Download Relations and Their Properties: Definition, Types, and Composition and more Lecture notes Calculus in PDF only on Docsity!
CHAPTER 7
Introduction to Relations
- Relations and Their Properties
1.1. Definition of a Relation. Definition: A binary relation from a set A to a set B is a subset
R ⊆ A × B.
If (a, b) ∈ R we say a is related to b by R.
A is the domain of R, and
B is the codomain of R.
If A = B, R is called a binary relation on the set A.
Notation:
- If (a, b) ∈ R, then we write aRb.
- If (a, b) 6 ∈ R, then we write a R b 6.
Discussion
Notice that a relation is simply a subset of A × B. If (a, b) ∈ R, where R is some relation from A to B, we think of a as being assigned to b. In these senses students often associate relations with functions. In fact, a function is a special case of a relation as you will see in Example 1.2.4. Be warned, however, that a relation may differ from a function in two possible ways. If R is an arbitrary relation from A to B, then
- it is possible to have both (a, b) ∈ R and (a, b′) ∈ R, where b′^6 = b; that is, an element in A could be related to any number of elements of B; or
- it is possible to have some element a in A that is not related to any element in B at all. 204
Often the relations in our examples do have special properties, but be careful not to assume that a given relation must have any of these properties.
1.2. Examples. Example 1.2.1. Let A = {a, b, c} and B = { 1 , 2 , 3 , 4 }, and let R 1 = {(a, 1), (a, 2), (c, 4)}. Example 1.2.2. Let R 2 ⊂ N × N be defined by (m, n) ∈ R 2 if and only if m|n. Example 1.2.3. Let A be the set of all FSU students, and B the set of all courses offered at FSU. Define R 3 as a relation from A to B by (s, c) ∈ R 3 if and only if s is enrolled in c this term.
Discussion
There are many different types of examples of relations. The previous examples give three very different types of examples. Let’s look a little more closely at these examples.
Example 1.2.1. This is a completely abstract relation. There is no obvious reason for a to be related to 1 and 2. It just is. This kind of relation, while not having any obvious application, is often useful to demonstrate properties of relations.
Example 1.2.2. This relation is one you will see more frequently. The set R 2 is an infinite set, so it is impossible to list all the elements of R 2 , but here are some elements of R 2 : (2, 6), (4, 8), (5, 5), (5, 0), (6, 0), (6, 18), (2, 18).
Equivalently, we could also write
2 R 26 , 4 R 28 , 5 R 25 , 5 R 20 , 6 R 20 , 6 R 218 , 2 R 218.
Here are some elements of N × N that are not elements of R 2 :
(6, 2), (8, 4), (2, 5), (0, 5), (0, 6), (18, 6), (6, 8), (8, 6).
Example 1.2.3. Here is an element of R 3 : (you, MAD2104).
Example 1.2.4. Let A and B be sets and let f : A → B be a function. The graph of f , defined by graph(f ) = {(x, f (x))|x ∈ A}, is a relation from A to B.
Notice the previous example illustrates that any function has a relation that is associated with it. However, not all relations have functions associated with them.
Exercise 1.2.1. Suppose f : R → R is defined by f (x) = bx/ 2 c.
(1) Find 5 elements of the relation graph(f ).
A digraph for R 2 in Example 1.2.2 would be difficult to illustrate (and impossible to draw completely), since it would require infinitely many vertices and edges. We could draw a digraph for some finite subset of R 2. It is possible to indicate what the graph of some infinite relations might look like, but this one would be particularly difficult.
Example 1.3.1. Let R 5 be the relation from { 0 , 1 , 2 , 3 , 4 , 5 , 6 } defined by mR 5 n if and only if m ≡ n(mod 3). The digraph that represents R 5 is
0
(^3 )
1
4
2
5
1.4. Inverse Relation.
Definition 1.4.1. Let R be a relation from A to B. Then R−^1 = {(b, a)|(a, b) ∈ R} is a relation from B to A.
R−^1 is called the inverse of the relation R.
Discussion
The inverse of a relation R is simply the relation obtained by reversing the ordered pairs of R. The inverse relation is also called the converse relation.
Example 1.4.1. Recall Example 1.2.1 A = {a, b, c} and B = { 1 , 2 , 3 , 4 } and R 1 = {(a, 1), (a, 2), (c, 4)}. Then R−^1 = {(1, a), (2, a), (4, c)}.
Exercise 1.4.1. Recall Example 1.2.4. A and B are sets and f : A → B is a function. The graph of f , graph(f ) = {(x, f (x))|x ∈ A} is a relation from A to B.
