Reduction of Quasi-Lattices to Lattices

TL;DR

This paper explores the conditions under which quasi-lattices can be transformed into lattices, providing a fundamental homomorphism theorem that clarifies their structural relationship.

Contribution

It introduces a fundamental theorem of homomorphism for quasi-lattices, showing how they can be mapped onto lattices under specific conditions.

Findings

Quasi-lattices become lattices when associativity and modularity conditions are met.

A homomorphism theorem for quasi-lattices is established.

Conditions for mapping quasi-lattices onto lattices are identified.

Abstract

Quasi-lattices are introduced in terms of 'join' and 'meet' operations. It is observed that quasi-lattices become lattices when these operations are associative and when these operations satisfy 'modularity' conditions. A fundamental theorem of homomorphism proved in this article states that a quasi-lattice can be mapped onto a lattice when some conditions are satisfied.

Full text

a* Department of Mathematics, Alagappa University, Karaikudi-630 004,

Tamil Nadu, India.

b* Department of Mathematics, Vidhyaa Giri College of Arts and Science, Puduvayal-630 108, Tamil Nadu, India.*

E-mail: ganesamoorthyc@gmail.com (C. Ganesa Moorthy)

 sgkarpa@gmail.com (SG.Karpagavalli)

Key words: Minimal upper bound, Congruence relation, Partition.

AMS Subject Classification(2010): 06B10,18B35,06C05.

1 Introduction

The concept of a minimal upper bound is not widely known. A lattice is a partially ordered set (poset) in which any two elements have a least upper bound and a greatest lower bound. A quasi-lattice is a poset in which any two elements have a minimal upper bound and a maximal lower bound. Every quasi-lattice is a lattice. This article tries to establish fundamental facts about quasi-lattices. But, it finds that associativity of $`$meet’ and $`$join’ operations of quasi-lattices is a unique property of lattices. Similarly it is established that $`$modularity’ is also a unique property of lattices. A fundamental theorem of homomorphism found in this article also reduces quasi-lattices into lattices. The books [3] and [2] are referred to fundamental definitions and properties for posets and lattices. Although there are many recent articles (see, for example [4, 5, 6]) the results of these articles will not be extended to quasi-lattices, because quasi-lattices reduce to lattices when some fundamental properties are assumed.

A partial order $\leq$ on a non empty set $P$ is a relation that is reflexive, anti-symmetric and transitive. A poset $(P,\leq)$ is a non empty set $P$ with a partial order $\leq$. An element $a$ in a partially ordered set $(P,\leq)$ is a maximal lower bound of a non empty subset $A$ of $P$ if $a\leq x$, $\forall x\in A$, and if there is no element $d$ in $P$ such that $a<d\leq x$, $\forall x\in A$.Dually a minimal upper bound is defined. A partially ordered set $(P,\leq)$ is called quasi-lattice, if any two elements of $P$ have a minimal upper bound and a maximal lower bound. However, two elements in a quasi-lattice may have more than one maximal lower bound and may have more than one minimal upper bound. Let us use the notations $x\wedge y$ and $x\vee y$ to denote some (particular) maximal lower bound and some minimal upper bound of $x$ and $y$, respectively, in a partially ordered set.

Example 1.1

The Hasse diagram given in the Figure 1 represents a quasi-lattice. In this diagram the point $x\vee(y\vee z)$ represents another minimal upper bound of {x,y} in addition to $x\vee y$. So, it is not a lattice. In this quasi-lattice, $(x\vee y)\vee z$ can never take $``$ the form” $x\vee(y\vee z)$. So, associativity fails to be true.

yzx0(x\vee y)\vee z$$x\vee(y\vee z)$$y\vee z$$x\vee yFigure 1

2 Associative quasi-lattices

It would be difficult to derive many results for quasi-lattices, when associativity is not assumed.

Definition 2.1

A quasi-lattice $(P,\leq)$ is called an associative lattice, if

1. (i)

$a\vee(b\vee c)=(a\vee b)\vee c$, and 2. (ii)

$a\wedge(b\wedge c)=(a\wedge b)\wedge c$* hold for every $a,b,c\in P$.*

Here (i) means that if $a_{1}$ is a minimal upper bound of {$b,c$} and if $a_{2}$ is a minimal upper bound of {$a,a_{1}$}, then there is a minimal upper bound $a_{3}$ of {$a,b$} such that $a_{2}$ is a minimal upper bound of{$a_{3},c$} and similarly; if $b_{1}$ is a minimal upper bound of {$a,b$} and $b_{2}$ is minimal upper bound of {$b_{1},c$}, then there is a minimal upper bound $b_{3}$ of {$b,c$} such that $b_{2}$ is a minimal upper bound of {$a,b_{3}$}. This interpretation clarifies the meanings for the present notation. When this is followed, the meaning of the following proposition is unambigious.

