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Question about the defining equivalence relations on sets
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Suppose I have an equivalence relation $sim$ on $S=e,f,g,h,i$ such that $e sim f, f sim g$ and $e nsim i$. I’m trying to find the number of such relations that can be defined on $S$. I know that $e,f,g$ will always be an equivalence class and that $i$ will also always be an equivalence class. The questions therefore is equivalent to asking how many different equivalence classes can $h$ belong to and the answer is obviously $3$ since it can belong to its own equivalence class $h$, $i$ or $e,f,g$. However I’m not sure if it’s possible that $h$ does not belong to any equivalence class, i.e. the set of equivalence classes for the relations would be $e,f,g,i$. I think the answer is no because the set of equivalence classes has to partition $S$ but I’m not 100% sure.
elementary-set-theory relations equivalence-relations
asked 1 hour ago
Reinhild Van Rosenú
689719
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up vote
2
down vote
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Suppose I have an equivalence relation $sim$ on $S=e,f,g,h,i$ such that $e sim f, f sim g$ and $e nsim i$. I’m trying to find the number of such relations that can be defined on $S$. I know that $e,f,g$ will always be an equivalence class and that $i$ will also always be an equivalence class. The questions therefore is equivalent to asking how many different equivalence classes can $h$ belong to and the answer is obviously $3$ since it can belong to its own equivalence class $h$, $i$ or $e,f,g$. However I’m not sure if it’s possible that $h$ does not belong to any equivalence class, i.e. the set of equivalence classes for the relations would be $e,f,g,i$. I think the answer is no because the set of equivalence classes has to partition $S$ but I’m not 100% sure.
elementary-set-theory relations equivalence-relations
asked 1 hour ago
Reinhild Van Rosenú
689719
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose I have an equivalence relation $sim$ on $S=e,f,g,h,i$ such that $e sim f, f sim g$ and $e nsim i$. I’m trying to find the number of such relations that can be defined on $S$. I know that $e,f,g$ will always be an equivalence class and that $i$ will also always be an equivalence class. The questions therefore is equivalent to asking how many different equivalence classes can $h$ belong to and the answer is obviously $3$ since it can belong to its own equivalence class $h$, $i$ or $e,f,g$. However I’m not sure if it’s possible that $h$ does not belong to any equivalence class, i.e. the set of equivalence classes for the relations would be $e,f,g,i$. I think the answer is no because the set of equivalence classes has to partition $S$ but I’m not 100% sure.
elementary-set-theory relations equivalence-relations
asked 1 hour ago
Reinhild Van Rosenú
689719
Suppose I have an equivalence relation $sim$ on $S=e,f,g,h,i$ such that $e sim f, f sim g$ and $e nsim i$. I’m trying to find the number of such relations that can be defined on $S$. I know that $e,f,g$ will always be an equivalence class and that $i$ will also always be an equivalence class. The questions therefore is equivalent to asking how many different equivalence classes can $h$ belong to and the answer is obviously $3$ since it can belong to its own equivalence class $h$, $i$ or $e,f,g$. However I’m not sure if it’s possible that $h$ does not belong to any equivalence class, i.e. the set of equivalence classes for the relations would be $e,f,g,i$. I think the answer is no because the set of equivalence classes has to partition $S$ but I’m not 100% sure.
elementary-set-theory relations equivalence-relations
elementary-set-theory relations equivalence-relations
asked 1 hour ago
Reinhild Van Rosenú
689719
asked 1 hour ago
Reinhild Van Rosenú
689719
asked 1 hour ago
Reinhild Van Rosenú
689719
asked 1 hour ago
Reinhild Van Rosenú
689719
asked 1 hour ago
Reinhild Van Rosenú
689719
689719
add a comment |Â
add a comment |Â
2 Answers
2
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up vote
2
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accepted
An equivalence relation most certainly has to completely partition the set. It follows from the reflexivity requirement: at the very least, each element must be equivalent to itself, therefore constituting an equivalence class of its own.
answered 1 hour ago
Drinkwater
43529
add a comment |Â
up vote
3
down vote
There is a bijection between the set of equivalence relations on a set, and the partitions of that set. This allows us to switch between these notions fluidly.
