Show Summary Details
More options …

Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

IMPACT FACTOR 2018: 0.726
5-year IMPACT FACTOR: 0.869

CiteScore 2018: 0.90

SCImago Journal Rank (SJR) 2018: 0.323
Source Normalized Impact per Paper (SNIP) 2018: 0.821

Mathematical Citation Quotient (MCQ) 2018: 0.34

ICV 2018: 152.31

Open Access
Online
ISSN
2391-5455
See all formats and pricing
More options …

Classification lattices are geometric for complete atomistic lattices

Hua Mao
Published Online: 2017-07-13 | DOI: https://doi.org/10.1515/math-2017-0078

Abstract

We characterize complete atomistic lattices whose classification lattices are geometric. This implies an proper solution to a problem raised by S. Radeleczki in 2002.

MSC 2010: 06B05; 06B15; 06C10

1 Introduction and Preliminaries

Given a complete atomistic lattice L (see [1, p.251,Definition 285]), a subset C of L \ {0} is called a classification system of L if

(C1) cd = 0 for any two distinct elements c, dC, and

(C2) x = ∨{xc | cC} holds for each xL.

This concept, not only for the atomistic case, is due to Radeleczki [2], and it has motivations in the theory of concept lattices (see [2, 3]). Let Cls(L) stand for the collection of all classification systems of L. For C, DCls(L), we let CD if and only if for each cC there is a dD such that cd. By [2, Theorem 4.2], this relation turns Cls(L) into a complete lattice.

Radeleczki [2] provides two interesting and attractive open problems for classification systems. Mao [4] solves one of them completely. Another open problem asks to characterize those complete atomistic lattices whose classification lattices are geometric. Though this problem is partly solved by Mao [5], it is unsolved completely to date. This problem is related to the application of concept lattices to one of the main problems in group technology (see [2, 6-8]). Actually, the problem is also related with the study of CD-bases of a lattice (see [9,10]), or to the investigation of the decomposition systems of a closure system (cf. [11,12]). In other words, the solution of this problem may have useful applications in several related fields. Based on these discussions, we will reveal the answer to this problem in this paper.

1.1 Definitions and properties

To find the answer of the open problem, we first review some definitions and properties what we need later on. In what follows, for more basic concepts of lattice theory and classification systems, the reader is kindly referred to [1,13] and [2] respectively.

Some definitions

1. [1, p.1] A nonempty set equipped with a relation (reflexivity, antisymmetry, and transitivity) is called a poset.

2. [1, p.9] A poset (K, ≤) is a lattice if sup{a, b} and inf{a, b} exist for all a, bK.

3. [1, p.28] The lattices (K, ≤) and (K′, ≤) are isomorphic (in symbols, (K, ≤) ≅ (K′, ≤)) if the map ϕ : KK′ is a bijection such that ab in (K, ≤) if and only if ϕ(a) ≤ ϕ(b) in (K′, ≤).

4. [13, p.11] N5 = {a, b, c, d, e} is a lattice which satisfies the following conditions:

(p.1) e < x < d for any x ∈ {a, b, c}; (p.2) c < a; (p.3) a||b and c||b.

5. [1, p.35] An interval in a lattice K is a subset of K of the form [a, b] = {xK | axb} for some a, b, ∈ K with ab.

6. [13, p.5 & 1,p.4] The length of a finite chain n is defined to be n — 1; the length l(P) of a poset P is defined as the least upper bound of the lengths of the chains in P.

[1, p.4] The height of an element aK, denoted by h(a), is the length of the order {xK | xa}.

That a covers b in a poset P is defined in [1, p.6].

The diagram of a finite poset is defined in [1, p.6].

7. [1, p.50] A lattice K is called complete if ∨H and ∧H exist for any subset HK, in which ∨H and ∧H are seen [1, p.5].

8. [1, p.101] An element a of a lattice K is an atom if a covers 0 (i.e the least element in K if it exists). A lattice K is called atomistic if every element is a join of atoms.

9. [1, p.329] A lattice K is called semimodular if it satisfies for all a, bK, ababbc or ac = bc.

10. [1, p.52] Let K be a complete lattice and let a be an element of K. Then a is called compact if a ≤ ∨X, for any XK, implies that a ≤ ∨X1 for some finite X1X.

11. [1, p.342] A lattice K is called geometric if the lattice K is complete, K is atomistic, all atoms are compact, and K is semimodular.

Some notations

1. Let K be a lattice.

2. L denotes a complete atomistic lattice; 0 and 1 denote its least and greatest element, respectively.

Sx {x} ∪ {aA(L) | ax = 0} for any xL \ {0}; S0 = ∧{S |SCls(L)}.

Some properties

1. [1, p.342,Corollary 390] An interval of a geometric lattice is again a geometric lattice.

2. [13, p.13,Theorem 12] Up to isomorphism, any non-modular lattice contains the lattice N5 as a sublattice.

[1, p.109,Theorem 102] A lattice K is modular if and only if it does not contain a N5.

3. [2] S0Cls(L) and S1 = {1} ∈ Cls(L) hold, and SxCls(L) holds for any xL \ {0}.

4. [5] Cls(L) possesses the following properties.

(4.1) Let x, yL \ {0}. Then xySxSy. Further, x < ySx < Sy.

(4.2) A(Cls(L)) = {Sd | d ∈ ℱ2}.

1.2 Answer to the cases of |𝓕2| ≤ 1

This subsection will reveal the answers to the open problem if a complete atomistic lattice L satisfies |ℱ2| ≤ 1. In fact, |ℱ2| ≤ 1 means |ℱ2| = 0 or |ℱ2| = 1. We will search out the answers to the open problem correspondent to the above two cases respectively in this subsection.

Case I

|ℱ2| = 0.

Answer: Cls(L) is geometric if and only if h(L) = 1.

The reason is the following.

|ℱ2| = 0 means h(L) = 1 or h(L) = 0.

If h(L) = 1. Then we obtain L = {0, 1} such that 01 and A(L) = {1}. This implies Cls(L) = {S0} = {S1}.

So, Cls(L) is a geometric lattice.

If h(L) = 0. Then we obtain L = {0}. This implies Cls(L) to be empty. So, Cls(L) is not a lattice according to the definition of lattice in Subsection 1.1.

Conversely, if Cls(L) is geometric. According to Cls(L) ≠ ∅ and the above analysis, we may be assured h(L) = 1.

Case II

|ℱ2| = 1.

Answer: Cls(L) is geometric if and only if h(L) = 2.

The reason is the following.

From |ℱ2| = 1, we obtain h(L) ≥ 2.

If h(L) = 2. According to the atomic of L, we obtain L = {0, 1} ∪ A(L) such that 0 < 1 and A(L) ≠ ∅. From these results, we confirm Cls(L) = {S0, S1} with S0S1. So, Cls(L) is a geometric lattice.

If h(L) > 2. In view of |ℱ2| = 1, we may assume ℱ2 = {d}. Considering the atomic of L with |ℱ2| = 1 and h(1) = h(L), we may affirm that for any xL \ ({0, 1, d} ∪ A(d)), we receive xA(L) or d < x < 1. Thereby, we may decide Cls(L) = {S0, S1 Sd} ∪ {Sx | xL \ ({0, 1, d} ∪ A(d)) and d < x < 1} satisfying A(Cls(L)) = {Sd}. Furthermore, S1 is not a join of some atoms in Cls(L). Hence, Cls(L) is not geometric.