(1) What is the inverse of this relation? (2) Does f have to be invertible for the inverse of this relation to exist? (3) If f is invertible, find the inverse of the relation graph(f ) in terms of the inverse function f −^1.
1.5. Special Properties of Binary Relations. Definitions 1.5.1. Let A be a set, and let R be a binary relation on A.
(1) R is reflexive if ∀x[(x ∈ A) → ((x, x) ∈ R)]. (2) R is irreflexive if ∀x[(x ∈ A) → ((x, x) 6 ∈ R)]. (3) R is symmetric if ∀x∀y[((x, y) ∈ R) → ((y, x) ∈ R)]. (4) R is antisymmetric if ∀x∀y[([(x, y) ∈ R] ∧ [(y, x) ∈ R]) → (x = y)]. (5) R is asymmetric if ∀x∀y[((x, y) ∈ R) → ((y, x) 6 ∈ R)]. (6) R is transitive if ∀x∀y∀z[([(x, y) ∈ R] ∧ [(y, z) ∈ R]) → ((x, z) ∈ R)].
Discussion
Study the definitions of the definitions of the properties given above. You must know these properties, be able to recognize whether or not a relation has a particular property, and be able to prove that a relation has or does not have a particular property. Notice that the definitions of reflexive and irreflexive relations are not complementary. In fact, a relation on a set may be neither reflexive nor irreflexive. The same is true for the symmetric and antisymmetric properties, as well as the symmetric and asymmetric properties. Some texts use the term antireflexive for irreflexive. Exercise 1.5.1. Before reading further, find a relation on the set {a, b, c} that is neither
(a) reflexive nor irreflexive. (b) symmetric nor antisymmetric. (c) symmetric nor asymmetric.
1.6. Examples of Relations and Their Properties. Example 1.6.1. Suppose A is the set of FSU students and R is the relation given by aRb if students a and b have the same last name. This relation is...
- reflexive
- not irreflexive
- symmetric
- not antisymmetric
- not symmetric Proof. Both 2 and 3 are integers, 2 · 2 − 3 = 1, and 2 · 3 − 2 = 4 6 = 1. This shows 2R3, but 3 R 6 2; that is, ∀x∀y[(x, y) ∈ Z → (y, x) ∈ T ] is not true.
- antisymmetric Proof. Let m, n ∈ Z be such that (m, n) ∈ T and (n, m) ∈ T. By the definition of T , this implies both equations 2m − n = 1 and 2n − m = 1 must hold. We may use the first equation to solve for n, n = 2m − 1, and substitute this in for n in the second equation to get 2(2m − 1) − m = 1. We may use this equation to solve for m and we find m = 1. Now solve for n and we get n = 1. This shows that the only integers, m and n, such that both equations 2 m − n = 1 and 2n − m = 1 hold are m = n = 1. This shows that ∀m∀n[((m, n) ∈ T ∧ (n, m) ∈ T ) → m = n].
- not asymmetric Proof. 1 is an integer such that (1, 1) ∈ T. Thus ∀x∀y[((x, y) ∈ T → (b, a) 6 ∈ T ] is not true (counterexample is a = b = 1).
- not transitive Proof. 2, 3, and 5 are integers such that (2, 3) ∈ T , (3, 5) ∈ T , but (2, 5) 6 ∈ T. This shows ∀x∀y∀z[(x, y) ∈ T ∧ (y, z) ∈ T → (x, z) ∈ T ] is not true. Example 1.7.2. Recall Example 1.2.2: R 2 ⊂ N × N was defined by (m, n) ∈ R 2 if and only if m|n.
- reflexive Proof. Since n|n for all integers, n, we have nR 2 n for every integer. This shows R 2 is reflexive.
- not irreflexive Proof. 1 is an integer and clearly 1R 2 1. This shows R 2 is not irreflexive. (you could use any natural number to show R 2 is not irreflexive).
- not symmetric Proof. 2 and 4 are natural numbers with 2|4, but 4 6 |2, so 2R 2 4, but 4 R 6 2 2. This shows R 2 is not reflexive.
- antisymmetric Proof. Let n, m ∈ N be such that nR 2 m and mR 2 n. By the definition of R 2 this implies n|m and m|n. Hence we must have m = n. This shows R 2 is antisymmetric.
Proof. Let m = n be any natural number. Then nR 2 m and mR 2 n, which shows R 2 is not asymmetric. (You may use a particular number to show R 2 is not asymmetric.
Proof. Let p, q, r ∈ N and assume pR 2 q and qR 2 r. By the definition of R 2 this means p|q and q|r. We have proven in Integers and Division that this implies p|r, thus pR 2 r. This shows R 2 is transitive.
