Lecture 06 Relations PDF

Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Lecture 6: Relations Aaron Tan Part of the contents here is taken from AY2024...

Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Lecture 6: Relations Aaron Tan Part of the contents here is taken from AY2024/25 Semester 1 Dr Wong Tin Lok’s lecture notes. 1 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations 6. Relations 6.1 Relations on Sets Definition of relation; arrow diagram, inverse of a relation. Relation on a set; directed graph of a relation. Composition of relations. N-ary relations and relational databases. 6.2 Reflexivity, Symmetry and Transitivity Definitions of reflexivity, symmetry and transitivity. Transitive closure of a relation. 6.3 Equivalence Relations Partition of a set; the relation induced by a partition. Equivalence relation; equivalence classes. Congruence. Dividing a set by an equivalence relation. Summary Reference: Epp’s Chapter 8 Properties of Relations 2 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations 6. Relations 6.4 Partial Order Relations Antisymmetry. Partial order relations. Hasse diagrams. Comparability. Maximal/minimal/largest/smallest element. Linearization. Total order relations; well ordered sets. Reference: Epp’s Chapter 8 Properties of Relations 3 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations 6.1 Relations on Sets 4 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions 6.1.1 Definitions As the topic of relations is built on sets, definitions on sets, such as ordered pair, ordered 𝑛-tuple, Cartesian product, etc. (see Lecture 5 Set Theory) will be referred to here. Recall: The Cartesian product of sets 𝐴 and 𝐵, denoted 𝐴 × 𝐵, consists of all ordered pairs whose first element is in 𝐴 and second element in 𝐵: 𝐴 × 𝐵 = { 𝑥, 𝑦 ∶ 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵}. Example #1: Let 𝐴 = {0,1,2} and 𝐵 = 𝑎, 𝑏, 𝑐. Then 𝐴 × 𝐵 = { 0, 𝑎 , 0, 𝑏 , 0, 𝑐 , 1, 𝑎 , 1, 𝑏 , 1, 𝑐 , 2, 𝑎 , 2, 𝑏 , 2, 𝑐 }. 5 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions Definition: Relation Let 𝐴 and 𝐵 be sets. A (binary) relation from 𝑨 to 𝑩 is a subset of 𝑨 × 𝑩. Given an ordered pair (𝑥, 𝑦) in 𝐴 × 𝐵, 𝒙 is related to 𝒚 by 𝑹, or 𝒙 is 𝑹-related to 𝒚, written 𝒙 𝑹 𝒚, iff (𝒙, 𝒚) ∈ 𝑹. Symbolically, 𝑥 𝑅 𝑦 means (𝑥, 𝑦) ∈ 𝑅 𝑥 𝑅 𝑦 means (𝑥, 𝑦) ∉ 𝑅 We read 𝑥 𝑅 𝑦 as “𝑥 is 𝑅-related to 𝑦” or, if there is no risk of confusion, simply “𝑥 is related to 𝑦”. 6 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions Example #2: Let 𝐴 = {0,1,2} and 𝐵 = 1,2,3. Suppose we define the relation 𝑅 s.t. 𝑥𝑅𝑦 iff 𝑥 < 𝑦. Then 0𝑅1, 0𝑅2, 0𝑅3, 1𝑅2, 1𝑅3 and 2𝑅3, but 1𝑅1, 2𝑅1 and 2𝑅2. Example #3: Let 𝐴 = {1,2} and 𝐵 = 1,2,3. Define a relation 𝑅 from 𝐴 to 𝐵 as follows: 𝑥−𝑦 ∀ 𝑥, 𝑦 ∈ 𝐴 × 𝐵 𝑥, 𝑦 ∈ 𝑅 ⇔ ∈ℤ. 2 State explicitly which ordered pairs are in 𝐴 × 𝐵 and which are in 𝑅. 𝐴×𝐵 = 1,1 , 1,2 , 1,3 , 2,1 , 2,2 , 2,3. 𝑅 = { 1,1 , 1,3 , 2,2 }. 7 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Application of Relations An application: A simple database Let 𝑆 be the set of students, 𝑀 the set of modules, and 𝑅 the relation “is enrolled in” from 𝑆 to 𝑀. Student Module Ali CS1101S 𝑅 = { (Ali, CS1101S), (Aiken, CS1231S), Aiken CS1231S (Dueet, CS1231S) Dueet CS1231S (Bam Boo, MA1101R), Bam Boo MA1101R (Lian Eng, CS1101S), Lian Eng CS1101S (Manimaran, CS1231S), Manimaran CS1231S (James Tan, MA1101R), James Tan MA1101R …} : : 8 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions Definitions: Domain, Co-domain, Range Let 𝐴 and 𝐵 be sets and 𝑅 be a relation from 𝐴 to 𝐵. The domain of 𝑅, 𝐷𝑜𝑚 𝑅 , is the set {𝑎 ∈ 𝐴 ∶ 𝑎𝑅𝑏 for some 𝑏 ∈ 𝐵}. The co-domain of 𝑅, 𝑐𝑜𝐷𝑜𝑚(𝑅), is the set 𝐵. The range of 𝑅, 𝑅𝑎𝑛𝑔𝑒 𝑅 , is the set {𝑏 ∈ 𝐵: 𝑎𝑅𝑏 for some 𝑎 ∈ 𝐴}. Example #4: Let 𝐴 = {1,2,3} and 𝐵 = 2,4,9 , and define a relation 𝑅 from 𝐴 to 𝐵 as follows: ∀ 𝑥, 𝑦 ∈ 𝐴 × 𝐵, 𝑥, 𝑦 ∈ 𝑅 ⇔ 𝑥 2 = 𝑦. 𝐷𝑜𝑚 𝑅 = 2,3 𝑐𝑜𝐷𝑜𝑚 𝑅 = {2,4,9} 𝑅𝑎𝑛𝑔𝑒 𝑅 = {4,9} 9 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions Example #5: Define a relation 𝑅 from ℤ to ℤ as follows: ∀ 𝑥, 𝑦 ∈ ℤ × ℤ (𝑥𝑅𝑦 ⇔ 𝑥 − 𝑦 is even). a. Is 4𝑅0? Is 2𝑅6? Is 3𝑅(– 3)? Is 5𝑅2? Yes Yes Yes No b. List five integers that are related by 𝑅 to 1. Infinitely many possible answers. One answer: 1, 57, 12345, -203, -999. c. Prove that if 𝑎 is any odd integer, then 𝑎𝑅1. 1. Let 𝑎 be an odd integer. 2. Then 𝑎 = 2𝑘 + 1 for some integer 𝑘 (by the definition of odd). 3. Therefore, 𝑎 − 1 = 2𝑘 which is even (by the definition of even). 4. Hence 𝑎𝑅1 (by the definition of 𝑅). 10  Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Arrow Diagram A relation 𝑅 from set 𝐴 to set 𝐵 can be depicted as an arrow diagram: 1. Represent the elements of 𝐴 as points in one region and the elements of 𝐵 as points in another region. 