Equivalence relation and the quotient set
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Good day everyone,
let $sim$ be an equivalence relation on some set $E$, by definition we have $bar{x}={y in E mid x sim y}$ the equivalence class of x, and $E/sim$ the quotient set (set of all equivalence classes).
Wouldn't $E/sim$ contain redundancies since if $x sim y$ with $x,y in E$ then $bar{x}=bar{y}$? Which is a problem because sets are composed of different elements.
abstract-algebra
asked Nov 18 at 9:33
archaic
325
add a comment |
up vote
0
down vote
favorite
Good day everyone,
let $sim$ be an equivalence relation on some set $E$, by definition we have $bar{x}={y in E mid x sim y}$ the equivalence class of x, and $E/sim$ the quotient set (set of all equivalence classes).
Wouldn't $E/sim$ contain redundancies since if $x sim y$ with $x,y in E$ then $bar{x}=bar{y}$? Which is a problem because sets are composed of different elements.
abstract-algebra
asked Nov 18 at 9:33
archaic
325
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Good day everyone,
let $sim$ be an equivalence relation on some set $E$, by definition we have $bar{x}={y in E mid x sim y}$ the equivalence class of x, and $E/sim$ the quotient set (set of all equivalence classes).
Wouldn't $E/sim$ contain redundancies since if $x sim y$ with $x,y in E$ then $bar{x}=bar{y}$? Which is a problem because sets are composed of different elements.
abstract-algebra
asked Nov 18 at 9:33
archaic
325
Good day everyone,
let $sim$ be an equivalence relation on some set $E$, by definition we have $bar{x}={y in E mid x sim y}$ the equivalence class of x, and $E/sim$ the quotient set (set of all equivalence classes).
Wouldn't $E/sim$ contain redundancies since if $x sim y$ with $x,y in E$ then $bar{x}=bar{y}$? Which is a problem because sets are composed of different elements.
abstract-algebra
abstract-algebra
asked Nov 18 at 9:33
archaic
325
asked Nov 18 at 9:33
archaic
325
edited Nov 18 at 9:41
asked Nov 18 at 9:33
archaic
325
asked Nov 18 at 9:33
archaic
325
asked Nov 18 at 9:33
archaic
325
325
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
If $sim$ is an equivalence relation, then we know it is a reflexive, symmetric, and transitive relation. Only then can say that $sim$ partitions $E$ into equivalence classes. Let $x sim y$. Then for any $zin bar x$, we know that $xsim z$ and $y sim x$ (by the symmetry), and then $ysim z$ (transitivity). Therefore, if $x sim y$ then $bar x = bar y$.
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
Yes and therefore $bar{x}$ and $bar{y}$ intersect hence only one should be considered in $E/sim$
– archaic
Nov 18 at 9:49
Am I failing to notice something?
– archaic
Nov 18 at 9:50
Not only do they intersect, we showed that they must be the exact same set. In $E/sim$ this corresponds to one equivalence class that is equal to $bar x$ and to $bar y$
– Joey Kilpatrick
Nov 18 at 9:50
Alright I did a blunder here, thank you. I was thinking that since $x$ and $y$ are different elements of their own then each one should see its equivalent class in $E/sim$.
– archaic
Nov 18 at 9:52
Not just that, how to explain ..
– archaic
Nov 18 at 9:53
|
show 2 more comments
up vote
0
down vote
No, it doesn't. Given two sets $A,B$. Define $A=B$ iff $forall x [xin ALeftrightarrow xin B]$. Thus a set can be represented such that it contains an element multiple times and the ordering of elements in the set doesn't matter, e.g., ${1,2,3}={3,2,1} = {1,2,1,3,1}$.
answered Nov 18 at 9:37
Wuestenfux
2,6821410
math.stackexchange.com/questions/934378/…
– archaic
Nov 18 at 9:39
add a comment |
up vote
0
down vote
Equivalence classes are defined over equivalence relations so them are well defined.
Check This
answered Nov 18 at 9:38
Kato
534
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If $sim$ is an equivalence relation, then we know it is a reflexive, symmetric, and transitive relation. Only then can say that $sim$ partitions $E$ into equivalence classes. Let $x sim y$. Then for any $zin bar x$, we know that $xsim z$ and $y sim x$ (by the symmetry), and then $ysim z$ (transitivity). Therefore, if $x sim y$ then $bar x = bar y$.
