Partial Order Relations Ioan Despi despi@turing.une.edu.au - - PowerPoint PPT Presentation

Partial Order Relations

Ioan Despi

despi@turing.une.edu.au

University of New England

August 12, 2013

Outline

1 Partial Orderings 2 Totally Ordered Set 3 Special Elements 4 Hasse Diagrams

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Motivation

Partial oderings have important applications, e.g., the analysis of computer programs.

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Motivation

Partial oderings have important applications, e.g., the analysis of computer programs. Trees, that you met earlier, have an inherent partial order.

Ioan Despi – AMTH140 3 of 21

Motivation

Partial oderings have important applications, e.g., the analysis of computer programs. Trees, that you met earlier, have an inherent partial order. In addition, partial orders play an important role in aspects of pure mathematics.

Ioan Despi – AMTH140 3 of 21

Partial Orderings

Definition Let R be a binary relation on a set A. ◮ R is antisymmetric if for all x, y ∈ A, if xRy and yRx, then x = y.

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Partial Orderings

Definition Let R be a binary relation on a set A. ◮ R is antisymmetric if for all x, y ∈ A, if xRy and yRx, then x = y. ◮ R is a partial order relation if R is reflexive, antisymmetric and transitive.

Ioan Despi – AMTH140 4 of 21

Partial Orderings

Definition Let R be a binary relation on a set A. ◮ R is antisymmetric if for all x, y ∈ A, if xRy and yRx, then x = y. ◮ R is a partial order relation if R is reflexive, antisymmetric and transitive. ◮ A partially ordered set (poset) consists of a set together with a partial

• rder relation on it.

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Partial Orderings

Definition Let R be a binary relation on a set A. ◮ R is antisymmetric if for all x, y ∈ A, if xRy and yRx, then x = y. ◮ R is a partial order relation if R is reflexive, antisymmetric and transitive. ◮ A partially ordered set (poset) consists of a set together with a partial

• rder relation on it.

Ioan Despi – AMTH140 4 of 21

Partial Orderings

Definition Let R be a binary relation on a set A. ◮ R is antisymmetric if for all x, y ∈ A, if xRy and yRx, then x = y. ◮ R is a partial order relation if R is reflexive, antisymmetric and transitive. ◮ A partially ordered set (poset) consists of a set together with a partial

• rder relation on it.

In terms of the digraph of a binary relation R, the antisymmetry is tantamount to saying there are no arrows in opposite directions joining a pair of (different) vertices.

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Example

• 1. Let A = {0, 1, 2} and R = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} and

S = {(0, 0), (1, 1), (2, 2)} be two relations on A. Show that (i) R is a partial order relation. (ii) S is an equivalence relation.

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Example

• 1. Let A = {0, 1, 2} and R = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} and

S = {(0, 0), (1, 1), (2, 2)} be two relations on A. Show that (i) R is a partial order relation. (ii) S is an equivalence relation. Solution. We choose to use digraphs to make the explanations in this case. (i) The digraph for R on the right implies the relation R is Reflexive: loops on every vertex. Transitive: if you can travel from vertex v to vertex w along consecutive arrows of the same direction, then there is also a single arrow pointing from v to w. Antisymmetric: no ⇆ type of arrows.

1 2

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Example

• 1. Let A = {0, 1, 2} and R = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} and

S = {(0, 0), (1, 1), (2, 2)} be two relations on A. Show that (i) R is a partial order relation. (ii) S is an equivalence relation. Solution. We choose to use digraphs to make the explanations in this case. (i) The digraph for R on the right implies the relation R is Reflexive: loops on every vertex. Transitive: if you can travel from vertex v to vertex w along consecutive arrows of the same direction, then there is also a single arrow pointing from v to w. Antisymmetric: no ⇆ type of arrows.

1 2

(ii) The digraph for S on the right is reflexive due to loops on every vertex, symmetric and transitive because no no-loop arrows exist.

1 2

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Notes

In Example 1, R and S are built on A from “≤” and “=” respectively by R = {(x, y) : x, y ∈ A, x ≤ y} , S = {(x, y) : x, y ∈ A, x = y} .

