Page 229 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Equivalence Relations 215
F is a partition of A. On the other hand, the family G ={{1, 2}, {1, 3}, {4}} is
not pairwise disjoint, because {1, 2}∩{1, 3}={1} = ∅, so it is not a partition
of A. The family H = {∅, {2}, {1, 3}, {4}} is also not a partition of A, because
it fails on the third requirement in the definition.

Definition 4.6.3. Suppose R is an equivalence relation on a set A, and x ∈ A.
Then the equivalence class of x with respect to R is the set

[x] R ={y ∈ A | yRx}.
If R is clear from context, then we just write [x] instead of [x] R . The set of all
equivalence classes of elements of A is called A modulo R, and is denoted A/R.
Thus,

A/R ={[x] R | x ∈ A}={X ⊆ A |∃x ∈ A(X = [x] R )}.
In the case of the same-birthday relation B,if p is any person, then according
to Definition 4.6.3,
[p] B ={q ∈ P | qBp}
={q ∈ P | the person q has the same birthday as the person p}.

For example, if John was born on August 10, then
[John] B ={q ∈ P | the person q has the same birthday as John}
={q ∈ P | the person q was born on August 10}.

In the notation we introduced earlier, this is just the set P d , for d = August
10. In fact, it should be clear now that for any person p,ifwelet d be p’s
birthday, then [p] B = P d . This is in agreement with our earlier statement that
the sets P d are the equivalence classes for the equivalence relation B. Accord-
ing to Definition 4.6.3, the set of all of these equivalence classes is called P
modulo B:
P/B ={[p] B | p ∈ P}={P d | d ∈ D}.
You are asked to give a more careful proof of this equation in exercise 5. As
we observed before, this family is a partition of P.
Let’s consider one more example. Let S be the relation on R defined as
follows:
S ={(x, y) ∈ R × R | x − y ∈ Z}.

For example, (5.73, 2.73) ∈ S and (−1.27, 2.73) ∈ S, since 5.73 − 2.73 = 3 ∈
Z and −1.27 − 2.73 =−4 ∈ Z, but (1.27, 2.73) /∈ S, since 1.27 − 2.73 =
−1.46 ∈ Z. Clearly for any x ∈ R, x − x = 0 ∈ Z,so(x, x) ∈ S, and therefore