Why are partitions and equivalence relations the same thing?

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My lecturer omitted the proof in the lecture notes. From what I can gather, it's because equivalence classes partition always partition a set (the class can contain only that element or more and the elements in that class can only be in that one class so it acts as a partition). However, this, firstly, doesn't tell me why equivalence relations are the same thing as partitions and not equivalence classes, and to me it sounds like partitions cannot be arbitrary. From what I understood all they have to be are a collection of subsets $A_i$ forming some set $A$ such that:
- $A_i ne emptyset$
- $A_i cap A_j = emptyset , textif $j ne i$$
- $cup _i A_i = A$
I don't see any of these rules demanding subtly there be an equivalence relation, and doing so sounds like it puts restrictions on the creation of a partition that I don't see in its definition. How are they equivalent?
equivalence-relations set-partition
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up vote
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My lecturer omitted the proof in the lecture notes. From what I can gather, it's because equivalence classes partition always partition a set (the class can contain only that element or more and the elements in that class can only be in that one class so it acts as a partition). However, this, firstly, doesn't tell me why equivalence relations are the same thing as partitions and not equivalence classes, and to me it sounds like partitions cannot be arbitrary. From what I understood all they have to be are a collection of subsets $A_i$ forming some set $A$ such that:
- $A_i ne emptyset$
- $A_i cap A_j = emptyset , textif $j ne i$$
- $cup _i A_i = A$
I don't see any of these rules demanding subtly there be an equivalence relation, and doing so sounds like it puts restrictions on the creation of a partition that I don't see in its definition. How are they equivalent?
equivalence-relations set-partition
Do you mean $A_icap A_j=varnothing$ if $jne i$?
– Frpzzd
4 hours ago
@Frpzzd Good spot, thanks for pointing it out.
– sangstar
4 hours ago
3
Declare two things equivalent if they live in the same subset. That’s an equivalence relation.
– Randall
4 hours ago
@Randall Ah, that's a clever example. Can I use this to extend the general sameness of equivalence relations and partitions that I've been offered to accept however, or only if we consider the equivalence relation "if they live in the same subset"?
– sangstar
4 hours ago
They are equivalent: partitions and eq relations are the same concept via my remark. It’s a great learning exercise.
– Randall
4 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My lecturer omitted the proof in the lecture notes. From what I can gather, it's because equivalence classes partition always partition a set (the class can contain only that element or more and the elements in that class can only be in that one class so it acts as a partition). However, this, firstly, doesn't tell me why equivalence relations are the same thing as partitions and not equivalence classes, and to me it sounds like partitions cannot be arbitrary. From what I understood all they have to be are a collection of subsets $A_i$ forming some set $A$ such that:
- $A_i ne emptyset$
- $A_i cap A_j = emptyset , textif $j ne i$$
- $cup _i A_i = A$
I don't see any of these rules demanding subtly there be an equivalence relation, and doing so sounds like it puts restrictions on the creation of a partition that I don't see in its definition. How are they equivalent?
equivalence-relations set-partition
My lecturer omitted the proof in the lecture notes. From what I can gather, it's because equivalence classes partition always partition a set (the class can contain only that element or more and the elements in that class can only be in that one class so it acts as a partition). However, this, firstly, doesn't tell me why equivalence relations are the same thing as partitions and not equivalence classes, and to me it sounds like partitions cannot be arbitrary. From what I understood all they have to be are a collection of subsets $A_i$ forming some set $A$ such that:
- $A_i ne emptyset$
- $A_i cap A_j = emptyset , textif $j ne i$$
- $cup _i A_i = A$
I don't see any of these rules demanding subtly there be an equivalence relation, and doing so sounds like it puts restrictions on the creation of a partition that I don't see in its definition. How are they equivalent?
equivalence-relations set-partition
equivalence-relations set-partition
edited 4 hours ago
asked 4 hours ago
sangstar
786214
786214
Do you mean $A_icap A_j=varnothing$ if $jne i$?
– Frpzzd
4 hours ago
@Frpzzd Good spot, thanks for pointing it out.
– sangstar
4 hours ago
3
Declare two things equivalent if they live in the same subset. That’s an equivalence relation.
– Randall
4 hours ago
@Randall Ah, that's a clever example. Can I use this to extend the general sameness of equivalence relations and partitions that I've been offered to accept however, or only if we consider the equivalence relation "if they live in the same subset"?
