## 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.

Show Summary Details# Classification lattices are geometric for complete atomistic lattices

#### Open Access

## Abstract

## 1 Introduction and Preliminaries

## 1.1 Definitions and properties

## Some definitions

## Some notations

## Some properties

## 1.2 Answer to the cases of |𝓕^{2}| ≤ 1

#### Case I

#### Case II

## 1.3 Organization

## 2 Main results

#### Lemma 2.1

#### Proof

#### Corollary 2.2

#### Proof

## 2.1 Answer for ‘First status’

#### Theorem 2.3

#### Lemma 2.4

#### Proof

#### Step 1

#### Case 1

#### Case 2

#### Case 3

#### Step 2

#### Step 3

#### Part (I)

#### Case I1

#### Case I2

#### Case I3

#### Case I4

#### Part (II)

#### Case II1

#### Case II2

#### Case 2.3.1

#### Case 2.3.2

#### Corollary 2.5

#### Proof

#### Lemma 2.6

#### Proof

#### Step 1

#### Step 2

#### Part (I1)

#### Part (I2)

#### Step 3

#### Proof of Theorem 2.3

#### Example 2.7

#### Example 2.8

## 2.2 Answer for ‘Second status’

#### Theorem 2.9

#### Lemma 2.10

#### Proof

#### Case 1

#### Case 2

#### Case 3.1

#### Case 3.2

#### Lemma 2.11

#### Proof

#### Lemma 2.12

#### Proof

#### Case 1

#### Case 2

#### Lemma 2.13

#### Proof

#### Step 1

#### Step 2

#### Step 3

#### Proof of Theorem 2.9

#### Example 2.14

## 3 Conclusion

## Acknowledgement

## References

## About the article

More options …# Open Mathematics

### formerly Central European Journal of Mathematics

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

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

Keywords: Geometric; Classification system; Classification lattice; Complete atomistic lattice

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) *c* ∧ *d* = 0 for any two distinct elements *c*, *d* ∈ *C*, and

(C2) *x* = ∨{*x* ∧ *c* | *c* ∈ *C*} holds for each *x* ∈ *L*.

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 *Cl _{s}*(

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.

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.

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

*poset*.[1, p.9] A poset (

*K*, ≤) is a*lattice*if sup{*a*,*b*} and inf{*a*,*b*} exist for all*a*,*b*∈*K*.[1, p.28] The lattices (

*K*, ≤) and (*K′*, ≤) are*isomorphic*(in symbols, (*K*, ≤) ≅ (*K′*, ≤)) if the map*ϕ*:*K*→*K′*is a bijection such that*a*≤*b*in (*K*, ≤) if and only if*ϕ*(*a*) ≤*ϕ*(*b*) in (*K′*, ≤).[13, p.11]

*N*_{5}= {*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*.[1, p.35] An interval in a lattice

*K*is a subset of*K*of the form [*a*,*b*] = {*x*∈*K*|*a*≤*x*≤*b*} for some*a*,*b*, ∈*K*with*a*≤*b*.[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*a*∈*K*, denoted by*h*(*a*), is the length of the order {*x*∈*K*|*x*≤*a*}.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].[1, p.50] A lattice

*K*is called*complete*if ∨*H*and ∧*H*exist for any subset*H*⊆*K*, in which ∨*H*and ∧*H*are seen [1, p.5].[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.[1, p.329] A lattice

*K*is called*semimodular*if it satisfies for all*a*,*b*∈*K*,*a*≺*b*⇒*a*∨*b*≺*b*∨*c*or*a*∨*c*=*b*∨*c*.[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*X*⊆*K*, implies that*a*≤ ∨*X*_{1}for some finite*X*_{1}⊆*X*.[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.

Let

*K*be a lattice.*L*denotes a complete atomistic lattice;**0**and**1**denote its least and greatest element, respectively.*S*{_{x}*x*} ∪ {*a*∈*A*(*L*) |*a*∧*x*=**0**} for any*x*∈*L*\ {**0**};*S*_{0}= ∧{*S*|*S*∈*Cl*(_{s}*L*)}.

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

[13, p.13,Theorem 12] Up to isomorphism, any non-modular lattice contains the lattice

*N*_{5}as a sublattice.[1, p.109,Theorem 102] A lattice

*K*is modular if and only if it does not contain a*N*_{5}.[2]

*S*_{0}∈*Cl*_{s}(*L*) and*S*_{1}= {**1**} ∈*Cl*_{s}(*L*) hold, and*S*_{x}∈*Cl*_{s}(*L*) holds for any*x*∈*L*\ {**0**}.[5]

*Cl*_{s}(*L*) possesses the following properties.(4.1) Let

*x*,*y*∈*L*\ {**0**}. Then*x*≤*y*⇔*S*≤_{x}*S*. Further,_{y}*x*<*y*⇔*S*<_{x}*S*._{y}(4.2)

*A*(*Cl*(_{s}*L*)) = {*S*|_{d}*d*∈ ℱ^{2}}.

This subsection will reveal the answers to the open problem if a complete atomistic lattice *L* satisfies |ℱ^{2}| ≤ 1. In fact, |ℱ^{2}| ≤ 1 means |ℱ^{2}| = 0 or |ℱ^{2}| = 1. We will search out the answers to the open problem correspondent to the above two cases respectively in this subsection.

|ℱ^{2}| = 0.

Answer: *Cl _{s}*(

The reason is the following.

|ℱ^{2}| = 0 means *h*(*L*) = 1 or *h*(*L*) = 0.

If *h*(*L*) = 1. Then we obtain *L* = {**0**, **1**} such that **0** ≺ **1** and *A*(*L*) = {**1**}. This implies *Cl*_{s}(*L*) = {*S*_{0}} = {*S*_{1}}.

