Reflexive, symmetric, transitive, and antisymmetric
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Can there be a relation which is reflexive, symmetric, transitive, and antisymmetric at the same time? I tried to find so.
If $A = a,b,c $. Let $R$ be a relation which is reflexive, symmetric, transitive, and antisymmetric.
$R = (a,a), (b,b), (c,c) $
Is this correct? If I'm wrong, can you help me understand it?
Since if $(a, b)$ and $(b, c)$ are elements of $R$ by transitive there would be $(a, c)$, but then there should be $(b, a)$, $(c, b)$ and $(c, a)$ by symmetry, but then it would not be antisymmetric. If I'm not mistaken.
relations symmetric-functions
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
asked Sep 25 at 9:51
Shehan Tearz
654
add a comment |Â
up vote
5
down vote
favorite
Can there be a relation which is reflexive, symmetric, transitive, and antisymmetric at the same time? I tried to find so.
If $A = a,b,c $. Let $R$ be a relation which is reflexive, symmetric, transitive, and antisymmetric.
$R = (a,a), (b,b), (c,c) $
Is this correct? If I'm wrong, can you help me understand it?
Since if $(a, b)$ and $(b, c)$ are elements of $R$ by transitive there would be $(a, c)$, but then there should be $(b, a)$, $(c, b)$ and $(c, a)$ by symmetry, but then it would not be antisymmetric. If I'm not mistaken.
relations symmetric-functions
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
asked Sep 25 at 9:51
Shehan Tearz
654
2
Rather than simply telling you if you're right or where you're wrong, I would recommend you check methodically so you can be confident in the answer. To test transitivity, if you're concerned about missing something, you can write down all 9 pairs of elements or $R$ and see if they're of the form $(x,y)$ and $(y,z)$ (where some of $x,y,z$ can be the same) and if the corresponding $(x,z)$ is in $R$ too. For symmetry, look at all 3 elements of $R$. For reflectivity, look at all 3 elements of $A$. For antisymmetry, look at all 6 unordered pairs of elements of $R$ to look for $(x,y)$ and $(y,x)$.
– Mark S.
Sep 25 at 10:07
Symmetric and antisymmetric forces the relation to be a subset of the diagonal. Reflexive forces the diagonal to be a subset of the relation. Transitivity doesn't really play a role here, though it follows from the other properties.
– John Coleman
Sep 25 at 10:59
Any reason why none of the answers you received were accepted by you?
– 5xum
2 days ago
Sorry for the delay
– Shehan Tearz
2 days ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Can there be a relation which is reflexive, symmetric, transitive, and antisymmetric at the same time? I tried to find so.
If $A = a,b,c $. Let $R$ be a relation which is reflexive, symmetric, transitive, and antisymmetric.
$R = (a,a), (b,b), (c,c) $
Is this correct? If I'm wrong, can you help me understand it?
Since if $(a, b)$ and $(b, c)$ are elements of $R$ by transitive there would be $(a, c)$, but then there should be $(b, a)$, $(c, b)$ and $(c, a)$ by symmetry, but then it would not be antisymmetric. If I'm not mistaken.
relations symmetric-functions
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
asked Sep 25 at 9:51
Shehan Tearz
654
Can there be a relation which is reflexive, symmetric, transitive, and antisymmetric at the same time? I tried to find so.
If $A = a,b,c $. Let $R$ be a relation which is reflexive, symmetric, transitive, and antisymmetric.
$R = (a,a), (b,b), (c,c) $
Is this correct? If I'm wrong, can you help me understand it?
Since if $(a, b)$ and $(b, c)$ are elements of $R$ by transitive there would be $(a, c)$, but then there should be $(b, a)$, $(c, b)$ and $(c, a)$ by symmetry, but then it would not be antisymmetric. If I'm not mistaken.
relations symmetric-functions
relations symmetric-functions
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
asked Sep 25 at 9:51
Shehan Tearz
654
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
asked Sep 25 at 9:51
Shehan Tearz
654
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
edited Sep 25 at 17:13
Ng Chung Tak
13.2k31132
13.2k31132
asked Sep 25 at 9:51
Shehan Tearz
654
asked Sep 25 at 9:51
Shehan Tearz
654
asked Sep 25 at 9:51
Shehan Tearz
654
654
2
Rather than simply telling you if you're right or where you're wrong, I would recommend you check methodically so you can be confident in the answer. To test transitivity, if you're concerned about missing something, you can write down all 9 pairs of elements or $R$ and see if they're of the form $(x,y)$ and $(y,z)$ (where some of $x,y,z$ can be the same) and if the corresponding $(x,z)$ is in $R$ too. For symmetry, look at all 3 elements of $R$. For reflectivity, look at all 3 elements of $A$. For antisymmetry, look at all 6 unordered pairs of elements of $R$ to look for $(x,y)$ and $(y,x)$.