– sangstar
4 hours ago
They are equivalent: partitions and eq relations are the same concept via my remark. It’s a great learning exercise.
– Randall
4 hours ago
add a comment |
Do you mean $A_icap A_j=varnothing$ if $jne i$?
– Frpzzd
4 hours ago
@Frpzzd Good spot, thanks for pointing it out.
– sangstar
4 hours ago
3
Declare two things equivalent if they live in the same subset. That’s an equivalence relation.
– Randall
4 hours ago
@Randall Ah, that's a clever example. Can I use this to extend the general sameness of equivalence relations and partitions that I've been offered to accept however, or only if we consider the equivalence relation "if they live in the same subset"?
– sangstar
4 hours ago
They are equivalent: partitions and eq relations are the same concept via my remark. It’s a great learning exercise.
– Randall
4 hours ago
Do you mean $A_icap A_j=varnothing$ if $jne i$?
– Frpzzd
4 hours ago
Do you mean $A_icap A_j=varnothing$ if $jne i$?
– Frpzzd
4 hours ago
@Frpzzd Good spot, thanks for pointing it out.
– sangstar
4 hours ago
@Frpzzd Good spot, thanks for pointing it out.
– sangstar
4 hours ago
3
3
Declare two things equivalent if they live in the same subset. That’s an equivalence relation.
– Randall
4 hours ago
Declare two things equivalent if they live in the same subset. That’s an equivalence relation.
– Randall
4 hours ago
@Randall Ah, that's a clever example. Can I use this to extend the general sameness of equivalence relations and partitions that I've been offered to accept however, or only if we consider the equivalence relation "if they live in the same subset"?
– sangstar
4 hours ago
@Randall Ah, that's a clever example. Can I use this to extend the general sameness of equivalence relations and partitions that I've been offered to accept however, or only if we consider the equivalence relation "if they live in the same subset"?
– sangstar
4 hours ago
They are equivalent: partitions and eq relations are the same concept via my remark. It’s a great learning exercise.
– Randall
4 hours ago
They are equivalent: partitions and eq relations are the same concept via my remark. It’s a great learning exercise.
– Randall
4 hours ago
add a comment |
3 Answers
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up vote
5
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accepted
A partition and an equivalence relation are not the same thing; however, they can induce each other (as explained at the end of this answer). An equivalence relation $R$ on a set $A$ is a subset of $Atimes A$ satisfying the following properties:
$$(a,a)in Rspaceforall ain A$$
$$(a,b)in Rimplies (b,a)in Rspaceforall a,bin A$$
$$(a,b)in Rspacetextandspace (b,c)in Rimplies (a,c)in R spaceforall a,b,cin A$$
However, a partition $P$ of $A$ is a subset of $2^A$ satisfying the following two properties:
$$p_icap p_j=varnothingspaceforall p_i,p_jin 2^Aspacetextwithspace p_ine p_j$$
$$bigcup_i=1^p_i=A$$
We have that $Rsubset Atimes A$ and $Psubset 2^A$, so they're not even the same type of object. However, your professor probably meant that every equivalence relation on a set $A$ induces a partition of $A$, and vice versa.
More specifically, if $R$ is an equivalence relation on $A$, then the induced partition $P$ is
$$P=b:(a,b)in R:ain A$$
and if $P$ is a partition of $A$, then the induced equivalence relation $R$ is defined by
$$R=(a,b):exists p_iin Pspacetexts.t.space a,bin p_i$$
In plain words: If $R$ is a given equivalence relation, then the induced partition $P$ partitions $A$ into all sets of elements which are equivalent to each other under $R$. If $P$ is a given partition, then the induced equivalence relation $R$ is the relation for which $xsim y$ if and only if $x,y$ are in the same set of the partition P.
A bit ashamed to admit the notation looks a bit daunting to read and it's presenting as difficult to me. Is there a more wordsy explanation to this? Mainly the end of your answer addressing the induced partition and induced equivalence relation bit?
– sangstar
4 hours ago
@sangstar Sure, thanks for asking! I edited my question for you.
– Frpzzd
4 hours ago
Thanks! I think I hopefully understand now. Does my example illustrate this understanding? Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of that space. If we define an equivalence relation $R_W$ on $V$ by $u R_W v$ if $u-v in W$, then the set of elements which satisfy $u-v in W$ for arbitrary $u, v in V$ form a partition of $V$?