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Notes

In Example 1, R and S are built on A from “≤” and “=” respectively by R = {(x, y) : x, y ∈ A, x ≤ y} , S = {(x, y) : x, y ∈ A, x = y} . Hence, partial order relation and equivalence relation can be in general regarded as “generalisation” of “≤” and “=” respectively.

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Notes

In Example 1, R and S are built on A from “≤” and “=” respectively by R = {(x, y) : x, y ∈ A, x ≤ y} , S = {(x, y) : x, y ∈ A, x = y} . Hence, partial order relation and equivalence relation can be in general regarded as “generalisation” of “≤” and “=” respectively. For the same reasons, they are often denoted by

Ioan Despi – AMTH140 6 of 21

Notes

In Example 1, R and S are built on A from “≤” and “=” respectively by R = {(x, y) : x, y ∈ A, x ≤ y} , S = {(x, y) : x, y ∈ A, x = y} . Hence, partial order relation and equivalence relation can be in general regarded as “generalisation” of “≤” and “=” respectively. For the same reasons, they are often denoted by

◮ x ⪯ y if xR1y and R1 is a partial order relation,

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Notes

In Example 1, R and S are built on A from “≤” and “=” respectively by R = {(x, y) : x, y ∈ A, x ≤ y} , S = {(x, y) : x, y ∈ A, x = y} . Hence, partial order relation and equivalence relation can be in general regarded as “generalisation” of “≤” and “=” respectively. For the same reasons, they are often denoted by

◮ x ⪯ y if xR1y and R1 is a partial order relation, ◮ x ∼ y if xR2y and R2 is an equivalence relation.

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Totally Ordered Set

Let R be a partial order relation on a set A. ◮ For any elements a, b ∈ A, aRb can be alternatively denoted by a ⪯ b , meaning

element a precedes element b under the partial order relation R.

◮ Two elements a, b ∈ A are comparable if either

◮ aRb or bRa (i.e., either a ⪯ b or b ⪯ a), or if ◮ a = b.

◮ If all elements of A are comparable with each other, then the partially

• rdered set A (w.r.t. R) is said to be a totally ordered set, and the

relation R is also said to be a total order relation. A total order let us arrange the elements of set A in order as thought on a line.

◮ Hence a total ordering is also called a linear ordering. Ioan Despi – AMTH140 7 of 21

Special Elements

◮ An element a ∈ A is a maximal element of A if b ⪯ a holds for every b ∈ A whenever b and a are comparable.

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Special Elements

◮ An element a ∈ A is a maximal element of A if b ⪯ a holds for every b ∈ A whenever b and a are comparable. ◮ An element a ∈ A is a greatest element (top element) of A if b ⪯ a holds for all b ∈ A.

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Special Elements

◮ An element a ∈ A is a maximal element of A if b ⪯ a holds for every b ∈ A whenever b and a are comparable. ◮ An element a ∈ A is a greatest element (top element) of A if b ⪯ a holds for all b ∈ A. ◮ An element a ∈ A is a minimal element of A if a ⪯ b holds for every b ∈ A whenever b and a are comparable.

Ioan Despi – AMTH140 8 of 21

Special Elements

◮ An element a ∈ A is a maximal element of A if b ⪯ a holds for every b ∈ A whenever b and a are comparable. ◮ An element a ∈ A is a greatest element (top element) of A if b ⪯ a holds for all b ∈ A. ◮ An element a ∈ A is a minimal element of A if a ⪯ b holds for every b ∈ A whenever b and a are comparable. ◮ An element a ∈ A is a least element (bottom element) of A if a ⪯ b holds for all b ∈ A.

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Example

• 2. Let A be the set of all subsets of set {a, b, c}. Show the “subset” relation ⊆
• n A, i.e. ∀u, v ∈ A,

u ⪯ v or uRv, iff u ⊆ v, is a partial order relation. Find a minimal element and a greatest element.

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Example

• 2. Let A be the set of all subsets of set {a, b, c}. Show the “subset” relation ⊆
• n A, i.e. ∀u, v ∈ A,

u ⪯ v or uRv, iff u ⊆ v, is a partial order relation. Find a minimal element and a greatest element. Solution. It is easy to verify that “⊆” is a partial ordering. Since ∅ is a subset of any u ∈ A, i.e. ∅ ⪯ u, we see ∅ is not only a minimal element, it is also a least element of A. Since for any u ∈ A one has u ⊆ {a, b, c}, i.e., u ⪯ {a, b, c}, we see that {a, b, c} is a greatest element of A.