– Mark S.
Sep 25 at 10:07
Symmetric and antisymmetric forces the relation to be a subset of the diagonal. Reflexive forces the diagonal to be a subset of the relation. Transitivity doesn't really play a role here, though it follows from the other properties.
– John Coleman
Sep 25 at 10:59
Any reason why none of the answers you received were accepted by you?
– 5xum
2 days ago
Sorry for the delay
– Shehan Tearz
2 days ago
add a comment |Â
2
Rather than simply telling you if you're right or where you're wrong, I would recommend you check methodically so you can be confident in the answer. To test transitivity, if you're concerned about missing something, you can write down all 9 pairs of elements or $R$ and see if they're of the form $(x,y)$ and $(y,z)$ (where some of $x,y,z$ can be the same) and if the corresponding $(x,z)$ is in $R$ too. For symmetry, look at all 3 elements of $R$. For reflectivity, look at all 3 elements of $A$. For antisymmetry, look at all 6 unordered pairs of elements of $R$ to look for $(x,y)$ and $(y,x)$.
– Mark S.
Sep 25 at 10:07
Symmetric and antisymmetric forces the relation to be a subset of the diagonal. Reflexive forces the diagonal to be a subset of the relation. Transitivity doesn't really play a role here, though it follows from the other properties.
– John Coleman
Sep 25 at 10:59
Any reason why none of the answers you received were accepted by you?
– 5xum
2 days ago
Sorry for the delay
– Shehan Tearz
2 days ago
2
2
Rather than simply telling you if you're right or where you're wrong, I would recommend you check methodically so you can be confident in the answer. To test transitivity, if you're concerned about missing something, you can write down all 9 pairs of elements or $R$ and see if they're of the form $(x,y)$ and $(y,z)$ (where some of $x,y,z$ can be the same) and if the corresponding $(x,z)$ is in $R$ too. For symmetry, look at all 3 elements of $R$. For reflectivity, look at all 3 elements of $A$. For antisymmetry, look at all 6 unordered pairs of elements of $R$ to look for $(x,y)$ and $(y,x)$.
– Mark S.
Sep 25 at 10:07
Rather than simply telling you if you're right or where you're wrong, I would recommend you check methodically so you can be confident in the answer. To test transitivity, if you're concerned about missing something, you can write down all 9 pairs of elements or $R$ and see if they're of the form $(x,y)$ and $(y,z)$ (where some of $x,y,z$ can be the same) and if the corresponding $(x,z)$ is in $R$ too. For symmetry, look at all 3 elements of $R$. For reflectivity, look at all 3 elements of $A$. For antisymmetry, look at all 6 unordered pairs of elements of $R$ to look for $(x,y)$ and $(y,x)$.
– Mark S.
Sep 25 at 10:07
Symmetric and antisymmetric forces the relation to be a subset of the diagonal. Reflexive forces the diagonal to be a subset of the relation. Transitivity doesn't really play a role here, though it follows from the other properties.
– John Coleman
Sep 25 at 10:59
Symmetric and antisymmetric forces the relation to be a subset of the diagonal. Reflexive forces the diagonal to be a subset of the relation. Transitivity doesn't really play a role here, though it follows from the other properties.
– John Coleman
Sep 25 at 10:59
Any reason why none of the answers you received were accepted by you?
– 5xum
2 days ago
Any reason why none of the answers you received were accepted by you?
– 5xum
2 days ago
Sorry for the delay
– Shehan Tearz
2 days ago
Sorry for the delay
– Shehan Tearz
2 days ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
12
down vote
accepted
For any set $A$, there exists only one relation which is both reflexive, symmetric and assymetric, and that is the relation $R=(a,a)$.
You can easily see that any reflexive relation must include all elements of $R$, and that any relation that is symmetric and antisymmetric cannot include any pair $(a,b)$ where $aneq b$. So already, $R$ is your only candidate for a reflexive, symmetric, transitive and antisymmetric relation.
Since $R$ is also transitive, we conclude that $R$ is the only reflexive, symmetric, transitive and antisymmetric relation.
answered Sep 25 at 9:55
5xum
84.4k388152
add a comment |Â
up vote
3
down vote
Your answer is correct and you can easily generalize it to a set with more elements
Apparently the only solution to your question is the diagonal relation, $$R=(x,x)$$ for any set A.
