College Algebra Exam Review 118

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

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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 ŒxŁ D f y 2 X W x ± y g : Note that x 2 ŒxŁ 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, ŒxŁ D ŒyŁ . Proof. If ŒxŁ D ŒyŁ , then x 2 ŒxŁ D ŒyŁ , so x ± y . For the converse, suppose that x ± y (and hence y ± x , by symmetry of the equivalence relation). If z 2 ŒxŁ , then z ± x . Since x ± y by assumption, transitivity of the equivalence relation implies that z ± y (i.e., z 2 ŒyŁ ). This shows that ŒxŁ ² ŒyŁ . Similarly, ŒyŁ ² ŒxŁ . Therefore, ŒxŁ D ŒyŁ . n Corollary 2.6.6. Let ± be an equivalence relation on X , and let x;y 2 X . Then either ŒxŁ \ ŒyŁ D ; or ŒxŁ D ŒyŁ . Proof. We have to show that if ŒxŁ \ ŒyŁ ¤ ; , then ŒxŁ D ŒyŁ . But if z 2 ŒxŁ \ ŒyŁ , then ŒxŁ D ŒzŁ D ŒyŁ , 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 x . 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 dene an relation on X by x P y if, and only if, x and y are in the same subset of the par-tition. Lets 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 denition 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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