chapter 6. order relations and structure

Chapter 6. Order Relations and Structure

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： A309

mailto:[email protected]

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6.1. Partially Ordered Sets 6.2. Extremal Elements of Partially Ordered Sets 6.3. Lattices 6.4. Finite Boolean Algebras 6.5. Functions on Boolean Algebras 6.6. Circuit Design

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Chapter six: Order Relations and Structures

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Partial Order A relation R on a set A is called a partial order if R is reflexive,

antisymmetric and transitive. The set A together with the partial order R is called a partially ordered set, or simply a poset, denoted by (A, R)

For instance,

1.Let A be a collection of subsets of a set S. The relation ⊆ of set inclusion is a partial order on A, so (A, ) is a poset. ⊆

2.Let Z+ be the set of positive integers. The usual relation ≤

is a partial order on Z+, as is “≥”

6.1 Partially Ordered Sets

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Example 6 Let R be a partial order on a set A, and let R-1 be the

inverse relation of R. Then R-1 is also a partial order.Proof:

(a)Reflexive Δ ⊆ R Δ=Δ-1 ⊆ R-1

(b)Antisymmetric R ∩ R-1 ⊆ Δ R-1 ∩ R ⊆ Δ

(c)Transitive R2 ⊆ R (R-1)2 ⊆ R-1

Thus, R-1 is also a partial order.

The poset (A, R-1) is called the dual of the poset (A, R).

whenever (A, ≤) is a poset, we use “≥” for the partial order ≤-1


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Comparable If (A, ≤) is a poset, elements a and b of A are comparable if

a ≤ b or b ≤ a

In some poset, e.g. the relation of divisibility (a R b iff a | b), some pairs of elements are not comparable

2 | 7 and 7 | 2

Note: if every pair of elements in a poset A is comparable, we say that A is linear ordered set, and the partial order is called a linear order. We also say that A is a chain.


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Theorem 1 If (A, ≤ ) and (B, ≤ ) are posets, then (A×B, ≤) is a poset, with

partial order ≤ defined by

(a, b) ≤ (a’, b’) if a ≤ a’ in A and b≤b’ in B

Note: the ≤ is used to denote three different partial orders.

Proof:

(a) Reflexive

support (a, b) in A×B, then

(a, b) ≤ (a, b) since a ≤ a in A and b ≤ b in B ((A, ≤) and (B, ≤) are posets)


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(b) Antisymmetry

support (a, b) ≤ (a’, b’) and (a’, b’) ≤ (a, b), then

a ≤ a’ and a’ ≤ a in A ; b ≤ b’ and b’ ≤ b in B

since A and B are posets, a=a’, b=b’ (antisymmetry property in A and B, respectively), which means that (a, b)=(a’, b’) and thus ≤ satisfies the antisymmetry property in A×B

(c) Transitive

support (a, b) ≤ (a’, b’) and (a’, b’) ≤ (a’’, b’’), then

a ≤ a’ and a’ ≤ a’’ in A ; b ≤ b’ and b’ ≤ b’’ in B ,

since A and B are posets, a ≤ a’’ and b ≤ b’’ (transitive property in A and B, respectively), which means that

(a, b) ≤ (a’’, b’’)


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Product partial order The partial order ≤ defined on the Cartesian product A× B is called the

Product partial order

The symbol < If (A, ≤ ) is a poset, we say a<b if a ≤ b but a ≠ b

Lexicographic (dictionary) order

Another useful partial order on A×B, denoted by ≺, is defined as (a, b) ≺ (a’, b’) if a<a’ or a=a’ and b ≤ b’

why ≺ is a partial order?


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Example 8

Let A=R, with the usual order ≤. Then the plane R2=R×R may be given lexicographic order


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xx=x2x=x1

p1

p2

p3

P1 ≺ P2P1 ≺ P3 (x1 ≤ x2, x1 ≠x2) P2 ≺ P3 (x2=x2, y2 ≤ y3)

P1=(x1,y1)

P2=(x2,y2) P3=(x2,y3)

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Lexicographic ordering is easily extended to Cartesian products A1×A2 …×An as follows:

(a1, a2, …, an) ≺ (a’1, a’2, …, a’n) if and only if

a1 < a’1 or

a1 = a’1and a2 < a’2 or

a1 = a’1and a2 = a’2 and a3 < a’3 or …

a1 = a’1and a2 = a’2 … , an-1 = a’n-1, an ≤ a’n


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Why?