Proposition 2.2

The following identities are true in a quasi-lattice $(P,\leq)$. (A1): $a\vee a=a$; (A2): $a\wedge a=a$; (A3): $a\vee b=b\vee a$; (A4):$a\wedge b=b\wedge a$; (A5): $a\vee(a\wedge b)=a=(a\wedge b)\vee a$ ; (A6):$a\wedge(a\vee b)=a=(a\vee b)\wedge a$; $\forall a,b\in P$.

Proof: Let us verify $a\vee(a\wedge b)=a$. Let $a_{1}$ be a maximal lower bound of {$a,b$}, and $a_{2}$ be a minimal upper bound of {$a_{1},a$}. Then $a_{2}=a$ because $a_{1}\leq a$. Other relations can also be verified in this way.

If $(P,\leq)$ is an associative quasi-lattice, then it further has the properties: ( A7) : $a\vee(b\vee c)=(a\vee b)\vee c$ and ( A8) : $a\wedge(b\wedge c)=(a\wedge b)\wedge c$; $\forall a,b,c\in P$. It is known that the relations ( A1) to (A8) characterize a lattice, when $a\vee b$ and $a\wedge b$ are unique elements ( see Theorem 1 in Section 1 in Chapter 1 in [2] ). It is to be proved that an associative quasi-lattice should be a lattice. For this purpose, let us introduce some changes in applications of the notations $\vee$ and $\wedge$. For a given poset $(P,\leq),A\subseteq P$ and $B\subseteq P$, let $A\vee B$ (respectively, $A\wedge B$) denote the collection of all elements of the form $a\vee b$ (respectively, $a\wedge b$) with $a\in A$ and $b\in B$. So, for example, the relation $a\wedge(a\vee b)=a=(a\vee b)\wedge a$ will mean $\{a\}\wedge(\{a\}\vee\{b\})=\{a\}=(\{a\}\vee\{b\})\wedge\{a\}$. Thus a poset $(P,\leq)$ is a quasi-lattice if and only if $\{a\}\vee\{b\}$ and $\{a\}\wedge\{b\}$ are non empty subsets of $P$, for any $a,b\in P$. It is a lattice if and only if $\{a\}\vee\{b\}$ and $\{a\}\wedge\{b\}$ are singleton subsets of $P$, for any $a,b\in P$.

Theorem 2.3

A quasi-lattice $(P,\leq)$ is associative if and only if it is a lattice.

Proof: Suppose $(P,\leq)$ is an associative quasi-lattice. Let $x,y\in P$ and $a,b\in\{x\}\wedge\{y\}$.Then $a\leq y$, $a\leq x$ and $(\{x\}\wedge\{y\})\wedge\{a\}=\{x\}\wedge(\{y\}\wedge\ \{a\})=\{x\}\wedge\{a\}=\{a\}$, when $(\{x\}\wedge\{y\})\wedge\{a\}\supseteq\{a,b\}\wedge\{a\}=\{a\}\cup(\{b\}\wedge\{a\})$.Thus $\{a\}\wedge\{b\}=\{a\}$ so that $a\leq b$. Similarly $b\leq a$ so that $a=b$.Thus $\{x\}\wedge\{y\}$ contains a unique element. Dually, $\{x\}\vee\{y\}$ contains a unique element. This proves that $(P,\leq)$ is a lattice.

3 Modular quasi-lattices

Definition 3.1

A quasi-lattice $(P,\leq)$ is said to be modular if $\{x\}\vee(\{y\}\wedge\{z\})=(\{x\}\vee\{y\})\wedge\{z\}$ whenever $x,y,z\in P$ and $x\leq z$.

Theorem 3.2

A modular quasi-lattice $(P,\leq)$ is a lattice.

Proof: Fix $x,y$ in the given modular lattice $(P,\leq)$. Let $a,b\in\{x\}\wedge\{y\}$. Then $a\leq x,a\leq y,b\leq x$, and $b\leq y$. So, $(\{x\}\wedge\{y\})\vee\{a\}=\{a\}\vee(\{y\}\wedge\{x\})=(\{a\}\vee\{y\})\wedge\{x\}=\{y\}\wedge\{x\}=\{x\}\wedge\{y\}$, when $\{a,b\}\subseteq\{x\}\wedge\{y\}$. So $\{a,b\}\vee\{a\}\subseteq\{x\}\wedge\{y\}$ and hence $\{a,a\vee b\}\subseteq\{x\}\wedge\{y\}$.Thus $a\vee b\in\{x\}\wedge\{y\}$, when $a\vee b\geq a,a\vee b\geq b$ , $a\in\{x\}\wedge\{y\}$ and $b\in\{x\}\wedge\{y\}$. So,the maximality of $a$ and $b$ implies that $a=a\vee b=b$. In particular, $\{x\}\wedge\{y\}$ contains atmost one point. Dually $\{x\}\vee\{y\}$ contains atmost one point. This proves the theorem.

