Page 232 - HOW TO PROVE IT: A Structured Approach, Second Edition

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218 Relations
Proof of Theorem 4.6.4. To prove that A/R is a partition of A, we must prove the
three properties in Definition 4.6.2. For the first, we must show that ∪(A/R) =
A, or in other words that ∪ x∈A [x] = A. Now every equivalence class in A/Ris
a subset of A, so it should be clear that their union is also a subset of A. Thus,
∪(A/R) ⊆ A, so all we need to show to finish the proof is that A ⊆∪(A/R).
To prove this, suppose x ∈ A. Then by Lemma 4.6.5, x ∈ [x], and of course
[x] ∈ A/R,so x ∈∪(A/R). Thus, ∪(A/R) = A.
To see that A/R is pairwise disjoint, suppose that X and Y are two elements of
A/R, and X ∩ Y = ∅. By definition of A/R, X and Y are equivalence classes,
so we must have X = [x] and Y = [y] for some x, y ∈ A. Since X ∩ Y = ∅,
we can choose some z such that z ∈ X ∩ Y = [x] ∩ [y]. Now by Lemma 4.6.5,
since z ∈ [x] and z ∈ [y], it follows that [x] = [z] = [y]. Thus, X = Y. This
shows that if X = Y then X ∩ Y = ∅,so A/R is pairwise disjoint.
Finally, for the last clause of the definition of partition, suppose X ∈ A/R.
As before, this means that X = [x] for some x ∈ A. Now by Lemma 4.6.5,
x ∈ [x] = X,so X = ∅, as required.

Commentary. We have given an intuitive reason why ∪(A/R) ⊆ A, but if
you’re not sure why this is correct, you should write out a formal proof.
(You might also want to look at exercise 16 in Section 3.3.) The proof that
A ⊆∪(A/R) is straightforward.
The definition of pairwise disjoint suggests that to prove that A/R is pair-
wise disjoint we should let X and Y be arbitrary elements of A/R and then
prove X = Y → X ∩ Y = ∅. Recall that the statement that a set is empty is
really a negative statement, so both the antecedent and the consequent of this
conditional are negative. This suggests that it will probably be easier to prove
the contrapositive, so we assume X ∩ Y = ∅ and prove X = Y. The givens
X ∈ A/R, Y ∈ A/R, and X ∩ Y = ∅ are all existential statements, so we use
them to introduce the variables x, y, and z. Lemma 4.6.5 now takes care of the
proof that X = Y as well as the proof of the final clause in the definition of
partition.

Theorem 4.6.4 shows that if R is an equivalence relation on A then A/R is a
partition of A. In fact, it turns out that every partition of A arises in this way.

Theorem 4.6.6. Suppose A is a set and F is a partition of A. Then there is an
equivalence relation R on A such that A/R = F.

Before proving this theorem, it might be worthwhile to discuss the strategy
for the proof briefly. Because the conclusion of the theorem is an existential
statement, we should try to find an equivalence relation R such that A/R = F.