– sangstar
4 hours ago
1
@sangstar That's right! If you're familiar with a bit of elementary group theory, the sets in this induced partition are called the cosets of the subspace $W$.
– Frpzzd
4 hours ago
Ah, okay. Thank you!
– sangstar
4 hours ago
add a comment |
up vote
4
down vote
They are "the same thing" in the sense that given an equivalence relation there is a natural way to construct a partition, and given a partition there is a natural way to construct an equivalence relation, and these two natural ways invert one another. That's useful because whenever you encounter one of these objects you are free to reason about the other if that makes your argument easier.
You are right when you say that it is the equivalence classes of an equivalence relation that form a partition (that is in fact the natural thing to look at), not the equivalence relation itself.
The natural way to construct an equivalence relation from a partition is to define two elements to be equivalent just when they are in the same block of the partition.
I see. What if there is a defined equivalence relation $R$ that isn't defined as "is in the same partition as". Can we not then equate equivalence classes and partitions in this case? Just trying to make sure I understand you.
– sangstar
4 hours ago
@sangstar Every equivalent relation can be expressed as "is in the same partition as" for some partition; that is, every equivalence relation has an induced partition just as every partition has an induced equivalence relation.
– Frpzzd
4 hours ago
@Frpzzd I see. I'm afraid the notation in your answer is a bit difficult to read for my less experienced eyes. Is there a more wordy explanation for your answer?
– sangstar
4 hours ago
My answer says in words just what @Frpzzd 's answer says with notation.
– Ethan Bolker
3 hours ago
add a comment |
up vote
0
down vote
Here's a visual explanation.

Note that the nodes in the first three panels should have had edges to themselves as well (from reflexivity of equivalence relations).
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
A partition and an equivalence relation are not the same thing; however, they can induce each other (as explained at the end of this answer). An equivalence relation $R$ on a set $A$ is a subset of $Atimes A$ satisfying the following properties:
$$(a,a)in Rspaceforall ain A$$
$$(a,b)in Rimplies (b,a)in Rspaceforall a,bin A$$
$$(a,b)in Rspacetextandspace (b,c)in Rimplies (a,c)in R spaceforall a,b,cin A$$
However, a partition $P$ of $A$ is a subset of $2^A$ satisfying the following two properties:
$$p_icap p_j=varnothingspaceforall p_i,p_jin 2^Aspacetextwithspace p_ine p_j$$
$$bigcup_i=1^p_i=A$$
We have that $Rsubset Atimes A$ and $Psubset 2^A$, so they're not even the same type of object. However, your professor probably meant that every equivalence relation on a set $A$ induces a partition of $A$, and vice versa.
More specifically, if $R$ is an equivalence relation on $A$, then the induced partition $P$ is
$$P=b:(a,b)in R:ain A$$
and if $P$ is a partition of $A$, then the induced equivalence relation $R$ is defined by
$$R=(a,b):exists p_iin Pspacetexts.t.space a,bin p_i$$
In plain words: If $R$ is a given equivalence relation, then the induced partition $P$ partitions $A$ into all sets of elements which are equivalent to each other under $R$. If $P$ is a given partition, then the induced equivalence relation $R$ is the relation for which $xsim y$ if and only if $x,y$ are in the same set of the partition P.
A bit ashamed to admit the notation looks a bit daunting to read and it's presenting as difficult to me. Is there a more wordsy explanation to this? Mainly the end of your answer addressing the induced partition and induced equivalence relation bit?
– sangstar
4 hours ago
@sangstar Sure, thanks for asking! I edited my question for you.
– Frpzzd
4 hours ago
Thanks! I think I hopefully understand now. Does my example illustrate this understanding? Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of that space. If we define an equivalence relation $R_W$ on $V$ by $u R_W v$ if $u-v in W$, then the set of elements which satisfy $u-v in W$ for arbitrary $u, v in V$ form a partition of $V$?
– sangstar
4 hours ago
1
@sangstar That's right! If you're familiar with a bit of elementary group theory, the sets in this induced partition are called the cosets of the subspace $W$.
– Frpzzd
4 hours ago
Ah, okay. Thank you!
