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Mathematics Thesis and Dissertations

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CLOSURE OPERATORS ON

COMPLETE ALMOST DISTRIBUTIVE LATTICES

MIHRET, ANTENEH

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CLOSURE OPERATORS ON COMPLETE ALMOST

DISTRIBUTIVE LATTICES

BY

MIHRET ANTENEH

DEPARTMENT OF MATHEMATICS COLLEGE OF SCIENCE BAHIR DAR UNIVERSITY

SEPTEMBER, 2016

CLOSURE OPERATORS ON COMPLETE ALMOST

DISTRIBUTIVE LATTICES

A project

Submitted in Partial Fulfillment of the Requirements for the Degree of

Master of Science in Mathematics

By

MIHRET ANTENEH

ADVISOR: BERHANU ASSAYE (PhD)

DEPARTMENT OF MATHEMATICSCOLLEGE OF SCIENCE

SEPTEMBER, 2016

The project entitled “CLOSURE OPERATORS ON COMPLETE

ALMOST DISTRIBUTIVE LATTICES” by Mihret Anteneh, is approved

for the degree of “master of science in mathematics”.

Board of Examiners Name Signature

Advisor: Berhanu Assaye (PhD) _______ Examiner 1: Jejaw Demamu (PhD) ________

Examiner2: Mihret Alamneh (PhD) ________

Date:_____________

ACKNOWLEDGEMENTSFirst, I would like to thank the almighty God, who is always with me and helped me to be where I am today.Second, my deepest gratitude goes to my advisor Dr. Berhanu Assaye for introducing me with a new project topic and giving me his valuable support and constructive comments about the whole work. Last but not least, I would like to thank my husband Dr. Ebabu Tefera, my children, and all who helped me directly and indirectly to complete this project.

Table of contentsAbstract-------------------------------------------------------------------------------------------------iChapter one--------------------------------------------------------------------------------------------11. Introduction and preliminaries-------------------------------------------------------------------1 1.1 Introduction --------------------------------------------------------------------------------1 1.2 Preliminaries--------------------------------------------------------------------------------2Chapter Two------------------------------------------------------------------------------------------82. Closure operators on complete almost distributive lattice ----------------------------------8References-------------------------------------------------------------------------------------------18

Abstract

In this project, we discussed the concept of complete Almost Distributive Lattice. We pr

oved all closure operators of a complete Almost Distributive Lattice L with fixed maximal

element m is dual atomistic. We defined the concept of a complete meet-irreducible

element in a complete and prove the necessary and sufficient condition for a dual atom

of complemented.

CHAPTER ONE

INTRODUCTION AND PRELIMINARIES

1.1 INTRODUCTION

In [19] Swamy and Rao introduced the concept of an Almost Distributive Lattice () as a

common abstraction of almost all the existing ring theoretic generalizations of a

Boolean algebra like p-rings [13], regular rings[12], biregular rings [17], associate rings

[11], p1-rings [14], triple systems [16], baer rings [1], m-domain rings [15] and-rings [2]

on one hand and the class of distributive lattices on the other. Thus, a study of any

concept in the class of will yield results in all the classes of algebras mentioned above.

In [19], they also observed that the set PI(L) of all principal ideals of an ADL (L, 0, m)

with a maximal element m forms a bounded distributive lattice. Through this

distributive lattice PI(L), many existing concepts of lattice theory were extended to the

class of ADLs [4, 5, 7, 20].

In mathematics, closure operators play important role in topology, algebra and

logic and in theoretical computer sciences, closure operators have been widely used in

the semantic area, notably in domain theory, in program semantics and in the theory of

semantics approximation by abstract interpretation. In view of the rich applications of

complete lattices and the closure operators in different fields, we introduced the

concept of a complete ADL [8] and the concept of a closure operator of a complete [9,1

0] and derived some important properties on closure operators.

In this project, we define the concept of a completely meet irreducible element in

a complete ADL (L, 0, m) and establish a relation between completely meet irreducible

elements in a complete ADL L and dual atoms of the lattice ((L)) of all closure

operators of L. We derive necessary and sufficient conditions for dual atoms in the

lattice ((L)) to have complements.