Conversely, if Cls(L) is geometric. According to the above analysis for h(L) > 2, we may be assured h(L) = 2.

Combining the above answers for |ℱ2| ≤ 1, we determine that in what follows, we will consider |ℱ2| > 1 to discover the answer for the open problem in [2].

1.3 Organization

The rest of the paper is organized as follows. Section 2 will give the answer to the open problem for the case of |ℱ2| > 1. The final part of this paper, Section 3, concludes this paper and leaves room for our future work.

2 Main results

This section answers the open problem for a complete atomistic lattice L which satisfies |ℱ2| > 1.

Before we provide the answer, we present some properties with regards to L which we will use to find the answer to the open problem.

Lemma 2.1

(1) Sa = S0 if and only if aA(L).

(2) Let x, yL \ (A(L) ∪ 0) and xy. Then x < ySx < Sy.

Furthermore, for any x, yL\ (A(L) ∪ 0), we obtain xySxSy.

Proof

Firstly, we prove item (1).

Since L is atomistic, we may easily obtain that A(L) ⊆ L\ {0} holds and A(L) satisfies both of (C1) and (C2). That is to say, A(L) ∈ Cls(L) holds.

Let SCls(L) \ A(L). Then, we affirm that S contains an element x satisfying xA(L) according to SL \ {0} and SA(L). Thus, by the definition of classification system and the atomic of L, we obtain A(L) < S in Cls(L).

Therefore, S0 = A(L) holds.

Since Sa = {a} ∪ {xA(L) | xa} for any aA(L), we may easily obtain Sa = A(L). Hence, we obtain Sa = S0.

According to Sx = {x} ∪ {aA(L) | aA(x)} for any xL \ {0} with the above A(L) = S0, we may easily attain that if Sx = S0 holds, then xA(L) holds.

Secondly, we prove item (2).

It will be finished by routine verification of the property (4.1) in Subsection 1.1.

Let x, yA(L) and xy. Then we may easily see that Sx = Sy = S0 = A(L) since Lemma 2.1 (1). Even though, in the following, we still write S0 not A(L) when we consider Cls(L). The reason is that we hope to keep the coherence since SzCls(L) holds for any zL\ (A(L) ∪ 0) in view of property (3) in Subsection 1.1.□

Taking Lemma 2.1 (2) and Lemma 3.1(3.1) in [10], we will obtain the following corollary.

Corollary 2.2

(1) Let x, yL\ (A(L) ∪ 0).

Then Sx = Syx = y. In addition, x||ySx||Sy.

(2) Let x, yL\ {0}. Then

xySxSy;   x < ySx < Sy.

Proof

At first, we prove Sx = Syx = y in item (1).

If x = y, then it is easy to see Sx = Sy by the definitions of Sx and Sy.

Conversely, let Sx = Sy. According to Sx = {x}∪{aA(L) | ax = 0}, Sy = {y}∪{bA(L) | by = 0} and xA(L) ∪ 0, we confirm x ∉ {bA(L) | by = 0}. Considering Sx = Sy, we decide x ∈ {y}. Hence, we obtain x = y.

Secondly, we prove x||ySx||Sy in item (1).

Considering Lemma 2.1 (1), we may be assured Sx, SyCls(L)\{S0}.

x||y implies the true of xy, xy and yx.

Using the above and xy, we find SxSy. Using xy (yx) and Lemma 2.1 (2), we find SxSy (SySx). Therefore, we know Sx||Sy.

Conversely, if Sx||Sy. This means SxSy, SxSy and SySx. Applying the ‘first part’ above and SxSy, we find xy. Applying SxSy (SySx) and Lemma 2.1 (2), we find xy (yx). Therefore, we know x||y.

Next, we will prove item (2).

If x = y. Then we easily know Sx = Sy.

If x, yL \ (A(L) ∪ 0). Then by Lemma 2.1 (2) and item (1) above, we obtain xySxSy, and x < ySx < Sy.

If x, yA(L) and xy hold, or yA(L) and xA(L) ∪ 0 hold. Then it is easy to see that both of x < y and xy will not happen. Thus, we do not consider the needed results.

If xA(L) and yA(L) ∪ 0. Then xy will mean xA(y). Let bA(L) \ A(y). We may easily see that by = 0 since L is atomistic. This implies bSy according to Sy = {y} ∪ {aA(L)| ay = 0}. In addition, Sx = A(L) = S0 holds in light of Lemma 2.1 (1). Hence, Sx < Sy holds. That is to say, SxSy is correct.

Actually, the three of xA(L), yA(L) ∪ 0 and xy imply x < y. So, we easily obtain S0 < Sy. That is to say, Sx < Sy holds since Sx = S0 holds by Lemma 2.1 (1). ☐

Let L be a complete atomistic lattice with at least two elements of height 2. Based on the Subsection 1.2 and the above discussions, in answering the problem for L we will consider two statuses:

First status: there is xy0 for some x, yL \ (A(L) ∪ 01) and x||y.

Second status: there are xy = 0 for any x, yL \ (A(L) ∪ 01) and x||y.

In fact, Since |ℱ2| > 1, we obtain |L \ (A(L) ∪ 01)| > 1. Thus, we may easily find that only one of the above statuses will happen for L.

In this subsection, we will give an answer to the open problem for the complete atomistic lattices that satisfy the conditions in First status.

Through this subsection, LCls(L) denotes L \ A(L). We may easily see that LCls(L) with the same order in L is a sublattice of L. We still denote this sublattice as LCls(L).

The main result in this subsection is the following Theorem 2.3, that is, an answer to the open problem for ‘First status’.

Theorem 2.3

Let L satisfy the conditions in ‘First status’. Then Cls(L) is geometric if and only if LCls(L) satisfies the following conditions.

1. For any m, nL \ (A(L) ∪ 01) with m||n, mn0 holds.

2. LCls(L) is geometric.

Before giving the proof of Theorem 2.3, we present two lemmas as preparations.

Lemma 2.4

Let L satisfy the conditions in ‘First status’. If Cls(L) is geometric. Then the following statements hold.

(1) Cls(L) = {Sx | xL \ {0}}.

(2) (2.1) Let x, yA(L) and xy. Then

SxSy = S0 < Sxy;

SxSy = S0Sxy.

(2.2) Let xA(L) and yL \ (A(L) ∪ 0). Then

SxSy = Sxy and SxSy = Sx if x < y;

SxSy < Sxy and SxSy = S0Sxy if x||y.

(2.3) Let x, yL \ (A(L) ∪ 0) and xy. Then

Sxy = SxSy;

S0 = SxSy if xy0;

Sxy = SxSy if xy0.

Proof

We prove item (1).

Otherwise, there is SCls(L) \ {Sx | xL \ {0}}. Using Lemma 2.1 (1), Sx = S0 holds for any xA(L). Thus, there must exist y, zS \ (A(L) ∪ 0) satisfying yz and y||z according to the definition of Sx (∀xL \ (A(L) ∪ 0)) in Subsection 1.1. Hence, we may determine yz = 0 since S satisfies (C1).