2. For each 𝑥 ∈ 𝐴 and 𝑦 ∈ 𝐵, draw an arrow from 𝑥 to 𝑦 iff 𝑥𝑅𝑦. Example #6: Let 𝐴 = {1,2,3} and 𝐵 = 1,3,5. Define relations 𝑆 and 𝑇 from 𝐴 to 𝐵 as follows: ∀ 𝑥, 𝑦 ∈ 𝐴 × 𝐵, 𝑥, 𝑦 ∈ 𝑆 ⇔ 𝑥 < 𝑦 𝑇 = (2,1 , (2,5)}. 11 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Inverse of a Relation 6.1.2 The Inverse of a Relation If 𝑅 is a relation from 𝐴 to 𝐵, then a relation 𝑅 −1 from 𝐵 to 𝐴 can be defined by interchanging the elements of all the ordered pairs of 𝑅. Definition: Inverse of a Relation Let 𝑅 be a relation from 𝐴 to 𝐵. Define the inverse relation 𝑅−1 from 𝐵 to 𝐴 as follows: 𝑅 −1 = { 𝑦, 𝑥 ∈ 𝐵 × 𝐴 ∶ (𝑥, 𝑦) ∈ 𝑅}. This definition can be written operationally as follows: ∀𝑥 ∈ 𝐴, ∀𝑦 ∈ 𝐵 𝑦, 𝑥 ∈ 𝑅−1 ⇔ 𝑥, 𝑦 ∈ 𝑅. 12 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Inverse of a Relation If 𝑛, 𝑑 ∈ ℤ: 𝑑 | 𝑛 ⟺ ∃𝑘 ∈ ℤ such that 𝑛 = 𝑑𝑘. Example #7: Let 𝐴 = {2, 3, 4} and 𝐵 = {2, 6, 8} and let 𝑅 be the “divides” relation from A to 𝐵: ∀ 𝑥, 𝑦 ∈ 𝐴 × 𝐵 𝑥𝑅𝑦 ⇔ 𝑥 𝑦 a. State explicitly which ordered pairs are in 𝑅 and 𝑅−1 , and draw arrow diagrams for 𝑅 and 𝑅−1. 𝑅 = {(2,2), (2,6), (2,8), (3,6), (4,8)} 𝑅−1 = {(2,2), (6,2), (8,2), (6,3), (8,4)} b. Describe 𝑅−1. ∀ 𝑦, 𝑥 ∈ 𝐵𝐴 (𝑦𝑅−1 𝑥 ⇔ 𝑦 = 𝑘𝑥) for some integer 𝑘. 13  Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Directed Graph of a Relation 6.1.3 Directed Graph of a Relation Definition: Relation on a Set A relation on a set 𝑨 is a relation from 𝐴 to 𝐴. In other words, a relation on a set 𝐴 is a subset of 𝐴 × 𝐴. We may write 𝐴2 for 𝐴 × 𝐴. In general, we may write 𝐴𝑛 for 𝐴 × ⋯ × 𝐴 (𝑛 times). The arrow diagram of such a relation can be modified so that it becomes a directed graph. Instead of representing 𝐴 as two separate sets of points, represent 𝐴 only once, and draw an arrow from each point of 𝐴 to its related point. If a point is related to itself, a loop is drawn that extends out from the point and goes back to it. 14 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Directed Graph of a Relation Example #8: Let 𝐴 = {3, 4, 5, 6, 7, 8} and define a relation 𝑅 on 𝐴 as follows: ∀𝑥, 𝑦 ∈ 𝐴, 𝑥𝑅𝑦 ⇔ 2 | (𝑥 – 𝑦). Draw the directed graph of 𝑅. 15 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Composition of Relations 6.1.4 Composition of Relations Definition: Composition of Relations Let 𝐴, 𝐵 and 𝐶 be sets. Let 𝑅 ⊆ 𝐴 × 𝐵 be a relation. Let 𝑆 ⊆ 𝐵 × 𝐶 be a relation. The composition of 𝑅 with 𝑆, denoted 𝑆 ∘ 𝑅, is the relation from 𝐴 to 𝐶 such that: ∀𝑥 ∈ 𝐴, ∀𝑧 ∈ 𝐶 𝑥 𝑆 ∘ 𝑅 𝑧 ⇔ ∃𝑦 ∈ 𝐵 𝑥𝑅𝑦 ∧ 𝑦𝑆𝑧. In other words, 𝑥 ∈ 𝐴 and 𝑧 ∈ 𝐶 are “𝑆 ∘ 𝑅”-related iff there is a “path” from 𝑥 to 𝑧 via some intermediate element 𝑦 ∈ 𝐵 in the arrow diagram. 16 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Composition of Relations Students Module Venue takes held in Ann CS1010 LT15 Bryan CS1231 ICube IS1103 Candy MA1101 SR1 Danny CS2100 Students Venue “held in ∘ takes” = “go to” Ann LT15 Bryan ICube Candy Danny SR1 17 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Composition of Relations Proposition: Composition is Associative Let 𝐴, 𝐵, 𝐶, 𝐷 be sets. Let 𝑅 ⊆ 𝐴 × 𝐵, 𝑆 ⊆ 𝐵 × 𝐶 and 𝑇 ⊆ 𝐶 × 𝐷 be relations. 𝑇∘ 𝑆∘𝑅 = 𝑇∘𝑆 ∘𝑅 =𝑇∘𝑆∘𝑅 Proposition: Inverse of Composition Let 𝐴, 𝐵 and 𝐶 be sets. Let 𝑅 ⊆ 𝐴 × 𝐵 and 𝑆 ⊆ 𝐵 × 𝐶 be relations. (𝑆 ∘ 𝑅)−1 = 𝑅−1 ∘ 𝑆 −1 18 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations N-ary Relations and Relational Databases 6.1.5 N-ary Relations and Relational Databases A relation involving two sets is called binary relation. We can generalize a relation to involve more than two sets. Definition: 𝑛-ary Relation Given 𝑛 sets 𝐴1 , 𝐴2 , ⋯ , 𝐴𝑛 , an 𝒏-ary relation 𝑹 on 𝐴1 × 𝐴2 × ⋯ × 𝐴𝑛 is a subset of 𝐴1 × 𝐴2 × ⋯ × 𝐴𝑛. The special cases of 2-ary, 3-ary and 4-ary relations are called binary, ternary and quaternary relations respectively. 19 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations N-ary Relations and Relational Databases Example #9: (The following is a radically simplified version of a database that might be used in a hospital.) Let 𝐴1 be a set of positive integers, 𝐴2 and 𝐴4 be sets of alphabetic character strings, and 𝐴3 be a set of numeric character strings. Define a quaternary relation 𝑅 on 𝐴1 × 𝐴2 × 𝐴3 × 𝐴4 as follows: (𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ) ∈ 𝑅 ⇔ a patient with ID number 𝑎1 , named 𝑎2 , was admitted on date 𝑎3 with primary diagnosis 𝑎4. At a particular hospital, (011985, John Schmidt, 020710, asthma) this relation might (574329, Tak Kurosawa, 011410, pneumonia) contain these 4-tuples: (466581, Mary Lazars, 010310, appendicitis) (008352, Joan Kaplan, 112409, gastritis) (011985, John Schmidt, 021710, pneumonia) (244388, Sarah Wu, 010310, broken leg) (778400, Jamal Baskers, 122709, appendicitis) 20 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations N-ary Relations and Relational Databases For example, in the database language SQL, if the above database is denoted as 𝑆, the result of the query SELECT Patient_ID#, Name FROM 𝑆 WHERE Admission_Date=010310 would yield a list of the ID numbers and names of all patients admitted on 01-03-10: 466581 Mary Lazars 244388 Sarah Wu This is obtained by taking the intersection of the set 𝐴1 × 𝐴2 × {010310} × 𝐴4 with the database and then projecting onto the first two fields. 