Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

Equivalence Relations, Classes (and bijective maps) watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    Hi,

    Question:

    Consider the unit interval [0, 1] ⊂ R and the relations x ∼ x for all x ∈ [0, 1],plus 0 ∼ 1 and 1 ∼ 0.
    (a) Show that ∼ defines an equivalence relation on [0, 1].
    (b) Identify the equivalence classes of ∼ in [0, 1].
    (c) Construct a bijective map g : [0, 1]/∼ → S1from the quotient set [0, 1]/∼to the circle S1 (unit circle) .

    My attempt:

    a) The relation is reflexive because of the definition x ∼ x for all x ∈ [0, 1]
    The relation is also symmetric because of the definition that 0 ∼ 1 and 1 ∼ 0.
    Finally, the relation is also transitive, take 0 ∼ 1 and 1 ∼ 0 then 0 ~ 0 which holds as 0 ∈ [0, 1] . Is this right?

    b) I understand that the equivalence class is say you have a set S with equivalence relation R, then the set of elements in S that are equivalent to one another is called an equivalence class.
    My attempt for the this equivalence relation is:
    {[0, 1] ⊂ R | for all x ∈ [0, 1], x ~ x, also 0 ~ 1 AND 1 ~ 0}
    I'm not too certain with this...

    c) Not really considered c) yet. More concerned with understanding a) and b) right now.

    Thanks for all help given!
    Offline

    11
    ReputationRep:
    (Original post by randlemcmurphy)
    Hi,

    Question:

    Consider the unit interval [0, 1] ⊂ R and the relations x ∼ x for all x ∈ [0, 1],plus 0 ∼ 1 and 1 ∼ 0.
    (a) Show that ∼ defines an equivalence relation on [0, 1].
    (b) Identify the equivalence classes of ∼ in [0, 1].
    (c) Construct a bijective map g : [0, 1]/∼ → S1from the quotient set [0, 1]/∼to the circle S1 (unit circle) .

    My attempt:

    a) The relation is reflexive because of the definition x ∼ x for all x ∈ [0, 1]
    The relation is also symmetric because of the definition that 0 ∼ 1 and 1 ∼ 0.
    Finally, the relation is also transitive, take 0 ∼ 1 and 1 ∼ 0 then 0 ~ 0 which holds as 0 ∈ [0, 1] . Is this right?

    b) I understand that the equivalence class is say you have a set S with equivalence relation R, then the set of elements in S that are equivalent to one another is called an equivalence class.
    My attempt for the this equivalence relation is:
    {[0, 1] ⊂ R | for all x ∈ [0, 1], x ~ x, also 0 ~ 1 AND 1 ~ 0}
    I'm not too certain with this...

    c) Not really considered c) yet. More concerned with understanding a) and b) right now.

    Thanks for all help given!
    I'm not the best person for this but:

    1. I don't think that you have quite shown clearly that the definitions for an equivalence relation are satisfied (e.g. what about transitivity for 1?)

    2. I don't understand your notation for equivalence classes. I would expect to see e.g.

    [x] = \{ a \in [0,1] : a=?\}, [0] = \{ \cdots \}, [1] = \{ \cdots \}

    where [x] denotes the equivalence class for x.

    3. Note that this relation effectively identifies 0 with 1 i.e. it sticks the ends of the unit interval together in some sense. For the bijective map, you will therefore need a function where f(0)=f(1) - that suggests using some kind of periodicity.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by atsruser)
    I'm not the best person for this but:

    1. I don't think that you have quite shown clearly that the definitions for an equivalence relation are satisfied (e.g. what about transitivity for 1?)

    2. I don't understand your notation for equivalence classes. I would expect to see e.g.

    [x] = \{ a \in [0,1] : a=?\}, [0] = \{ \cdots \}, [1] = \{ \cdots \}

    where [x] denotes the equivalence class for x.

    3. Note that this relation effectively identifies 0 with 1 i.e. it sticks the ends of the unit interval together in some sense. For the bijective map, you will therefore need a function where f(0)=f(1) - that suggests using some kind of periodicity.
    I'm going to speak to my lecturer about b) as I am slightly confused by it.

    So for a) could I have a hint about how I would show the transitivity of 1?

    I realised for c) that 1 and 0 would be glued together to make the unit circle, and that f(0)=f(1).
    This can be achieved by the function:
    f : [0,1] -> S1 : a -> (cos(2pi*a), sin(2pi*a))
    Offline

    11
    ReputationRep:
    (Original post by randlemcmurphy)
    I'm going to speak to my lecturer about b) as I am slightly confused by it.

    So for a) could I have a hint about how I would show the transitivity of 1?

    I realised for c) that 1 and 0 would be glued together to make the unit circle, and that f(0)=f(1).
    This can be achieved by the function:
    f : [0,1] -> S1 : a -> (cos(2pi*a), sin(2pi*a))
    For a) just do what you did for 0 \in [0,1], only for 1 instead.

    As for b) the equivalence class of a given element x is, as you said, the set of y such that x \sim y.
    So for each x \in [0,1], you simply need to write down the set of such y.
    You have been given this information in the question.
    Offline

    10
    ReputationRep:
    (Original post by randlemcmurphy)
    I'm going to speak to my lecturer about b) as I am slightly confused by it.

    So for a) could I have a hint about how I would show the transitivity of 1?

    I realised for c) that 1 and 0 would be glued together to make the unit circle, and that f(0)=f(1).
    This can be achieved by the function:
    f : [0,1] -> S1 : a -> (cos(2pi*a), sin(2pi*a))
    Note that to show symmetry you need to show x~y implies y~x for all x,y in [0,1]; not just when x = 1, y = 0. The same thing for transitivity; you need to show if x~y and y~z then x~z for all x,y,z in [0,1].