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Special Elements

A greatest (top) element is always a maximal element, but a maximal element needs not be a greatest, not even if the maximal element is unique.

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Special Elements

A greatest (top) element is always a maximal element, but a maximal element needs not be a greatest, not even if the maximal element is unique. However, an unique maximal element in a finite partial ordered set is a top.

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Special Elements

A greatest (top) element is always a maximal element, but a maximal element needs not be a greatest, not even if the maximal element is unique. However, an unique maximal element in a finite partial ordered set is a top. Mutatis mutandis for bottom: A least (bottom) element is always a minimal element, but a minimal element needs not be a least, not even if the minimal element is unique.

Ioan Despi – AMTH140 10 of 21

Special Elements

A greatest (top) element is always a maximal element, but a maximal element needs not be a greatest, not even if the maximal element is unique. However, an unique maximal element in a finite partial ordered set is a top. Mutatis mutandis for bottom: A least (bottom) element is always a minimal element, but a minimal element needs not be a least, not even if the minimal element is unique. However, an unique minimal element in a finite partial ordered set is a bottom.

Ioan Despi – AMTH140 10 of 21

Special Elements

A greatest (top) element is always a maximal element, but a maximal element needs not be a greatest, not even if the maximal element is unique. However, an unique maximal element in a finite partial ordered set is a top. Mutatis mutandis for bottom: A least (bottom) element is always a minimal element, but a minimal element needs not be a least, not even if the minimal element is unique. However, an unique minimal element in a finite partial ordered set is a bottom.

Ioan Despi – AMTH140 10 of 21

Special Elements

A greatest (top) element is always a maximal element, but a maximal element needs not be a greatest, not even if the maximal element is unique. However, an unique maximal element in a finite partial ordered set is a top. Mutatis mutandis for bottom: A least (bottom) element is always a minimal element, but a minimal element needs not be a least, not even if the minimal element is unique. However, an unique minimal element in a finite partial ordered set is a bottom. Theorem A poset can have at most one top.

Ioan Despi – AMTH140 10 of 21

Special Elements

A greatest (top) element is always a maximal element, but a maximal element needs not be a greatest, not even if the maximal element is unique. However, an unique maximal element in a finite partial ordered set is a top. Mutatis mutandis for bottom: A least (bottom) element is always a minimal element, but a minimal element needs not be a least, not even if the minimal element is unique. However, an unique minimal element in a finite partial ordered set is a bottom. Theorem A poset can have at most one top. Proof. Let (A, R) be a poset. If x and y are tops then yRx and xRy so x = y (by antisymmetry of R). Same for bottoms, of course.

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Hasse Diagrams

Hasse diagrams are meant to present partial order relations in equivalent but somewhat simpler forms by removing certain deducible “noncritical” parts of the relations.

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Hasse Diagrams

Hasse diagrams are meant to present partial order relations in equivalent but somewhat simpler forms by removing certain deducible “noncritical” parts of the relations. For better motivation and understanding, we’ll introduce them through the following examples.

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Hasse Diagrams

Hasse diagrams are meant to present partial order relations in equivalent but somewhat simpler forms by removing certain deducible “noncritical” parts of the relations. For better motivation and understanding, we’ll introduce them through the following examples.

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Hasse Diagrams

Hasse diagrams are meant to present partial order relations in equivalent but somewhat simpler forms by removing certain deducible “noncritical” parts of the relations. For better motivation and understanding, we’ll introduce them through the following examples. Example

• 3. The relation in Example 2 can be drawn as

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

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Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

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Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops,

Ioan Despi – AMTH140 12 of 21

Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops, ◮ omit all arrows that can be inferred from transitivity,

Ioan Despi – AMTH140 12 of 21

Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops, ◮ omit all arrows that can be inferred from transitivity, ◮ draw arrows without heads,