So, *Cl*_{s}(*L*) is a geometric lattice.

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

Conversely, if *Cl*_{s}(*L*) is geometric. According to *Cl*_{s}(*L*) ≠ ∅ and the above analysis, we may be assured *h*(*L*) = 1.

|ℱ^{2}| = 1.

Answer: *Cl*_{s}(*L*) is geometric if and only if *h*(*L*) = 2.

The reason is the following.

From |ℱ^{2}| = 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 *Cl*_{s}(*L*) = {*S*_{0}, *S*_{1}} with *S*_{0} ≺ *S*_{1}. So, *Cl*_{s}(*L*) is a geometric lattice.

If *h*(*L*) > 2. In view of |ℱ^{2}| = 1, we may assume ℱ^{2} = {*d*}. Considering the atomic of *L* with |ℱ^{2}| = 1 and *h*(**1**) = *h*(*L*), we may affirm that for any *x* ∈ *L* \ ({**0**, **1**, *d*} ∪ *A*(*d*)), we receive *x* ∈ *A*(*L*) or *d* < *x* < 1. Thereby, we may decide *Cl*_{s}(*L*) = {*S*_{0}, *S*_{1} *S _{d}*} ∪ {

Conversely, if *Cl*_{s}(*L*) is geometric. According to the above analysis for *h*(*L*) > 2, we may be assured *h*(*L*) = 2.

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

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

This section answers the open problem for a complete atomistic lattice *L* which satisfies |ℱ^{2}| > 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.

*(1) S _{a}* =

*(2) Let x*, *y* ∈ *L* \ (*A*(*L*) ∪ **0**) *and x* ≠ *y*. *Then x* < *y* ⇔ *S*_{x} < *S*_{y}.

*Furthermore*, *for any x*, *y* ∈ *L*\ (*A*(*L*) ∪ 0), *we obtain x* ≤ *y* ⇔ *S*_{x} ≤ *S*_{y}.

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*) ∈ *Cl*_{s}(*L*) holds.

Let *S* ∈ *Cl*_{s}(*L*) \ *A*(*L*). Then, we affirm that *S* contains an element *x* satisfying *x* ∉ *A*(*L*) according to *S* ⊆ *L* \ {**0**} and *S* ≠ *A*(*L*). Thus, by the definition of classification system and the atomic of *L*, we obtain *A*(*L*) < *S* in *Cl*_{s}(*L*).

Therefore, *S*_{0} = *A*(*L*) holds.

Since *S*_{a} = {*a*} ∪ {*x* ∈ *A*(*L*) | *x* ≠ *a*} for any *a* ∈ *A*(*L*), we may easily obtain *S*_{a} = *A*(*L*). Hence, we obtain *S _{a}* =

According to *S*_{x} = {*x*} ∪ {*a* ∈ *A*(*L*) | *a* ≰ *A*(*x*)} for any *x* ∈ *L* \ {**0**} with the above *A*(*L*) = *S*_{0}, we may easily attain that if *S*_{x} = *S*_{0} holds, then *x* ∈ *A*(*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*, *y* ∈ *A*(*L*) and *x* ≠ *y*. Then we may easily see that *S*_{x} = *S*_{y} = *S*_{0} = *A*(*L*) since Lemma 2.1 (1). Even though, in the following, we still write *S*_{0} not *A*(*L*) when we consider *Cl*_{s}(*L*). The reason is that we hope to keep the coherence since *S*_{z} ∈ *Cl*_{s}(*L*) holds for any *z* ∈ *L*\ (*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.

*(1) Let x*, *y* ∈ *L*\ (*A*(*L*) ∪ **0**).

*Then S*_{x} = *S*_{y} ⇔ *x* = *y*. *In addition*, *x*||*y* ⇔ *S _{x}*||

*(2) Let x*, *y* ∈ *L*\ {**0**}. *Then*

*x* ≤ *y* ⇒ *S _{x}* ≤

At first, we prove *S*_{x} = *S*_{y} ⇔ *x* = *y* in item (1).

If *x* = *y*, then it is easy to see *S*_{x} = *S*_{y} by the definitions of *S*_{x} and *S*_{y}.

Conversely, let *S*_{x} = *S*_{y}. According to *S*_{x} = {*x*}∪{*a* ∈ *A*(*L*) | *a*∧*x* = **0**}, *S*_{y} = {*y*}∪{*b* ∈ *A*(*L*) | *b*∧*y* = **0**} and *x* ∉ *A*(*L*) ∪ **0**, we confirm *x* ∉ {*b* ∈ *A*(*L*) | *b* ∧ *y* = **0**}. Considering *S*_{x} = *S*_{y}, we decide *x* ∈ {*y*}. Hence, we obtain *x* = *y*.

Secondly, we prove *x*||*y* ⇔ *S*_{x}||*S*_{y} in item (1).

Considering Lemma 2.1 (1), we may be assured *S*_{x}, *S*_{y} ∈ *Cl*_{s}(*L*)\{*S*_{0}}.

*x*||*y* implies the true of *x* ≠ *y*, *x* ≮ *y* and *y* ≮ *x*.

Using the above and *x* ≠ *y*, we find *S*_{x} ≠ *S*_{y}. Using *x* ≮ *y* (*y* ≮ *x*) and Lemma 2.1 (2), we find *S*_{x} ≮ *S*_{y} (*S*_{y} ≮ *S*_{x}). Therefore, we know *S*_{x}||*S*_{y}.

Conversely, if *S _{x}*||

Next, we will prove item (2).