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Theorem 2 The digraph of a partial order has no cycle of length larger than 1

Proof: support that the digraph of the partial order ≤ on the set A contains a cycle of length n>=2. Then there exist distinct elements a1,a2,…,an in A such that

a1 ≤a2, a2 ≤a3,…, an-1 ≤an, an ≤a1

by the transitivity of the partial order, used n-1 times,

a1 ≤an

by antisymmetry, an ≤a1and a1 ≤an then a1=an

(Contradiction)


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Hasse Diagrams Just a reduced version of the diagram of the partial order of

the poset.

a) Reflexive

Every vertex has a cycle of length 1 (delete all cycles)


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Transitive a ≤ b, and b ≤c, then a ≤c (delete the edge from a to c)


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b

a

c

Vertex dot

a

b

c

Remove arrow (all edges pointing upward)

a

b

c

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Example 11

Let A={1,2,3,4,12}. Consider the partial order of divisibility on A. Draw the corresponding Hasse diagram.


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12

4

2

1

3

12

4

2

1

3

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Example 12 Let S={a,b,c} and A=P(S). Draw the Hasse diagram of

the poset A with the partial order ⊆


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{a,b,c}

{b,c} {a,b} {a,c}

{b}{c}

{a}

ф

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The Hasse diagram of a finite linearly order set

Let A={a1,a2,…,an} be a finite set, and ai ≤aj if i ≤j


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a1

a2

an-1

an

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Example 13 Fig. a shows the Hasse diagram of a poset (A, ≤), where

A={a, b, c, d, e, f}

Fig. b shows the Hasse diagram of the dual poset (A, ≥)


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f

d e

ba c

ba c

d e

f

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Topological Sorting If A is a poset with partial order ≤, we sometimes need

to find a linear order ≺ for the set A that will merely be an extension of the given partial order in the sense that if a ≤ b, then a ≺b. The process of constructing a linear order such as is called topological sorting. ≺

(refer to p.229 for the details of the algorithm)


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Example 14

Give a topological sorting for the poset whose Hasse diagram as follows


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f

e

b d

a

c

g

a

b

c

d

e

g

f

a

c

g

b

d

e

f

a

d

b

e

c

g

f

Usually, the topological sorting is not unique.

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Isomorphism Let (A, ≤) and (A’, ≤’) be posets and let f: AA’ be a one to

one correspondence between A and A’. The function f is called an isomorphism from (A, ≤) to (A’, ≤’) if, for any a and b in A,

a ≤ b if and only if f(a) ≤’ f(b)

If f: AA’ is an isomorphism, we say that (A, ≤ ) and (A’, ≤’) are isomorphic posets.


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Example 15

Let A be the set Z+ of positive integers, and let ≤ be the usual partial order on A. Let A’ be the set of positive even integers, and let ≤’ be the usual partial order on A’. The function f: AA’ given by

f(a) = 2 a

is an isomorphism form (A, ≤) to (A’, ≤’)

Proof: First, it is very to show that f is one to one, everywhere defined and onto (one to one correspondence).

Finally, if a and b are elements of A, then it is clear that

a ≤ b if and only if 2a ≤’ 2b. Thus f is an isomorphism.


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Theorem 3 Suppose that f : AA’ is an isomorphism from a poset (A,

≤) to a poset (A’, ≤’). Suppose also that B is a subset of A, and B’=f(B) is the corresponding subset of A’. The following principle must hold.

If the elements of B have any property relating to one another or to other elements of A, and if this property can be defined entirely in terms of the relation ≤, then the elements of B’ must possess exactly the same property, defined in terms of ≤’.


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Example 16 Let (A, ≤ ) be the poset whose Hasse diagram is shown below,

and suppose that f is an isomorphism from (A, ≤) to some other poset (A’, ≤’). Note d ≤ x for any x in A, then the corresponding element f(d) in A’ must satisfy the property f(d) ≤y for all y in A’.


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d

c

a b

As another example, a ≤ b and b ≤ a. Such a pair is called incomparable in A, then f(a) and f(b) are also incomparable in A’

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Let (A, ≤ ) and (A’, ≤ ’) be finite posets, let f: AA’ be a one-to-one correspondence, and let H be any Hasse diagram of (A, ≤ ). Then

If f is an isomorphism and each label a of H is replaced by f(a), then H will become a Hasse diagram for (A, ≤ ’)

Conversely

If H becomes a Hasse diagram for (A’, ≤ ’), whenever each label a is replaced by f(a), then f is an isomorphism.

