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Equivalence Relations

- A relation R on AA 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 PP.
- Is R an equivalence relation if:
- aRb when a is the brother of b?
- aRb when a is in the same family as b?

- Is R an equivalence relation if:
- R N N.
- xRy when x has the same remainder as y when they are divided by 5?

Examples?

- Let R2 = RR 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.)

- E1 = {((x1,y1),(x2,y2)) | (x1,y1) & (x2,y2) are on the same horizontal line}?

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)
- SiSj (S), SiSj = (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 SS 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)
- SiSj (S), SiSj = (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, xEx(reflexive):

- Since (S) is a partition, everyx is in some part.
- Every element x of S is in the same part as itself: xEx.
(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.

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