If *x* = *y*. Then we easily know *S*_{x} = *S*_{y}.

If *x*, *y* ∈ *L* \ (*A*(*L*) ∪ **0**). Then by Lemma 2.1 (2) and item (1) above, we obtain *x* ≤ *y* ⇔ *S*_{x} ≤ *S*_{y}, and *x* < *y* ⇔ *S*_{x} < *S*_{y}.

If *x*, *y* ∈ *A*(*L*) and *x* ≠ *y* hold, or *y* ∈ *A*(*L*) and *x* ∉ *A*(*L*) ∪ **0** hold. Then it is easy to see that both of *x* < *y* and *x* ≤ *y* will not happen. Thus, we do not consider the needed results.

If *x* ∈ *A*(*L*) and *y* ∉ *A*(*L*) ∪ **0**. Then *x* ≤ *y* will mean *x* ∈ *A*(*y*). Let *b* ∈ *A*(*L*) \ *A*(*y*). We may easily see that *b* ∧ *y* = **0** since *L* is atomistic. This implies *b* ∈ *S*_{y} according to *S*_{y} = {*y*} ∪ {*a* ∈ *A*(*L*)| *a* ∧ *y* = **0**}. In addition, *S _{x}* =

Actually, the three of *x* ∈ *A*(*L*), *y* ∉ *A*(*L*) ∪ **0** and *x* ≤ *y* imply *x* < *y*. So, we easily obtain *S*_{0} < *S*_{y}. That is to say, *S*_{x} < *S*_{y} holds since *S*_{x} = *S*_{0} 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 *x* ∧ *y* ≠ **0** for some *x*, *y* ∈ *L* \ (*A*(*L*) ∪ **0** ∪ **1**) and *x*||*y*.

**Second status**: there are *x* ∧ *y* = **0** for any *x*, *y* ∈ *L* \ (*A*(*L*) ∪ **0** ∪ **1**) and *x*||*y*.

In fact, Since |ℱ^{2}| > 1, we obtain |*L* \ (*A*(*L*) ∪ **0** ∪ **1**)| > 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, *L*_{Cls(L)} denotes *L* \ *A*(*L*). We may easily see that *L*_{Cls(L)} with the same order in *L* is a sublattice of *L*. We still denote this sublattice as *L*_{Cls(L)}.

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

*Let L satisfy the conditions in* **‘First status’**. *Then Cl _{s}*(

*For any m*,*n*∈*L*\ (*A*(*L*) ∪**0**∪**1**)*with m*||*n*,*m*∧*n*≠**0***holds*.*L*_{Cls(L)}*is geometric*.

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

*Let L satisfy the conditions in* **‘First status’**. *If Cl _{s}*(

*(1) Cl*_{s}(*L*) = {*S _{x}* |

*(2) (2.1) Let x*, *y* ∈ *A*(*L*) *and x* ≠ *y*. *Then*

*S _{x}* ∨

*S _{x}* ∧

*(2.2) Let x* ∈ *A*(*L*) *and y* ∈ *L* \ (*A*(*L*) ∪ **0**). *Then*

*S _{x}* ∨

*S _{x}* ∨

*(2.3) Let x*, *y* ∈ *L* \ (*A*(*L*) ∪ **0**) *and x* ≠ *y*. *Then*

*S*_{x∨y} = *S _{x}* ∨

*S*_{0} = *S*_{x}∧*S _{y}*

*S*_{x∧y} = *S*_{x} ∧ *S*_{y} *if x* ∧ *y* ≠ **0**.

We prove item (1).

Otherwise, there is *S* ∈ *Cl _{s}*(

From *y*, *z* ∈ *S* \ (*A*(*L*) ∪ **0**), we obtain *h*(*y*, *h*(*z*) ≥ 2 where *h* is the height function of *L*. So, we believe ℱ^{2}(*y*) ≠ ∅ and ℱ^{2}(*z*) ≠ ∅. According to *y* ∧ *z* = **0**, we decide ℱ^{2}(*y*) ∩ ℱ^{2}(*z*) = ∅.

We will continue the discussion in the following two steps. Let *d* ∈ ℱ^{2}(*y*) and *e* ∈ ℱ^{2}(*z*).

Let *S _{de}* = {

We may easily gain *d* ∧ *e* = **0** and *d*||*e* since **0** ≤ *d* ∧ *e* ≤ *y* ∧ *z* = **0** and *y*||*z*. We also obtain *d* ∧ *p* = *e* ∧ *p* = **0** since *p* ∈ *S _{de}* satisfies

We distinguish three cases to demonstrate *S _{de}* to satisfy (C2). Let

*g* ≤ *d*.

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

*g* ≤ *e*.

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

*g* ≰ *d* and *g* ≰ *e*.

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

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

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

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

Moreover, *g* = (*g* ∧ *d*) ∨ (*g* ∧ *e*) ∨ $(\underset{q\in A(L)\mathrm{\setminus}(A(d)\cup A(e))}{\bigvee}(g\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}q))$ is accepted. Summing up Cases 1, 2 and 3, we can decide that *S _{de}* satisfies (C2). Therefore,

From the definition of *S _{de}* in Step 1, we may easily determine

In addition, we also obtain *S*_{0} < *S _{d}* <

According to the conditions in **‘First status’**, we may confirm that there exist *m*, *n* ∈ *L* \ (*A*(*L*) ∪ **0** ∪ **1**) such that *m*||*n* and *m* ∧ *n* ≠ 0.

We will analyze the relationships among *S _{m}*,

If both of *S _{m}* and

Then we may easily find *S _{m}* ≤

*m* ≤ *d* and *n* ≤ *e*.

