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College Algebra Exam Review 118

# College Algebra Exam Review 118 - X The equivalence classes...

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128 2. BASIC THEORY OF GROUPS An equivalence relation on a set X always gives rise to a distinguished family of subsets of X : Definition 2.6.4. If is an equivalence relation on X , then for each x 2 X , the equivalence class of x is the set OExŁ D f y 2 X W x y g : Note that x 2 OExŁ because of reflexivity; in particular, the equivalence classes are nonempty subsets of X . Proposition 2.6.5. Let be an equivalence relation on X . For x; y 2 X , x y if, and only if, OExŁ D OEyŁ . Proof. If OExŁ D OEyŁ , then x 2 OExŁ D OEyŁ , so x y . For the converse, suppose that x y (and hence y x , by symmetry of the equivalence relation). If z 2 OExŁ , then z x . Since x y by assumption, transitivity of the equivalence relation implies that z y (i.e., z 2 OEyŁ ). This shows that OExŁ OEyŁ . Similarly, OEyŁ OExŁ . Therefore, OExŁ D OEyŁ . n Corollary 2.6.6. Let be an equivalence relation on X , and let x; y 2 X . Then either OExŁ \ OEyŁ D ; or OExŁ D OEyŁ . Proof. We have to show that if OExŁ \ OEyŁ ¤ ; , then OExŁ D OEyŁ . But if z 2 OExŁ \ OEyŁ , then OExŁ D OEzŁ D OEyŁ , by the proposition. n Consider an equivalence relation on a set
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Unformatted text preview: X . The equivalence classes of ± are nonempty and have union equal to X , since for each x 2 X , x 2 ŒxŁ . Furthermore, the equivalence classes are mutually disjoint ; this means that any two distinct equivalence classes have empty intersection. So the collection of equivalence classes is a partition of the set X . Any equivalence relation on a set X gives rise to a partition of X by equivalence classes. On the other hand, given a partition P of X , we can deﬁne an relation on X by x ± P y if, and only if, x and y are in the same subset of the par-tition. Let’s check that this is an equivalence relation. Write the partition P as f X i W i 2 I g . We have X i ¤ ; for all i 2 I , X i \ X j D ; if i ¤ j , and [ i 2 I X i D X . Our deﬁnition of the relation is x ± P y if, and only if, there exists i 2 I such that both x and y are elements of X i ....
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