– sangstar
4 hours ago
add a comment |
up vote
5
down vote
accepted
A partition and an equivalence relation are not the same thing; however, they can induce each other (as explained at the end of this answer). An equivalence relation $R$ on a set $A$ is a subset of $Atimes A$ satisfying the following properties:
$$(a,a)in Rspaceforall ain A$$
$$(a,b)in Rimplies (b,a)in Rspaceforall a,bin A$$
$$(a,b)in Rspacetextandspace (b,c)in Rimplies (a,c)in R spaceforall a,b,cin A$$
However, a partition $P$ of $A$ is a subset of $2^A$ satisfying the following two properties:
$$p_icap p_j=varnothingspaceforall p_i,p_jin 2^Aspacetextwithspace p_ine p_j$$
$$bigcup_i=1^p_i=A$$
We have that $Rsubset Atimes A$ and $Psubset 2^A$, so they're not even the same type of object. However, your professor probably meant that every equivalence relation on a set $A$ induces a partition of $A$, and vice versa.
More specifically, if $R$ is an equivalence relation on $A$, then the induced partition $P$ is
$$P=b:(a,b)in R:ain A$$
and if $P$ is a partition of $A$, then the induced equivalence relation $R$ is defined by
$$R=(a,b):exists p_iin Pspacetexts.t.space a,bin p_i$$
In plain words: If $R$ is a given equivalence relation, then the induced partition $P$ partitions $A$ into all sets of elements which are equivalent to each other under $R$. If $P$ is a given partition, then the induced equivalence relation $R$ is the relation for which $xsim y$ if and only if $x,y$ are in the same set of the partition P.
A bit ashamed to admit the notation looks a bit daunting to read and it's presenting as difficult to me. Is there a more wordsy explanation to this? Mainly the end of your answer addressing the induced partition and induced equivalence relation bit?
– sangstar
4 hours ago
@sangstar Sure, thanks for asking! I edited my question for you.
– Frpzzd
4 hours ago
Thanks! I think I hopefully understand now. Does my example illustrate this understanding? Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of that space. If we define an equivalence relation $R_W$ on $V$ by $u R_W v$ if $u-v in W$, then the set of elements which satisfy $u-v in W$ for arbitrary $u, v in V$ form a partition of $V$?
– sangstar
4 hours ago
1
@sangstar That's right! If you're familiar with a bit of elementary group theory, the sets in this induced partition are called the cosets of the subspace $W$.
– Frpzzd
4 hours ago
Ah, okay. Thank you!
– sangstar
4 hours ago
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
A partition and an equivalence relation are not the same thing; however, they can induce each other (as explained at the end of this answer). An equivalence relation $R$ on a set $A$ is a subset of $Atimes A$ satisfying the following properties:
$$(a,a)in Rspaceforall ain A$$
$$(a,b)in Rimplies (b,a)in Rspaceforall a,bin A$$
$$(a,b)in Rspacetextandspace (b,c)in Rimplies (a,c)in R spaceforall a,b,cin A$$
However, a partition $P$ of $A$ is a subset of $2^A$ satisfying the following two properties:
$$p_icap p_j=varnothingspaceforall p_i,p_jin 2^Aspacetextwithspace p_ine p_j$$
$$bigcup_i=1^p_i=A$$
We have that $Rsubset Atimes A$ and $Psubset 2^A$, so they're not even the same type of object. However, your professor probably meant that every equivalence relation on a set $A$ induces a partition of $A$, and vice versa.
More specifically, if $R$ is an equivalence relation on $A$, then the induced partition $P$ is
$$P=b:(a,b)in R:ain A$$
and if $P$ is a partition of $A$, then the induced equivalence relation $R$ is defined by
$$R=(a,b):exists p_iin Pspacetexts.t.space a,bin p_i$$
In plain words: If $R$ is a given equivalence relation, then the induced partition $P$ partitions $A$ into all sets of elements which are equivalent to each other under $R$. If $P$ is a given partition, then the induced equivalence relation $R$ is the relation for which $xsim y$ if and only if $x,y$ are in the same set of the partition P.