1. preliminary

The following definitions, remarks, lemmas, corollary and theorems are fundamental

and essential for the discussion of the next chapter. And are frequently refers when

they are necessary.

Definition 1. 2.1. [3] An algebra (L,) of type is called lattice if for all and c L the

following conditions hold:

1. (Idempotent law of

2. (Commutative law of

3. associative law of

4. ⋁

( absorption law of and respectively)

Note:(a) ( “join” ) are both binary operations which means that they can be applied to

a pair of elements a, b of L to yield again an elements of L.

(b) In (2, 2), 2 represents binary operation.

Definition 1.2. 2. [18] If L is a lattice and L’ is a subset of L such that for every pairs of

elements in L’ both and are in L’, where the lattice operation of L, then we say that L’ is

the sub lattice of L.

Definition 1.2.3. [18] An algebra (L,) of type (2, 2) is called distributive lattice if, for any

, satisfying any one of the following four operations.

1. (Left distributive of over )

2. (Right distributive of over

3. (Left distributive of over )

4.. (Right distributive of over)

Definition 1.2.4. [19] An algebra (L, 0) of type (2, 2, 0) is called an Almost distributive

lattice () if, for any, the following conditions hold:

(1)

(2) ).

(3)

(4)

(5)

(6)

Definition 1. 2.5 . [18] A partially ordered set () is a non empty set L together with a

binary relation which satisfy the following condition, for any

1. (reflexivity)

2. If and then ( anti- symmetric )

3. and then ( transitivity)

Definition 1. 2.6. [18] A (L, ) is said form a lattice if for every both and exist in L, in

that case we write and

Remark 1. 2.7. [18] The two definition of a lattice are equivalent.

Definition 1. 2.8. [17] If (L 0) is an ADL, for any, defineif and only if or equivalently,

thenis a partial ordering on L.

Definition 1. 2.9. [18] A lattice L is complete if for any subset R of L bothand the join exi

sts in L.

Definition 1. 2.10. [18] A L’ of a complete lattice L is called a complete of L if for every

subset A of L’ the element in L, are actually in L’.

Definition 1. 2.11. A completely distributive lattice is a complete lattice in which

arbitrary joins distribute over arbitrary meets.(WIKIPEDIA, the free Encyclopedia)

Lemma 1.2.12. [19] If (L 0) is an ADL, for any L, we have the following:

(1)

(2)

(3) Whenever

(4)is associative in L

(5)

(6)

(7)

(8)

(9) and

(10) and

(11) and

(12) and

(13) , then and

(14) .

Definition 1.2.13. [19] Let L be an ADL and , then the interval

[] = {} is a bounded distributive lattice.

Definition 1.2.14. [4] An element m maximal if it is maximal in the partial ordered set

(L,That is, for any , mm

Theorem 1.2.15 . [19] Let (L,) be an ADL. Then, for any m L, the following are

equivalent:

(1) m is maximal

(2) mx = m for all x L

(3) m x = x for all x L

Definition 1.2.16. [19] If (L,) is an and m is fixed maximal element of L, then

we say that (L,m) is an ADL with a maximal element m.

Definition 1.2.17. [19] A non-empty subset I of an L is called an ideal of L if

(i) I and

(ii) For any I and xL

Definition 1.2.18 . [19] For any non-empty subset S of an ADL L,

(S] = { n is a positive integer} is the smallest ideal of L containing S.

Definition 1.2.19. [19] For any x , (x] = ({x}] = {x t: t L} and (x] is called the principal

ideal generated by x.

Definition 1.2.20.[19] The set of all ideals of an is closed under arbitrary intersection an

d contains L. Thus (I(L), ) is a complete lattice where I J ={x y : xI, y J} and I J = IJ for

any I, J I(L). Since, for any x, y L, (x] (y] = (x y] and (x] (y] = (x y], the set PI(L) of all

principal ideals of L is a sub lattice of.

Definition 1.2.21.[8] An L = (L , m) with a maximal element m is called a complete if is

a complete sub lattice of), or equivalently, [0, m] is a complete distributive lattice.