Two finite isomorphic posets have the same Hasse diagrams


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Example 17 Let A={1,2,3,6} and let ≤ be the relation | .

Let A’= {ф, {a}, {b}, {a, b}} and let ≤’ be set containment, . ⊆ If f(1)= ф, f(2)={a}, f(3)={b}, f(6)={a, b}, then f is an

isomorphism. They have the same Hasse diagrams.


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1

32

6

ф

{b}{a}

{a, b}

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Homework Ex. 2, Ex. 14, Ex. 16, Ex. 18, Ex. 19 , Ex. 30, Ex. 32


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Consider a poset (A, ≤ ) Maximal Element An element a in A is called a maximal element of A if

there is no element c in A such that a<c.

Minimal Element An element b in A is called a minimal element of A if

there is no element c in A such that c<b.

an element a in A is a maximal (minimal) element of (A, ≥ ) if and only if a is a minimal (maximal) element of (A, ≤ )

6.2 Extremal Elements of Partially Ordered Sets

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Example 1

Find the maximal and minimal elements in the following Hasse diagram


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a1 a2

a3

b1 b2

b3

Maximal elements

Minimal element

Note: a1, a2, a3 are incomparable b1, b2, b3 are incomparable

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Example 2 Let A be the poset of nonnegative real number with the

usual partial order ≤ . Then 0 is a minimal element of A. There are no maximal elements of A

Example 3 The poset Z with the usual partial order ≤ has no

maximal elements and has no minimal elements


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Theorem 1 Let A be a finite nonempty poset with partial order ≤.

Then A has at least one maximal element and at least one minimal element.

Proof: Let a be any element of A. If a is not maximal, we can find an element a1 in A such that a<a1. If a1 is not maximal, we can find an element a2 in A such that a1<a2. This argument can not be continued indefinitely, since A is a finite set. Thus we eventually obtain the finite chain

a <a1<a2<…<ak-1<ak

which cannot be extended. Hence we cannot have ak < b for any b in A, so Ak is a maximal element of (A, ≤).


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Algorithm For finding a topological sorting of a finite poset (A ≤).

Step 1: Choose a minimal element a of A

Step 2: Make a the next entry of SORT and replace A with A-{a}

Step 3: Repeat step 1 and 2 until A={ }.


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Example 4


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a b

c

d e

SORT:

a b

c

e

d

a b

c

e

a b

c

a

b a

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Greatest element An element a in A is called a greatest element of A if

x ≤ a for all x in A.

Least element An element a in A is called a least element of A if

a ≤ x for all x in A.

Note: an element a of (A, ≤ ) is a greatest (or least) element if and only if it is a least (or greatest) element of (A, ≥ )


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Example 5

Let A be the poset of nonnegative real number with the usual partial order ≤ . Then 0 is a least element of A. There are no greatest elements of A

Example 7

The poset Z with usual partial order has neither a least nor a greatest element.


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Theorem 2 A poset has at most one greatest element and at most one least

element. Proof: Support that a and b are greatest elements of a poset A. since

b is a greatest element, we have a ≤ b;

since a is a greatest element, we have b ≤ a; thus

a=b by the antisymmetry property. so, if a poset has a greatest element, it only has one such element.

This is true for all posets, the dual poset (A, ≥) has at most one greatest element, so (A, ≤) also has at most one least element.


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Unit element The greatest element of a poset, if it exists, is denoted

by I and is often called the unit element.

Zero element

The least element of a poset, if it exists, is denoted by 0 and is often called the zero element.

Q: does unit/zero element exist for a finite nonempty poset?


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Consider a poset A and a subset B of A Upper bound An element a in A is called an upper bound of B if b ≤ a

for all b in B

Lower bound

An element a in A is called a lower bound of B if a ≤ b for all b in B


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Example 8 Find all upper and lower bounds of the following subset

of A: (a) B1={a, b}; B2={c, d, e}


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h

f g

d e

c

a b

B1 has no lower bounds; The upper bounds of B1 are c, d, e, f, g and h

The lower bounds of B2 are c, a and bThe upper bounds of B2 are f, g and h

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Let A be a poset and B a subset of A, Least upper bound

An element a in A is called a least upper bound of B, denoted by (LUB(B)), if a is an upper bound of B and a ≤a’, whenever a’ is an upper bound of B.