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
Yes and therefore $bar{x}$ and $bar{y}$ intersect hence only one should be considered in $E/sim$
– archaic
Nov 18 at 9:49
Am I failing to notice something?
– archaic
Nov 18 at 9:50
Not only do they intersect, we showed that they must be the exact same set. In $E/sim$ this corresponds to one equivalence class that is equal to $bar x$ and to $bar y$
– Joey Kilpatrick
Nov 18 at 9:50
Alright I did a blunder here, thank you. I was thinking that since $x$ and $y$ are different elements of their own then each one should see its equivalent class in $E/sim$.
– archaic
Nov 18 at 9:52
Not just that, how to explain ..
– archaic
Nov 18 at 9:53
|
show 2 more comments
up vote
0
down vote
accepted
If $sim$ is an equivalence relation, then we know it is a reflexive, symmetric, and transitive relation. Only then can say that $sim$ partitions $E$ into equivalence classes. Let $x sim y$. Then for any $zin bar x$, we know that $xsim z$ and $y sim x$ (by the symmetry), and then $ysim z$ (transitivity). Therefore, if $x sim y$ then $bar x = bar y$.
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
Yes and therefore $bar{x}$ and $bar{y}$ intersect hence only one should be considered in $E/sim$
– archaic
Nov 18 at 9:49
Am I failing to notice something?
– archaic
Nov 18 at 9:50
Not only do they intersect, we showed that they must be the exact same set. In $E/sim$ this corresponds to one equivalence class that is equal to $bar x$ and to $bar y$
– Joey Kilpatrick
Nov 18 at 9:50
Alright I did a blunder here, thank you. I was thinking that since $x$ and $y$ are different elements of their own then each one should see its equivalent class in $E/sim$.
– archaic
Nov 18 at 9:52
Not just that, how to explain ..
– archaic
Nov 18 at 9:53
|
show 2 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If $sim$ is an equivalence relation, then we know it is a reflexive, symmetric, and transitive relation. Only then can say that $sim$ partitions $E$ into equivalence classes. Let $x sim y$. Then for any $zin bar x$, we know that $xsim z$ and $y sim x$ (by the symmetry), and then $ysim z$ (transitivity). Therefore, if $x sim y$ then $bar x = bar y$.
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
If $sim$ is an equivalence relation, then we know it is a reflexive, symmetric, and transitive relation. Only then can say that $sim$ partitions $E$ into equivalence classes. Let $x sim y$. Then for any $zin bar x$, we know that $xsim z$ and $y sim x$ (by the symmetry), and then $ysim z$ (transitivity). Therefore, if $x sim y$ then $bar x = bar y$.
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
answered Nov 18 at 9:39
Joey Kilpatrick
1,183422
1,183422
Yes and therefore $bar{x}$ and $bar{y}$ intersect hence only one should be considered in $E/sim$
– archaic
Nov 18 at 9:49
Am I failing to notice something?
– archaic
Nov 18 at 9:50
Not only do they intersect, we showed that they must be the exact same set. In $E/sim$ this corresponds to one equivalence class that is equal to $bar x$ and to $bar y$
– Joey Kilpatrick
Nov 18 at 9:50
Alright I did a blunder here, thank you. I was thinking that since $x$ and $y$ are different elements of their own then each one should see its equivalent class in $E/sim$.
– archaic
Nov 18 at 9:52
Not just that, how to explain ..
– archaic
Nov 18 at 9:53
|
show 2 more comments
Yes and therefore $bar{x}$ and $bar{y}$ intersect hence only one should be considered in $E/sim$
– archaic
Nov 18 at 9:49
Am I failing to notice something?
– archaic
Nov 18 at 9:50
Not only do they intersect, we showed that they must be the exact same set. In $E/sim$ this corresponds to one equivalence class that is equal to $bar x$ and to $bar y$
– Joey Kilpatrick
Nov 18 at 9:50
Alright I did a blunder here, thank you. I was thinking that since $x$ and $y$ are different elements of their own then each one should see its equivalent class in $E/sim$.
– archaic
Nov 18 at 9:52
Not just that, how to explain ..
– archaic
Nov 18 at 9:53
Yes and therefore $bar{x}$ and $bar{y}$ intersect hence only one should be considered in $E/sim$
– archaic
Nov 18 at 9:49
Yes and therefore $bar{x}$ and $bar{y}$ intersect hence only one should be considered in $E/sim$
– archaic
Nov 18 at 9:49
Am I failing to notice something?
