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 Ioan Despi – AMTH140 2 of 21

Motivation Partial oderings have important applications, e.g., the analysis of computer programs. 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. 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 . 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. 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 order 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 order 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 order 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. Ioan Despi – AMTH140 4 of 21

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. Ioan Despi – AMTH140 5 of 21

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 1 0 v to vertex w along consecutive arrows of the same direction, then there is also a single arrow pointing from v to w . 2 Antisymmetric : no ⇆ type of arrows. Ioan Despi – AMTH140 5 of 21

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 1 0 v to vertex w along consecutive arrows of the same direction, then there is also a single arrow pointing from v to w . 2 Antisymmetric : no ⇆ type of arrows. (ii) The digraph for S on the right is reflexive due to loops on every vertex, 2 symmetric and transitive because no 0 no-loop arrows exist. 1 Ioan Despi – AMTH140 5 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 } . 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. 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 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 xR 1 y and R 1 is a partial order relation, 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 xR 1 y and R 1 is a partial order relation, ◮ x ∼ y if xR 2 y and R 2 is an equivalence relation. Ioan Despi – AMTH140 6 of 21

Totally Ordered Set Let R be a partial order relation on a set A . ◮ For any elements a, b ∈ A , can be alternatively denoted by a ⪯ b , meaning aRb 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 ordered 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. 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 . 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. 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 . Ioan Despi – AMTH140 8 of 21

Example 2. Let A be the set of all subsets of set { a, b, c } . Show the “subset” relation ⊆ on 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. Ioan Despi – AMTH140 9 of 21

Example 2. Let A be the set of all subsets of set { a, b, c } . Show the “subset” relation ⊆ on 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 . Ioan Despi – AMTH140 9 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. Ioan Despi – AMTH140 10 of 21