*m* ≤ *e* and *n* ≤ *d*.

*m* ≤ *d* and *n* ≤ *d*.

*m* ≤ *e* and *n* ≤ *e*.

Suppose that Case I1 happens. Then, we obtain **0** ≤ *m* ∧ *n* ≤ *d* ∧ *e* = **0**. So, we receive *m* ∧ *n* = **0**. This is a contradiction to *m* ∧ *n* ≠ 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 *n* ≤ *d* follows *n* ≤ *m*. This is a contradiction to *m*||*n*. Thus, we determine *m* < *d*. Analogously, *n* < *d* holds.

According to *d* ∈ ℱ^{2}(*y*) and *m* < *d*, we obtain *h*(*m*) < 2. This implies *m* ∈ *A*(*L*)∪**0**. But this is a contradiction to *m* ∈ *L* \ (*A*(*L*) ∪ **0** ∪ **1**). Thus, we can say *m* ≮ *d*.

Analogously, we can also find *n* ≮ *d*. 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 *S _{m}* ∈ [

If *S _{m}* ∉ [

*S _{m}* ∉ [

*S _{de}* <

We find *S _{d}*,

In view of the property (4.2) in Subsection 1.1 and *d*, *e* ∈ ℱ^{2}, we find *S _{d}*,

Let *S′* = {*d*, *e*} ∪ {*g* | *g* ∈ 𝒞} ∪ {*a* ∈ *A*(*L*) | *a* ∉ *A*(*d*) and *a* ∉ *A*(*e*), and *a* ∉ *A*(*g*) for any *g* ∈ 𝒞}. We may determine *S′* ∈ *Cl _{s}*(

Additionally, let *M* ∈ *A*(*Cl _{s}*(

Considering the property (4.2) in Subsection 1.1, we may confirm that *S _{m}* is not a join of atoms in

*S _{de}*||

If *m* ≤ *d* (or *m* ≤ *e*; or *d*, *e* ≤ *m*). Then, we decide *S _{m}* ≤

If *d*||*m* and *e*||*m*. We may easily decide *S _{d}*,

If *d*||*m* and *e* ∦*m*. Actually, we find *m* ≰ *e*. This implies *e* < *m* since *e* ∦*m*. Therefore, we confirm *e* ∈ ℱ^{2}(*m*) and *d* ∉ ℱ^{2}(*m*). Moreover, ℱ^{2}(*m*) ≠ ∅ holds. We attain *S _{e}* <

The above Case II1 and Case II2 show that if *S _{m}* ∉[

Analogously, if *S _{n}* ∉ [

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 *S* ∈ *Cl _{s}*(

Furthermore, we demonstrate *Cl _{s}*(

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

Let *x*, *y* ∈ *A*(*L*). This implies *x* ∧ *y* = **0** and 2 ≤ *h*(*x* ∨ *y*). By Lemma 2.1 (1), we indicate *S _{x}* =

In addition, *S*_{x∧y} does not exist since *S*_{x∧y} = {**0**} does not satisfy (C2). Therefore, we can decide the correctness of (2.1).

We prove (2.2).

*x* ∈ *A*(*L*) implies *S _{x}* =

If *x* < *y* holds. Then this implies *x* ∨ *y* = *y* and *x* ∧ *y* = *x*. Thus, we can receive *S _{x}* ∨

If *x*||*y* holds. Then this implies *x* ∧ *y* = **0** and *x*, *y* < *x*∨*y* with *h*(*y*) < *h*(*x*∨*y*). Considering Lemma 2.1 (2), we find *S*_{0} = *S _{x}* <

We prove (2.3).

We may easily find *x*, *y* ≤ *x* ∨ *y* ∈ *L*\ (*A*(*L*) ∪ **0**). This implies *S _{x}*,

Since *Cl _{s}*(

Additionally, we may find *x*, *y* ≤ *x*∨*y* since *L* is a lattice. Using Lemma 2.1 (2), we obtain *S _{x}*,

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

*S _{x}* ∧

When *x* ∧ *y* = **0**. If *c* ∉ *A*(*L*), then according to Lemma 2.1 (2), we determine *c* ≤ *x*, *y*. So, *c* ≤ *x* ∧ *y* holds. This means 2 ≤ *h*(*c*) ≤ *h*(*x*∧*y*). Thus, we obtain *x*∧*y* ≠ **0**. This is a contradiction to the supposition of *x*∧*y* = **0**. Moreover, we determine *c* ∈ *A*(*L*). Considering Lemma 2.1 (1), we indicate *S _{x}* ∧

When *x* ∧ *y* ≠ **0**. This means *S*_{x∧y} ∈ *Cl _{s}*(

If *S*_{x∧y} = *S _{x}* ∧

If *S*_{x∧y} < *S _{x}* ∧

*x* ∧ *y* ∉ *A*(*L*).

The hypothesis implies *S*_{0} < *S*_{x∧y} since Lemma 2.1 (1). Furthermore, we obtain *c* ∈ *L* \ (*A*(*L*) ∪ **0**) since *S*_{0} < *S*_{x∧y} < *S _{c}* and Lemma 2.1 (1). Taking

*x*∧*y* ∈ *A*(*L*).

The hypothesis implies *S*_{0} = *S*_{x∧y} since Lemma 2.1 (1). Furthermore, we obtain *c* ∈ *L* \ (*A*(*L*) ∪ **0**) since *S*_{0} < *S _{c}* and Lemma 2.1 (1). Combining

The above Case 2.3.1 and Case 2.3.2 show that *S*_{x∧y} < *S _{x}* ∧

Therefore, we confirm *S*_{x∧y} = *S _{x}* ∧

*Let L be defined in Lemma 2*.*4*. *If* *Cl _{s}*(

Define a map *π* : *L*_{Cls(L)} → *Cl _{s}*(

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*, *y* ∈ *L*_{Cls(L)} and *x* ≠ *y*.

