Equivalence Relations

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# Equivalence Relations - PowerPoint PPT Presentation

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

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?
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?
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.)

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.
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.
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.
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.

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.

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.

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.
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

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.