Ioan Despi – AMTH140 12 of 21

Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops, ◮ omit all arrows that can be inferred from transitivity, ◮ draw arrows without heads, ◮ understand that all arrows point upwards,

Ioan Despi – AMTH140 12 of 21

Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops, ◮ omit all arrows that can be inferred from transitivity, ◮ draw arrows without heads, ◮ understand that all arrows point upwards,

Ioan Despi – AMTH140 12 of 21

Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops, ◮ omit all arrows that can be inferred from transitivity, ◮ draw arrows without heads, ◮ understand that all arrows point upwards, then the above graph simplifies to

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

Ioan Despi – AMTH140 12 of 21

Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops, ◮ omit all arrows that can be inferred from transitivity, ◮ draw arrows without heads, ◮ understand that all arrows point upwards, then the above graph simplifies to

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

This type of graph is called a Hasse diagram, it is often used to represent a partially ordered set.

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Hasse Diagrams

It is somewhat “messy” and some arrows can be derived from transitivity

• anyway. If we

◮ omit all loops, ◮ omit all arrows that can be inferred from transitivity, ◮ draw arrows without heads, ◮ understand that all arrows point upwards, then the above graph simplifies to

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

This type of graph is called a Hasse diagram, it is often used to represent a partially ordered set. If a digraph of a partial order relation is simplified according to the steps under the four “◮” symbols in the above, then the graph becomes a Hasse diagram.

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Example

• 4. Let A = {1, 2, 3, 9, 18} and for any a, b ∈ A, a ⪯ b

iff a | b. Draw the Hasse diagram of the relation. Give a maximal element and a greatest element if any.

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Example

• 4. Let A = {1, 2, 3, 9, 18} and for any a, b ∈ A, a ⪯ b

iff a | b. Draw the Hasse diagram of the relation. Give a maximal element and a greatest element if any. Solution. The directed graph of relation ⪯ is drawn on the right. First it is easy to verify that the re- lation ⪯ defined above is a partial

• rdering. By removing all the loops,

deleting all the edges that can be de- rived transitively, and making sure all the retained edges will point up- wards implicitly, we obtain the fol- lowing Hasse diagram: 1 2 3 9 18

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1 2 3 9 18

This is because the edge from vertex 1 to vertex 18, for instance, can be derived from the edge from 1 to 2 and the edge from 2 to 18.

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1 2 3 9 18

This is because the edge from vertex 1 to vertex 18, for instance, can be derived from the edge from 1 to 2 and the edge from 2 to 18. Hence the edge from 1 to 18 shouldn’t be kept in the Hasse diagram as it can be induced from the transitivity.

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1 2 3 9 18

This is because the edge from vertex 1 to vertex 18, for instance, can be derived from the edge from 1 to 2 and the edge from 2 to 18. Hence the edge from 1 to 18 shouldn’t be kept in the Hasse diagram as it can be induced from the transitivity. Since 18 divides no elements in the set A

• ther than itself, no elements in A is preceded

by 18 other than 18 itself because of the definition of the partial order relation ⪯.

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1 2 3 9 18

This is because the edge from vertex 1 to vertex 18, for instance, can be derived from the edge from 1 to 2 and the edge from 2 to 18. Hence the edge from 1 to 18 shouldn’t be kept in the Hasse diagram as it can be induced from the transitivity. Since 18 divides no elements in the set A

• ther than itself, no elements in A is preceded

by 18 other than 18 itself because of the definition of the partial order relation ⪯. Hence 18 is a maximal element.

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1 2 3 9 18

This is because the edge from vertex 1 to vertex 18, for instance, can be derived from the edge from 1 to 2 and the edge from 2 to 18. Hence the edge from 1 to 18 shouldn’t be kept in the Hasse diagram as it can be induced from the transitivity. Since 18 divides no elements in the set A

• ther than itself, no elements in A is preceded

by 18 other than 18 itself because of the definition of the partial order relation ⪯. Hence 18 is a maximal element. Since all elements in A divide 18 means a ⪯ 18 holds for all elements a ∈ A, 18 is also a greatest element.