21 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations 6.2 Reflexivity, Symmetry and Transitivity 22 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions of Reflexivity, Symmetry and Transitivity 6.2.1 Definitions of Reflexivity, Symmetry and Transitivity Example #10: Let 𝐴 = {2, 3, 4, 6, 7, 9} and define a relation 𝑅 on 𝐴 as follows: ∀𝑥, 𝑦𝐴 𝑥𝑅𝑦 ⇔ 3 𝑥 − 𝑦. The directed graph for 𝑅 is shown below: 1. Each point of the graph has an arrow looping around from it back to itself. 2. Wherever there is an arrow going from one point to another, there is also an arrow going from the second point back to the first. 3. Wherever there is an arrow going from one point to a second and from the second point to a third, there is also an If 𝑛, 𝑑 ∈ ℤ: arrow going from the first point to the 𝑑 | 𝑛 ⟺ ∃𝑘 ∈ ℤ such that 𝑛 = 𝑑𝑘. third. 23 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions of Reflexivity, Symmetry and Transitivity Properties (1), (2), and (3) correspond to properties of general relations called reflexivity, symmetry, and transitivity. Definitions: Reflexivity, Symmetry, Transitivity Let 𝑅 be a relation on a set 𝐴. 1. 𝑅 is reflexive iff ∀𝑥 ∈ 𝐴 (𝑥𝑅𝑥). 2. 𝑅 is symmetric iff ∀𝑥, 𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ⇒ 𝑦𝑅𝑥). 3. 𝑅 is transitive iff ∀𝑥, 𝑦, 𝑧 ∈ 𝐴 (𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧 ⇒ 𝑥𝑅𝑧). Reflexive Symmetric Transitive ∀ ∀ ∀ ∀ ∃ ∃ ∃ 24 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions of Reflexivity, Symmetry and Transitivity Example #11: Let 𝐴 = {0, 1, 2, 3} and define relations 𝑅, 𝑆, and 𝑇 on 𝐴 as follows: 𝑅 = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}, 𝑆 = {(0, 0), (0, 2), (0, 3), (2, 3)}, 𝑇 = {(0, 1), (2, 3)}. a. Is 𝑅 reflexive? symmetric? transitive? Yes Yes No b. Is 𝑆 reflexive? symmetric? transitive? No No Yes c. Is 𝑇 reflexive? symmetric? transitive? No No Yes 25  Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions of Reflexivity, Symmetry and Transitivity Common mistake: Talking about reflexivity. “The element 0 is reflexive.” “The element 0 is related to itself.” “The element 1 is not reflexive.” “The element 1 is not related to itself.” “The element 2 is not reflexive.” “The element 2 is not related to itself.” Reflexivity, symmetry and transitivity are properties of a relation, not properties of members of the set. We say a relation is reflexive or not reflexive. 26 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions of Reflexivity, Symmetry and Transitivity Example #12: Define a relation 𝑅 on ℤ as follows: ∀𝑥, 𝑦 ∈ ℤ 𝑥 𝑅 𝑦 ⇔ 3 | 𝑥 − 𝑦. This relation is called congruence modulo 3. Is 𝑅 reflexive, symmetric, transitive? 𝑅 is reflexive. Proof of Reflexivity: 1. Let 𝑎 be an arbitrarily chosen integer. 2. Now 𝑎 − 𝑎 = 0. 3. But 3 | 0 (since 0 = 3 ∙ 0), hence, 3 |(𝑎 − 𝑎). 4. Therefore 𝑎 𝑅 𝑎 (by the definition of 𝑅). 27 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions of Reflexivity, Symmetry and Transitivity Example #12: Define a relation 𝑅 on ℤ as follows: ∀𝑥, 𝑦 ∈ ℤ 𝑥 𝑅 𝑦 ⇔ 3 | 𝑥 − 𝑦. This relation is called congruence modulo 3. Is 𝑅 reflexive, symmetric, transitive? 𝑅 is symmetric. Proof of Symmetry: 1. Let 𝑎 and 𝑏 be arbitrarily chosen integers that satisfy 𝑎 𝑅 𝑏. 2. Then 3|(𝑎 − 𝑏) (by the definition of 𝑅), hence 𝑎 − 𝑏 = 3𝑘 for some integer 𝑘 (by the definition of divisibility). 3. Multiplying both sides by −1 gives 𝑏 − 𝑎 = 3 −𝑘. 4. Since – 𝑘 is an integer, 3|(𝑏 − 𝑎) (by definition of divisibility). 5. Therefore 𝑏 𝑅 𝑎 (by the definition of 𝑅). 28 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definitions of Reflexivity, Symmetry and Transitivity Example #12: Define a relation 𝑅 on ℤ as follows: ∀𝑥, 𝑦 ∈ ℤ 𝑥 𝑅 𝑦 ⇔ 3 | 𝑥 − 𝑦. This relation is called congruence modulo 3. Is 𝑅 reflexive, symmetric, transitive? 𝑅 is transitive. Proof of Transitivity: 1. Let 𝑎, 𝑏 and 𝑐 be arbitrarily chosen integers that satisfy 𝑎 𝑅 𝑏 and 𝑏 𝑅 𝑐. 2. Then 3|(𝑎 − 𝑏) and 3|(𝑏 − 𝑐) (by the definition of 𝑅), hence 𝑎 − 𝑏 = 3𝑟 and 𝑏 − 𝑐 = 3𝑠 for some integers 𝑟 and 𝑠 (by the definition of divisibility). 