From y, zS \ (A(L) ∪ 0), we obtain h(y, h(z) ≥ 2 where h is the height function of L. So, we believe ℱ2(y) ≠ ∅ and ℱ2(z) ≠ ∅. According to yz = 0, we decide ℱ2(y) ∩ ℱ2(z) = ∅.

We will continue the discussion in the following two steps. Let d ∈ ℱ2(y) and e ∈ ℱ2(z).

Step 1

Let Sde = {d, e} ∪ {pA(L) | pA(d) and pA(e)}. We will prove SdeCls(L).

We may easily gain de = 0 and d||e since 0deyz = 0 and y||z. We also obtain dp = ep = 0 since pSde satisfies pA(d) and pA(e) with $d\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{a\in A\left(d\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}e\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{q\in A\left(e\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q$. This implies that Sde satisfies (C1).

We distinguish three cases to demonstrate Sde to satisfy (C2). Let gL \ {0}.

Case 1

gd.

This implies ge = gp = 0 for any pA(L) \ (A(d) ∪ A(e)) since de = 0 and pA(d). Thus, we may obtain g = (gd) ∨ (ge) ∨ $\left(\underset{p\in {S}_{de}\mathrm{\setminus }\left\{d,e\right\}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(g\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}p\right)\right)$.

Case 2

ge.

Similarly to Case 1, we can obtain $g=\left(g\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}e\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\vee \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\underset{q\in {S}_{de}\mathrm{\setminus }\left\{e\right\}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(g\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}q\right)\right)$.

Case 3

gd and ge.

We may easily find A(g) = (A(g) ∩ A(d)) ∪ (A(g) ∩ A(e)) ∪ (A(g) ∩ {qA(L) | qA(d) and qA(e)} and g = ∨A(g) = $\left(\underset{u\in A\left(g\right)\cap A\left(d\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}u\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\vee \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\underset{w\in A\left(g\right)\cap A\left(e\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}w\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\vee \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\underset{q\in A\left(L\right)\mathrm{\setminus }\left(A\left(d\right)\cup A\left(e\right)\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q\right)$.

If gd. Then it is easy to see g||d or d < g.

In fact, g||d means $\underset{u\in A\left(g\right)\cap A\left(d\right)}{\bigvee }u\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}<\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g$ means A(d) ⊂ A(g). Furthermore, we obtain$\underset{u\in A\left(g\right)\cap A\left(d\right)}{\bigvee }u=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{u\in A\left(d\right)}{\bigvee }u=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d$.

Similarly, we can affirm $\underset{w\in A\left(g\right)\cap A\left(e\right)}{\bigvee }w=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}e$.

Moreover, g = (gd) ∨ (ge) ∨ $\left(\underset{q\in A\left(L\right)\mathrm{\setminus }\left(A\left(d\right)\cup A\left(e\right)\right)}{\bigvee }\left(g\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}q\right)\right)$ is accepted. Summing up Cases 1, 2 and 3, we can decide that Sde satisfies (C2). Therefore, SdeCls (L) holds.

Step 2

From the definition of Sde in Step 1, we may easily determine Sde ∉ {Sx | xL \ {0}} since d, eA(L) ∪ 0.

In addition, we also obtain S0 < Sd < Sde and S0 < Se < SdeS.

Step 3

According to the conditions in ‘First status’, we may confirm that there exist m, nL \ (A(L) ∪ 01) such that m||n and mn ≠ 0.

We will analyze the relationships among Sm, Sn and [S0, Sde] in the following Parts (I) and (II).

Part (I)

If both of Sm and Sn belong to [S0, Sde].

Then we may easily find SmSde and SnSde. By the definition of “≤” in Cls(L), we receive md or me by Lemma 2.1 (2). At the same time, we also receive nd or ne. Thus, for clarity, we will distinguish four cases from Case I1 to Case I4 to continue our discussion. In fact, there are only the following four cases for m and n according to the above discussions.

md and ne.

me and nd.

md and nd.

Case I4

me and ne.

Suppose that Case I1 happens. Then, we obtain 0mnde = 0. So, we receive mn = 0. This is a contradiction to mn ≠ 0. Thus, we determine that Case I1 will not happen.

Similarly to the discussion for Case I1, we obtain that Case I2 will not happen. Suppose that Case I3 happens. If m = d. Then nd follows nm. This is a contradiction to m||n. Thus, we determine m < d. Analogously, n < d holds.

According to d ∈ ℱ2(y) and m < d, we obtain h(m) < 2. This implies mA(L)∪0. But this is a contradiction to mL \ (A(L) ∪ 01). Thus, we can say md.

Analogously, we can also find nd. Shortly, Case I3 can not happen.

Similarly to the discussion for Case I3, we obtain that Case I4 can not happen.

These discussions imply that if Sm ∈ [S0, Sde] (Sn ∈ [S0, Sde]), then Sn ∉ [S0, Sde] (Sm ∉ [S0, Sde]).

Part (II)

If Sm ∉ [S0, Sde] but Sn ∈ [S0, Sde].

Sm ∉ [S0, Sde] means SmSde. Thus, we obtain Sde < Sm or Sde||Sm. For clarity, we will use the following Cases II1 and II2 to discuss.

Case II1

Sde < Sm.

We find Sd, Se < Sm since we may easily see that Sd, Se < Sde. This implies d < m and e < m according to Lemma 2.1(2).

In view of the property (4.2) in Subsection 1.1 and d, e ∈ ℱ2, we find Sd, SeA(Cls(L)). However, we easily find out SdSe = Sde. Thus, from the geometric of Cls(L) and the property (4.2) in Subsection 1.1, we confirm that there exists a c ∈ ℱ2 satisfying Sc < Sm and Sc||Sde since Sde < Sm. This follows d||c and e||c. Hence, we obtain 𝒞 ≠ ∅ in which 𝒞 = ℱ2(m) \ {d,e}. We also easily find SxSde for any x ∈ 𝒞.

Let S′ = {d, e} ∪ {g | g ∈ 𝒞} ∪ {aA(L) | aA(d) and aA(e), and aA(g) for any g ∈ 𝒞}. We may determine S′Cls(L) according to the definition of Cls(L). In addition, we also determine S′ < Sm since d, e, g ∈ ℱ2(m) with d, e < m and gm for any g ∈ 𝒞.

Additionally, let MA(Cls(L)) and MSm. According to the property (4.2) in Subsection 1.1, we may decide M = Sx0 for some x0 ∈ ℱ2. Owing to the item (2) in Lemma 2.1, the property (4.2) in Subsection 1.1 and Sd < Sm, we may decide m ∉ ℱ2. Thus, we obtain Sx0Sm. So, Sx0 < Sm holds. In view of Lemma 2.1 (2), we can indicate x0 < m. This implies x0 ∈ ℱ2(m). Hence, we find x0 ∈ {d, e} ∪ {g | g ∈ 𝒞}. Therefore, we can infer to Sx0 < S′.

Considering the property (4.2) in Subsection 1.1, we may confirm that Sm is not a join of atoms in Cls(L). This is a contradiction to the geometric of Cls(L).

Case II2

Sde||Sm.

If md (or me; or d, em). Then, we decide SmSdSde (or SmSe; or SdeSm) according to Lemma 2.1 (2). No matter which results happens, it is a contradiction to Sde||Sm.

