The union is not disjoint in the example from the book
$begingroup$
I am reading 'Measure, Integral and Probability' book. There is this following para in the book:
An equivalence relation $sim$ on $E$ partitions $E$ into disjoint
equivalence classes: given $x in E$, write $[x] = {z : z sim x}$
for the equivalence class of $x$, i.e. the set of all elements of $E$
that are equivalent to $x$. Thus $x in [x]$, hence $E = cup_{x in
E} [x]$. This is a disjoint union: if $[x] cap [y] neq
emptyset$, then there is $z in E$ with $x sim z$ and $z sim y$,
hence $x sim y$, so that $[x] = [y]$. We shall denote the set of all
equivalence classes so obtained by $E/sim$.
First of all, I do not see how $ cup_{x in E} [x]$ is a disjoint union. Take, for example, a set
$$S = {1, 2, 3, 4, 5 }$$ with the following equivalence on it:
$$sim = {(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1) }$$
then
$[1] = {1,2,3}$, $[2]={1,2,3}$, $[3]={1,2,3}$, $[4]={4}$, $[5]={5}$. Already, you have that sert $[1]$ is the same as $[2]$. So that the union $cup_{x in E} [x] = {1,2,3} cup {1,2,3}$ and that's not disjoint. Unless I do not understand what disjoint union means.
Also, what is $E/sim$?
elementary-set-theory equivalence-relations
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
$endgroup$
add a comment |
$begingroup$
I am reading 'Measure, Integral and Probability' book. There is this following para in the book:
An equivalence relation $sim$ on $E$ partitions $E$ into disjoint
equivalence classes: given $x in E$, write $[x] = {z : z sim x}$
for the equivalence class of $x$, i.e. the set of all elements of $E$
that are equivalent to $x$. Thus $x in [x]$, hence $E = cup_{x in
E} [x]$. This is a disjoint union: if $[x] cap [y] neq
emptyset$, then there is $z in E$ with $x sim z$ and $z sim y$,
hence $x sim y$, so that $[x] = [y]$. We shall denote the set of all
equivalence classes so obtained by $E/sim$.
First of all, I do not see how $ cup_{x in E} [x]$ is a disjoint union. Take, for example, a set
$$S = {1, 2, 3, 4, 5 }$$ with the following equivalence on it:
$$sim = {(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1) }$$
then
$[1] = {1,2,3}$, $[2]={1,2,3}$, $[3]={1,2,3}$, $[4]={4}$, $[5]={5}$. Already, you have that sert $[1]$ is the same as $[2]$. So that the union $cup_{x in E} [x] = {1,2,3} cup {1,2,3}$ and that's not disjoint. Unless I do not understand what disjoint union means.
Also, what is $E/sim$?
elementary-set-theory equivalence-relations
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
$endgroup$
$begingroup$
It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union.
$endgroup$
– Rellek
Dec 13 '18 at 22:18
$begingroup$
what about $E/sim$ is that just $cup_{x in E} [x]$?
$endgroup$
– i squared - Keep it Real
Dec 13 '18 at 22:24
1
$begingroup$
No it would just consist of the representatives for each equivalence class. That is, $E / sim = { [x] | x in E }$. In your example, $E/ sim = { [1] , [4], [5] }$. Of course you could use $[2]$ and $[3]$ for representatives of the class of $[1]$ instead, but they are the same under the relation $sim$.
$endgroup$
– Rellek
Dec 13 '18 at 22:28
add a comment |
$begingroup$
I am reading 'Measure, Integral and Probability' book. There is this following para in the book:
An equivalence relation $sim$ on $E$ partitions $E$ into disjoint
equivalence classes: given $x in E$, write $[x] = {z : z sim x}$
for the equivalence class of $x$, i.e. the set of all elements of $E$
that are equivalent to $x$. Thus $x in [x]$, hence $E = cup_{x in
E} [x]$. This is a disjoint union: if $[x] cap [y] neq
emptyset$, then there is $z in E$ with $x sim z$ and $z sim y$,
hence $x sim y$, so that $[x] = [y]$. We shall denote the set of all
equivalence classes so obtained by $E/sim$.
First of all, I do not see how $ cup_{x in E} [x]$ is a disjoint union. Take, for example, a set
$$S = {1, 2, 3, 4, 5 }$$ with the following equivalence on it:
$$sim = {(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1) }$$
then
$[1] = {1,2,3}$, $[2]={1,2,3}$, $[3]={1,2,3}$, $[4]={4}$, $[5]={5}$. Already, you have that sert $[1]$ is the same as $[2]$. So that the union $cup_{x in E} [x] = {1,2,3} cup {1,2,3}$ and that's not disjoint. Unless I do not understand what disjoint union means.