3. Adding both equations gives 𝑎 − 𝑐 = 3 𝑟 + 𝑠. 4. Since 𝑟 + 𝑠 is an integer, 3|(𝑎 − 𝑐) (by definition of divisibility). 5. Therefore 𝑎 𝑅 𝑐 (by the definition of 𝑅). 29 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Transitive Closure of a Relation 6.2.2 The Transitive Closure of a Relation Generally speaking, a relation fails to be transitive because it fails to contain certain ordered pairs. For example, if (1, 3) and (3, 4) are in a relation R, then the pair (1, 4) must be in R for R to be transitive. To obtain a transitive relation from one that is not transitive, it is necessary to add ordered pairs. Roughly speaking, the relation obtained by adding the least number of ordered pairs to ensure transitivity is called the transitive closure of the relation. 30 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Transitive Closure of a Relation In a sense made precise by the formal definition, the transitive closure of a relation is the smallest transitive relation that contains the relation. Definition: Transitive Closure Let 𝐴 be a set and 𝑅 a relation on 𝐴. The transitive closure of 𝑅 is the relation 𝑅𝑡 on 𝐴 that satisfies the following three properties: 1. 𝑅𝑡 is transitive. 2. 𝑅 ⊆ 𝑅𝑡. 3. If 𝑆 is any other transitive relation that contains 𝑅, then 𝑅𝑡 ⊆ 𝑆. 31 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Transitive Closure of a Relation Example #13: Let 𝐴 = {0, 1, 2, 3} and consider the relation 𝑅 defined on 𝐴 as follows: 𝑅 = {(0, 1), (1, 2), (2, 3)}. Find the transitive closure of 𝑅. Since there are arrows from 0 to 1 and from 1 to 2, 𝑅𝑡 must have an arrow from 0 to 2. Hence (0,2) ∈ 𝑅𝑡. Then (0,2) ∈ 𝑅𝑡 and (2,3) ∈ 𝑅𝑡 , so (0,3) ∈ 𝑅𝑡. Also, since (1,2) ∈ 𝑅𝑡 and (2,3) ∈ 𝑅𝑡 , so (1,3) ∈ 𝑅𝑡. Directed graph of 𝑅𝑡 : 32 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations 6.3 Equivalence Relations 33 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Relation Induced by a Partition 6.3.1 The Relation Induced by a Partition A partition of a set 𝐴 is a finite or infinite collection of nonempty, mutually disjoint subsets whose union is 𝐴. The diagram below illustrates a partition of a set 𝐴 = {𝑏, 𝑒, 𝑓, 𝑘, 𝑚, 𝑝} by subsets 𝑏, 𝑝 , 𝑓, 𝑚 , 𝑘 , {𝑒}. 𝑏 𝑏𝑝 𝑘 𝑝 𝑘 𝑓 𝑚 𝑒 𝑓𝑚 𝑒 34 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Relation Induced by a Partition Definition: Partition C is a partition of a set 𝐴 if the following hold: (1) C is a set of which all elements are non-empty subsets of 𝐴, i.e., ∅ ≠ 𝑆 ⊆ 𝐴 for all 𝑆 ∈ C. (2) Every element of 𝐴 is in exactly one element of C, i.e., ∀𝑥 ∈ 𝐴 ∃𝑆 ∈ C (𝑥 ∈ 𝑆) and ∀𝑥 ∈ 𝐴 ∀𝑆1 , 𝑆2 ∈ C (𝑥 ∈ 𝑆1 ∧ 𝑥 ∈ 𝑆2 ⇒ 𝑆1 = 𝑆2 ). Elements of a partition are called components of the partition. 𝑏 𝑏𝑝 𝑘 𝑝 𝑘 𝑓 𝑚 𝑒 𝑓𝑚 𝑒 𝐴 = {𝑏, 𝑒, 𝑓, 𝑘, 𝑚, 𝑝} C= 𝑏, 𝑝 , 𝑓, 𝑚 , 𝑘 , 𝑒 35 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Relation Induced by a Partition Definition: Partition C is a partition of a set 𝐴 if the following hold: (1) C is a set of which all elements are non-empty subsets of 𝐴, i.e., ∅ ≠ 𝑆 ⊆ 𝐴 for all 𝑆 ∈ C. (2) Every element of 𝐴 is in exactly one element of C, i.e., ∀𝑥 ∈ 𝐴 ∃𝑆 ∈ C (𝑥 ∈ 𝑆) and ∀𝑥 ∈ 𝐴 ∀𝑆1 , 𝑆2 ∈ C (𝑥 ∈ 𝑆1 ∧ 𝑥 ∈ 𝑆2 ⇒ 𝑆1 = 𝑆2 ). Elements of a partition are called components of the partition. Definition (shorter): Partition A partition of set 𝐴 is a set C of non-empty subsets of 𝐴 such that ∀𝑥 ∈ 𝐴 ∃! 𝑆 ∈ C 𝑥 ∈ 𝑆. (Recall: ∃! means “there exists a unique”.) 36 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Relation Induced by a Partition Partitions as relations We may view a partition as a “is in the same component as” relation. Let 𝑅 be “in the same component as” relation. 𝑏 𝑝 𝑏𝑅𝑏 𝑓𝑅𝑓 𝑝𝑅𝑝 𝑚𝑅𝑚 𝑏𝑅𝑝 𝑓𝑅𝑚 𝑏𝑝 𝑓 𝑚 𝑝𝑅𝑏 𝑚𝑅𝑓 𝑘 𝑘𝑅𝑘 𝑒𝑅𝑒 𝑓𝑚 𝑘 𝑒 𝑒 37 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Relation Induced by a Partition Definition: Relation Induced by a Partition Given a partition C of a set 𝐴, the relation 𝑅 induced by the partition is defined on 𝐴 as follows: ∀𝑥, 𝑦 ∈ 𝐴, 𝑥𝑅𝑦 ⇔ ∃ a component 𝑆 of C s.t. 𝑥, 𝑦 ∈ 𝑆. Example #14: Let 𝐴 = {0, 1, 2, 3, 4} and consider this partition of 𝐴: 0, 3, 4 , 1 , 2. Find the relation 𝑅 induced by this partition. {0,3,4} is a component of the partition  0R0, 0R3, 0R4, 3R0, 3R3, 3R4, 4R0, 4R3 and 4R4. {1} is a component of the partition  1R1. {2} is a component of the partition  2R2. Therefore, R = {(0,0), (0,3), (0,4), (1,1), (2,2), (3,0), (3,3), (3,4), (4,0), (4,3), (4,4)}. 38 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations The Relation Induced by a Partition The fact is that a relation induced by a partition of a set satisfies all three properties: reflexivity, symmetry, and transitivity. Theorem 8.3.1 Relation Induced by a Partition Let 𝐴 be a set with a partition and let 𝑅 be the relation induced by the partition. Then 𝑅 is reflexive, symmetric, and transitive. 