If d||m and e||m. We may easily decide Sd, Se, Sx < Su in which x ∈ ℱ2(m) and u = dem. Analogously to the discussion for Case II1, we will find that Su is not a join of atoms in Clx(L) since S″ < Su and the property (4.2) in Subsection 1.1 in which S″ = {d, e} ∪ {x | x ∈ ℱ2(m)} ∪ {y | y ∈ ℱ2(u) and y ∉ {d, e} ∪ {x | x ∈ ℱ2(m)} ∪ {aA(L) | aA(d), and aA(e), and in addition, aA(m) ∈ Cls(L). This is a contradiction to the geometric of Cls(L).

If d||m and em. Actually, we find me. This implies e < m since em. Therefore, we confirm e ∈ ℱ2(m) and d ∉ ℱ2(m). Moreover, ℱ2(m) ≠ ∅ holds. We attain Se < Sm since Lemma 2.1 (2) and the property (4.2) in Subsection 1.1. We may easily find ${S}_{d\vee \left(\underset{x\in {\mathcal{F}}^{2}\left(m\right)}{\bigvee }x\right)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\in C{l}_{s}\left(L\right)$. Similarly to the discussion for Case II1, we can state that ${S}_{d\vee \left(\underset{x\in {\mathcal{F}}^{2}\left(m\right)}{\bigvee }x\right)}$ is not a join of atoms in Cls(L) since S‴ < ${S}_{d\vee \left(\underset{x\in {\mathcal{F}}^{2}\left(m\right)}{\bigvee }x\right)}$ and the property (4.2) in Subsection 1.1 in which ${S}^{‴}=\left\{d\right\}\cup \left\{x|x\in {\mathcal{F}}^{2}\left(m\right)\right\}\cup \left\{y|y\in {\mathcal{F}}^{2}\left(d\phantom{\rule{thinmathspace}{0ex}}\vee \left(\underset{x\in {\mathcal{F}}^{2}\left(m\right)}{\bigvee }x\right)\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}y\notin \left\{d\right\}\cup \left\{x|x\in {\mathcal{F}}^{2}\left(m\right)\right\}\right\}\cup \left\{a\in A\left(L\right)|a\nleqq A\left(d\phantom{\rule{thinmathspace}{0ex}}\vee \phantom{\rule{thinmathspace}{0ex}}\left(\underset{x\in {F}^{2}\left(m\right)}{\bigvee }x\right)\right)\right\}\in C{l}_{s}\left(L\right)$. This is a contradiction to the geometric of Cls(L).

The above Case II1 and Case II2 show that if Sm ∉[S0, Sde] and Sn ∈ [S0, Sde] hold, then Cls(L) is not geometric.

Analogously, if Sn ∉ [S0, Sde] and Sm ∈ [S0, Sde] hold, then Cls(L) is not geometric.

Combining the discussions in Parts (I) and (II), we may state that neither m nor n exist. This is a contradiction to the conditions in ‘First status’.

Therefore, we may be assured that the supposition of SCls(L) \ {Sx | xL \ {0}} does not hold.

Furthermore, we demonstrate Cls(L) = {Sx | xL \ {0}}.

We prove item (2). We prove (2.1).

Let x, yA(L). This implies xy = 0 and 2 ≤ h(xy). By Lemma 2.1 (1), we indicate Sx = Sy = S0. 2 ≤ h(xy) implies S0 < Sxy.

In addition, Sxy does not exist since Sxy = {0} does not satisfy (C2). Therefore, we can decide the correctness of (2.1).

We prove (2.2).

xA(L) implies Sx = S0 by Lemma 2.1 (1). yL \ (A(L) ∪ 0) implies 2 ≤ h(y) and SyCls(L) \ {S0} by Lemma 2.1 (1). Thus, we may easily find Sx < Sy. So, we obtain SxSy = Sy and SxSy = Sx.

If x < y holds. Then this implies xy = y and xy = x. Thus, we can receive SxSy = Sy = Sxy and SxSy = Sx = Sxy.

If x||y holds. Then this implies xy = 0 and x, y < xy with h(y) < h(xy). Considering Lemma 2.1 (2), we find S0 = Sx < Sxy and Sy < Sxy. However, SxSy = S0Sy = Sy and SxSy = S0Sy = S0 = Sx. Considering S0Cls(L), we may obtain SxSy < Sxy and SxSy = S0Sxy.

We prove (2.3).

We may easily find x, yxyL\ (A(L) ∪ 0). This implies Sx, Sy, SxyCls(L) \ {S0}.

Since Cls(L) is a lattice, we may easily obtain Sx, SySxSyCls(L) \ {S0}. In addition, considering the above item (1), we obtain SxSy = Sb for some bL \ (A(L) ∪ 0). Thus, we may decide xb and yb since Sx, SySb and Lemma 2.1 (2). Furthermore, we obtain xyb. So, we get SxySb.

Additionally, we may find x, yxy since L is a lattice. Using Lemma 2.1 (2), we obtain Sx, SySxy. So, we may be assured SxSySxy since Cls(L) is a lattice. Therefore, we have demonstrated SxSy = Sxy.

Next, we will consider the other correspondent statements in item (2.3).

SxSyCls(L) holds since Cls(L) is a lattice. According to item (1), we confirm SxSy = Sc for some cL \ {0}. In addition, we easily find S0SxSySx, Sy.

When xy = 0. If cA(L), then according to Lemma 2.1 (2), we determine cx, y. So, cxy holds. This means 2 ≤ h(c) ≤ h(xy). Thus, we obtain xy0. This is a contradiction to the supposition of xy = 0. Moreover, we determine cA(L). Considering Lemma 2.1 (1), we indicate SxSy = S0.

When xy0. This means SxyCls(L). From xyx, y and Corollary 2.2, we may find SxySx, Sy. So, we decide SxySxSy.

If Sxy = SxSy holds. Then we obtain the needed result.

If Sxy < SxSy happens. We will distinguish two cases 2.3.1 and 2.3.2 to discuss.

Case 2.3.1

xyA(L).

The hypothesis implies S0 < Sxy since Lemma 2.1 (1). Furthermore, we obtain cL \ (A(L) ∪ 0) since S0 < Sxy < Sc and Lemma 2.1 (1). Taking Sc = SxSySx, Sy with Lemma 2.1 (2) together, we find cx and cy. So, we obtain cxy. Furthermore, we obtain ScSxy since Lemma 2.1 (2). This is a contradiction to Sxy < SxSy.

Case 2.3.2

xyA(L).

The hypothesis implies S0 = Sxy since Lemma 2.1 (1). Furthermore, we obtain cL \ (A(L) ∪ 0) since S0 < Sc and Lemma 2.1 (1). Combining x, y, cL \ (A(L) ∪ 0) with Corollary 2.2 and Sc = SxSySx, Sy, we may find cx, y. So, we believe cxy. Therefore, cA(L) ∪ 0 holds since xyA(L). This means Sc = S0 according to Lemma 2.1 (1). This is a contradiction to S0 < Sc.

The above Case 2.3.1 and Case 2.3.2 show that Sxy < SxSy will not happen.