Definition 1.2.22.[18] Let A be any set, A mapping C: su(A)(A) is called a closure

operator on A if it satisfies the following for any x, y p(A)

1. x C(x) (extensive)

2. C(C(x)) C(x) (idempotent)

3. xyC(x)

⊆

C(y) (isotone)

Definition 1.2.23. [6] Let (L,

∨

,

∧

, 0) be an ADL and A mapping

∇

: L → L is called a

closure operator on L if, for any x, y L, the following conditions hold:

1. xx

2. ∇

x

3. x y

∇

x ≤

∇

y

Definition 1.2.24. [9] Let L be a complete ADL with a maximal element m. Then a mappin

g: L L is said to be a closure operator of L if, for any x, y L, the following conditions hold:

(1) (x) m

(2) (x) x = x

(3) If x y, then x) (y)

(4)(xy) = (y x)

(5) ((x)) = (x).

Definition 1.2.25. [9] Let L be a complete ADL with a maximal element m, and a

closure operator of L. Then an element x L said to be closed under, if (x) = x.

Lemma 1.2.26. [9] Let L be a complete ADL with a maximal element m, then m is

closed under every closure operator of L.

Lemma 1.2.27.[9] Let L be a complete ADL with a maximal element m, be a closure

operator of L and { J} be a family of elements of L closed under in L. Then (is also an

element of L closed under in L.

Remark 1.2.28 [9] Let L be a complete ADL with a maximal element m, define

t and: L L by t(x) = x m and (x) = m for all x L, then t, are closure operators of L. Where t

and are called the least and greatest elements of all closure operators defined in L

respectively.

Theorem 1.2.29. [9] Let (L) be the set of all closure operators of L and for any ,L), defin

e if and only if (x) (x) for all x L.Then ((L),) is a complete lattice in which the greatest

element is and least element is t.

Lemma 1.2.30. [9] Let L be a complete ADL with a maximal element m, L such that m m

and define: L L by (x) = , if = x and (x) = m, if x for all x L, then is a closure operator of

L.

Theorem 1.2.31.[9] Let L be a complete ADL with a maximal element m and (L) be the

set of all closure operators of L. Then we have the following:

(1) If {:(L) and =, then, for any x L,(x) = x if and only if (x) = x for all.

(2) If such that, then is a dual atom of (L).

(3) Every dual atom of (L) is of the form b for some b L such that b < m.

(4) For ,(L), if and only if, for any x L, (x) = x implies (x)= x.

CHAPTER TWO

CLOSURE OPERATORS ON COMPLETE ALMOST DISTRIBUTIVE LATTICES

In this chapter, we deal with the dual atoms of the lattice (L),), where (L) is the set of

all closure operators of a complete ADL. We define the concept of a completely meet-

irreducible element in a complete ADL and we prove a necessary and sufficient condition

for a dual atom (where such that) to have a complement in the lattice ((L),).

Definition 2.1. Let L be complete ADL with a maximal element m and a closure operator of L. Define =

{ x L:(x) = x} . That is, is the set of elements of L closed under.

Lemma 2.2. Let L be a complete ADL with a maximal element m, [0, m] such that M’ M

for all M’ M and for each x L, define : L L by (x) = I, where ={ y M: y x = x}.Then is a

closure operator of L.

Proof: (1) Clearly (x) m for all x L, (since (x)

∈

M for all x

∈

L and M

⊆

[0, m]

(2) By our assumption, we get that (x) M for all x L and hence(x) x = x.

(3) Let L such that. Let= {y M: y x = x}.

Suppose y . Then y = and hence y= y = = (because 1.2.8 ) y . ThereforeThus.()().

(4) Let z. Then z xy = xy and hence z xy x = xyx.

Therefore z y x = y x. Hence z . Thus. ……..

Conversely, let zThenand hence Therefore z x y = x y. Hence z. Thus …Therefore (By

and .

(5) BY 2(x)x = x (x) (x) ((x))…...

We have, (x) (x) = (x) (by 2), we get that (x). Hence (x). Therefore ((x)) (x). …... Thus ((x)

) = (x) (By and .

Therefore is a closure operator of L.