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1 2 3 9 18

This is because the edge from vertex 1 to vertex 18, for instance, can be derived from the edge from 1 to 2 and the edge from 2 to 18. Hence the edge from 1 to 18 shouldn’t be kept in the Hasse diagram as it can be induced from the transitivity. Since 18 divides no elements in the set A

• ther than itself, no elements in A is preceded

by 18 other than 18 itself because of the definition of the partial order relation ⪯. Hence 18 is a maximal element. Since all elements in A divide 18 means a ⪯ 18 holds for all elements a ∈ A, 18 is also a greatest element.

Finally we remark that the vertices in the digraphs in this and the previous examples happen to be conveniently placed for creating the corresponding Hasse diagrams, in the sense that all arrows are already pointing upwards.

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1 2 3 9 18

This is because the edge from vertex 1 to vertex 18, for instance, can be derived from the edge from 1 to 2 and the edge from 2 to 18. Hence the edge from 1 to 18 shouldn’t be kept in the Hasse diagram as it can be induced from the transitivity. Since 18 divides no elements in the set A

• ther than itself, no elements in A is preceded

by 18 other than 18 itself because of the definition of the partial order relation ⪯. Hence 18 is a maximal element. Since all elements in A divide 18 means a ⪯ 18 holds for all elements a ∈ A, 18 is also a greatest element.

Finally we remark that the vertices in the digraphs in this and the previous examples happen to be conveniently placed for creating the corresponding Hasse diagrams, in the sense that all arrows are already pointing upwards. If this is not the case, then one may have to ”twist” the digraph to make it so, ready for the transformation into a Hasse diagram.

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Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements.

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Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements.

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Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements. Two elements are comparable if there is a path connecting them and that the walk along the path does not change direction vertically on any edge

• f the path.

Ioan Despi – AMTH140 15 of 21

Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements. Two elements are comparable if there is a path connecting them and that the walk along the path does not change direction vertically on any edge

• f the path.

In terms of a Hasse diagram

Ioan Despi – AMTH140 15 of 21

Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements. Two elements are comparable if there is a path connecting them and that the walk along the path does not change direction vertically on any edge

• f the path.

In terms of a Hasse diagram

◮ A ∈ A is a maximal element of A if there is no y ∈ A such that y is above a. Ioan Despi – AMTH140 15 of 21

Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements. Two elements are comparable if there is a path connecting them and that the walk along the path does not change direction vertically on any edge

• f the path.

In terms of a Hasse diagram

◮ A ∈ A is a maximal element of A if there is no y ∈ A such that y is above a. ◮ A ∈ A is a minimal element of A if there is no y ∈ A such that y is below a. Ioan Despi – AMTH140 15 of 21

Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements. Two elements are comparable if there is a path connecting them and that the walk along the path does not change direction vertically on any edge

• f the path.

In terms of a Hasse diagram

◮ A ∈ A is a maximal element of A if there is no y ∈ A such that y is above a. ◮ A ∈ A is a minimal element of A if there is no y ∈ A such that y is below a. ◮ A ∈ A is the greatest (top) element of A if a is above every other element

• f A.

Ioan Despi – AMTH140 15 of 21

Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements. Two elements are comparable if there is a path connecting them and that the walk along the path does not change direction vertically on any edge

• f the path.

In terms of a Hasse diagram

◮ A ∈ A is a maximal element of A if there is no y ∈ A such that y is above a. ◮ A ∈ A is a minimal element of A if there is no y ∈ A such that y is below a. ◮ A ∈ A is the greatest (top) element of A if a is above every other element

• f A.

◮ A ∈ A is the least (bottom) element of A if a is below every other element

• f A.

Ioan Despi – AMTH140 15 of 21

Remember

In a partial order relation, all greatest elements are also maximal elements and all least elements are also minimal elements. In terms of a Hasse diagram, all local top vertices are maximal elements, and all local bottom vertices are minimal elements. Two elements are comparable if there is a path connecting them and that the walk along the path does not change direction vertically on any edge

• f the path.

In terms of a Hasse diagram

◮ A ∈ A is a maximal element of A if there is no y ∈ A such that y is above a. ◮ A ∈ A is a minimal element of A if there is no y ∈ A such that y is below a. ◮ A ∈ A is the greatest (top) element of A if a is above every other element

• f A.