– archaic
Nov 18 at 9:50
Am I failing to notice something?
– archaic
Nov 18 at 9:50
Not only do they intersect, we showed that they must be the exact same set. In $E/sim$ this corresponds to one equivalence class that is equal to $bar x$ and to $bar y$
– Joey Kilpatrick
Nov 18 at 9:50
Not only do they intersect, we showed that they must be the exact same set. In $E/sim$ this corresponds to one equivalence class that is equal to $bar x$ and to $bar y$
– Joey Kilpatrick
Nov 18 at 9:50
Alright I did a blunder here, thank you. I was thinking that since $x$ and $y$ are different elements of their own then each one should see its equivalent class in $E/sim$.
– archaic
Nov 18 at 9:52
Alright I did a blunder here, thank you. I was thinking that since $x$ and $y$ are different elements of their own then each one should see its equivalent class in $E/sim$.
– archaic
Nov 18 at 9:52
Not just that, how to explain ..
– archaic
Nov 18 at 9:53
Not just that, how to explain ..
– archaic
Nov 18 at 9:53
|
show 2 more comments
up vote
0
down vote
No, it doesn't. Given two sets $A,B$. Define $A=B$ iff $forall x [xin ALeftrightarrow xin B]$. Thus a set can be represented such that it contains an element multiple times and the ordering of elements in the set doesn't matter, e.g., ${1,2,3}={3,2,1} = {1,2,1,3,1}$.
answered Nov 18 at 9:37
Wuestenfux
2,6821410
math.stackexchange.com/questions/934378/…
– archaic
Nov 18 at 9:39
add a comment |
up vote
0
down vote
No, it doesn't. Given two sets $A,B$. Define $A=B$ iff $forall x [xin ALeftrightarrow xin B]$. Thus a set can be represented such that it contains an element multiple times and the ordering of elements in the set doesn't matter, e.g., ${1,2,3}={3,2,1} = {1,2,1,3,1}$.
answered Nov 18 at 9:37
Wuestenfux
2,6821410
math.stackexchange.com/questions/934378/…
– archaic
Nov 18 at 9:39
add a comment |
up vote
0
down vote
up vote
0
down vote
No, it doesn't. Given two sets $A,B$. Define $A=B$ iff $forall x [xin ALeftrightarrow xin B]$. Thus a set can be represented such that it contains an element multiple times and the ordering of elements in the set doesn't matter, e.g., ${1,2,3}={3,2,1} = {1,2,1,3,1}$.
answered Nov 18 at 9:37
Wuestenfux
2,6821410
No, it doesn't. Given two sets $A,B$. Define $A=B$ iff $forall x [xin ALeftrightarrow xin B]$. Thus a set can be represented such that it contains an element multiple times and the ordering of elements in the set doesn't matter, e.g., ${1,2,3}={3,2,1} = {1,2,1,3,1}$.
answered Nov 18 at 9:37
Wuestenfux
2,6821410
answered Nov 18 at 9:37
Wuestenfux
2,6821410
answered Nov 18 at 9:37
Wuestenfux
2,6821410
answered Nov 18 at 9:37
Wuestenfux
2,6821410
2,6821410
math.stackexchange.com/questions/934378/…
– archaic
Nov 18 at 9:39
add a comment |
math.stackexchange.com/questions/934378/…
– archaic
Nov 18 at 9:39
math.stackexchange.com/questions/934378/…
– archaic
Nov 18 at 9:39
math.stackexchange.com/questions/934378/…
– archaic
Nov 18 at 9:39
add a comment |
up vote
0
down vote
Equivalence classes are defined over equivalence relations so them are well defined.
Check This
answered Nov 18 at 9:38
Kato
534
add a comment |
up vote
0
down vote
Equivalence classes are defined over equivalence relations so them are well defined.
Check This
answered Nov 18 at 9:38
Kato
534
add a comment |
up vote
0
down vote
up vote
0
down vote
Equivalence classes are defined over equivalence relations so them are well defined.
Check This
answered Nov 18 at 9:38
Kato
534
Equivalence classes are defined over equivalence relations so them are well defined.
Check This
answered Nov 18 at 9:38
Kato
534
answered Nov 18 at 9:38
Kato
534
answered Nov 18 at 9:38
Kato
534
answered Nov 18 at 9:38
Kato
534
534
add a comment |
add a comment |
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