Page 234 - HOW TO PROVE IT: A Structured Approach, Second Edition

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220 Relations
Lemma 4.6.7. Suppose A is a set and F is a partition of A. Let R =
∪ X∈F (X × X). Then R is an equivalence relation on A. We will call R the
equivalence relation determined by F.

Proof. We’ll prove that R is reflexive and leave the rest for you to do in
exercise 7. Let x be an arbitrary element of A. Since F is a partition of
A, ∪F = A,so x ∈∪F. Thus, we can choose some X ∈ F such that x ∈ X.
But then (x, x) ∈ X × X,so(x, x) ∈∪ X∈F (X × X) = R. Therefore, R is
reflexive.

Commentary. After letting x be an arbitrary element of A, we must prove
(x, x) ∈ R. Because R =∪ X∈F (X × X), this means we must prove ∃X ∈
F((x, x) ∈ X × X), or in other words ∃X ∈ F(x ∈ X). But this just means
x ∈∪F, so this suggests using the first clause in the definition of partition,
which says that ∪F = A.

Lemma 4.6.8. Suppose A is a set and F is a partition of A. Let R be the equiv-
alence relation determined by F. Suppose X ∈ F and x ∈ X. Then [x] R = X.

Proof. Suppose y ∈ [x] R . Then (y, x) ∈ R, so by the definition of R there must
be some Y ∈ F such that (y, x) ∈ Y × Y, and therefore y ∈ Y and x ∈ Y. Since
x ∈ X and x ∈ Y, X ∩ Y = ∅, and since F is pairwise disjoint it follows that
X = Y. Thus, since y ∈ Y, y ∈ X. Since y was an arbitrary element of [x] R ,
we can conclude that [x] R ⊆ X.
Now suppose y ∈ X. Then (y, x) ∈ X × X,so(y, x) ∈ R and therefore
y ∈ [x] R . Thus X ⊆ [x] R ,so[x] R = X.

Commentary. To prove [x] R = X we prove [x] R ⊆ X and X ⊆ [x] R . For the
first we start with an arbitrary y ∈ [x] R and prove y ∈ X. Writing out the defini-
tion of [x] R we get (y, x) ∈ R, and since R was defined to be ∪ Y∈F (Y × Y), this
means ∃Y ∈ F((y, x) ∈ Y × Y). Of course, since this is an existential state-
ment we immediately introduce the new variable Y by existential instantiation.
Since this gives us y ∈ Y and our goal is y ∈ X, it is not surprising that the
proof is completed by proving Y = X.
The proof that X ⊆ [x] R also uses the definitions of [x] R and R, but is more
straightforward.

Proof of Theorem 4.6.6. Let R =∪ X∈F (X × X). We have already seen that
R is an equivalence relation, so we need only check that A/R = F. To see
this, suppose X ∈ A/R. This means that X = [x] for some x ∈ A. Since F is a
partition, we know that ∪F = A,so x ∈∪F, and therefore we can choose some