Associative quasi-lattices are lattices and modular quasi-lattices are lattices. So it is difficult to derive new results for quasi-lattices, because quasi-lattices with additional fundamental properties become lattices. However, one can derive fundamental results for ideals.

Definition 3.3

A subset $I$ $(\mathscr{F},respectively)$ of a quasi-lattice $(P,\leq)$ is called an ideal (a filter, respectively), if

1. (i)

$a,b\in I\Rightarrow\{a\}\vee\{b\}\subseteq I$** 2. ((i)

$a,b\in\mathscr{F}\Rightarrow\{a\}\wedge\{b\}\subseteq\mathscr{F}$, respectively) and 3. (ii)

$a\in I,b\in P,b\leq a\Rightarrow b\in I$** 4. ((ii)

$a\in\mathscr{F},b\in P,b\geq a\Rightarrow b\in\mathscr{F}$, respectively).

An arbitrary intersection of ideals (filters) in a quasi-lattice is an ideal (a filter). The intersection of a filter and an ideal is sub quasi-lattice. Here, a sub quasi-lattice $(Q,\leq)$ of a quasi-lattice $(P,\leq)$ means that $\{x\}\vee\{y\}\subseteq Q$, and $\{x\}\wedge\{y\}\subseteq Q$, whenever $x,y\in Q$. The intersection of a filter with an ideal is a convex subset in view of the following (usual) definition.

Definition 3.4

A subset $C$ of a quasi-lattice $(P,\leq)$ is said to be convex, if $a\in C$, whenever $x,y\in C$, $a\in P$ and $x\leq a\leq y$.

Notation 3.5

To each $A\subseteq P$, a quasi-lattice, let $(A]$ and $[A)$ denote the smallest ideal and the smallest filter, respectively, containing $A$. They exist in view of the previous remark.

Proposition 3.6

Let $(P,\leq)$ be a quasi-lattice. Let $I(P)$ (respectively, $F(P)$) be the collection of all ideals (respectively, filters) of $(P,\leq)$. Then $I(P)$ (respectively, $F(P)$) is a complete lattice under the inclusion relation (respectively, inverse inclusion relation).

Proof: Let $(I_{\lambda})_{\lambda\in A}$ be a collection of ideals in $P$, Then $\cap\{I_{\lambda}:\lambda\in A\}$ and $(\cup\{I_{\lambda}:\lambda\in A\}]$ are ideals which are the greatest lower bound and the least upper bound of the given collection. A similar argument is applicable for filters.

4 Congruence relations

Ideals are associated with inverse image of a least element for a lattice homomorphism. A lattice homomorphism is associated with a congruence. Let us first define a congruence relation for a quasi-lattice.

Definition 4.1

Let $(P,\leq)$ be a quasi-order lattice. An equivalence relation $\theta$ on $P$ is denoted by $x\equiv y$(mod $\theta$) when $x$ and $y$ are related in $P$ by $\theta$. Moreover, for subsets $A$,$B$ of $P$, the identity $A\equiv B$ (mod $\theta$) means the following:

1. (i)

to each $a\in A$, there is a $b\in B$ such that $a\equiv b$ (mod $\theta$), and 2. (ii)

to each $b\in B$, there is an $a\in B$ such that $a\equiv b$ (mod $\theta$).

*The equivalence relation $\theta$ on $P$ is called a congruence relation, if $\{x_{1}\}\wedge\{y_{1}\}\equiv\{x_{2}\}\wedge\{y_{2}\}$(mod $\theta$), and $\{x_{1}\}\vee\{y_{1}\}\equiv\{x_{2}\}\vee\{y_{2}\}$(mod $\theta$), whenever $x_{1}\equiv x_{2}$(mod $\theta$) and $y_{1}\equiv y_{2}$(mod $\theta$) in $P$, and if $\{x\}\wedge\{y\}\subseteq[z]$, when $z\in\{x\}\wedge\{y\}$ and $\{x\}\vee\{y\}\subseteq[z]$, when $z\in\{x\}\vee\{y\}$, for x,y,z in P, when $[z]$ refers to the equivalence class containing $z$, determined by $\theta$.