Greatest lower bound An element a in A is called a greatest lower bound of B,

denoted by (GLB(B)), if a is a lower bound of B and a’ ≤ a, whenever a’ is a lower bound of B.


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Some properties of dual of poset The upper bounds in (A, ≤ ) correspond to lower

bounds in (A, ≥) (for the same set of elements)

The lower bounds in (A, ≤ ) correspond to upper bounds in (A, ≥) (for the same set of elements)

Similar statements hold for greatest lower bounds and least upper bounds.


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Example 9 Find all least upper bounds and all greatest lower

bounds of (a) B1={a, b} (b) B2={c, d, e}


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h

f g

d e

c

a b

(a) Since B1 has no lower bounds, it has no greatest lower bounds; However, LUB(B1)=c

(b)Since the lower bounds of B2 are c, a and b, we find that GLB(B2)=c The upper bounds of B2 are f, g and h. Since f and g are not comparable, we conclude that B2 has no least upper bound.

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Theorem 3 Let (A, ≤) be a poset. Then a subset B of A has at most

one LUB and at most one GLB

Please refer to the proof of Theorem 2.


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Example 10 Let A={1,2,3,…,11} be the poset whose Hasse diagram

is shown below. Find the LUB and GLB of B={6,7,10}, if they exist.


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1

23

4

5 6 7 8

109

11 The upper bounds of B are 10, 11, and LUB(B) is 10 (the first vertex that can be Reached from {6,7,10} by upward paths )

The lower bounds of B are 1,4, and GLB(B) is 4 (the first vertex that can be Reached from {6,7,10} by downward paths )

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Theorem 4 Suppose that (A, ≤) and (A, ≤’) are isomorphic posets under

the isomorphic f: AA’

1. If a is a maximal (minimal) element of (A, ≤), then f(a) is a maximal (minimal) element of (A’, ≤’)

2. If a is the greatest (least) element of (A, ≤), then f(a) is the greatest (least) element of (A’,≤’)

3. If a is an upper (lower, least upper, greatest lower) bound of a subset B of A, then f(a) is an upper (lower, least upper, greatest lower) bound for subset f(B) of A’

4. If every subset of (A, ≤) has a LUB (GLB), then every subset of (A’, ≤’) has a LUB (GLB)


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Example 11

Show that the posets (A, ≤) and (A’, ≤’), whose Hasse diagrams are shown below are not isomorphic.


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a

b c

a’ b’

c’

(A, ≤) has a greatest element a, while (A’, ≤’) does not have a greatest element

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Homework

Ex. 2, Ex. 18, Ex. 22, Ex. 28, Ex. 34, Ex. 37


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Lattice A lattice is a poset (L, ≤) in which every subset {a, b}

consisting of two elements has a least upper bound and a greatest lower bound. we denote

LUB({a, b}) by a ∨ b (the join of a and b)

GLB({a, b}) by a ∧b (the meet of a and b)

6.3 Lattices

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Example 1 Let S be a set and let L=P(S). As we have seen, ⊆ , containment, is

a partial order on L. Let A and B belong to the poset (L, ⊆ ). Then

a ∨ b =A U B & a ∧b = A ∩ B

Why?

Assuming C is a upper bound of {a, b}, then

A ⊆ C and B ⊆ C thus A U B ⊆ C

Assuming C is a lower bound of {a, b}, then

C ⊆ A and C ⊆ B thus C ⊆ A ∩ B

6.3 Lattices

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Example 2 Consider the poset (Z+, ≤), where for a and b in Z+, a ≤

b if and only if a | b , then

a∨ b = LCM(a,b)

a∧ b = GCD(a,b)

LCM: least common multiple

GCD: greatest common divisor

6.3 Lattices

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Example 3

Let n be a positive integer and Dn be the set of all positive divisors of n. Then Dn is a lattice under the relation of divisibility. For instance,

D20= {1,2,4,5,10,20} D30= {1,2,3,5,6,10,15,20}

6.3 Lattices

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1

2 5

4 10

20

1

2 5

6 15

30

3

10

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Example 4 Which of the Hasse diagrams represent lattices?