◮ A ∈ A is the least (bottom) element of A if a is below every other element

• f A.

If a Hasse diagram has only finite number of vertices, then a maximal element is also a greatest element if there is only 1 maximal element, and a minimal element is also a least element if there is only 1 minimal element.

Ioan Despi – AMTH140 15 of 21

Example 5. (a) Let set A be given by A = {3, 4, 5, 6, 10, 12} and a binary relation R on A be defined by (x, y) ∈ R if and only if x divides y (x, y) ∈ R iff x | y Give R explicitly in terms of its elements and draw the corresponding Hasse diagram. List all the maximal, minimal, greatest, and least elements. (b) Let a new binary relation R′ on the set A given in (a) be defined by (x, y) ∈ R′ if and only if either x | y or y | x and R′′ be the transitive closure of R′. Use directed graphs to represent R, R′ and R′′ respectively. Which of the three relations R, R′ and R′′ is an equivalence relation? For the equivalence relation, give all the distinct equivalence classes.

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(a) Take 3 ∈ A = {3, 4, 5, 6, 10, 12}, then 3 ∈ A divides 3, 6 and 12. Hence, by the definition of the relation R, we conclude (3, 3), (3, 6) and (3, 12) are all elements of the relation R. Likewise, we can show that (4, 4), (4, 12), (5, 5), (5, 10), (6, 6), (6, 12), (10, 10), (12, 12) are all elements of R. In fact we have

R = { (3, 3), (3, 6), (3, 12), (4, 4), (4, 12), (5, 5), (5, 10), (6, 6), (6, 12), (10, 10), (12, 12) }

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(a) Take 3 ∈ A = {3, 4, 5, 6, 10, 12}, then 3 ∈ A divides 3, 6 and 12. Hence, by the definition of the relation R, we conclude (3, 3), (3, 6) and (3, 12) are all elements of the relation R. Likewise, we can show that (4, 4), (4, 12), (5, 5), (5, 10), (6, 6), (6, 12), (10, 10), (12, 12) are all elements of R. In fact we have

R = { (3, 3), (3, 6), (3, 12), (4, 4), (4, 12), (5, 5), (5, 10), (6, 6), (6, 12), (10, 10), (12, 12) }

Hence the digraph for R is

4 5 10 6 12 3

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(a) Take 3 ∈ A = {3, 4, 5, 6, 10, 12}, then 3 ∈ A divides 3, 6 and 12. Hence, by the definition of the relation R, we conclude (3, 3), (3, 6) and (3, 12) are all elements of the relation R. Likewise, we can show that (4, 4), (4, 12), (5, 5), (5, 10), (6, 6), (6, 12), (10, 10), (12, 12) are all elements of R. In fact we have

R = { (3, 3), (3, 6), (3, 12), (4, 4), (4, 12), (5, 5), (5, 10), (6, 6), (6, 12), (10, 10), (12, 12) }

Hence the digraph for R is

4 5 10 6 12 3

which induces the following Hasse diagram

3 4 5 10 6 12

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From the Hasse diagram, we observe that there are 2 local top vertices, 12 and 10, and 3 local bottom vertices, 3, 4 and 5.

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From the Hasse diagram, we observe that there are 2 local top vertices, 12 and 10, and 3 local bottom vertices, 3, 4 and 5. Hence 12 and 10 are maximal elements, and 3, 4 and 5 are minimal elements.

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From the Hasse diagram, we observe that there are 2 local top vertices, 12 and 10, and 3 local bottom vertices, 3, 4 and 5. Hence 12 and 10 are maximal elements, and 3, 4 and 5 are minimal elements. Since the Hasse diagram has only finite number of vertices, the fact that the Hasse diagram has more than 1 local top vertices means there are no greatest elements.

Ioan Despi – AMTH140 18 of 21

From the Hasse diagram, we observe that there are 2 local top vertices, 12 and 10, and 3 local bottom vertices, 3, 4 and 5. Hence 12 and 10 are maximal elements, and 3, 4 and 5 are minimal elements. Since the Hasse diagram has only finite number of vertices, the fact that the Hasse diagram has more than 1 local top vertices means there are no greatest elements. Likewise there are no least elements either.

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(b) The digraphs for R is already given in (a).