Also, what is $E/sim$?
elementary-set-theory equivalence-relations
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
$endgroup$
I am reading 'Measure, Integral and Probability' book. There is this following para in the book:
An equivalence relation $sim$ on $E$ partitions $E$ into disjoint
equivalence classes: given $x in E$, write $[x] = {z : z sim x}$
for the equivalence class of $x$, i.e. the set of all elements of $E$
that are equivalent to $x$. Thus $x in [x]$, hence $E = cup_{x in
E} [x]$. This is a disjoint union: if $[x] cap [y] neq
emptyset$, then there is $z in E$ with $x sim z$ and $z sim y$,
hence $x sim y$, so that $[x] = [y]$. We shall denote the set of all
equivalence classes so obtained by $E/sim$.
First of all, I do not see how $ cup_{x in E} [x]$ is a disjoint union. Take, for example, a set
$$S = {1, 2, 3, 4, 5 }$$ with the following equivalence on it:
$$sim = {(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1) }$$
then
$[1] = {1,2,3}$, $[2]={1,2,3}$, $[3]={1,2,3}$, $[4]={4}$, $[5]={5}$. Already, you have that sert $[1]$ is the same as $[2]$. So that the union $cup_{x in E} [x] = {1,2,3} cup {1,2,3}$ and that's not disjoint. Unless I do not understand what disjoint union means.
Also, what is $E/sim$?
elementary-set-theory equivalence-relations
elementary-set-theory equivalence-relations
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
edited Dec 14 '18 at 0:04
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
asked Dec 13 '18 at 22:12
i squared - Keep it Reali squared - Keep it Real
1,62511128
1,62511128
$begingroup$
It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union.
$endgroup$
– Rellek
Dec 13 '18 at 22:18
$begingroup$
what about $E/sim$ is that just $cup_{x in E} [x]$?
$endgroup$
– i squared - Keep it Real
Dec 13 '18 at 22:24
1
$begingroup$
No it would just consist of the representatives for each equivalence class. That is, $E / sim = { [x] | x in E }$. In your example, $E/ sim = { [1] , [4], [5] }$. Of course you could use $[2]$ and $[3]$ for representatives of the class of $[1]$ instead, but they are the same under the relation $sim$.
$endgroup$
– Rellek
Dec 13 '18 at 22:28
add a comment |
$begingroup$
It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union.
$endgroup$
– Rellek
Dec 13 '18 at 22:18
$begingroup$
what about $E/sim$ is that just $cup_{x in E} [x]$?
$endgroup$
– i squared - Keep it Real
Dec 13 '18 at 22:24
1
$begingroup$
No it would just consist of the representatives for each equivalence class. That is, $E / sim = { [x] | x in E }$. In your example, $E/ sim = { [1] , [4], [5] }$. Of course you could use $[2]$ and $[3]$ for representatives of the class of $[1]$ instead, but they are the same under the relation $sim$.
$endgroup$
– Rellek
Dec 13 '18 at 22:28
$begingroup$
It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union.
$endgroup$
– Rellek
Dec 13 '18 at 22:18
$begingroup$
It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union.
$endgroup$
– Rellek
Dec 13 '18 at 22:18
$begingroup$
what about $E/sim$ is that just $cup_{x in E} [x]$?
$endgroup$
– i squared - Keep it Real
Dec 13 '18 at 22:24
$begingroup$
what about $E/sim$ is that just $cup_{x in E} [x]$?
$endgroup$
– i squared - Keep it Real
Dec 13 '18 at 22:24
1
1
$begingroup$
No it would just consist of the representatives for each equivalence class. That is, $E / sim = { [x] | x in E }$. In your example, $E/ sim = { [1] , [4], [5] }$. Of course you could use $[2]$ and $[3]$ for representatives of the class of $[1]$ instead, but they are the same under the relation $sim$.
$endgroup$
– Rellek
Dec 13 '18 at 22:28
$begingroup$
No it would just consist of the representatives for each equivalence class. That is, $E / sim = { [x] | x in E }$. In your example, $E/ sim = { [1] , [4], [5] }$. Of course you could use $[2]$ and $[3]$ for representatives of the class of $[1]$ instead, but they are the same under the relation $sim$.
$endgroup$
– Rellek
Dec 13 '18 at 22:28
add a comment |
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$begingroup$
It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union.
$endgroup$
– Rellek
Dec 13 '18 at 22:18
$begingroup$
what about $E/sim$ is that just $cup_{x in E} [x]$?
$endgroup$
– i squared - Keep it Real
Dec 13 '18 at 22:24
1
$begingroup$
No it would just consist of the representatives for each equivalence class. That is, $E / sim = { [x] | x in E }$. In your example, $E/ sim = { [1] , [4], [5] }$. Of course you could use $[2]$ and $[3]$ for representatives of the class of $[1]$ instead, but they are the same under the relation $sim$.
$endgroup$
– Rellek
Dec 13 '18 at 22:28