Page 315 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Closures Again 301
Base case: When n = 1wehave
m
R m+1 = R ◦ R (by definition of R m+1 )
1
m
= R ◦ R 1 (by definition of R ).
Induction step: Let n be an arbitrary positive integer and suppose R m+n =
m
n
R ◦ R . Then
R m+(n+1) = R (m+n)+1
= R m+n ◦ R (by definition of R (m+n)+1 )
m
n
= (R ◦ R ) ◦ R (by inductive hypothesis)
n
m
= R ◦ (R ◦ R) (by associativity of composition)
m
= R ◦ R n+1 (by definition of R n+1 ).
We can now say precisely how to form the transitive closure of R.
n
Theorem 6.5.2. The transitive closure of R is ∪ n∈Z R .
+
1
n
Proof. Let S =∪ n∈Z R . Clearly R = R ⊆ S. To see that S is transitive,
+
suppose (x, y) ∈ S and (y, z) ∈ S. Then by the definition of S, we can choose
n
m
positive integers n and m such that (x, y) ∈ R and (y, z) ∈ R . But then by
n
m
n
Lemma 6.5.1, (x, z) ∈ R ◦ R = R m+n ,so(x, z) ∈∪ n∈Z R = S. Thus S is
+
transitive.
Finally, suppose R ⊆ T ⊆ A × A and T is transitive. We must show that
S ⊆ T , and clearly by the definition of S it suffices to show that ∀n ∈
n
+
Z (R ⊆ T ). We prove this by induction on n.
n
1
We have assumed R ⊆ T , so when n = 1wehave R = R = R ⊆ T .For
n
the induction step, suppose n is a positive integer and R ⊆ T . Now suppose
(x, y) ∈ R n+1 . Then by definition of R n+1 we can choose some z ∈ A such
n
that (x, z) ∈ R and (z, y) ∈ R . By assumption R ⊆ T , and by inductive hy-
n
pothesis R ⊆ T . Therefore, (x, z) ∈ T and (z, y) ∈ T , so since T is transitive,
(x, y) ∈ T . Since (x, y) was an arbitrary element of R n+1 , this shows that
R n+1 ⊆ T .
n
Commentary . Because the proof must refer to the set ∪ n∈Z R often, it is
+
convenient to give this set a name right at the beginning of the proof. According
to the definition of transitive closure we must prove three things: R ⊆ S, S is
transitive, and for all T,if R ⊆ T ⊆ A × A and T is transitive, then S ⊆ T .Of
course, we prove them one at a time.
Theproofofthefirstofthesegoalsisnotspelledout.Asusual,ifyoudon’tsee
why it is true you should work out the details of the proof yourself. The second
goal is to prove that S is transitive, and the proof is based on the definition of