Prove there is a bijection φ : R(X) → P(X)
$begingroup$
For a set $X$, let $R(X)$ be the set of equivalence relations on $X$ and let $P(X)$ be the set of partitions of $X$. Prove there is a bijection $varphi : R(X) to P(X)$.
Stuck on how to proceed with this. I understand that equivalence relations and partitions are in essence the same. But in terms of writing a proof, I am really lost how to proceed. Any help will be appreciated.
equivalence-relations set-partition
edited Dec 11 '18 at 22:51
platty
3,370320
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
$endgroup$
add a comment |
$begingroup$
For a set $X$, let $R(X)$ be the set of equivalence relations on $X$ and let $P(X)$ be the set of partitions of $X$. Prove there is a bijection $varphi : R(X) to P(X)$.
Stuck on how to proceed with this. I understand that equivalence relations and partitions are in essence the same. But in terms of writing a proof, I am really lost how to proceed. Any help will be appreciated.
equivalence-relations set-partition
edited Dec 11 '18 at 22:51
platty
3,370320
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
$endgroup$
2
$begingroup$
You say that they are in essence the same -- can you show in what way this holds? That is, if I give you an equivalence relation, can you come up with its corresponding partition and vice versa? This would define such a $varphi$, it then remains to show that it is a bijection.
$endgroup$
– platty
Dec 11 '18 at 22:52
$begingroup$
In addition to platty's advice, it might help to write down exactly what an element of $R(X)$ and an element of $P(X)$ might look like.
$endgroup$
– AlexanderJ93
Dec 11 '18 at 23:09
$begingroup$
See this and this.
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:17
add a comment |
$begingroup$
For a set $X$, let $R(X)$ be the set of equivalence relations on $X$ and let $P(X)$ be the set of partitions of $X$. Prove there is a bijection $varphi : R(X) to P(X)$.
Stuck on how to proceed with this. I understand that equivalence relations and partitions are in essence the same. But in terms of writing a proof, I am really lost how to proceed. Any help will be appreciated.
equivalence-relations set-partition
edited Dec 11 '18 at 22:51
platty
3,370320
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
$endgroup$
For a set $X$, let $R(X)$ be the set of equivalence relations on $X$ and let $P(X)$ be the set of partitions of $X$. Prove there is a bijection $varphi : R(X) to P(X)$.
Stuck on how to proceed with this. I understand that equivalence relations and partitions are in essence the same. But in terms of writing a proof, I am really lost how to proceed. Any help will be appreciated.
equivalence-relations set-partition
equivalence-relations set-partition
edited Dec 11 '18 at 22:51
platty
3,370320
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
edited Dec 11 '18 at 22:51
platty
3,370320
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
edited Dec 11 '18 at 22:51
platty
3,370320
edited Dec 11 '18 at 22:51
platty
3,370320
edited Dec 11 '18 at 22:51
platty
3,370320
3,370320
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
asked Dec 11 '18 at 22:47
Nihar GuptaNihar Gupta
61
61
2
$begingroup$
You say that they are in essence the same -- can you show in what way this holds? That is, if I give you an equivalence relation, can you come up with its corresponding partition and vice versa? This would define such a $varphi$, it then remains to show that it is a bijection.
$endgroup$
– platty
Dec 11 '18 at 22:52
$begingroup$
In addition to platty's advice, it might help to write down exactly what an element of $R(X)$ and an element of $P(X)$ might look like.
$endgroup$
– AlexanderJ93
Dec 11 '18 at 23:09
$begingroup$
See this and this.
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:17
add a comment |
2
$begingroup$
You say that they are in essence the same -- can you show in what way this holds? That is, if I give you an equivalence relation, can you come up with its corresponding partition and vice versa? This would define such a $varphi$, it then remains to show that it is a bijection.
$endgroup$
– platty
Dec 11 '18 at 22:52
$begingroup$
In addition to platty's advice, it might help to write down exactly what an element of $R(X)$ and an element of $P(X)$ might look like.
$endgroup$
– AlexanderJ93
Dec 11 '18 at 23:09
$begingroup$
See this and this.
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:17
2
2
$begingroup$
You say that they are in essence the same -- can you show in what way this holds? That is, if I give you an equivalence relation, can you come up with its corresponding partition and vice versa? This would define such a $varphi$, it then remains to show that it is a bijection.
$endgroup$
– platty
Dec 11 '18 at 22:52
$begingroup$
You say that they are in essence the same -- can you show in what way this holds? That is, if I give you an equivalence relation, can you come up with its corresponding partition and vice versa? This would define such a $varphi$, it then remains to show that it is a bijection.
$endgroup$
– platty
Dec 11 '18 at 22:52
$begingroup$
In addition to platty's advice, it might help to write down exactly what an element of $R(X)$ and an element of $P(X)$ might look like.
$endgroup$
– AlexanderJ93
Dec 11 '18 at 23:09
$begingroup$
In addition to platty's advice, it might help to write down exactly what an element of $R(X)$ and an element of $P(X)$ might look like.
$endgroup$
– AlexanderJ93
Dec 11 '18 at 23:09
$begingroup$
See this and this.
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:17
$begingroup$
See this and this.
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:17
add a comment |
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$begingroup$
You say that they are in essence the same -- can you show in what way this holds? That is, if I give you an equivalence relation, can you come up with its corresponding partition and vice versa? This would define such a $varphi$, it then remains to show that it is a bijection.
$endgroup$
– platty
Dec 11 '18 at 22:52
$begingroup$
In addition to platty's advice, it might help to write down exactly what an element of $R(X)$ and an element of $P(X)$ might look like.
$endgroup$
– AlexanderJ93
Dec 11 '18 at 23:09
$begingroup$
See this and this.
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:17