6.3 Lattices

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a

b

c

d

a

b c

d

e

f g

a

b

c

e

d

a

b c

e

d

a

b c

d

a

b c

d e

f

a

c db

a

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Example 6 Let S be a set and L =P(S). Then (L, ⊆ ) is a lattice, and

its dual lattice is (L, ), where “⊇ ⊆ ” is “contained in”, and “⊇ ” is “contains”. Then, in the poset (L, ⊇ )

join: A B=A∩B , ∨ meet: A B=A B.∧ ∪

6.3 Lattices

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Theorem 1 If (L1, ≤ ) and (L2, ≤ ) are lattices, then (L, ≤ ) is a lattices,

where L= L1 ×L2, and the partial order ≤ of L is the product partial order.

Proof: we denote

the join and meet in is L1by ∨ 1 and ∧ 1

the join and meet in is L2by ∨ 2 and ∧ 2

We know that L is a poset (Theorem 1 in p.219)

for (a1,b1) and (a2,b2) in L. then

(a1,b1) ∨ (a2,b2) = (a1 ∨1 a2, b1 ∨2 b2 ) in L

(a1,b1) ∧ (a2,b2) = (a1 ∧ 1 a2, b1 ∧ 2 b2 ) in L

6.3 Lattices

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Example 7

6.3 Lattices

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× =01

I1

L1

ba

02

I2

L2

(I1 , I2)

(I1 ,a) (I1 ,b)(01 ,I2)

(01 ,a)

(01 ,02 )

(01 ,b )(I1 ,02 )

L1×L2

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Sublattice Let (L, ≤) be a lattice. A nonempty subset S of L is

called a sublattice of L if a ∨ b in S and a ∧ b in S whenever a and b in S

For instance

Example 3 is one of sublattices of Example 2

6.3 Lattices

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Example 9

6.3 Lattices

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0

I

e f

ca b

0

I

e f

a b

I

e f

a b 0

cba

c

Not a lattice

a lattice, not a Sublattice

a sublattice

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Isomorphic Lattices If f: L1L2 is an isomorphism form the poset (L1, ≤1 )

to the poset (L2, ≤ 2) , then L1 is a lattice if and only if L2 is a lattice. In fact, if a and b are elements of L1, then

f(a ∨ b)= f(a) ∨ f(b) & f (a ∧ b)=f(a) ∧ f(b).

If two lattices are isomorphic, as posets, we say they are isomorphic lattices.

6.3 Lattices

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Example 10 (P.225 Ex.17) Let A={1,2,3,6} and let ≤ be the relation | .

Let A’= {ф, {a}, {b}, {a, b}} and let ≤’ be set containment, . ⊆ If f(1)= ф, f(2)={a}, f(3)={b}, f(6)={a, b}, then f is an

isomorphism. They have the same Hasse diagrams.

6.3 Lattices

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1

32

6

ф

{b}{a}

{a, b}

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a ∨ b (LUB{a, b}) 1. a ≤ a∨ b and b ≤ a∨ b; a∨ b is an upper bound of a and b

2. If a ≤ c and b ≤ c, then a ∨ b ≤ c ; a ∨ b is the least upper bound of a and b

a ∧ b (GLB{a, b}) 3. a∧ b ≤ a and a ∧ b ≤ b; a ∧ b is a lower bound of a and b

4. If c≤ a and c ≤ b, then c≤ a ∧ b ; a ∧ b is the greatest lower bound of a and b

6.3 Lattices

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Theorem 2 Let L be a lattice. Then for every a and b in L

(a) a ∨ b =b if and only if a ≤ b

(b) a ∧ b =a if and only if a ≤ b

(c) a ∧ b =a if and only if a ∨ b =b

Proof:

(a) if a∨ b =b, since a≤ a∨ b, thus a ≤ b

if a ≤ b, since b ≤ b , thus b is a upper bound of a and b, by definition of least upper bound we have a∨ b ≤ b. since a∨ b is an upper bound of a and b, b ≤ a∨ b, so a∨ b =b

(b) Similar to (a); (c) the proof follows from (a) & (b)

6.3 Lattices

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Example 12 Let L be a linearly ordered set. If a and b in L, then

either a ≤ b or b ≤ a. It follows form Theorem 2 that L is a lattice, since every pair of elements has a least upper bound and a greatest lower bound.