Therefore, we confirm Sxy = SxSy since we have proved SxySxSy. □

Corollary 2.5

Let L be defined in Lemma 2.4. If Cls(L) is geometric, then Cls(L) is isomorphic to LCls(L).

Proof

Define a map π : LCls(L)Cls(L) as $π(x)=Sx,forx∈LCls(L)∖{0}S0,forx=0.$ We will demonstrate that π is a lattice isomorphism between two lattices LCls(L) and Cls(L).

By item (1) in Lemma 2.4, we find that π is a surjection. In light of both of items (1) and (2) in Lemma 2.1 with Corollary 2.2, we find that π is an injection.

Let x, yLCls(L) and xy.

If x = y = 0. Then we indicate π(x) = π(y) = π(xy) = π(xy) = S0. So, both of π(x) ∧ π(y) = π(xy) and π(x) ∨ π(y) = π(xy) hold.

If x = 0 and y0. Then we indicate xy = 0, xy = y, π(x) = S0 and π(y) = Sy. Furthermore, we obtain π(xy) = π(y) = Sy = S0Sy = π(x) ∨ π(y) and π(xy) = π(0) = S0 = S0Sy = π(x) ∧ π(y).

If x0 and y0. Then we indicate x, yL \ (A(L) ∪ 0). By item (2.3) in Lemma 2.4, we easily obtain π(xy) = π(x) ∨ π(y).

If xy = 0, then we obtain S0 = SxSy by item (2.3) in Lemma 2.4. Thus, π(xy) = SxSy = π(x)∧π(y) is accepted.

If xy0, then we obtain Sxy = SxSy by item (2.3) in Lemma 2.4. Thus π(xy) = Sxy = SxSy = π(x) ∧ π(y) is accepted.

Therefore, we have demonstrated π(xy) = π(x)∧π(y).

Summarizing the above, we affirm that π is a lattice isomorphism between LCls(L) and Cls(L) □

Lemma 2.6

Let L satisfy the conditions in ‘First status’. If L satisfies both of the following conditions, then Cls(L) is geometric.

1. Let m, nL \ (A(L) ∪ 01) and m||n. Then mn0 holds.

2. LCls(L) is geometric.

Proof

We will prove the needed results by the following three steps.

Step 1

We prove Cls(L) = {Sx | xL \ (A(L) ∪ 0)}.

Using Lemma 2.1 (1), we find {Sx | xL \ (A(L) ∪ 0)} ∪ {S0} = {Sx | xL \ {0}}. We may easily obtain {Sx | xL \ {0}} ⊆ Cls(L). Let SCls(L) \ {Sx | xL \ {0}}. Then, considering Lemma 2.1 (1) and the definition of Cls(L), we find that there are at least two elements d, eSL \ (A(L) ∪ 01) with d||e. This implies de = 0 by (C1). But this is a contradiction to the given condition (1). Thus, we decide Cls(L) = {Sx | xL \ {0}}.

Step 2

We prove Cls(L) ≅ LCls(L) as

Define a map π : LCls(L)Cls(L) as $π(x)=Sx,forx∈LCls(L)∖{0}S0,forx=0.$ We may easily find that π is well defined by Lemma 2.1 and Corollary 2.2.

We will prove that π is a lattice isomorphism. According to Step 1, we may confirm that π is a surjection.

We may easily decide that 0 < x in Cls(L) means S0 < Sx in Cls(L) by Lemma 2.1 (1) and Lemma 2.1 (2). Combining Step 1, we obtain that in Cls(L), S0 < S follows S = Sz for some zLCls(L) \ {0}.

Let x, yLCls(L) \ {0} with xy. According to Corollary 2.2 (2) and Lemma 2.1 (2), we obtain $x≤yinLCls(L)⇔π(x)≤π(y)inCls(L).$(*) and also obtain x < y in LCls(L)π(x) < π(y) in Cls(L)\{S0}. Therefore, we may state that π is an injection.

Let m, nL. We will prove that π keeps the two operators-meet ∧ and join ∨ in lattices by the following Parts (I1) and (I2).

Part (I1)

Let m = 0 and nLCls(L). This implies mn = n since LCls(L) is a lattice with 0 as the least element.

If n = 0. Then it is ease to see π(mn) = π(mn) = π(0) = S0 = π(m) ∨ π(n) = π(m) ∧ π(n).

If n0. By Step 1 and Lemma 2.1, the formula π(n) = Sn > S0 is accepted. Thus, we obtain π(mn) = π(n) = Sn = S0Sn = π(m) ∨ π(n) and π(mn) = π(0) = S0 = S0Sn = π(m) ∧ π(n).

Part (I2)

Let m, nLCls(L) \ {0}.

Since m, nmn in LCls(L), we receive Sm, SnSmn. This implies SmSnSmn.

Additionally, from Step 1 and SmSnCls(L), we affirm that there is cLCls(L) satisfying SmSn = Sc. According to SmSc and the definition of LCls(L), we confirm mc and nc. So, cLCls(L) \ {0} holds. Furthermore, we gain mnc. Hence, we decide SmnSc since the expression (٭). This means SmnSmSn. Therefore, we determine Smn = SmSn. Thus, we can express π(mn) = Smn = SmSn = π(m) ∨ π(n).

We will prove π(mn) = π(m) ∧ π(n).

When mn0. We obtain Smn, Sm, SnCls(L). From mnm, n, we may easily find SmnSm, Sn. Moreover, SmnSmSn holds. So, there is bLCls(L) \ {0} satisfying Sb = SmSn. Thus, we obtain bm, n since SbSm, Sn. Meanwhile, we receive mnb since SmnSb. Furthermore, we find b = mn. Hence, Smn = SmSn holds. That is to say, π(mn) = π(m) ∧ π(n) is correct.

When mn = 0.

If mn. Then according to m, nLCls(L) \ {0}, we determine mn = m0. This is a contradiction to mn = 0.

If m||n. Then according to m, nLCls(L) \ {0} and m||n with the given condition (1), we determine mn0. This is a contradiction to mn = 0. Therefore, up to now, we have demonstrated π(mn) = Smn = SmSn = π(m) ∧ π(n).

Summing up, we determine that π is a lattice isomorphism. That is to say, Cls(L) ≅ LCls(L) holds.

Step 3

We prove that Cls(L) is geometric.

Considering the given condition (2) with Step 2, we may state that Cls(L) is geometric. Combining Lemmas 2.4 and 2.6, we may demonstrate the true of Theorem 2.3 as follows. ☐

Proof of Theorem 2.3

(⇒) If the condition (1) is not true then there are two elements x, yL \ (A(L) ∪ 01) with x||y satisfying xy = 0. Hence, we obtain S = {x, y} ∪ {zA(L) | zA(x) and zA(y)} ∈ Cls(L).

However, considering Lemma 2.4(1), we confirm SCls(L). This is a contradiction to the above SCls(L). Therefore, the condition (1) is accepted. Using Corollary 2.5 and the geometric of Cls(L), we accept the condition (2).

(⇐) Applying Lemma 2.6, the needed result is followed. ☐

The following examples will express the correctness of Theorem 2.3.

Example 2.7

Let L1 be shown in Figure 1. It is easy to see that L1 is complete and atomistic. In addition, we also easily explore that L1 satisfies the conditions forFirst status’. LCls(L) is shown as Figure 2. According to the definition of geometric lattice in Subsection 1.1, LCls(L1) is geometric.