A partition and an equivalence relation are not the same thing; however, they can induce each other (as explained at the end of this answer). An equivalence relation $R$ on a set $A$ is a subset of $Atimes A$ satisfying the following properties:
$$(a,a)in Rspaceforall ain A$$
$$(a,b)in Rimplies (b,a)in Rspaceforall a,bin A$$
$$(a,b)in Rspacetextandspace (b,c)in Rimplies (a,c)in R spaceforall a,b,cin A$$
However, a partition $P$ of $A$ is a subset of $2^A$ satisfying the following two properties:
$$p_icap p_j=varnothingspaceforall p_i,p_jin 2^Aspacetextwithspace p_ine p_j$$
$$bigcup_i=1^p_i=A$$
We have that $Rsubset Atimes A$ and $Psubset 2^A$, so they're not even the same type of object. However, your professor probably meant that every equivalence relation on a set $A$ induces a partition of $A$, and vice versa.
More specifically, if $R$ is an equivalence relation on $A$, then the induced partition $P$ is
$$P=b:(a,b)in R:ain A$$
and if $P$ is a partition of $A$, then the induced equivalence relation $R$ is defined by
$$R=(a,b):exists p_iin Pspacetexts.t.space a,bin p_i$$
In plain words: If $R$ is a given equivalence relation, then the induced partition $P$ partitions $A$ into all sets of elements which are equivalent to each other under $R$. If $P$ is a given partition, then the induced equivalence relation $R$ is the relation for which $xsim y$ if and only if $x,y$ are in the same set of the partition P.
edited 4 hours ago
answered 4 hours ago
Frpzzd
19k63798
19k63798
A bit ashamed to admit the notation looks a bit daunting to read and it's presenting as difficult to me. Is there a more wordsy explanation to this? Mainly the end of your answer addressing the induced partition and induced equivalence relation bit?
– sangstar
4 hours ago
@sangstar Sure, thanks for asking! I edited my question for you.
– Frpzzd
4 hours ago
Thanks! I think I hopefully understand now. Does my example illustrate this understanding? Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of that space. If we define an equivalence relation $R_W$ on $V$ by $u R_W v$ if $u-v in W$, then the set of elements which satisfy $u-v in W$ for arbitrary $u, v in V$ form a partition of $V$?
– sangstar
4 hours ago
1
@sangstar That's right! If you're familiar with a bit of elementary group theory, the sets in this induced partition are called the cosets of the subspace $W$.
– Frpzzd
4 hours ago
Ah, okay. Thank you!
– sangstar
4 hours ago
add a comment |
A bit ashamed to admit the notation looks a bit daunting to read and it's presenting as difficult to me. Is there a more wordsy explanation to this? Mainly the end of your answer addressing the induced partition and induced equivalence relation bit?
– sangstar
4 hours ago
@sangstar Sure, thanks for asking! I edited my question for you.
– Frpzzd
4 hours ago
Thanks! I think I hopefully understand now. Does my example illustrate this understanding? Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of that space. If we define an equivalence relation $R_W$ on $V$ by $u R_W v$ if $u-v in W$, then the set of elements which satisfy $u-v in W$ for arbitrary $u, v in V$ form a partition of $V$?
– sangstar
4 hours ago
1
@sangstar That's right! If you're familiar with a bit of elementary group theory, the sets in this induced partition are called the cosets of the subspace $W$.
– Frpzzd
4 hours ago
Ah, okay. Thank you!
– sangstar
4 hours ago
A bit ashamed to admit the notation looks a bit daunting to read and it's presenting as difficult to me. Is there a more wordsy explanation to this? Mainly the end of your answer addressing the induced partition and induced equivalence relation bit?
– sangstar
4 hours ago
A bit ashamed to admit the notation looks a bit daunting to read and it's presenting as difficult to me. Is there a more wordsy explanation to this? Mainly the end of your answer addressing the induced partition and induced equivalence relation bit?
– sangstar
4 hours ago
@sangstar Sure, thanks for asking! I edited my question for you.
– Frpzzd
4 hours ago
@sangstar Sure, thanks for asking! I edited my question for you.
– Frpzzd
4 hours ago
Thanks! I think I hopefully understand now. Does my example illustrate this understanding? Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of that space. If we define an equivalence relation $R_W$ on $V$ by $u R_W v$ if $u-v in W$, then the set of elements which satisfy $u-v in W$ for arbitrary $u, v in V$ form a partition of $V$?
