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Equivalence Relations

Equivalence Relations. Equivalence Relations. A relation R on A A is an equivalence relation when R satisfies 3 conditions:  x  A, xRx ( reflexive ).  x, y  A, xRy  yRx ( symmetric ).  x, y, z  A, (xRy  yRz)  xRz ( transitive ).

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Equivalence Relations

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  1. Equivalence Relations

  2. Equivalence Relations • A relation R on AA is an equivalence relation when R satisfies 3 conditions: • x  A, xRx (reflexive). • x, y  A, xRy  yRx (symmetric). • x, y, z  A, (xRy  yRz)  xRz (transitive). • How are the properties of an equivalence relation reflected in its graph representation?

  3. Examples? • Let P be the set of all human beings; R  PP. • Is R an equivalence relation if: • aRb when a is the brother of b? • aRb when a is in the same family as b? • R  N N. • xRy when x has the same remainder as y when they are divided by 5?

  4. Examples? • Let R2 = RR be the set of points in the plane. • Is E  R2 R2 when • E1 = {((x1,y1),(x2,y2)) | (x1,y1) & (x2,y2) are on the same horizontal line}? (E1 “partitions” the plane into horizontal lines.) • E2 = {((x1,y1),(x2,y2)) | (x1,y1) & (x2,y2) are equidistant from the origin}? (E2 “partitions” the plane into concentric circles.)

  5. Partitions • Let S be a set. A partition, (S), of S is a set of nonempty subsets of S such that: • Si  (S) Si = S (the parts cover S) • SiSj  (S), SiSj =  (the parts are disjoint) • Example: (check 2 conditions of partition): • “The same remainder when divided by 5” partitions N into 5 parts. • E1 partitions the plane into horizontal lines. • E2 partitions the plane into concentric circles.

  6. Partitions as Equivalence Relations • Let E  SS be an equivalence relation, and a S. • The equivalence class determined by a is: [a] = {b S | aEb}: the set of all elements of S equivalent to a. • Let P be the set of equivalence classes under E.

  7. aEb  [a] = [b] I.e., any member of [a] can name the class. Assume[a][b]: Without loss of generality, c [b] and c[a] (draw a Venn diagram) • aEb, (given); • bEc, (by assumption) • aEc, (E is transitive) • c [a] (definition of equivalence class). • Therefore, [a][b] is false.

  8. Equivalence classes partition S • To prove this, we must show that: (i) the union of all equivalence classes equals S; (ii) if a is not equivalent to b, then [a] [b] = . (i): Since E is reflexive, a S, there is some equivalence class that contains a: [a]. Therefore, a  S [a] = S.

  9. Equivalence classes partition S ... (ii): To show: For [a][b], [a] [b] =  • Assume not:  c  [a] [b]. • c  [a]  aEc which implies [a] = [c]; • c  [b]  bEc which implies [b] = [c]; • Therefore, [a] = [b], a contradiction. • Therefore, [a] [b] = . The set of equivalence classes partitions S.

  10. A partition defines an equivalence relation • Let (S) be a partition of S: • Si  (S) Si = S (the parts “cover” S) • SiSj  (S), SiSj =  (the parts are disjoint) • Define E = {(a,b) | a, b Si  (S)}. • Illustrate on blackboard. • Claim: E is an equivalence relation: E is reflexive, symmetric, & transitive.

  11. E is an equivalence relation (i): x S, xEx(reflexive): • Since (S) is a partition, everyx is in some part. • Every element x of S is in the same part as itself: xEx. (ii):x, y S, x Ey  y Ex (symmetric). • If x is in the same part as y, then y is in the same part as x.

  12. E is an equivalence relation ... (iii): x, y, z S, (x Ey  y Ez)  x Ez (transitive): • If x is in the same part as y and y is in the same part as z, then x is in the same part as z. E is an equivalence relation

  13. Equivalence relations: summary • Partitioning a set S is the same thing as defining an equivalence relation over S. • If E is an equivalence relation of S, the associated partition is called the quotient set of S relative to E and is denoted S/E.

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