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208 Relations
difficult to find. For the proofs we are concerned with we will need to be able to
find, for any given relation R, relations that satisfy the definitions of symmetric
and transitive closure for R. This will actually be the most difficult part of
figuring out these proofs. Once we’ve found the right relations, verifying that
they satisfy the definitions of symmetric and transitive closure will be somewhat
long, but not difficult.
The case of the symmetric closure is the easier of the two. Suppose R is a
relation on a set A, and we want to find the symmetric closure of R. Looking at
the earlier examples, it appears that all we need to do is to add to R all ordered
pairs (x, y) such that (y, x) ∈ R. In other words, it looks like the symmetric
−1
closure of R will be R ∪ R .
Theorem 4.5.5. Suppose R is a relation on a set A. Then R has a symmetric
closure.
−1
Proof. Let S = R ∪ R . We will show that S is the symmetric closure of R.
Clearly R ⊆ S, so the first clause of the definition of symmetric closure
is satisfied. For the second, suppose (x, y) ∈ S. Then by the definition of S,
−1
either (x, y) ∈ R or (x, y) ∈ R .If(x, y) ∈ R, then (y, x) ∈ R −1 ⊆ S.If
(x, y) ∈ R −1 then (y, x) ∈ R ⊆ S. Thus, we can conclude that if (x, y) ∈ S
then (y, x) ∈ S. Since (x, y) was arbitrary, this shows that S is symmetric.
Finally, for the third clause in the definition of symmetric closure, sup-
pose R ⊆ T ⊆ A × A and T is symmetric. Suppose (x, y) ∈ S. As before, this
−1
means that either (x, y) ∈ R or (x, y) ∈ R , so we consider these two possi-
bilities separately. If (x, y) ∈ R then (x, y) ∈ T , since R ⊆ T .If(x, y) ∈ R −1
then (y, x) ∈ R, so since R ⊆ T, (y, x) ∈ T . But then since T is symmetric, it
follows that (x, y) ∈ T . Since (x, y) was arbitrary, we have shown that S ⊆ T ,
as required.

Commentary. The overall form of this proof is quite similar to the form of the
proof of Theorem 4.5.2, but some of the details are a little trickier. We start
by specifying a relation S that will be the symmetric closure of R, and then
we prove the three statements in the definition of symmetric closure one at a
time. The logical form of the definition of symmetric suggests that to prove
that S is symmetric we should start with an arbitrary ordered pair (x, y) ∈ S
−1
and prove (y, x) ∈ S. Because we defined S to be R ∪ R , the assumption that
−1
(x, y) ∈ S means (x, y) ∈ R ∨ (x, y) ∈ R , and since this is a disjunction it
suggests the use of proof by cases.
As in the proof of Theorem 4.5.2, to prove the third statement in the definition
of symmetric closure we let T be an arbitrary relation on A such that R ⊆ T
and T is symmetric, and we prove S ⊆ T .Toprove S ⊆ T we let (x, y)be
an arbitrary element of S and prove (x, y) ∈ T . As before, the assumption that
(x, y) ∈ S is a disjunction, so it leads to a proof by cases.