Theorem 2.3. Let L be a complete with a maximal element m and [0, m]. Then there is

a closure operator of L such that M = if and only if M’ M for all M’M.

Proof: Suppose is a closure operator of L and M =. Let M’M. Suppose x = M’. Since

every element of M’ is closed under and by Lemma 1.2.27, the of closed elements

under is again closed under, we get that x M. then, M’ M

Conversely, suppose that M’ M for all M’M. Now, we prove that there exists a closure

operator of L such that M = For each x L, define: L L by (x) = {y M : y x =x}. Then by

Lemma 2:2, we get that is a closure operator of L. Let x Then (x) = x. Now, {y M: y x =

x} M implies that (x) = {y M: y x = x} M, by our assumption. Thus (x) M. That is, x M. He

nce M …..

Now, suppose x M.So that (x)x. Thus (x) = (x) x = x. Therefore x . Hence M

⊆

Thus M =. (B

y and

Lemma 2.4. Let L be a complete ADL with a maximal element m. Let

∈

(L). Then if and

only if

⊆

.

Proof: -

⟹

Let

∈

(L). Suppose Then) for each x

∈

L. (by theorem 1.2.29). Let x

∈

. Then

and hence. So that. Therefore x

∈

. Thus

⊆

Conversely, suppose that

⊆

. Now, we prove that. Let L such that. Then x

∈

. Hence x

∈ . There fore Thus (by Theorem 1.2.31(4)).

Lemma 2.5. Let L be a complete ADL with a maximal element m. Let

). Then =

Proof: Let x L. Then J (by Theorem 1.2.31(1)x for all α

∈

J

⟺

x

∈

.Thus =

Definition 2.6 . An element b of a complete lattice L is called a dual atom if b m, where

m is a maximal element of L. And L is called a dually atomic lattice if there can be found

for any element a (m) a dual atom y such that ay.

Definition 2.7 Let L be a complete lattice. An elementL is said to be dual atomistic, if it

is the of set of all dual atoms above it.

Definition 2.8. A closure operator of a complete ADL L is called dual atom if for any

closure operator of L, then either

Theorem 2.9. Let L be a complete ADL with a maximal element m. Then the lattice is

dual atomistic.

Proof: Let and. Write =. Choose y L such that (y) m. Write x = (y).Then. Therefore x . He

nce Also, by Theorem 1.2.31 (2). is a dual atom of for all x . Now, we prove that

Let x and y L such that (y) = y and hence) or y = m. If , then = = x = y (since x ). If ,

then = m. That is, = y. Thereforeall x. (by Theorem 1.2.31(4)) Hence Let such that = y.

If, then y.Now, ( Hence (= ( y =y. Thus ………..

(by Theorem 1.2.31(4)). Therefore (by and

Let B = { is a dual atom and Let C = {: x }. Let x . Then and is dual atom. Therefore B

and hence. Thus ϕ Therefore ϕ. Thus is the infimum of set of all dual atoms about it.

Hence is dual atomistic.

Note: For any closure operator of a complete ADL L and by condition (4) of Definition 1

2.24.

Definition 2.10. Let L be complete with a maximal element m and be any member of,

by a complement of is any satisfying t and

Then is called the complemented element of.

Lemma 2.11. Let L be complete with a maximal element m and L such that . If is a

complemented element of and if is the complement of , then ().

Proof: Since, we get that suppose () =. Also, by Lemma 1.2.30 we have () =. Then, by Th

eorem 1.2.31 (1), we get that

m = () = = which is a contradiction. Therefore

Definition 2.12 . Let L be a complete ADL with a maximal element m. Let L such that m m

Then m is said to be meet-irreducible, if

∧

m = b

∧

c

∧

m, then either or.

Definition 2.13. Let L be a complete ADL with a maximal element m and x L such that x

m m. Then x m is said to be completely meet-irreducible, if where then for some J.

Theorem 2.14. Let L be a complete with at least two elements and let such that ≠ m.

Then is complemented element of (L) if and only if is completely meet- irreducible

element of L.

Proof: Let such that. Suppose (L) is complemented element of (L) and suppose. Let { α

J}

⊆

Lsuch that. We prove that for some We have for all. Supposefor all . Then.