Figure 1

diagram of L1

Figure 2

diagram of LCls(L1)

We may easily see that the diagram of Cls(L) is in Figure 3, in which S0 = {a, b, c, d}, S1 = {1}, S(bc) = {(bc), a, d}, S(ab) = {(ab), c, d}, S(cd) = {(cd, a, b} and S(ab)(cd) = {(ab, (cd)}. From the definition of geometric lattice in Subsection 1.1, we decide that Cls(L1) is not geometric. In fact, (ab), (cd) ∈ L1 \ (A(L1) ∪ 0) satisfies (ab) ∧ (cd) = 0. In other words, L1 does not satisfy the given condition (1) in Theorem 2.3.

Figure 3

diagram of Cls(L1)

Example 2.8

Let L2 be shown as Figure 4. We may easily demonstrate that L2 satisfies the conditions inFirst statusand the condition (1) in Theorem 2.3.

Figure 4

diagram of L2

The diagram of LCls(L2) is in Figure 5. We may confirm the geometry of LCls(L2). Hence, using Theorem 2.3, Cls(L2) is geometric.

Figure 5

diagram of LCls(L2)

The diagram of Cls(L2) is in Figure 6, in which S(ab) = {(ab), c}, S(ac) = {(ac), b}, S(bc) = {(bc), a}, S0 = {a, b, c} and S1 = {1}. We may easily prove the geometry of Cls(L2). This is the same to the result from Theorem 2.3.

Figure 6

diagram of Cls(L2)

This subsection will discover the answer to the open problem for the complete atomistic lattices satisfying the conditions in ‘Second status’.

The main result in this subsection is the following theorem.

Theorem 2.9

Let L satisfy the conditions inSecond status’. Then, Cls(L) is geometric if and only if L satisfies the following conditions.

1. L is modular.

2. L is compact.

3. $x=\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }$ d holds for any xL \ (A(L) ∪ 0).

To get Theorem 2.9, we need the following series of lemmas.

Lemma 2.10

Let L satisfy the conditions inSecond status’. If Cls(L) is geometric, then the following statements hold.

1. L is modular.

2. L is compact.

3. $x=\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }$ d holds for any xL \ (A(L) ∪ 0).

Proof

To prove item (1). Suppose that L is not modular. Then in view of the property (2) in Subsection 1.1, L contains a sublattice L′ which is isomorphic to N5. Let L′ = {ab, a, b, c, ab} with ab < c < a < ab, ab < b < ab, a||b and c||b. Thus, we obtain h′(a) ≥ 2 and h′(ab) ≥ 3 in which h′ is the height function of L′. Certainly, we may affirm h(a) ≥ 2 and h(ab) ≥ 3 since h′(a) ≥ 2 and h′(ab) ≥ 3 in which h is the height function in L. We will distinguish the following Cases 1 and 2 to continue the discussions.

Case 1

ab = 0.

Suppose that in L, xA(L) holds for any x ∈ [0, ab] with x||a. Then we obtain [S0, Sab] ⊆ Cls(L) to be {S0, Sa} ∪ {Sy | a < y < ab} ∪ {Sz | 0 < z < a and zA(L)} ∪ {Sab}. From the definition of Cls(L) and Lemma 2.1, we may easily find S0 < St < Ss for any t, s ∈ [0, ab] \ (A(L) ∪ 0) and t < s. Therefore, we obtain |A([S0, Sab])| = 1 since the property (4.2) in Subsection 1.1 and h′(a) ≥ 2. We may assume A([S0, Sab]) = {SA}. Then, Sab is not a join of atoms in [S0, Sab] since SA < Sab. This implies that [S0, Sab] is not geometric. This is a contradiction to the property (2) in Subsection 1.1 since Cls(L) is geometric.

Suppose that there is d ∈ [0, ab] satisfying d ||a and dA(L). This implies h′(d) ≥ 2. Moreover, we decide S0 < Sa and S0 < Sd, and besides, we find Sa, Sd < SaSd = Sad = {a, d}∪{xA(L) | xA(a) and xA(d)} < SadSab.

Let {xt, t ∈ 𝕋} = {y | y ∈ [0, ab], yab, h(y) ≥ 2}. Then we may be assured |𝕋| ≥ 1 since h′(a) ≥ 2 and h′(d) ≥ 2 follows h(a) ≥ 2 and h(d) ≥ 2 respectively. Additionally, for any x, z ∈ {xt, t ∈ 𝕋} and xz, x||z holds since the definitions of {xt, t ∈ 𝕋}. In addition, we confirm xz = 0 since the conditions in ‘Second status’.

Let Sxt, t∈𝕋 = {xt, t ∈ 𝕋} ∪ {xA(L) | xA(a) and xA(xt) for any t ∈ 𝕋}. For any S ∈ [S0, Sab] \ {Sab}, we may easily find SSxt,t∈𝕋 < Sab. This implies SaSxt,t∈𝕋 < Sab for any SaA([S0, Sab]). Thus, we decide $\underset{{S}^{a}\in A\left(\left[{S}_{0,}{S}_{a\vee b}\right]\right)}{\bigvee }{S}^{a}\le {S}_{{x}_{t},t\in \mathbb{T}}.$ Hence, we may state that Sab is not a join of atoms in [S0, Sab].

Furthermore, we confirm that the interval [S0, Sab] ⊆ Cls(L) is not geometric. This is a contradiction to the property (2) in Subsection 1.1 since Cls(L) is geometric.

Case 2

ab0.

The supposition of ab ≠ 0 and ab < b taken together implies h(b) ≥ 2. Considering the conditions in ‘Second status’ with a||b and a, b < ab1, we find ab = 0. This is a contradiction to the supposition of ab0.

To summarize the whole discussion in Case 1 and Case 2, we can express that L′ does not exist. Furthermore, L is modular according to the property (2) in Subsection 1.1.

To prove item (2).

Otherwise, we can suppose that there is aL \ {0} and a ≤ ∨i∈𝕀xi for some xiL \ {0}, (i ∈ 𝕀) such that for any finite subset J ⊆ 𝕀, a ≰ ∨jJxj holds. Thus, we may find ∨jJxj < ∨i∈𝕀xi. Thereby, we confirm |𝕀| ≮ ∞.

In fact, if xsxt holds for some s, t ∈ 𝕀, then ∨i∈𝕀xi = ∨i∈𝕀\{s}xi is accepted. Hence, we can suppose that for any s, t ∈ 𝕀, s||t holds.

Let x1, x2 ∈ {xi, i ∈ 𝕀} satisfy x1x2. Since a ≰ ∨jJxj holds for any finite subset J ⊆ 𝕀, we may be assured that there is x3 ∈ {xi, i ∈ 𝕀} \ {x1, x2} satisfying x3||x1x2. Thus, we obtain that the sublattice {0, x1, x3, x1x2, x1x2x3} in L is isomorphic to N5. Therefore, L is not a modular. This is a contradiction to item (1).

Moreover, L must be compact.

To prove item (3).

We fulfill the proof with the following Cases 3.1 and 3.2.

Case 3.1

x ∈ ℱ2. The result is trivial.