That is, we know that $e,f,g$ are in the same equivalence class, which is one that is different from $i$. So, it is a question of where you place $h$, as you mentioned. The answer to your last part , is that it has to be part of its own equivalence class, since it is related to itself, and therefore must appear in one of the classes. Alternately, a partition covers every element, so must cover $h$. This gives us just the three possibilities, of $1.[e,f,g,h],[i]$ , $2 . [e,f,g],[h],[i]$, and $3. [e,f,g],[h,i]$.
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
An equivalence relation most certainly has to completely partition the set. It follows from the reflexivity requirement: at the very least, each element must be equivalent to itself, therefore constituting an equivalence class of its own.
answered 1 hour ago
Drinkwater
43529
add a comment |Â
up vote
2
down vote
accepted
An equivalence relation most certainly has to completely partition the set. It follows from the reflexivity requirement: at the very least, each element must be equivalent to itself, therefore constituting an equivalence class of its own.
answered 1 hour ago
Drinkwater
43529
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
An equivalence relation most certainly has to completely partition the set. It follows from the reflexivity requirement: at the very least, each element must be equivalent to itself, therefore constituting an equivalence class of its own.
answered 1 hour ago
Drinkwater
43529
An equivalence relation most certainly has to completely partition the set. It follows from the reflexivity requirement: at the very least, each element must be equivalent to itself, therefore constituting an equivalence class of its own.
answered 1 hour ago
Drinkwater
43529
answered 1 hour ago
Drinkwater
43529
answered 1 hour ago
Drinkwater
43529
answered 1 hour ago
Drinkwater
43529
43529
add a comment |Â
add a comment |Â
up vote
3
down vote
There is a bijection between the set of equivalence relations on a set, and the partitions of that set. This allows us to switch between these notions fluidly.
That is, we know that $e,f,g$ are in the same equivalence class, which is one that is different from $i$. So, it is a question of where you place $h$, as you mentioned. The answer to your last part , is that it has to be part of its own equivalence class, since it is related to itself, and therefore must appear in one of the classes. Alternately, a partition covers every element, so must cover $h$. This gives us just the three possibilities, of $1.[e,f,g,h],[i]$ , $2 . [e,f,g],[h],[i]$, and $3. [e,f,g],[h,i]$.
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
add a comment |Â
up vote
3
down vote
There is a bijection between the set of equivalence relations on a set, and the partitions of that set. This allows us to switch between these notions fluidly.
That is, we know that $e,f,g$ are in the same equivalence class, which is one that is different from $i$. So, it is a question of where you place $h$, as you mentioned. The answer to your last part , is that it has to be part of its own equivalence class, since it is related to itself, and therefore must appear in one of the classes. Alternately, a partition covers every element, so must cover $h$. This gives us just the three possibilities, of $1.[e,f,g,h],[i]$ , $2 . [e,f,g],[h],[i]$, and $3. [e,f,g],[h,i]$.
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
add a comment |Â
up vote
3
down vote
up vote
3
down vote
There is a bijection between the set of equivalence relations on a set, and the partitions of that set. This allows us to switch between these notions fluidly.
That is, we know that $e,f,g$ are in the same equivalence class, which is one that is different from $i$. So, it is a question of where you place $h$, as you mentioned. The answer to your last part , is that it has to be part of its own equivalence class, since it is related to itself, and therefore must appear in one of the classes. Alternately, a partition covers every element, so must cover $h$. This gives us just the three possibilities, of $1.[e,f,g,h],[i]$ , $2 . [e,f,g],[h],[i]$, and $3. [e,f,g],[h,i]$.
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
There is a bijection between the set of equivalence relations on a set, and the partitions of that set. This allows us to switch between these notions fluidly.
That is, we know that $e,f,g$ are in the same equivalence class, which is one that is different from $i$. So, it is a question of where you place $h$, as you mentioned. The answer to your last part , is that it has to be part of its own equivalence class, since it is related to itself, and therefore must appear in one of the classes. Alternately, a partition covers every element, so must cover $h$. This gives us just the three possibilities, of $1.[e,f,g,h],[i]$ , $2 . [e,f,g],[h],[i]$, and $3. [e,f,g],[h,i]$.
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35k33274
35k33274
add a comment |Â
add a comment |Â
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