39 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definition of an Equivalence Relation 6.3.2 Definition of an Equivalence Relation A relation on a set that satisfies the three properties of reflexivity, symmetry, and transitivity is called an equivalence relation. Definition: Equivalence Relation Let 𝐴 be a set and 𝑅 a relation on 𝐴. 𝑅 is an equivalence relation iff 𝑅 is reflexive, symmetric and transitive. Note: The symbol ~ is commonly used to denote an equivalence relation. 40 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definition of an Equivalence Relation Example #15: Let 𝑋 be the set of all nonempty subsets of {1, 2, 3}. Then 𝑋 = {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}. Define a relation R on 𝑋 as follows: For all 𝐴, 𝐵 ∈ 𝑋, 𝐴 R 𝐵 ⇔ the least element of 𝐴 equals the least element of 𝐵. Prove that R is an equivalence relation on 𝑋. R is reflexive: Suppose 𝐴 is a nonempty subset of {1, 2, 3}. It is true to say that the least element of 𝐴 equals the least element of 𝐴. Thus, 𝐴 R 𝐴. R is symmetric: Suppose 𝐴 and 𝐵 are nonempty subsets of {1, 2, 3} and 𝐴 R 𝐵. Since 𝐴 R 𝐵, the least element of 𝐴 equals the least element of 𝐵. But this means that the least element of 𝐵 equals the least element of 𝐴, and so by definition of R, 𝐵 R 𝐴. 41 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Definition of an Equivalence Relation Example #15: Let 𝑋 be the set of all nonempty subsets of {1, 2, 3}. Then 𝑋 = {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}. Define a relation R on 𝑋 as follows: For all 𝐴, 𝐵 ∈ 𝑋, 𝐴 R 𝐵 ⇔ the least element of 𝐴 equals the least element of 𝐵. Prove that R is an equivalence relation on 𝑋. R is transitive: Suppose 𝐴, 𝐵 and 𝐶 are nonempty subsets of {1, 2, 3}, 𝐴 R 𝐵 and 𝐵 R 𝐶. Since 𝐴 R 𝐵, the least element of 𝐴 equals the least element of 𝐵 and since 𝐵 R 𝐶, the least element of 𝐵 equals the least element of 𝐶. Thus the least element of 𝐴 equals the least element of 𝐶, and so, by definition of R, 𝐴 R 𝐶. 42 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation 6.3.3 Equivalence Classes of an Equivalence Relation Suppose there is an equivalence relation on a certain set. If 𝑎 is any particular element of the set, then one can ask, “What are the elements that are related to 𝑎?” This set of elements is called the equivalence class of 𝑎. Definition: Equivalence Class Suppose 𝐴 is a set and ~ is an equivalence relation on 𝐴. For each 𝑎 ∈ 𝐴, the equivalence class of 𝑎, denoted [𝒂] and called the class of 𝒂 for short, is the set of all elements 𝑥 ∈ 𝐴 s.t. 𝑎 is ~-related to 𝑥. Symbolically, [𝑎]~ = {𝑥 ∈ 𝐴 ∶ 𝑎~𝑥 } 43 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation The procedural version of this definition is ∀𝑥 ∈ 𝐴 𝑥 ∈ [𝑎]~ ⇔ 𝑎~𝑥. 𝑎 [𝑎]~ (When there is no risk of confusion, we may drop the subscript ~ and write [𝑎].) 44 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation Example #16: Let 𝐴 = {0, 1, 2, 3, 4} and define a relation 𝑅 on 𝐴 as follows: 𝑅 = {(0, 0), (0, 4), (1, 1), (1, 3), (2, 2), (3, 1), (3, 3), (4, 0), (4, 4)}. The directed graph for 𝑅 is as shown below. As can be seen by inspection, 𝑅 is an equivalence relation on 𝐴. Find the distinct equivalence classes of 𝑅. 45 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation First find the equivalence class of every element of 𝐴. 0 = {𝑥 ∈ 𝐴 ∶ 0 𝑅 𝑥} = 0,4 1 = {𝑥 ∈ 𝐴 ∶ 1 𝑅 𝑥} = {1,3} 2 = {𝑥 ∈ 𝐴 ∶ 2 𝑅 𝑥} = {2} 3 = {𝑥 ∈ 𝐴 ∶ 3 𝑅 𝑥} = {1,3} 4 = {𝑥 ∈ 𝐴 ∶ 4 𝑅 𝑥} = {0,4} Note that = and =. Thus the distinct equivalence classes of the relation are {0, 4}, {1, 3}, and {2}. 46 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation Lemma Rel.1 Equivalence Classes We prove this by proving: Let ~ be an equivalence relation on a set 𝐴. The following are equivalent for all 𝑥, 𝑦 ∈ 𝐴. (i) (i) 𝑥~𝑦. (ii) 𝑥 = 𝑦. (iii) 𝑥 ∩ [𝑦] ≠ ∅. (iii) ⇐ (ii) Proof Definition: 1. ((i) ⇒ (ii)) [𝑎]~ = {𝑥 ∈ 𝐴 ∶ 𝑎~𝑥 } 1.1. Suppose 𝑥~𝑦. 1.2. Then 𝑦~𝑥 by symmetry. 1.3. For every 𝑧 ∈ [𝑥], 1.3.1. 𝑥~𝑧 by the definition of [𝑥]; 1.3.2. ∴ 𝑦~𝑧 by transitivity, as 𝑦~𝑥; 1.3.3. ∴ 𝑧 ∈ [𝑦] by the definition of [𝑦]. 1.4. This shows 𝑥 ⊆ 𝑦. 1.5. Switching the roles of 𝑥 and 𝑦, we see also that 𝑦 ⊆ 𝑥. 1.6. Therefore, [𝑥] = [𝑦]. 47 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation Lemma Rel.1 Equivalence Classes We prove this by proving: Let ~ be an equivalence relation on a set 𝐴. The following are equivalent for all 𝑥, 𝑦 ∈ 𝐴. (i) (i) 𝑥~𝑦. (ii) 𝑥 = 𝑦. (iii) 𝑥 ∩ [𝑦] ≠ ∅. (iii) ⇐ (ii) Proof Definition: 2. ((ii) ⇒ (iii)) [𝑎]~ = {𝑥 ∈ 𝐴 ∶ 𝑎~𝑥 } 2.1. Suppose [𝑥] = [𝑦]. 2.2. Then 𝑥 ∩ 𝑦 = [𝑥] by the Idempotent Law for ∩. 