Hence, by Lemma 2.11, we get that. Now, Since, we get that () = m. Hence, Now, = (

. Hence=, which is a contradiction. Therefore, there exists α

∈

J such that. Thus is comp

letely meet-irreducible. Conversely, assume that is completely meet-irreducible. Let B

= {. Since m, we get that. Hence Let = .Now, we prove that is a complement of in the

lattice ((L),

∨

,

∧

). Let x L. If , then m (since t(x). If, then and hence. Now, (Thus = =

t(x) for all x L. Therefore = t. Now, we prove that Let b B. Then bm and m. So that ( fo

r all b B. Since is completely meet-irreducible, we get that

()≠

∧

m. Thus) . Let Then = x and

hence .

Suppose m. We have x

∈. So that (x) = x and hence Therefore () =

∧

m. This is not true. Hence. Thus = {m}. So

that = (since ={x L:(x) = x} = {m}). Hence, by Lemma 2.5, = and, by Lemma 2.4,

∨

φ= ω.

Therefore is the complement of in the lattice

Theorem 2.15 . If L is a complete with at least two elements and such that, then is

complemented if and only if ), where B ={

Proof: Let such that. Then, by Theorem 1.2.31 (2) is a dual atom of L. Suppose is

complemented. Let. Then from the proof of the Theorem 2.14, we get that is a

complement of and () >. Hence (a)= . Thus) >. Conversely, assume the condition.

Now, we prove that is complemented. It is enough to prove that is completely meet-

irreducible element in L. Let {J}

⊆

L such that () we prove that there exists J such that. Su

ppose for allα

∈

J and hence for all α

∈

J. Therefore {: α

∈

J }

⊆

B. Now, = (since

∧

m

∧

a

∧

m = Thus, which is a contradiction. Therefore there exists α

∈

J such that. Hence

is completely meet-irreducible. Therefore is complemented.

Theorem 2.16. If is a dual atom of (L), then there is at most one complement of

Proof: Suppose 0 is not completely meet-irreducible, then by Theorem 2.14, we get that

is not complemented. Suppose 0 is completely meet-irreducible, and then by the proof

of the Theorem 2.14, is complemented. Also =, where B =} is the complement ofand (0

0. Now, we prove that has at most one complement. Suppose is another complement o

f. Let. If, then and hence, which is a contradiction. Therefore, We have =, so that (x)

=

(x) = . Hence (x) = (x)

∧

x

∧

m = x m, by condition (2) of definition 1.2.24. We have x m

(x) for all x L (by theorem 2.11). If x = 0, then 0 = 0m (0). If 0, then xm (x).Therefore (x)

0 for all x L and hence (x) L – {0} = B. Hence ((x)) = ()((x)) (x) ( (x) ) = (x)

∧

m = (x) , since(x) (x) = (x). Since x

∧

m ≤ (x) for all x L, we get that (x ) (x) ) (x) for all x L. Hence…... Let L-{0}. Then = = since, so that = m = Suppose (x) = x for x L-{0} = B. Case (i) = x.

Then x = . Hence (x) = = = = x. Case (ii) Then . Hence (x) = (m) = m x. Thus we have proved that ( ) for all x B. By Theorem 1.2.31 (4), we get that fo

r all x B. Hence=. Therefore Thus

.( by and .

Lemma 2.17 Let L be a complete ADL. Then, for any (L) and x L we have

(i)

(ii)

Proof: i Let (L) and x L. Then (by lemma 1.2.12 (10))

and hence

Therefore

ii. Since and hence

Corollary 2.18 . Let L be a complete ADL. Then, for any (L) and xL we have

(i)

(ii)

Proof: i. Let (L) and x L. Then (by lemma 1.2.12 (10))

and hence

Therefore

ii. Since and hence

Theorem 2.19. is complemented, then there is at least one complement of preceding .

Proof: be a complement of. Let x L. Then. Hence t. Now, we prove that

Case (i) Then By above corollary 2.18 (ii), we get that ()((( Therefore (

Case (ii) ifAgain, by above Lemma 2.17(i), we get Therefore Thus

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