If *x* = *y* = **0**. Then we indicate *π*(*x*) = *π*(*y*) = *π*(*x* ∧ *y*) = *π*(*x* ∨ *y*) = *S*_{0}. So, both of *π*(*x*) ∧ *π*(*y*) = *π*(*x* ∧ *y*) and *π*(*x*) ∨ *π*(*y*) = *π*(*x* ∨ *y*) hold.

If *x* = **0** and *y* ≠ **0**. Then we indicate *x* ∧ *y* = **0**, *x* ∨ *y* = *y*, *π*(*x*) = *S*_{0} and *π*(*y*) = *S _{y}*. Furthermore, we obtain

If *x* ≠ **0** and *y* ≠ **0**. Then we indicate *x*, *y* ∈ *L* \ (*A*(*L*) ∪ **0**). By item (2.3) in Lemma 2.4, we easily obtain *π*(*x* ∧ *y*) = *π*(*x*) ∨ *π*(*y*).

If *x* ∧ *y* = **0**, then we obtain *S*_{0} = *S _{x}*∧

If *x* ∧ *y* ≠ **0**, then we obtain *S*_{x∧y} = *S _{x}* ∧

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

Summarizing the above, we affirm that *π* is a lattice isomorphism between *L*_{Cls(L)} and *Cl*_{s}(*L*) □

*Let L satisfy the conditions in* **‘First status’**. *If L satisfies both of the following conditions*, *then Cl _{s}*(

*Let m*,*n*∈*L*\ (*A*(*L*) ∪**0**∪**1**)*and m*||*n*.*Then m*∧*n*≠**0***holds*.*L*_{Cls}(*L*)*is geometric*.

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

We prove *Cl _{s}*(

Using Lemma 2.1 (1), we find {*S _{x}* |

We prove *Cl _{s}*(

Define a map *π* : *L*_{Cls(L)} → *Cl _{s}*(

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 *Cl _{s}*(

*Let x*, *y* ∈ *L*_{Cls(L)} \ {**0**} with *x* ≠ *y*. According to Corollary 2.2 (2) and Lemma 2.1 (2), we obtain
$$x\le y\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{L}_{C{l}_{s}(L)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\iff \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\pi (x)\le \pi (y)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}C{l}_{s}(L).$$(*)
and also obtain *x* < *y* in *L*_{Cls(L)} ⇔ *π*(*x*) < *π*(*y*) in *Cl _{s}*(

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

Let *m* = 0 and *n* ∈ *L*_{Cls(L)}. This implies *m* ∨ *n* = *n* since *L*_{Cls(L)} is a lattice with **0** as the least element.

If *n* = **0**. Then it is ease to see *π*(*m* ∨ *n*) = *π*(*m* ∧ *n*) = *π*(**0**) = *S*_{0} = *π*(*m*) ∨ *π*(*n*) = *π*(*m*) ∧ *π*(*n*).

If *n* ≠ **0**. By Step 1 and Lemma 2.1, the formula *π*(*n*) = *S _{n}* >

Let *m*, *n* ∈ *L*_{Cls(L)} \ {**0**}.

Since *m*, *n* ≤ *m* ∨ *n* in *L*_{Cls(L)}, we receive *S _{m}*,

Additionally, from Step 1 and *S*_{m} ∨ *S*_{n} ∈ *Cl*_{s}(*L*), we affirm that there is *c* ∈ *L*_{Cls(L)} satisfying *S*_{m} ∨ *S*_{n} = *S*_{c}. According to *S*_{m} ≤ *S*_{c} and the definition of *L*_{Cls(L)}, we confirm *m* ≤ *c* and *n* ≤ *c*. So, *c* ∈ *L*_{Cls(L)} \ {**0**} holds. Furthermore, we gain *m* ∨ *n* ≤ *c*. Hence, we decide *S*_{m∨n} ≤ *S _{c}* since the expression (٭). This means

We will prove *π*(*m* ∧ *n*) = *π*(*m*) ∧ *π*(*n*).

When *m* ∧ *n* ≠ **0**. We obtain *S*_{m∧n}, *S*_{m}, *S*_{n} ∈ *Cl*_{s}(*L*). From *m* ∧ *n* ≤ *m*, *n*, we may easily find *S*_{m∧n} ≤ *S _{m}*,

When *m* ∧ *n* = **0**.

If *m* ≤ *n*. Then according to *m*, *n* ∈ *L*_{Cls(L)} \ {**0**}, we determine *m* ∧ *n* = *m* ≠ **0**. This is a contradiction to *m* ∧ *n* = **0**.

If *m*||*n*. Then according to *m*, *n* ∈ *L*_{Cls(L)} \ {**0**} and *m*||*n* with the given condition (1), we determine *m* ∧ *n* ≠ **0**. This is a contradiction to *m* ∧ *n* = **0**. Therefore, up to now, we have demonstrated *π*(*m* ∧ *n*) = *S*_{m∧n} = *S*_{m} ∧ *S*_{n} = *π*(*m*) ∧ *π*(*n*).

Summing up, we determine that *π* is a lattice isomorphism. That is to say, *Cl*_{s}(*L*) ≅ *L*_{Cls(L)} holds.

We prove that *Cl*_{s}(*L*) is geometric.

