Page 231 - 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 217
fact that the equivalence classes determined by an equivalence relation always
form a partition.

Theorem 4.6.4. Suppose R is an equivalence relation on a set A. Then A/Ris
a partition of A.

The proof of Theorem 4.6.4 will be easier to understand if we first prove
a few facts about equivalence classes. Facts that are proven primarily for the
purpose of using them to prove a theorem are usually called lemmas.

Lemma 4.6.5. Suppose R is an equivalence relation on A. Then:

1. For every x ∈ A, x ∈ [x].
2. For every x ∈ A and y ∈ A, y ∈ [x] iff [y] = [x].

Proof.

1. Let x ∈ A be arbitrary. Since R is reflexive, xRx. Therefore, by the definition
of equivalence class, x ∈ [x].
2. (→) Suppose y ∈ [x]. Then by the definition of equivalence class, yRx.
Now suppose z ∈ [y]. Then zRy. Since zRy and yRx, by transitivity of R
we can conclude that zRx,so z ∈ [x]. Since z was arbitrary, this shows that
[y] ⊆ [x].
Now suppose z ∈ [x], so zRx. We already know yRx, and since R is
symmetric we can conclude that xRy. Applying transitivity to zRx and xRy,
we can conclude that zRy,so z ∈ [y]. Therefore [x] ⊆ [y], so [x] = [y].
(←) Suppose [y] = [x]. By part 1 we know that y ∈ [y], so since [y] =
[x], it follows that y ∈ [x].

Commentary

1. According to the definition of equivalence classes, x ∈ [x] means xRx.
This is what leads us to apply the fact that R is reflexive.
2. Of course, the iff form of the goal leads us to prove both directions separately.
For the → direction, the goal is [y] = [x], and, since [y] and [x] are sets, we
can prove this by proving [y] ⊆ [x] and [x] ⊆ [y]. We prove each of these
statements by the usual method of taking an arbitrary element of one set and
proving that it is in the other. Throughout the proof we use the definition of
equivalence classes repeatedly, as we did in the proof of statement 1.