2.3. However, we know 𝑥~𝑥 by the reflexivity of ~. 2.4. This shows 𝑥 ∈ 𝑥 = 𝑥 ∩ 𝑦 by the definition of [x] and line 2.2. 2.5. Therefore, 𝑥 ∩ [𝑦] ≠ ∅. 48 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation Lemma Rel.1 Equivalence Classes We prove this by proving: Let ~ be an equivalence relation on a set 𝐴. The following are equivalent for all 𝑥, 𝑦 ∈ 𝐴. (i) (i) 𝑥~𝑦. (ii) 𝑥 = 𝑦. (iii) 𝑥 ∩ [𝑦] ≠ ∅. (iii) ⇐ (ii) Proof Definition: 3. ((iii) ⇒ (i)) [𝑎]~ = {𝑥 ∈ 𝐴 ∶ 𝑎~𝑥 } 3.1. Suppose 𝑥 ∩ [𝑦] ≠ ∅. 3.2. Take 𝑧 ∈ 𝑥 ∩ 𝑦. 3.3. Then 𝑧 ∈ 𝑥 and 𝑧 ∈ 𝑦 by the definition of ∩. 3.4. Then 𝑥~𝑧 and 𝑦~𝑧. by the definition of [𝑥] and [𝑦]. 3.5. 𝑦~𝑧 implies 𝑧~𝑦 by symmetry. 3.6. Therefore, 𝑥~𝑦 by transitivity. 49 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation Theorem 8.3.4 The Partition Induced by an Equivalence Relation If 𝐴 is a set and 𝑅 is an equivalence relation on 𝐴, then the distinct equivalence classes of 𝑅 form a partition of 𝐴; that is, the union of the equivalence classes is all of 𝐴, and the intersection of any two distinct classes is empty. 50 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation Revisit Example #12: Define a relation 𝑅 on ℤ as follows: ∀𝑥, 𝑦 ∈ ℤ 𝑥 𝑅 𝑦 ⇔ 3 | 𝑥 − 𝑦. This relation is called congruence modulo 3. It has been shown that 𝑅 is an equivalence relation. What are the distinct equivalence classes of 𝑅? The distinct equivalent classes of 𝑅 are:  {3𝑘 ∶ 𝑘 ∈ ℤ}, {…,-9,-6,-3,0,3,6,9,…}  {3𝑘 + 1 ∶ 𝑘 ∈ ℤ}, and {…,-8,-5,-2,1,4,7,10,…}  3𝑘 + 2 ∶ 𝑘 ∈ ℤ. {…,-7,-4,-1,2,5,8,11,…} Observe that {{…,-9,-6,-3,0,3,6,9,…}, {…,-8,-5,-2,1,4,7,10,…}, {…,-7,-4,-1,2,5,8,11,…}} is a partition of ℤ. 51 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Equivalence Classes of an Equivalence Relation Congruence modulo 𝑛 (congruence-mod-𝑛) relation: ∀𝑥, 𝑦 ∈ ℤ 𝑥 𝑅 𝑦 ⇔ 𝑛 | 𝑥 − 𝑦. Congruence modulo 2 Congruence modulo 3 Congruence modulo 4 : : ℤ : : : ℤ : : : : ℤ -4 -3 -6 -5 -4 -8 -7 -6 -5 -2 -1 -3 -2 -1 -4 -3 -2 -1 0 1 0 1 2 0 1 2 3 2 3 3 4 5 4 5 6 7 4 5 6 7 8 8 9 10 11 : : : : : : : : : Partition of ℤ: Partition of ℤ: Partition of ℤ: 2𝑘 ∶ 𝑘 ∈ ℤ , 3𝑘 ∶ 𝑘 ∈ ℤ , 4𝑘 ∶ 𝑘 ∈ ℤ , 2𝑘 + 1 ∶ 𝑘 ∈ ℤ 3𝑘 + 1 ∶ 𝑘 ∈ ℤ , 4𝑘 + 1 ∶ 𝑘 ∈ ℤ , 3𝑘 + 2 ∶ 𝑘 ∈ ℤ 4𝑘 + 2 ∶ 𝑘 ∈ ℤ , 4𝑘 + 3 ∶ 𝑘 ∈ ℤ 52 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Congruence 6.3.4 Congruence Definition: Divisibility Let 𝑛, 𝑑 ∈ ℤ. Then 𝑑 | 𝑛 ⇔ 𝑛 = 𝑑𝑘 for some 𝑘 ∈ ℤ. Definition: Congruence Let 𝑎, 𝑏 ∈ ℤ and 𝑛 ∈ ℤ+. Then 𝑎 is congruent to 𝑏 modulo 𝑛 iff 𝑎 − 𝑏 = 𝑛𝑘 for some 𝑘 ∈ ℤ. In other words, 𝑛 | 𝑎 − 𝑏. In this case, we write 𝑎 ≡ 𝑏 (mod 𝑛). Example #17: Are the following true? (a) 7 ≡ 1 (mod 2) Yes, because 7 − 1 = 6 = 2 × 3. 𝑘 = 3. (b) −3 ≡ 12 (mod 5) Yes, because −3 − 12 = −15 = 5 × (−3). 𝑘 = −3. (c) −4 ≡ 5 (mod 7) No, because −4 − 5 = −9 ≠ 7𝑘 for any 𝑘 ∈ ℤ. 53  Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Congruence Proposition Congruence-mod 𝑛 is an equivalence relation on ℤ for every 𝑛 ∈ ℤ+. Proof: A relation 𝑅 on a set 𝐴 is 1. (Reflexivity) For all 𝑎 ∈ ℤ, reflexive: ∀𝑥 ∈ 𝐴 (𝑥 𝑅 𝑥); 1.1. 𝑎 − 𝑎 = 0 = 𝑛 × 0. symmetric: 1.2. So 𝑎 ≡ 𝑎 (mod 𝑛) by the defn of congruence. ∀𝑥, 𝑦 ∈ 𝐴 (𝑥 𝑅 𝑦 ⇒ 𝑦 𝑅 𝑥); transitive: 2. (Symmetry) ∀𝑥, 𝑦, 𝑧 ∈ 𝐴 2.1. Let 𝑎, 𝑏 ∈ ℤ such that 𝑎 ≡ 𝑏 (mod 𝑛). 𝑥𝑅𝑦∧𝑦𝑅𝑧 ⇒𝑥𝑅𝑧. 2.2. Then there is a 𝑘 ∈ ℤ such that 𝑎 − 𝑏 = 𝑛𝑘. 2.3. Then 𝑏 − 𝑎 = − 𝑎 − 𝑏 = −𝑛𝑘 = 𝑛(−𝑘). 2.4. −𝑘 ∈ ℤ (by closure of integers under ×), so 𝑏 ≡ 𝑎 (mod 𝑛) by the definition of congruence. 3. (Transitivity) 3.1. Let 𝑎, 𝑏, 𝑐 ∈ ℤ such that 𝑎 ≡ 𝑏 (mod 𝑛) and 𝑏 ≡ 𝑐 (mod 𝑛). 3.2. Then there are 𝑘, 𝑙 ∈ ℤ such that 𝑎 − 𝑏 = 𝑛𝑘 and 𝑏 − 𝑐 = 𝑛l. 3.3. Then 𝑎 − 𝑐 = 𝑎 − 𝑏 + 𝑏 − 𝑐 = 𝑛𝑘 + 𝑛𝑙 = 𝑛(𝑘 + 𝑙). 3.4. 𝑘 + 𝑙 ∈ ℤ (by closure of integers under +), so 𝑎 ≡ 𝑐 (mod 𝑛) by the definition of congruence. 54 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Congruence Definition: Equivalence Class Suppose 𝐴 is a set and ~ is an equivalence relation on 𝐴. The equivalence class of 𝑎 ∈ 𝐴, is [𝑎]~ = {𝑥 ∈ 𝐴 ∶ 𝑎~𝑥 }. Congruence modulo 4 Congruence: Equivalence classes : : : : ℤ + Let 𝑛 ∈ ℤ. The equivalence classes w.r.t. the -8 -7 -6 -5 congruence-mod-𝑛 relation on are of the form: -4 -3 -2 -1 0 1 2 3 𝑥 = 𝑦 ∈ ℤ ∶ 𝑥 ≡ 𝑦 mod 𝑛 4 5 6 7 = {𝑦 ∈ ℤ ∶ 𝑥 − 𝑦 = 𝑛𝑘 for some 𝑘 ∈ ℤ} 8 9 10 11 : : : : = 𝑥 + 𝑛𝑘 ∶ 𝑘 ∈ ℤ = … , 𝑥 − 2𝑛, 𝑥 − 𝑛, 𝑥, 𝑥 + 𝑛, 𝑥 + 2𝑛, … where 𝑥 ∈ ℤ. Note that for all 𝑥 ∈ ℤ, 𝑥 + 𝑛 = … , 𝑥 − 𝑛, 𝑥, 𝑥 + 𝑛, 𝑥 + 2𝑛, 𝑥 + 3𝑛, … = 𝑥. For example, if 𝑛 = 4, then ⋯ = −8 = −4 = 0 = 4 = ⋯ and ⋯ = −7 = −3 = 1 = 5 = ⋯ and so on. 55 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Dividing a Set by an Equivalence Relation 6.3.5 Dividing a Set by an Equivalence Relation Definition: Set of equivalence classes Let 𝐴 be a set and ~ be an equivalence relation on 𝐴. Denote by 𝐴/~ the set of all equivalence classes with respect to ~, i.e., 𝐴/~ = 𝑥 ~ ∶ 𝑥 ∈ 𝐴. We may read 𝐴/~ as “the quotient of 𝐴 by ~”. Example #18: Let 𝑛 ∈ ℤ+. If ~𝑛 denotes the congruence-mod-𝑛 relation on ℤ, then ℤ/~𝑛 = { 𝑥 ∶ 𝑥 ∈ ℤ} = 𝑛𝑘 ∶ 𝑘 ∈ ℤ , 𝑛𝑘 + 1 ∶ 𝑘 ∈ ℤ , ⋯ , 𝑛𝑘 + 𝑛 − 1 ∶ 𝑘 ∈ ℤ. 56 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Dividing a Set by an Equivalence Relation Theorem Rel.2 Equivalence classes form a partition 𝐴/~ = 𝑥 ~ ∶𝑥∈𝐴. C is a partition of a set 𝐴 if: Let ~ be an equivalence relation on a set 𝐴. (1) C is a set of which all Then 𝐴/~ is a partition of 𝐴. elements are nonempty subsets of 𝐴. Proof: (2) Every element of 𝐴 is in 1. 