    Your example for c) is good, just notationally you should write f:[0,1]/~ -> S^1.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by metaltron)
    Note that to show symmetry you need to show x~y implies y~x for all x,y in [0,1]; not just when x = 1, y = 0. The same thing for transitivity; you need to show if x~y and y~z then x~z for all x,y,z in [0,1].

    Your example for c) is good, just notationally you should write f:[0,1]/~ -> S^1.
    Thanks for your input,

    How would I go about showing it, I could easily say for x,y in [0,1] then x~y implies y~x, but that isn't really showing it...

    Thanks for pointing that out for me in c)!
    Online

    17
    ReputationRep:
    (Original post by randlemcmurphy)
    Thanks for your input,

    How would I go about showing it, I could easily say for x,y in [0,1] then x~y implies y~x, but that isn't really showing it..
    Given x ~ y, we wish to show y ~ x.

    But if x ~ y, by definition of ~ we have 3 possibilities:

    (1) x = y, and then y ~ x is immediate.
    (2) x = 0, y = 1, then by definition of ~ we have 1 ~ 0, so y ~ x.
    (3) x = 1, y = 0, then by definition of ~ we have 0 ~ 1, so y ~ x.

    So in all cases we can deduce y~x and so we are done.
    Offline

    10
    ReputationRep:
    (Original post by randlemcmurphy)
    Thanks for your input,

    How would I go about showing it, I could easily say for x,y in [0,1] then x~y implies y~x, but that isn't really showing it...

    Thanks for pointing that out for me in c)!
    You'll have to split it into a few cases ie if x is not 0 or 1 then if x ~ y then y = x so...
    If x = 0 then ...
    If x = 1 then...

    EDIT DFranklin has already provided a perfectly good response!
    Offline

    17
    ReputationRep:
    (1) and (3) have been pretty well covered, so I'll attack (2):

    An equivalence relation is fundamentally just a way of taking a set, and saying which elements of it we want to treat as the same: you should be able to see, with a bit of thought why this intuition forces the axioms given above to hold. Now, what we mean by an equivalence class is just one of the points that's left over after the squashing together stuff. Formally, that is: we want, for every point in there, the set of all points that we are saying are the same as it (obviously, there will be some overlap if our relation isn't trivial, so we'll only take each of these sets once).
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by DFranklin)
    Given x ~ y, we wish to show y ~ x.

    But if x ~ y, by definition of ~ we have 3 possibilities:

    (1) x = y, and then y ~ x is immediate.
    (2) x = 0, y = 1, then by definition of ~ we have 1 ~ 0, so y ~ x.
    (3) x = 1, y = 0, then by definition of ~ we have 0 ~ 1, so y ~ x.

    So in all cases we can deduce y~x and so we are done.
    How is this to show the relation is transitive?:

    Given x~y and y~z we wish to show x~z

    The possibilities:

    (1) x=y=z in which case x~z
    (2) x=0, y=1, z=0, then we have 0~0 which by definition of ~ holds as 0 ∈ [0, 1]
    (3) x=1, y=0, z=1 then we have 1~1 which by definition of ~ holds as 1 ∈ [0, 1]
    Online

    17
    ReputationRep:
    (Original post by randlemcmurphy;60629109

    The possibilities:

    (1) x=y=z in which case x~z
    (2) x=0, y=1, z=0, then we have 0~0 which by definition of ~ holds as 0 ∈ [0, 1
    )

    (3) x=1, y=0, z=1 then we have 1~1 which by definition of ~ holds as 1 ∈ [0, 1]
    You also have the cases x = y or y = z.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by DFranklin)
    You also have the cases x = y or y = z.
    Sorry I'm a bit confused, if x=y then wouldn't z also have to be equal to x and y?
    • Thread Starter
    Offline

    10
    ReputationRep:
    One other question, if I do the function for c) as f : [0,1]/~ -> S1 : a -> (cos(2pi*a), sin(2pi*a)) . Then this isn't bijective as it isn't injective (f(0)=f(1)) :s

    EDIT: Or because we have said that 0 and 1 are equivalent, it is still injective?
    Online

    17
    ReputationRep:
    (Original post by randlemcmurphy)
    Sorry I'm a bit confused, if x=y then wouldn't z also have to be equal to x and y?
    You could have
    x = y= 0, z = 1

    or x = 0, y=z=1, (this is the y=z case, not the x=y case)

    etc.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by DFranklin)
    You could have
    x = y= 0, z = 1

    or x = 0, y=z=1, (this is the y=z case, not the x=y case)

    etc.
    Oh, silly me! I get it now.

    One last question if you don't mind...

    If I do the function for c) as f : [0,1]/~ -> S1 : a -> (cos(2pi*a), sin(2pi*a)) . Then this isn't bijective as it isn't injective (f(0)=f(1)) :s
    Or because we have said that 0 and 1 are equivalent, it is still injective?

    Thanks for all your help! (And everybody else who has helped me )
    Online

    17
    ReputationRep:
    (Original post by randlemcmurphy)
    If I do the function for c) as f : [0,1]/~ -> S1 : a -> (cos(2pi*a), sin(2pi*a)) . Then this isn't bijective as it isn't injective (f(0)=f(1)) :s
    Or because we have said that 0 and 1 are equivalent, it is still injective?
    It's bijective because in the quotient set {0,1} is a single element (since the quotient set is made from equivalence classes).
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 15, 2015
Poll
Do you like carrot cake?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.