6.3 Lattices

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Theorem 3Let L be a lattice. Then

1. Idempotent properties: a∨ a =a ; a∧ a =a

2. Commutative properties: a∨ b= b∨ a; a∧ b = b∧ a

3. Associative properties: (a) (a∨ b)∨ c= a∨ (b∨ c )

(b) (a∧ b) ∧ c= a∧ (b∧ c)

4. Absorption properties: (a) a ∨ (a ∧ b) =a

(b) a ∧ (a ∨ b) =a

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Proof: 3. (a) (a∨ b)∨ c= a∨ (b∨ c)

a≤ a∨ (b∨ c) & b∨ c≤ a∨ (b∨ c)

b≤ b∨ c & c ≤ b∨ c (definition of LUB)

b≤ b∨ c & c ≤ b∨ c & b∨ c≤ a∨ (b∨ c)

b≤ a∨ (b∨ c) & c ≤ a∨ (b∨ c) (transitivity)

a≤ a∨ (b∨ c) & b≤ a∨ (b∨ c) a∨ (b∨ c) is a upper of a and b

then we have a∨ b ≤ a∨ (b∨ c) (why?)

a∨ b ≤ a∨ (b∨ c) & c ≤ a∨ (b∨ c) a∨ (b∨ c) is a upper of a∨ b and c

then we have (a∨ b)∨ c ≤ a∨ (b∨ c)

Similarly, a∨ (b∨ c) ≤ (a∨ b)∨ c

Therefore (a∨ b)∨ c = a∨ (b∨ c) (why?)

6.3 Lattices

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(a ∨ b) ∨ c = a ∨ (b ∨ c) = a ∨ b ∨ c

(a ∧ b) ∧ c = a ∧ (b ∧ c) = a ∧ b ∧ c

LUB({a1,a2,…,an})= a1 ∨ a2 ∨… ∨ an

GLB({a1,a2,…,an}) =a1 ∧ a2 ∧… ∧ an

6.3 Lattices

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Theorem 4 Let L be a lattice. Then, for every a, b and c in L

1. If a ≤ b, then

(a) a ∨ c ≤ b ∨ c

(b) a ∧c ≤ b ∧ c

2. a ≤ c and b ≤ c if and only if a ∨ b ≤ c

3. c ≤ a and c ≤ b if and only if c ≤ a ∧ b

4. If a ≤b and c ≤d, then

(a) a∨ c ≤ b∨ d

(b) a ∧ c ≤ b∧ d

6.3 Lattices

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Proof

1. (a) If a ≤ b, then a c ≤ b c ∨ ∨ c ≤ b c ; b≤ b c ∨ ∨ (definition of LUB)

a ≤ b ; b≤ b c ∨ a≤ b c ∨ (transitivity)

therefore,

b c is a upper bound of a and c , which means ∨ a c ≤ b c ∨ ∨ (why? )

The proofs for others left as exercises.

6.3 Lattices

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Bounded

A lattice L is said to be bounded if it has a greatest element I and a least element 0

For instance:

Example 15: The lattice P(S) of all subsets of a set S, with the relation containment is bounded. The greatest element is S and the least element is empty set.

Example 13: The lattice Z+ under the partial order of divisibility is not bounded, since it has a least element 1, but no greatest element.

6.3 Lattices

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If L is a bounded lattice, then for all a in A

0 ≤ a ≤ I

a ∨ 0 = a, a ∨ I = I

a ∧ 0 = 0 , a ∧ I = a

Note: I (0) and a are comparable, for all a in A.

6.3 Lattices

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Theorem 5

Let L={a1,a2,…,an} be a finite lattice. Then L is bounded.

Proof:

The greatest element of L is a1 ∨ a2 ∨… ∨ an, and the least element of L is a1 ∧ a2 ∧… ∧ an

6.3 Lattices

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Distributive

A lattice L is called distributive if for any elements a, b and c in L we have the following distributive properties:

1. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)

2. a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c)

If L is not distributive, we say that L is nondistributive.

Note: the distributive property holds when

a. any two of the elements a, b and c are equal or

b. when any one of the elements is 0 or I.

6.3 Lattices

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Example 16 For a set S, the lattice P(S) is distributive, since union

and intersection each satisfy the distributive property.

Example 17 The lattice whose Hasse diagram shown as follows is

distributive.

6.3 Lattices

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0

a c

b d

I

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Example 18 Show that the lattices as follows are nondistributive.