It is known that the collection of all partitions is a complete lattice under the $``$refinement” relation. The collection of all congruences on a lattice is a (complete) sublattice of the lattice of all partitions. In the same way(see the proof of theorem 3.9 in [1]), one can verify that the collection of all congruences on a quasi-lattice is a complete lattice and a sublattice of the lattice of all partitions.*

Lemma 4.2

Let $(P,\leq)$ be a quasi-lattice, and $\theta$ be a congruence relation on $P$. If $u\equiv v$ (mod $\theta$), $a\in\{u\}\wedge\{v\}$, $b\in\{u\}\vee\{v\}$, and if $a\leq x\leq b$ , then $u\equiv x$ (mod $\theta$).

Proof: Under the assumptions, we have $\{x\}=\{x\}\vee\{a\}\equiv\{x\}\vee(\{u\}\wedge\{v\})\equiv\{x\}\vee(\{u\}\wedge\{u\})\equiv(\{x\}\vee\{u\})$(mod $\theta$). Dually, we have $\{x\}=\{x\}\wedge\{b\}\equiv\{x\}\wedge(\{u\}\vee\{v\})\equiv\{x\}\wedge(\{u\}\vee\{u\})\equiv\{x\}\wedge\{u\}$(mod $\theta$). So, we have $\{u\}=\{u\}\wedge(\{u\}\vee\{x\})=\{u\}\wedge\{x\}\equiv\{x\}$(mod $\theta$). This proves the lemma.

Definition 4.3

Let $T:P_{1}\to P_{2}$ be a mapping from a quasi-lattice $P_{1}$ into a quasi-lattice $P_{2}$. It is said to be a q-lattice homomorphism, if $T(\{x\}\vee\{y\})=\{T(x)\}\vee\{T(y)\}$ and $T(\{x\}\wedge\{y\})=\{T(x)\}\wedge\{T(y)\},\forall x,y\in P$

Definition 4.4

Let $\theta$ be an equivalence relation on a quasi-lattice $(P,\leq)$. Let $[x]$ denote the equivalence class containing $x$. Let us say that $\theta$ satisfies the condition () if the following are true in $P$ :*

1. (i)

If $[x]\neq[y]$, $x\leq z$ and $y\leq z$, then there are elements $a\in[x]$ and $b\in[y]$ and there is an element $d\in\{a\}\vee\{b\}$ such that $d\leq z$. 2. (ii)

If $[x]\neq[y]$, $x\geq z$ and $y\geq z$, then there are elements $a\in[x]$ and $b\in[y]$ and there is an element $d\in\{a\}\wedge\{b\}$ such that $d\geq z$.

Let us now state a fundamental theorem of homomorphism.

Theorem 4.5

Let $(P,\leq)$ be a quasi-lattice. Let $\theta$ be a congruence relation on $P$ that satisfies () of definition 4.4. Let $P/\theta$ be the collection of all equivalence classes. Let $[x]$ denote the equivalence class containing $x$. Then $P/\theta$ is a lattice in which we have $[x]\wedge[y]=[x\wedge y]$ and $[x]\vee[y]=[x\vee y]$, for any elements $x\wedge y$ and $x\vee y$ in $\{x\}\wedge\{y\}$ and $\{x\}\vee\{y\}$, respectively. Also, the quotient mapping $\pi:P\to P/\theta$ defined by $\pi(x)=[x]$, $x\in P$, is a surjective q-lattice homomorphism. On the other hand if $T:P\to L$ is a surjective q-lattice homomorphism from a quasi-lattice $P$ onto a lattice $L$, then $\{T^{-1}(a):a\in L\}$ defines a partition that leads to a congruence relation satisfying () of definition 4.4.