Case 3.2

x ∉ ℱ2. Then we confirm $\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d\le \phantom{\rule{thinmathspace}{0ex}}x$ and ℱ2(x) ≠ ∅ since xL \ (A(L) ∪ 0).

If $\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d=\phantom{\rule{thinmathspace}{0ex}}x$ holds. This is the needed result.

If $\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d<\phantom{\rule{thinmathspace}{0ex}}x$ holds.

Since L is atomistic, we decide $x=\left(\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d\right)\vee \left(\underset{{a}_{i}\in A\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\text{a}\text{n}\text{d}\phantom{\rule{thinmathspace}{0ex}}{a}_{i}\nleqq d\phantom{\rule{thinmathspace}{0ex}}\text{f}\text{o}\text{r}\phantom{\rule{thinmathspace}{0ex}}\text{a}\text{n}\text{y}\phantom{\rule{thinmathspace}{0ex}}d\in {\mathcal{F}}^{2},\phantom{\rule{thinmathspace}{0ex}}i\in \phantom{\rule{thinmathspace}{0ex}}I}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{a}_{i}\right)$ satisfying {ai | aiA(x) and aid for any d ∈ ℱ2, iI} ≠ ∅. According to item (2), we obtain that there are some dj ∈ ℱ2(x)(j = 1,...,n < ∞) and at ∈ {ai | aiA(x) and aid for any d ∈ ℱ2; iI}(t = 1,...,m < ∞) satisfying $\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d={d}_{1}\vee ...\vee {d}_{n}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\underset{{a}_{i}\in A\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\text{a}\text{n}\text{d}\phantom{\rule{thinmathspace}{0ex}}{a}_{i}\nleqq d\phantom{\rule{thinmathspace}{0ex}}\text{f}\text{o}\text{r}\phantom{\rule{thinmathspace}{0ex}}\text{a}\text{n}\text{y}\phantom{\rule{thinmathspace}{0ex}}d\in {\mathcal{F}}^{2},\phantom{\rule{thinmathspace}{0ex}}i\in \phantom{\rule{thinmathspace}{0ex}}I}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{a}_{i}={a}_{1}\vee ...\vee {a}_{m}.$

In addition, owing to d1 ∈ ℱ2, we obtain that there is bA(x) satisfying bd1. Thus, we infer that the sublattice {x, d1, b, a1, 0} in L is isomorphic to N5. Hence, L is not modular. This is a contradiction to item (1).

Therefore, we demonstrate $\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d=x.$

We will reveal the construction of Cls(L).

Lemma 2.11

Let L satisfy the conditions inSecond status’. If L satisfies the following statements (1) and (2), then Cls(L) = {Sx | xL \ {0}} holds.

1. L is modular.

2. $x=\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d$holds for any xL \ (A(L) ∪ 0).

Proof

According to Lemma 2.1 (1), we can obtain {Sx | xL \ (A(L) ∪ 0)} ∪ {S0} = {Sx | xL \ {0}}. It is easy to see {Sx | xL \ {0}} ⊆ Cls(L).

Let SCls(L) \ {Sx | xL \ {0}}. Then, by SS0 and Lemma 2.1 (1), we affirm that there exists yS such that yA(L) ∪ 0.

Let {yj, jJ} = {y | yS \ (A(L) ∪ 0)}. Owing to (C1), we obtain yα||yβ for any α, βJ and αβ. According to S ∉ {Sx | xL \ (A(L) ∪ 0)} ∪ {S0} and Lemma 2.1 (1), we obtain |J| ≥ 2. In addition, we may easily receive Syj < S for any jJ in view of the definition of Cls(L).

Since |J| ≥ 2, we can select y1, y2S\(A(L)∪0) with y1y2. Thus, we confirm y1 || y2 and h(y1), h(y2) ≥ 2 since yjA(L) ∪ 0 follows h(yj) ≥ 2 (j = 1, 2) in which h is the height function in L.

Combining SCls(L) and (C1), we obtain y1y2 = 0. Considering the atomic of L with h(y1), h(y2) ≥ 2 and y1 || y2 and y1y2 = 0, we obtain A(y1) \ A(y2) ≠ ∅ and A(y2) \ A(y1) ≠ ∅. Let a1A(y1) \ A(y2) and a2A(y2) \ A(y1). Then we decide that the sublattice {y1y2, y1, a1, a2, 0} of L is isomorphic to N5. This is a contradiction to the given condition (1) according to property (2) in Subsection 1.1.

Therefore, we determine Cls(L) = {Sx | xL \ {0}}. ☐

Lemma 2.12

Let L satisfy the conditions inSecond statusand the statements

1. L is modular.

2. L is compact.

3. $x=\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}d$ holds for any xL \ (A(L) ∪ 0).

Then $\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}={S}_{{\vee }_{y\in \mathfrak{Y}}y}$ holds for any {y | y ∈ 𝔜} ⊆ L\ (A(L) ∪ 0).

Proof

Since L is complete, we obtain ∨y∈𝔜yL. Let b = ∨y∈𝔜y. Then, we confirm Sy∈𝔜yCls(L).

From SyCls(L) (∀y ∈ 𝔜) and the completion of Cls(L), we receive $\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}\in C{l}_{s}\left(L\right).$ Combining Lemma 2.11, we decide $\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}={S}_{z}$ for some zL \ (A(L) ∪ 0) or $\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}={S}_{0}.$

We distinguish two cases to continue the proof.

Case 1

$\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}={S}_{z}$ for some zL \ (A(L) ∪ 0).

Considering ySy∈𝔜y for any y ∈ 𝔜 with Lemma 2.1 (2) and Corollary 2.2 (2), we decide SySy∈𝔜y. Hence, we affirm $\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}\le {S}_{{\vee }_{y\in \mathfrak{Y}}y}.$ That is to say, SzSb holds. Using Lemma 2.1 (2) and Corollary 2.2, we obtain zb.

Since L is compact, we may be assured b = ∨y∈𝔜y = y1 ∨ ... ∨ yn for some yj ∈ 𝔜 with n < ∞, (j = 1,..., n).

In view of the selection of z, we obtain SySz for any y ∈ 𝔜. Combining Lemma 2.1 (2) and Corollary 2.2 (2), we decide yz for any y ∈ 𝔜. Hence, we get yjz, (j = 1,..., n). So, we also get y1 ∨ ... ∨ ynz. That is, bz holds. Therefore, we produce SbSz according to Lemma 2.1 (2).

Thereby, we determine Sb = Sz.

Case 2

$\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}={S}_{0}.$

Since Sy$\underset{y\in \mathfrak{Y}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{y}={S}_{0}$ and Lemma 2.1 (1), we obtain yA(L). So, we believe {y | y ∈ 𝔜} ⊆ A(L). This is a contradiction to {y | y ∈ 𝔜} ⊆ L \ (A(L) ∪ 0). ☐

By extension of Lemmas 2.11 and 2.12, we will obtain the following result.

Lemma 2.13

Let L satisfy the conditions inSecond statusand the following conditions.

1. L is modular.

2. L is compact.

3. $x=\underset{d\in {\mathcal{F}}^{2}\left(x\right)}{\bigvee }$ d holds for any xL \ (A(L) ∪ 0).

Then, Cls(L) is geometric.

Proof

From the given three conditions, we confirm that L is geometric in view of the definition of geometric lattice in Subsection 1.1.

