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Appendix 1: Solutions to Selected Exercises 347
(c) Imitating the first half of the proof in part (b), we can use exercise 11
to show that S 1 ∪ S 2 ⊆ S. However, the answer to exercise 13(c) of
Section 4.3 was no, so we can’t imitate the second half of the proof.
In fact, the example given in the solution to exercise 13(c) works as an
example for which S 1 ∪ S 2 = S.
15. Hint: Let S = R ∪ R −1 ∪ i A .
18. (a) We have R ⊆ Q and Q ⊆ S,so R ⊆ S. By definition of symmetric
closure, Q is symmetric, and therefore, by exercise 17, S is symmetric.
By definition of transitive closure, S is also transitive. Now suppose
that T ⊆ A × A, R ⊆ T , and T is both symmetric and transitive. Since
Q is the smallest symmetric relation on A containing R, Q ⊆ T . But
then since S is the smallest transitive relation on A containing Q,
S ⊆ T .
(b) Since R ⊆ Q, Q is the transitive closure of R, and S is the transitive

closure of Q, by exercise 11, Q ⊆ S. Since S is symmetric and S is
the smallest symmetric relation on A containing Q , S ⊆ S.

(c) No. Counterexample: A ={1, 2, 3}, R ={(1, 2), (3, 2)}.
20. (a) One example is {(San Francisco, Chicago), (Chicago, Dallas), (Dallas,
New York), (New York, Washington, D.C.), (Washington, D.C., San
Francisco)}.
(b) No.

Section 4.6
1. Here is a list of all partitions:

{{1, 2, 3}}
{{1, 2}, {3}}
{{1, 3}, {2}}
{{2, 3}, {1}}
{{1}, {2}, {3}}

3. (a) R is an equivalence relation. There are 26 equivalence classes – one
for each letter of the alphabet. The equivalence classes are: the set of
all words that start with a, the set of all words that start with b,...,
the set of all words that start with z.
(b) S is not an equivalence relation, because it is not transitive.
(c) T is an equivalence relation. The equivalence classes are: the set of
all one-letter words, the set of all two-letter words, ..., the set of all
n-letter words, where n is the length of the longest English word.