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

(⇒) If the condition (1) is not true then there are two elements *x*, *y* ∈ *L* \ (*A*(*L*) ∪ **0** ∪ **1**) with *x*||*y* satisfying *x* ∧ *y* = **0**. Hence, we obtain *S* = {*x*, *y*} ∪ {*z* ∈ *A*(*L*) | *z* ∉ *A*(*x*) and *z* ∉ *A*(*y*)} ∈ *Cl*_{s}(*L*).

However, considering Lemma 2.4(1), we confirm *S* ∉ *Cl*_{s}(*L*). This is a contradiction to the above *S* ∈ *Cl*_{s}(*L*). Therefore, the condition (1) is accepted. Using Corollary 2.5 and the geometric of *Cl*_{s}(*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.

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

*We may easily see that the diagram of Cl*_{s}(*L*) *is in Figure 3, in which S*_{0} = {*a*, *b*, *c*, *d*}, *S*_{1} = {**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 Cl*_{s}(*L*_{1}) *is not geometric*. *In fact*, (*ab*), (*cd*) ∈ *L*_{1} \ (*A*(*L*_{1}) ∪ **0**) *satisfies* (*ab*) ∧ (*cd*) = **0**. *In other words*, *L*_{1} *does not satisfy the given condition (1) in Theorem 2.3*.

*Let L*_{2} *be shown as Figure 4. We may easily demonstrate that L*_{2} *satisfies the conditions in* ‘**First status**’ *and the condition (1) in Theorem 2.3*.

*The diagram of L*_{Cls(L2)} *is in Figure 5. We may confirm the geometry of L*_{Cls(L2)}. *Hence, using* Theorem 2.3, *Cl*_{s}(*L*_{2}) *is geometric*.

*The diagram of Cl*_{s}(*L*_{2}) *is in Figure 6, in which S*_{(ab)} = {(*ab*), *c*}, *S*_{(ac)} = {(*ac*), *b*}, *S*_{(bc)} = {(*bc*), *a*}, *S*_{0} = {*a*, *b*, *c*} *and S*_{1} = {**1**}. *We may easily prove the geometry of Cl*_{s(L2)}. *This is the same to the result from Theorem 2.3*.

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.

*Let L satisfy the conditions in* ‘**Second status**’. *Then, Cl*_{s}(*L*) *is geometric if and only if L satisfies the following conditions*.

*L is modular*.*L is compact*.$x=\underset{d\in {\mathcal{F}}^{2}(x)}{\bigvee}$

*d holds for any x*∈*L*\ (*A*(*L*) ∪**0**).

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

*Let L satisfy the conditions in* ‘**Second status**’. *If Cl*_{s}(*L*) *is geometric, then the following statements hold*.

*L is modular*.*L is compact*.$x=\underset{d\in {\mathcal{F}}^{2}(x)}{\bigvee}$

*d holds for any x*∈*L*\ (*A*(*L*) ∪**0**).

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 *N*_{5}. Let *L′* = {*a* ∧ *b*, *a*, *b*, *c*, *a* ∨ *b*} with *a* ∧ *b* < *c* < *a* < *a* ∨ *b*, *a* ∧ *b* < *b* < *a* ∨ *b*, *a*||*b* and *c*||*b*. Thus, we obtain *h′*(*a*) ≥ 2 and *h′*(*a* ∨ *b*) ≥ 3 in which *h′* is the height function of *L′*. Certainly, we may affirm *h*(*a*) ≥ 2 and *h*(*a* ∨ *b*) ≥ 3 since *h′*(*a*) ≥ 2 and *h′*(*a* ∨ *b*) ≥ 3 in which *h* is the height function in *L*. We will distinguish the following Cases 1 and 2 to continue the discussions.

*a* ∧ *b* = **0**.

Suppose that in *L*, *x* ∈ *A*(*L*) holds for any *x* ∈ [**0**, *a* ∨ *b*] with *x*||*a*. Then we obtain [*S*_{0}, *S*_{a∨b}] ⊆ *Cl*_{s}(*L*) to be {*S*_{0}, *S*_{a}} ∪ {*S*_{y} | *a* < *y* < *a* ∨ *b*} ∪ {*S*_{z} | **0** < *z* < *a* and *z* ∉ *A*(*L*)} ∪ {*S*_{a∨b}}. From the definition of *Cl*_{s}(*L*) and Lemma 2.1, we may easily find *S*_{0} < *S*_{t} < *S*_{s} for any *t*, *s* ∈ [**0**, *a* ∨ *b*] \ (*A*(*L*) ∪ **0**) and *t* < *s*. Therefore, we obtain |*A*([*S*_{0}, *S*_{a∨b}])| = 1 since the property (4.2) in Subsection 1.1 and *h′*(*a*) ≥ 2. We may assume *A*([*S*_{0}, *S*_{a∨b}]) = {*S*_{A}}. Then, *S*_{a∨b} is not a join of atoms in [*S*_{0}, *S*_{a∨b}] since *S*_{A} < *S*_{a∨b}. This implies that [*S*_{0}, *S*_{a∨b}] is not geometric. This is a contradiction to the property (2) in Subsection 1.1 since *Cl*_{s}(*L*) is geometric.

Suppose that there is *d* ∈ [**0**, *a* ∨ *b*] satisfying *d* ||*a* and *d* ∉ *A*(*L*). This implies *h′*(*d*) ≥ 2. Moreover, we decide *S*_{0} < *S*_{a} and *S*_{0} < *S*_{d}, and besides, we find *S*_{a}, *S*_{d} < *S*_{a}∨*S*_{d} = *S*_{ad} = {*a*, *d*}∪{*x* ∈ *A*(*L*) | *x* ≰ *A*(*a*) and *x* ≰ *A*(*d*)} < *S*_{a∨d} ≤ *S*_{a∨b}.

