover of L. It is easy to prove that C contains 1L . Hence the cover {1L } is a refinement of every cover of L, and therefore ord({1L }) = 0. Thus, dim(L) = 0. □ Theorem 4. If a finite lattice L has only one atom x, then dim(L) = max{|C | − 1 | C is a minimal cover of L}. Proof. If a finite lattice L has only one atom x, then Fig. 3 is the order diagram of L. Let C be a minimal cover of L. Since ⋀ x is the only one atom of L, we have a ≥ x for every a ∈ L\{0L }. Also since C is a minimal cover of L, we have 0L ̸∈ C . Thus, C ≥ x ̸ = 0L , and therefore ord(C ) = |C | − 1. By Corollary 1, dim(L) = max{|C | − 1 : C is a minimal cover of L}. □ Theorem 5. Let (L1 , ≤1 ), (L2 , ≤2 ), . . . , (Ls , ≤s ) be finite lattices with the property that Li ∩ Lj = ∅, where i ̸ = j. If L = (L1 \{1L1 }) ∪ (L2 \{1L2 }) ∪ · · · ∪ (L(s−1) \{1L(s−1) }) ∪ Ls and the relation ≤ on L is defined as follows (see Fig. 4): a, b ∈ L i a ∈ Li , b ∈ Lj
{ a ≤ b ⇐⇒
and and
a≤i b i < j,
then L is a finite lattice and dim(L) = max{|C | − 1 : C is a minimal cover of L}. Proof. It is easy to see that 0Lk ≥ (Lk−1 \{1Lk−1 }) for every k ∈ {2, . . . , s} in L. By checking the existence of meets and joins, L is a finite lattice. Let C be a minimal cover of L. If C ̸ ⊆ Ls , then there exists a ∈ C such that a ̸ ∈ Ls .
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Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Fig. 4. The lattice L.
Fig. 5. The lattices L1 , L2 and their lexicographic product L1 ◦ L2 .
It is easy to prove that C \{a} is a cover of L. And obviously C \{a} is a refinement of C . Since C is a minimal cover of L, this is a contradiction. Thus, C ⊆⋀ Ls . ⋀ In Fig. 4 we observe that Ls = xs ̸ = 0L . So C ̸ = 0L , and therefore ord(C ) = |C | − 1. By Corollary 1, we have dim(L) = max{|C | − 1 : C is a minimal cover of L}.
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In [4], the authors posed another open question as follows: Question 2. Does the relation dim(L1 ◦ L2 ) ≤ dim(L1 ) + dim(L2 ) + 1 hold for all finite lattices L1 and L2 ? Now, we give a negative answer to Question 2 by the following example: Example 2. Consider the finite lattices L1 and L2 shown on the left in Fig. 5. Consider their lexicographic product L1 ◦ L2 , shown on the right in Fig. 5. Let C = {(1L1 , x1 ), (1L1 , x2 ), (1L1 , x3 )}. We observe that C is a minimal cover of L1 ◦ L2 and ord(C ) = 2. Thus, dim(L1 ◦ L2 ) ≥ 2. We also observe that dim(L1 ) = dim(L2 ) = 0. Thus, dim(L1 ◦ L2 ) ̸ ≤ dim(L1 ) + dim(L2 ) + 1. Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019
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Garrett Birkhoff, Lattice theory, Mathematical Gazette (34) (1948). Charalambous, G. M, Dimension theory for sigma-frames, J. Lond. Math. Soc. (1974) 149–160. B.A. Davey, H.A. Priestley, Introduction to Lattices and Order, Cambridge University Press, 2002. Themba Dube, Dimitris N. Georgiou, Athanasios C. Megaritis, Seithuti P. Moshokoa, A study of covering dimension for the class of finite lattices, Discrete Math. 338 (7) (2015) 1096–1110. [5] Ben Dushnik, E.W. Miller, Partially ordered sets, Amer. J. Math. 63 (1941) 600–610. [6] Jorge Picado, Ales Pultr, Frames and Locales, Springer, Basel, 2012.
Please cite this article in press as: H.-f. Zhang, et al., Answer to some open questions on covering dimension for finite lattices, Discrete Mathematics (2016), http://dx.doi.org/10.1016/j.disc.2016.10.019