We prove that the complete lattice Cls(L) is geometric with the following steps.

Step 1

We will prove that Cls(L) is atomistic.

In virtue of Lemma 2.11, we find Cls(L) = {Sx | xL \ {0}}. With the given condition (3), we obtain Sy = Sd∈ℱ2(y)d for any yL \ (A(L) ∪ 0). Combining Lemma 2.12, we receive ${S}_{{\vee }_{d\in {\mathcal{F}}^{2}\left(y\right)}d}=\underset{d\in {\mathcal{F}}^{2}\left(y\right)}{\bigvee }{S}_{d}$

By the property (4.2) in Subsection 1.1, we decide {Sd | d ∈ ℱ2(y)} ⊆ A(Cls(L)) for any yL\(A(L)∪0).

Hence, we may state that Sy is the join of some atoms in Cls(L) for any yL \ (A(L) ∪ 0). Therefore, we indicate that Cls(L) is atomistic.

Step 2

We will prove that all atoms in Cls(L) are compact.

Let SA(Cls(L)). Then by the property (4.2) in Subsection 1.1, we obtain S = Sd for some d ∈ ℱ2.

Suppose ${S}_{d}\le \underset{j\in \mathbb{J}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{{x}_{j}}$ for some xjL(j ∈ 𝕁). According to Lemma 2.12, we determine $\underset{j\in \mathbb{J}}{\bigvee }\phantom{\rule{thinmathspace}{0ex}}{S}_{{x}_{j}}={S}_{{\vee }_{j\in \mathbb{J}}{x}_{j}}.$ These imply SdSj∈𝕁xj. By Lemma 2.1 (2), we obtain d ≤ ∨j∈𝕁xj. Since L is compact, we obtain dx1 ∨ ... ∨ xn for some {x1,...., xn} ⊆ {xj, j ∈ 𝕁} and n < ∞. Thereby, combining Lemma 2.1 (2) and Lemma 2.12, we confirm SdSx1∨...∨xn = Sx1 ∨ ... ∨ Sxn. Thus, we can express that Sd is compact.

Therefore, every atom in Cls(L) is compact.

Step 3

We will prove that Cls(L) is modular.

Suppose that Cls(L) is not modular. Then in Cls(L), there is a sublattice {A, B, C, AC, AC} which is isomorphic to N5, where A||C, B||C, AC < A < B < AC and AC < C < AC.

According to Lemma 2.1, Lemma 2.11 and AC < A, B with ACCls(L), we own Sa = A, Sb = B and Sc = C for some a, b, cL \ (A(L) ∪ 0). In light of Lemma 2.12, we may be assured AC = SaSc = Sac.

Additionally, combining A||C, B||C, A < B and Lemma 2.1 with Corollary 2.2, we affirm a||c and b||c; a < b; b, c < ac.

From a||c and a, cL \(A(L)∪0), we confirm ac = 0 since L satisfies the conditions in ‘Second status’.

Taking Lemma 2.1 (2) and Corollary 2.2, we may be assured that the sublattice {0, ac, a, b, c} in L satisfies a||c and b||c, and 0 < a < b < ac and 0 < c < ac. This implies that the sublattice {0, ac, a, b, c} in L is isomorphic to N5. Thus, we find that L is not modular. This is a contradiction to the given condition (1). ☐

The proof of Theorem 2.9 follows.

Proof of Theorem 2.9

Combining all the results in Lemmas 2.10 and 2.13, we can express that Theorem 2.9 holds. ☐

The correctness of Theorem 2.9 has been proved. Hence, the following example will show that if a complete atomistic lattice L3 satisfies the conditions in ‘Second status’, but does not satisfy the conditions in Theorem 2.9, then Cls(L3) will not be geometric. This will show that the importance of Theorem 2.9 to decide the geometric of Cls(L) for a complete atomistic lattice L.

Example 2.14

Let L3 be defined as Figure 7. We may easily reveal that L3 satisfies all of conditions inSecond statusand the conditions (2) and (3) in Theorem 2.9.

Figure 7

diagram of L3

Since {1, (ab), b, 0, c} is N5 up to isomorphism. This means that L3 is not modular in view of property (2) in Subsection 1.1. By Theorem 2.9, we demonstrate that Cls(L3) is not geometric.

Actually, the diagram of Cls(L3) is shown as Figure 8, in which S(ab) = {(ab), c, d}, S(cd) = {(cd), a, b}, S(ab)(cd) = {(ab), (cd)}, S0 = {a, b, c, d} and S1 = {1}. We may easily find the non-geometry of Cls(L3) since S1 is not the join of atoms in Cls(L3). This is the same consequence of that by Theorem 2.9.

Figure 8

diagram of Cls(L3)

3 Conclusion

Utilizing the results in Subsection 1.2 with Theorems 2.3 and 2.9, we completely respond to Radeleczki’s open problem which hope to characterize complete atomistic lattice whose classification lattice is geometric in [2]. For the answer of the open problem, we will discover the applied fields, such as concept lattice theory, in the future.

Acknowledgement

This research is supported by National Nature Science Foundation of China (61572011).

References

• [1]

Grätzer G., Lattice Theory: Foundation, Birkhäuser-Verlag, Basel, 2011 Google Scholar

• [2]

Radeleczki S., Classification systems and their lattice, Dis. Math. Gene. Alge. and Appl. 2002, 22, 167-181 Google Scholar

• [3]

Wille R., Subdirect decomposition of concept lattices, Alge. Univ. 1983, 17, 275-287

• [4]

Mao H., Complete atomistic lattices are classification lattices, Alge. Univ., 2012, 68, 293-294

• [5]

Mao H., Characterizations of atomistic complete finite lattices relative to geometric, Miskolc Math. Note., 2016, 17(1), 421-440

• [6]

Ganter B., Körei A., Radeleczki S., Classification systems and context extension, http://www.math.u-szeged.hu/~czedli/otka-t049433, Cited 15 January 2007

• [7]

Ganter B., Körei A., Radeleczki S., Extent partitions and context extensions, Math. Slovaca, 2013, 63(4), 693-706

• [8]

Körei A., Radeleczki S., Box elements in a concept lattice, Lect. Notes in Comp. Sci., 2006, 3874, 41-56 Google Scholar

• [9]

Czedli G., Hartmann M., Schmidt E.T., CD-independent subsets in distributive lattices, Publ. Math. Debr., 2009, 74(1-2), 127-134 Google Scholar

• [10]

Horvath E.K., Radeleczki S., Notes on CD-independent subsets, Acta Sci. Math.(Szeged), 2012, 78(1), 3-24 Google Scholar

• [11]

Foldes S., Radeleczki S., On interval decomposition lattices, Dis. Math., Gene. Alge. and Appl., 2004, 24, 95-114 Google Scholar

• [12]

Mohring R.H., Algorithmic aspects of the substitution decomposition in optimization over relations, sets systems and Boolean functions, Anna. of Oper. Rese., 1985, 4(6), 195-225

• [13]

Birkhoff G., Lattice Theory, 3rd. edition, Amer. Math. Soc, Provindence, 1967 Google Scholar

Accepted: 2017-05-25

Published Online: 2017-07-13

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 959–973, ISSN (Online) 2391-5455,

Export Citation