◮ As for R′ we observe that, according to the definition of the relation R′, if

(x, y) (e.g. (3, 6) ) is in R so will (y, x) (thus e.g. (6, 3) ).

Ioan Despi – AMTH140 19 of 21

(b) The digraphs for R is already given in (a).

◮ As for R′ we observe that, according to the definition of the relation R′, if

(x, y) (e.g. (3, 6) ) is in R so will (y, x) (thus e.g. (6, 3) ).

◮ Hence we can obtain R′ by adding to R the symmetric pairs like (6, 3),

(12, 3) etc.

Ioan Despi – AMTH140 19 of 21

(b) The digraphs for R is already given in (a).

◮ As for R′ we observe that, according to the definition of the relation R′, if

(x, y) (e.g. (3, 6) ) is in R so will (y, x) (thus e.g. (6, 3) ).

◮ Hence we can obtain R′ by adding to R the symmetric pairs like (6, 3),

(12, 3) etc.

◮ In terms of the digraph, such addition of elements is equivalent to drawing

an opposite arrows to each existing (non-loop) arrows.

Ioan Despi – AMTH140 19 of 21

(b) The digraphs for R is already given in (a).

◮ As for R′ we observe that, according to the definition of the relation R′, if

(x, y) (e.g. (3, 6) ) is in R so will (y, x) (thus e.g. (6, 3) ).

◮ Hence we can obtain R′ by adding to R the symmetric pairs like (6, 3),

(12, 3) etc.

◮ In terms of the digraph, such addition of elements is equivalent to drawing

an opposite arrows to each existing (non-loop) arrows.

◮ The digraph of R′ thus takes the following form Ioan Despi – AMTH140 19 of 21

(b) The digraphs for R is already given in (a).

◮ As for R′ we observe that, according to the definition of the relation R′, if

(x, y) (e.g. (3, 6) ) is in R so will (y, x) (thus e.g. (6, 3) ).

◮ Hence we can obtain R′ by adding to R the symmetric pairs like (6, 3),

(12, 3) etc.

◮ In terms of the digraph, such addition of elements is equivalent to drawing

an opposite arrows to each existing (non-loop) arrows.

◮ The digraph of R′ thus takes the following form Ioan Despi – AMTH140 19 of 21

(b) The digraphs for R is already given in (a).

◮ As for R′ we observe that, according to the definition of the relation R′, if

(x, y) (e.g. (3, 6) ) is in R so will (y, x) (thus e.g. (6, 3) ).

◮ Hence we can obtain R′ by adding to R the symmetric pairs like (6, 3),

(12, 3) etc.

◮ In terms of the digraph, such addition of elements is equivalent to drawing

an opposite arrows to each existing (non-loop) arrows.

◮ The digraph of R′ thus takes the following form

4 5 10 6 12 3

Ioan Despi – AMTH140 19 of 21

Since relation R′′ is the transitive closure of R′, we can derive R′′ from R′ by connecting 3 to 4 and 4 to 6 and connecting 4 to 3 and 6 to 4. 4 5 10 6 12 3

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Since relation R′′ is the transitive closure of R′, we can derive R′′ from R′ by connecting 3 to 4 and 4 to 6 and connecting 4 to 3 and 6 to 4. We connect 3 to 4 via an direct arrow because we can travel from 3 to 12 and then from 12 to 4 all along the arrows, and we connect 4 to 3 because we can travel from 4 to 12 then to 3 all along the arrows too. 4 5 10 6 12 3

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Since relation R′′ is the transitive closure of R′, we can derive R′′ from R′ by connecting 3 to 4 and 4 to 6 and connecting 4 to 3 and 6 to 4. We connect 3 to 4 via an direct arrow because we can travel from 3 to 12 and then from 12 to 4 all along the arrows, and we connect 4 to 3 because we can travel from 4 to 12 then to 3 all along the arrows too. Similar reasons are applicable for the arrows from 4 to 6 and 6 to 4. 4 5 10 6 12 3

Ioan Despi – AMTH140 20 of 21

Since relation R′′ is the transitive closure of R′, we can derive R′′ from R′ by connecting 3 to 4 and 4 to 6 and connecting 4 to 3 and 6 to 4. We connect 3 to 4 via an direct arrow because we can travel from 3 to 12 and then from 12 to 4 all along the arrows, and we connect 4 to 3 because we can travel from 4 to 12 then to 3 all along the arrows too. Similar reasons are applicable for the arrows from 4 to 6 and 6 to 4. Hence R′′ can be drawn as 4 5 10 6 12 3

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Since there is no arrow from element 12 to element 3 in the digraph of R despite the existence of an arrow from 3 to 12, relation R is not symmetric hence is not an equivalence relation.

