Page 233 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: PIG/ P2: OYK/
0521861241c04 CB996/Velleman October 20, 2005 2:54 0 521 86124 1 Char Count= 0

Equivalence Relations 219
Clearly for different choices of F we will need to choose R differently, so
the definition of R should depend on F in some way. Looking back at the
same-birthday example may help you see how to proceed. Recall that in that
example the equivalence relation B consisted of all pairs of people (p, q) such
that p and q were in the same set in the partition {P d | d ∈ D}. In fact, we
found that we could also express this by saying that B =∪ d∈D (P d × P d ). This
suggests that in the proof of Theorem 4.6.6 we should let R be the set of all
pairs (x, y) ∈ A × A such that x and y are in the same set in the partition F.
An alternative way to write this would be R =∪ X∈F (X × X).
For example, consider again the example of a partition given after Definition
4.6.2. In that example we had A ={1, 2, 3, 4} and F ={{2}, {1, 3}, {4}}.Now
let’s define a relation R on A as suggested in the last paragraph. This gives us:
R = ∪ (X × X)
X∈F
= ({2}×{2}) ∪ ({1, 3}×{1, 3}) ∪ ({4}×{4})
={(2, 2)}∪{(1, 1), (1, 3), (3, 1), (3, 3)}∪{(4, 4)}
={(2, 2), (1, 1), (1, 3), (3, 1), (3, 3), (4, 4)}.
The directed graph for this relation is shown in Figure 1. We will let you check
that R is an equivalence relation and that the equivalence classes are
[2] ={2}, [1] = [3] ={1, 3}, [4] ={4}.
Thus, the set of all equivalence classes is A/R ={{2}, {1, 3}, {4}}, which is
precisely the same as the partition F we started with.

Figure 1
Of course, the reasoning that led us to the formula R =∪ X∈F (X × X) will
not be part of the proof of Theorem 4.6.6. When we write the proof, we can
simply define R in this way and then verify that it is an equivalence relation on
A and that A/R = F. It may make the proof easier to follow if we once again
prove some lemmas first.