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# Equivalence Relations, Classes (and bijective maps) watch

1. 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!
2. (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.

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.
3. (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.
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))
4. (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 , only for instead.

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.
5. (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.
6. (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)!
7. (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.
8. (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!
9. (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).
10. (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]
11. (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.
12. (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?
13. 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?
14. (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.
15. (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 )
16. (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).

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