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Equivalence Relations

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1 Equivalence Relations
Lecture 45 Section 10.3 Fri, Apr 8, 2005

2 Equivalence Relations
An equivalence relation on a set A is a relation on A that is reflexive, symmetric, and transitive. We often use the symbol ~ as a generic symbol for an equivalence relation.

3 Examples of Equivalence Relations
Which of the following are equivalence relations? a  b, on Z+. gcd(a, b) > 1, on Z+. A  B, on (U). p  q, on a set of statements. p  q, on a set of statements. a  b (mod 10), on Z.

4 Examples of Equivalence Relations
Which of the following are equivalence relations? p  q = p, on a set of statements. gcd(a, b) = 1, on Z+. gcd(a, b) = a, on Z+. A  B = , on (U). A = B, on (U).

5 Examples of Equivalence Relations
Which of the following are equivalence relations? R  R, on R. , on R.

6 Equivalence Classes Let ~ be an equivalence relation on a set A and let a  A. The equivalence class of a is [a] = {x  A  x ~ a}.

7 Examples: Equivalence Classes
Describe the equivalence classes of each of the following equivalence relations. a  b (mod 10), on Z. A = B, on (U). p  q, on a set of statements. R  R, on R.

8 Equivalence Classes and Partitions
Theorem: Let ~ be an equivalence relation on a set A. The equivalence classes of ~ form a partition of A. Proof: We must show that The equivalence classes are pairwise disjoint, The union of the equivalence classes equals A.

9 Equivalence Classes and Partitions
Proof that the equivalence classes are pairwise disjoint. Let [a] and [b] be two distinct equivalence classes. Suppose [a]  [b]  . Let x  [a]  [b]. Then x ~ a and x ~ b. Therefore, a ~ x and x ~ b.

10 Equivalence Classes and Partitions
By transitivity, a ~ b. Now let y  [a]. Then y ~ a. By transitivity, y ~ b. So y  [b]. Therefore, [a]  [b]. By a similar argument, [b]  [a].

11 Equivalence Classes and Partitions
Thus, [a] = [b], which is a contradiction Therefore, [a]  [b] = . Thus, the equivalence classes are pairwise disjoint.

12 Equivalence Classes and Partitions
Proof that the union of the equivalence classes is A. Let a  A. Then a  [a] since a ~ a. Therefore, a is in the union of the equivalence classes. So, A is a subset of the union of the equivalence classes.

13 Equivalence Classes and Partitions
On the other hand, every equivalence class is a subset of A. Therefore, the union of the equivalence classes is a subset of A. Therefore, the union of the equivalence classes equals A. Therefore, the equivalence classes form a partition of A.

14 Example Let F be the set of all functions f : R  R.
For f, g  F, define f ~ g to mean that f is (g).

15 Example Theorem: ~ is an equivalence relation on F. Proof: Reflexivity
Obviously, f ~ f for all f  F.

16 M1g(x)  f(x)  M2g(x),
Example Symmetry Suppose that f ~ g for some f, g  F. Then f(x) is (g(x)). There exist positive constants M1, M2, and x0 such that M1g(x)  f(x)  M2g(x), for all x > x0.

17 (1/M2)f(x)  g(x)  (1/M1)f(x),
Example It follows that (1/M2)f(x)  g(x)  (1/M1)f(x), for all x > x0. Therefore, g(x) is (f(x)).

18 Example Transitivity Let f, g, h  F and suppose that f ~ g and g ~ h.
Then there exist constants M1 and x1 and M2 and x2 such that f(x) M1g(x) for all x  x1 and

19 Example g(x) M2h(x) for all x  x2. Let x0 = max(x1, x2).
Then for all x  x0, f(x) M1g(x)  M1  M2h(x) Therefore, f(x) is O(h(x)).

20 Example Similarly, we can show that h(x) is O(f(x)).
Therefore, f(x) is (h(x)). Therefore, f ~ h. Therefore, ~ is an equivalence relation on F.

21 Example The equivalence class of f is the set [f] of all functions with the same growth rate as f. The most important equivalence classes are [xa], a  R, a > 0. [ax], a  R, a > 1. [xa logb x], a  R, a > 0, b > 1.

22 Example Furthermore, However, [xa]  [xb] if a  b.
[ax]  [bx] if a  b. However, [loga x] = [logb x] for all a, b > 1.

23 The Equivalence Relation Induced by a Partition
Let A be a set and let {Ai}iI be a partition of A. Define a relation ~ on A as x ~ y  x, y  Ai for some i  I.

24 The Equivalence Relation Induced by a Partition
Theorem: The relation ~ defined above is an equivalence relation on A.

25 The Equivalence Relation Induced by a Partition
Proof: We must prove that ~ is reflexive, symmetric, and transitive. Proof that ~ is reflexive. Let a  A. Then a is in Ai for some i  I. So a ~ a.

26 The Equivalence Relation Induced by a Partition
Proof that ~ is symmetric. Let a, b  A and suppose that a ~ b. Then a, b  Ai for some i  I. So b, a  Ai for some i  I. Therefore b ~ a.

27 The Equivalence Relation Induced by a Partition
Proof that ~ is transitive. Let a, b, c  A and suppose a ~ b and b ~ c. Then a, b  Ai for some i  I and b, c  Aj for some j  I. That means that b  Ai  Aj. This is possible only if Ai = Aj. Therefore, a, c  Ai. So, a ~ c.

28 Example Consider the set P of all computer programs.
Partition P into subsets by putting in the same subset any two programs that always produce identical output for the same input.

29 Example This partition determines an equivalence relation  on P.
Let p1 and p2 be two computer programs. Then p1  p2 if p1 and p2 always produce identical output for the same input.

30 Example Let A be the set of all people on Earth.
Let R be the relation defined by x R y if x and y have ever shaken hands. Is R reflexive? Symmetric? Transitive? Let R* be the reflexive-transitive closure of R. Is R* an equivalence relation? What are the equivalence classes of R*?


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