Page 224 - HOW TO PROVE IT: A Structured Approach, Second Edition

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210 Relations
Finally, for the third clause, suppose T is a relation on A, R ⊆ T , and T
is transitive. Then T ∈ F, and by exercise 9 of Section 3.3, it follows that
S =∩F ⊆ T .

Commentary. Once again we start by defining S, but this time the definition
S =∩F doesn’t make sense unless we know F = ∅, so we must prove this
first. Because F = ∅ means ∃Q(Q ∈ F), we prove it by giving an example
of an element of F. The example is A × A, so we must prove A × A ∈ F,
and according to the definition of F this means A × A ⊆ A × A, R ⊆ A × A,
and A × A is transitive. The statement in the proof that “you should be able to
check” that these statements are true really does mean that you should do the
checking. In particular, you should verify that A × A is transitive by assuming
that (x, y) ∈ A × A and (y, z) ∈ A × A and then proving that (x, z) ∈ A × A.
As in Theorems 4.5.2 and 4.5.5, we must now prove the three statements
in the definition of transitive closure. To prove the first statement, R ⊆ S,we
let (x, y) be an arbitrary element of R and prove (x, y) ∈ S. Since S =∩F,
the goal (x, y) ∈ S means ∀T ∈ F((x, y) ∈ T ), so to prove it we let T be an
arbitrary element of F and prove (x, y) ∈ T . To prove that S is transitive we
assume (x, y) ∈ S and (y, z) ∈ S, and prove (x, z) ∈ S. Once again, by the
definition of S this goal means ∀T ∈ F((x, z) ∈ T ), so we let T be an arbitrary
element of F and prove (x, z) ∈ T .

Exercises

1. Find the reflexive, symmetric, and transitive closures of the following
∗
relations.
(a) A ={a, b, c}, R ={(a, a), (a, b), (b, c), (c, b)}.
(b) R ={(x, y) ∈ R × R | x < y}.
(c) D r , as defined in part 3 of Example 4.3.1, for any positive real
number r.
2. Find the reflexive, symmetric, and transitive closures of the relations in
exercise 4 of Section 4.3.
∗
3. Suppose R is a relation on A. R is called asymmetric if ∀x ∈ A∀y ∈
A((x, y) ∈ R → (y, x) /∈ R).
(a) Show that if R is asymmetric then R is antisymmetric.
(b) Show that if R is a strict partial order, then R is asymmetric. Note that
it follows by part (a) that it is also antisymmetric.
4. Suppose R is a strict partial order on A. Let S be the reflexive closure of R.
(a) Show that S is a partial order on A.