Page 223 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Closures 209
Before moving on to the proof that all relations have transitive closures, we
would like to describe an alternative method of proving Theorem 4.5.5. If R
is a relation on a set A, then we know that the symmetric closure of R,ifit
exists, must be the smallest element of the family F ={T ⊆ A × A | R ⊆ T
and T is symmetric}. Now according to exercise 20 in the last section, the
smallest element of a set is also always the greatest lower bound of the set, and
by Theorem 4.4.11, the g.l.b. of a nonempty family of sets F is always ∩F.
Thus, if the symmetric closure of R exists, then it must be equal to ∩F. This
suggests an alternative approach to proving Theorem 4.5.5. We could define
−1
the family F, prove that F = ∅, and then let S =∩F (instead of R ∪ R )
and prove that S satisfies the definition of symmetric closure. Recall that we
only defined ∩F for F = ∅, so we must check that F = ∅ before we can
form ∩F. You are asked in exercise 10 to work out the details of this alternative
proof.
There are also two ways we could prove that any relation R on a set A
has a transitive closure. One possibility would be to try to form the transitive
closure by starting with R and then adding extra ordered pairs to try to create
a transitive relation, as we did in the earlier examples. Although this can be
done, a careful treatment of the details of this proof would require the method
of mathematical induction, which we have not yet discussed. We will present
this proof in Chapter 6 after we’ve discussed mathematical induction. For now,
we’ll use the alternative method suggested by the last paragraph of letting
F ={T ⊆ A × A | R ⊆ T and T is transitive} and then showing that F = ∅
and ∩F is the transitive closure of R.

Theorem 4.5.6. Suppose R is a relation on a set A. Then R has a transitive
closure.
Proof. Let F ={T ⊆ A × A | R ⊆ T and T is transitive}. First of all, you
should be able to check that R ⊆ A × A and A × A is a transitive relation on
A,so A × A ∈ F, and therefore F = ∅. Thus, we can let S =∩F. We will
show that S is the transitive closure of R.
To prove the first clause in the definition of transitive closure, suppose
(x, y) ∈ R. Let T be an arbitrary element of F. Then by the definition of
F, R ⊆ T ,so(x, y) ∈ T . Since T was arbitrary, this shows that ∀T ∈
F((x, y) ∈ T ), so (x, y) ∈∩F = S. Thus, R ⊆ S.
Forthesecondclause,suppose(x, y) ∈ S and(y, z) ∈ S,andagainletT bean
arbitrary element of F. Then since (x, y) ∈ S =∩F, (x, y) ∈ T , and similarly
(y, z) ∈ T . But since T ∈ F, T is transitive, so it follows that (x, z) ∈ T . Since
T wasarbitrary,wecanconcludethat∀T ∈ F((x, z) ∈ T ),so(x, z) ∈∩F = S.
Thus, we have shown that if (x, y) ∈ S and (y, z) ∈ S then (x, z) ∈ S,so S is
transitive.