Discussion
When proving a relation, R, on a set A has a particular property, the property must be shown to hold for all possible members of the set. For example, if you wish to prove that a given relation, R, on A is reflexive, you must take an arbitrary element x from A and show that xRx. Some properties, such as the symmetric property, are defined using implications. For example, if you are asked to show that a relation, R, on A is symmetric, you would suppose that x and y are arbitrary elements of A such that xRy, and then try to prove that yRx. It is possible that a property defined by an implication holds vacuously or trivially.
Exercise 1.7.1. Let R be the relation on the set of real numbers given by xRy if and only if x < y. Prove R is antisymmetric.
When proving R does not have a property, it is enough to give a counterexample. Recall ¬[∀x∀yP (x, y)] ⇔ ∃x∃y¬P (x, y).
Exercise 1.7.2. Prove whether or not each of the properties in Section 1.5 holds for the relation in Example 1.6.1.
Exercise 1.7.3. Prove whether or not each of the properties in Section 1.5 holds for the relation in Example 1.6.3.
1.8. Combining Relations. Important Question: Suppose property P is one of the properties listed in Section 1.5, and suppose R and S are relations on a set A, each having property P. Then the following questions naturally arise.
(1) Does R (necessarily) have property P? (2) Does R ∪ S have property P? (3) Does R ∩ S have property P? (4) Does R − S have property P?
Then the composition of R 1 with R 2 , denoted R 2 ◦ R 1 , is the relation from A to C defined by the following property: (x, z) ∈ R 2 ◦ R 1 if and only if there is a y ∈ B such that (x, y) ∈ R 1 and (y, z) ∈ R 2. (2) Let R be a binary relation on A. Then Rn^ is defined recursively as follows: Basis: R^1 = R Recurrence: Rn+1^ = Rn^ ◦ R, if n ≥ 1.
Discussion
The composition of two relations can be thought of as a generalization of the composition of two functions, as the following exercise shows.
Exercise 1.10.1. Prove: If f : A → B and g : B → C are functions, then graph(g ◦ f ) = graph(g) ◦ graph(f ).
Exercise 1.10.2. Prove the composition of relations is an associative operation.
Exercise 1.10.3. Let R be a relation on A. Prove Rn^ ◦ R = R ◦ Rn^ using the previous exercise and induction.
Exercise 1.10.4. Prove an ordered pair (x, y) ∈ Rn^ if and only if, in the digraph D of R, there is a directed path of length n from x to y.
Notice that if there is no element of B such that (a, b) ∈ R 1 and (b, c) ∈ R 2 for some a ∈ A and c ∈ C, then the composition is empty.
1.11. Example of Composition.
Example 1.11.1. Let A = {a, b, c}, B = { 1 , 2 , 3 , 4 }, and C = {I, II, III, IV }.
- R 1 = {(a, 4), (b, 1)}
- R 2 = {(1, II), (1, IV ), (2, I)}
- Then R 2 ◦ R 1 = {(b, II), (b, IV )}
Discussion
It can help to consider the following type of diagram when discussing composition of relations, such as the ones in Example 1.11.1 as shown here.
a
b
c
I
II
III
IV
Example 1.11.2. If R and S are transitive binary relations on A, is R ◦ S tran- sitive?
Solution: No. Here is a counterexample: Let R = {(1, 2), (3, 4)}, and S = {(2, 3), (4, 1)}.
Then both R and S are transitive (vacuously). However,
R ◦ S = {(2, 4), (4, 2)}
is not transitive. (Why?)
Example 1.11.3. Suppose R is the relation on Z defined by aRb if and only if a|b. Then R^2 = R.
Exercise 1.11.1. Let R be the relation on the set of real numbers given by xRy
if and only if
x y
(1) Describe the relation R^2. (2) Describe the relation Rn. Exercise 1.11.2. Let P be a property given below and let R and S be relations on A satisfying property P. When does the relation obtained by combining R and S using the operation given satisfy property P?
(1) P is the reflexive property. (a) R ∪ S (b) R ∩ S (c) R ⊕ S (d) R − S (e) R ◦ S (f ) R−^1 (g) Rn (2) P is the symmetric property.
Theorem 1.12.1 gives an important theorem characterizing the transitivity rela- tion. Notice that, since the statement of the theorem was a property that was to be proven for all positive integers, induction was a natural choice for the proof.
Exercise 1.12.1. Prove that a relation R on a set A is transitive if and only if R^2 ⊆ R. [Hint: Examine not only the statement, but the proof of Theorem 1.12.1.]