Let {*x _{t}*,

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

Furthermore, we confirm that the interval [*S*_{0}, *S*_{a∨b}] ⊆ *Cl*_{s}(*L*) is not geometric. This is a contradiction to the property (2) in Subsection 1.1 since *Cl*_{s}(*L*) is geometric.

*a* ∧ *b* ≠ **0**.

The supposition of *a* ∧ *b* ≠ 0 and *a* ∧ *b* < *b* taken together implies *h*(*b*) ≥ 2. Considering the conditions in ‘**Second status**’ with *a*||*b* and *a*, *b* < *a* ∨ *b* ≤ **1**, we find *a* ∧ *b* = **0**. This is a contradiction to the supposition of *a* ∧ *b* ≠ **0**.

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 *a* ∈ *L* \ {**0**} and *a* ≤ ∨_{i∈𝕀}*x _{i}* for some

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

Let *x*_{1}, *x*_{2} ∈ {*x*_{i}, *i* ∈ 𝕀} satisfy *x*_{1} ≠ *x*_{2}. Since *a* ≰ ∨_{j∈J}*x _{j}* holds for any finite subset

Moreover, *L* must be compact.

To prove item (3).

*x* ∈ ℱ^{2}. The result is trivial.

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

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

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

Since *L* is atomistic, we decide $x=(\underset{d\in {\mathcal{F}}^{2}(x)}{\bigvee}\phantom{\rule{thinmathspace}{0ex}}d)\vee (\underset{{a}_{i}\in A(x)\phantom{\rule{thinmathspace}{0ex}}\text{a}\text{n}\text{d}\phantom{\rule{thinmathspace}{0ex}}{a}_{i}\nleqq d\phantom{\rule{thinmathspace}{0ex}}\text{f}\text{o}\text{r}\phantom{\rule{thinmathspace}{0ex}}\text{a}\text{n}\text{y}\phantom{\rule{thinmathspace}{0ex}}d\in {\mathcal{F}}^{2},\phantom{\rule{thinmathspace}{0ex}}i\in \phantom{\rule{thinmathspace}{0ex}}I}{\bigvee}\phantom{\rule{thinmathspace}{0ex}}{a}_{i})$ satisfying {*a _{i}* |

In addition, owing to *d*_{1} ∈ ℱ^{2}, we obtain that there is *b* ∈ *A*(*x*) satisfying *b* ≺ *d*_{1}. Thus, we infer that the sublattice {*x*, *d*_{1}, *b*, *a*_{1}, **0**} in *L* is isomorphic to *N*_{5}. Hence, *L* is not modular. This is a contradiction to item (1).

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

We will reveal the construction of *Cl*_{s}(*L*).

*Let L satisfy the conditions in* ‘**Second status**’. *If L satisfies the following statements (1) and (2), then Cl*_{s}(*L*) = {*S*_{x} | *x* ∈ *L* \ {**0**}} *holds*.

*L is modular*.$x=\underset{d\in {\mathcal{F}}^{2}(x)}{\bigvee}\phantom{\rule{thinmathspace}{0ex}}d$

*holds for any x*∈*L*\ (*A*(*L*) ∪**0**).

According to Lemma 2.1 (1), we can obtain {*S*_{x} | *x* ∈ *L* \ (*A*(*L*) ∪ **0**)} ∪ {*S*_{0}} = {*S*_{x} | *x* ∈ *L* \ {**0**}}. It is easy to see {*S*_{x} | *x* ∈ *L* \ {**0**}} ⊆ *Cl*_{s}(*L*).

Let *S* ∈ *Cl*_{s}(*L*) \ {*S*_{x} | *x* ∈ *L* \ {**0**}}. Then, by *S* ≠ *S*_{0} and Lemma 2.1 (1), we affirm that there exists *y* ∈ *S* such that *y* ∉ *A*(*L*) ∪ **0**.

Let {*y _{j}*,

Since |*J*| ≥ 2, we can select *y*_{1}, *y*_{2} ∈ *S*\(*A*(*L*)∪**0**) with *y*_{1} ≠ *y*_{2}. Thus, we confirm *y*_{1} || *y*_{2} and *h*(*y*_{1}), *h*(*y*_{2}) ≥ 2 since *y _{j}* ∉

Combining *S* ∈ *Cl*_{s}(*L*) and (C1), we obtain *y*_{1} ∧ *y*_{2} = **0**. Considering the atomic of *L* with *h*(*y*_{1}), *h*(*y*_{2}) ≥ 2 and *y*_{1} || *y*_{2} and *y*_{1} ∧ *y*_{2} = **0**, we obtain *A*(*y*_{1}) \ *A*(*y*_{2}) ≠ ∅ and *A*(*y*_{2}) \ *A*(*y*_{1}) ≠ ∅. Let *a*_{1} ∈ *A*(*y*_{1}) \ *A*(*y*_{2}) and *a*_{2} ∈ *A*(*y*_{2}) \ *A*(*y*_{1}). Then we decide that the sublattice {*y*_{1} ∨ *y*_{2}, *y*_{1}, *a*_{1}, *a*_{2}, **0**} of *L* is isomorphic to *N*_{5}. This is a contradiction to the given condition (1) according to property (2) in Subsection 1.1.

Therefore, we determine *Cl*_{s}(*L*) = {*S*_{x} | *x* ∈ *L* \ {**0**}}. ☐

*Let L satisfy the conditions in* ‘**Second status**’ *and the statements*

*L is modular*.*L is compact*.$x=\underset{d\in {\mathcal{F}}^{2}(x)}{\bigvee}\phantom{\rule{thinmathspace}{0ex}}d$

*holds for any x*∈*L*\ (*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**).

