What is the irreflexive closure of an irreflexive relation?
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I am working on a problem that states the following:
When is it possible to define the irreflexive closure of a relation R, that is, a relation that contains R, is irreflexive, and is contained in every irreflexive relation that contains R?
Attempt
First note that a relation $R$ defined on a set $A$ is irreflexive if $(a,a) notin R$ for all $a in A$.
Suppose $R$ is a relation defined in a set $A$. Let $S$ be an irreflexive relation on $A$ that contains $R$. Then $(a,a)notin S$ for all $a in A$ and $R subseteq S$ which implies that $(a,a) notin R$ for all $a in A$. Therefore, $R$ is irreflexive on $A$. Hence, unless $R$ is irreflexive, we can't define irreflexive closure of $R$.
Finally, note that when $R$ is irreflexive, the irreflexive closure of $R$ is $R$ itself.
Question is my approach correct?
relations
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up vote
2
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favorite
I am working on a problem that states the following:
When is it possible to define the irreflexive closure of a relation R, that is, a relation that contains R, is irreflexive, and is contained in every irreflexive relation that contains R?
Attempt
First note that a relation $R$ defined on a set $A$ is irreflexive if $(a,a) notin R$ for all $a in A$.
Suppose $R$ is a relation defined in a set $A$. Let $S$ be an irreflexive relation on $A$ that contains $R$. Then $(a,a)notin S$ for all $a in A$ and $R subseteq S$ which implies that $(a,a) notin R$ for all $a in A$. Therefore, $R$ is irreflexive on $A$. Hence, unless $R$ is irreflexive, we can't define irreflexive closure of $R$.
Finally, note that when $R$ is irreflexive, the irreflexive closure of $R$ is $R$ itself.
Question is my approach correct?
relations
3
seems sound to me
– Exodd
Jun 10 '15 at 16:23
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am working on a problem that states the following:
When is it possible to define the irreflexive closure of a relation R, that is, a relation that contains R, is irreflexive, and is contained in every irreflexive relation that contains R?
Attempt
First note that a relation $R$ defined on a set $A$ is irreflexive if $(a,a) notin R$ for all $a in A$.
Suppose $R$ is a relation defined in a set $A$. Let $S$ be an irreflexive relation on $A$ that contains $R$. Then $(a,a)notin S$ for all $a in A$ and $R subseteq S$ which implies that $(a,a) notin R$ for all $a in A$. Therefore, $R$ is irreflexive on $A$. Hence, unless $R$ is irreflexive, we can't define irreflexive closure of $R$.
Finally, note that when $R$ is irreflexive, the irreflexive closure of $R$ is $R$ itself.
Question is my approach correct?
relations
I am working on a problem that states the following:
When is it possible to define the irreflexive closure of a relation R, that is, a relation that contains R, is irreflexive, and is contained in every irreflexive relation that contains R?
Attempt
First note that a relation $R$ defined on a set $A$ is irreflexive if $(a,a) notin R$ for all $a in A$.
Suppose $R$ is a relation defined in a set $A$. Let $S$ be an irreflexive relation on $A$ that contains $R$. Then $(a,a)notin S$ for all $a in A$ and $R subseteq S$ which implies that $(a,a) notin R$ for all $a in A$. Therefore, $R$ is irreflexive on $A$. Hence, unless $R$ is irreflexive, we can't define irreflexive closure of $R$.
Finally, note that when $R$ is irreflexive, the irreflexive closure of $R$ is $R$ itself.
Question is my approach correct?
relations
relations
edited Jun 10 '15 at 16:35
Andrés E. Caicedo
64.2k8157243
64.2k8157243
asked Jun 10 '15 at 16:21
Camilo Celis Guzman
1585
1585
3
seems sound to me
– Exodd
Jun 10 '15 at 16:23
add a comment |
3
seems sound to me
– Exodd
Jun 10 '15 at 16:23
3
3
seems sound to me
– Exodd
Jun 10 '15 at 16:23
seems sound to me
– Exodd
Jun 10 '15 at 16:23
add a comment |
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3
seems sound to me
– Exodd
Jun 10 '15 at 16:23