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
For any set $A$, there exists only one relation which is both reflexive, symmetric and assymetric, and that is the relation $R=(a,a)$.
You can easily see that any reflexive relation must include all elements of $R$, and that any relation that is symmetric and antisymmetric cannot include any pair $(a,b)$ where $aneq b$. So already, $R$ is your only candidate for a reflexive, symmetric, transitive and antisymmetric relation.
Since $R$ is also transitive, we conclude that $R$ is the only reflexive, symmetric, transitive and antisymmetric relation.
answered Sep 25 at 9:55
5xum
84.4k388152
add a comment |Â
up vote
12
down vote
accepted
For any set $A$, there exists only one relation which is both reflexive, symmetric and assymetric, and that is the relation $R=(a,a)$.
You can easily see that any reflexive relation must include all elements of $R$, and that any relation that is symmetric and antisymmetric cannot include any pair $(a,b)$ where $aneq b$. So already, $R$ is your only candidate for a reflexive, symmetric, transitive and antisymmetric relation.
Since $R$ is also transitive, we conclude that $R$ is the only reflexive, symmetric, transitive and antisymmetric relation.
answered Sep 25 at 9:55
5xum
84.4k388152
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
For any set $A$, there exists only one relation which is both reflexive, symmetric and assymetric, and that is the relation $R=(a,a)$.
You can easily see that any reflexive relation must include all elements of $R$, and that any relation that is symmetric and antisymmetric cannot include any pair $(a,b)$ where $aneq b$. So already, $R$ is your only candidate for a reflexive, symmetric, transitive and antisymmetric relation.
Since $R$ is also transitive, we conclude that $R$ is the only reflexive, symmetric, transitive and antisymmetric relation.
answered Sep 25 at 9:55
5xum
84.4k388152
For any set $A$, there exists only one relation which is both reflexive, symmetric and assymetric, and that is the relation $R=(a,a)$.
You can easily see that any reflexive relation must include all elements of $R$, and that any relation that is symmetric and antisymmetric cannot include any pair $(a,b)$ where $aneq b$. So already, $R$ is your only candidate for a reflexive, symmetric, transitive and antisymmetric relation.
Since $R$ is also transitive, we conclude that $R$ is the only reflexive, symmetric, transitive and antisymmetric relation.
answered Sep 25 at 9:55
5xum
84.4k388152
edited Sep 26 at 5:51
answered Sep 25 at 9:55
5xum
84.4k388152
answered Sep 25 at 9:55
5xum
84.4k388152
answered Sep 25 at 9:55
5xum
84.4k388152
84.4k388152
add a comment |Â
add a comment |Â
up vote
3
down vote
Your answer is correct and you can easily generalize it to a set with more elements
Apparently the only solution to your question is the diagonal relation, $$R=(x,x)$$ for any set A.
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
add a comment |Â
up vote
3
down vote
Your answer is correct and you can easily generalize it to a set with more elements
Apparently the only solution to your question is the diagonal relation, $$R=(x,x)$$ for any set A.
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Your answer is correct and you can easily generalize it to a set with more elements
Apparently the only solution to your question is the diagonal relation, $$R=(x,x)$$ for any set A.
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
Your answer is correct and you can easily generalize it to a set with more elements
Apparently the only solution to your question is the diagonal relation, $$R=(x,x)$$ for any set A.
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
answered Sep 25 at 10:02
Mohammad Riazi-Kermani
33.4k41855
33.4k41855
add a comment |Â
add a comment |Â
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Rather than simply telling you if you're right or where you're wrong, I would recommend you check methodically so you can be confident in the answer. To test transitivity, if you're concerned about missing something, you can write down all 9 pairs of elements or $R$ and see if they're of the form $(x,y)$ and $(y,z)$ (where some of $x,y,z$ can be the same) and if the corresponding $(x,z)$ is in $R$ too. For symmetry, look at all 3 elements of $R$. For reflectivity, look at all 3 elements of $A$. For antisymmetry, look at all 6 unordered pairs of elements of $R$ to look for $(x,y)$ and $(y,x)$.
– Mark S.
Sep 25 at 10:07
Symmetric and antisymmetric forces the relation to be a subset of the diagonal. Reflexive forces the diagonal to be a subset of the relation. Transitivity doesn't really play a role here, though it follows from the other properties.
– John Coleman
Sep 25 at 10:59
Any reason why none of the answers you received were accepted by you?
– 5xum
2 days ago
Sorry for the delay
– Shehan Tearz
2 days ago