6.3 Lattices

72

0

b

a

I

c

0

a b

I

c

a ( b c) = a I = a∧ ∨ ∧(a ∧ b) (a c) = b 0 = b∨ ∧ ∨

a ( b c) = a I = a∧ ∨ ∧(a ∧ b) (a c) = 0 0 = 0∨ ∧ ∨

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Theorem 6 A lattice L is nondistributive if and only if it contains a

sublattice that is isomorphic to one of the lattices whose Hasse diagrams are as show.

6.3 Lattices

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0

b

a

I

c

0

a b

I

c

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Complement Let L be bounded lattice with greatest element I and

least element 0, and let a in L. An element a’ in L is called a complement of a if

a a’ = I and a a’ =0 ∨ ∧ Note that 0’=I and I’=0

6.3 Lattices

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Example 19

The lattice L=P(S) is such that every element has a complement, since if A in L, then its set complement A has the properties A ∨ A = S and A ∧ A=ф. That is, the set complement is also the complement in L.

Example 20

6.3 Lattices

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0

b

a

I

c

0

a b

I

c

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Example 21

6.3 Lattices

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1

2 5

4 10

20

1

2 5

6 15

30

3

10

D20 D30

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Theorem 7

Let L be a bounded distributive lattice. If a complement exists, it is unique. Proof: Let a’ and a’’ be complements of the element a in L, then

a ∨ a’ = I, a ∨ a’’= I ; a ∧ a’ = 0, a ∧ a’’ =0

using the distributive laws, we obtain

a’= a’ ∨ 0 = a’ ∨(a ∧ a’’ ) = (a’ ∨ a) ∧ (a’ ∨ a’’)

= I ∧ (a’ ∨ a’’) = a’ ∨ a’’

Also

a’’= a’’ ∨ 0 = a’’ ∨(a ∧ a’ ) = (a’’ ∨ a) ∧ (a’’ ∨ a’)

= I ∧ (a’ ∨ a’’) = a’ ∨ a’’

Hence a’=a’’

6.3 Lattices

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Complemented

A lattice L is called complemented if it is bounded and if every element in L has a complement.

6.3 Lattices

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Example 22

The lattice L=P(S) is complemented. Observe that in this case each element of L has a unique complement, which can be seen directly or is implied by Theorem 7.

Example 23

6.3 Lattices

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0

b

a

I

c

0

a b

I

c

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Homework Ex. 6, Ex. 7, Ex. 12, Ex. 19, Ex. 24, Ex. 32, Ex. 40

6.3 Lattices

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Theorem 1 If S1={x1,x2,…,xn} and S2={y1,y2,…,yn} are any two finite

sets with n elements, then the lattices (P(S1), ⊆ ) and (P(S2), ⊆ ) are isomorphic. Consequently, the Hasse diagrams of these lattices may be drawn identically.

6.4 Finite Boolean Algebras

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S1: x1 x2 x3 … xn

S2: y1 y2 y3 … yn

S1: x1 x2 x3 … xn

S2: y1 y2 y3 … yn

A

f(A)

Arrange the elements in S1 and S2

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Example 1: S={a, b, c} and T={2,3,5}. Consider the Hasse diagrams of

the two lattices (P(S), ) and (P(T), ). ⊆ ⊆


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{a,b,c}

{a,b} {b,c}{a,c}

{a}

ф

{c}{b}

{2,3,5}

{2,3} {3,5}{2,5}

{2}

ф

{5}{3}

Note : the lattice depends only on the number of elements in set, not on the elements.

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Label the subsets Let a set S={a1,a2,…,an}, then P(S) has 2n subsets. We

label subsets by sequences of 0’s and 1’s of length n.

For instance,

{a1,a2} 1 1 0 0 …0

{a1,an } 1 0 0 0 …1

ф 0 0 0 0 …0

{a1,a2,…,an} 1 1 1 1 …1

…


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Get the unique Hasse Diagram


84

{2,3,5}

{2,3} {3,5}{2,5}

{2}

ф

{5}{3}

{a,b,c}

{a,b} {b,c}{a,c}

{a}

ф

{c}{b}

111

110 011101

100

000

001010

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Lattice Bn

If the Hasse diagram of the lattice corresponding to a set with n elements is labeled by sequences of 0’s and 1’s of length n, the resulting lattice is named Bn. The properties of the partial order on Bn can be described directly as follows. If x=a1a2…an and y=b1b2…bn are two element of Bn, then

1. x ≤ y iff ak ≤ bk (as numbers 0 or 1) for k=1,2,…,n

2. x ∧ y=c1c2…cn, where ck= min{ak,bk}

3. x ∨ y=c1c2…cn, where ck= max{ak,bk}

4. x has a complement x’=z1z2…zn, where zk=1 if xk=0 and zk=0 if xk=1


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Boolean algebra

A finite lattice is called a Boolean algebra if it is isomorphic with Bn for some nonnegative integer n.