– sangstar
4 hours ago
Thanks! I think I hopefully understand now. Does my example illustrate this understanding? Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of that space. If we define an equivalence relation $R_W$ on $V$ by $u R_W v$ if $u-v in W$, then the set of elements which satisfy $u-v in W$ for arbitrary $u, v in V$ form a partition of $V$?
– sangstar
4 hours ago
1
1
@sangstar That's right! If you're familiar with a bit of elementary group theory, the sets in this induced partition are called the cosets of the subspace $W$.
– Frpzzd
4 hours ago
@sangstar That's right! If you're familiar with a bit of elementary group theory, the sets in this induced partition are called the cosets of the subspace $W$.
– Frpzzd
4 hours ago
Ah, okay. Thank you!
– sangstar
4 hours ago
Ah, okay. Thank you!
– sangstar
4 hours ago
add a comment |
up vote
4
down vote
They are "the same thing" in the sense that given an equivalence relation there is a natural way to construct a partition, and given a partition there is a natural way to construct an equivalence relation, and these two natural ways invert one another. That's useful because whenever you encounter one of these objects you are free to reason about the other if that makes your argument easier.
You are right when you say that it is the equivalence classes of an equivalence relation that form a partition (that is in fact the natural thing to look at), not the equivalence relation itself.
The natural way to construct an equivalence relation from a partition is to define two elements to be equivalent just when they are in the same block of the partition.
I see. What if there is a defined equivalence relation $R$ that isn't defined as "is in the same partition as". Can we not then equate equivalence classes and partitions in this case? Just trying to make sure I understand you.
– sangstar
4 hours ago
@sangstar Every equivalent relation can be expressed as "is in the same partition as" for some partition; that is, every equivalence relation has an induced partition just as every partition has an induced equivalence relation.
– Frpzzd
4 hours ago
@Frpzzd I see. I'm afraid the notation in your answer is a bit difficult to read for my less experienced eyes. Is there a more wordy explanation for your answer?
– sangstar
4 hours ago
My answer says in words just what @Frpzzd 's answer says with notation.
– Ethan Bolker
3 hours ago
add a comment |
up vote
4
down vote
They are "the same thing" in the sense that given an equivalence relation there is a natural way to construct a partition, and given a partition there is a natural way to construct an equivalence relation, and these two natural ways invert one another. That's useful because whenever you encounter one of these objects you are free to reason about the other if that makes your argument easier.
You are right when you say that it is the equivalence classes of an equivalence relation that form a partition (that is in fact the natural thing to look at), not the equivalence relation itself.
The natural way to construct an equivalence relation from a partition is to define two elements to be equivalent just when they are in the same block of the partition.
I see. What if there is a defined equivalence relation $R$ that isn't defined as "is in the same partition as". Can we not then equate equivalence classes and partitions in this case? Just trying to make sure I understand you.
– sangstar
4 hours ago
@sangstar Every equivalent relation can be expressed as "is in the same partition as" for some partition; that is, every equivalence relation has an induced partition just as every partition has an induced equivalence relation.
– Frpzzd
4 hours ago
@Frpzzd I see. I'm afraid the notation in your answer is a bit difficult to read for my less experienced eyes. Is there a more wordy explanation for your answer?
– sangstar
4 hours ago
My answer says in words just what @Frpzzd 's answer says with notation.
– Ethan Bolker
3 hours ago
add a comment |
up vote
4
down vote
up vote
4
down vote
They are "the same thing" in the sense that given an equivalence relation there is a natural way to construct a partition, and given a partition there is a natural way to construct an equivalence relation, and these two natural ways invert one another. That's useful because whenever you encounter one of these objects you are free to reason about the other if that makes your argument easier.
You are right when you say that it is the equivalence classes of an equivalence relation that form a partition (that is in fact the natural thing to look at), not the equivalence relation itself.
The natural way to construct an equivalence relation from a partition is to define two elements to be equivalent just when they are in the same block of the partition.
They are "the same thing" in the sense that given an equivalence relation there is a natural way to construct a partition, and given a partition there is a natural way to construct an equivalence relation, and these two natural ways invert one another. That's useful because whenever you encounter one of these objects you are free to reason about the other if that makes your argument easier.
You are right when you say that it is the equivalence classes of an equivalence relation that form a partition (that is in fact the natural thing to look at), not the equivalence relation itself.
