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!

randlemcmurphy
 Follow
 6 followers
 10 badges
 Send a private message to randlemcmurphy
 Thread Starter
Offline10ReputationRep: Follow
 1
 14112015 13:41

 Follow
 2
 14112015 14:34
(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!
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.
where denotes the equivalence class for .
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  that suggests using some kind of periodicity. 
randlemcmurphy
 Follow
 6 followers
 10 badges
 Send a private message to randlemcmurphy
 Thread Starter
Offline10ReputationRep: Follow
 3
 14112015 15:54
(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.
where denotes the equivalence class for .
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  that suggests using some kind of periodicity.
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)) 
 Follow
 4
 14112015 16:19
(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))
As for b) the equivalence class of a given element is, as you said, the set of such that .
So for each , you simply need to write down the set of such .
You have been given this information in the question. 
metaltron
 Follow
 4 followers
 10 badges
 Send a private message to metaltron
Offline10ReputationRep: Follow
 5
 14112015 17:39
(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))
Your example for c) is good, just notationally you should write f:[0,1]/~ > S^1. 
randlemcmurphy
 Follow
 6 followers
 10 badges
 Send a private message to randlemcmurphy
 Thread Starter
Offline10ReputationRep: Follow
 6
 14112015 19:08
(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.
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)! 
DFranklin
 Follow
 60 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 7
 14112015 19:50
(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..
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. 
metaltron
 Follow
 4 followers
 10 badges
 Send a private message to metaltron
Offline10ReputationRep: Follow
 8
 14112015 19:51
(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)!
If x = 0 then ...
If x = 1 then...
EDIT DFranklin has already provided a perfectly good response!Last edited by metaltron; 14112015 at 19:53. 
 Follow
 9
 15112015 02:24
(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). 
randlemcmurphy
 Follow
 6 followers
 10 badges
 Send a private message to randlemcmurphy
 Thread Starter
Offline10ReputationRep: Follow
 10
 15112015 12:40
(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.
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] 
DFranklin
 Follow
 60 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 11
 15112015 12:48
(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] 
randlemcmurphy
 Follow
 6 followers
 10 badges
 Send a private message to randlemcmurphy
 Thread Starter
Offline10ReputationRep: Follow
 12
 15112015 15:11
(Original post by DFranklin)
You also have the cases x = y or y = z. 
randlemcmurphy
 Follow
 6 followers
 10 badges
 Send a private message to randlemcmurphy
 Thread Starter
Offline10ReputationRep: Follow
 13
 15112015 15:18
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?Last edited by randlemcmurphy; 15112015 at 15:44. 
DFranklin
 Follow
 60 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 14
 15112015 15:46
(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?
x = y= 0, z = 1
or x = 0, y=z=1, (this is the y=z case, not the x=y case)
etc. 
randlemcmurphy
 Follow
 6 followers
 10 badges
 Send a private message to randlemcmurphy
 Thread Starter
Offline10ReputationRep: Follow
 15
 15112015 15:50
(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.
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 ) 
DFranklin
 Follow
 60 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 16
 15112015 15:51
(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?

Cardiff University

Business Management and Mathematics
Keele University

Coventry University

Aberystwyth University

Mathematics for Finance and Management
University of Portsmouth

Initial Year for Extended Degree in Science  Mathematics
University of Hertfordshire

Durham University

Mathematics (with Foundation Year)
University of the West of England

University of Oxford

Mathematics/Business (Joint Honours)
University of Hertfordshire
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 SherlockHolmes
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 rayquaza17
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Labrador99
 EmilySarah00