86

0

1

n=1

0110

00

11

n=2

111

110 011101

100

000

001010

n=3n=0

|Bn| =2n

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(P(S), ) ⊆ Each x and y in Bn correspond to subsets A and B of S. Then

x ≤ y, x ∧ y, x ∨ y and x’ correspond to A B, A ∩ B, A ⊆U B and A. Therefore,

(P(S), ) is isomorphic with Bn, where n=|S| ⊆

Example 3

Consider the lattice D6 consisting of all positive integer divisors of 6 under the partial order of divisibility.


871

32

6

00

0110

11

D6 is a Boolean algebras

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Example 4

Consider the lattices D20 and D30 of all positive integer divisors of 20 and 30, respectively.


88

1

2 5

4 10

20

1

2 5

6 15

30

3

10

D20 is not a Boolean algebra(why? 6 is not 2n )

D30 is a Boolean algebra,D30 B3

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Theorem 2 Let n=p1p2…pk, where the pi are distinct primes. The Dn is a

Boolean algebra.

Proof: Let S={p1 , p2 , … , pk}. If T S and a⊆ T is the product of the primes

in T, then aT | n. Any divisor of n must be of the form aT for some subset T of S (let aф=1) .

If V and T are subsets of S, V T if and only if a⊆ V | aT

aV ∩ T = aV a∧ T = GCD(aV,aT) &

aV U T= aV a∨ T = LCM(aV,aT)

Thus, the function f: P(S) Dn given by f(T)=aT is a isomorphism form P(S) to Dn. Since P(S) is a Boolean algebra, so is Dn.


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Example

Let S={2,3,5}, show the Hasse diagrams of (P(S), ) ⊆and D30 as follows.


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1

2 5

6 15

30

3

10

ф

{2} {5}

{2,3}{3,5}

{2,3,5}

{3}

{2,5}

2×3×5=30

3×5=15

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Example 5

Since 210=2×3×5×7, 66=2×3×11 and 646=2×17×19, then D210, D66 D646 are all Boolean algebras.

Example 9

Since 40=23×5, and 75=3×52, neither D40 and D75 are Boolean algebras.

Note: If n is positive integer and p2 | n, where p is a prime number, then Dn is not a Boolean algebra.


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Theorem 3 (Substitution rule for Boolean algebra) Any formula involving U or ∩ that holds for arbitrary subsets of a set S

will continue to hold for arbitrary elements of a Boolean algebra L if is ∧ substituted for ∩ and for U. ∨

Example 6 If L is any Boolean algebra and x,y and z are in L, then the following three properties hold.

1. (x’)’=x 2. (x∧ y)’ = x’ y’ 3. (x y)’ = x’∨ ∨ ∧ y’

This is true by theorem 3,

1. (A)=A 2. (A∩B) =A U B 3. (A U B ) = A ∩ B

hold for arbitrary subsets A and B of a set S.

More properties can be found in p. 247, 1 ~12


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Example 7 Show the lattice whose Hasse diagram shown below is

not a Boolean algebra.


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0

c

f

I

a

e

b

d

a and e are both complements of c

However, based on the 11. Every element x has a unique complement x’ Every element A has a unique complement x

Theorem 3 (e.g. properties 1~14) is usually used to show that a lattice L is not a Boolean algebra.

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Denote the Boolean algebra B1 simply as B. Thus B contains only the two elements 0 and 1. It is a fact that any of the Boolean algebras Bn can be described in terms of B. The following theorem gives this description.

Theorem 4 For any n>=1, Bn is the product B×B×…×B of B, n factors,

where B×B×…×B is given the product partial order.


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Homework Ex. 6, Ex. 10, Ex. 12, Ex. 20, Ex. 24, Ex. 26, Ex. 29


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chapter 6. order relations and structure

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