The natural way to construct an equivalence relation from a partition is to define two elements to be equivalent just when they are in the same block of the partition.
answered 4 hours ago
Ethan Bolker
38.7k543102
38.7k543102
I see. What if there is a defined equivalence relation $R$ that isn't defined as "is in the same partition as". Can we not then equate equivalence classes and partitions in this case? Just trying to make sure I understand you.
– sangstar
4 hours ago
@sangstar Every equivalent relation can be expressed as "is in the same partition as" for some partition; that is, every equivalence relation has an induced partition just as every partition has an induced equivalence relation.
– Frpzzd
4 hours ago
@Frpzzd I see. I'm afraid the notation in your answer is a bit difficult to read for my less experienced eyes. Is there a more wordy explanation for your answer?
– sangstar
4 hours ago
My answer says in words just what @Frpzzd 's answer says with notation.
– Ethan Bolker
3 hours ago
add a comment |
I see. What if there is a defined equivalence relation $R$ that isn't defined as "is in the same partition as". Can we not then equate equivalence classes and partitions in this case? Just trying to make sure I understand you.
– sangstar
4 hours ago
@sangstar Every equivalent relation can be expressed as "is in the same partition as" for some partition; that is, every equivalence relation has an induced partition just as every partition has an induced equivalence relation.
– Frpzzd
4 hours ago
@Frpzzd I see. I'm afraid the notation in your answer is a bit difficult to read for my less experienced eyes. Is there a more wordy explanation for your answer?
– sangstar
4 hours ago
My answer says in words just what @Frpzzd 's answer says with notation.
– Ethan Bolker
3 hours ago
I see. What if there is a defined equivalence relation $R$ that isn't defined as "is in the same partition as". Can we not then equate equivalence classes and partitions in this case? Just trying to make sure I understand you.
– sangstar
4 hours ago
I see. What if there is a defined equivalence relation $R$ that isn't defined as "is in the same partition as". Can we not then equate equivalence classes and partitions in this case? Just trying to make sure I understand you.
– sangstar
4 hours ago
@sangstar Every equivalent relation can be expressed as "is in the same partition as" for some partition; that is, every equivalence relation has an induced partition just as every partition has an induced equivalence relation.
– Frpzzd
4 hours ago
@sangstar Every equivalent relation can be expressed as "is in the same partition as" for some partition; that is, every equivalence relation has an induced partition just as every partition has an induced equivalence relation.
– Frpzzd
4 hours ago
@Frpzzd I see. I'm afraid the notation in your answer is a bit difficult to read for my less experienced eyes. Is there a more wordy explanation for your answer?
– sangstar
4 hours ago
@Frpzzd I see. I'm afraid the notation in your answer is a bit difficult to read for my less experienced eyes. Is there a more wordy explanation for your answer?
– sangstar
4 hours ago
My answer says in words just what @Frpzzd 's answer says with notation.
– Ethan Bolker
3 hours ago
My answer says in words just what @Frpzzd 's answer says with notation.
– Ethan Bolker
3 hours ago
add a comment |
up vote
0
down vote
Here's a visual explanation.

Note that the nodes in the first three panels should have had edges to themselves as well (from reflexivity of equivalence relations).
add a comment |
up vote
0
down vote
Here's a visual explanation.

Note that the nodes in the first three panels should have had edges to themselves as well (from reflexivity of equivalence relations).
add a comment |
up vote
0
down vote
up vote
0
down vote
Here's a visual explanation.

Note that the nodes in the first three panels should have had edges to themselves as well (from reflexivity of equivalence relations).
Here's a visual explanation.

Note that the nodes in the first three panels should have had edges to themselves as well (from reflexivity of equivalence relations).
answered 24 mins ago
rlms
2591317
2591317
add a comment |
add a comment |
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Do you mean $A_icap A_j=varnothing$ if $jne i$?
– Frpzzd
4 hours ago
@Frpzzd Good spot, thanks for pointing it out.
– sangstar
4 hours ago
3
Declare two things equivalent if they live in the same subset. That’s an equivalence relation.
– Randall
4 hours ago
@Randall Ah, that's a clever example. Can I use this to extend the general sameness of equivalence relations and partitions that I've been offered to accept however, or only if we consider the equivalence relation "if they live in the same subset"?
– sangstar
4 hours ago
They are equivalent: partitions and eq relations are the same concept via my remark. It’s a great learning exercise.
– Randall
4 hours ago