**Proof:

First Part:** Define $[x]\leq[y]$ if and only if $a\leq b$ for some $a\in[x]$ and some $b\in[y]$. Suppose $a_{1}\in[x]$ and $b_{1}\in[y]$ such that $a_{1}\leq b_{1}$. If $a_{2}\in[x]$, then $a_{1}\equiv a_{2}$ (mod $\theta$), $a_{2}\leq b_{1}\vee a_{2}$ (for any element of this type) and $\{b_{1}\}\vee\{a_{2}\}\equiv\{b_{1}\}\vee\{a_{1}\}\equiv\{b_{1}\}$(mod $\theta$). If $b_{2}\in[y]$, then $b_{1}\equiv b_{2}$(mod $\theta$), $a_{1}\wedge b_{2}\leq b_{2}$, and $\{a_{1}\}\wedge\{b_{2}\}\equiv\{a_{1}\}\wedge\{b_{1}\}\equiv\{a_{1}\}$(mod $\theta$). Thus, if $[x]\leq[y]$, then for any $a_{1}\in[x]$, there is a $b_{1}\in[y]$ such that $a_{1}\leq b_{1}$ and for any $b_{2}\in[y]$ there is an $a_{2}\in[x]$ such that $a_{2}\leq b_{2}$. Now let us verify that this relation in $P/\theta$ is a partial order relation. Since $x\leq x$, we have $[x]\leq[x],\forall x\in P$. To prove anti-symmetricity, assume that $[x]\leq[y]$ and $[y]\leq[x]$ for two elements $x,y\in P$. Then there is an element $y_{1}\in[y]$ such that $x\leq y_{1}$; and there is an element $x_{1}\in[x]$ such that $y_{1}\leq x_{1}$. Thus $x\leq y_{1}\leq x_{1}$ and $x_{1}\equiv x$ (mod $\theta$). By the previous lemma 4.2 it is concluded that $y_{1}\equiv x$ (mod $\theta$). This proves that $\leq$ is anti-symmetric in $P/\theta$. To prove transitivity, assume that $[x]\leq[y]$ and $[y]\leq[z]$ for some $x,y,z\in P$. Then there is an element $y_{1}\in[y]$ and there is an element $z_{1}\in[z]$ satisfying $x\leq y_{1}\leq z_{1}$ so that $x\leq z_{1}$, So $(P/\theta,\leq)$ is a poset. To prove that $P/\theta$ is a lattice, consider an element $a\in\{x\}\wedge\{y\}$, for some fixed elements $x,y$. Then $a\leq x$ and $a\leq y$. So $[a]\leq[x]$ and $[a]\leq[y]$. Suppose $[b]\leq[x]$ and $[b]\leq[y]$ for some element $b$ of $P$, and assume that $[a]\leq[b]$. Then there is an element $b_{1}\in[b]$ such that $a\leq b_{1}$. There are elements $c_{1}\in[x]$ and $c_{2}\in[y]$ such that $b_{1}\leq c_{1}$ and $b_{1}\leq c_{2}$. By the condition (*) satisfied, there are elements $a_{1}\in[c_{1}]$ and $a_{2}\in[c_{2}]$ and there is an element $d\in\{a_{1}\}\wedge\{a_{2}\}$ such that $b_{1}\leq d$. Since $\{a\}\equiv\{x\}\wedge\{y\}\equiv\{c_{1}\}\wedge\{c_{2}\}\equiv\{a_{1}\}\wedge\{a_{2}\}\equiv\{d\}$(mod $\theta$), we have the relation $[b]\leq[a]$. Thus $[a]=[b]$. This proves that $[x]\wedge[y]=[x\wedge y]$ for any element $x\wedge y,\forall x,y\in P$. Dually, one can prove that $[x]\vee[y]=[x\vee y]$, for any element $x\vee y,\forall x,y\in P$. So, $P/\theta$ is a lattice. Other sub divisions of the first part are trivial.

Second Part: Let $\theta$ be the equivalence relation induced by the partition $\{T^{-1}(a):a\in L\}$. The condition (*) of definition 4.4 has to be checked to complete the proof as the other sub divisions are trivial. If $T^{-1}(a)\neq T^{-1}(b)$, $a_{1}\in T^{-1}(a),b_{1}\in T^{-1}(b),d_{1}\geq a_{1},d_{1}\geq b_{1}$ and $T(d_{1})=d$, then $d\geq a\vee b,\{a_{1}\}\vee\{b_{1}\}\subseteq T^{-1}(a\vee b)$, and $[a_{1}\vee b_{1}]\leq[d_{1}]$ (in view of the order relation introduced in first part) so that there are $a_{2}\in T^{-1}(a),b_{2}\in T^{-1}(b)$ such that $d_{1}\geq a_{2}\vee b_{2}\equiv a_{1}\vee b_{1}$(mod $\theta$). Similarly, if $T^{-1}(a)\neq T^{-1}(b)$, $a_{1}\in T^{-1}(a),b_{1}\in T^{-1}(b)$, $d_{1}\leq a_{1}$ and $d_{1}\leq b_{1}$ then there are $a_{2}\in T^{-1}(a),b_{2}\in T^{-1}(b)$, such that $d_{1}\leq a_{2}\wedge b_{2}\equiv a_{1}\wedge b_{1}$ (mod $\theta$). This completes the proof of the theorem.