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.



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    seems sound to me
    – Exodd
    Jun 10 '15 at 16:23















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?










share|cite|improve this question




















  • 3




    seems sound to me
    – Exodd
    Jun 10 '15 at 16:23













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?










share|cite|improve this question















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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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














  • 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















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