𝐴/~ is by definition a set. exactly one element of C. 2. We show that every element of 𝐴/~ is a nonempty subset of 𝐴. 2.1. Let 𝑆 ∈ 𝐴/~. 2.2. Use the definition of 𝐴/~ to find 𝑥 ∈ 𝐴 such that 𝑆 = [𝑥]. 2.3. Then 𝑆 = [𝑥] ⊆ 𝐴 in view of the definition of equivalence classes. 2.4. 𝑥~𝑥 by the reflexivity of ~. 2.5. Hence 𝑥 ∈ 𝑥 = 𝑆 by the definition of [x]. 2.6. In particular, we know 𝑆 is nonempty. 3. We show that every element of 𝐴 is in at least one element of 𝐴/~. 4. We show that every element of 𝐴 is in at most one element of 𝐴/~. 57 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Dividing a Set by an Equivalence Relation Theorem Rel.2 Equivalence classes form a partition 𝐴/~ = 𝑥 ~ ∶𝑥∈𝐴. C is a partition of a set 𝐴 if: Let ~ be an equivalence relation on a set 𝐴. (1) C is a set of which all Then 𝐴/~ is a partition of 𝐴. elements are nonempty subsets of 𝐴. Proof: (2) Every element of 𝐴 is in 1. 𝐴/~ is by definition a set. exactly one element of C. 2. We show that every element of 𝐴/~ is a nonempty subset of 𝐴. 3. We show that every element of 𝐴 is in at least one element of 𝐴/~. 3.1. Let 𝑥 ∈ 𝐴. 3.2. 𝑥~𝑥 by the reflexivity of ~. 3.3. So 𝑥 ∈ 𝑥 ∈ 𝐴/~. 4. We show that every element of 𝐴 is in at most one element of 𝐴/~. 58 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Dividing a Set by an Equivalence Relation Theorem Rel.2 Equivalence classes form a partition 𝐴/~ = 𝑥 ~ ∶𝑥∈𝐴. C is a partition of a set 𝐴 if: Let ~ be an equivalence relation on a set 𝐴. (1) C is a set of which all Then 𝐴/~ is a partition of 𝐴. elements are nonempty subsets of 𝐴. Proof: (2) Every element of 𝐴 is in 1. 𝐴/~ is by definition a set. exactly one element of C. 2. We show that every element of 𝐴/~ is a nonempty subset of 𝐴. 3. We show that every element of 𝐴 is in at least one element of 𝐴/~. 4. We show that every element of 𝐴 is in at most one element of 𝐴/~. 4.1. Let 𝑥 ∈ 𝐴 that is in two elements of 𝐴/~, say 𝑆1 and 𝑆2. 4.2. Use the definition of 𝐴/~ to find 𝑦1 , 𝑦2 ∈ 𝐴 such that 𝑆1 = [𝑦1 ] and 𝑆2 = [𝑦2 ]. 4.3. 𝑥 ∈ [𝑦1 ] ∩ [𝑦2 ] by lines 4.1 and 4.2. 4.4. So [𝑦1 ] ∩ [𝑦2 ] ≠ ∅. 4.5. Therefore 𝑆1 = 𝑦1 = 𝑦2 = 𝑆2 by lemma: equivalence classes. Lemma Rel.1 Equivalence Classes Let ~ be an equivalence relation on a set 𝐴. The following are equivalent for all 𝑥, 𝑦 ∈ 𝐴. (i) 𝑥~𝑦; (ii) 𝑥 = 𝑦 ; (iii) 𝑥 ∩ [𝑦] ≠ ∅. 59 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Summary 6.3.6 Summary Definition: A relation on set 𝐴 is a subset of 𝐴2. Definition: If 𝑅 is a relation on a set 𝐴, then we write 𝑥 𝑅 𝑦 for (𝑥, 𝑦) ∈ 𝑅. Definition: A partition of a set 𝐴 is a set C of non-empty subsets of 𝐴 such that ∀𝑥 ∈ 𝐴 ∃! 𝑆 ∈ C 𝑥 ∈ 𝑆. Definition: A relation 𝑅 on 𝐴 is an equivalence relation if  (reflexivity) ∀𝑥 ∈ 𝐴 (𝑥 𝑅 𝑥);  (symmetry) ∀𝑥, 𝑦 ∈ 𝐴 (𝑥 𝑅 𝑦 ⇒ 𝑦 𝑅 𝑥); and  (transitivity) ∀𝑥, 𝑦, 𝑧 ∈ 𝐴 (𝑥 𝑅 𝑦 ∧ 𝑦𝑅𝑧 ⇒ 𝑦 𝑅 𝑧). Definition: Let ~ be an equivalence relation on 𝐴. Then the set of equivalence classes is denoted by 𝐴/~ = 𝑥 ~ ∶ 𝑥 ∈ 𝐴 , where 𝑥 ~ = {𝑦 ∈ 𝐴 ∶ 𝑥~𝑦}. Proposition: The same-component relation w.r.t. a partition is an equivalence relation. Theorem Rel.2: If ~ is an equivalence relation on 𝐴, then 𝐴/~ is a partition of 𝐴. 60 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Summary Informal descriptions of the terms 𝑏 𝑘 𝑝 𝑓 𝑚 𝑒 1. underlying set 𝐴 the set to be “partitioned” 2. components 𝑆 subsets of 𝐴, mutually disjoint, 𝐴 = {𝑏, 𝑒, 𝑓, 𝑘, 𝑚, 𝑝} together union to 𝐴 3. partition C the set of all components 4. same-component relation ~ equivalence relation 𝑏𝑝 𝑘 1. underlying set 𝐴 the set of all vertices 2. relation 𝑅 the set of all arrows 𝑓𝑚 𝑒 3. equivalence relation ~ if ignoring directions of arrows one can walk from 𝑥 to 𝑦, then C = 𝑏, 𝑝 , 𝑓, 𝑚 , 𝑘 , 𝑒 there is an arrow from 𝑥 to 𝑦 4. equivalence classes [𝑥] connected components 5. quotient 𝐴/~ the set of all connected components 𝑏 𝑝 𝑓 𝑚 𝑘 𝑒 61 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations 6.4 Partial Order Relations 62 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Antisymmetry 6.4.1 Antisymmetry Definition: Antisymmetry Let 𝑅 be a relation on a set 𝐴. 