Equivalence relations
Download
1 / 13

Equivalence Relations - PowerPoint PPT Presentation


  • 154 Views
  • Uploaded on

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 ).

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Equivalence Relations' - ginger


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Equivalence relations1
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
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?


Examples1
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
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
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
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
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 s1
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
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
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 relation1
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
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.


ad