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Since there is no arrow from element 12 to element 3 in the digraph of R despite the existence of an arrow from 3 to 12, relation R is not symmetric hence is not an equivalence relation. Since relation R′ is not transitive (because its transitive closure R′′ is not the same as R′ itself), relation R′ is not an equivalence relation either.

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Since there is no arrow from element 12 to element 3 in the digraph of R despite the existence of an arrow from 3 to 12, relation R is not symmetric hence is not an equivalence relation. Since relation R′ is not transitive (because its transitive closure R′′ is not the same as R′ itself), relation R′ is not an equivalence relation either. As for the relation R′′ , it is obviously reflexive, symmetric and transitive. Hence R′′ is an equivalence relation.

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Since there is no arrow from element 12 to element 3 in the digraph of R despite the existence of an arrow from 3 to 12, relation R is not symmetric hence is not an equivalence relation. Since relation R′ is not transitive (because its transitive closure R′′ is not the same as R′ itself), relation R′ is not an equivalence relation either. As for the relation R′′ , it is obviously reflexive, symmetric and transitive. Hence R′′ is an equivalence relation. Since elements 3, 4, 6 and 12 are all related (connected) to each other through the arrows of the digraph R′′ and none of these 4 elements are related to any other elements, they must form a single equivalence class.

Ioan Despi – AMTH140 21 of 21

Since there is no arrow from element 12 to element 3 in the digraph of R despite the existence of an arrow from 3 to 12, relation R is not symmetric hence is not an equivalence relation. Since relation R′ is not transitive (because its transitive closure R′′ is not the same as R′ itself), relation R′ is not an equivalence relation either. As for the relation R′′ , it is obviously reflexive, symmetric and transitive. Hence R′′ is an equivalence relation. Since elements 3, 4, 6 and 12 are all related (connected) to each other through the arrows of the digraph R′′ and none of these 4 elements are related to any other elements, they must form a single equivalence class. Hence we have [3] = { 3, 4, 6, 12 } .

Ioan Despi – AMTH140 21 of 21

Since there is no arrow from element 12 to element 3 in the digraph of R despite the existence of an arrow from 3 to 12, relation R is not symmetric hence is not an equivalence relation. Since relation R′ is not transitive (because its transitive closure R′′ is not the same as R′ itself), relation R′ is not an equivalence relation either. As for the relation R′′ , it is obviously reflexive, symmetric and transitive. Hence R′′ is an equivalence relation. Since elements 3, 4, 6 and 12 are all related (connected) to each other through the arrows of the digraph R′′ and none of these 4 elements are related to any other elements, they must form a single equivalence class. Hence we have [3] = { 3, 4, 6, 12 } . Likewise we can derive another equivalence class [5] = { 5, 10 } .

Ioan Despi – AMTH140 21 of 21

Since there is no arrow from element 12 to element 3 in the digraph of R despite the existence of an arrow from 3 to 12, relation R is not symmetric hence is not an equivalence relation. Since relation R′ is not transitive (because its transitive closure R′′ is not the same as R′ itself), relation R′ is not an equivalence relation either. As for the relation R′′ , it is obviously reflexive, symmetric and transitive. Hence R′′ is an equivalence relation. Since elements 3, 4, 6 and 12 are all related (connected) to each other through the arrows of the digraph R′′ and none of these 4 elements are related to any other elements, they must form a single equivalence class. Hence we have [3] = { 3, 4, 6, 12 } . Likewise we can derive another equivalence class [5] = { 5, 10 } . Because any element of A is either in the equivalence class [3] or in the equivalence class [5], these two classes are all the distinct equivalence classes.

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