Page 199 - HOW TO PROVE IT: A Structured Approach, Second Edition

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More About Relations 185
because the statement ∀x ∈ B∀y ∈ B∀z ∈ B((x ⊆ y ∧ y ⊆ z) → x ⊆ z)is
true.
For any set A the identity relation i A will be reflexive, symmetric, and
transitive, because the statements ∀x ∈ A(x = x), ∀x ∈ A∀y ∈ A(x = y →
y = x), and ∀x ∈ A∀y ∈ A∀z ∈ A((x = y ∧ y = z) → x = z) are all clearly
true. Finally, suppose r is a positive real number and consider the relation D r .
For any real number x, |x − x|= 0 < r,so(x, x) ∈ D r . Thus, D r is reflexive.
Also, for any real numbers x and y, |x − y|=|y − x|,soif |x − y| < r then
|y − x| < r. Therefore, if (x, y) ∈ D r then (y, x) ∈ D r ,so D r is sym-
metric. But D r is not transitive. To see why, let x be any real number.
Let y = x + 2r/3 and z = y + 2r/3 = x + 4r/3. Then |x − y|= 2r/3 < r
and |y − z|= 2r/3 < r,but |x − z|= 4r/3 > r. Thus, (x, y) ∈ D r and
(y, z) ∈ D r , but (x, z) /∈ D r .

Perhaps you’ve already guessed that the properties of relations defined in
Definition 4.3.2 are related to the operations defined in Definition 4.2.3. To
say that a relation R is symmetric involves reversing the roles of two variables
−1
in a way that may remind you of the definition of R . The definition of
transitivity of a relation involves stringing together two ordered pairs, just as
the definition of composition of relations does. The following theorem spells
these connections out more carefully.

Theorem 4.3.4. Suppose R is a relation on a set A.
1. R is reflexive iff i A ⊆ R, where as before i A is the identity relation on A.
−1
2. R is symmetric iff R = R .
3. R is transitive iff R ◦ R ⊆ R.
Proof. We will prove 2 and leave the proofs of 1 and 3 as exercises (see exer-
cises 7 and 8).
2. (→) Suppose R is symmetric. Let (x, y) be an arbitrary element of R. Then
xRy, so since R is symmetric, yRx. Thus, (y, x) ∈ R, so by the definition of
−1
−1
−1
R , (x, y) ∈ R . Since (x, y) was arbitrary, it follows that R ⊆ R .
−1
Now suppose (x, y) ∈ R . Then (y, x) ∈ R, so since R is symmetric,
−1
(x, y) ∈ R. Thus, R −1 ⊆ R,so R = R .
−1
(←) Suppose R = R , and let x and y be arbitrary elements of A. Suppose
−1
−1
xRy. Then (x, y) ∈ R, so since R = R , (x, y) ∈ R . By the definition of
R −1 this means (y, x) ∈ R,so yRx. Thus, ∀x ∈ A∀y ∈ A(xRy → yRx), so
R is symmetric.
Commentary. This proof is fairly straightforward. The statement to be proven
is an iff statement, so we prove both directions separately. In the → half we