Since *L* is complete, we obtain ∨_{y∈𝔜}*y* ∈ *L*. Let *b* = ∨_{y∈𝔜}*y*. Then, we confirm *S*_{∨y∈𝔜y} ∈ *Cl*_{s}(*L*).

From *S _{y}* ∈

We distinguish two cases to continue the proof.

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

Considering *y* ≤ *S*_{∨y∈𝔜y} for any *y* ∈ 𝔜 with Lemma 2.1 (2) and Corollary 2.2 (2), we decide *S*_{y} ≤ *S*_{∨y∈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, *S*_{z} ≤ *S _{b}* holds. Using Lemma 2.1 (2) and Corollary 2.2, we obtain

Since *L* is compact, we may be assured *b* = ∨_{y∈𝔜}*y* = *y*_{1} ∨ ... ∨ *y*_{n} for some *y _{j}* ∈ 𝔜 with

In view of the selection of *z*, we obtain *S _{y}* ≤

Thereby, we determine *S _{b}* =

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

Since *S _{y}* ≤ $\underset{y\in \mathfrak{Y}}{\bigvee}\phantom{\rule{thinmathspace}{0ex}}{S}_{y}={S}_{0}$ and Lemma 2.1 (1), we obtain

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

*Let L satisfy the conditions in* ‘**Second status**’ *and the following conditions*.

*L is modular*.*L is compact*.$x=\underset{d\in {\mathcal{F}}^{2}(x)}{\bigvee}$

*d holds for any x*∈*L*\ (*A*(*L*) ∪**0**).

*Then, Cl*_{s}(*L*) *is geometric*.

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 *Cl*_{s}(*L*) is geometric with the following steps.

We will prove that *Cl*_{s}(*L*) is atomistic.

In virtue of Lemma 2.11, we find *Cl*_{s}(*L*) = {*S*_{x} | *x* ∈ *L* \ {**0**}}. With the given condition (3), we obtain *S _{y}* =

By the property (4.2) in Subsection 1.1, we decide {*S*_{d} | *d* ∈ ℱ^{2}(*y*)} ⊆ *A*(*Cl*_{s}(*L*)) for any *y* ∈ *L*\(*A*(*L*)∪**0**).

Hence, we may state that *S*_{y} is the join of some atoms in *Cl*_{s}(*L*) for any *y* ∈ *L* \ (*A*(*L*) ∪ **0**). Therefore, we indicate that *Cl*_{s}(*L*) is atomistic.

We will prove that all atoms in *Cl*_{s}(*L*) are compact.

Let *S* ∈ *A*(*Cl*_{s}(*L*)). Then by the property (4.2) in Subsection 1.1, we obtain *S* = *S*_{d} for some *d* ∈ ℱ^{2}.

Suppose ${S}_{d}\le \underset{j\in \mathbb{J}}{\bigvee}\phantom{\rule{thinmathspace}{0ex}}{S}_{{x}_{j}}$ for some *x _{j}* ∈

Therefore, every atom in *Cl*_{s}(*L*) is compact.

We will prove that *Cl*_{s}(*L*) is modular.

Suppose that *Cl*_{s}(*L*) is not modular. Then in *Cl*_{s}(*L*), there is a sublattice {*A*, *B*, *C*, *A* ∧ *C*, *A* ∨ *C*} which is isomorphic to *N*_{5}, where *A*||*C*, *B*||*C*, *A* ∧ *C* < *A* < *B* < *A* ∨ *C* and *A* ∧ *C* < *C* < *A* ∨ *C*.

According to Lemma 2.1, Lemma 2.11 and *A* ∧ *C* < *A*, *B* with *A* ∧ *C* ∈ *Cl*_{s}(*L*), we own *S*_{a} = *A*, *S*_{b} = *B* and *S*_{c} = *C* for some *a*, *b*, *c* ∈ *L* \ (*A*(*L*) ∪ **0**). In light of Lemma 2.12, we may be assured *A* ∨ *C* = *S*_{a} ∨ *S*_{c} = *S*_{a∨c}.

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* < *a* ∨ *c*.

From *a*||*c* and *a*, *c* ∈ *L* \(*A*(*L*)∪**0**), we confirm *a* ∧ *c* = **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**, *a* ∨ *c*, *a*, *b*, *c*} in *L* satisfies *a*||*c* and *b*||*c*, and **0** < *a* < *b* < *a* ∨ *c* and **0** < *c* < *a* ∨ *c*. This implies that the sublattice {**0**, *a* ∨ *c*, *a*, *b*, *c*} in *L* is isomorphic to *N*_{5}. Thus, we find that *L* is not modular. This is a contradiction to the given condition (1). ☐

The proof of Theorem 2.9 follows.

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 *L*_{3} satisfies the conditions in ‘**Second status**’, but does not satisfy the conditions in Theorem 2.9, then *Cl _{s}*(

*Let L*_{3} *be defined as Figure 7. We may easily reveal that L*_{3} *satisfies all of conditions in* ‘**Second status**’ *and the conditions (2) and (3) in Theorem 2.9*.

*Since* {**1**, (*ab*), *b*, **0**, *c*} *is N*_{5} *up to isomorphism. This means that L _{3} is not modular in view of property (2) in Subsection 1.1. By Theorem 2.9, we demonstrate that Cl_{s}*(

*Actually, the diagram of Cl _{s}*(

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.

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

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**Received**: 2016-09-10

**Accepted**: 2017-05-25

**Published Online**: 2017-07-13

**Citation Information: **Open Mathematics, Volume 15, Issue 1, Pages 959–973, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0078.

© 2017 Mao. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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