Page 217 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Closures 203
Example 4.4.7, some families of sets don’t have smallest elements, so it might
not be clear at first whether every relation has a reflexive closure. In fact, every
relation does have a reflexive closure.

Theorem 4.5.2. Suppose R is a relation on A. Then R has a reflexive closure.
Proof. Let S = R ∪ i A , where as usual i A is the identity relation on A.We
will show that S is the reflexive closure of A. Thus, we must show that S
has the three properties listed in Definition 4.5.1. The first property is ob-
viously true, since clearly R ⊆ R ∪ i A = S. For the second and third, we use
Theorem 4.3.4. Clearly i A ⊆ R ∪ i A = S, so by part 1 of Theorem 4.3.4, S is re-
flexive. Finally, to prove the third property, suppose T is a relation on A, R ⊆ T ,
and T is reflexive. Then by Theorem 4.3.4, since T is reflexive, i A ⊆ T . Com-
bining this with the fact that R ⊆ T , we can conclude that S = R ∪ i A ⊆ T ,as
required.

Commentary. Our goal is the existential statement ∃S(S is the reflexive closure
of R), so we start by specifying a value for S. Our earlier discussion suggests that
to get S we should start with R and then add just those ordered pairs that must
be added to create a reflexive relation. The ordered pairs that must be added
are the elements of i A that are not already in R,sowelet S = R ∪ i A . Our goal
now is to prove that S is the reflexive closure of R, and by definition this means
that we must prove statements 1–3 of Definition 4.5.1. We prove these one at a
time. Statements 1 and 2 are easy, and the logical form of statement 3 suggests
the strategy of letting T be an arbitrary relation on A, assuming R ⊆ T and T
is reflexive, and then proving S ⊆ T .

For another example of a reflexive closure, let A be any set and consider
the relation P ={(x, y) ∈ P (A) × P (A) | x ⊆ y and x = y}. Thus, if x and y
are any two subsets of A, then xPy means that x ⊆ y and x = y.If xPy, then
we will say that x is a proper subset of y, which is written x ⊂ y. The reflexive
closure of P would be the relation

P ∪ i P (A) ={(x, y) ∈ P (A) × P (A) | (x, y) ∈ P or (x, y) ∈ i P (A) }
={(x, y) ∈ P (A) × P (A) | x ⊂ y or x = y}
={(x, y) ∈ P (A) × P (A) | x ⊆ y}.

Thus, the reflexive closure of the proper subset relation is the subset relation.
The relations M and P in these examples are similar to partial orders except
that they are not reflexive. Rather than expressing a sense in which one object
can be “at least as large as” another, these relations seem to represent a sense