𝑅 is antisymmetric iff ∀𝑥, 𝑦 ∈ 𝐴 (𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑥 ⇒ 𝑥 = 𝑦).  ∀ or ∀ By taking the negation of the definition, you can see that a relation 𝑅 is not antisymmetric iff ∃𝑥, 𝑦 ∈ 𝐴 (𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑥 ∧ 𝑥 ≠ 𝑦). 63 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Antisymmetry The big question: Is antisymmetry ≡ ~(symmetry)? Let 𝑅 be a relation on a set 𝐴. 𝑅 is symmetric iff ∀𝑥, 𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ⇒ 𝑦𝑅𝑥). 𝑅 is not symmetric iff ∃𝑥, 𝑦 ∈ 𝐴 𝑥𝑅𝑦 ∧ 𝑦𝑅𝑥.  𝑅 is antisymmetric ≡ iff ∀𝑥, 𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∧ 𝑦𝑅𝑥 ⇒ 𝑥 = 𝑦). ? 64  Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Antisymmetry Example #19: Let 𝑅1 be the “divides” relation on the set of all positive integers, and let 𝑅2 be the “divides” relation on the set of all integers. ∀𝑎, 𝑏 ∈ ℤ+ , 𝑎𝑅1 𝑏 ⇔ 𝑎|𝑏. ∀𝑎, 𝑏 ∈ ℤ, 𝑎𝑅2 𝑏 ⇔ 𝑎|𝑏. a. Is 𝑅1 antisymmetric? Prove or give a counterexample. 𝑅1 is antisymmetric: 1. Suppose 𝑎, 𝑏 ∈ ℤ+ such that 𝑎𝑅1 𝑏 and 𝑏𝑅1 𝑎. 2. Then 𝑏 = 𝑟𝑎 and 𝑎 = 𝑠𝑏 for some integers 𝑟 and 𝑠 (by definition of “divides”). It follows that 𝑏 = 𝑟𝑎 = 𝑟 𝑠𝑏. 3. Dividing both sides by 𝑏 gives 1 = 𝑟𝑠. 4. The only product of two positive integers that equals 1 is 1 ∙ 1. 5. Thus 𝑟 = 𝑠 = 1, and so 𝑎 = 𝑠𝑏 = 1 ∙ 𝑏 = 𝑏. 65 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Antisymmetry Example #19: Let 𝑅1 be the “divides” relation on the set of all positive integers, and let 𝑅2 be the “divides” relation on the set of all integers. ∀𝑎, 𝑏 ∈ ℤ+ , 𝑎𝑅1 𝑏 ⇔ 𝑎|𝑏. ∀𝑎, 𝑏 ∈ ℤ, 𝑎𝑅2 𝑏 ⇔ 𝑎|𝑏. a. Is 𝑅1 antisymmetric? Prove or give a counterexample. Alternatively, we may use Theorem 4.3.1 (see lecture #4): For all 𝑎, 𝑏 ∈ ℤ+ , if 𝑎 | 𝑏 then 𝑎 ≤ 𝑏. 𝑅1 is antisymmetric: 1. Suppose 𝑎, 𝑏 ∈ ℤ+ such that 𝑎𝑅1 𝑏 and 𝑏𝑅1 𝑎. 2. Then 𝑎 ≤ 𝑏 and 𝑏 ≤ 𝑎 by Theorem 4.3.1. 3. So 𝑎 = 𝑏. 66 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Antisymmetry Example #19: Let 𝑅1 be the “divides” relation on the set of all positive integers, and let 𝑅2 be the “divides” relation on the set of all integers. ∀𝑎, 𝑏 ∈ ℤ+ , 𝑎𝑅1 𝑏 ⇔ 𝑎|𝑏. ∀𝑎, 𝑏 ∈ ℤ, 𝑎𝑅2 𝑏 ⇔ 𝑎|𝑏. b. Is 𝑅2 antisymmetric? Prove or give a counterexample. 𝑅2 is not antisymmetric. Counterexample: Let 𝑎 = 2 and 𝑏 = −2. Then 𝑎|𝑏 and 𝑏|𝑎. Hence 𝑎𝑅2 𝑏 and 𝑏𝑅2 𝑎 but 𝑎 ≠ 𝑏. 67  Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Partial Order Relations 6.4.2 Partial Order Relations Definition: Partial Order Relation Let 𝑅 be a relation on a set 𝐴. Then 𝑅 is a partial order relation (or simply partial order) iff 𝑅 is reflexive, antisymmetric and transitive. Two fundamental partial order relations are the “less than or equal to (≤)” relation on a set of real numbers and the “subset (⊆)” relation on a set of sets. Definition: Partially Ordered Set A set 𝐴 is called a partially ordered set (or poset) with respect to a partial order relation 𝑅 on 𝐴, denoted by (𝐴, 𝑅). 68 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Partial Order Relations Example #20: Let | be the “divides” relation on a set 𝐴 of positive integers. That is, ∀𝑎, 𝑏 ∈ 𝐴, 𝑎|𝑏 ⇔ 𝑏 = 𝑘𝑎 for some integer 𝑘. Prove that | is a partial order relation on 𝐴. | is reflexive: Suppose 𝑎 ∈ 𝐴. Then 𝑎 = 1𝑎, so 𝑎|𝑎 (by the definition of divisibility). | is antisymmetric: To show that ∀𝑎, 𝑏 ∈ 𝐴, 𝑎 𝑏 ∧ 𝑏 𝑎 → 𝑎 = 𝑏. Proof identical to Exercise #19a. | is transitive: To show that for ∀𝑎, 𝑏, 𝑐 ∈ 𝐴, 𝑎 𝑏 ∧ 𝑏 𝑐 → 𝑎|𝑐. This is Theorem 4.3.3 (5th: 4.4.3). Theorem 4.3.3 (5th: 4.4.3) Transitivity of Divisibility For all integers 𝑎, 𝑏 and 𝑐, if 𝑎 | 𝑏 and 𝑏 | 𝑐, then 𝑎 | 𝑐. 69 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Partial Order Relations Theorem 4.3.3 (5th: 4.4.3) Transitivity of Divisibility For all integers 𝑎, 𝑏 and 𝑐, if 𝑎 | 𝑏 and 𝑏 | 𝑐, then 𝑎 | 𝑐. Proof: 1. Suppose 𝑎, 𝑏, 𝑐 are integers such that 𝑎 | 𝑏 and 𝑏 | 𝑐. 2. Then 𝑏 = 𝑎𝑟 and 𝑐 = 𝑏𝑠 for some integers 𝑟 and 𝑠 by the definition of divisibility. 3. 𝑐 = 𝑏𝑠 = (𝑎𝑟)𝑠 = 𝑎(𝑟𝑠) by basic algebra. 4. Therefore 𝑎 | 𝑐 since 𝑟𝑠 is an integer (by closure of integers under ×). 70 Relations on Sets Reflexivity, Symmetry and Transitivity Equivalence Relations Partial Order Relations Partial Order Relations Exercise: Let ≤ be the “less than or equal to” relation on ℚ. Show that ≤ is a partial order. Notation Because of the special paradigmatic role played by the ≤ relation in the study of partial order relations, the symbol ≼ is often used to refer to a general partial order, and the notation 𝑥 ≼ 𝑦 is read “𝑥 is curly less than or equal to 𝑦”. Question: The “less than” relation on ℚ is denoted as